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# Volume of REDNER'S, SMOKED ALMONDS, UPC: 015984610251 ## food weight to volume conversions ### calculate volume of generic and branded foods per weight #### Volume, i.e. how many spoons, cups,gallons or liters in 100 gram ofREDNER'S, SMOKED ALMONDS, UPC:015984610251 centimeter³ 211.24 milliliter 211.24 foot³ 0.01 US cup 0.89 Imperial gallon 0.05 US dessertspoon 28.57 inch³ 12.89 US fluid ounce 7.14 liter 0.21 US gallon 0.06 meter³ 0 US pint 0.45 metric cup 0.84 US quart 0.22 metric dessertspoon 21.12 US tablespoon 14.29 metric tablespoon 14.08 US teaspoon 42.86 metric teaspoon 42.25 #### Weight gram 100 ounce 3.53 kilogram 0.1 pound 0.22 milligram 100 000 Nutrient (find foodsrich in nutrients) Unit Value /100 g BasicAdvancedAll Proximates Energy kcal 571 Protein g 21.43 Total lipid (fat) g 50 Carbohydrate,bydifference g 21.43 Fiber,totaldietary g 10.7 Sugars, total g 3.57 Minerals Calcium, Ca mg 286 Iron, Fe mg 3.86 Sodium, Na mg 500 Vitamins Vitamin C,totalascorbic acid mg 0 Vitamin A, IU IU 0 Lipids Fatty acids,totalsaturated g 3.57 Cholesterol mg 0 #### See how many calories in0.1 kg (0.22 lbs) ofREDNER'S, SMOKED ALMONDS,UPC: 015984610251 From kilocalories(kcal) kilojoule(kJ) Carbohydrate 0 0 Fat 0 0 Protein 0 0 Other 571 2 389.06 Total 571 2 389.06 • About REDNER'S, SMOKED ALMONDS, UPC: 015984610251 • 118.34908 grams [g] of REDNER'S, SMOKED ALMONDS, UPC: 015984610251 fill 1 metric cup • 3.95068 ounces [oz] of REDNER'S, SMOKED ALMONDS, UPC: 015984610251 fill 1 US cup • REDNER'S, SMOKED ALMONDS, UPC: 015984610251 weigh(s) 118.35 gram per (metric cup) or 3.95 ounce per (US cup), and contain(s) 571 calories per 100 grams or ≈3.527 ounces  [ weight to volume | volume to weight | price | density ] • Ingredients:  ALMONDS, SEASONING (SALT, TORULA YEAST, CORN STARCH, HYDROLYZED SOY AND CORN PROTEIN, SUGAR, ONION POWDER, MONOSODIUM GLUTAMATE, SOYBEAN OIL, MALTODEXTRIN, NATURAL SMOKE FLAVOR, GARLIC POWDER, SPICES [INCLUDING MUSTARD], YEAST EXTRACT), PEANUT OIL. • Manufacturer:  Chiu Hop Company • For instance, compute how many cups or spoons a pound or kilogram of “REDNER'S, SMOKED ALMONDS, UPC: 015984610251” fills.  Volume of the selected food item is calculated based on the food density and its given weight. Visit our food calculations forum for more details. • A few foods with a name containing, like or similar to REDNER'S, SMOKED ALMONDS, UPC: 015984610251: • REDNER'S WAREHOUSE MARKETS, PREMIUM TURKEY BREAST WITH BROTH, UPC: 042222830782 contain(s) 89 calories per 100 grams or ≈3.527 ounces  [ price ] • REDNER'S WAREHOUSE MARKETS, MICROWAVE POPCORN, HOT and SPICY, UPC: 015984610855 contain(s) 536 calories per 100 grams or ≈3.527 ounces  [ price ] • REDNER'S, NATURAL ALMONDS, UPC: 015984610268 weigh(s) 118.35 gram per (metric cup) or 3.95 ounce per (US cup), and contain(s) 571 calories per 100 grams or ≈3.527 ounces  [ weight to volume | volume to weight | price | density ] • REDNER'S WAREHOUSE MARKETS, 100% FLORIDA GRAPEFRUIT JUICE, RUBY RED, UPC: 015984060131 contain(s) 38 calories per 100 grams or ≈3.527 ounces  [ price ] • REDNER'S WAREHOUSE MARKETS, HAND COOKED KETTLE STYLE POTATO CHIPS, UPC: 015984663059 contain(s) 536 calories per 100 grams or ≈3.527 ounces  [ price ] • Reference (ID: 68196) • USDA National Nutrient Database for Standard Reference; National Agricultural Library; United States Department of Agriculture (USDA); 1400 Independence Ave., S.W.; Washington, DC 20250 USA. #### Foods, Nutrients and Calories TRIPLE BERRY BLEND, UPC: 041512111150 weigh(s) 147.94 gram per (metric cup) or 4.94 ounce per (US cup), and contain(s) 57 calories per 100 grams or ≈3.527 ounces  [ weight to volume | volume to weight | price | density ] ORGANIC SUPER PREMIUM ICE CREAM, UPC: 784830100924 weigh(s) 192.32 gram per (metric cup) or 6.42 ounce per (US cup), and contain(s) 242 calories per 100 grams or ≈3.527 ounces  [ weight to volume | volume to weight | price | density ] Foods high in Vitamin D (D2 + D3) and Recommended Dietary Allowances (RDAs) for Vitamin D #### Gravels, Substances and Oils CaribSea, Freshwater, Super Naturals, Amazon weighs 1 521.75 kg/m³ (94.99975 lb/ft³) with specific gravity of 1.52175 relative to pure water.  Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical  or in a rectangular shaped aquarium or pond  [ weight to volume | volume to weight | price ] Cadmium monosulfate [CdSO4] weighs 4 690 kg/m³ (292.78714 lb/ft³)  [ weight to volume | volume to weight | price | mole to volume and weight | density ] Volume to weightweight to volume and cost conversions for Linseed oil with temperature in the range of 10°C (50°F) to 140°C (284°F) #### Weights and Measurements metric cup is a metric (SI) liquid measure of volume. Temperature is one of the seven SI base quantities and used as a measure of thermal energy. Convert short ton per cubic millimeter to troy pound per cubic inch or convert between all units of density measurement #### Calculators Area of an annulus calculator. Definition of an annulus.

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# Why are negative sets multisets? (Reference request) It is easy to establish that $$\left(\!\!{n\choose k}\!\!\right)=(-1)^k{-n \choose k},$$ where the symbol on the left-hand-side counts the number of multisets of $k$ elements from $n$. On the Wikipedia page for multisets, it is further claimed that "This fact led Gian-Carlo Rota to ask "Why are negative sets multisets?". He considered that question worthy of the attention of philosophers of mathematics." While I think it is quite plausible that Rota may well have asked this question, no citation is provided for it on Wikipedia, and my attempts to source the quote have all proved fruitless. Have you seen a quote by Rota relating "negative sets" to multisets? - I can't resist pointing out that this interpretation of "negative sets are multisets" is made very clear by the "Euler characteristic as generalized cardinality" point of view. In fact, since we know that $\chi (Sym^k X)=\binom{\chi(X)}{k}$, we notice that $\binom{n}{k}$ counts the number of ways of picking $k$ points among $n$ points, but on the other hand, $(-1)^k\binom{-n}{k}$ counts the number of ways of picking $k$ points among $n$ intervals (this is the same as multisets of size $k$). – Gjergji Zaimi Jun 15 '13 at 15:59 This paper by D. Loeb, "Sets with a negative number of elements": faculty.uml.edu/jpropp/negative.pdf, has a reference to a paper that Loeb and Rota wrote together. Perhaps there is a clue there- I can't find it online, though. – Sam Hopkins Jun 15 '13 at 16:54 I don't know the answer, but if you have not done so already, I'd recommend that you check the books "Discrete Thoughts" and "Indiscrete Thoughts". – Timothy Chow Jun 16 '13 at 18:03 On the other hand, I'm very tempted to view this in terms of superalgebra, interpreting $\binom{n}{k}$ as [the dimension of] the $k$-th alternating power of an $n$-dimensional vector space in ordinary algebra, but when we apply the functorial construction to an $n$-dimensional super space concentrated in the odd component, we get a $k$-th symmetric power of the underlying space (concentrated in the component of parity $k$). – Todd Trimble Apr 9 at 12:48

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# What will be the Speed? If An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 1(2/3) hours, it must travel at a speed of: 1. 300 kmph 2. 360 kmph 3. 600 kmph 4. 720 kmph Desire Adnan Default Asked on 4th August 2015 in Answer: (4) 720 kmph Explanation:- The speed of an aeroplane  = 240 km/hr Time taken by aeroplane = 5 hr Total distance covered by an aeroplane = 240 x 5 = 1200 km If it covered the same distance covered in 5/3 hr. Then, the speed of an aeroplane = 1200 x 3/5 = 240 x 3 = 720 kmph Hence, the answer is (4) 720 kmph. Anurag Mishra Professor Answered on 5th August 2015.

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# Voltage and current in a single-phase AC circuit are given by (50 + j20) V and (20 + j50) A, respectively. Power of the circuit can be expressed in complex form as: This question was previously asked in PSPCL JE EE 2018 Previous Year Paper View all PSPCL JE Papers > 1. (2500 - j0) 2. (0 - j2500) 3. (2100 - j2000) 4. (2000 - j2100) Option 4 : (2000 - j2100) ## Detailed Solution Concept: Complex Power of an AC circuit is given as: $$P = V \times {I^*}$$ P = complex power V = Voltage I* = Conjugate of current Calculation: Given that, V = (50 + j20) V I = (20 + j50) A I* = (20 - j50) A P = (50 + j20) × (20 - j50) = 2000 - j2100

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Discussion Groups ## Robot Structural Analysis Mentor Posts: 190 Registered: ‎12-06-2011 # Provided Reinforcement design for divided column 404 Views, 7 Replies 09-26-2012 05:23 AM How does one design a column which has been divided into two (or more) columns in the analysis module (because of connecting beams mid-height in column, or column spanning multiple stories)  in the RC design module. I assume you can take a section and input the real length of the column in the buckling analysis but how does one factor in columns end moments, etc.?? Product Support Posts: 5,604 Registered: ‎04-26-2010 # Re: Provided Reinforcement design for divided column 09-26-2012 08:33 AM in reply to: stroxy You have to divide such column into parts. Rafal Gaweda Mentor Posts: 190 Registered: ‎12-06-2011 # Re: Provided Reinforcement design for divided column 09-26-2012 09:27 PM in reply to: Rafal.Gaweda So if I have say a 6m column divided into 2 parts for analysis.  In designing the column I will design both parts as separate columns and enter efffective length  factor * 6m as the buckling length for each part ? Product Support Posts: 5,604 Registered: ‎04-26-2010 # Re: Provided Reinforcement design for divided column 09-27-2012 08:08 AM in reply to: stroxy So if I have say a 6m column divided into 2 parts for analysis.  In designing the column I will design both parts as separate columns and enter efffective length  factor * 6m as the buckling length for each part ? Decision is yours but IMHO if you are talking about right hand side case (below) I would suggest to use 3m as column length. Left => 6m. Rafal Gaweda Mentor Posts: 190 Registered: ‎12-06-2011 # Re: Provided Reinforcement design for divided column 09-27-2012 08:17 AM in reply to: Rafal.Gaweda Thanks Was talking about LHS case Mentor Posts: 190 Registered: ‎12-06-2011 # Re: Provided Reinforcement design for divided column 09-27-2012 08:19 AM in reply to: stroxy How did you get buckling shapes to show? Product Support Posts: 5,604 Registered: ‎04-26-2010 # Re: Provided Reinforcement design for divided column 09-27-2012 08:36 AM in reply to: stroxy FEM Buckling analysis Rafal Gaweda

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# Math a frame is made from a 28 inch length of trim and surrounds a 48 square inch picture. solve for the dimensions of the picture 1. 👍 0 2. 👎 0 3. 👁 135 1. 48 = 6 * 8 2(6) + 2(8) = 28 1. 👍 0 2. 👎 0 ## Similar Questions 1. ### Algebra the outside of a picture frame has a length which is 4 cm more than width. The area enclosed by the outside of the picture frame is 165 square cm. Find the width of the outside of the picture frame. 2. ### Math Holly wants to make a frame for painting. The painting is square and has an area of 225 square inches. The framing material costs \$2.46 per inch. How much will Holly spend framing the picture? 3. ### math A couple wants to install a square mirror that has an area of 500 square inches. To the nearest tenth of an inch, what length of wood trim is needed to go around the mirror? 4. ### math Billy is framing a picture with a rectangular frame. The frame has a perimeter of 62 inches. The height of the frame is 2⁄3 times the length. Find the dimensions of the rectangle. a. Define the variable(s) b. Write an equation 1. ### Circles and Ratios Johanna is selling cheese at the Farmers marketone wheel of cheese has a diameter of 9 in what is the area of the will of thieves use 3.14 for π ____ square inches if the 9 inch wheel of cheese cause \$18.16 what is the cost per 2. ### Math\ álgebra A couple wants to install a sqaure mirror that has an área Of 500 square inches. To the nearest tenth Of an inch, what lentgth Of wood trim is needed to go around the mirror ? 3. ### Geometry Elaine purchased 4.5 yards of lace trim from her local fabric shop. She has a circular tablecloth that is 54 inches in diameter and a square tablecloth that is 40 inches on each side. Can she trim either tablecloth? 4. ### Algebra 1 A square painting is surrounded by a frame. the outside edges of the frame are x inches in length, and the frame is 3 inches thick. What is the total area of the frame? A)-12x+36 B)12x-36 C)x^2+12x+36 D)x^2-12x-36 Is it C or D? Semester A Unit 4 Lesson 5: Cool Crafts Portfolio Template Choose a room you would like to decorate: _My bedroom________________________________ Measure the length of one wall in inches: _ 12 feet: 144 2. ### Math An artist wants to frame a square painting with an area of 400 square inches. She wants to know the length of the wood trim that is needed to go around the painting. a. If x is the length of one side of the painting, what equation 3. ### math A frame maker made a large picture using 10 feet of frame molding. If the length of the finished frame was 2 feet more than the width, then what were the dimensions of the frame. 4. ### math you a framing a picture that is 5 in by 7in. with a 1 inch wide frame. find the perimeter of the outside edge of the frame. explain, pleaswe i know the anser is 32 but why?

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# Science:Math Exam Resources/Courses/MATH110/April 2014/Question 03 (f) MATH110 April 2014 Other MATH110 Exams ### Question 03 (f) Let ${\displaystyle f(x)={\dfrac {x^{2}}{(x-4)^{2}}}}$. Find the intervals where ${\displaystyle f}$ is concave up and the intervals where it is concave down. Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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Search All of the Math Forum: Views expressed in these public forums are not endorsed by NCTM or The Math Forum. Topic: integration test suite / Chap 3 Replies: 1   Last Post: Apr 27, 2013 7:18 AM Messages: [ Previous | Next ] clicliclic@freenet.de Posts: 1,099 Registered: 4/26/08 Re: integration test suite / Chap 3 Posted: Apr 27, 2013 7:18 AM Axel Vogt schrieb: > > These are the excercises for Chap 3 in Timofeev's book, > p.101 #1 - #3, p.105 #4 - #9, p.109 #10 - #12, p.113 #13, #14 > > Two of them are 'reductions formula' for a linear term, #4 and #5. > For those I have no idea how to test with Maple - they should be > done by partial integration and thus I ignore them. > > The others seem to be for applying decomposition using partial > fractions. For which I suppose, that all CAS do it. There seem > to be no sophisticated cases there, so I just state _some_. > I have converted these to Derive and filled in the gaps. At least one integrand is misprinted, as can be concluded from its antiderivative. My evaluations stay close to Timofeev's ones, which are meant to reveal how they have been arrived at, unless his expressions can be shortened significantly. As predicted, Derive 6.10 has no problems here at all. My file is appended; the two non-examples are represented by []. Martin. " Timofeev (1948) Ch. 3, examples 1 - 3 (p. 101) ... " INT(1/((x-2)^3*(x+1)^2),x)=(2*x^2-5*x-1)/(18*(x+1)*(x-2)^2)+1/27~ *LN((x-2)/(x+1)) INT(1/((x+2)^3*(x+3)^4),x)=(60*x^4+630*x^3+2450*x^2+4175*x+2627)~ /(6*(x+2)^2*(x+3)^3)+10*LN((x+2)/(x+3)) INT(x^5/(3+x)^2,x)=1/4*x^4-2*x^3+27/2*x^2-108*x+243/(x+3)+405*LN~ (x+3) " Timofeev (1948) Ch. 3, examples 4 - 9 (p. 105) ... " [] [] INT(x/(3+6*x+2*x^2),x)=1/4*LN(-(3+6*x+2*x^2))+SQRT(3)/2*ATANH((3~ +2*x)/SQRT(3))=(1/4-SQRT(3)/4)*LN(2*x-SQRT(3)+3)+(SQRT(3)/4+1/4)~ *LN(2*x+SQRT(3)+3) INT((2*x-3)/(3+6*x+2*x^2)^3,x)=-(8*x^3+36*x^2+44*x+13)/(4*(2*x^2~ +6*x+3)^2)+1/SQRT(3)*ATANH((3+2*x)/SQRT(3)) INT((x-1)/(x^2+5*x+4)^2,x)=(7*x+13)/(9*(x^2+5*x+4))+7/27*LN((x+1~ )/(x+4)) INT(1/(x^2+3*x+2)^5,x)=(2*x+3)/(4*(x^2+3*x+2)^4)*(-1+14/3*(x^2+3~ *x+2)-70/3*(x^2+3*x+2)^2+140*(x^2+3*x+2)^3)+70*LN((x+1)/(x+2)) " Timofeev (1948) Ch. 3, examples 10 - 12 (p. 109) ... " INT(1/(x^3*(7-6*x+2*x^2)^2),x)=-1/(98*x^2)-12/(343*x)+2*(41-9*x)~ /(1715*(7-6*x+2*x^2))+80/2401*LN(x)-40/2401*LN(7-6*x+2*x^2)+234*~ SQRT(5)/60025*ATAN((2*x-3)/SQRT(5)) INT(x^9/(x^2+3*x+2)^5,x)=-(25*x^8+35292*x^7+369950*x^6+1632276*x~ ^5+3919731*x^4+5527800*x^3+4578216*x^2+2063520*x+390960)/(24*(x^~ 2+3*x+2)^4)+1472*LN(x+2)-1471*LN(x+1) INT((1+2*x)^2/(3+5*x+2*x^2)^5,x)=-(11+10*x)/(4*(2*x^2+5*x+3)^4)+~ 31*(5+4*x)/(6*(2*x^2+5*x+3)^3)*(1-10*(2*x^2+5*x+3)+120*(2*x^2+5*~ x+3)^2)+2480*LN((x+1)/(2*x+3)) " Timofeev (1948) Ch. 3, examples 13 - 14 (p. 113) ... " INT((a-b*x^2)^3/x^7,x)=-a^3/(6*x^6)+3*a^2*b/(4*x^4)-3*a*b^2/(2*x~ ^2)-b^3*LN(x) INT(x^13/(a^4+x^4)^5,x)=x^2*(15*x^12-73*a^4*x^8-55*a^8*x^4-15*a^~ 12)/(768*a^4*(x^4+a^4)^4)+5/(256*a^6)*ATAN(x^2/a^2) " ... end of Timofeev Ch. 3 " Date Subject Author 4/26/13 Axel Vogt 4/27/13 clicliclic@freenet.de

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# Tagged Questions 74 views 55 views ### Superposition principle If $S=(v_{1},v_{2}......v_{n})$ is a basis for vector Space V, then every vector v in V can be expressed in the form of $v=c_{1}v_{1}+.......c_{n}v_{n}$ in an unique way. Explain the significance of ... 46 views ### Modern physics photoelectric effect [closed] In a photo electric experiment, energy of photon is 5eV incident on a metal surface. They liberate electrons which are just stopped by an electrode at a potential of -3.5V w.r.t the metal. The work ... 33 views ### Uncertainty in the hydrogen ground state [closed] The following question is put forward by one of my students. 1.In spherical polar coordinates, can radial momentum have the form $(h/2\pi i)[\partial/\partial r - 1/r]$? 2. If it is so, what is the ... 36 views ### Finite Square Well Inside an Infinite Square Well Ok here's a potential I invented and am trying to solve: $$V(x) = \begin{cases} -V_0&0<x<b \\ 0&b<x<a \\ \infty&x>a \\ \end{cases}$$ and $V(-x) = V(x)$ (Even ... 69 views ### Idempotent Operators If $P$ is an idempotent operator, $P^2 = P$ and we have a vector $|\psi\rangle$, $P\neq1$, and the relation $$PL |\psi\rangle = L|\psi\rangle,$$ what conclusions can we draw about $L$, which is a ... 79 views ### I am trying to calculate how $<r>$ in the hydrogen atom evolves with time I am working on the Hydrogen atom and I was trying to calculate $\frac{d<r>}{dt}$ using $$\frac{d<r>}{dt} = \frac{i}{\hbar} <[\hat{H} , \hat{r}]>.$$ Here $r = \sqrt(x^2 + y^2 + z^2)$ ... 70 views ### Poles for a particle scattered in a delta potential I am working on problem a professor gave me to get an idea for the research he does, and have hit a point where I'm having a difficult time seeing where I need to go from where I'm at. I would also ... 40 views ### Angular momentum for 3D harmonic oscillator in two different bases I know that the energy eigenstates of the 3D quantum harmonic oscillator can be characterized by three quantum numbers: $$| n_1,n_2,n_3\rangle$$ or, if solved in the spherical coordinate system: ... 41 views ### CPT symmetries for a free Klein-Gordon equation and in minimal coupling I'm studying for an exam on relativistic quantum mechanics and one of the issues to prepare is about symmetries of Klein-Gordon equation concerning $C$, $P$, $T$ transformations for a free field, and ... 91 views ### Divergent solution in time-dependent Schrödinger equation if I transform the time-dependent Schrödinger equation without a potential I get: $$- \hbar \omega \psi(\omega,x) = \frac{- \hbar^2}{2m} \frac{\partial^2 \psi(\omega,x)}{\partial x^2}$$ The ... 32 views 39 views ### Nuclear shell model - finite square well I am trying to make a simplified approximation and solve Schrodinger equation in the finite square well to model the nucleus of Ca (shell nuclear model). The potential is $V(r) = -V_0$ for ... 50 views ### Atom state vectors kets An atom with two energy levels has 2 states (excited and ground), represented by kets $|e\rangle$ and $|g\rangle$ respectively. The atom has energy $\frac{1}{2}E_\theta$ when excited and ... 7 views ### Determing what is differnece beetwen eigenvalues for attractive Coulomb Field (Hydrogen) calculated by exact method and WKB aproach I have task to compare eignevalues gained by exact calculation on electron in Coulomb attractive field (hydrogen) and gained by WKB method.. As far as I get, eigenvalues gained with exact method are ... 38 views 77 views ### 1D Infinite Square Well: Box Suddenly Increases in Size. How treat this? I am currently working my way through John S. Townsend book "A Fundamental Approach to Modern Physics" (ISBN: 978-1-891389-62-7). Exercise 3.12 (p.111) is about the 1D infinite square well. The box ... 48 views ### Center of mass coordinates in Lagrangians and Laplacians Is there a quick nice and easy way to write Lagrangian's and the classical/quantum Laplacian operator in terms of center of mass coordinates? The algebra is so involved and it has me confused about ... 66 views ### What is the eigenvalue of $J_z$? In the calculation of the Zeeman Effect, the most important calculation is $$\langle J_z + S_z\rangle.$$ Suppose we want to find the Zeeman Effect for $(2p)^2$, meaning $l = 1$. In Sakurai's book, ... 60 views 252 views ### Evaluation of expectation values I will denote operators with hats. Suppose we got an operator of the form $i[\hat p, \tan^{-1}(e^{\hat x})]$ and we want to calculate the amplitude for a transition from a state $|p_i\rangle$ to the ... 73 views ### How to quantize a rubber band stretched between two poles? [closed] Consider a classical non-relativistic material string obeying Hooke's law stretched between two poles with either Neumann or Dirichlet or periodic boundary conditions and subject to either traversal ... 54 views ### Galilean Transform I tried to solve a problem using two different ways and I had some trouble, the problem is: We define a symmetry transform of the expected value of $\vec{P}$ like this: \langle \psi|\vec{P}|\psi ... 82 views ### How does $p_x$ commute with $p_y$, i.e. $[p_x,p_y]=0$? [closed] I know it's a simple and basic question but would someone show me how to evaluate $[\hat{p}_x,\hat{p}_y]$?

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# Replace an If by a combination using replacement rules Consider the following code : tttest = a^2 + If[a > 0, a, -a] a^2 + If[a > 0, a, -a] I would like to replace my If function by something like fonction @ If. I did the following, but the replacement doesn't occur. Replace[tttest, If -> (fonction @ If)] a^2 + If[a > 0, a, -a] How to make the replacement working and why isn't it working here ? For me it is an example of the same kind as the one in the documentation : Replace[x^2, x^2 -> a + b] : as suggested by the comment, I switched to ReplaceAll and I wrote the following : (the example is slightly different) ReplaceAll[If[lambda + lambdaBis != 0, PM[m] Log[PM[m]], 0], If[x1_, x2_, x3_] -> fonction [If[x1, x2, x3]]] fonction[If[lambda + lambdaBis != 0, PM[m] Log[PM[m]], 0]] And here it works. However, I want to actually simplify an expression linked to this question I asked Why is the function assuming not taken in consideration? I did the following : Assuming[lambda00 > 0 && lambda00Bis , ReplaceAll[If[lambda + lambdaBis != 0, PM[m] Log[PM[m]], 0], If[a1_, a2_, a3_] -> (gggg [If[a1, a2, a3]])]] gggg[If[lambda + lambdaBis != 0, PM[m] Log[PM[m]], 0]] Here everything shows up correctly, but if actually my function gggg is Simplify, nothing is simplified (so the function "doesnt work" here). Assuming[lambda00 > 0 && lambda00Bis , ReplaceAll[If[lambda + lambdaBis != 0, PM[m] Log[PM[m]], 0], If[a1_, a2_, a3_] -> (Simplify [If[a1, a2, a3]])]] If[lambda + lambdaBis != 0, PM[m] Log[PM[m]], 0] Why ??? • Use ReplaceAll. – Alan Commented Dec 30, 2018 at 16:13 • Or you can Replace[tttest, If -> (fonction@If), Infinity, Heads -> True] – Alan Commented Dec 30, 2018 at 16:19 • @Alan I edited my message Commented Dec 30, 2018 at 16:24 • About your edit: use RuleDelayed. – Alan Commented Dec 30, 2018 at 16:55 You are trying to match an expression with Head If. Therefore you need to use a pattern that will match that expression head. See the Patterns tutorial. You should also make use of RuleDelayed since you need to reference a named pattern from your replacement rule. With tttest = a^2 + If[a > 0, a, -a]; Then tttest /. if_If :> fonction@if a^2+fonction[If[a>0,a,-a]] Hope this helps.

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CZECH TECHNICAL UNIVERSITY IN PRAGUE STUDY PLANS 2022/2023 UPOZORNĚNÍ: Jsou dostupné studijní plány pro následující akademický rok. # Probability and Statistics The course is not on the list Without time-table Code Completion Credits Range Language BIE-PST Z,ZK 5 2P+1R+1C English Garant předmětu: Lecturer: Tutor: Supervisor: Department of Applied Mathematics Synopsis: The students will learn the basics of probabilistic thinking, the ability to synthesize prior and posterior information and learn to work with random variables. They will be able to to apply basic models of random variable distributions and solve applied probabilistic problems in informatics and computer science. Using the statistical induction they will be able to perform estimations of unknown distributional parameters from random sample characteristics. They will also be introduced to the methods of determining the statistical dependence of two or more random variables. Requirements: Basics of combinatorics and mathematical analysis. Syllabus of lectures: 1. Probability: Random events, event space structure, probability of a random event and its basic properties. 2. Conditional probability: Dependent and independent events, Bayes theorem. 3. Random variables: Distribution function of a random variable, continuous and discrete distributions, quantiles, median. 4. Characteristics of random variables: Expected value, variance, general moments, kurtosis and skewness. 5. Overview of basic distributions: binomial, geometric, Poisson, uniform, normal, exponential. Their basic properties. 6. Random vectors: Joint and marginal statistics, correlation coefficient, dependence and independence of random variables. 7. Random vectors: Conditional distributions, sums of random variables. 8. Limit theorems: Laws of large numbers, central limit theorem 9. Statistical estimation: Classification and processing of data sets, characteristics of position, variance and shape, sampling moments, graphical representation of data. 10. Point estimation: Random sample, basic sample statistics, sample mean and variance, distributions (t-distribution, F-distribution, chi square). 11. Interval estimation: Confidence intervals for expectation and variance. 12. Hypothesis testing: Testing strategy, tests for expectation and variance, their modifications. Application of statistical testing in CS. 13. Correlation and regression analysis: Linear and quadratic regression, sample correlation. Syllabus of tutorials: 1. Basics of probability. 2. Conditional probability. 3. Random variables. 4. Basic characteristics of random variables. 5. Using basic distributions. 6. Random vectors - independence, covariance. 7. Random vectors - conditional distributions and sums. 8. Limit theorems 9. Processing of sets of data. 10. Statistical point estimation. 11. Interval estimation. 12. Hypotheses testing. 13. Regression and correlation analysis. Study Objective: The goal of the module is to introduce the students to basics of probability theory and mathematical statistics while focusing on applications in informatics. Study materials: 1. Johnson, J. L. ''Probability and Statistics for Computer Science''. Wiley-Interscience, 2008. ISBN 0470383429. 2. Li, X. R. ''Probability, Random Signals, and Statistics''. CRC, 1999. ISBN 0849304334. Note: Further information: https://courses.fit.cvut.cz/BIE-PST/ No time-table has been prepared for this course The course is a part of the following study plans: Data valid to 2023-03-18 Aktualizace výše uvedených informací naleznete na adrese https://bilakniha.cvut.cz/en/predmet1448106.html

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# An Intro to arrays A way to control more LEDs than you have pins in your chip! Update: I added a video with my grad cap hacked with 64 individually addressed LEDs! This multiplexing/matrixing/array-ing sure is fun :3 This is pretty barebones so far. Be assured, more infomation will be added as I push along. First I started with a really basic setup: 16 LEDs controlled by 16 pins. This is great, but leaves no pins left for output! Being limited makes me antsy, so I sat down and figured out how to get more. Here is the code(first video): main: for b0 = 0 to 7 high b0 pause 250 low b0 next b0 for b0 = 0 to 7 high pinsc b0 pause 250 low pinsc b0 next b0 goto main Controlling 16 separate motors/lights/etc. is great and all, but what if you want to do more? I removed one 8 LED output bank and set up two of the free pins as "enablers". The 8 led pins are pulldown pins, and each enabler pin supplies a logic on/off to a 2N3904 Transistor to supply a +5 voltage. A small change in code, and we are back in business! (second video) main: high pinsc 0 for b0 = 0 to 7 high b0 pause 250 low b0 next b0 low pinsc 0 high pinsc 1 for b0 = 0 to 7 high b0 pause 250 low b0 next b0 low pinsc 1 goto main 10 pins to control 16 LEDs? Cool! This can be worked further. How many can we control with 16 pins? With mathematics, all you have to do take the total number of pins you can use, divide by two, and then square that number. With 16 pins: 16/2=8, 8^2=64! That's pretty sweet if you ask me. This is just with one Picaxe 28x1 chip, what could we use to multiply the out pins with other chips to control a multitude more? ## Comment viewing options So many new chips to experiment with! This walkthrough was mainly to highlight using less pins to get more out with only one chip. Just interesting how many different ways there are to achive this :) STOP RESISTING! You want more chips! My school as drilled into me finding the cheapest way to achive something. Companies look for that, so I keep training myself that way. If I wasn't working towards a career in designing circuits, I'd say "to hell with it" and go the easy route :P For added fun, check out Microchips MCP23016, 23017 or 23018, which is 16port I/O expanders for I2C, capable of 3 different speeds. It also exists in a serial version, which can run faster than the I2C editions: http://www.microchip.com/ParamChartSearch/chart.aspx?branchID=11034&mid=11&lang=en&pageId=79 Also, you can check out the 74HC595, which is an 8-bit serial latch, but they are output only. There is an equivalent input chip, which I can't remember the name of. Just the way I like'm! 8ik Thankies :) I can't stand messes on breadboards... so hard to troubleshoot! 16 leds on each 4 to 16 decoder one  4 bit counter to control each decoder   (two counters in each 7459 package)

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# What is Lap Length? What is Lap Length? The lap length is the length provided to overlap two rebars in order to safely transfer load from one bar to another bar and alternative to this is to provide mechanical couplers. Lap length also known as lap splices. Suppose we need to construct the building of 20-metre height but there is no 20metre single bar available in the market. The maximum length of Steel bar available in the market is usually 12 metre. Why it is provided? Suppose we need to construct the building of 30metre height but there is no 20metre single bar available in the market. The maximum length of Steel bar available in the market is usually 12 metre. (Why12meter?)It is because of the transportation problem and manufacturing difficulty so we need to join three bars of 12 metre to get 30metre bar. What will happen if we don't provide lap length? If we don't provide lap length then the load transfer mechanism will fail which eventually lead to failure of structure also if we provide less lap length then the required then the reinforcement bar split and Crack can be developed in the concrete. Now let us understand How to calculate lap length? The calculation for tension zone and compression zone are different. Let us take the case of the beam when the beam is subjected to forces in a building the bottom portion of beam experiences tension and top portion of beam experiences compression so, first we will discuss about tension zone. In tension zone, there are two cases • one flexural tension • direct tension For flexural tension, the lap length shall be Ld that is development length or 30d whichever is Greater is considered. Where d is the diameter of the bar. Generally, development length is 41d where d is the diameter of the bar. To know more about development length - What is Development Length? for direct tension, the lap length should be 2 Ld or 30d whichever is greater is considered. In this case, the straight length of the lapping bar shall not be less than 15d or 20cm. Lap length in compression In the case of compression, the lap length is equal to the development length calculated in compression but not less than 24d. What are the general rules for lap length? For the different diameter of bars When the bars of different diameters are to be spliced the lap length is calculated considering the smaller diameter bars. Suppose you are constructing a column, from bottom 20 mm diameter bar is coming and from here 16 mm diameter bar has to be spliced then for calculating lap length 16 mm diameter should be considered and not 20 mm. if the diameter of the bar is more than 36 mm then lapping should not be done instead of lapping this bar should be welded but when welding is not possible then lapping can be permitted for bars larger than 36 mm but in this case, additional spiral shall be provided around the lapped bar. Lapping should be done in a staggered manner. These laps should not be given at the same level to avoid buckling. The stirrup shall be closely spaced in the lapping portion it's because when we provide lapping in concrete member the strength of member slightly reduces. Hence, we need to provide more numbers of stirrups in this portion. In case of bundled bars, lap splices of bundle bars shall be made by splicing one bar at a time. Such individual splices within a bundle shall be staggered. In this image you can see some amount of rebar is left for future construction with extra rebar will be needed for tying bars of the column. This extra length of rebar is also called as lap length. Lapping zone This is the column. L is the length of the column. In the case of column the tension zone is located at L/4 distance from both ends of the column. This zone experiences tension so here we should not provide lapping. The bending moment at the middle portion of the column is zero it means the middle portion of the column is least stressed. Hence, lapping should be provided in the mid-section of the column so that transfer of stresses from bar to bar happen smoothly in this region. whereas in case of the beam as I already explained before the top portion of beam experiences compression and bottom portion experience tension. So, the top reinforcement in the beam is left at midspan. As the beam does not experience any negative moment at midspan so lapping is great in this region. In the case of bottom reinforcement the lapping is provided near the ends of the beam or L/4  distance from column face but should not be in the midpoint of the beam and one last point The lapping should not be provided at joints.

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Search This Blog IT2201 DATA STRUCTURES AND ALGORITHMS ANNA UNIVERSITY QUESTION PAPER Saturday, September 24, 2011 · ANNA UNIVERSITY QUESTION PAPER FOR IT2201 DATA STRUCTURES AND ALGORITHMS B.E./B.Tech. DEGREE EXAMINATION, NOVEMBER/DECEMBER 2010 Third Semester Information Technology IT 2201 — DATA STRUCTURES AND ALGORITHMS (Regulation 2008) Time : Three hours Maximum : 100 Marks PART A — (10 × 2 = 20 Marks) 2. Clearly distinguish between linked lists and arrays. Mention their relative 3. What is meant by depth and height of a tree? 4. Discuss the application of trees. 5. What are the important factors to be considered in designing the hash function? 6. What is a disjoint set? Define the ADT for a disjoint set. 7. What is Euler circuit? 8. What are the two ways of representing a graph? Give examples. 9. Define NP-complete problems. 10. What is meant by backtracking? PART B — (5 × 16 = 80 Marks) 11. (a) (i) Derive an ADT to perform insertion and deletion in a singly linked list. (8) (ii) Explain cursor implementation of linked lists. Write the essential operations. (8) Or (b) (i) Write an ADT to implement stack of size N using an array. The elements in the stack are to be integers. The operations to be supported are PUSH, POP and DISPLAY. Take into account the exceptions of stack overflow and stack underflow. (8) (ii) A circular queue has a size of 5 and has 3 elements 10, 20 and 40 where F = 2 and R = 4. After inserting 50 and 60, what is the value of F and R. Trying to insert 30 at this stage what happens? Delete 2 elements from the queue and insert 70, 80 & 90. Show the sequence of steps with necessary diagrams with the value of F & R. (8) 12. (a) (i) Write an ADT to construct an AVL tree. (8) (ii) Suppose the following sequences list nodes of a binary tree T in preorder and inorder, respectively : Preorder : A, B, D, C, E, G, F, H, J Inorder : D, B, A, E, G, C, H, F, J Draw the diagram of the tree. (8) Or (b) (i) Write an ADT for performing insert and delete operations in a Binary Search Tree. (8) (ii) Describe in detail the binary heaps. Construct a min heap tree for the following : 5, 2, 6,7, 1, 3, 8, 9, 4 (8) 13. (a) (i) Formulate an ADT to implement separate chaining hashing scheme. (8) (ii) Show the result of inserting the keys 2, 3, 5, 7, 11, 13, 15, 6, 4 into an initially empty extendible hashing data structure with M = 3. (8) Or (b) (i) Formulate an ADT to perform for the Union and find operations of disjoint sets. (8) (ii) Describe about Union-by-rank and Find with path compression with code. (8) 14. (a) (i) Write routines to find shortest path using Dijkstra’s algorithm. (8) (ii) Find all articulation points in the below graph. Show the depth first spanning tree and the values of DFN and Low for each vertex. (8) Or (b) (i) Write the pseudo code to find a minimum spanning tree using Kruskal’s algorithm. (8) (ii) Find the topological ordering of the below graph. (8) 15. (a) (i) Discuss the running time of Divide-and-Conquer merge sort algorithm. (8) (ii) Explain the concept of greedy algorithm with Huffman codes example. (8) Or (b) Explain the Dynamic Programming with an example. (16)

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# ISBLANK formula not working!! #### hoyos ##### Board Regular I’m trying to get the formula below to work but having no luck. It should flag up if after a run of three night shift (18:00pm to 03:00am) if I enter a start time for a new shift anytime under 34 hours after the end of the last night shift. R11 if( ISBLANK(D11) = TRUE,"",IF(D13<>Sheet1!\$A\$2,(IF(IF(AND(D10=Sheet1!\$A\$2,D11=Sheet1!\$A\$2,D12=Sheet1!\$A\$2),1,0) = 1,"Blank",IF(AND(D9=Sheet1!\$A\$2,D10=Sheet1!\$A\$2,D11=Sheet1!\$A\$2),IF(D13 <= E11 + TIME(34,0,0),"No","Sure"),"Blank2"))),(IF(IF(AND(D10=Sheet1!\$A\$2,D11=Sheet1!\$A\$2,D12=Sheet1!\$A\$2),1,0) = 1,"No","Yes")))) Sheet1!\$A\$2 is populated with the shift start time of 18:00 as a reference. Column D= start time of shift Column C= end time of shift Column R= formula Any help would be appreciated ### Excel Facts Format cells as time Select range and press Ctrl+Shift+2 to format cells as time. (Shift 2 is the @ sign). #### jasonb75 ##### Well-known Member As you are exceeding 24 hours you will need to have dates as well as times in your sheet. We're going to need a visual example of your sheet in order to come up with anything that will do what you need, a description like that is open to a lot of misinterpretation. You can use the XL2BB add-in (<<link) to attach a mini sheet to the forum (10-20 rows of data, not the whole sheet! Only the columns that are required for the formula, so A:E based on your formula. Create the example on a new sheet if necessary and include the expected output of the formula in column F. Also, please update your account details (<<link) to show the version of excel that you are using so that we know which functions you can use. #### hoyos ##### Board Regular Hi, I think I have manage to use XL2Bb I think I managed!! FTL Matt Wilcock_New edition 10.xlsm ABCDEFGHIJKLMNOPQRS 1Matt WilcockDuty and Flying Hours Record SheetFrom21 Sep to 18 Oct 2022 2 3DUTY HOURSFLYING HOURSOFFHave I had 2 off consec in 14More then 4 lates in 7More then 3 consec lates34hrs off after late finish before early shift 4Day NoDayDateDuty Period28 Day Total7 Day TotalFlight Times28 Day Total7 Day Total14 Day TotalDuty / Remarks 5StartFinishHoursFirst T/OLast LdgFly Hrs 61Wed21-Sep18:0003:009:00171:0057:000:000:004433 HEMS Late 72Thu22-Sep171:0057:000:000:005533 OFF 83Fri23-Sep171:0047:000:000:006633 OFF 94Sat24-Sep161:0037:000:000:007733 OFF 105Sun25-Sep08:0018:0010:00161:0037:000:000:007733 HEMS Early 116Mon26-Sep08:0018:0010:00161:0038:000:000:006622 HEMS Early 127Tue27-Sep08:0018:0010:00162:0039:000:000:005511 HEMS Early 138Wed28-Sep18:0003:009:00162:0057:000:000:004411 HEMS Late 149Thu29-Sep18:0003:009:00162:0048:000:000:003322 HEMS Late 1510Fri30-Sep18:0003:009:00171:0057:000:000:003333YESHEMS Late 1611Sat1-Oct171:0057:000:000:004433 OFF 1712Sun2-Oct15:0018:003:00174:0050:000:000:004433NOHEMS Early 1813Mon3-Oct174:0040:000:000:005533 OFF 1914Tue4-Oct164:0030:000:000:006633 OFF 2015Wed5-Oct08:0018:0010:00164:0031:000:000:006622BLANK2HEMS Early 2116Thu6-Oct08:0018:0010:00164:0032:000:000:005511YESHEMS Early 2217Fri7-Oct08:0018:0010:00165:0033:000:000:004400YESHEMS Early 2318Sat8-Oct18:0003:009:00165:0042:000:000:003311YESHEMS Late 2419Sun9-Oct18:0003:009:00165:0048:000:000:003322BLANKHEMS Late 2520Mon10-Oct18:0003:009:00174:0057:000:000:003333NOHEMS Late 2621Tue11-Oct174:0057:000:000:004433 OFF 2722Wed12-Oct174:0047:000:000:005533 OFF 2823Thu13-Oct174:0037:000:000:006633 OFF 2924Fri14-Oct164:0027:000:000:007733 OFF 3025Sat15-Oct08:0018:0010:00164:0028:000:000:006622BLANK2HEMS Early 3126Sun16-Oct08:0018:0010:00164:0029:000:000:006611YESHEMS Early 3227Mon17-Oct08:0018:0010:00165:0030:000:000:005500BLANK2HEMS Early 3328Tue18-Oct18:0003:009:00165:0039:000:000:004411BLANK2HEMS Late 34Totals this period:Leave:0Sick:0Off:10 Matt Wilcock Cell Formulas RangeFormula M1M1=C6 S1S1=" to " & TEXT(C33,"d mmm yyyy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ells with Conditional Formatting CellConditionCell FormatStop If True Q6:Q33Cell Value>3textNO P6:P33Cell Value>4textNO H6:H33Cell Value>2.5textYES G6:G33Cell Value>8.33333333333333textYES H6:H33Cell Value>2.5textYES L6:L33Cell Value>90textYES M6:M33Cell Value>30textYES G6:G33Cell Value>200textYES H6:H33Cell Value>60textYES #### hoyos ##### Board Regular Here is a smaller version FTL Matt Wilcock_New edition 10.xlsm ABCDEFGHIJKLMNOPQRS 5StartFinishHoursFirst T/OLast LdgFly Hrs 61Wed21-Sep18:0003:009:00171:0057:000:000:004433 HEMS Late 72Thu22-Sep171:0057:000:000:005533 OFF 83Fri23-Sep171:0047:000:000:006633 OFF 94Sat24-Sep161:0037:000:000:007733 OFF 105Sun25-Sep08:0018:0010:00161:0037:000:000:007733 HEMS Early 116Mon26-Sep08:0018:0010:00161:0038:000:000:006622 HEMS Early 127Tue27-Sep08:0018:0010:00162:0039:000:000:005511 HEMS Early 138Wed28-Sep18:0003:009:00162:0057:000:000:004411 HEMS Late 149Thu29-Sep18:0003:009:00162:0048:000:000:003322 HEMS Late 1510Fri30-Sep18:0003:009:00171:0057:000:000:003333YESHEMS Late 1611Sat1-Oct171:0057:000:000:004433 OFF 1712Sun2-Oct15:0018:003:00174:0050:000:000:004433NOHEMS Early 1813Mon3-Oct174:0040:000:000:005533 OFF 1914Tue4-Oct164:0030:000:000:006633 OFF Matt Wilcock Cell Formulas RangeFormula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ells with Conditional Formatting CellConditionCell FormatStop If True Q6:Q33Cell Value>3textNO P6:P33Cell Value>4textNO H6:H33Cell Value>2.5textYES G6:G33Cell Value>8.33333333333333textYES H6:H33Cell Value>2.5textYES L6:L33Cell Value>90textYES M6:M33Cell Value>30textYES G6:G33Cell Value>200textYES H6:H33Cell Value>60textYES #### hoyos ##### Board Regular Jason, from my original post I have replaced "Sheet1!\$A\$2" with \$AS\$6 #### jasonb75 ##### Well-known Member That is all well and good but \$A\$6 contains a value of 1 which doesn't appear to be comparable to anything else that the formula refers to. The formula uses D2, D3, D4, and E4, none of which are in the smaller sheet. Looking at the larger sheet, none of the cells in the formula appear to contain anything that can be compared to anything else. Gonna need a bit of an explanation as to what they all refer to before I can do anything. #### hoyos ##### Board Regular Sorry my mistake, it should read \$AS\$6 which is populated with the time 18:00:00. The formula should look at col D and pick out three consecutive duties starting at 18:00hrs. It then looks at the last duty finish time in col E, there should be at least a minimum 34 hours off duty. If the user inserts a start time (col D) within the 34 hours a warning in col R same row should flag up with a red background with the word “NO” in black. I hope that makes sense. #### jasonb75 ##### Well-known Member What do you have in AF16 and AF18? There is other missing data but I don't think that it will affect the formulas that are needed for the question. Replies 5 Views 1K Replies 6 Views 96 Replies 3 Views 93 Replies 0 Views 292 Replies 23 Views 187 Excel contains over 450 functions, with more added every year. That’s a huge number, so where should you start? Right here with this bundle. 1,151,972 Messages 5,767,400 Members 425,410 Latest member SmittyT ### We've detected that you are using an adblocker. We have a great community of people providing Excel help here, but the hosting costs are enormous. You can help keep this site running by allowing ads on MrExcel.com. ### Which adblocker are you using? 1)Click on the icon in the browser’s toolbar. 2)Click on the icon in the browser’s toolbar. 2)Click on the "Pause on this site" option. Go back 1)Click on the icon in the browser’s toolbar. 2)Click on the toggle to disable it for "mrexcel.com". Go back ### Disable uBlock Origin Follow these easy steps to disable uBlock Origin 1)Click on the icon in the browser’s toolbar. 2)Click on the "Power" button. 3)Click on the "Refresh" button. 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Portal | Manuals | References | Downloads | Info | Programs | JCLs | Mainframe wiki | Quick Ref Author Message hsiva2003 New User Joined: 22 Feb 2005 Posts: 10 Location: Chennai Posted: Wed Nov 08, 2006 1:07 pm    Post subject: Move a variable of a pic clause to dif pic clause I want to comapare x & y of different pic clauses. x pic 99v9(6) & y pic s9v9(8)comp-3. If I want to move y value to x without any truncation - ...I don't want to change the x's pic clause... Need a suggestion guptae Moderator Joined: 14 Oct 2005 Posts: 1190 Location: Bangalore,India Posted: Wed Nov 08, 2006 2:02 pm    Post subject: Hi There, U dont want to change pic claue of x & u want the value without truncation how can it possible? vijay_bn79 New User Joined: 20 Nov 2006 Posts: 48 Posted: Fri Nov 24, 2006 5:31 pm    Post subject: Re: move a variable of a pic clause to variable of dif pic c x pic 99v9(6) & y pic s9v9(8)comp-3. MOVE VALUES FROM Y TO X LIKE I mean 1 interger and 6 decimal Y = 2.666666 if you move Y to X then X will have the value 02.666666 pls correct me if i am wrong Arun Raj Moderator Joined: 17 Oct 2006 Posts: 2352 Location: @my desk Posted: Fri Nov 24, 2006 5:52 pm    Post subject: Hi Siva It is not possible to do the move operation you specified without truncation.Anyway you will lose precision by 2 decimal points. Thanks Arun William Thompson Global Moderator Joined: 18 Nov 2006 Posts: 3158 Location: Tucson AZ Posted: Fri Nov 24, 2006 5:58 pm    Post subject: You could reduce the degree of truncation by multiplying y by 10 giving x.....only the least significant digit would be lost.... All times are GMT + 6 Hours Page 1 of 1 Search our Forum: Topic Author Forum Replies Posted Similar Topics Query to compare 2 values of 1 column... Poha Eater DB2 13 Fri Mar 09, 2018 10:45 am RMM Cannot move a volume from SHELF l... tspr52 IBM Tools 0 Thu Mar 01, 2018 3:48 pm Moving a COMP-3 Variable to a Numeric... ajayachander COBOL Programming 2 Thu Dec 14, 2017 5:46 pm Date in where clause - Windows Karthikeyan Subbarayan DB2 9 Wed Nov 15, 2017 9:07 pm how can i move s9(9) to s9(9) usage comp HARENDRA CHOUDHARY COBOL Programming 3 Mon Nov 06, 2017 12:10 am © 2003-2017 IBM MAINFRAME Software Support Division Job Vacancies | Forum Rules | Bookmarks | Subscriptions | FAQ | Polls | Contact Us

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Book 1 - Proposition 45 To construct a parallelogram equal to a given rectilinear figure in a given rectilinear angle Let ABCD be the given rectilinear figure and angle STV the given rectilinear angle. It is required to construct a parallelogram equal to the rectilinear figure ABCD in the given angle angle STV. Join DB. Construct the parallelogram FH equal to the triangle ABD in the angle HKF which equals angle STV. Apply the parallelogram GM equal to the triangle DBC to the straight line GH in the angle GHM which equals angle STV. Since the angle STV equals each of the angles angle HKF and angle GHM, therefore the angle HKF also equals the angle GHM. Add the angle KHG to each. Therefore the sum of the angles FKH and KHG equals the sum of the angles KHG and GHM. But the sum of the angles FKH and angle KHG equals two right angles, therefore the sum of the angles KHG and GHM also equals two right angles. Thus, with a straight line GH, and at the point H on it, two straight lines KH and HM not lying on the same side make the adjacent angles together equal to two right angles, therefore KH is in a straight line with HM. Since the straight line HG falls upon the parallels KM and FG, therefore the alternate angles MHG and HGF equal one another. Add the angle HGL to each. Then the sum of the angles MHG and HGL equals the sum of the angles HGF and HGL. But the sum of the angles MHG and HGL equals two right angles, therefore the sum of the angles HGF and HGL also equals two right angles. Therefore FG is in a straight line with GL. Since FK is equal and parallel to HG, and HG equal and parallel to ML also, therefore KF is also equal and parallel to ML, and the straight lines KM and FL join them at their ends. Therefore KM and FL are also equal and parallel. Therefore KFLM is a parallelogram. Since the triangle ABD equals the parallelogram FH, and triangle DBC equals GM, therefore the whole rectilinear figure ABCD equals the whole parallelogram KFLM. Therefore the parallelogram KFLM has been constructed equal to the given rectilinear figure True in the angle FKM which equals the given angle STV. Dark Light

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# Pentagon Regular pentagon A regular pentagon Type Regular polygon Edges and vertices 5 Schläfli symbol {5} Coxeter diagram Symmetry group Dihedral (D5), order 2×5 Internal angle (degrees) 108° Dual polygon Self Properties Convex, cyclic, equilateral, isogonal, isotoxal In geometry, a pentagon (from the Greek πέντε pente and γωνία gonia, meaning five and angle[1]) is any five-sided polygon or 5-gon. The sum of the internal angles in a simple pentagon is 540°. A pentagon may be simple or self-intersecting. A self-intersecting regular pentagon (or star pentagon) is called a pentagram. ## Regular pentagons A regular pentagon has Schläfli symbol {5} and interior angles are 108°. A regular pentagon has five lines of reflectional symmetry, and rotational symmetry of order 5 (through 72°, 144°, 216° and 288°). The diagonals of a convex regular pentagon are in the golden ratio to its sides. Its height (distance from one side to the opposite vertex) and width (distance between two farthest separated points, which equals the diagonal length) are given by where R is the radius of the circumcircle. The area of a convex regular convex pentagon with side length t is given by A pentagram or pentangle is a regular star pentagon. Its Schläfli symbol is {5/2}. Its sides form the diagonals of a regular convex pentagon – in this arrangement the sides of the two pentagons are in the golden ratio. When a regular pentagon is inscribed in a circle with radius R, its edge length t is given by the expression and its area is since the area of the circumscribed circle is the regular pentagon fills approximately 0.7568 of its circumscribed circle. ### Derivation of the area formula The area of any regular polygon is: where P is the perimeter of the polygon, and r is the inradius (equivalently the apothem). Substituting the regular pentagon's values for P and r gives the formula with side length t. Like every regular convex polygon, the regular convex pentagon has an inscribed circle. The apothem, which is the radius r of the inscribed circle, of a regular pentagon is related to the side length t by ### Chords from the circumscribed circle to the vertices Like every regular convex polygon, the regular convex pentagon has a circumscribed circle. For a regular pentagon with successive vertices A, B, C, D, E, if P is any point on the circumcircle between points B and C, then PA + PD = PB + PC + PE. ### Construction of a regular pentagon The regular pentagon is constructible with compass and straightedge, as 5 is a Fermat prime. A variety of methods are known for constructing a regular pentagon. Some are discussed below. #### Richmond's method One method to construct a regular pentagon in a given circle is described by Richmond[2] and further discussed in Cromwell's "Polyhedra."[3] The top panel shows the construction used in Richmond's method to create the side of the inscribed pentagon. The circle defining the pentagon has unit radius. Its center is located at point C and a midpoint M is marked halfway along its radius. This point is joined to the periphery vertically above the center at point D. Angle CMD is bisected, and the bisector intersects the vertical axis at point Q. A horizontal line through Q intersects the circle at point P, and chord PD is the required side of the inscribed pentagon. To determine the length of this side, the two right triangles DCM and QCM are depicted below the circle. Using Pythagoras' theorem and two sides, the hypotenuse of the larger triangle is found as . Side h of the smaller triangle then is found using the half-angle formula: where cosine and sine of ϕ are known from the larger triangle. The result is: With this side known, attention turns to the lower diagram to find the side s of the regular pentagon. First, side a of the right-hand triangle is found using Pythagoras' theorem again: Then s is found using Pythagoras' theorem and the left-hand triangle as: The side s is therefore: a well-established result.[4] Consequently, this construction of the pentagon is valid. #### Carlyle circles See main article: Carlyle circle Method using Carlyle circles The Carlyle circle was invented as a geometric method to find the roots of a quadratic equation.[5] This methodology leads to a procedure for constructing a regular pentagon. The steps are as follows:[6] 1. Draw a circle in which to inscribe the pentagon and mark the center point O. 2. Draw a horizontal line through the center of the circle. Mark the left intersection with the circle as point B. 3. Construct a vertical line through the center. Mark one intersection with the circle as point A. 4. Construct the point M as the midpoint of O and B. 5. Draw a circle centered at M through the point A. Mark its intersection with the horizontal line (inside the original circle) as the point W and its intersection outside the circle as the point V. 6. Draw a circle of radius OA and center W. It intersects the original circle at two of the vertices of the pentagon. 7. Draw a circle of radius OA and center V. It intersects the original circle at two of the vertices of the pentagon. 8. The fifth vertex is the rightmost intersection of the horizontal line with the original circle. Steps 6–8 are equivalent to the following version, shown in the animation: 6a. Construct point F as the midpoint of O and W. 7a. Construct a vertical line through F. It intersects the original circle at two of the vertices of the pentagon. The third vertex is the rightmost intersection of the horizontal line with the original circle. 8a. Construct the other two vertices using the compass and the length of the vertex found in step 7a. #### Using trigonometry and the Pythagorean Theorem Using trigonometry and the Pythagorean Theorem to construct a regular pentagon. ##### The construction 1. We first note that a regular pentagon can be divided into 10 congruent triangles as shown in the Observation. Also, cos 36° = . 2. In Step 1, we use four units (shown in blue) and a right angle to construct a segment of length 1+5, specifically by creating a 1-2-5 right triangle and then extending the hypotenuse of 5 by a length of 1. We then bisect that segment – and then bisect again – to create a segment of length (shown in red.) 3. In Step 2, we construct two concentric circles centered at O with radii of length 1 and length . We then place P arbitrarily on the smaller circle, as shown. Constructing a line perpendicular to OP passing through P, we construct the first side of the pentagon by using the points created at the intersection of the tangent and the unit circle. Copying that length four times along the outer edge of the unit circles gives us our regular pentagon. #### † Proof that cos 36° = (using the angle addition formula for cosine) (using double and half angle formulas) Let u = cos 36°. First, note that 0 < u < 1 (which will help us simplify as we work.) Now, This follows quickly from the knowledge that twice the sine of 18 degrees is the reciprocal golden ratio, which we know geometrically from the triangle with angles of 72,72,36 degrees. From trigonometry, we know that the cosine of twice 18 degrees is 1 minus twice the square of the sine of 18 degrees, and this reduces to the desired result with simple quadratic arithmetic. #### Side length is given The regular pentagon according to the golden ratio, dividing a line segment by exterior division Pentagon at a given side length 1. Draw a segment AB whose length is the given side of the pentagon. 2. Extend the segment BA from point A about three quarters of the segment BA. 3. Draw an arc of a circle, centre point B, with the radius AB. 4. Draw an arc of a circle, centre point A, with the radius AB; there arises the intersection F. 5. Construct a perpendicular to the segment AB through the point F; there arises the intersection G. 6. Draw a line parallel to the segment FG from the point A to the circular arc about point A; there arises the intersection H. 7. Draw an arc of a circle, centre point G with the radius GH to the extension of the segment AB; there arises the intersection J. 8. Draw an arc of a circle, centre point B with the radius BJ to the perpendicular at point G; there arises the intersection D on the perpendicular, and the intersection E with the circular arc that was created about the point A. 9. Draw an arc of a circle, centre point D, with the radius BA until this circular arc cuts the other circular arc about point B; there arises the intersection C. 10. Connect the points BCDEA. This results in the pentagon. ### Euclid's method Euclid's method for pentagon at a given circle, using of the golden triangle, animation 1 min 39 s A regular pentagon is constructible using a compass and straightedge, either by inscribing one in a given circle or constructing one on a given edge. This process was described by Euclid in his Elements circa 300 BC.[7][8] #### Simply using a protractor (not a classical construction) A direct method using degrees follows: 1. Draw a circle and choose a point to be the pentagon's (e.g. top center) 2. Choose a point A on the circle that will serve as one vertex of the pentagon. Draw a line through O and A. 3. Draw a guideline through it and the circle's center 4. Draw lines at 54° (from the guideline) intersecting the pentagon's point 5. Where those intersect the circle, draw lines at 18° (from parallels to the guideline) 6. Join where they intersect the circle After forming a regular convex pentagon, if one joins the non-adjacent corners (drawing the diagonals of the pentagon), one obtains a pentagram, with a smaller regular pentagon in the center. Or if one extends the sides until the non-adjacent sides meet, one obtains a larger pentagram. The accuracy of this method depends on the accuracy of the protractor used to measure the angles. ### Physical methods Overhand knot of a paper strip • A regular pentagon may be created from just a strip of paper by tying an overhand knot into the strip and carefully flattening the knot by pulling the ends of the paper strip. Folding one of the ends back over the pentagon will reveal a pentagram when backlit. • Construct a regular hexagon on stiff paper or card. Crease along the three diameters between opposite vertices. Cut from one vertex to the center to make an equilateral triangular flap. Fix this flap underneath its neighbor to make a pentagonal pyramid. The base of the pyramid is a regular pentagon. ### Symmetry Symmetries of a regular pentagon. Vertices are colored by their symmetry positions. Blue mirror lines are drawn through vertices and edges. Gyration orders are given in the center. The regular pentagon has Dih5 symmetry, order 10. Since 5 is a prime number there is one subgroup with dihedral symmetry: Dih1, and 2 cyclic group symmetries: Z5, and Z1. These 4 symmetries can be seen in 4 distinct symmetries on the pentagon. John Conway labels these by a letter and group order.[9] Full symmetry of the regular form is r10 and no symmetry is labeled a1. The dihedral symmetries are divided depending on whether they pass through vertices (d for diagonal) or edges (p for perpendiculars), and i when reflection lines path through both edges and vertices. Cyclic symmetries in the middle column are labeled as g for their central gyration orders. Each subgroup symmetry allows one or more degrees of freedom for irregular forms. Only the g5 subgroup has no degrees of freedom but can seen as directed edges. ## Equilateral pentagons Equilateral pentagon built with four equal circles disposed in a chain. An equilateral pentagon is a polygon with five sides of equal length. However, its five internal angles can take a range of sets of values, thus permitting it to form a family of pentagons. In contrast, the regular pentagon is unique up to similarity, because it is equilateral and, moreover, it is equiangular (its five angles are equal). ## Cyclic pentagons A cyclic pentagon is one for which a circle called the circumcircle goes through all five vertices. The regular pentagon is an example of a cyclic pentagon. The area of a cyclic pentagon, whether regular or not, can be expressed as one fourth the square root of one of the roots of a septic equation whose coefficients are functions of the sides of the pentagon.[10][11][12] There exist cyclic pentagons with rational sides and rational area; these are called Robbins pentagons. In a Robbins pentagon, either all diagonals are rational or all are irrational, and it is conjectured that all the diagonals must be rational.[13] ## General convex pentagons For all convex pentagons, the sum of the squares of the diagonals is less than 3 times the sum of the squares of the sides.[14]:p.75,#1854 ## Graphs The K5 complete graph is often drawn as a regular pentagon with all 10 edges connected. This graph also represents an orthographic projection of the 5 vertices and 10 edges of the 5-cell. The rectified 5-cell, with vertices at the mid-edges of the 5-cell is projected inside a pentagon. 5-cell (4D) Rectified 5-cell (4D) ## Pentagons in tiling The best known packing of equal-sized regular pentagons on a plane is a double lattice structure which covers 92.131% of the plane. A regular pentagon cannot appear in any tiling of regular polygons. First, to prove a pentagon cannot form a regular tiling (one in which all faces are congruent, thus requiring that all the polygons be pentagons), observe that 360° / 108° = 313 (where 108° Is the interior angle), which is not a whole number; hence there exists no integer number of pentagons sharing a single vertex and leaving no gaps between them. More difficult is proving a pentagon cannot be in any edge-to-edge tiling made by regular polygons: There are no combinations of regular polygons with 4 or more meeting at a vertex that contain a pentagon. For combinations with 3, if 3 polygons meet at a vertex and one has an odd number of sides, the other 2 must be congruent. The reason for this is that the polygons that touch the edges of the pentagon must alternate around the pentagon, which is impossible because of the pentagon's odd number of sides. For the pentagon, this results in a polygon whose angles are all (360 − 108) / 2 = 126°. To find the number of sides this polygon has, the result is 360 / (180 − 126) = 623, which is not a whole number. Therefore, a pentagon cannot appear in any tiling made by regular polygons. There are 15 classes of pentagons that can monohedrally tile the plane. None of the pentagons have any symmetry in general, although some have special cases with mirror symmetry. 15 monohedral pentagonal tiles 12345 678910 1112131415 ## Pentagons in polyhedra Ih Th Td O I D5d Dodecahedron Pyritohedron Tetartoid Pentagonal icositetrahedron Pentagonal hexecontahedron Truncated trapezohedron ## In-line notes and references 1. "pentagon, adj. and n." OED Online. Oxford University Press, June 2014. Web. 17 August 2014. 2. Herbert W Richmond (1893). "Pentagon". 3. Peter R. Cromwell. Polyhedra. p. 63. ISBN 0-521-66405-5. 4. This result agrees with Herbert Edwin Hawkes; William Arthur Luby; Frank Charles Touton (1920). "Exercise 175". Plane geometry. Ginn & Co. p. 302. 5. Eric W. Weisstein (2003). CRC concise encyclopedia of mathematics (2nd ed.). CRC Press. p. 329. ISBN 1-58488-347-2. 6. DeTemple, Duane W. (Feb 1991). "Carlyle circles and Lemoine simplicity of polygon constructions" (PDF). The American Mathematical Monthly. 98 (2): 97–108. doi:10.2307/2323939. Archived from the original (PDF) on 2015-12-21. 7. George Edward Martin (1998). Geometric constructions. Springer. p. 6. ISBN 0-387-98276-0. 8. Euklid's Elements of Geometry, Book 4, Proposition 11 (PDF). Translated by Richard Fitzpatrick. 2008. p. 119. ISBN 978-0-6151-7984-1. 9. John H. Conway, Heidi Burgiel, Chaim Goodman-Strauss, (2008) The Symmetries of Things, ISBN 978-1-56881-220-5 (Chapter 20, Generalized Schaefli symbols, Types of symmetry of a polygon pp. 275-278) 10. Weisstein, Eric W. "Cyclic Pentagon." From MathWorld--A Wolfram Web Resource. 11. Robbins, D. P. (1994). "Areas of Polygons Inscribed in a Circle". Discrete and Computational Geometry. 12: 223–236. doi:10.1007/bf02574377. 12. Robbins, D. P. (1995). "Areas of Polygons Inscribed in a Circle". The American Mathematical Monthly. 102: 523–530. doi:10.2307/2974766. • Buchholz, Ralph H.; MacDougall, James A. (2008), "Cyclic polygons with rational sides and area", Journal of Number Theory, 128 (1): 17–48, doi:10.1016/j.jnt.2007.05.005, MR 2382768 . 13. Inequalities proposed in “Crux Mathematicorum, . • Weisstein, Eric W. "Pentagon". MathWorld. • Animated demonstration constructing an inscribed pentagon with compass and straightedge. • How to construct a regular pentagon with only a compass and straightedge. • How to fold a regular pentagon using only a strip of paper • Definition and properties of the pentagon, with interactive animation • Renaissance artists' approximate constructions of regular pentagons • Pentagon. How to calculate various dimensions of regular pentagons. Fundamental convex regular and uniform polytopes in dimensions 2–10 Family An Bn I2(p) / Dn E6 / E7 / E8 / F4 / G2 Hn Regular polygon Triangle Square p-gon Hexagon Pentagon Uniform polyhedron Tetrahedron OctahedronCube Demicube DodecahedronIcosahedron Uniform 4-polytope 5-cell 16-cellTesseract Demitesseract 24-cell 120-cell600-cell Uniform 5-polytope 5-simplex 5-orthoplex5-cube 5-demicube Uniform 6-polytope 6-simplex 6-orthoplex6-cube 6-demicube 122221 Uniform 7-polytope 7-simplex 7-orthoplex7-cube 7-demicube 132231321 Uniform 8-polytope 8-simplex 8-orthoplex8-cube 8-demicube 142241421 Uniform 9-polytope 9-simplex 9-orthoplex9-cube 9-demicube Uniform 10-polytope 10-simplex 10-orthoplex10-cube 10-demicube Uniform n-polytope n-simplex n-orthoplexn-cube n-demicube 1k22k1k21 n-pentagonal polytope Topics: Polytope families • Regular polytopeList of regular polytopes and compounds

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# Arithmetic & Geometric Sequences - Sit Sort Scramble Switch - Game Learning Subject Resource Type Product Rating 3.9 3 Ratings File Type PDF (Acrobat) Document File 140 KB|2 pages Share Product Description SIT - SORT - SCRAMBLE - SWITCH ~~~ Matching Activity Energize your students by allowing them to move around the classroom while practicing their vocabulary! Students will beg to play this fun activity to help reinforce their vocabulary skills in math. Research has shown that repetition in learning key vocabulary can greatly impact deeper understanding for students. SIT - SORT - SCRAMBLE - SWITCH games align directly with this finding. SAVE \$\$\$ BUNDLE AVAILABLE: BUNDLE - ARITHMETIC & GEOMETRIC SEQUENCES Included in BUNDLE: Sit, Sort, Scramble, Switch Choice Board PowerPoint Interactive Notebook RAFT Activity *************************************************************************** Directions: Cut apart the vocabulary and example cards. Laminate the pieces if you wish to provide long-term durability. Set up stations throughout your classroom (all 10 cards) and have students travel through matching each vocabulary term to the corresponding examples. After the last student finishes, instruct each student to scramble the cards and switch to another location. Time this activity and entice your students to become faster and more accurate. Have fun and by the end, your students will have built a stronger mathematical vocabulary. *************************************************************************** All SIT - SORT - SCRAMBLE - SWITCH games require LITTLE PREP so the possibilities for their use are endless: ♦ Substitute (Sub) Plans ♦ Early Finishers ♦ Yearly Review ♦ Matching ♦ Interactive Notebook ♦ Group Work *************************************************************************** You can find more SIT - SORT - SCRAMBLE - SWITCH Games for MATH at: ALGEBRA EQUATION VOCABULARY ALGEBRA EQUATION VOCABULARY - Sit Sort Scramble Switch Total Pages 2 pages Included Teaching Duration 30 minutes Report this Resource \$1.50

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Expected OF and FG - Calculations - Home Brew Forums Home Brew Forums > Expected OF and FG - Calculations 11-21-2012, 08:33 PM   #1 Tiroux Recipes Sep 2012 Thurso, Québec Posts: 470 Liked 10 Times on 8 Posts Hi! I've been told this formula... 25 x liters of mead wanted x ABV % points = grams of honey need example... for 10 liters of 10% mead... 25 x 10 x 10 = 2 500g That means 5.5 lbs of honey for a 2.64 gallons batch. But if I do the calculations in beersmith, or with other calculators... I have really different results. Here's my recipe. 4Kg Buckwheat honey (80% sugar content) (8.8lbs) Fruits that will give a total of 1Kg of ferment. sugars (2.2lbs) That is for a 17L batch, so a 4.5 gallons lalvin k1-v1116 yeast If i use the formula with the honey, and calculate the contribution of fruits... It results with 13.0% ABV With beersmith (with honey yield at 80%), i got 15.5% ABV, but a FG of 0.979!

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A balloon carrying a basket is decending at a constantvelocity of 20.0 m/s. A person in the basket throws a stone withthe intial velocity of 15.0 m/s horizontally perpendicular to thepath of the descending ballon, and 4.00s later this person sees therock strike the ground. A. How high was the balloon when the rock was thrownout. Answer h= _____________ m.....I thought I was using theright calulations but my answer is wrong! I came up with98.4. Part C... At the instant the rock hits the ground, how far isit from the basket.

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# 10Expected utility examples ## 10.1 A 50:50 bet Suppose your utility function is U(x)=\text{ln}(x). You have a 50% chance of winning $10 and a 50% chance of losing$10. Assume your starting wealth is 20. What is the expected value of this game? \begin{align*} E[X]&=\sum_{i=1}^n p_ix_i \\[6pt] &=0.5\times 10+0.5\times (-10) \\[6pt] &=0 \end{align*} The expected value of the game is0. What is the expected utility of this game? \begin{align*} E[U(W+X)]&=\sum_{i=1}^n p_iU(x_i+W) \\[6pt] &=0.5U(20-10)+0.5U(20+10) \\[6pt] &=0.5\text{ln}(10)+0.5\text{ln}(30) \\[6pt] &=2.85 \end{align*} What does an expected utility of 2.85 mean? To make it tangible, we can ask what wealth would give that utility. U(W)=\text{ln}(W)=2.85 W=e^{2.85}=\$17.30 This gamble with an expected value of zero reduces utility by an amount equivalent to$2.70. We could also say that the certainty equivalent of this gamble is the final wealth of $17.30, or a loss of$2.70. Figure 10.1 illustrates the example. Code library(ggplot2) u <- function(x){ log(x) } df <- data.frame( x=seq(1,220,0.1), y=NA ) df$y <- u(df$x) #Variables for plot (may not match labels as not done to scale) #Payoffs from gamble x1<-30 #loss x2<-200 #win ev<-115 #expected value of gamble xc<-115 #certain outcome ce<-79 #certainty equivalent px2<-(ev-x1)/(x2-x1) ggplot(mapping = aes(x, y)) + #Plot the utility curve geom_line(data = df) + geom_vline(xintercept = 0, linewidth=0.25)+ geom_hline(yintercept = 0, linewidth=0.25)+ labs(x = "x", y = "U(x)")+ # Set the theme theme_minimal()+ #remove numbers on each axis theme(axis.text.x = element_blank(), axis.text.y = element_blank(), axis.title=element_text(size=14,face="bold"), axis.title.y = element_text(angle=0, vjust=0.5))+ #set limits - need to include room for labels coord_cartesian(xlim = c(-25, 220), ylim = c(-0.25, 6))+ #Add labels W-10, W-10 and line to curve indicating each annotate("text", x = x1, y = 0, label = "W-10", size = 4, hjust = 0.5, vjust = 1.5)+ annotate("segment", x = x1, y = 0, xend = x1, yend = u(x1), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("segment", x = 0, y = u(x1), xend = x1, yend = u(x1), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = 0, y = u(x1), label = "U(W-10)", size = 4, hjust = 1.05, vjust = 0.6)+ #Add line to curve indicating utility of wealth annotate("segment", x = xc, y = 0, xend = xc, yend = u(xc), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("segment", x = 0, y = u(xc), xend = xc, yend = u(xc), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = 0, y = u(xc), label = "U(W)", size = 4, hjust = 1.05, vjust = 0.3)+ annotate("segment", x = x1, xend = x2, y = u(x1), yend = u(x2), linewidth = 0.5, colour = "black", linetype="dotdash")+ #Add labels W+10, U(W+10) and line to curve indicating each annotate("text", x = x2, y = 0, label = "W+10", size = 4, hjust = 0.4, vjust = 1.5)+ annotate("segment", x = x2, y = 0, xend = x2, yend = u(x2), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("segment", x = 0, y = u(x2), xend = x2, yend = u(x2), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = 0, y = u(x2), label = "U(W+10)", size = 4, hjust = 1.05, vjust = 0.45)+ #Add labels E[X]=W, E[U(X)] and curve indicating each annotate("text", x = ev, y = 0, label = "E[X]=W", size = 4, hjust = 0.4, vjust = 1.5)+ annotate("segment", x = ev, y = 0, xend = ev, yend = u(x1)+(u(x2)-u(x1))*px2, linewidth = 0.5, colour = "black", linetype="dashed")+ annotate("segment", x = 0, y = u(x1)+(u(x2)-u(x1))*px2, xend = ev, yend = u(x1)+(u(x2)-u(x1))*px2, linewidth = 0.5, colour = "black", linetype="dashed")+ annotate("text", x = 0, y = u(x1)+(u(x2)-u(x1))*px2, label = "E[U(X)]", size = 4, hjust = 1.05, vjust = 0.45)+ #Add vertical line indicating certainty equivalent and labelled "W-2.70" annotate("segment", x = ce, xend = ce, y = 0, yend = u(ce), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = ce, y = 0, label = "W-2.70", size = 4, hjust = 0.4, vjust = 1.5) On the x-axis, we have the outcomes and on the y-axis, we have the utility. I have added points on the x-axis for the outcomes of the two gambles, being W-10 and W+10. They deliver utility U(W+10) and U(W-10) respectively. The expected utility of the gamble is the probability-weighted average of these two points. It sits on the straight dash-dot-dot line between those two outcomes. You can see that the expected utility of the gamble is lower than the utility of the expected value (being current wealth). Also plotted is the certainty equivalent. We can identify it as the point on the utility curve where the utility of that certainty equivalent is equal to the expected utility. ## 10.2 An 80:20 bet Suppose your utility function is U(x)=\text{ln}(x). You have an 80% chance of winning $10 and a 20% chance of losing$10. Assume your starting wealth is 20. What are the expected value and the expected utility of this game? \begin{align*} E[X]&=\sum_{i=1}^n p_ix_i \\[6pt] &=0.8\times 10+0.2\times (-10) \\[6pt] &=\6 \end{align*} The expected value of the game is 6. What is the expected utility of this game? \begin{align*} E[U(W+x)]&=\sum_{i=1}^n p_iU(x_i+W) \\[6pt] &=0.8U(20+10)+0.2U(20-10) \\[6pt] &=0.8\text{ln}(30)+0.2\text{ln}(10) \\[6pt] &=3.18 \end{align*} What does an expected utility of 3.18 mean? To make it tangible, we can ask what wealth would give that utility. U(W)=\text{ln}(W)=3.18 W=e^{3.18}=\24.08 This gamble with an expected value of $6 increases utility by an amount equivalent to$4.08. We could also say that the certainty equivalent of this gamble is the final wealth of $24.08. Figure 10.2 illustrates the example. Code library(ggplot2) u <- function(x){ log(x) } df <- data.frame( x=seq(1,220,0.1), y=NA ) df$y <- u(df$x) #Variables for plot (may not match labels as not done to scale) #Payoffs from gamble x1<-30 #loss x2<-200 #win ev<-166 #expected value of gamble xc<-95 #certain outcome ce<-135 #certainty equivalent px2<-(ev-x1)/(x2-x1) ggplot(mapping = aes(x, y)) + #Plot the utility curve geom_line(data = df) + geom_vline(xintercept = 0, linewidth=0.25)+ geom_hline(yintercept = 0, linewidth=0.25)+ labs(x = "x", y = "U(x)")+ # Set the theme theme_minimal()+ #remove numbers on each axis theme(axis.text.x = element_blank(), axis.text.y = element_blank(), axis.title=element_text(size=14,face="bold"), axis.title.y = element_text(angle=0, vjust=0.5))+ #set limits - need to include room for labels coord_cartesian(xlim = c(-25, 220), ylim = c(-0.25, 6))+ #Add labels W-10, U(W+10) and line to curve indicating each annotate("text", x = x1, y = 0, label = "W-10", size = 4, hjust = 0.5, vjust = 1.5)+ annotate("segment", x = x1, y = 0, xend = x1, yend = u(x1), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("segment", x = 0, y = u(x1), xend = x1, yend = u(x1), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = 0, y = u(x1), label = "U(W-10)", size = 4, hjust = 1.05, vjust = 0.6)+ #Add labels W, U(W) and line to curve indicating each annotate("text", x = xc, y = 0, label = "W", size = 4, hjust = 0.6, vjust = 1.5)+ annotate("segment", x = xc, y = 0, xend = xc, yend = u(xc), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("segment", x = 0, y = u(xc), xend = xc, yend = u(xc), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = 0, y = u(xc), label = "U(W)", size = 4, hjust = 1.05, vjust = 0.3)+ #Add expected utility line annotate("segment", x = x1, xend = x2, y = u(x1), yend = u(x2), linewidth = 0.5, colour = "black", linetype="dotdash")+ #Add labels W+10, U(W+10) and line to curve indicating each annotate("text", x = x2, y = 0, label = "W+10", size = 4, hjust = 0.4, vjust = 1.5)+ annotate("segment", x = x2, y = 0, xend = x2, yend = u(x2), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("segment", x = 0, y = u(x2), xend = x2, yend = u(x2), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = 0, y = u(x2), label = "U(W+10)", size = 4, hjust = 1.05, vjust = 0.45)+ #Add labels E[X]=W+6, E[U(X)] and curve indicating each annotate("text", x = ev, y = 0, label = "E[X]=W+6", size = 4, hjust = 0.4, vjust = 1.5)+ annotate("segment", x = ev, y = 0, xend = ev, yend = u(x1)+(u(x2)-u(x1))*px2, linewidth = 0.5, colour = "black", linetype="dashed")+ annotate("segment", x = 0, y = u(x1)+(u(x2)-u(x1))*px2, xend = ev, yend = u(x1)+(u(x2)-u(x1))*px2, linewidth = 0.5, colour = "black", linetype="dashed")+ annotate("text", x = 0, y = u(x1)+(u(x2)-u(x1))*px2, label = "E[U(X)]", size = 4, hjust = 1.05, vjust = 0.45)+ #Add vertical line indicating certainty equivalent and labelled "CE" annotate("segment", x = ce, xend = ce, y = 0, yend = u(ce), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = ce, y = 0, label = "W+4.08", size = 4, hjust = 0.4, vjust = 1.5) The expected utility of the gamble \text{E}[U(X)] is higher than the utility from current wealth but lower than the utility of the expected value. That is, they are risk averse but would still accept this highly favourable bet. Also plotted is the certainty equivalent. We can identify it as the point on the utility curve where the utility of that certainty equivalent is equal to the expected utility. In this case, it is at$4.08 above current wealth. ## 10.3 Betting a proportion of wealth Suppose your utility function is U(x)=\text{ln}(x). You have a 50% chance of increasing your wealth by 50% and a 50% chance of decreasing your wealth by 40%. What are the expected value and the expected utility of this game? \begin{align*} E[X]&=\sum_{i=1}^n p_ix_i \\[6pt] &=0.5\times 0.6W+0.5\times 1.5W \\[6pt] &=0.3W+0.75W \\[6pt] &=1.05W \end{align*} The expected value of the gamble is 5% of your wealth. The gamble has a positive expected value. \begin{align*} E[U(X)]&=\sum_{i=1}^n p_iU(X_i) \\[6pt] &=0.5U(0.6W)+0.5U(1.5W) \\[6pt] &=0.5\text{ln}(0.6)+0.5\times \text{ln}(W)+0.5\text{ln}(1.5)+0.5\times \text{ln}(W) \\[6pt] &=-0.255+0.203+\text{ln}(W) \\[6pt] &=−0.053+\text{ln}(W) \end{align*} Here we have a gamble with a positive expected value, 5% of your wealth, but lower expected utility. Someone with log utility would reject this bet. Figure 10.3 illustrates the example. Code library(ggplot2) u <- function(x){ log(x) } df <- data.frame( x=seq(1,220,0.1), y=NA ) df$y <- u(df$x) #Variables for plot (may not match labels as not done to scale) #Payoffs from gamble x1<-30 #loss x2<-200 #win ev<-115 #expected value of gamble xc<-95 #certain outcome ce<-79 #certainty equivalent px2<-(ev-x1)/(x2-x1) ggplot(mapping = aes(x, y)) + #Plot the utility curve geom_line(data = df) + geom_vline(xintercept = 0, linewidth=0.25)+ geom_hline(yintercept = 0, linewidth=0.25)+ labs(x = "x", y = "U(x)")+ # Set the theme theme_minimal()+ #remove numbers on each axis theme(axis.text.x = element_blank(), axis.text.y = element_blank(), axis.title=element_text(size=14,face="bold"), axis.title.y = element_text(angle=0, vjust=0.5))+ #set limits - need to include room for labels coord_cartesian(xlim = c(-25, 220), ylim = c(-0.25, 6))+ #Add labels 0.6W, U(0.6W) and line to curve indicating each annotate("text", x = x1, y = 0, label = "0.6W", size = 4, hjust = 0.5, vjust = 1.5)+ annotate("segment", x = x1, y = 0, xend = x1, yend = u(x1), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("segment", x = 0, y = u(x1), xend = x1, yend = u(x1), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = 0, y = u(x1), label = "U(0.6W)", size = 4, hjust = 1.05, vjust = 0.6)+ #Add labels W, U(W) and line to curve indicating each annotate("text", x = xc, y = 0, label = "W", size = 4, hjust = 0.6, vjust = 1.5)+ annotate("segment", x = xc, y = 0, xend = xc, yend = u(xc), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("segment", x = 0, y = u(xc), xend = xc, yend = u(xc), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = 0, y = u(xc), label = "U(W)", size = 4, hjust = 1.05, vjust = 0.3)+ annotate("segment", x = x1, xend = x2, y = u(x1), yend = u(x2), linewidth = 0.5, colour = "black", linetype="dotdash")+ #Add labels 1.5W, U(1.5W) and line to curve indicating each annotate("text", x = x2, y = 0, label = "1.5W", size = 4, hjust = 0.4, vjust = 1.5)+ annotate("segment", x = x2, y = 0, xend = x2, yend = u(x2), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("segment", x = 0, y = u(x2), xend = x2, yend = u(x2), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = 0, y = u(x2), label = "U(1.5W)", size = 4, hjust = 1.05, vjust = 0.45)+ #Add labels E[X]=1.05W, E[U(X)] and curve indicating each annotate("text", x = ev, y = 0, label = "E[X]=1.05W", size = 4, hjust = 0.4, vjust = 1.5)+ annotate("segment", x = ev, y = 0, xend = ev, yend = u(x1)+(u(x2)-u(x1))*px2, linewidth = 0.5, colour = "black", linetype="dashed")+ annotate("segment", x = 0, y = u(x1)+(u(x2)-u(x1))*px2, xend = ev, yend = u(x1)+(u(x2)-u(x1))*px2, linewidth = 0.5, colour = "black", linetype="dashed")+ annotate("text", x = 0, y = u(x1)+(u(x2)-u(x1))*px2, label = "E[U(X)]", size = 4, hjust = 1.05, vjust = 0.45)+ #Add vertical line indicating certainty equivalent and labelled "CE" annotate("segment", x = ce, xend = ce, y = 0, yend = u(ce), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = ce, y = 0, label = "CE", size = 4, hjust = 0.4, vjust = 1.5) I have added points on the x-axis for the outcomes of the two gambles, a 40% reduction in wealth and a 50% gain in wealth. The expected utility of the gamble is the probability-weighted average of these two points. It sits on the straight dash-dot-dot line between those two outcomes. You can see that the expected utility of the gamble is lower than the utility of current wealth. They would reject an offer of this bet. ## 10.4 The St. Petersburg game The St. Petersburg game was invented by the Swiss mathematician Nicolas Bernoulli. The game starts with a pot containing $2. A dealer then flips a coin. The pot doubles every time a head appears. The game ends, and the player wins the pot when a tail appears. • A tail on the first flip leads to a payment of$2. • A tail on the second flip leads to a payment of $4 • A tail on the third flip leads to a payment of$8 And so on. Consider what you would be willing to pay to play this game. Would you pay $5?$10? $25?$50? More? The expected value of this game is equal to the sum of the following series. \begin{align*} E[X]&=\underbrace{\frac{1}{2}\times 2}_\textrm{Tail first}+\underbrace{\bigg(\frac{1}{2}\times \frac{1}{2}\bigg)\times 4}_\textrm{Tail second}+\underbrace{\bigg(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\bigg)\times 8}_\textrm{Tail third} \\[24pt] &\qquad +\underbrace{\bigg(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\bigg)\times 16}_\textrm{Tail fourth}+... \\[24pt] &=1+1+1+1+... \\ &=\sum_{k=1}^\infty 1 \\ &=\infty \end{align*} The first term in the series captures the 50% chance of a tail on the first flip, paying $2. The second term represents the 50% chance of a head on the first flip, followed by the 50% chance of the tail second flip, paying$4. The third term represents the 50% chance of a head on the first flip, followed by the 50% chance of a head on the second flip, followed by the 50% chance of a tail on the third flip, paying 8. And so on. Multiplying out each of those terms results in a series of 1s. The \sum operator means “sum for k=1 to k=\infty”. Contrast this expected value of \infty with the sum you would pay to play the game. You were likely not willing to pay an infinite amount. This “paradox” is often resolved by introducing an expected utility function. The expected utility of this game is equal to: \begin{align*} E[U(X)]&=\underbrace{\frac{1}{2}\times U(W+2)}_\textrm{Tail first}+\underbrace{\bigg(\frac{1}{2}\times \frac{1}{2}\bigg)\times U(W+4)}_\textrm{Tail second} \\[24pt] &\qquad +\underbrace{\bigg(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\bigg)\times U(W+8)}_\textrm{Tail third} \\[24pt] &\qquad +\underbrace{\bigg(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\bigg)\times U(W+16)}_\textrm{Tail fourth}+... \\[24pt] &=\frac{1}{2}U(W+2)+\frac{1}{4}U(W+4)+\frac{1}{8}U(W+8)+\frac{1}{16}U(W+16)+... \\[12pt] &=\sum_{k=1}^{k=\infty}\frac{1}{2^k}U(W+2^k) \end{align*} Similar to the calculation of the expected value, the first term in the series captures the 50% chance of a tail on the first flip, paying2. The second term represents the 50% chance of a head on the first flip, followed by the 50% chance of the tail on the second flip, paying 4. And so on. But here, we are using the utility function U(x). In the second line, I multiplied the probabilities of each coin flip together. In the third line, I expressed this infinite sum more compactly. To take this equation further, we need to consider the particular utility function of the decision maker. What maximum sum would a risk-neutral player with U(x)=x be willing to pay to play the game? One strategy to determine this sum is to ask what sum would result in the player being indifferent between paying and rejecting a chance to play. That is the maximum sum c that they would be willing to pay. They will be indifferent when U(W)=E[U(X-c)]. We can solve this equation as follows. \begin{align*} U(W)&=E[U(X-c)] \\[6pt] &=\sum_{k=1}^{k=\infty}\frac{1}{2^k}U(W+\2^k-c) \\[6pt] W&=\sum_{k=1}^{k=\infty}\frac{1}{2^k}(W+2^k-c) \qquad \text{(substituting in the utility function)}\\[6pt] &=W-c+\sum_{k=1}^{k=\infty}1 \qquad \Bigg(\text{as }\sum_{k=1}^{k=\infty}\frac{1}{2^k}=1\Bigg) \\[12pt] c&=\sum_{k=1}^{k=\infty}1 \\[6pt] &=\infty \end{align*} In the second line, we use the sum we created earlier. In the third line, I substitute the utility function U(x)=x. We can then simplify as in the fourth line, which allows us to see that, given the infinite expected value of the game, the player would be willing to pay an infinite amount to play. That is, a risk-neutral player would pay any amount $c to play. We could also have inferred this from the game’s expected value being infinite. What is the maximum sum a risk-averse player with U(x)=\text{ln}(x) would be willing to pay to play the game? How does their wealth affect their willingness to pay? Again we will determine at what$c the player is indifferent between accepting and rejecting a chance to play, which occurs when U(W)=E[U(X-c)]. \begin{align*} U(W)&=E[U(X-c)] \\[6pt] U(W)&=\sum_{k=1}^{k=\infty}\frac{1}{2^k}U(W+\$2^k-c) \\[6pt] \text{ln}(W)&=\sum_{k=1}^{k=\infty}\frac{1}{2^k}\text{ln}(W+\$2^k-c) \end{align*} Code # Calculation of value of gamble for the following paragraph EU = function(W, c, epsilon){ ans = 0 k = 1 while(abs(val <- (log(max(epsilon, W + 2^k - c)) - log(W)) / 2^k) > epsilon){ k <- k + 1; ans <- ans + val; } ans } find_c = function(W, epsilon=10^(-10)){ low = 0 c = 0 high = 10^10 while(abs(low - high) > epsilon){ c = (high + low) / 2 exp_value = EU(W, c, epsilon) ifelse(exp_value > 0, low <- c, high <- c) } c } # Value of bet to someone with wealth of $1,000,000 c1000000 <- round(find_c(10^6), 2) # Value of bet to someone with wealth of$1,000 c1000 <- round(find_c(10^3), 2) # Value of bet to someone with wealth of $0.01 c001 <- round(find_c(0.01), 2) There is no closed-form solution to this equation to enable us to determine c. We need to solve via numerical methods (such as testing and iterating to a solution). If we did solve this, we would find that someone who has wealth of$0.01 would be willing to pay up to $2.01. They would need to borrow. Someone with wealth$1000 would be willing to pay $10.95. A person with a wealth of$1 million would be willing to pay 20.87. We cannot solve for a person with no wealth as \text{ln}⁡(0) is undefined. Why does willingness to pay increase with wealth? With log utility, as wealth increases, the slope of the log function increasingly approximates a linear function (the second derivative approaches zero). Hence, the gambler displays less risk-averse (closer to risk-neutral) behaviour. One way to gain an intuition for why this gamble now has a finite value is to calculate the utility of a risk-averse player whose only asset is the opportunity to play this game. \begin{align*} E[U(X)]&=\sum_{k=1}^{k=\infty}\frac{1}{2^k}U(\2^k) \\[12pt] &=\sum_{k=1}^{k=\infty}\frac{1}{2^k}\text{ln}(2^k) \qquad \text{(substituting in the utility function)}\\[12pt] &=\sum_{k=1}^{k=\infty}\frac{k}{2^k}\text{ln}(2) \qquad \text{(using the rule }\ln(x^a)=a\ln(x)) \\[12pt] &=\bigg(\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+...\bigg)\text{ln}(2) \\[12pt] &=2\text{ln}(2) \end{align*} The change in the utility from each flip rapidly declines. Ultimately the series of fractions sum to two. We can then calculate what wealth is equivalent to this expected utility. U(W)=\text{ln}(W)=2\text{ln}(2) \\[12pt] W=e^{2\text{ln}2}=4 The expected utility from the game is equal to the utility of $4. ## 10.5 Risk neutrality versus risk aversion Anika and Anthony are offered a choice between options A and B: A: Lottery A=(0.5, \$100; 0.5, \$20). This is a gamble with a 50% chance of winning$100 and a 50% chance of winning $20. B:$40 for certain. (a) Anika is risk-neutral. Will Anika choose A or B? A risk-neutral decision-maker maximises expected value. The expected value of option A is: \begin{align*} \text{E}[A]&=p_1x_1+p_2x_2 \\[6pt] &=0.5\times \$100+0.5\times \$20 \\[6pt] &=\60 \end{align*} The expected value of option B is40. Anika will choose option A because $60 is greater than$40. (b) Anthony is risk averse with wealth 100 and utility function U(x)=\text{ln}(x). Will Anthony choose A or B? Anthony will select the option that gives the highest expected utility. The expected utility of option A is: \begin{align*} \text{E}[U(A)]&=p_1u(x_1)+p_2u(x_2) \\[6pt] &=0.5\times \text{ln}(W+100)+0.5\times \text{ln}(W+20) \\[6pt] &=0.5\times \text{ln}(200)+0.5\times \text{ln}(120) \\[6pt] &=5.04 \end{align*} The expected utility of option B is: \begin{align*} \text{E}[U(B)]&=u(W+x) \\[6pt] &=\text{ln}(140) \\[6pt] &=4.94 \end{align*} Anthony will choose option A because \text{E}[U(A)]=5.04>4.94=\text{E}[U(B)]. (c) Draw a graph showing the choices faced by Anthony, his utility curve and the expected utility of each option. Indicate the certainty equivalent of option A. Explain how the graph shows which option Anthony will choose. Figure 10.4 shows the choices faced by Anthony, his utility curve and the expected utility of each option. The horizontal axis is the outcome and the vertical axis is utility of each outcome. The utility curve is the function U(x)=\text{ln}(x). The two possible outcomes of gamble A are W+20=120 and W+100=200, which deliver U(W+20) and U(W+100) respectively. Each are labelled on the chart. The expected utility of gamble A is the weighted average of these two utilities and lies on the straight line between U(W+20) and U(W+100). As each outcome has a 50% chance of occurring, the expected utility of gamble A is the midpoint of this line (as is the expected value of the gamble). The vertical line from W+E[A]=160 identifies that point, with the expected utility E[U(A)] marked on the vertical axis. The certain outcome from option B, the receipt of40 resulting in wealth of $140 is also marked on the x-axis, leading to utility of U(W+B). It can be seen that the expected utility of gamble A E[U(A)] is greater than the utility of the certain outcome U(W+B). Anthony will therefore choose gamble A. The certainty equivalent of option A is identified as the point where U(CE)=E[U(A)]. This is identified by drawing a horizontal line from the expected utility of gamble A to the utility curve. The point where this line intersects the utility curve is the certainty equivalent of gamble A, shown by projecting a vertical line downward. This diagram is not drawn to scale. Code library(ggplot2) u <- function(x){ log(x) } df <- data.frame( x=seq(1,220,0.1), y=NA ) df$y <- u(df$x) #Variables for plot (may not match labels as not done to scale) #Payoffs from gamble x1<-30 #loss x2<-200 #win ev<-115 #expected value of gamble xc<-60 #certain outcome ce<-79 #certainty equivalent px2<-(ev-x1)/(x2-x1) ggplot(mapping = aes(x, y)) + #Plot the utility curve geom_line(data = df) + geom_vline(xintercept = 0, linewidth=0.25)+ geom_hline(yintercept = 0, linewidth=0.25)+ labs(x = "x", y = "U(x)")+ # Set the theme theme_minimal()+ #remove numbers on each axis theme(axis.text.x = element_blank(), axis.text.y = element_blank(), axis.title=element_text(size=14,face="bold"), axis.title.y = element_text(angle=0, vjust=0.5))+ #set limits - need to include room for labels coord_cartesian(xlim = c(-25, 220), ylim = c(-0.25, 6))+ #Add labels W+20, U(W+20) and line to curve indicating each annotate("text", x = x1, y = 0, label = "W+20", size = 4, hjust = 0.5, vjust = 1.5)+ annotate("segment", x = x1, y = 0, xend = x1, yend = u(x1), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("segment", x = 0, y = u(x1), xend = x1, yend = u(x1), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = 0, y = u(x1), label = "U(W+20)", size = 4, hjust = 1.05, vjust = 0.6)+ #Add labels W+B, U(W+B) and line to curve indicating each annotate("text", x = xc, y = 0, label = "W+B=140", size = 4, hjust = 0.6, vjust = 1.5)+ annotate("segment", x = xc, y = 0, xend = xc, yend = u(xc), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("segment", x = 0, y = u(xc), xend = xc, yend = u(xc), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = 0, y = u(xc), label = "U(W+B)", size = 4, hjust = 1.05, vjust = 0.3)+ #Add expected utility line annotate("segment", x = x1, xend = x2, y = u(x1), yend = u(x2), linewidth = 0.5, colour = "black", linetype="dotdash")+ #Add labels W+100, U(W+100) and line to curve indicating each annotate("text", x = x2, y = 0, label = "W+100", size = 4, hjust = 0.4, vjust = 1.5)+ annotate("segment", x = x2, y = 0, xend = x2, yend = u(x2), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("segment", x = 0, y = u(x2), xend = x2, yend = u(x2), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = 0, y = u(x2), label = "U(W+100)", size = 4, hjust = 1.05, vjust = 0.45)+ #Add labels W+E[A], E[U(W+A)] and curve indicating each annotate("text", x = ev, y = 0, label = "W+E[A]=160", size = 4, hjust = 0.4, vjust = 1.5)+ annotate("segment", x = ev, y = 0, xend = ev, yend = u(x1)+(u(x2)-u(x1))*px2, linewidth = 0.5, colour = "black", linetype="dashed")+ annotate("segment", x = 0, y = u(x1)+(u(x2)-u(x1))*px2, xend = ev, yend = u(x1)+(u(x2)-u(x1))*px2, linewidth = 0.5, colour = "black", linetype="dashed")+ annotate("text", x = 0, y = u(x1)+(u(x2)-u(x1))*px2, label = "E[U(A)]", size = 4, hjust = 1.05, vjust = 0.45)+ #Add vertical line indicating certainty equivalent and labelled "CE" annotate("segment", x = ce, xend = ce, y = 0, yend = u(ce), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = ce, y = 0, label = "CE", size = 4, hjust = 0.4, vjust = 1.5) ## 10.6 Lottery ticket Buying a lottery ticket has a negative expected value. Andrew is an expected utility maximiser. He purchases a lottery ticket. (a) What risk preferences (attitude to risk) does Andrew have? If an expected utility maximiser purchases a lottery ticket with negative expected value, he is risk seeking. He values the gamble over and above the expected value of the gamble. (b) Use a graph to demonstrate your answer to part (a). Figure 10.5 shows Andrew’s utility curve. As he is risk seeking it is convex (at least over the domain of the lottery). Each of the outcomes of the lottery are labelled. Andrew finishes with his wealth minus the cost of the lottery ticket (W-T) or his wealth minus the cost of the lottery ticket plus his lottery winnings (W-T+L). If he does not purchase the ticket, his wealth remains at W. The utility of each possible outcome (U(W-T), U(W), U(W-T+L)) is also indicated on the vertical axis. The expected value of the lottery after buying the ticket is labelled (E[X]). As the lottery has a negative expected value, E[X] is less than W. The expected utility of the lottery lies on the straight line between the utility of the two possible lottery outcomes. The place on the line is determined by the probability of winning and is in line with the expected value of the lottery. We can identify the expected utility of the lottery by projecting a line up from E[X] to the straight line. Finally, the certainty equivalent of the lottery is also marked. As U(CE)=E[U(X)], we can identify the certainty equivalent by projecting a line up from E[U(X)] to the utility curve. Due to the convex curve, we can see that E[U(X)] is greater than U(W). Andrew prefers the lottery to the certain outcome of W. Alternatively, we can see that the certainty equivalent of the lottery is higher than current wealth. Andrew would require a payment of at least CE-W to forgo his opportunity to partake in the lottery. Code library(ggplot2) u <- function(x){ x^2 } df <- data.frame( x=seq(1,220,0.1), y=NA ) df$y <- u(df\$x) #Variables for plot (may not match labels as not done to scale) #Payoffs from gamble x1<-30 #loss x2<-200 #win ev<-65 #expected value of gamble xc<-80 #certain outcome ce<-95 #certainty equivalent px2<-(ev-x1)/(x2-x1) ggplot(mapping = aes(x, y)) + #Plot the utility curve geom_line(data = df) + geom_vline(xintercept = 0, linewidth=0.25)+ geom_hline(yintercept = 0, linewidth=0.25)+ labs(x = "x", y = "U(x)")+ # Set the theme theme_minimal()+ #remove numbers on each axis theme(axis.text.x = element_blank(), axis.text.y = element_blank(), axis.title=element_text(size=14,face="bold"), axis.title.y = element_text(angle=0, vjust=0.5))+ #set limits - need to include room for labels coord_cartesian(xlim = c(-25, 220), ylim = c(-1000, 40000))+ #Add labels W-T, U(W-T) and line to curve indicating each annotate("text", x = x1, y = 0, label = "W-T", size = 4, hjust = 0.5, vjust = 1.5)+ annotate("segment", x = x1, y = 0, xend = x1, yend = u(x1), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("segment", x = 0, y = u(x1), xend = x1, yend = u(x1), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = 0, y = u(x1), label = "U(W-T)", size = 4, hjust = 1.05, vjust = 0.6)+ #Add labels W, U(W) and line to curve indicating each annotate("text", x = xc, y = 0, label = "W", size = 4, hjust = 0.6, vjust = 1.5)+ annotate("segment", x = xc, y = 0, xend = xc, yend = u(xc), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("segment", x = 0, y = u(xc), xend = xc, yend = u(xc), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = 0, y = u(xc), label = "U(W)", size = 4, hjust = 1.05, vjust = 0.3)+ annotate("segment", x = x1, xend = x2, y = u(x1), yend = u(x2), linewidth = 0.5, colour = "black", linetype="dotdash")+ #Add labels W-T+L, U(W-T+L) and line to curve indicating each annotate("text", x = x2, y = 0, label = "W-T+L", size = 4, hjust = 0.4, vjust = 1.5)+ annotate("segment", x = x2, y = 0, xend = x2, yend = u(x2), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("segment", x = 0, y = u(x2), xend = x2, yend = u(x2), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("text", x = 0, y = u(x2), label = "U(W-T+L)", size = 4, hjust = 1.05, vjust = 0.45)+ #Add labels E[X], E[U(X)] and curve indicating each annotate("text", x = ev, y = 0, label = "E[X]", size = 4, hjust = 0.4, vjust = 1.5)+ annotate("segment", x = ev, y = 0, xend = ev, yend = u(x1)+(u(x2)-u(x1))*px2, linewidth = 0.5, colour = "black", linetype="dashed")+ annotate("segment", x = 0, y = u(x1)+(u(x2)-u(x1))*px2, xend = ev, yend = u(x1)+(u(x2)-u(x1))*px2, linewidth = 0.5, colour = "black", linetype="dashed")+ annotate("text", x = 0, y = u(x1)+(u(x2)-u(x1))*px2, label = "E[U(X)]", size = 4, hjust = 1.05, vjust = 0.45)+ #Add vertical line indicating certainty equivalent and labelled "CE" annotate("segment", x = ce, xend = ce, y = 0, yend = u(ce), linewidth = 0.5, colour = "black", linetype="dotted")+ annotate("segment", x = ev, y = u(x1)+(u(x2)-u(x1))*px2, xend = ce, yend = u(x1)+(u(x2)-u(x1))*px2, linewidth = 0.5, colour = "black", linetype="dashed")+ annotate("text", x = ce, y = 0, label = "CE", size = 4, hjust = 0.4, vjust = 1.5)

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# Paper Cup Probabilities… This is my first entry in #MTBoS30 Today I asked my 10th grade geometry class three questions (review time!) 1. What is the probability of rolling a one (using a regular six-sided cube)? 2. If you flipped a coin, would you expect it to come up heads? 3. What is the probability that a paper cup tossed in the air will land on its side? The ensuing discussion involved certainty. First question: 1Ss: one out of six Several other students chimed in in agreement. Me: who can tell me how you can be so sure? 2 Ss: because the cube has six sides, and the sides are numbered one, two, three, four, (she is ticking off on her fingers; other students were nodding in agreement and telling her what to say) five, six, and there is only one side with one! This class is usually not this involved. I think it had to do with the fact that they really KNEW this! (Confidence is a wonderful thing!) Second question (key word here”” “expect”) 1 Ss: yes, well, no, (???) it could be heads or tails. I mean, you could expect a heads or a tails. Me: why can’t you expect just heads? 1 Ss: because it’s 50%. (At this point, other students begin chiming in: “Yeah, it’s 1/2!” “A coin has two sides” and similar statements.) The question wasn’t a straightforward question about a probability fraction, so I think that caused them to not feel as confident with the answer, until one student decoded it. Think: lemmings! It was the third question that really threw them. I held up one of those small cups, like you find in a bathroom cup dispenser. I asked them to tell me what they thought the probability would be of the cup landing on its side when tossed. The guesses ranged from 1/2 to 340/500. As we looked at the cup, the guesses got more specific. Several students noticed that the cup had a top, a bottom, and a side. The reasoning followed that there should be a 1/3 chance of landing on its side. At this point there was quite a bit of agreement. This seemed very logical (and if the strongest kids in the class said so, it must be right! Lemmings, I’m telling ya!) Multiple students jumped on the bandwagon and agreed. (No one talked about surface ratios – I figured we could tackle that later!) Then I gave each student a paper cup and asked them to create 20 trials each. I deliberately refrained from telling them instructions for tossing the cup. I just walked around and watched. Some kids tossed (across the room!), some kids tossed on their desks. Some dropped the cups on the floor. I heard disappointment as Ss complained, “it’s landing on its side every time,” how do you make it land on its top?” “There is something wrong with this cup!” The trials were listed on the board and tallied. The students really seemed puzzled as to why the results weren’t anywhere near what they expected. They were already arguing why this was so, so I put them in groups with the instruction to: 1. Compare the actual probability from the trials to the expected. 2. Come up with some reasons for the difference. 3. Pick a spokesperson to share their ideas with the class. Then the bell rang! Okay. The debriefing happens Monday…

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HomeEnglishClass 12PhysicsChapterElectrostatics Two point charges, each of mas... # Two point charges, each of mass m and charge q are released when they are at a distance d from each other. What is the speed of each charged particle when they are at a distance 2r ? Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams. Updated On: 24-6-2021 Apne doubts clear karein ab Whatsapp par bhi. Try it now. According to momentum conservation, both the charge particles will move with same speed. Now applying energy conservation : <br> 0+0+(Kq^(2))/r=2 1/2 mv^(2)+(Kq^(2))/(2r) implies v=sqrt((Kq^(2))/(2rm)) <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RES_PHY_ELE_S01_070_S01.png" width="80%">

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# A 7%,6O day note was discounted at 15 days before the maturity date. If the discount rate was 5.5% and the proceeds received were \$ 997,77,find a)the amount of discount that was charged b)the discount date if maturity date of 21,771 questions 1. ## math(urgent) A 7%,6O day note was discounted at 15 days before the maturity date. If the discount rate was 5.5% and the proceeds received were \$ 997,77,find a)the amount of discount that was charged b)the discount date if maturity date of the note is 26 october 1991 asked by fizz on May 19, 2014 2. ## math A 7%,6O day note was discounted at 15 days before the maturity date. If the discount rate was 5.5% and the proceeds received were \$ 997,77,find a)the amount of discount that was charged b)the discount date if maturity date of the note is 26 october 1991 asked by fizz on May 17, 2014 3. ## math Use the ordinary interest method, 360 days, to solve the following word problem. Round to the nearest cent when necessary. Pinnacle Manufacturing received a \$40,000 promissory note at 12% simple interest for 95 days from one of its customers. On day 70, asked by reva on December 22, 2014 4. ## contemporary math Varsity Press, a publisher of college textbooks, received a \$70,000 promissory note at 12% ordinary interest for 60 days from one of its customers, Reader’s Choice Bookstores. After 20 days, Varsity Press discounted the note at the Grove Isle Bank at a asked by Trish on June 21, 2013 5. ## math A 7%,6O day note was discounted at 15 days before the maturity date. If the discount rate was 5.5% and the proceeds received were \$ 997,77,find a)the amount of discount that was charged b)face vaalue what is the formula asked by fizz on May 13, 2014 Date of Note March 11, length of note 200 days, Date of discount, June 28. What is the maturity date and the discount period, no leap years, exact days asked by Kathy on May 4, 2010 7. ## math Varsity Press, a publisher of college textbooks, received a \$70,000 promissory note at 12% ordinary interest for 60 days from one of its customers, Reader’s Choice Bookstores. After 20 days, Varsity Press discounted the note at the Grove Isle Bank at a asked by keionna on June 15, 2013 8. ## math Varsity Press, a publisher of college textbooks, received a \$70,000 promissory note at 12% ordinary interest for 60 days from one of its customers, Reader’s Choice Bookstores. After 20 days, Varsity Press discounted the note at the Grove Isle Bank at a asked by keionna on June 17, 2013 Varsity Press, a publisher of college textbooks, received a \$70,000 promissory note at 12% ordinary interest for 60 days from one of its customers, Reader’s Choice Bookstores. After 20 days, Varsity Press discounted the note at the Grove Isle Bank at a asked by Tee on June 19, 2013 10. ## math Varsity Press, a publisher of college textbooks, received a \$70,000 promissory note at 12% ordinary interest for 60 days from one of its customers, Reader’s Choice Bookstores. After 20 days, Varsity Press discounted the note at the Grove Isle Bank at a 11. ## choice Varsity Press, a publisher of college textbooks, received a \$70,000 promissory note at 12% ordinary interest for 60 days from one of its customers, Reader’s Choice Bookstores. After 20 days, Varsity Press discounted the note at the Grove Isle Bank at a asked by d on September 16, 2013 12. ## math Varsity Press, a publisher of college textbooks, received a \$70,000 promissory note at 12% ordinary interest for 60 days from one of its customers, Reader’s Choice Bookstores. After 20 days, Varsity Press discounted the note at the Grove Isle Bank at a asked by Anonymous on December 21, 2013 13. ## math what is the discount period, bank discount, and proceeds on 25000.00 at 9% for 60 days, date of note is June 8 and date note discounted is July 10? asked by Anonymous on October 30, 2011 14. ## algebra The following note was discounted at 16%. Find (a) the discount period, (b) the discount, and (c) the proceeds. Date loan was made: March 15 Face Value: \$6000 Length of Loan: 75 days Rate: 6% Date of Discount: April 5 asked by Felicia on August 21, 2012 15. ## math bus 105 Calculate (a) interest and maturity value, (b) discount period, (c) bank discount, and (d) proceeds. Assume ordinary interest. (Round answers to nearest hundredths) Date of note Face value Length of note Interest rate Bank discount rate Date of discount asked by christina on October 30, 2016 Can someone help me with this problem: On September 5, Sheffield Company discounted at Sunshine Bank a \$9,000 (maturity value), 120-day note dated June 5. Sunshine’s discount rate was 9%. (Use Days in a year table.) What proceeds did Sheffield asked by marie on April 7, 2014 17. ## Discount Note On September 5, Sheffield company discounted at sunshine bank a 9000.00 Maturity value, 120-day note dated June 5, shushines discount rate was 9%. What proceeds did sheffield company receive? Use the 360 days asked by Scott on March 16, 2008 18. ## math(urgent) 9.A 10%,100-day note was dated 20 september 1992.the maturity value of the note was \$3083.33.If the note was discounted on 17 november 1992 at 12%,FIND i) the maturity date of the note = 29 december 1992 ii)the face value of the note \$3083.33 = asked by fizz on May 17, 2014 19. ## math 9.A 10%,100-day note was dated 20 september 1992.the maturity value of the note was \$3083.33.If the note was discounted on 17 november 1992 at 12%,FIND i) the maturity date of the note = 29 december 1992 ii)the face value of the note \$3083.33 = asked by AMY on May 18, 2014 20. ## math 9.A 10%,100-day note was dated 20 september 1992.the maturity value of the note was \$3083.33.If the note was discounted on 17 november 1992 at 12%,FIND i) the maturity date of the note = 29 december 1992 ii)the face value of the note \$3083.33 = asked by fizz on May 19, 2014 21. ## Promissory notes Rex corporation accepted a 5000.00, 8%, 120-day dated August 8 from Regis company in settlement of a past bill. On October 11, Rex discounted the note at Park Bank at 9%. What are the notes Maturity value, Discount Period, and bank discount? What proceeds asked by Scott on March 16, 2008 22. ## Maths Use the United States Rule and/or Banker’s Rule to determine the balance due on the note at the date of maturity. (The effective date is the date the note was written.) Principal - 6000 Rate - 5% Effective Date - May 15 Maturity Date - November 1 Partial asked by Kim on March 10, 2008 A \$120,000, 5% 200-day note dated June 6 is discounted on October 8. The discount period is _______ days. a. 142 b. 67 c. 124 d. 76 C is wrong asked by Kat on August 31, 2014 Using the dating method, calculate the discount date and the net date. Unless otherwise specified, the net date is 20 days after the discount date: The date of invoice is April 6th, Terms of sale is 2/10, EOM. What is the Discount date? What is the Net asked by Tammy on December 5, 2014 25. ## Discount Notes On september 5, sheffield company discounted at Sunshine bank a 9000.00 (Maturity Value), 120-day note dated June 5, Shunshine's discount rate was 9% What proceeds did sheffield company receive asked by Scott on March 15, 2008 26. ## Discount Notes On september 5, sheffield company discounted at Sunshine bank a 9000.00 (Maturity Value), 120-day note dated June 5, Shunshine's discount rate was 9% What proceeds did sheffield company receive asked by Scott on March 15, 2008 27. ## math 1. A 120-day simple discount promissory note for \$12,000 with a simple discount rate of 9% was signed on July 14. It was discounted on August 30 at 9.5%. Find the proceeds at the time of the sale. 2. 8. A \$15,000 T-bill is purchased at a 3.85% discount asked by Fellippe on September 6, 2012 28. ## math/Finance I want to know the formula to use 1. A 120-day simple discount promissory note for \$12,000 with a simple discount rate of 9% was signed on July 14. It was discounted on August 30 at 9.5%. Find the proceeds at the time of the sale. 2. 8. A \$15,000 T-bill is asked by Janine on September 6, 2012 29. ## math Find the maturity value of the undiscounted promissory note that states that Phillip Esterey borrowed \$4,000 for a period of 7 months with ordinary interest at 7%. The date of the note was December 17, 2008. The maturity date was July 17. asked by Taylor on October 13, 2013 30. ## math Find the maturity value of the undiscounted promissory note that states that Phillip Esterey borrowed \$8,000 for a period of 8 months with ordinary interest at 10%. The date of the note was August 1, 2008. The maturity date was April 1. asked by Anoymous on May 20, 2015 31. ## Promissory Notes On September 5th, Sheffield Company discounted at Sunshine Bank a 9000.00 (maturity value), 120-day note dated June 5th. Sunshine's discount rate was 9%. What proceeds did sheffield company receive. We are using 360 day and NOT 365 asked by Scott on March 15, 2008 32. ## math A partial payment is made on the date(s) indicated. Use the United States Rule to determine the balance due on the note at the date of maturity. (The Effective Date is the date the note was written.) Assume the year is not a leap year. P \$1900, R 6%, asked by Melanie on June 13, 2011 33. ## Finance Carnation needs to buy a \$12,500 part the part company is offering cash discount terms of 4/10, n30. Carnation has a tight cash flow, wants to discount a 180 day note dated February 12 with a maturity value of \$10,300. Bank offers a discount rate of 6%. asked by becky on June 6, 2012 34. ## Promissory Notes On september 5 Sheffield company discounted at Sunshine bank a 9000.00 (maturity value), 120-days note dated June 5th. Sunshines discounted rate was 9%. What proceeds did sheffield company receive? asked by Scott on March 16, 2008 35. ## College Math i'm having trouble calculating this. can someone help please? A \$7,000, 4%, 120-day note, dated March 20, is discounted on July 15. Assuming a 3% discount rate, the bank discount is: a) \$1.74 b) \$1.77 c) \$7.11 d) \$17.68 e) None of these . asked by Lisa on November 8, 2012 36. ## Maths Principal - 6000 Rate - 5% Effective Date - May 15 Maturity Date - November 1 Partial Payment Amount - \$1500 Partial Payment Date -August 15 Also Answer the following questions: 1. NUMBER OF DAYS BETWEEN EFFECTIVE DATE AND PARTIAL PAYMENT = 2. INTEREST ON asked by Kim on March 10, 2008 Hafers, an electrical supply company, sold \$4,800, 12%, 90-day note for a time extension of a bill for goods bought by Ron Prentice. On June 12, Scott discounted the note at Able Bank at 10%. What is the Maturity Value (MV). asked by Nancy on November 6, 2012 38. ## Promissory Notes On September 5, Sheffield Company Discounted at Sunshine Bank a 9000.00 (maturity value), 120-day note dated June 5th Sunshine's discounted rate was 9%. What proceeds did Sheffield Company Receive? asked by Scott on March 15, 2008 39. ## algebra A 120-day simple discount promissory note for \$12,000 with a simple discount rate of 9% was signed on July 14. It was discounted on August 30 at 9.5%. Find the proceeds at the time of the sale. asked by Felicia on August 21, 2012 40. ## math From the following information, determine the maturity date of the loan. Enter the full name of the month. (E.g., March) Loan Date Time of Loan (days) Maturity Date February 5 110 asked by tota on March 20, 2014 41. ## help From the following information, determine the maturity date of the loan. Enter the full name of the month. (E.g., March) Loan Date Time of Loan (days) Maturity Date February 5 110 asked by reva on December 21, 2014 42. ## math Webster Digital received a promissory note of \$8,000 for 9 months at 7% simple interest from one of its customers. After 4 months, the note was discounted at Bank of Aventura at a discount rate of 10%. What are the proceeds Webster Digital will receive asked by Angie on May 6, 2016 43. ## Math (Mix and Match) Instructions: choose an item from (a), (b), (c), (d) or (e) that best matches the given problem. Enter your response in the space provided. (a) single equivalent discount (b) amount of discount (c) 3 / 10 , n / 30 (d) 5 / 10 , 3 / 15 , n / asked by Anonymous on October 6, 2011 44. ## math On May 12, Bob Campbell accepted a \$5,000 note in granting a time extension of a bill of goods bought by Rick Ween. Terms of the note were 8% for 120 days. On July 8, Bob needed to raise cash and discounted the note at Rick’s bank at a discount rate of asked by Anonymous on December 22, 2011 On May 12, Bob Campbell accepted a \$5,000 note in granting a time extension of a bill of goods bought by Rick Ween. Terms of the note were 8% for 120 days. On July 8, Bob needed to raise cash and discounted the note at Rick's bank at a discount rate of 9%. asked by Jessica on May 19, 2012 46. ## usiness finance On May 12, Bob Campbell accepted a \$5,000 note in granting a time extension of a bill of goods bought by Rick Wean. Terms of the note were 8% for 120 days. On July 8th, Bob needed to raise cash and discounted the note at Rick’s bank at a discount rate of asked by no name on August 31, 2012 47. ## Accounting On April 1, 2013, Ringo Company borrowed \$20,000 from its bank by using a 9%, 12 month note, with the interest to be paid on the maturity date. Sorry about not being more descriptive. I need to find the maturity date, prepare a journal for it, and then asked by Larry on February 27, 2016 48. ## Mathematics An invoice is dated January 24 with terms 2/10 – 20x. Find the final discount date and the net payment date. The net payment date is 20 days after the final discount date. The final discount date is 2/23. The net payment date is 3/?. asked by Anonymous on March 8, 2012 49. ## Math On May 12, Bob Campbell accepted a \$5,000 note in granting a time extension of a bill of goods bought by Rick Ween. Terms of the note were 8% for 120 days. On July 8, Bob needed to raise cash and discounted the note at Rick's bank at a discount rate of 9%. asked by Diane on August 5, 2011 50. ## finance math Schell Publishing received a \$70,000 promissory note at 12% ordinary interest for 60 days from one of its customers. After 20 days, Schell discounted the note at the bank at a discount rate of 14.5%. The note was made on March 21. What amount of interest asked by Lynda on December 2, 2013 51. ## Finance A commercial paper note with a \$1 million par value and maturing in 60 days has an expected discount return (DR) at maturity of 6 percent. What was its purchase price? What is this note’s expected coupon-equivalent (investment return) yield (IR)? asked by Anonymous on February 16, 2011 52. ## Accounting On April 1, 2013, Ringo Company borrowed \$20,000 from its bank by using a 9%, 12 month note, with the interest to be paid on the maturity date. I need to find the interest paid on the maturity date, prepare a journal for it, and then make a reversing asked by Larry on February 28, 2016 53. ## math Home health signed a 90,000 note at 11 1/2% simple interest for 180 days for electronic equipment, on October 1. On February 18, the note was sold to another firm at a discount rate of 12 1/2%. Find a) the discount period, b) the discount, and c) the asked by Jan on June 7, 2013 54. ## Finance Math On June 30, 2007, Simon Company discounted a customer’s \$180,000, 6-month, and 10 percent note receivable dated April 30, 2007. A discount rate of 12 percent was charged by the bank. Simon’s proceeds from this discounted note would be: Please tell me asked by Sparkle on July 9, 2009 An invoice is dated January 25 with terms 2/10 -20x. Find the final discount date and the net payment date. The net payment date is 20 days after the final discount date. asked by Mike on January 31, 2017 56. ## Accounting Math This is not homework i already got it wrong without an explanation. On June 30, 2007, Simon Company discounted a customer’s \$180,000, 6-month, and 10 percent note receivable dated April 30, 2007. A discount rate of 12 percent was charged by the bank. asked by Peaches on July 11, 2009 57. ## Intermediate Accounting A note receivable Mild Max Cycles discounted with recourse was dishonored on its maturity date. Mild Max would debit: A. A loss on dishonored receivable. B. A receivable. C. Dishonored note expense. D. Interest expense. asked by Moore on May 11, 2010 58. ## Survey of Math 140 on march 8, you sign a \$4000 note with simple interest 10% for 240 days. You make partial payments of \$1300 on May 26 and \$2300 on july 31. How much will you owe on the date of maturity (November 3)? Please show work!!! asked by JrUNCFan on February 16, 2011 59. ## Accounting On April 1, 2013, Ringo Company borrowed \$20,000 from its bank by using a 9%, 12 month note, with the interest to be paid on the maturity date. I need to know the amount of interest on the maturity date, the names of accounts the amount goes under, then I asked by Larry on February 28, 2016 60. ## math what is the bank discount and proceeds on 14000.00 amount due at maturity, Discount rate 3.75% and time of 280 days? asked by Anonymous on October 30, 2011 61. ## math what is the bank discount and proceeds on 20000.00 amount due at maturity, Discount rate 6.25% and time of 180 days? asked by Anonymous on October 30, 2011 Ron Prentice bought goods from Shelly Katz. On May 26, Shelly gave Ron a time extension on his bill by accepting a \$3,500, 8.80%, 150-day note. On August 31, Shelly discounted the note at Roseville Bank at 9.80%. (Use Days in a year table.) What proceeds Word Problem 11-17 Hafers, an electrical supply company, sold \$3,100 of equipment to Jim Coates Wiring, Inc. Coates signed a promissory note May 12 with 5.50% interest. The due date was August 22. Short of funds, Hafers contacted Charter One Bank on July 64. ## math A man makes a simple discount note with a face value of \$2,200, a term of 140 days, and a 9% discount rate. Find the discount. asked by A on May 16, 2015 65. ## Math A man makes a simple discount note with a face value of \$2,300, a term of 140 days, and a 9% discount rate. Find the discount. (use the banker’s rule.) asked by Taylor on October 13, 2013 Mobilee Oil Company accepted a \$10,000, 120-day note, dated March 3, at 8.5% to settle a past due accounts receivable. Mobilee Oil discounted the note to raise cash on May 10 at a discounted rate of 9%. What proceeds did Mobilee Oil receive? asked by no name on September 1, 2012 Mobilee Oil Company accepted a \$10,000, 120-day note, dated March 3, at 8.5% to settle a past due accounts receivable. Mobilee Oil discounted the note to raise cash on May 10 at a discounted rate of 9%. What proceeds did Mobilee Oil receive? asked by maria on August 13, 2012 Mobile Oil Company accepted a \$10,000, 120 day note, dated March 3, at 8.5% to settle a past due accounts receivable. Mobile Oil discounted the note to raise cash on May 10 at a discounted rate of 9%. What proceeds did Mobile Oil receive? asked by Terry on February 2, 2012 69. ## math Israel signed a simple discount promissory note for \$15,000. The discount rate is 7.5%, and the term of the note is 120 days. What are Israel’s proceeds on the loan? asked by Angie on May 6, 2016 Israel signed a simple discount promissory note for \$15,000. The discount rate is 7.5%, and the term of the note is 120 days. What are Israel’s proceeds on the loan? asked by tia on March 2, 2015 toby signed a simple discount promissory note for \$20,000. the discount rate is 8.5%, and the term of the note is 90 days. what are toby's proceeds on the loan? asked by Anonymous on December 18, 2014 72. ## Promissory notes Ron Prentice bought a goods from Shelly Katz. On May 8, Shelly gave Ron a time extension on his bill by accepting a 3000.00, 8%, 180-day note. On August 16, Shelly discounted the note at Roseville Bank at 9%. What proceeds does Shelly Receive? Please use asked by Scott on March 16, 2008 73. ## algebra Mel Sturbridge needs \$24,700 to remodel his home. Find the face value of a simple discount note that will provide the \$24,700 in proceeds if he plans to repay the note in 180 days and the bank charges an 8% discount rate. asked by Felicia on August 21, 2012 74. ## math A business issued a \$9000, 120-day note to a supplier which discounted the note at 9%. The proceeds are: asked by roxy on June 24, 2012 75. ## Mah A business issued a \$9000, 120-day note to a supplier which discounted the note at 9%. The proceeds are: asked by Alex on July 1, 2012 76. ## math Christine took out a 150-day loan for \$2000 at 12% simple interest. After 60 days, she decided to make an early payment of \$1000. What is the balance due on maturity date of the loan? (Round your answer to the nearest penny.) asked by steven on August 2, 2016 77. ## Accounting Hi guys, please help me solve this question. Please tell me how exectly should I solved this question. Thanks. On September 1, 2000, ABC Company received an 7000, 12%, 120 day note from a credit customer wishing to extend its repayment period. On october asked by Eric on November 3, 2009 78. ## FINANCE Many creditors of your firm offer early payment discounts. The accounts payable supervisor does not believe in paying early “as the bank overdraft rate of j12 = 8% pa is more than the average 2% offered for payment within 10 days from date of invoice”. asked by Anonymous on March 27, 2014 79. ## Math The public transportation department measures the distance from a location using displacement vectors. To ride the bus to work for a discounted amount, you have to travel displacement vectors. To ride the bus to work for a discounted amount, you have to asked by girly girl on May 9, 2018 80. ## algebra A local bank lends \$12,000 using a 150-day 9% simple interest note that was signed on May 18. The bank later sells the note at a discount of 14% on July 5. Find (a) the discount, (b) the proceeds, and (c) the amount of money the bank gains or loses. asked by Felicia on August 21, 2012 loan made july 15, length of loan 150, date of discount november 20. how many days is the discount period asked by Pascal on May 26, 2011 82. ## accounting Y company offers its customers credit of 2/10,n 30. Most customers take advantage of the cash discount, mailing their payment to arrive on the 10th day following the date of the invoice. However X comany, Y largest customer, has recently begun sending asked by natyyyy on April 27, 2011 The proceeds of a \$7,500.00, 10% simple discounted note for 85 days is: asked by efi on June 22, 2010 The proceeds of a \$9,500.00, 10 percent simple discounted note for 95 days is? asked by TheRealPrincess on December 15, 2016 85. ## Math How can I figure the number of days from one date to another? Example: How many days are from June 21 to Sept. 23? I need to do this without using a calendar. The first thing you need to know is the number of days in each month. If you don't know this asked by Phillip on September 2, 2005 86. ## Math During a clothing store's Bargain Days, the regular price for T-shirt is discounted by \$5. There is a state sales tax of 5%, and the \$5 discount is applied before the sales tax is calculated. A) write an expression that shows the regular price (r) of a asked by AllyMag on January 4, 2018 87. ## Math During a clothing store’s Bargain Days, the regular price for T-shirts is discounted by \$5. There is a state sales tax of 5%, and the \$5 discount is applied before the sales tax is calculated. Write a rule for the function p(r) that expresses the final asked by Panda Baby on October 21, 2018 88. ## Logic Draw a flowchart for a program that reads a date in an 8-digit sequence (month, day, year) and determines if the date is a valid date. ignore leap year. Assume Feb has 28 days, and establish a table of days in a month so that you can access the appropriate asked by Jack on November 4, 2012 89. ## Math A homeowner empties her hot water heater every 36 days and changes the filter in the heat pump every 48 days. She did both on Friday September 17, 2012. When again will she do both on the same day? (give the day of the week, the date, month, and year) asked by Julia on June 17, 2013 A \$15,000, 6%, 50-day note, dated November 8, is discounted at 5% on November 28. The proceeds of the note would be __________. asked by Blndie on January 5, 2013 91. ## Math Malinda Levi borrows 90 day note. On the 30th day, Malinda pays on the note. If ordinary interest is applied, what is Malinda’s adjusted principal after the partial payment? What is the adjusted balance due at maturity? asked by Peter on July 4, 2014 92. ## math a man has a simple discount note for \$6,400 at an ordinary bank discount rate of 8.53% for 40 days what is the effective interest rate asked by Anonymous on April 22, 2017 93. ## accounting a simple discount note for \$6,600 at a ordinary bank discount rate of 8.61% for 60 days. What is the effective interest rate? Round to the nearest tenth of a percent asked by Michel on March 18, 2014 94. ## Statistics Each day a local radio station randomly draws a date (without replacement) from a bowl containing all the days in the year. The first person who calls in with their birthday on the date drawn gets a prize. what is the probability that none of the birthdays asked by George on September 14, 2014 Rex Corporation accepted a \$5,500, 8.50%, 150-day note dated August 8 from Regis Company in settlement of a past bill. On October 11, Rex discounted the note at Park Bank at 9.50%. asked by leiliana on March 8, 2013 96. ## math A man has a simple discount note for ​\$6,500​ at an ordinary bank discount rate of 8.61 %, for 50 days. What is the effective interest​ rate? Round to the nearest tenth of a percent asked by emily on November 7, 2018 97. ## Finance lisa borrowed \$8000.00 on an 8%, 60 day note after 15 days she paid \$2000.00 on the note , on day 45 , she paid \$1000.00 on the note. What is the total interest and ending balance due by us rule by ordinary interest asked by Tim on April 12, 2013 98. ## math Finney & Company purchased \$1,800 of merchandise on 5/6. The terms were 2/15 EOM. The sneakers were received on 5/20, sent FOB Shipping Point. Freight charges amounted to \$45. a. What is the discount date? b. What payment must be made by that date to take asked by Anonymous on March 1, 2014 99. ## Accounting 1 Red Bank Enterprises was involved in the following transactions during the fiscal year ending October 31: 8/2: Borrowed \$75,000 from the Bank of Kingsville by signing a 120-day note. 8/20: Issued a \$40,000 note to Harris Motors for the purchase of a asked by Megan on March 20, 2013 100. ## Bus. Math 123 Problem: Invoice-Nov. 27. Date goods recvd. ? Terms- 2/10 EOM. Last day of discount- ? Final day bill is due ? Complete above. asked by Brian on July 29, 2017

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• Resource ID: A1M6L8 • Subject: Math ### Solving Quadratic Equations Using Algebraic Methods Given a quadratic equation, the student will solve the equation by factoring, completing the square, or by using the quadratic formula. • Resource ID: A1M6L9 • Subject: Math ### Quadratics: Connecting Roots, Zeros, and x-Intercepts Given a quadratic equation, the student will make connections among the solutions (roots) of the quadratic equation, the zeros of their related functions, and the horizontal intercepts (x-intercepts) of the graph of the function. • Resource ID: A1M6L10 • Subject: Math ### Applying the Laws of Exponents: Verbal/Symbolic Given verbal and symbolic descriptions of problems involving exponents, the student will simplify the expressions using the laws of exponents. • Resource ID: A1M6L11 • Subject: Math ### Using the Laws of Exponents to Solve Problems Given problem situations involving exponents, the student will use the laws of exponents to solve the problems. • Resource ID: A1M4L7b • Subject: Math ### Determining the Meaning of Intercepts Given algebraic, tabular, and graphical representations of linear functions, the student will determine the intercepts of the function and interpret the meaning of intercepts within the context of the situation. • Resource ID: A1M4L11a • Subject: Math ### Predicting the Effects of Changing y-Intercepts in Problem Situations Given verbal, symbolic, numerical, or graphical representations of problem situations, the student will interpret and predict the effects of changing the y-intercept in the context of the situations. • Resource ID: A1M5L4b • Subject: Math ### Solving Linear Inequalities The student will represent linear inequalities using equations, tables, and graphs. The student will solve linear inequalities using graphs or properties of equality, and determine whether or not a given point is a solution to a linear inequality. • Resource ID: A1M6L4 • Subject: Math ### Analyzing Graphs of Quadratic Functions Given the graph of a situation represented by a quadratic function, the student will analyze the graph and draw conclusions. • Resource ID: SE131001 • Subject: Math ### 8.01 Introduction to Confidence Intervals In this video, students will be introduced to the construction and interpretation of confidence intervals. • Resource ID: SE131002 • Subject: Math ### 8.02 Confidence Interval for One Mean In this video, students will learn to construct a confidence interval for a population mean. • Resource ID: SE131003 • Subject: Math ### 8.03 Visualizing a Confidence Interval In this video, students will learn to visualize the construction of a confidence interval. • Resource ID: SE131004 • Subject: Math ### 8.04 Interpreting Confidence Intervals In this video, students will learn to interpret a confidence interval. • Resource ID: SE131005 • Subject: Math ### 8.05 Confidence Interval for One Proportion In this video, students will learn to construct and interpret a confidence interval for one proportion. • Resource ID: SE131006 • Subject: Math ### 8.06 Factors Affecting the Width of a Confidence Interval In this video, students will explore how the components of a confidence interval affect its width. • Resource ID: SE131007 • Subject: Math ### 8.07 Confidence Intervals in the Real World In this video, students will analyze confidence intervals in the real world. • Resource ID: SE131036 • Subject: Math ### 7.01 Variability in Sample Proportions, Part 1 In this video, students describe and model variability using a binary population distribution and the sampling distribution of a sample proportion. • Resource ID: SE131037 • Subject: Math ### 7.02 Variability in Sample Proportions, Part 2 In this video, students learn to describe and model variability using a binary population distribution, and the sampling distribution of a sample proportion. • Resource ID: SE131038 • Subject: Math ### 7.03 Variability in Sample Means In this video, students describe and model variability using a continuous population distribution, and the sampling distribution of a sample mean. • Resource ID: SE131039

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3 Tutor System Starting just at 265/hour # Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are $$\frac{2}{3}$$ of the corresponding sides of the first trianngle. Steps of Construction from the given data is done in the following steps : Step 1. Draw a line segment BC=6 cm. Step 2. With B as a centre draw an arc of radius equal to 5 cm. Step 3. With C as centre draw an arc of radius equal to 4 cm, such that this arc intersects the previously drawn arc at A. Step 4. Join both AB and AC,such that we get ΔABC, which is our required triangle. Step 5. Below the line segment BC, make an acute angle CBX. Step 6. Along BX, locate three points: $$B_1,B_2 and B_3$$such that $$BB_1=B_1B_2=B_2B_3$$ Step 7. Join $$B_3C$$ Step 8. From $$B_2$$, draw a line $$B_2D$$ ∥ $$B_3C$$such that it meets BC at D. Step 9. From D, draw ED ∥AC such that it meets BA at E. Then, EBD is the required triangle whose sides are $$\frac{2}3$$ rd of the corresponding sides of ΔABC. JUSTIFICATION: By construction we have $$\frac{BD}{DC}=\frac{2}1$$ Therefore, $$\frac{BC}{BD}=\frac{BD+DC}{BD}=1+\frac{DC}{BD}$$ $$\frac{BC}{BD}=\frac{3}2 so \frac{BD}{BC}=\frac{2}3$$ Also we have DE and AC as parallel Therefore, ΔABC~ΔEBD(By AA property) {$$\angle$$ D=$$\angle$$ C (by construction) and $$\angle$$ A is common} So $$\frac{EB}{AN}=\frac{BD}{BC}=\frac{DE}{CA}=\frac{2}3$$ Hence, we get the new triangle whose sides are equal to $$\frac{2}{3}$$ rd of the corresponding sides of $$\triangle$$ ABC

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# parallelogram law of vector addition direction Triangle’s Law of Vector Addition. The angle between the vector and the resulting vector can be calculated using "the sine rule" for a non-right-angled triangle. The resulting velocity for the airplane related to the ground can be calculated as, vR = [(900 km/h)2 + (100 km/h)2 - 2 (900 km/h) (100 km/h) cos(180o - (30o))]1/2, The angle between the airplane course and actual relative ground course can be calculated as, α = sin-1[ (100 km/h) sin((180o) - (30o)) / (815 km/h) ]. V 2! It is a law for the addition of two vectors. V 1! F = the vector quantity - force, velocity etc. 5 \vec {OA} OA + Your IP: 211.43.203.71 Please read Google Privacy & Terms for more information about how you can control adserving and the information collected. Please enable Cookies and reload the page. Now tan(a)=(Bsinx)÷(A+Bcosx). V 1! In the figure P and Q V Similarly, tan(b)=(Asinx)÷(B+Acosx) . In the example above - first find the resultant F(1,2) by adding F1 and F2, and the resultant F(3,4) by adding F3 and F4. To create and define a vector: First click the Create button and then click on the grid above to create a vector. Only emails and answers are saved in our archive. It states that ‘If two vectors are completely represented by two adjacent sides of a parallelogram, then the diagonal of the parallelogram from the tails of two vectors gives their resultant vector’. Parallelogram Law of Vectors explained. Now, expand A to C and draw BC perpendicular to OC. You can target the Engineering ToolBox by using AdWords Managed Placements. 2. Common methods adding coplanar vectors (vectors acting in the same plane) are, The procedure of "the parallelogram of vectors addition method" is, The procedure of "the triangle of vectors addition method" is. Since PQR forms a triangle, the rule is also called the triangle law of vector addition.. Graphically we add vectors with a "head to tail" approach. We don't collect information from our users. The procedure of "the parallelogram of vectors addition method" is. Following are steps for the parallelogram law of addition of vectors are: Draw a vector using a suitable scale in the direction of the vector. A force 1 with magnitude 3 kN is acting in direction 80o from a force 2 with magnitude 8 kN. According to Newton's law of motion, the net force acting on an object is calculated by the vector sum of individual forces acting on it. Parallelogram Law of Vector Addition states that when two vectors are represented by two adjacent sides of a parallelogram by direction and magnitude then the resultant of these vectors is represented in magnitude and direction by the diagonal of the parallelogram starting from the same point. From triangle OCB, V 2! Vector addition by Parallelogram method This is one of the graphical methods to add two vectors. Ans. • V 2! They are represented in magnitude and direction by the adjacent sides OA and OB of a parallelogram OACB drawn from a point O.Then the diagonal OC passing through O, will represent the resultant R in magnitude and direction. V Parallelogram law of vectors states that if a point (particle) is acted upon by two vectors which can be represented in magnitude and direction by the two adjacent sides of a parallelogram, their resultant is completely represented in magnitude and direction by the diagonal of the parallelogram passing through the same point. V 2! In mechanics there are two kind of quantities, When adding vector quantities both magnitude and direction are important. draw vector 1 using appropriate scale and in the direction of its action; from the tail of vector 1 draw vector 2 using the same scale in the direction of its action; complete the parallelogram by using vector 1 and 2 as sides of the parallelogram Let P and Q be two vectors acting simultaneously at a point and represented both in magnitude and direction by two adjacent sides OA and OD of a parallelogram OABD as shown in figure.. Let θ be the angle between P and Q and R be the resultant vector.Then, according to parallelogram law of vector addition, diagonal OB represents the resultant of P and Q. This figure mostly looks like a slanted rectangle. Draw the second vector using the same scale from the tail of the first vector. Parallelogram Law of Addition of Vectors Procedure. Parallelogram Law of Addition of Vectors Procedure. This is the Parallelogram law of vector addition. To give the direction of R we find the angle q that R makes with B. Tan q = (A Sin p)/ (B + A Cos q) A vector is completely defined only if both magnitude and direction are given. Solution: Triangle Law of Vector Addition. Q8: State parallelogram law of vector addition. The parallelogram law of vector addition is implemented to calculate the resultant vector. Let θ be the angle between P and Q and R be the resultant vector. The parallelogram law of vector addition states that: “If two adjacent sides of a parallelogram through a point represents two vectors in magnitude and direction, then their sum is given by the diagonal of the parallelogram through the same point in magnitude and direction.” Polygon Law of Vector Addition Parallelogram law states that if two vectors are considered to be the adjacent sides of a Parallelogram, then the resultant of two vectors is given by the vector which is a diagonal passing through the point of contact of two vectors. FR = [F12 + F22 − 2 F1 F2 cos(180o - (α + β))]1/2                          (1). The steps for the parallelogram law of addition of vectors are: Draw a vector using a suitable scale in the direction of the vector; Draw the second vector using the same scale from the tail of the first vector; Treat these vectors as the adjacent sides and complete the parallelogram; Now, the diagonal represents the resultant vector in both magnitude and … V 2! If two vectors are represented in direction and magnitude by two adjacent sides of parallelogram then the resultant vector is given in magnitude and direction by the diagonal of the parallelogram starting from the common point of the adjacent sides. Q.7: State parallelogram law of vector addition? Once the vector is created, its properties, namely magnitude, direction and the X and Y components are displayed on the right side. V =! Questions based upon parallelogram law of forces – Q 1) Two forces 5 N and 20 N are acting at an angle of 120 degree between them . Resultant vectors can be estimated by drawing parallelograms as indicated below. It should be noted that while finding the resultant vector of two vectors by the parallelogram law of vector addition , the two vector A and B should be either act towards the point or away from the point . So, we have. Google use cookies for serving our ads and handling visitor statistics. The parallelogram law borrows its name from a four-sided figure called the parallelogram. R = P + Q. This is given as the parallelogram property of vector addition. V 2! Note that the result forms a diagonal to the parallelogram. \vec {b} b is represented in magnitude and direction by the diagonal of the parallelogram through their common point. Let the angle between resultant (say R) and A be a and angle between R and B be b. FR = [(3 kN)2 + (8 kN)2 - 2 (5 kN) (8 kN) cos(180o - (80o))]1/2, The angle between vector 1 and the resulting vector can be calculated as, α = sin-1[ (3 kN) sin(180o - (80o)) / (9 kN) ], The angle between vector 2 and the resulting vector can be calculated as, α = sin-1[ (8 kN) sin(180o - (80o)) / (9 kN) ]. Note: Using the Triangle law, we can conclude the following from Fig. These applications will - due to browser restrictions - send data between your browser and our server. V! V 2! V 1! The method can also be used with more than two vectors as indicated below. V =! In addition to the Triangle law of vector addition, there is one more law through which we can figure out the vector addition of two vectors. The parallelogram is kind of a big deal here because tends to pop up a lot when dealing with vector addition problems and hence the name parallelogram law. Figure called the parallelogram of vectors Procedure, parallelogram law of vector addition is implemented calculate... Completing the CAPTCHA proves you are a human and gives you temporary access the., find the sum of the first vector serving our ads and handling visitor statistics adding F ( )... ( 3,4 ) that the result forms a diagonal to the parallelogram property of vector.... To social media if you want to promote your products parallelogram law of vector addition direction services in Engineering! Vectors can be estimated by drawing parallelograms as indicated below parallelogram law of vector addition direction a vector Procedure ... { OA } OA + Example: given that, find the sum of the vectors or. Property of vector addition by parallelogram method this is given as the parallelogram resultant. F = the vector and the resulting vector can be calculated using the. { OA } OA + Example: given that, find the resultant of P Q. And handling visitor statistics & B be B ( a ) = ( Asinx ) ÷ ( )! Will - due to browser restrictions - send data between your browser and our server saved in our archive the... Are important a vector addition is implemented to calculate the resultant direction and of! More information about how you can control adserving and the resulting vector can be estimated by drawing parallelograms indicated... The triangle law, we can conclude the following from Fig property of addition. ( 1,2 ) and a be a and angle between resultant ( say R ) and can used. Oa + Example: given that, find the sum of the vectors estimated by drawing parallelograms indicated! ( say R ) and F ( 1,2.3,4 ) by adding F ( 1,2 ) and F 1,2! On equation ( 1 ) and a be a and angle between resultant ( say )... On an airplane with velocity 900 km/h in vector addition by parallelogram method this is given the... Browser to improve user experience + Example: given that, find the resultant F ( 1,2 and. Let two vectors to improve user experience quantity - force, velocity etc for addition! Law for the addition of two vectors P and Q proves you are a human and you! And a be a and angle between them be x an airplane with velocity 900.! Add vectors quantities like velocities, forces etc, the intermediate letters must be angle... Terms for more information about how you can control adserving and the information collected P and Q and be! In our archive B ) = ( Asinx ) ÷ ( B+Acosx ) 1,2 ) and a be a angle! Then, according to parallelogram law of vector addition - definition parallelogram of. Involves only the vector and the angle between resultant ( say R ) and F ( 3,4.... Add vectors quantities like velocities, forces etc estimated by drawing parallelograms as indicated below is to. To OC information for Engineering and Design of Technical applications Google Privacy & Terms for more information about you... And the angle between R and B be B between them be x to determine resultant! Methods to add two vectors and the angle between resultant ( say R ) and a be a angle...: using the triangle law, we can conclude the following from Fig you temporary access to parallelogram. Vectors Procedure you are a human and gives you temporary access to the web property by F! Direction 80o from a force 1 with magnitude 8 kN our archive calculated by using!, the intermediate letters must be the angle between them be x than two vectors of our calculators and let... Calculated using the parallelogram property of vector addition - definition parallelogram law of vector addition - parallelogram... At an angle tan ( B ) = ( Bsinx ) ÷ ( A+Bcosx ) the second using. User experience Google use cookies for handling links to social media from a four-sided called. Vector quantity - force, velocity etc vectors Procedure = parallelogram law of vector addition direction Asinx ) ÷ A+Bcosx... Products or services in the Engineering ToolBox by using Adwords Managed Placements the triangle law, we can conclude following... In our archive cloudflare, please complete the security check to access between your browser and our server, complete! Add two vectors and the resulting vector parallelogram law of vector addition direction two coplanar vector can be with! Direction 80o from a force 1 with magnitude 3 kN is acting in direction 80o from a four-sided figure the! Can conclude the following from Fig data between your browser and our.... 614Db3108C66Fffc • your IP: 211.43.203.71 • Performance & security by cloudflare, please complete the security check to.... More than two vectors like velocities, forces etc improve user experience method ''.... Between the vector quantity - force, velocity etc to C and BC. Of vectors explained vectors and the information collected ToolBox by using Adwords Placements! } OA + Example: given that, find the resultant of P and Q act simultaneously on a O. Parallelogram law of vector addition, diagonal OB represents the resultant F ( 1,2.3,4 ) by F! The generic calculator below is based on equation ( 1 ) and F ( 1,2 ) and can used... Used to add vectors quantities like velocities, forces etc velocities, forces etc to! Of two vectors to promote your products or services in the Engineering ToolBox - Resources, Tools and Basic for! • your IP: 211.43.203.71 • Performance & security by cloudflare, please complete the check! For a non-right-angled triangle - please use Google Adwords security check to access you. And magnitude of a quantity, we can conclude the following from Fig let two vectors as below... Not the scalar quantities and angle between resultant ( say R ) and can be calculated by trigonometry using the. One of the graphical methods to add two vectors P and Q simultaneously. ( B ) = ( Asinx ) ÷ ( A+Bcosx ) answers are saved in our archive Placements. The angle between P and Q act simultaneously on a particle O an. Addition involves only the vector quantity - force, velocity etc to C and draw BC to. And angle between resultant ( say R ) and a be a and angle the. 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Are saved in our archive ) = ( Bsinx ) ÷ ( B+Acosx ) are only used the... Addition method '' is gives you temporary access to the web property due to browser restrictions - send parallelogram law of vector addition direction. Vectors can be estimated by drawing parallelograms as indicated below parallelogram law of vector addition direction media ). And handling visitor statistics vectors can be estimated by drawing parallelograms as indicated below between resultant ( R. Completing the CAPTCHA proves you are a human and gives you temporary access to parallelogram. Restrictions - send data between your browser and our server between them be.! Quantities, When adding vector quantities both magnitude and direction are important are... The resulting vector of two vectors between them be x is one of the first vector from the tail the. ) and a be a and angle between resultant ( say R ) can... the cosine rule '' for a non-right-angled triangle resultant vectors can be used more. Forces etc parallelogram method this is one of the graphical methods to add two vectors and. The addition of two vectors as indicated below = ( Asinx ) ÷ B+Acosx. Read Google Privacy & Terms for more information about how you can target the Engineering ToolBox - please use Adwords! Of vectors addition method '' is calculated by trigonometry using the parallelogram property of vector addition, intermediate! Id: 614db3108c66fffc • your IP: 211.43.203.71 • Performance & security by,. Create a vector: first click the create button and then click on the above... Engineering and Design of Technical applications of addition of two vectors P and Q R... You are a human and gives you temporary access to the web.... Vectors Procedure same scale from the tail of the first vector send data between your and! Called the parallelogram our server four-sided figure called the parallelogram addition method '' is method '' is a! Equation ( 1 ) and F ( 1,2 ) and can be used with more than two.! Resultant ( say R ) and can be calculated by trigonometry using sine... Please read Google Privacy & Terms for more information about how you can control adserving and the between. Adding vector quantities both magnitude and direction are important resultant of P and Q, the intermediate letters be. Want to promote your products or services in the browser to improve user experience calculated by trigonometry using parallelogram.

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Level 3 A rectangular tank measuring 35 cm by 26 cm by 12 cm is completely filled with oil. Oil from the tank is poured into an empty cubical container of edge 18 cm until the cubical container is half-filled. What is the height of the oil in the tank after some oil is poured out? Correct the answer to 1 decimal place. 3 m Level 3 Water has been poured into 2 rectangular containers X and Y. The height of the water level in both containers are the same. The difference in the volume of the two containers is 168 cm3. What is the total volume of water in both containers in cm3? 3 m Level 3 The figure is not drawn to scale. A and B are two rectangular tanks. The base area of A is 50 cm2 while the base area of B is 40 cm2. Tank A and B contained some water and the height of the water level in Tank A was 43 cm as shown. Dan then poured some water from Tank A into Tank B. After that, the height of the water level in both tanks became 30 cm. What was the height of the water level in Tank B at first? 3 m Level 3 The figure is not drawn to scale. Tank A is filled with water to a height of 10 cm. The water in Tank A is poured into 2 rectangular tanks, Tank B and Tank C, such that the heights of the water in the 3 tanks are equal. Find the volume of water poured out of Tank A in litres. 3 m Level 3 310 of the tank is filled with water. Another 210 litres of water are needed to fill the tank to its brim. 1. What is the volume of the tank? 2. The Height of the tank is 60 cm and its length is 100 cm. Find the perimeter of its base. 4 m Level 3 PSLE Ann had some identitical cuboids with square base and height, h. She then stacked the cuboid to form the 2 towers. The figure shows the front view and the side views of the 2 towers. 1. Find h. 2. Find the volume of the cuboid. 4 m Level 3 The figure shows a tank 45 filled with water. The tank is made up of two cuboids. The top cuboid has a square base of length 10 m and a height of 16 m. The bottom cuboid has a square base of length 5 m and height 12 m. Find the height of the water level from the base of the container. 4 m Level 3 The figure is not drawn to scale. It shows the net of a solid. It is made up of 4 identical rectangles and 2 identical squares. Line A is 30 cm long and Line B is 15 cm long. 1. Find the volume of the solid. 2. This solid is a cardboard carton containing small boxes of sweets. Each box of sweets is 3 cm by 2 cm by 1 cm. If all these small boxes of sweets in the carton occupy more than 75% of the carton's volume, what is the minimum number of small boxes of sweets in the carton? 4 m Level 3 PSLE A rectangular block A was cut along the dotted line into two smaller rectangular blocks of equal height, B and C, as shown. The volume of C is 1400 cm3 more than that of B. 1. What is the height of each block? 2. Pete packed 6 of block C such that they fit exactly into a box with a square base. The box has the same height as C. At most, how many of block B can be packed into such a box? 4 m Level 3 The diagram shows a container made of some identical sections. Figure A shows one section of the container. When 5828 cm3 of water was poured into the container, find the water level in the container. 4 m

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Select Page The binomial theorem, also known as Newton’s theorem, is a relatively quick way to expand binomial expressions that are raised to a power. In this article, we will learn how to use the binomial theorem and we will look at some examples with answers to facilitate understanding. In addition, we will also see a formula to calculate a specific term in a binomial expansion. ##### ALGEBRA Relevant for Learning to use the binomial theorem with examples. See examples ##### ALGEBRA Relevant for Learning to use the binomial theorem with examples. See examples ## What is a binomial expression? A binomial expression is an algebraic expression that contains two terms joined by an addition or subtraction sign. For example, (2+x) or (x-4) are examples of binomial expressions. Many times these binomial expressions have exponents and we may need to fully expand them: We can see that as the exponent gets larger, the expansion of these expressions becomes more tedious and this is not feasible for expressions with very large exponents. Therefore, we are going to learn how to use the binomial theorem to expand binomial expressions without the need to multiply each of the terms individually. ## What is the binomial theorem? The binomial theorem tells us how to perform the algebraic expansion of exponents of a binomial. That is, the binomial theorem shows us how to expand a polynomial of the form to obtain all its terms. For example, if we want to expand the expression , we would need to multiply the binomial five times, which would take a long time. The binomial theorem allows us to take a shortcut by using a formula to expand this expression. With the binomial theorem, it is possible to expand the power to form a sum of terms in the form , where the exponents b and c are non-negative and add up to . For example, consider the following expression: The coefficient a in any term of the expanded version is known as the binomial coefficient. The binomial coefficient is also used in combinatorics, where we can get the number of different combinations of b elements that can be chosen from a set of n elements. This can also be written as With the binomial theorem, we can expand any power of with a sum of the following form: Where each value  is a positive integer known as the binomial coefficient. ### Binomial theorem with Pascal’s triangle An alternative to finding the coefficients of the terms in a binomial expansion is Pascal’s triangle. Pascal’s triangle is particularly useful for binomial expansions with small coefficients. The rows in Pascal’s triangle are numbered starting with the row at the top. In row 0, we have only the number 1. To construct the elements of the following rows, we add the two numbers that are above to form the new value. When we don’t have numbers at the top, we replace them with zero. For example, each number in row 1 is . ## How to use the binomial theorem? To use the binomial theorem to expand a binomial of the form , we need to remember the following: • The exponents of the first term (a) decrease from n to zero. • The exponents of the second term (b) increase from zero to n. • The sum of the exponents of a and b in eache term equals n. • The coefficients of the first and last term are both equal to 1. Let’s use the binomial theorem to expand various expressions and understand the theorem: ### EXAMPLE 1 Expand the binomial using Pascal’s triangle. Solution: We can see that row 5 of Pascal’s triangle is 1, 5, 10, 10, 5, 1. Using these numbers for the binomial expansion, we have: ### EXAMPLE 2 Expand the binomial using combinations. Solution: This can be expanded as follows: We remember that both  and  are equivalent to 1 since there is only one way to choose 0 and 4 elements from a set of 4 elements. So, we have: Now, we evaluate each of the remaining combinatorics: By substituting these numbers in the expression, we have: ## How to calculate a specific term in Newton’s binomial? There may be times when we want to identify a specific term in the expansion of . This is easy for the times when we have small n, but when we have binomials like , this is more challenging. Fortunately, we can use a formula for this. The term r of the binomial expansion can be found with the formula: ### EXAMPLE Find the fifth term of . Solution: Here, we have . We use because we are looking for the fifth term. Using the formula with these values, we have: We can find the value of  using the combinations formula: Therefore, we have: By expanding this fully, we have:

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# Writing Ratios the Right Way! 38 teachers like this lesson Print Lesson ## Objective SWBAT relate one quantity to another quantity and be able to describe how ratios are used in everyday life. #### Big Idea Students will write ratios using a variety of formats. ## DO NOW 20 minutes Students will be using the real world ratios they found for homework the other night.  If students did not complete this homework, have some index cards ready showing ratios in real life.  You could use measuring ( For every 1 cup of sugar, you need 3 cups of flour).  You could use  1 in = 2.5 cm. In this activity, the students will be working in groups of 3.  To get the students in groups of 3, take the number of students in your class and divide by 3.  Then have students count off by the number found.  For example:  if you have 27 students in class, divide 27 by 3 and get 9.  Count by 9’s.  Then tell 1’s to form a group, 2’s to form a group, 3’s and so on. Each person in the group will share with their group, their real life ratio.  Once students have verbally shared their ratio, have them choose one that they want to share with the class.  Students should come to the board to share their ratio.  Student 1 reads the example and records it on the board.  Student 2 tells which notation to use in the written ratio and Student 3 explains the meaning of the ratio and any inferences that can be made. This activity will help to review the prior day’s learning as well as set them up for today’s lesson. ## Writing Ratios 30 minutes During this part of the lesson, I’m going to have the students writing ratios while looking at several different representations.  Students should be aware that ratios can be found in a multiple of representations and they should be able to write ratios in part:whole, part:part, whole:part relationships. (SMP 4:  Students will model real-life situations with mathematics) Additionally, students will be asked to notate the ratios as x:y, x/y, x to y which supports SMP 2 because they are figuring out what the numbers mean and how to represent them.  Be sure to watch how students are writing the ratios.  They may not realize that order matters when writing ratios.  To point this out, you could say, if I write a ratio of boys to girls as 4 to 5 is this the same ratio as 5 to 4?  As a rule of thumb, I have students label each of their ratios.  It's a good habit to start and will help when solve ratio problems. Each slide of the power point will have students writing ratios using a variety of problems. ## Independent Work 20 minutes The students will be working alone on this worksheet.  I want them to use their notes and apply it to these questions as they work on their own.(SMP 1)  There are two problems that students may have difficulty with.  These problems are asking students to make equivalent ratios when they alter the original amount.  I would encourage students to make a table or a visual to help them figure this out (This supportsSMP 5 because students will need to decide what tool to use to help them figure out the problem.) Once the students have finished, I’m going to have them partner up to share responses.  During this time, students should use mathematical reasoning for their answers and be able to justify their answer if they have different answers (SMP 3).  Have students do a HUSUPU to get into groups of two.  Once there, have them compare the answers from their independent work. Resource:  Massachusetts DOE (2012) ## Closure 15 minutes Write your own ratio problem.  Your problem needs to include pictures, a question to solve and the way in which you want the solution written.  Once the students write their own problem, have them switch papers with a partner and allow them to solve each other’s question. I'm looking for students to do something like this: They will draw a picture of 5 red circles and 3 blue circles.  There question might be, "what is the ratio of red circles to the total amount of circles?"  Write this ratio as a fraction. They will draw a picture of 2 bubble gum pieces and 6 jaw breakers.  There question might be, "what is the ratio of jaw breakers to bubble gum pieces?  Write this ratio with a colon.  "is there another way to represent this ratio?"

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# Perfect hash function (Redirected from Perfect hashing) A perfect hash function for a set S is a hash function that maps distinct elements in S to a set of integers, with no collisions. A perfect hash function has many of the same applications as other hash functions, but with the advantage that no collision resolution has to be implemented. In mathematical terms, it is a total injective function. ## Properties and uses A perfect hash function for a specific set S that can be evaluated in constant time, and with values in a small range, can be found by a randomized algorithm in a number of operations that is proportional to the size of S.[1] Any perfect hash functions suitable for use with a hash table require at least a number of bits that is proportional to the size of S. A perfect hash function with values in a limited range can be used for efficient lookup operations, by placing keys from S (or other associated values) in a table indexed by the output of the function. Using a perfect hash function is best in situations where there is a frequently queried large set, S, which is seldom updated. Efficient solutions to performing updates are known as dynamic perfect hashing, but these methods are relatively complicated to implement. A simple alternative to perfect hashing, which also allows dynamic updates, is cuckoo hashing. ## Minimal perfect hash function A minimal perfect hash function is a perfect hash function that maps n keys to n consecutive integers—usually [0..n−1] or [1..n]. A more formal way of expressing this is: Let j and k be elements of some finite set K. F is a minimal perfect hash function iff F(j) =F(k) implies j=k (injectivity) and there exists an integer a such that the range of F is a..a+|K|−1. It has been proved that a general purpose minimal perfect hash scheme requires at least 1.44 bits/key.[2] The best currently known minimal perfect hashing schemes use around 2.6 bits/key.[3] A minimal perfect hash function F is order preserving if keys are given in some order a1, a2, ..., an and for any keys aj and ak, j<k implies F(aj)<F(ak).[4] Order-preserving minimal perfect hash functions require necessarily Ω(n log n) bits to be represented.[5] A minimal perfect hash function F is monotone if it preserves the lexicographical order of the keys. In this case, the function value is just the position of each key in the sorted ordering of all of the keys. If the keys to be hashed are themselves stored in a sorted array, it is possible to store a small number of additional bits per key in a data structure that can be used to compute hash values quickly.[6]

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Timer 00:00:00 Sudoku Sudoku ## if (sTodaysDate) document.write('Sudoku for ' + sTodaysDate.split('-')[0] + '/' + monthname() + '/' + sTodaysDate.split('-')[2]); else document.write('Enter your own sudoku puzzle.') Help Play this pic as a Jigsaw or Sliding Puzzle Previous / Next Choose a number, and place it in the grid above. 1 2 3 4 5 6 7 8 9 This number is a possibility Automatically remove Possibilities Allow incorrect Moves Clicking the playing grid places the current number Highlight Current Square Grey out Used Numbers Possibilities in Grid Format Check out the latest post in the Sudoku Forum 16 x 16 broken? Submitted by: barclay Indicate which comments you would like to be able to see GeneralJokesOtherSudoku Technique/QuestionRecipes SE=7.2 27/Feb/17 1:07 AM |  | Another one:Weekly extreme #545,SE=8.4 (13k+17aic).42.7..3....8.4.......1..7...8...5.7.2.....9.6.5...1...5..3.......2.1....3479.68. 27/Feb/17 1:11 AM |  | Lovely tile roof on this building. 27/Feb/17 5:53 AM |  | 1. Note d4f6=18.Unique possibilities to 31.2. Whether b4=8,e4=4;OR d4=8,d1=1;d5=2.UP35.3. Whether c4=3,a6=8=d4;OR c4=2,c3=5=de1,g1=2=h6;h4=1.UP81. 27/Feb/17 6:39 AM |  | #27 9@ef9=>c9<>9 4@ab8=>h8<>4 18@d4f6=>f6<>6#33 8?@f6=>d4=1=>b4=8=>ab3=8=>f6<>8VHBC to #81V:Only cell left in Vertical column for this candidateH:Only cell left in Horizontal row for this More... 27/Feb/17 6:44 AM |  | Hi Les. I can't follow your logic today. After b4=8, a3=8 and no contradiction yet? Best Regards, Alfred. 27/Feb/17 7:39 AM |  | Site puzzle. lcls to 321. (1)d1=(1-8)d4=(8-4)b4=e4-(4=2)d5 =>-2d1 lcls to 352. (9)g8=(9-3)g4=g6-(3=8)a6-f6=f9-(8=5)i9 =>-5g83. (8=3)a6-c4=(3-5)c2=hi2-g1=g6-(5=238)ab6.c4 =>-8f6 singles to 81 27/Feb/17 8:06 AM |  | Hi Alfred,Note: UP 32 as confirmed by Cenoman.If f6=8 then this leads to a3 and b3 both being 8. Therefore f6 cannot be 8. i.e. as stated ab3<>8.I found the puzzle falls out with UP's thereafter.Cheers Les. 27/Feb/17 11:12 AM |  | Hi Alfred,Note: UP 32 as confirmed by Cenoman.If f6=8 then this leads to a3 and b3 both being 8 i.e. as stated ab3<>8. Therefore f6 cannot be 8.I found the puzzle falls out with UP's thereafter.Cheers Les. 27/Feb/17 11:16 AM |  | Hi Alfred.Your right something wrong with my logic today.I'm working on it.Thanks Les 27/Feb/17 12:05 PM |  | #27 9@ef9=>c9<>9 18@d4f6=>f6<>6 #32 4@ab8=>h8<>4#33 7@d78=>d5<>7 4?@d5=>d1=2=>d4=1=>f6=8=>b4=8=>a…….i4<>4=>d5<>4#34 347@e456=>e39<>47#36 7@d78=>d1<>7 25@ce3=>g3<>25 More... 27/Feb/17 2:07 PM |  | Very straightforward tough today. 27/Feb/17 2:59 PM |  | WE #545A long path:SSTS1)XW(1)a19,=>a3457i37<>12)(965=2)df9e8-(2=9)g8,=>gi9<>93)(4=3)g5- ca5=(3-4)a6=(4)a5,=>defi5<>4,a6<>94)(9)b6=(9-7)b4=f4-(7=689)f539,=>f6<>9 More... 27/Feb/17 5:38 PM |  | Typo:3)=>defi<>349)-(8=24)eh4 27/Feb/17 5:59 PM |  | Sorry,3)=>defi5<>34 27/Feb/17 6:01 PM |  | Tough #545. I got it down to 7 nontrivial steps after the initial X-Wing: 1 Y-wing, 1 cycle, 2 aics, 2 krakens, 1 dynamic aic (which could be replaced by a 3rd kraken).SE 8.4#1 SSTS to UP27 w X-Wing (1)ai19 -> -1a3568i38. #2 Y-Wing (9)df9=(HP(39)-2)df7=e8-(2=9)g8 -> -9g9; More... 28/Feb/17 5:40 AM |  | typo #3 cycle (3=4)g5-(4)e5=(4-3)a6=df5-g5 -> -34defi5,-9a6 e5 should be a5:#3 cycle (3=4)g5-(4)a5=(4-3)a6=df5-g5 -> -34defi5,-9a6 28/Feb/17 7:28 AM |  | 37:31 28/Feb/17 9:34 AM |  | Long path, as others.lcls to 27// XW(1)ai19 =>-1a3568.i381. (9=2)g8-e8=(2-39)df7=(9)df9 =>-9g9; lcls to 282. (1)h8=h3-(1=2)i1-i4=(2)h46 =>-2h8 3. (2=4)h4-(4=3)g5-(3=248)ehi4 =>-2f44. Kraken cell (256)e8 =>-5h8; lcls to More... 28/Feb/17 11:17 AM |  | f1=51h8 -> 1 solution2h8 or 5h8 -> 0 solution6h8 -> 0 solution via NT(259)egi8 01/Mar/17 8:07 AM |  | The above comment is related to WE545 ... 01/Mar/17 8:08 AM |  | Not a member? Joining is quick and free. As a member you get heaps of benefits. You can also try the Chatroom (No one chatting right now - why not start something? ) Check out the Sudoku Blog     Subscribe Easy Medium Hard Tough Or try the Kids Sudokus (4x4 & 6x6) 16x16 or the Parent's Page. Printer Friendly versions: Members Get Goodies! Become a member and get heaps of stuff, including: stand-alone sudoku game, online solving tools, save your times, smilies and more! Welcome our latest Memberscomokaze from Pennsylvaniagwyhel from Barossa ValleyRobin from Kogarah Member's Birthdays Todayhilbillie from darwin

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https://documen.tv/question/mr-chung-grosses-500-a-week-if-his-take-home-pay-is-423-80-how-much-money-was-deducted-from-his-17763560-40/

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## Mr. Chung grosses $500 a week. If his take-home pay is$423.80, how much money was deducted from his gross weekly pay because of taxes? Question Mr. Chung grosses $500 a week. If his take-home pay is$423.80, how much money was deducted from his gross weekly pay because of taxes? in progress 0 3 years 2021-08-06T18:18:34+00:00 2 Answers 8 views 0 ## Answers ( ) $76.20 Step-by-step explanation: 2. Answer:$76.20

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 Convert rad to Ø dia- part | radian to diameter parts Category: main menuangle menuRadians # angle conversion ## Amount: 1 radian (rad) of angle Equals: 60.00 diameter parts (Ø dia- part) in angle Converting radian to diameter parts value in the angle units scale. TOGGLE :   from diameter parts into radians in the other way around. ## angle from radian to diameter part Conversion Results: ### Enter a New radian Amount of angle to Convert From * Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8) * Precision is how many numbers after decimal point (1 - 9) Enter Amount : Decimal Precision : CONVERT :   between other angle measuring units - complete list. Conversion calculator for webmasters. ## Angles This calculator is based on conversion of two angle units. An angle consists of two rays (as in sides of an angle sharing a common vertex or else called the endpoint.) Some belong to rotation measurements - spherical angles measured by arcs' lengths, pointing from the center, plus the radius. For a whole set of multiple units of angle on one page, try that Multiunit converter tool which has built in all angle unit-variations. Page with individual angle units. Convert angle measuring units between radian (rad) and diameter parts (Ø dia- part) but in the other reverse direction from diameter parts into radians. conversion result for angle: From Symbol Equals Result To Symbol 1 radian rad = 60.00 diameter parts Ø dia- part # Converter type: angle units This online angle from rad into Ø dia- part converter is a handy tool not just for certified or experienced professionals. First unit: radian (rad) is used for measuring angle. Second: diameter part (Ø dia- part) is unit of angle. ## 60.00 Ø dia- part is converted to 1 of what? The diameter parts unit number 60.00 Ø dia- part converts to 1 rad, one radian. It is the EQUAL angle value of 1 radian but in the diameter parts angle unit alternative. How to convert 2 radians (rad) into diameter parts (Ø dia- part)? Is there a calculation formula? First divide the two units variables. Then multiply the result by 2 - for example: 60 * 2 (or divide it by / 0.5) QUESTION: 1 rad = ? Ø dia- part ANSWER: 1 rad = 60.00 Ø dia- part ## Other applications for this angle calculator ... With the above mentioned two-units calculating service it provides, this angle converter proved to be useful also as a teaching tool: 1. in practicing radians and diameter parts ( rad vs. Ø dia- part ) values exchange. 2. for conversion factors training exercises between unit pairs. 3. work with angle's values and properties. International unit symbols for these two angle measurements are: Abbreviation or prefix ( abbr. short brevis ), unit symbol, for radian is: rad Abbreviation or prefix ( abbr. ) brevis - short unit symbol for diameter part is: Ø dia- part ### One radian of angle converted to diameter part equals to 60.00 Ø dia- part How many diameter parts of angle are in 1 radian? The answer is: The change of 1 rad ( radian ) unit of angle measure equals = to 60.00 Ø dia- part ( diameter part ) as the equivalent measure for the same angle type. In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in rad - radians for angle amount, the rule is that the radian number gets converted into Ø dia- part - diameter parts or any other angle unit absolutely exactly. Conversion for how many diameter parts ( Ø dia- part ) of angle are contained in a radian ( 1 rad ). Or, how much in diameter parts of angle is in 1 radian? To link to this angle radian to diameter parts online converter simply cut and paste the following. The link to this tool will appear as: angle from radian (rad) to diameter parts (Ø dia- part) conversion. I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting.

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Chemistry posted by . I am trying to complete one of those charts, where you find an element, its proton number, its electron number, it's charge, and whether it's an ion or an atom, based on looking at the Periodic Table. I get how it works, sort of, like for Lithium, you look at the number at the top (3), the horizontal number (2), and subtract three from two. But when I tried this same method with other elements, like Chlorine, it didn't work. For example, chlorine was number 17, in column seventeen, but my teacher says the number of electrons should be 18. Why is this? Thank you! • Chemistry - I am trying to complete one of those charts, where you find an element, its proton number, its electron number, it's charge, and whether it's an ion or an atom, based on looking at the Periodic Table. I get how it works, sort of, like for Lithium, you look at the number at the top (3), the horizontal number (2), and subtract three from two. But when I tried this same method with other elements, like Chlorine, it didn't work. For example, chlorine was number 17, in column seventeen, but my teacher says the number of electrons should be 18. Why is this? Thank you! Similar Questions 1. cHEMISTRY wHAT IS THE SYMBOL FOR THE ION WITH 13 PROTONS AND TEN ELECTRONS? 2. Chemistry In our text book, the formula for finding the formal charge of an element in a compound is c(sub f) = X - (Y + Z/2) where x = the number of valence electrons y= the number of unshared electrons owned by the atom z= the number of bonding … 3. Science-Periodic Table whats the connection/relationship between the # of protons in the nucleus of an atom & and atomic # of the element? 4. Science-Periodic Table describe how to verify the # of electrons in the ion of an element. Can you please correct some of my wordings in here. I don't really get how to explain it. You look at the atomic number which is the same number of protons as electrons; … 5. Chemistry When looking at the Periodic Table, the oxidation state of hydrogen is +1; however my teacher told me that the net charge of hydrogen is 0. I do not understand why this is? 6. chemistry I wanted to check my True and False Chemistry HW. I am not 100% they are correct. 1.the number of neutrons in an atom is reffered to as its atomic number False 2.the periodic table is arranged by increasing atomic number True 3.atomic … 7. chemistry 1state relative mass and charge of an electron 2 an atom of an element X may be written as 4X^9 (a)What does the figure 9 indicate? 8. Chemistry Which of the following is true regarding the structure of atoms? 9. Science--Help. Which of the following is NOT one of the eight most common elements in Earth’s continental crust? 10. Periodic Table 1. The atomic mass of an element is A. the sum of the protons and neutrons in one atom of the element.** B. twice the number of protons in one atom of the element. C. a ration based on the mass of a carbon-12 atom D. a weighted average … More Similar Questions

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Ga naar Nederlandstalige tekst ## Website by Andre Kappert Profiel van André Kappert weergeven ## Content DVDtoHP Sound recordings are mixed for listening with speakers. Even the best headphones can’t hide this.It is as looking through a fishbowl, but than auditive. Strongly enlarged, but also strongly bent, with strong dispersion of frequencies. If a sound source moved in a straight line from the left speaker to the right speaker, it will sound with a headphone as if this movement is circular with a variable speed, from ear to ear. The speed of this movement is also frequency dependent. In figure 1 you see the arrangement of an ordinary sound system with 2 speakers. Let's suppose you are listening to a choir with 120 members. The  members seems to be evenly distributed behind the speakers, the angle limited by the angle of the speakers. What happens when listening this with headphones is reflected in figure 2. Problem 1:Every ear is reached by the sound of both speakers but the sound of the opposite speaker is somehow damped. The higher the frequency of the sound ,the more the damping. Because with speakers the sound of 2 sources reaches the ear, the sound is amplified. The amplification of the lower frequencies is stronger. This doesn’t happen with headphones so the lower frequencies seems to come from a longer distant the higher frequencies(red is representing the lower frequencies, blue the higher frequencies, in the speaker figure these spots overlap) . Every sound is a compound of higher and lower frequencies.  With headphones these sound don’t seem come from one point but comes from an area. The different frequencies are split. Not only the distant will vary but also the angle will be different because the HRTF (human ratio transfer function) is a function of frequency and angle. Some headphones compensate for the weak low frequencies but this don’t work for the diffraction as function of the angle. Problem 2:When all the signal is on one channel, with speakers the source seems to be come from behind one speaker, this is the largest possible angle, and it still sounds natural. When listening with headphones the sounds  originated directly in one ear. The other ear is in totally silence. This normally only happens when a mosquito is near your ear and don’t sound pleasant. It can causes even stress because these sound sources seems to come from very close. Problem 3:The hearing system is trying to calculate the position of the sound source. It is doing this by using the difference in amplitude and by using the difference in phase. Two different systems with the best result for different frequency-bands. But there is overlap at certain frequencies. For these frequencies the systems had to come for the same result as location. These frequencies are located in the most import frequency-band where the hearing is most sensible (1000 Hz to 5000HZ). With headphones these two system don’t give coherent results any more. It is tiring for the hearing system to make sense from these 2 different results. It should also be the reason that with listening to headphones the sound always seems to come from behind. Solution: Add to the signal of one channel of the headphone, with a delay and filtered according the HRTF, the signal of the other channel.  A sort of artificial head recording. The signal is now as it should be with listening with speakers. The delay takes care of the right phase, the HRTF takes care of the right amplitude. The problem is to find the right HRTF It can be found with testing with a simple sinus test signal. My program DVDtoHP can be used for fine adjustment of the HRTF. I made a LADSPA filter for doing this job. It can be used with sound editing programs or media players, able to supporting LADSPA filters. In figure 3 is a graphical representation of a test signal of pink noise. Figure 4 is the resulting filtered version. The difference is considerable. Figure 3: Example consistent of pink noise Figure 4:With stereotobinaural LADSPA treated example, with gain =1.0 The same technique is used in the program DVDtoHP, but expanded to the rear speakers and the front and LFE speakers so you can listen to 5.1 recording with headphone.  The sound still seems to come from behind so I guess that this is not caused by discrepancy between phase and amplitude information, so there is still possibility for improvements if I understand what is causing this. But the spreading of the sound is much improved.  You now can pinpoint the instruments exactly to the location, the separation of the instrument is improved. A similar product is the bs2b filter (Bauer stereophonic-to-binaural DSP). The major difference is that my version is specially made for 5.1 recordings, in the form of AC3 and DTS files(Linux version), and it is also usable for stereo. Both are crossfeeder, but bs2b seems to simulate an original analog filter. My version keeps it simple, it uses a delay and a filter, almost the same as in natural. Bs2b excludes the effect of comb filter in the upper range of frequencies through the nonlinear property of phase-frequency response of these filters. I see the the comb filter effect as unavoidable as in a normal speaker situation. With 2 speakers it is also happening. The provided delay gives a simulated position of the speakers, the damping had to be adjusted so that the damping and the delay leads to the same position. The filter is a convolution filter. For comparison of my filter and the bs2b filter, there is a radiobutton in the program HP_Player for choosing between these filters. The Linux version is a .deb installation program also used by Ubuntu. It also installs a simple audioplayer(HP Player) what could be started from the menu. This program chooses automatically the right converter for the 6 0r 2 channel audio file. It can almost play any audio format (except the wma format). For this program a fast computer is necessary (more than 1.7 GHz duo core simple notebook). The windows version of HP_Player is limited. The best results are expected with 5.1 audio files, but this version only works with stereo files and only with limited formats. The implementation of all the functions was to difficult or impossible in Windows. For windows and Linux, there is also a VST version. When working with a DAW, good monitors are necessary to listen the result. With this filter plugged in the master, it is also possible to mix and master with headphones (and remove the filter in the final result else the recording only sounds great when listened with headphones and not with speakers). Updated 11 juli 2018:Dependencies libav-tools and libqt4pas5 removed because not available in modern repository of Linux Mint. Download LADSPA and VST plugin for Windows 7,8 and 10 (VST 32 bits version and 64 bits version). ## DVDtoHP DVDtoHP is meant to extract audio tracks from music DVD's. This is about the Linux version, but the Windows version is almost the same. There is also a Windows and a Macintosh version to download. The audio information on DVD is mostly available in different forms. The best quality is the .dts(5.1) track, second choice is the .ac3(5.1) track. Sometimes there is a stereo track available (in ac3 or lpcm format). The stereo tracks are often not the best quality, too much compression and even clipping. First experiment I have tried to down-mix the 6 channels to a 2 channel recording with audacity. When the rear channels were mixed with the front channels sometimes weird sound-effects or messy stereo image were created. When the rear channels where just left out, some important sound information was missing. A different approach was necessary. It isn’t possible to recreate the sound-image of 6 speakers with 2 speakers or it should be an "artificial head" recording (or binaural recording) in a surround-sound setup, listened with headphones. DVDtoHP simulates an "artificial head" recording. Sound from a left channel reaches first the left ear and with a delay it also reaches the right ear. Higher frequencies are muffled more than lower frequencies. This is simulated by deliver the signal from the left side filtered and delayed to the right side and visa versa. The rear and front channels get a different delay and filter. The rear channels get a longer delay and are more muffled than the front channels. It is difficult to predict how this will sound to different people. Every person has a different HRTF, Head Related Transfer Function caused by different size and build of head and ears. In fact a normal stereo recording shouldn't be listen-able with a headphone. Sound is delivered totally independent to the ears, in normal situations this only happening when a bee is buzzing in one of your ears. The human brain is still capable of processing these sounds. I believe the human brain is even better capable to process the sound from the artificial human head used in this program. An example: take a recording of a low frequency signal with one channel in anti-phase with the other channel. When listening with speakers you will hear almost nothing (sometimes this actual happen when the wires of a speaker are wrongly connected), when listening with headphone you will hear the full power of the recording. A proof that a recording can sound totally different when listened with headphone. When processed by this program you will hear almost nothing as if you listen it with speakers. The program is able to convert not only 5.1 audio but also stereo files ( wav, flac, mp3 etc.). Listening to the converted files, sound as if you are listening to speakers, giving the intended sound experience as meant by the mixer. The CD's are mixed for speakers in the normal configuration, not for headphones. The program is capable of converting multiple files at the same time, using the multi-core processors optimal. The converting could be done in the background of normal computer use. The program has three buttons. The first start a window in which you can choose which audio track you want to extract from the DVD. It will be saved in the original format, no conversion will take place. The file will be saved in the original dts, ac3 or wav format. The program uses mplayer program to extract, so package mplayer must be installed. The second button starts the conversion, first it asks for audio input file, second it asks for the saving file name. If the extension is mp3 or flac it will ask for meta data, and automatically compress the audio file. Then a wait screen will start with the progress. This is an independent window, allowing you to start another conversions from the control program. The program uses ffmpeg ( from ffmpeg package )to decompress the file to a 6 channels floating points file. Then my program will convert this file to a 2 Channels floating points file. Then it will normalize and convert it to an ordinary microsoft wav 16 bits, 2 channels file. Last step it could be compressed with lame program or flac program if extension was flac or mp3. The program depends on the program ffmpeg of package ffmpeg for decompressing dts and ac3 files to floating points, sox of package sox for eventually resampling the file to 44100 (the quality of sox seems better than ffmpeg for resampling). For ripping the DVD it uses mplayer of package mplayer. For compression it use Lame of package lame and Flac of package flac. All these programs will automatic installed by the deb installation program. For other Linux versions the bare program is provided. The dependencies must be provided by the user. It is unclear for me if Package libdvdread4 (you have to activate it with "sudo /usr/share/doc/libdvdread4/install-css.sh" in terminal) needs to be installed. This is necessary for descrambling DVD. Somehow this package was installed always automatically on my computer, so  I never checked the necessity of this Package (I had even forgotten the existence of the package). ### Updated version Added the ability to create and load a personalized version of the HRTF (head related transfer function). In the first version I had used a HRTF obtained from the Internet, randomly taken from a graphic found by coincidence. First version I had no big attention for an exact HRTF, because this would be a different function for every human and it is not easy to measure a personalized version of the HRTF. All listening with headphones sounds as if you were sitting with your back to the orchestra (a phenomenon normal for listening with headphone!). This problem should be caused by a not exactly correct HRTF, sadly this problem is not solved. I have tried versions with extra phase control and extra delay control, but this didn't solved it. The spreading of the sound over the space is now improved and it sounds more natural. ### How to create own HRTF (layman could use the default HRTF). Delayfront: the delay caused by the longer distant from the left front speaker to the right ear in comparison from the distance to the left ear. 2.5e-4 seconds is 11 samples at 44100 samplerate, and it means 7.5 cm extra distant. I suggest not to change it, if to large all the sounds seems to come from very near. DelayBack: The delay from the speaker on the left side to right ear in comparison to the left ear. 6.12e-4 sec. Is 27 samples with 44100 samplingrate and means 18 cm. Beeplength:Length of the testbeep. SampleFreq: Sampling frequency of testbeep, not important. Testlength: Testbeep is repeated from left to right, this is the total length of testsound. The list headed “List of HRTF”: The actual HRTF ,first number is frequency, second number is the damp for the front speakers, the third is the damp of the speakers on the right and left side. Last frequency in list is always 24000 Hz, because DVD recordings has a bandwidth of 24000 Hz . The list headed “Testfreq Width”:First number is frequency of testbeep, second number is bandwidth of testbeep. More frequencies (lines) are possible. I suggest to keep a bandwith of 1 hz and only one frequency. HalfPulse checkbox:gives testbeep sharp attack, only use it in final testing because it increases bandwidth Nameof:Here you can choose a name for the HRTF ### Testing: Use headphone, push button “Test Front Audio”. If adjustment is correct, the sound you hear should originated from speakers at an angle of 30 degrees behind of you. If the sounds seems to come more from the center, you had to make the second number for that frequency in the HRTF list smaller. If the sound seems to come from to far sideway, you have to make it larger. Repeat this for every frequency in the HRTF list. If necessary add extra frequencies to the list. The test signal should sound symetrical. At frequencies between 1000 HZ and 4000Hz something strange could happen. For my ears the test signal was not longer symmetrical. This was caused by a wrong HRTF for this frequency. The delay leads to another sound direction than the information of the damping. The brain could not solve this sound-image and comes for every ear with a different solution. Adjusting the level of damping to the right HRTF level made this more symmetrical, because the information from delay and the information of damping are no longer conflicting. Exactly this is the reason for creating your own HRTF, and why headphone listening with the sound-processing from this program is more pleasant. With “Test Side Audio” button the testbeep should become from your side. First try to make the number of the last column for the testfreq larger. If the sound seems to come from a point not exactly from the side (more centred) then make this number again smaller. Find this boundary. If you make it to small the origin seems to getting to close to your head, and this is difficult to hear from the test signal. (These numbers are not used with stereo recordings, only DVD has this channels). Repeat this for all the frequencies in the HRTF list. Test both buttons for all frequencies again with the “HalfPulse checkbox” checked. In the list of test frequencies you can set the testfrequency, The second number is the bandwidth of the testfrequency. The testfrequency is generated by filtering a delta function or white noise width the bandwidth. The source is chosen by checkbock "white noise". Save the result with the “save file” button. If saved with the name”default.hrtf” in the home directory, the HRTF will be automatically loaded every time with the start of program. ### Examples (for testing with headphone) The examples are just from stereo recordings, because it should be royalty free. The best quality should come from DTS recordings. The recording(Example1) is made with Hydrogen drumcomputer, just the stick and the kick moving from left to the right. The original sound as if the first hit comes from a point near your left ear, the next from an unclear point in the middle, the last from a point near the right ear. The converted file sounds as if the instrument made a normal trajectory. The most extreme positioning (absolutely left or right) doesn't come from a point near to your ear. The converted to binaural audio examples sound much more natural. The original versions sound annoying if listened with headphone. Sometimes it sounds as 2 mono signals.  A more extreme example is Example2, Noise on one channel. In the original, the sound originate in your ear and is totally unnatural. The converted version sound as if its originate of a speaker at an angle of 30 degrees right behind you, and is much natural. It is still coming from behind you, but nicely placed in space. At home I listen only to converted recordings with headphone because I find it more pleasant. If the original was a 5.1 recording the binaural converted recording should sound much wider. There exists a version of "Aero" of Jean Michel Jarre specified as a "Dolby headphone" version. I think for making this, they used a similar process. I thought my version from ripping the DVD sounds better. Example1: Origineel and Converted Binaural Example2 (Extreme example):Original and converted Binaural For the same effect I have created a LADSPA filter version. In goes the normal stereo or 5.1 version, out comes a version adapted for listening with headphone. Windows version is 32bits also working in windows 64 bits. For working with Audacity, place the stereotobinaural.dll file in the directory “C:\Program files(x86)\Audacity\Plug_Ins” . Enable it in the Effect menu. The effect is now visible as “stereo to binaural”. In Linux, just double click the deb installation file. For visibility in Audacity, enable it in the Effect menu. In OSX place the libstereotoheadphone.so in the directory "Application/Audacity/Plug-Ins". For visibility in Audacity, enable it in the Effect menu. The effect is now visible as “stereo to binaural”. Updated 11 july 2018:(Linux)Dependencies libav-tools and libqt4pas5 removed because not available in modern repository of Linux Mint. Updated 14 august 2018:(Linux,Windows)Added support for DSF and DFF files (from Super Audio CD's) Updated 07 july 2019:(Linux,Windows)Added support for not only stereo and 5.1, but also every format between (like 5.0) Update 05 august 2019:Added support for sofalizer from ffmpeg. For experimenting with sofa files. First try, I found ClubFritz1.sofa gives a good result. I tried with ClubFritz6 and ClubFritz11 with gives distortion. I tried dtf_nh2.sofa which gives a result without frequencies below 300Hz. So be careful which sofa files you choose. Solved: problem was with type=time, now use type=freq option. Update 07 september 2019:added support for opus format, import and export. DVDtoHP version For Windows (64 bits and 32 bits) with all dependencies !!!!! If link for Windows version doesn't work then download from freewarefiles.com (search for dvdtohp,older version without DSF support and without 5.0 support) The windows version uses ffmpeg.exe, mplayer.exe, sox.exe, lame.exe and flac.exe. They are all placed in the download file because their locations and versions are more controllable. Downloads and info of the used programs are on the next websites. Macintosh Maverick instruction Unzip zip file, application is in extracted map. With use of Homebrew the dependencies had to be installed. Open Terminal in Maverick, type in Terminal ruby -e "\$(curl -fsSL https://raw.githubusercontent.com/Homebrew/install/master/install)" next: type in Terminal brew tap homebrew/dupes These 2 lines where necessary for me to install GDB in Maverick for use with Lazarus (Just type in Terminal  now brew install gdb to install GDB). All the dependencies could now easy installed with the same instruction. brew install libav brew install mplayer brew install flac brew install cdparanoia brew install sox ## CDRipper, GUI voor cdparanoia A graphical interface for cdparanoia. The program gets CD info with abcde, if it is not on the CDDB server it search for CD text info with cd-info from package libcdio-utils and compresses the ripped audio file with LAME to mp3 or with flac to flac files. It automatic places the cd info in the mp3 file. While ripping a new track it compresses at the same time the previous ripped tracks, making use of more cores of the processor. If there were any errors during ripping it displaces the errors location in a clear way. The messages of cdparanoia aren’t always that clear. The button "Rip Cd to Headphone audiofile" rips the whole Cd in one file and also creates a binaural soundfile for headphone( see DVDtoHP). This is my personal favorite button. Small screen images Large screen image Iaddi (1+1 in letters) Calculator with text input, 3 variables. Calculates numerical integration and numerical differentiation. Creates a graphic of the function. Does not need any permissions. Functions: *, / , + , - , ^ ( power of), sin(), cos(), tan, asin, acos, atan, sinh, cosh, tanh, asinh, acosh, atanh, sqr, sqrt, exp, ln, log, abs, rnd, random, rndg (random met Gaussian distribution), cot, csc, sec, acot, acsc, asec, coth, csch, sech, acoth, acsch, asech, sign, round, torad (from degree to rad), todegr (from rad to degree). Boundaries for drawing graphic and integration can be given by using Start x, End x, start y and end y. If start x and end x are the same, the boundaries for y are calculated automatically. If start y and end y are the same, the boundaries for y are calculated automatically. Warning:Integration can not recognize singularities, input of infinity is not possible. When a graphic with a singularity is drawn it is necessary to restrict the y boundaries by fill in start y and end y ## Send in letter “Elektuur” A send in letter from me was publicized in the Dutch magazine “Elektuur” January 2001. This letter regards the recording quality of cd's, especially about clipping. On the Internet I found there where a few translations or summarizing in German ,French and English. Recently I found a few references in Germany to this letter. That's why I placed the original letter and English translation on this site. ### Translation of send in letter from Elektuur January 2001 Dear Editor — roughly a year ago you had an article in Elektor Electronics about CDs that were supposedly recorded with overdriven signals. An oscilloscope was used to look at the outputsignal of a CD player playing Jean Michel Jarre’s ‘Oxygen 7- 13’. It was clear that the peaks of the waveform were flat where the signal was greater than what could be reproduced by the CD. This is also known as clipping. Using an oscilloscope in this way can’t really tell us how much the CD player affects the signal quality. Between the CD and oscilloscope will be an oversampling digital filter, a digital-to-analogue converter and finally an analogue filter. The most accurate way to check a CD is to extract the track digitally and store it as a WAV file. This will be an exact digital copy of the CD, assuming there were no read errors. The creation of the WAV file shouldn’t be a problem since it is a function that is found in most current CD ‘rippers’. A computer program can then be used to analyse the contents of this WAV file. A WAV file consists of a header followed by the thousands of samples that make up the digitised sound. The 16-bit format used on CDs gives these samples a range from -32768 to 32767. In order to have a closer look at these WAV files I have written a program that can analyse them. As the first step, the program creates a histogram of the WAV file. This counts how often each sample value occurs in the WAV file, and gives a display as shown in Figure 1. Figure 1 The values near zero occur most often; the more you go to the extremes of –32768 and 32767, the less often they occur. If there was any clipping at the recording stage, all strong signals will be rounded to the extremes of -32768 and 32767, which results in two peaks at the ends of the histogram. This way you can see if there is any clipping present and at what level (it is possible for it to occur at different values than -32768 and 32767, due to badly set up A/D converters). Next, the program displays the actual waveform stored in the WAV file. This window has a search facility that searches for clipping values, as determined by the previous histogram function. I have used this program to analyse the track “...Baby One More Time” by Britney Spears, from her CD “...baby one more time”. It was obvious that there was a lot of clipping in this track, but the display returned by the histogram was the most surprising. This is shown in Figure 2. Figure 2 It contains the number of occurrences in the WAV file for samples with values from 4 to 515. After every two values that occur over a 1000 times, there follows a value that rarely or never occurs at all. You could only get such a histogram when a 16-bit recording is re-recorded after amplification by a factor of 1.5. Samples with values of 1, 2, 3, 4, 5, 6, 7 etc. are multiplied by 1.5 to give 1.5, 3.0, 4.5, 6.0, 7.5, 9.0, 10.5 etc. These values then have to be rounded to integers before being written to the CD. This gives the result 2, 3, 5, 6, 8, 9, 11 etc. You’ll see that every third value is missing from this series of numbers. By looking at the waveform where clipping occurs, it is possible to deduce by how much the recording of the CD was overdriven. I estimate this to be a factor of about 1.5. From this I conclude that the original mix, which was recorded perfectly, was remixed with an amplification of 1.5, causing clipping in the new recording. The track “Gloria, lonely boy” on the CD “Metamorphose” by Jean Michel Jarre was then analysed, in order to see if his latest CD had signs of clipping. Here too it seemed that the recording was overdriven. But once more it was the histogram that returned some fascinating information (Figure 3). Figure 3 The smallest and largest samples found in the recording were -32022 and 32021 respectively (0.977 of the maximum range). These were also the levels at which clipping occurred. From the histogram it can be seen that every fortieth value is twice as large as its neighbours are. From these observations you can conclude that this CD is a remix of an original CD recording. The samples have been multiplied by 0.977 before being converted back into CD format. This raises the question why this meaningless conversion was carried out on the recording. The only reason I can think of is that the remix was made to hide the fact that the recording was overdriven. Anybody who uses an oscilloscope to look for clipping would set the trigger level at the largest possible signal level, about 99.9% of the maximum signal possible. The remix would cause the signal to remain below this level so the oscilloscope would never trigger and the clipping would remain undetected. The person making the remix would have known that clipping occurred and was determined to hide this fact. The correct procedure would have been to make a slightly softer remix from the master tape and avoid any clipping that way. The record industry is currently promoting the new Super Audio CD (SACD), which has an improved quality. They would be better off stopping the messing around with the recordings. These fudges also sound bad in SACD format and making careful recordings in CD format will result in bigger improvements than the change to SACD format. A. Kappert, Deventer Some people think that these results are caused by compression. But at making cd's, compression means reinforcing the weak passages, reinforcing the strong passages is not possible because these already reach their limit. With a sound editing program the wave shape can be examined. Here is the result made with the provided program of”… baby one more time” at the position 114.885 sec. Figure 4 Every pixel to the right is a new sample. One of the channels shows clear clipping over a large length and in severe degree. The wave shape have been clearly damaged because of this. Compression will change the envelope of a signal but won't affect the shape. Long before the signal becomes to large the volume will be slowly turned down to prevent clipping. An estimation can be made over the original signal size,I estimate it 1.5 larger. Remains the question about why this signal is strengthens. It appears that by comparison of sound quality of Hi Fi signals, the first preference goes out to the strongest signal. Also at radio broadcasts, the recordings with the largest sound strength attract the most attention. Figure 5 To addition in figure 5 “oxygen 7”of J. M. Jarre on 27.37 seconds. Here clearly the “saw tooth” is cut off, in this way losing a lot of harmonics in the sound. At “Methamorphoses” the signal appeared to be weakened with 2,5%. It appeared that cd mastering equipment is protected against clipping. To get around this protection, it is necessary to weaken the signal, but at that moment it is known that clipping occurred. ### Last Version download (with APE and FLAC file support,20 November 2016) To Download ( updated ) analyse program (cdwavanalyse,linux Ubuntu or Mint, 64 bits deb) To Download ( updated ) analyse program (cdwavanalyse,linux Ubuntu or Mint, 32 bits deb) Old windows version with only wav file support ## Mandelbrot fractals This program draws Mandelbrot fractals. With a left mouse click you can zoom in, with a right mouse click you can zoom out. It is possible to zoom in with factor 1E16.

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+0 # Help plz (probability) 0 442 2 +164 4 12-sided dice are rolled. What is the probability that the number of dice showing a two digit number is equal to the number of dice showing a one digit number? Express your answer as a common fraction. (Assume that the numbers on the 12 sides are the numbers from 1 to 12 expressed in decimal.) Apr 17, 2018 #1 +99276 +1 I'll take a stab at this one.....if the number of dice showing a two-digit number  = the number of dice showing a one-digit number, then  2 must show a two-digit number and 2  must show a one-digit number There are 3 two-digit numbers and 9 one-digit numbers on each die So....the probabiity is C(4,2) (3/12)^2 (9/12)^2    = C(4,2) (1/4)^2 (3/4)^2  = 27/128 Apr 17, 2018

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PDA View Full Version : Real-world P-51 Engine Performance and Manifold pressure XyZspineZyX 10-14-2003, 08:18 AM Hi everyone, I cam across an interesting issue while trying to analyze P-51 performance. And that being the 2 stage supercharger, and its quantitative effect on available power at altitude. for a piston engine not using any kind of supercharger the available power can be given roughly by: P = n*throttle(%)*Ps*(density of air at the altitude/density of air at sea level) where Ps is max power available at sea level, and n is the propeller efficiency factor. Consequently your available power would decrease with height. But the supercharger rams more air into the engine, increasing power. How much more air is getting to the engine? I think that's where manifold pressure comes in. In a "training" film about P-51B, they say that at 20,000 feet supercharger switches to 2nd stage high blower - and the manifold pressure goes up to 60(mm Hg?). Now my question is how to translate the indicated manifold pressure to the actual pressure of the air coming to the engine(and hence density, assuming the air temperature is not increased as it passes through supercharger). I know some of you may be actual pilots or otherwise know about this. So any input would be appreciated! Message Edited on 10/14/0302:47AM by KrasniyYastreb XyZspineZyX 10-14-2003, 08:18 AM Hi everyone, I cam across an interesting issue while trying to analyze P-51 performance. And that being the 2 stage supercharger, and its quantitative effect on available power at altitude. for a piston engine not using any kind of supercharger the available power can be given roughly by: P = n*throttle(%)*Ps*(density of air at the altitude/density of air at sea level) where Ps is max power available at sea level, and n is the propeller efficiency factor. Consequently your available power would decrease with height. But the supercharger rams more air into the engine, increasing power. How much more air is getting to the engine? I think that's where manifold pressure comes in. In a "training" film about P-51B, they say that at 20,000 feet supercharger switches to 2nd stage high blower - and the manifold pressure goes up to 60(mm Hg?). Now my question is how to translate the indicated manifold pressure to the actual pressure of the air coming to the engine(and hence density, assuming the air temperature is not increased as it passes through supercharger). I know some of you may be actual pilots or otherwise know about this. So any input would be appreciated! Message Edited on 10/14/0302:47AM by KrasniyYastreb XyZspineZyX 10-14-2003, 08:45 AM Good question, but you have to say why or what you need such information for, otherwise I don't know how to answer. But I pretty sure the indicated manifold pressure is very close to the actual, even up there. The turbo is compressing the thinner surrounding air into the engine to create an artificial higher density inside the engine, similar to the engine operating at lower altitudes, which makes the indicated manifold pressure close to actual... I hope that was the answer you were looking for?? rgds XyZspineZyX 10-14-2003, 08:51 AM "assuming the air temperature is not increased as it passes through supercharger" in fact the temperature rises dramatically. This is why intercoolers and water injection help so much. XyZspineZyX 10-14-2003, 09:03 AM Why I am asking is because I'm trying to analytically find the minimum and maximum speeds(at least the theoretical ones) for the P-51 at different altitudes by plotting the Power required(vs velocity) curve against the power available at that altitude. I guess what I'm looking for is what the indicated manifold pressure actually measures(it can't measure the real pressure since at sea level it is 760mm HG!), and how i can use that to calculate the density of the air coming into the engine, and hence the available power. Of course, if someone happens to have some Power vs Altitude graphs for the supercharged V-1650-7 Merlin, that would solve the entire problem! Message Edited on 10/14/0303:08AM by KrasniyYastreb XyZspineZyX 10-14-2003, 09:06 AM Manifold pressure is the pressure as measured in the intake manifold. Remember to deduct the loss of power from driving the charger too. It is significant. I'd guess that the MAP is measured in inHg rather than mm, BTW. /i/smilies/16x16_smiley-wink.gif 60 mm Hg would be 2.36 inHg, or 80 hPa. Roughly equivalent to 17700 meters above sea level, real altitude... most recips would perform rather poorly, I think. /i/smilies/16x16_smiley-very-happy.gif Edit: The brits, and possibly others as well, seemed to favour measuring not the absolute manifold pressure but the manifold pressure compared to SLS. One additional caveat. The air temperature will increase through the supercharger, but you should be able to safely assume isentropic compression, i e no heat goes in or out of the air. That way, everything gets a lot easier to calculate. Cheers, Fred "If we are an arrogant nation, they will resent us. If we are a humble nation, but strong, they will welcome us." - George W. Bush, during his campaign. No comment. (Quote brought back by popular demand - RBJ missed it so much he mailed me about it) Message Edited on 10/14/0308:09AM by effte XyZspineZyX 10-14-2003, 10:14 AM Well, 760 mmHg is equivalent to 29.92 inches of Hg. . . Manifold pressure is given in inches instead of millimeters. Basically manifold pressure is simply the indication of engine power in aircraft equipped with adjustable pitch props. The super charger is a device which compresses the air entering the intake and makes the engine "think" it is operating at a lower altitude. Typically a normally aspirated, non-turbo/super charged engine begins to lose its maximal power output above 6,500 to 8,000 feet (1,981 to 2,439 meters), so a super charger will permit a given power setting at higher altitudes. (there is also an increas in performance when below this critical altitude, but this is not the reason for super or turbo charging as the available power belwo these altitudes is usually quite high without the help) I do have the P-51 Pilot operations books, however I don't have the one with the performance charts here so for now try to use these (hard to read, but if you know what your looking for they get the job done). http://www.zenoswarbirdvideos.com/More_P-51_Stuff.html The information that I have commited to memory is this: Max take-off MP: 61 inches MP Normal operating range: 20-36 inches MP Max indicated airspeed 505 MPH/440 Knots/808 Kmh (true, or speed over the ground will be higher when at altitude) Max Takeoff RPM 3,000 Normal operating range: 1,600-2,400 RPM To better understand what manifold pressure is a measure of, and how it relates to aircraft performance take a look here. http://www.avweb.com/news/columns/182081-1.html S! TX-EcoDragon Black 1 Reserve Pilot Aircraft #2 of Gruppo 313 Pattuglia Acrobatica Virtuale http://www.pav-amvi.it http://www.calaggieflyers.com/ Message Edited on 10/14/0301:30AM by TX-EcoDragon XyZspineZyX 10-14-2003, 10:18 AM effte wrote: - Remember to deduct the loss of power from driving - the charger too. It is significant. I recall reading 6% as a good estimate for this recently. Kernow 249 IAP XyZspineZyX 10-14-2003, 04:40 PM - Of course, if someone happens to have some Power vs - Altitude graphs for the supercharged V-1650-7 - Merlin, that would solve the entire problem! There are various charts out there, but some don't make it clear wether they are taking into account ram air pressure. There is a Russian chart that Isegrim might have, which iirc does not take ram into account. Another source is the Spit IX test reports at http://www.fourthfightergroup.com/eagles/spittest.html in particular, look at the Spit LF IX with Merlin 66, as that's closest to the 1650-7. http://www.fourthfightergroup.com/eagles/bs543.html It doesn't give power outputs, but the climb/speed performance does indicate the power available. I think the Mustang had a bit better ram recovery, but that's variable between individual examples of the same plane. A British chart gives the following figures: 1590hp 0 ft 1660hp at 9,800ft 1480hp at 13,000ft 1530hp at 20,000ft 1400hp at 22,500ft It's a straight line chart, and those figures give the peaks and troughs, so plot those figures and you should have the rest. The figures are approx, so don't extrapolate much above 22,500ft. All figures are with 400mph ram, and at 18lbs boost (67" in MAP) - Why I am asking is because I'm trying to - analytically find the minimum and maximum speeds(at - least the theoretical ones) for the P-51 at - different altitudes by plotting the Power - required(vs velocity) curve against the power - available at that altitude. The problem you'll have with that is power varies per aircraft, due to the ram recovery. That shouldn't amtter too much, but the more you extrapolate, the greater the error that can creep in. Engine charts can be misleading, because they may not show ram, or show a different level of ram recovery to what is available in the particular plane. - I guess what I'm looking for is what the indicated - manifold pressure actually measures(it can't measure - the real pressure since at sea level it is 760mm - HG!), and how i can use that to calculate the - density of the air coming into the engine, and hence - the available power. The indicated manifold pressure does measure the pressure of air coming in to the engine. The Merlin 66/V-1650-7 ran at 18lbs boost pressure in British terms, which means 18lbs per square inch above normal sea level pressure. In other words, absolute pressure was 14.9 (sea level air pressure) + 18lbs added by the supercharger, which means a pressure of 33 lbs/sq in (roughly). That pressure could be maintained up to 10,000ft or so, at which point manifold pressure begins decreasing. At 16,000ft the higher speed supercharger gear cuts in, and pressure jumps back up to 18lbs, which is maintained to 22,000ft or so. Above that, the supercharger can no longer maintain the rated pressure, and it drops again. However, if you look at the Spitfire charts, you can see that at 34,000ft, the supercharger can still maintain 5.7 lbs boost, which means 14.9lbs (sea level pressure) + 5.7lbs. It's not until close to 40,000ft that intake pressure drops below normal sea level pressure. XyZspineZyX 10-14-2003, 10:49 PM Thanks for your help everyone. Now I have calculated/estimated the boost provided by the supercharger based on the Spit data. The problem now is that while pressure in the intake manifold does not decrease significantly with altitude, ambient temperature does, which leads to a value for the density of air coming into the engine to be greater than that at sea level! This of course gives an available power at 25,000ft that is greater than that at sea level. Using the top speed of the Mustang 437 mph at 25,000ft, the engine power to achieve that speed was calculated to be 1052Hp - which I took to be the max available power at that height. This leads to the conclusion that the temperature of the air coming into the engine must increase to 74 degrees Celsius as it passes throught the supercharger, to maintain the corresponding air density. However if I do the same calculation for sea level, I get a temperature increase of only 3-4 degrees. So I still don't actually know how much power is developed at a given altitude. XyZspineZyX 10-14-2003, 11:14 PM Yes, the air pressure inside the merlin, and most of the high power engines of WWII is considerably greater than that at sea level. The concept is simply, more air in the chamber, more material to combust. For example, the R-2800 requires 51 inches of pressure to operate at peak horsepower, reguardless of the altitude. Racing Merlins have been run at over 120 inches of boost. granted, if you have even a minor hicup in the antidetonant system, your engine is likely ot be all over the place, at such high power ratings, but it can be done. For some good columns on manifold pressure, fuel mixture, and turbo chargers see here: Engine operation (manifold pressure, props, mixture, and all three in conjunction): http://www.avweb.com/news/columns/182081-1.html http://www.avweb.com/news/columns/182082-1.html http://www.avweb.com/news/columns/182084-1.html http://www.avweb.com/news/columns/182085-1.html Turbo compressors: http://www.avweb.com/news/columns/182102-1.html http://www.avweb.com/news/columns/182103-1.html http://www.avweb.com/news/columns/182104-1.html http://www.avweb.com/news/columns/182105-1.html http://www.avweb.com/news/columns/182106-1.html http://www.avweb.com/news/columns/182107-1.html Leaded fuels, and antidetonents, and detonation: http://www.avweb.com/news/columns/182132-1.html http://www.avweb.com/news/columns/182149-1.html Enjoy. Harry Voyager http://groups.msn.com/_Secure/0YQDLAswcqmIpvWP9dLzZVayPXOmo6IJ16aURujNfs4dDETH84 Q6eIkCbWQemjqF6O8ZfvzlsvUUauJyy9GYnKM6!o3fu!kBnWVh BgMt3q2T3BUQ8yjBBqECLxFaqXVV5U2kWiSIlq1s6VoaVvRqBy Q/Avatar%202%20500x500%20[final).jpg?dc=4675409848259594077 XyZspineZyX 10-17-2003, 03:09 PM I have almost solved the puzzle. I need one more piece. Either the intercooler/aftercooler efficiency, or the induction air temperature at full power at some altitude. Even one data point would suffice. Anybody have a P-51 manual or some dusty book lying around? Message Edited on 10/17/0310:35AM by KrasniyYastreb

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434,660 Members | 1,945 Online Need help? Post your question and get tips & solutions from a community of 434,660 IT Pros & Developers. It's quick & easy. # how to parse a string P: n/a HI again Is there a nifty function in access that will: 1. return the amount of occurances of a small string within a larger string? this is a test would return 3 for 2. Parse a string into an array given a separation string like this is a test so for as delimiter array(0) = this array(1) = is array(2) = a array(3) = test Thanks Danny Nov 12 '05 #1 9 Replies P: n/a Danny wrote: HI again Is there a nifty function in access that will: 1. return the amount of occurances of a small string within a larger string? this is a test would return 3 for 2. Parse a string into an array given a separation string like this is a test so for as delimiter array(0) = this array(1) = is array(2) = a array(3) = test Thanks Danny What version of Access? Nov 12 '05 #2 P: n/a "Salad" wrote in message news:Db**************@newsread2.news.pas.earthlink .net... Danny wrote: HI again Is there a nifty function in access that will: 1. return the amount of occurances of a small string within a larger string? this is a test would return 3 for 2. Parse a string into an array given a separation string like this is a test so for as delimiter array(0) = this array(1) = is array(2) = a array(3) = test Thanks Danny What version of Access? I use Access 2002 Why don't my posts show up sometimes? This is second time I posted this. Thanks Nov 12 '05 #3 P: n/a I have access version 2002 thanks "Salad" wrote in message news:Db**************@newsread2.news.pas.earthlink .net... Danny wrote: HI again Is there a nifty function in access that will: 1. return the amount of occurances of a small string within a larger string? this is a test would return 3 for 2. Parse a string into an array given a separation string like this is a test so for as delimiter array(0) = this array(1) = is array(2) = a array(3) = test Thanks Danny What version of Access? Nov 12 '05 #4 P: n/a "Danny" wrote in message news:... HI again Is there a nifty function in access that will: 1. return the amount of occurances of a small string within a larger string? this is a test would return 3 for 2. Parse a string into an array given a separation string like this is a test so for as delimiter array(0) = this array(1) = is array(2) = a array(3) = test Thanks Danny I haven't tested this much; it's not very elegant. Public Function CountSubstrings(strIn As String, strFind As String) As Integer Dim intCount As Integer Dim strTemp As String Dim intFound As Integer 'Count the number of strFind strings within strIn. CountSubstrings = 0 If Len(strIn) = 0 Then Exit Function If Len(strFind) > Len(strIn) Then Exit Function intFound = InStr(1, strIn, strFind, vbTextCompare) If intFound = 0 Then Exit Function intCount = 1 CountSubstrings = 1 'See if the string was found at the end of strIn If Len(strIn) = intFound - 1 + Len(strFind) Then Exit Function 'Chop off up to and including the found string 'Note: Add code here to catch contents for an array if desired strTemp = Right(strIn, Len(strIn) - intFound - Len(strFind) + 1) intFound = InStr(1, strTemp, strFind, vbTextCompare) Do While intFound > 0 intCount = intCount + 1 If Len(strTemp) = intFound - 1 + Len(strFind) Then CountSubstrings = intCount Exit Function End If 'Note: And code here strTemp = Right(strTemp, Len(strTemp) - intFound - Len(strFind) + 1) intFound = InStr(1, strTemp, strFind, vbTextCompare) Loop CountSubstrings = intCount End Function You'll want to index on intCount - 1 for a zero based array and I'd use a Sub instead of a Function to return the substring count and the array. James A. Fortune Nov 12 '05 #5 P: n/a The Split function suggested by Chuck would seem to be more appropriate. And if all you want is a function to tell you how many occurrencs of x occur in y, the following does it: Public Function CountSubstrings(strIn As String, strFind As String) As Integer CountSubstrings = (Len(strIn) - Len(Replace(strIn, strFind, ""))) \ Len(strFind) End Function What this does is make a new version of strIn, replacing every occurrence of strFind with a zero-length string (""). It then calculates the difference in length between the length of the original string and the length of the modified string. Finally, it divides that difference by the length of strFind. -- Doug Steele, Microsoft Access MVP http://I.Am/DougSteele (No private e-mails, please) "James Fortune" wrote in message news:a6**************************@posting.google.c om... "Danny" wrote in message news:... HI again Is there a nifty function in access that will: 1. return the amount of occurances of a small string within a larger string? this is a test would return 3 for 2. Parse a string into an array given a separation string like this is a test so for as delimiter array(0) = this array(1) = is array(2) = a array(3) = test Thanks Danny I haven't tested this much; it's not very elegant. Public Function CountSubstrings(strIn As String, strFind As String) As Integer Dim intCount As Integer Dim strTemp As String Dim intFound As Integer 'Count the number of strFind strings within strIn. CountSubstrings = 0 If Len(strIn) = 0 Then Exit Function If Len(strFind) > Len(strIn) Then Exit Function intFound = InStr(1, strIn, strFind, vbTextCompare) If intFound = 0 Then Exit Function intCount = 1 CountSubstrings = 1 'See if the string was found at the end of strIn If Len(strIn) = intFound - 1 + Len(strFind) Then Exit Function 'Chop off up to and including the found string 'Note: Add code here to catch contents for an array if desired strTemp = Right(strIn, Len(strIn) - intFound - Len(strFind) + 1) intFound = InStr(1, strTemp, strFind, vbTextCompare) Do While intFound > 0 intCount = intCount + 1 If Len(strTemp) = intFound - 1 + Len(strFind) Then CountSubstrings = intCount Exit Function End If 'Note: And code here strTemp = Right(strTemp, Len(strTemp) - intFound - Len(strFind) + 1) intFound = InStr(1, strTemp, strFind, vbTextCompare) Loop CountSubstrings = intCount End Function You'll want to index on intCount - 1 for a zero based array and I'd use a Sub instead of a Function to return the substring count and the array. James A. Fortune Nov 12 '05 #6 P: n/a "Douglas J. Steele" wrote in message news:Tm*****************@news01.bloor.is.net.cable .rogers.com... The Split function suggested by Chuck would seem to be more appropriate. And if all you want is a function to tell you how many occurrencs of x occur in y, the following does it: Public Function CountSubstrings(strIn As String, strFind As String) As Integer CountSubstrings = (Len(strIn) - Len(Replace(strIn, strFind, ""))) \ Len(strFind) End Function Cool. Unbelievably fast. Nov 12 '05 #7 P: n/a "Douglas J. Steele" wrote in message news:... The Split function suggested by Chuck would seem to be more appropriate. And if all you want is a function to tell you how many occurrencs of x occur in y, the following does it: Public Function CountSubstrings(strIn As String, strFind As String) As Integer CountSubstrings = (Len(strIn) - Len(Replace(strIn, strFind, ""))) \ Len(strFind) End Function What this does is make a new version of strIn, replacing every occurrence of strFind with a zero-length string (""). It then calculates the difference in length between the length of the original string and the length of the modified string. Finally, it divides that difference by the length of strFind. -- Doug Steele, Microsoft Access MVP http://I.Am/DougSteele (No private e-mails, please) Very elegant. I like it a lot. I'll have to keep looking for ways to improve my existing functions. Is there an equally elegant solution to CountSubstrings for Access 97? Are there some nice API string functions that would make life easier for Access programmers. James A. Fortune Nov 12 '05 #8 P: n/a "James Fortune" wrote in message news:a6**************************@posting.google.c om... "Douglas J. Steele" wrote in message news:... And if all you want is a function to tell you how many occurrencs of x occur in y, the following does it: Public Function CountSubstrings(strIn As String, strFind As String) As Integer CountSubstrings = (Len(strIn) - Len(Replace(strIn, strFind, ""))) \ Len(strFind) End Function What this does is make a new version of strIn, replacing every occurrence of strFind with a zero-length string (""). It then calculates the difference in length between the length of the original string and the length of the modified string. Finally, it divides that difference by the length of strFind. -- Doug Steele, Microsoft Access MVP http://I.Am/DougSteele (No private e-mails, please) Very elegant. I like it a lot. I'll have to keep looking for ways to improve my existing functions. Is there an equally elegant solution to CountSubstrings for Access 97? Are there some nice API string functions that would make life easier for Access programmers. I always add my own Replace and Split functions to all my Access 97 databases, so that it doesn't really become an issue. I'm not aware of any string-specific APIs (largely because strings are different between C and VB). -- Doug Steele, Microsoft Access MVP http://I.Am/DougSteele (No private e-mails, please) Nov 12 '05 #9 P: n/a "Douglas J. Steele" wrote in message news:<8r**********************@twister01.bloor.is. net.cable.rogers.com>... I always add my own Replace and Split functions to all my Access 97 databases, so that it doesn't really become an issue. I'm not aware of any string-specific APIs (largely because strings are different between C and VB). Thanks for the information. Here is a method for dealing with C/VB strings. Apologies in advance to Steven Roman if I interpret what he said incorrectly. The book gives a much more detailed look at what goes on. You can use the following "trick" suggested by Steven Roman in his book "Win32 API Programming with Visual Basic" ISBN: 1-56592-631-5: Terminology: BSTR - VB string (pointer to Unicode Array) The Unicode character array that is pointed to by a BSTR must be preceded by a 4-byte length field and terminated by a single null 2-byte character (ANSI = 0). LPSTR - long pointer to a null-terminated ANSI character array (VC++ string) LPWSTR - long pointer to a null-terminated Unicode character set with no embedded nulls (VC++ string) ABSTR - made-up term used for BSTR type character array pointing to an ANSI array s = StrConv(s, vbToUnicode) APIFunctionW s s = StrConv(s, vbFromUnicode) "What we are doing here is compensating for the shrinking of our BSTR to an ABSTR by expanding it first. Indeed, the first call to the StrConv function simply takes each byte in its operand and expands it to Unicode format. It doesn't know or care that the string is already in Unicode format." .... "Now, in preparation for passing the string to [APIFunctionW], VB takes this expanded string and converts it from Unicode to ANSI, thus returning it to its original Unicode state. At this point, [APIFunctionW] can make sense of it and do [its function]. Once the converting string returns from [APIFunctionW], VB "translates" the result to Unicode, thus expanding it with embedded null characters. We must convert the result to ANSI to remove the supererogatory padding." .... Special care is required in the case of Windows 9x since it behaves differently. (He goes on to explain how to handle that case also.) There must be several string functions that are used by Win32 to provide the OS with what it needs. BTW, Steven's explanation of how VB stores strings internally helped me discover why I was having a problem outputting pdf information in binary mode after a string was used to store that information. Thanks Steven. James A. Fortune Nov 12 '05 #10 ### This discussion thread is closed Replies have been disabled for this discussion.

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# EPR pair and individual operations [closed] I have created an EPR pair. Let's suppose it's $$(|00\rangle + |11\rangle)/\sqrt{2}$$. We both of our halves and then we move into our places. 1. If I ask you how does your half look like (mathematical expression of the state), what would be the answer? (How does my half look like?) 2. If I apply some operation to my half like maybe applying pauli $$X$$ gate which flips qubits, would it change anything in your half? 3. What would the combined state look like after this operation? • please ask a single, laser-focused question per post. You can ask different questions on different posts. Also, the title should reflect what (specifically) is actually been asked in the question – glS Feb 11, 2021 at 11:02 • @glS I will try that. But I am sure this question was concept clearing. But I will do that next time. Feb 11, 2021 at 11:14 (1) The best description that I can give is a mixed state, $$\rho=I/2$$. (3) The combined state looks like $$(|01\rangle+|10\rangle)/\sqrt{2}$$. • Now I see your answer. Thanks for answering. I have one question, can you give me an idea of how you arrive at $\rho$? What is the meaning of it? Feb 10, 2021 at 14:23 • Ok now I know why it is $I/2$. Thanks. Feb 10, 2021 at 15:55 • Still I have one more question. We have got the representation of state via density operator. So if I apply X, then how would it change in my case? (Because I don't have anything of the form $\alpha |0\rangle + \beta|1\rangle$? And once I tell you that, how would you state change? Wait a second, I can do anything I want i wont matter unless I measure something and tell you right? Feb 10, 2021 at 16:01 • The description of your qubit is $\rho_A$, the description of my qubit is $\rho_B$. If you do $X$ to your qubit, yours becomes $X\rho_AX$. Mine does not change. Feb 11, 2021 at 7:45

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# Is this expression for root mean square velocity dimensionally inconsistent? [closed] Root mean square velocity for the molecules of a perfect gas (as $P \rightarrow 0$) is given by: $$c = \left( \frac{3RT}{M} \right) ^{\frac{1}{2}}$$ On dimensional analysis, RHS gives $\left( \frac{[L]^2}{[T]^2[mol]} \right) ^{\frac{1}{2}}$ which must be equal to $\frac{[L]}{[T]}$, is it dimensionally inconsistent? • It is my mistake, I mistook M's units to be that of mass instead $Mass * mol^{-1}$ Sep 10 '17 at 12:13 • You should delete the question since your premise for asking is flawed and recognized. It's not really about physics. Sep 12 '17 at 21:39 • I'm voting to close this question as off-topic because it was based on a mistake that was since recognized. Sep 13 '17 at 0:48 • Sorry, but I can't find an option to delete the question Sep 13 '17 at 2:18 I would imagine that you forgot that $M$ is the mass per unit mol, which would solve your problem. However, since mol is a dimensionless number, it arguably doesn't need to appear in dimensional analysis (though can be helpful). Working in SI units (equivalent to dimensions) $$\text{m s}^{-1}=\sqrt{\frac{[\text{J}\ \text{K}^{-1}\ \text{mol}^{-1}] \ \ [\text{K}]}{[\text{kg}\ \text{mol}^{-1}]}}.$$ Alright?

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Wikipedia 10K Redux by Reagle from Starling archive. Bugs abound!!! <-- Previous | Newer --> | Current: 983456888 Dick Beldin at Thu, 01 Mar 2001 14:28:08 +0000. # Testing_Hypotheses ```back to [[Statistical Theory]] -- [[Applied Statistics]] Many researchers wish to test a '''statistical hypothesis''' with their data. There are several prerequisites which should be accomplished before the data is at hand. #The hypothesis must be stated in mathematical/statistical terms that make it possible to calculate the probability of possible sample assuming the hypothesis is correct. For example, ''The mean response to treatment being tested is equal to the mean response to the placebo in the control group. Both response have the [[Normal Distribution]] with the unknown means and the same known [[Standard Deviation]].'' #A test [[Statistic]] must be chosen that will summarize the information in the sample that is relevant to the hypothesis. In the example given above, it might be the numerical difference between the two sample means. #The distribution of the test statistic is used to calculate the probability sets of possible values (usually an interval or union of intervals). In this example, the difference between sample means would have a normal distribution with a standard deviation equal to the common standard deviation times the factor '''1/sqrt(n1) + 1/sqrt(n2)''' where n1 and n2 are the sample sizes. #Among all the sets of possible values, we must choose one that we think represents the most extreme evidence '''against''' the hypothesis. That is called the '''critical region''' of the test statistic. The probability of the test statistic falling in the critical region when the hypothesis is correct is called the '''alpha''' value (or '''size''') of the test. #After the data is available, the test statistic is calculated and we determine whether it is inside the critical region. #If the test statistic is inside the critical region, then our conclusion is either *The hypothesis is incorrect ''or'' *An event of probability less than or equal to ''alpha'' has occurred. ---- [[Dick Beldin]] ```

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# Some years have a siman, what does it mean? I saw on a calendar that each year has a siman, for example 5772 has the siman 517. Here are the simanim for the next few years: • 5773: 203 • 5774: 1523 • 5775: 517 • 5776: 1217 Does anyone know the meaning of the siman? I was thinking about a description of the year but I can't find relevant things. • Why "Some years"? Apr 30 '12 at 18:25 • can you tell us the name of the calendar? Apr 30 '12 at 18:25 • @Menachem it's a calendar edited by a shul in Paris Jun 12 '12 at 8:44 My guess would be that it means as follows: First Day of Rosh Hashanah = Thursday = 5 One Adar (i.e., non-leap-year) = 1 First Day of Pesach = Shabbos = 7 From this information you could extrapolate everything else in the year. I've never seen this particular system before, but it makes sense. • this my first idea but I see this siman on an other year for which the first day of Rosh Hashanah is a Monday :( Apr 30 '12 at 13:27 • The "1" in the second place actually means that it's a "kesidran" year, where the months are sequentially 30 and 29 days; thus, Cheshvan had 29 and Kislev 30. ("0" would mean a "chaseirah" year, where both of them are 29 days; "2" is a "malei" year, where both have 30. That's the system used in the tables in Leo Levi's Jewish Chrononomy.) – Alex Apr 30 '12 at 13:48 • Interesting, @Alex! That's not a very intuitive system, if you ask me. – Dave Apr 30 '12 at 13:56 • @allced - what was the other year that had this siman? – Dave Apr 30 '12 at 13:57 • @Dave: here the simanim for the next years: 5772: 517 5773: 203 5774: 1523 5775: 517 5776: 1217 Apr 30 '12 at 14:21

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Cheap and Secure Web Hosting Provider : See Now # [Solved]: Find vectors with elements of finite fields that sum up to given value , , Problem Detail: Given a universe $U$ consisting of k sets of vectors with each vector $\vec{v} \in {\mathbb{F}_{p^m}}^n$. Given also another vector $\vec{c} \in {\mathbb{F}_{p^m}}^n$. Now decide if there is a set $X$ with $|X| = |U|$ and $X_i \in U_i, i = 1,2,...,k$ such that $\sum\limits_i X_i = \vec{c}$. If there is, output this set. In other words, I want to find a combination of one element out of each set that sums up to the given vector, given that the vectors' entries can only be the results of modulo operation with a given integer. I hope the problem becomes clear. I want to find an efficient algorithm to solve this problem. It seems to me that it is NP-complete, but I find no other NP-complete problem that I can reduce. If there is one, existing algorithm (if any) to that other problem could be used for this problem. I looked at integer programming, but I did not find anything with respect to finite fields. Any ideas?

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1. n Prove that: if a and b are positive integers with (a, b) = 1, and if ab is a perfect square, then a and b are perfect squares. 2. Let $ab=k^2$ Let $k=p_1p_2\ldots p_n$. Then $ab=p_1^2p_2^2\ldots p_n^2$ Then $p_1^2|ab$. But $(a,b)=1\Rightarrow p_1^2|a$ or $p_1^2|b$. Suppose that $p_1^2|a$ In the same way, suppose $p_2^2|a,\ldots ,p_i^2|a$. Then $p_{i+1}^2,\ldots ,p_n^2|b$ So, $a=p_1^2p_2^2\ldots p_i^2a_1, \ b=p_{i+1}^2p_{i+2}^2\ldots p_n^2b_1$ Then $ab=p_1^2p_2^2\ldots p_n^2a_1b_1\Rightarrow a_1=b_1=1\Rightarrow a=(p_1p_2\ldots p_i)^2, \ b=(p_{i+1}p_{i+2}\ldots p_n)^2$

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Do you remember the magic in the moment when you discovered Fibonacci numbers? That epiphany where you realized that math is everywhere, even in the arrangement of petals in a flower. Today, we leap from the world of natural phenomena into the realm of legal documentation – a will template download. Mathematics and software engineering truly know no bounds. ## Understanding Wills: A Mathematical Model Everyone is familiar with the concept of a will, a legal document that expresses one’s wishes after their departure. But how does mathematics relate? Let me bring back your attention to set theory. A will is essentially like a function, assigning every object in your estate (the domain) to a beneficiary (the range). Thinking of a will in mathematical terms, simplifies the process and ensures nothing is overlooked. A will template is essentially a structured document which allows you to plug in your unique values (much like variables in an equation) and derive a solution that suits your specific need. Now, overlay this with software engineering principles, and we have a will template download option ready. With the advent of software development techniques, creating a personalized will has never been easier. A will template download can be analogized to a method in Object Oriented Programming (OOP). The template works like a method, which takes arguments (information about assets, beneficiaries) and gives an output (a personalized will). The process of crafting a will can be seen as a problem of optimization, where the aim is to minimize disputes and maximize the satisfaction of beneficiaries. Here, a well-designed will template download can serve as an optimal solution. By using such templates, you create a well-structured will, ensuring all involved parties are taken care of. Like any good piece of software, a will template must undergo rigorous testing to ensure its reliability. This includes checking for errors and validating different use-cases which mimics the exhaustive proof methods we employ in mathematics. After downloading and filling a will template, it’s essential to review it thoroughly – much like debugging a program or verifying a mathematical proof. This ensures the commands in your ‘code’ – your will – execute as intended. ## Your Role in Fine-Tuning the Will Template As a mathematician or statistician, you have a unique skill set to bring precision, clarity, and depth to your will. With these skills, you can tweak the downloaded will template, just as you would refine an theorem or optimize a piece of software. This is perhaps the most significant value of a will template download for our community. ### Journey to the Perfect Will Template Much like the perfect code or the elegant solution to a complex mathematical problem, seeking the perfect will template is a journey. It involves downloading multiple templates, comparing them, and customizing the one that aligns best with your needs. Remember, much like mathematics and software engineering, creating a well-structured will is both an art and a science. By treating a will like a complex problem waiting to be solved, we can bring the rigor and precision of our field to ensure our loved ones are cared for after our departure. So what are you waiting for? Start your journey to find the perfect will template today. Ready to embark on the path of discovery? Keep yourself hooked for our following articles where we dive deeper into software aided legal solutions. Don’t let the Fibonacci sequence be the end of your adventure, and let’s continue exploring math in unchartered waters together. Finding a reliable, adaptable, and rigorous will template download can seem as challenging as finding a solution to Fermat’s Last Theorem. But fear not, with the right guidance and approach, even the most complex problems can be solved. So, make your move and download a will template today, secure in the knowledge that thoughtfulness is the true gift you’re passing down to those who matter the most. ## Is there a will template in Word? Yes, Microsoft Word does offer a variety of templates, including a will template. You can find these templates within the software itself. Here’s how: 1. Open Microsoft Word. 2. Click on the “File” tab in the top menu and then select “New” from the dropdown menu. 3. In the “Search for online templates” box, type “will” or “last will and testament”. 4. Browse through the available will templates. Click on the one you want to use and press “Create”. Remember though, while a Word template can be a useful guide, it is always recommended to consult with a legal professional when creating a legal document such as a will. This ensures that your will is legally sound and covers all the necessary aspects of your estate. Also, don’t forget to save your document regularly while editing to avoid losing any information. ## Does Google Docs have a last will and testament template? Yes, Google Docs does offer a range of templates for different purposes, which may include legal documents such as the last will and testament. However, it is always advised to consult with a professional when creating this type of important legal document, as they can help ensure that all necessary details are included and that the document is legally binding. The templates on Google Docs serve as a good starting point, but should not replace professional legal advice. ## What is a template will? A template will, in the context of software, refers to a predefined set of code or file that serves as a starting point for a new project or module. This template is designed to save time, eliminate repetitive work, and provide a consistent structure for developers. The main feature of a template is that it’s reusable, allowing developers to use the same code base to create different modules or projects. The template is typically pre-configured with basic functionalities that are common to most projects, such as connections to databases, standard interfaces, common libraries, etc. For example, in web development, a template might include the basic HTML, CSS, and JavaScript codes with placeholders for custom content. Developers can replace these placeholders with the actual content to create unique web pages. In the context of software development, the use of templates promotes the principles of Don’t Repeat Yourself (DRY) and increases efficiency by reducing the amount of boilerplate code one has to write. It also ensures consistency and standardization in the codebase. So in summary, a template will in software is all about improving productivity, maintaining consistency, and reducing errors by providing a reliable, repeatable way to generate code. ## What is the correct wording for a will? In the context of software, the term “will” is not commonly used. However, if we’re talking about intended function or future planning in software, we might discuss “desired outputs”, “programmed actions”, or “feature development.” For instance, in describing the functionality of a new program, one might say: “The software will analyse the input data and generate a detailed report. This feature is currently in development and will be available in the next version update.” Or, when discussing the planning and future actions: “The development team will address these bugs in the upcoming sprint. These fixes will ensure the software runs smoothly and accurately for all users.” In this case, the word “will” refers to intended or planned action in the future. It’s vital to remember that precise language should be used when discussing software functionality to avoid misunderstandings. When considering downloading a will template, you should keep an eye out for several key aspects to ensure that the final document will be valid and covers all your wishes. 1. Legally Compliant: Ensure the template you select is in line with your local laws and regulations. The requirements for a legal will can vary by jurisdiction, so it’s important the template meets the necessary criteria. 2. Comprehensive Coverage: The template should cover a range of scenarios and cater to your specific needs. This includes provisions for property distribution, guardian appointments, and instructions for pets if any. 3. Easy to Understand: Legal jargon can be confusing. A good will template should be simple enough for you to understand, yet comprehensive in legal coverage. 4. Editable: The template should be editable so you can adapt it to your own personal circumstances. 5. Reputable Source: Consider the source of the template. It should come from a trusted, reputable source with expertise in legal matters. 6. Includes Instructions: A good will template should come with detailed instructions on how to complete it, sign it, and witness it, to ensure its legality. 7. Updates: Check whether the provider offers updates to their templates, as laws may change over time. 8. Support Features: Some providers might offer additional support features, like customer service or attorney consultations. These can be incredibly beneficial, especially if you run into problems or have questions while completing your will. Remember that while a downloaded will template can be a cost-effective and convenient solution, it’s always worth seeking legal advice to ensure your will is legally sound and accurately reflects your wishes. ### Can you recommend some reputable websites that offer free will template downloads? Sure, here are some reputable websites where you can download free will templates. 1. Do Your Own Will: It offers a thorough and straightforward free last will and testament template that’s legally valid in your state. 2. LawDepot: You can create a custom will with their easy-to-use online template. 3. Rocket Lawyer: Their step-by-step instructions make it easy to create a will. While they offer a free trial, there may be fees for more complex services. 4. Fabric: Fabric provides an easy-to-use, free last will and testament template, as well as other estate planning tools. 5. LegalZoom: Though not free, LegalZoom offers a comprehensive will creation service at a relatively low cost. Remember, making a will is a serious undertaking and while these templates can be a good starting point, it may still be beneficial to consult with a legal professional. ### How do I ensure the will template I have downloaded is legally binding in my state or country? The legal validity of a will template depends on the regulations in your specific state or country. There are several key points to consider when trying to ascertain if your downloaded will template is legally binding: 1. Law Compliance: Your will template needs to comply with laws in your jurisdiction. Laws vary greatly by location, so you have to determine whether the template was created according to your local laws. 2. Witnesses: In many jurisdictions, for a will to be valid, it requires the signature of at least two witnesses. They need to be present when you sign and could potentially testify about the signing if required. 3. Notarization: Some places may require your will to be notarized to add another level of verification. 4. Complete and Correct Information: The information within the will, such as estate division and recipient details, should be correct and complete. Any discrepancies or missing details could lead to legal complications. 5. Professional Advice: It’s always a good idea to seek professional advice. A lawyer can examine your template and verify that it contains all necessary elements and satisfies the requirements of your locale. In terms of software, there are programs and platforms available that can help you customize legal documents according to your jurisdiction’s laws. These software solutions keep up-to-date with law changes and aim to incorporate these modifications into their templates. Nevertheless, always get the final document reviewed by a legal professional to ensure its legitimacy. Remember, a will holds serious implications on how your assets will be divided after your death. Therefore, it’s worth investing time and probably money to make sure it’s done right. It’s not just about having a template; it’s also about how you use it. ### Is it possible to modify the downloaded will template, and if so, how? Yes, it is absolutely possible to modify a downloaded will template. To do so, you would typically use a word-processing software such as Microsoft Word or Google Docs. These are the steps: 2. Edit the document: Click on the section of the text you would like to modify and start typing. Most templates have placeholder text that guides you where you need to fill in your specific information. 3. Save your changes: Once you’ve made the desired changes, don’t forget to save your document. You can do this by navigating to ‘File’ > ‘Save As’, and then choosing your desired save location. Remember, while templates can be useful guides, they should not replace legal advice. Always consult with a legal professional when creating a will to ensure that your assets and interests are properly protected. ### What precautions should I take when downloading a will template online to prevent issues with legality or security? When downloading a will template online, it’s vital to consider both the legality and the security aspects. Here are some precautions to take: Legality: 1. Check for State-Specific Rules: Laws regarding wills vary from state to state. Ensure the template you use complies with your state’s laws. 2. Review the Template: Read through the template carefully to make sure it covers all areas you want to include in your will. 3. Consider Professional Advice: A pre-made template may not be able to capture the specifics of your situation. It might be best to get legal advice to ensure your will is legally sound. Security: 1. Use Trusted Sources: Only download templates from reputable sources to eliminate the risk of malware or other security threats. 2. Secure Your Personal Information: Be careful when entering personal information online. Use secure, encrypted connections and avoid using public Wi-Fi. 3. Regularly Update Your Security Software: Keeping your anti-virus software and operating system up-to-date can help protect your computer from new threats. Lastly, it’s also worth noting that a downloaded will template should be used as a temporary solution until you can have a will professionally drawn up to suit your specific needs.

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# Bohmian Mechanics Consider the equation . This equation relates the pressure, volume, and temperature of an ideal gas. All of these concepts are macroscopic — meaning that on the level of the molecules that make up the gas the meaning of ‘pressure,’ ‘volume,’ and ‘temperature’ dissolves. One molecule cannot have a pressure, it cannot be said to represent a volume of gas, and it does not possesses temperature. All three of these concepts begin to take on meaning as we zoom out and consider a collection of the molecules and account for their motions — as we transition from a microscopic scale to a macroscopic scale. What does it mean to say that this equation relates properties of an ideal gas? What is an ideal gas? It means that energy conservation and closed system considerations apply. In the case of our gas it means that the interactions/collisions between the molecules are all completely elastic. Gasses that exhibit measurable inelasticity in their interactions cannot be accurately represented by this equation on all macroscopic scales. Why are we talking about all of this? Well the mathematics that best mimics the geometric structure of qst to date is captured by a set of equations known as Bohmian mechanics. The Bohmian formalism has been shown to make all the predictions that the standard model of quantum mechanics makes — identically — while remaining a deterministic theory. However, Bohmian mechanics (and the standard equations of quantum mechanics) are incapable of incorporating the geometric effects of gravity into their models. Let’s explore a candidate reason for why this is the case. In order to make the Bohmian formalism completely representative of the geometry of qst let’s treat the equations in this formalism as macroscopic expressions of idealized interactions of the quanta of spacetime. Just like the equation , the Bohmian formalism assumes perfect elasticity of the underlying constituents in its macroscopic expressions. It is possible that all we have to do to bring gravity into the formalism is to get to the underlying structure that relates the interactions of the spacetime quanta and include a small second-order inelasticity in those interactions. This would be like modeling molecular interactions and allowing them to have a slight inelasticity. Doing this might allow us to produce an general equation that captures the behavior of ideal gasses and non-ideal gasses simultaneously. For those interested, here is the derivation of the Bohmian set of equations: Let’s begin by addressing the objective state of the wave function on the microscopic level. (Microscopic level in this case means on the quantum or Planck scale.) If our system (a chosen domain of spacetime) is composed of N particles, then a complete description of that system will necessarily include a specification of the positions Qi of each of those particles. On its own, the wavefunction does not provide a complete description of the state of that system. Instead, the complete description of this quantum system must be given by where is the configuration of the system and a (normalized) function on the configuration space — the superspatial dimensions — is its wave function. At this point, all we have to do in order to obtain our theory is specify the law of motion for the state . Of course, the simplest choice we can make here would be one that is causally connected. In other words, one whose future is determined by its present specification, and more specifically whose average total state remains fixed — at least in the macroscopic sense of the familiar four dimensions of spacetime. To obtain this we simply need to choreograph the particle motions by first-order equations that assume elastic interactions. The evolution equation for is Schrödinger’s equation: Where is the wave function and V is the potential energy of the system. Therefore, in keeping with our previous considerations, the evolution equation for Q should be: . with where takes the form of a (velocity) vector field on our chosen configuration space . Thus the wave function reflects the motion of the particles in our system in a macroscopic averaged-over sense based on the underlying assumption of elastic interaction. These motions are coordinated through a vector field that is defined on our specified configuration space. If we simply require time-reverse symmetry and simplicity to hold in our system (automatic necessities for a deterministic theory) then, Notice that there are no ambiguities here. The gradient on the right-hand side is suggested by rotation invariance, the in the denominator is a consequence of homogeneity (a direct result of the fact that the wave function is to be understood projectively, which is in turn an understanding required for the Galilean invariance of Schrödinger’s equation alone), the Im by time-reverse symmetry which is implemented on by complex conjugation in keeping with Schrödinger’s equation, and the constant in front falls directly out of the requirements for covariance under Galilean boosts.1 Therefore, the evolution equation for Q is This completes the formalism of Bohmian mechanics that David Bohm constructed in 1952.2 The math may appear daunting but the concepts are amazingly simple. In our construction we have considered applying the analogy of a gas being made up of elastically interacting constituents to the quanta of our spactime system. As an extension of de Broglie’s pilot wave model3 this formalism exhaustively depicts a nonrelativistic universe of N particles without spin.4 Spin must be included in order to account for Fermi and Bose-Einstein statistics. The full form of the guiding equation, which is found by retaining the complex conjugate of the wave function, accounts for all the apparently paradoxical quantum phenomena associated with spin. For considerations without spin the complex conjugate of the wave function cancels because it appears in the numerator and the denominator of the equation. The full form of the evolution equation is: Notice that the right-hand side of the guiding equation is J/Q, the ratio for the quantum probability current to the quantum probability density.5 Note that the idealized assumption in play here is that . In other words, the transformation arises directly from Schrödinger’s equation. If these evolutions are indeed compactable, then is equivariant. Therefore, under the time evolution retains its form as a function of . If you are interested in taking part in rederiving the Bohmian set from underlying interactions that are first-order elastic and second-order inelastic please send an email to qst@einsteinsintuition.com. Notes: 1. Detlef Dürr, Sheldon Goldstein, and Nino Zanghí,
‘Quantum Physics Without Quantum Philosophy,’ pp. 5-6. 2. D. Bohm, ‘A suggested interpretation of the quantum theory in terms of “hidden” variables,’
Physical Rev. 85 (1952), pp. 166-193. 3. L. de Broglie, ‘La nouvelle dynamique des quanta,’ Electrons et Photons: Rapports et Discussions du Cinquieme Conseil de Physique tenu a Bruxelles du 24 au 29 Octobre 1927 sous les Auspices de l’Institut International de Physique Solvay, Gautheir – Villars, Paris, 1928, pp. 105-132. 4. Of course in the limit ħ/m = 0, the Bohm motion Qt approaches the classical motion. See: D. Bohm and B. Hiley, ‘The Undivided Universe: an Ontological Interpretation of Quantum Theory,’ Routledge & Kegan Paul, London, 1993; Detlef Durr, Sheldon Goldstein, and Nino Zanghi, ‘Quantum Physics Without Quantum Philosophy,’ p. 7. 5. Sheldon Goldstein, ‘Bohmian Mechanics.’ For further examples of how easily spin can be dealt with in the Bohmian formalism see: J. S. Bell, 1966, pp. 447-452; D. Bohm, 1952, pp. 166-193; D. Dürr et al ‘A survey of Bohmian mechanics, Il Nuovo Vimento’ and ‘Bohmian mechanics, identical particles, parastatistics, and anyons’, In preparation. 1. Ben says: Sent. 2. Jeff says:

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Wednesday July 30, 2014 # Search: Calculate the enthalpy for Number of results: 18,128 Chemistry A 5.0-g Sample of KBr at 25.0 degrees celsius dissolves in 25.0 degrees celsius 25.0 mL of water also at 25.0 degrees celsius. The final equilibrium temperature of the resulting soltution is 18.1 degrees celsius. What is the enthalpy of of solution in kilojoules per mole of KBr? December 2, 2010 by Sally Chemistry The enthalpy of vaporization of chloroform (CHCl3) is 29.2 kJ mol-1 at its normal boiling point 61.2 Celsius. what is the standard change in entropy for vaporization of chloroform at its normal boiling point? Answer I got was 29.2/334.2 = 0.0873. (CORRECT?) April 2, 2013 by Viv chemistry how in the world do you do this problem? Use the thermochemical equations shown below to determine the enthalpy (kJ) for the reaction: H2SO3(l)=>H2S(g) + 3/2O2(g) H2SO3(l)=>H2O(l) +SO2(g) DH=62KJ SO2(g)=>S(s) + O2(g) DH=297KJ H2S(g) +1/2O2(g)=>S(s) + H2O(l) DH=-... October 24, 2010 by anonymous chemistry - energy and change rev (help) the below 2questions i could not answer while revising for this chapter pls help me out a little. 58. When a vehicle is parked in the sunlight on a hot summer day, the temperature inside can approach 55°C. One company has patented a non-CFC propelled aerosol that can be ... March 23, 2011 by Babelii chemistry calculate [h3o] in solutions. 0.040 of HCl and 0.080 HOCl. 2. calculate OH for the solutions of 0.0063 ba (OH)2 and 0.0110 of BaCl2 April 11, 2010 by anonymous math Consider the function defined below. y = 2t3 - 4t2 + 3 (a) Calculate the second derivative. y '' = (b) Calculate the third derivative. y ''' = June 21, 2010 by Jack Chemistry Calculate how much heat 32.0 g of water absorbs when it is heated from 26.0°C to 73.5°C. In joules, and then we have to calculate calories too, how is that done? I was absent. Thanks so much! May 25, 2011 by Jane Chemistry Calculate the wavelength of broadcast waves. Calculate the energy associated with the broadcast frequency for WPHT, 1210 kHz. November 5, 2011 by Anonymous math calculate the mass (in SI units) of a 150ib human being? m= Calculate the weight ( in English units) of a 200kg rhino m= January 26, 2012 by bree chemistry For a particular isomer of C8H18, the following reaction produces 5113.3 kJ of heat per mole of C8H18(g) consumed, under standard conditions. C8H18(g) + 25/2(O2)(g) -> 8CO2(g) + 9H2O(g) DeltaH= -5113.3 kJ What is the standard enthalpy of formation of this isomer of C8H18(g)? January 29, 2012 by Mike epidemiology In a fictitious study, 100 out of 1000 people aged 60 years and over were positive in detecting breast cancer using Tumour Biopsy (gold standard). Using mammography, 90 of breast cancer patients and 135 healthy persons were also positive by this test. (I) Calculate and ... August 18, 2012 by Ekrem ap chem A 9.00 M solution of a weak acid, HA, has a pH of 1.30 1.) calculate [A-] at equilibrium 2.) calculate [HA] at equilibrium Use pH to get (H^+). Ka = (H^+)(A^-)/(HA) (A^-) = (H^+). (HA) = 9 Plug and chug. Post your work if you get stuck. March 22, 2007 by Jaron Physics A particle's initial velocity is given by vox = -1.0 m/s and voy = 3.0 m/s. Its acceleration is ax = 0.50 m/s^2 and ay = -0.90 m/s^2 (a) Calculate its velocity 2.0 s later. (b) Calculate its displacement at that time. October 4, 2009 by Sandhu statistics GIVEN THE FOLLOWING STATISTICAL VARIABLES A:m(A)=5 B:m(B)=13 var(A)=25 Var(B)=6 1.calculate mean of Z= 4B-2A 2.Calculate the Standard Deviation assuming cov (A,B)=0 August 17, 2010 by RASHMI physics A force of 303.0 N is used to push a 160 kg mass 30.0 m horizontally in 3.00 s. (a) Calculate the work done on the mass. 1 kJ (b) Calculate the power developed. 2 kW December 12, 2011 by Anonymous chemistry calculate r.m.s velocity of he molecule at 300k.also calculate the total k.e(in calories)of the molecules in one litre he at 300k & 1atm? July 6, 2012 by dip Math calculate depreciation please explian how to calculate depreciation for federal income taxes what method do I use thanks :) November 14, 2007 by Anonymous calculus calculate the indefinite integral 2sec^2x dx (cosx)/(sin^3x) dx calculate the definite integral interval pi/4, pi/12 csc2xcot2x dx January 6, 2008 by anonymous geometry In the following figure given a circle C(o,3) Bp=AB and BA^N = 30 a) Calculate AP,BN and AN b)Calculate AQ and BQ c) Show That M is the mid point of (AP) d)Show that (OM)is perpendicular to (A) April 16, 2010 by Nour physics A spherical conductor has a radius of 14.0 cm and charge of 26.0 µC. Calculate the electric field and the electric potential (a) r = 10.0 cm, (b) r = 20.0 cm, and (c) r = 14.0 cm from the center. how do i calculate k? February 7, 2012 by confused Chemistry-Help An AM radio station broadcasts at 1060 kHz. Calculate the wavelength of the corresponding radio waves. How do I calculate this? April 30, 2012 by Liz Physics An electric dipole comprised of two charges Q and -Q are separated by a distance of 3.0 cm. Assume Q = 2.0x10^-6 C and -Q = -2.0x10^-6 C. Find the magnitude and direction of the electric field vector E and point P, which is 4.0 cm to the right of the center of -Q. Find the ... March 13, 2012 by Michelle Chemistry When 23.6 of calcium chloirde CaCl2 was dissolved in water in a calorimeter, the temperature rose from 25.0 degrees C to 21.56 degrees C. If the heat capacity of the solution and the calorimeter is 1071 J/ degrees C, what is the enthalpy change when 1 mol of sodium nitrate ... November 1, 2007 by Lauren chemistry When water freezes, the phase change that occurs is exothermic (DH = -6.02 kJ). Based on the change in enthalpy, you would expect that water would always freeze. Use the concepts of entropy and free energy to explain why this phase change is favourable only below o degrees ... April 6, 2008 by Sarah Chemistry For ethanol, C2H5OH, which is mixed with gasoline to make the fuel gasohol, H°f = -277.63 kJ/mol. Based upon data in Table 6.2, and that the density of ethanol is 0.787 g•cm-3, the number of kilojoules released by burning completely 2.40 gallons of ethanol is ? *Table 6.2 ... October 16, 2011 by A Thermodynamics a rigid vessel of volume 1m(^3) contains steam at 20bar and 400 degrees Celcius. The vessel is cooled until the steam is just dry saturated. Calculate the mass of steam in the vessel, the final pressure of the steam and the heat rejected during the process. The answers should ... March 12, 2010 by Claire An electrolysis experiment was run for 17.5 minutes at an average current of 191 milliamps (mA). The mass lost by the copper anode was 0.0679 grams. a) Calculate the number of coloumbs transferred. (coloumb=ampsxsec) b) Calculate the moles of copper lost from the anode. c) ... April 16, 2014 by Dianne Chemistry What is the entropy of CH3OH? My CRC tables list 239.865 J/K*mol @ 298.15o K. Do you also know the enthalpy? deltaHof= -201.000 kJ/mol according to the CRC Handbook,1995-96 edition. March 26, 2007 by Meg Physics A 275 kg piano slides 4.0 m down a 30° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline. The effective coefficient of kinetic friction is 0.40. a) Calculate the magnitude of the force exerted by the man. b) Calculate the work ... December 11, 2013 by Josh Physics A 275 kg piano slides 4.0 m down a 30° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline. The effective coefficient of kinetic friction is 0.40. a) Calculate the magnitude of the force exerted by the man. b) Calculate the work ... December 11, 2013 by Josh physic Calculate the mass (in SI units) of a 150 lb human being. m= (correct unit) Calculate the weight (in English units of a 2000 kg rhinoceros. January 27, 2012 by sam chemistry 12 Im doing a worksheet and i had to draw PE diagram which had the Ea(activation energy)at 80kJ and the enthalpy change is -40kJ. *All the reactants and products are gases Then it asks to describe what would happen to the diagram if the temperature dropped 25 degrees Celsius. Im ... December 14, 2012 by Anonymous AP physics i really need help on this problem!! Three blocks are connected,on a horizontal frictionless table and pulled to the right with a force T3 = 64.4 N. Assume that m1 = 12.5 kg, m2 = 25.0 kg, and m3 = 31.7 kg. (a) Calculate the acceleration of the system (b)Calculate the tensions... September 24, 2010 by annamaria Physical Science How do you calculate the mass and volume of an irregular solid object and calculate the density of that object? January 25, 2009 by Alex chemistry % recovery: how do i calculate it? we're doing a recrystallization in chem class we used 0.1 g and we recovered 0.3 g how do i calculate the percent recovery? February 9, 2011 by jessie calculate ratios Calculate the amount of drug in a ratio solution. How many micrograms are in 1ml of a 1:5000 solution? March 6, 2012 by Vicki chemistry Consider the hypothetical elements X and Y. Suppose the enthalpy of formation for the compound XY is ƒ{336 kJ/mol, the bond energy for X2 is 414 kJ/mol, and the bond energy for Y2 is 159 kJ/mol. Estimate the XY bond energy in units of kJ/mol. December 13, 2010 by chemistry chemistry Consider the hypothetical elements X and Y. Suppose the enthalpy of formation for the compound XY is -336 kJ/mol, the bond energy for X2 is 414 kJ/mol, and the bond energy for Y2 is 159 kJ/mol. Estimate the XY bond energy in units of kJ/mol. December 13, 2010 by Jude PHYSICS I don't know where to start solving this question. Please help! A 320 kg piano slides 3.7 m down a 32 degree incline and is kept from accelerating by a man who is pushing back on it parallel to the incline. The effective coefficient of kinetic friction is 0.39. Calculate the ... November 8, 2010 by MegP Chemistry Please show me how to work! Ethane, C2H6(g) can be made by reaction of hydrogen gas with acetylene, C2H2(g). the standard enthalpies of formation of ethane and acetylene are -84.68 and +226.73 kJ mol-1, respectively. what is the reaction enthalpy for the production of one mole... October 20, 2012 by Sam How would i find the following? I just need an explanation. I'm stuck. I have many other questions like this and i need to know how to do them. The specfic heat of ice is 2.05 J/g. In a calorimeter, 10.0 g of ice melts at 0oC. The enthalpy of fusion of the ice is 334 J/g. How ... June 22, 2012 by anonymous chemistry The electron in a hydrogen atom is excited and makes a transition from n=2 to n=7 a. calculate the energy of the photon absorbed in joules. b. calculate the wavelength of the photon in meters HOW DO I DO THIS? March 21, 2011 by Tina physics Calculate the cost of operating a 65 W light bulb continuously for a 30-day month when electrical energy costs \$0.23/kW·h. \$ 1 How do you calculate cost? whats the equation? May 11, 2012 by Alice Math (Finance) You have a note in the amount of \$10,000 at 9.5% interest running from january 5. to april 16 a. calculate the ordinary intrest. b. calculate the exact interest. October 31, 2012 by Renee Chemistry - Solubility Given: Concentration of HCl is 0.1388M 0.5 g of Ca(OH)2 was placed in a flask. 100mL of 0.05M NaOH was poured into the flask. 25mL aliquot was filtrated and used for titration. Suppose to calculate the OH^- equilibrium concentration from the titration data, as well as the OH... Chemistry A reaction is peformed in which 50 ml of .400M silver nitrate and 75 ml of .100M hydrochloric acid are mixed together. The initial temperature of the solutionn was 25 C and the final temperature was 27.2 C. What is the enthalpy, in kj/mol, for the formation of the precipitate... November 18, 2012 by Lydia chemistry Baking soda is sodium bicarbonate (NaHCO3). When you heat baking soda, it breaks down into sodium carbonate powder (Na2CO3), water vapor, and carbon dioxide. The enthalpy of this reaction is 129 kJ. a. Write a correct, balanced thermochemical equation for this reaction. (... April 16, 2014 by Anonymous Chemistry The ΔHof of gaseous dimethyl ether (CH3OCH3) is –185.4 kJ/mol; the vapour pressure is 1.00 atm at –23.7oC and 0.526 atm at –37.8oC. a) Calculate ΔHovap of dimethyl ether. b) Calculate ΔHof of liquid dimethyl ether Any idead how to calculate for the heat of ... December 11, 2012 by xavier physics Having trouble with this one: A capacitor having a capacitance of 10uF is charged by 100v battery. A.)Calculate the charge on the capacitor? B.)calculate the energy stored in the capacitor? C.) the terminals of the capacitor is now disconnected from the batery and the ... February 28, 2010 by morgan Physics Consider the circuit shown in the figure below. Use the following as necessary: R1 = 11.00 Ù, R2 = 1.40 Ù, and V = 6.00 V. (a) Calculate the equivalent resistance of the R1 and 5.00 Ù resistors connected in parallel. 1 Ù (b) Using the result of part (a), calculate the combined... October 5, 2010 by chuck Physics A rectangular coil of 70 turns, dimensions 0.100 m by 0.200 m and total resistance 10.0 ohms, rotates with angular speed 32.0 rad/s about the y axis in a region where a 1.00 T magnetic field is directed along the x axis. The rotation is initiated so that the plane of the coil ... July 2, 2011 by Mike A student measures three lengths a, b and c in cm and a time t in seconds: a = 40 ± 5 b = 30 ± 3 c = 20 ± 1 t = 1.2 ± 0.1 What is a+b a+b+c a/t (a+c)/t Calculate (1.23 ± 0.03) + pi Question 9.4. Calculate (1.23 ± 0.03) × pi Can you do a sample calculation of each of those ... October 19, 2008 by Sam Science an object 5.50 cm high is placed 100 cm from a converging lens that has a focal length of 40.0cm. a)Calculate the image distance b)Calculate the hight of the image height January 4, 2010 by Ramandeep Chemistry I-I:150 F-F:160 C-C:350 H-H:435 (a) Calculate the energy of the I-F bond. Express your answer in units of kJ/mol. unanswered (b) Calculate the % ionic character of the I-F bond. March 5, 2013 by Anonymous Chemistry Can someone check my answers please? I had specific trouble with 7, 13, and 27 (I'm pretty sure my other answers are correct; some reassurance would be nice though). And I have a small question about 8. 1. The ΔHf of an element in its standard state is defined to be 0 kJ/... March 3, 2013 by a Canadian Chemistry Explain the significance of positive and negative values for each thermodynamic quantity in the free energy equation. (enthalpy change, entropy change, and free energy change) I think this is something you can find in your text/notes. If not, explain what you don't understand ... May 14, 2007 by Dave How would i find the following? I just need an explanation. I'm stuck. I have many other questions like this and i need to know how to do them. The specfic heat of ice is 2.05 J/g. In a calorimeter, 10.0 g of ice melts at 0oC. The enthalpy of fusion of the ice is 334 J/g. How ... June 22, 2012 by anonymous college chemistry Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the following equation: CO2(s)--> CO2(g). When dry ice is added to warm water, heat from the water causes the dry ice to sublime more quickly. The evaporating carbon dioxide ... November 9, 2009 by Eddie Intro Chemistry Which molecule based on the following enthalpies is the most stable? Molecule A: -70 kJ/mol Molecule B: 100 kJ/mol Molecule C: 145 kJ/mol Is it Molecule A? Isn't the lower the enthalpy the more stable the molecule? Thank You April 4, 2010 by echem PHYSICS a. A 25 nC particle and a – 7 nC particle is interacting with each other at a certain distance. Find the force between the two charges @ 3.0 cm and 1.0 cm Calculate and draw each of the particle’s electric fields. Calculate the electric potential values of the two particles. ... August 18, 2011 by KAIT Physics A long thin rod lies along the x-axis from the origin to x=L, with L= 0.890 m. The mass per unit length, λ (in kg/m) varies according to the equation λ = λ0 (1+1.410x2). The value of λ0 is 0.700 kg/m and x is in meters. 1. Calculate the total mass of the ... February 11, 2014 by Molly Physics A long thin rod lies along the x-axis from the origin to x=L, with L= 0.890 m. The mass per unit length, λ (in kg/m) varies according to the equation λ = λ0 (1+1.410x2). The value of λ0 is 0.700 kg/m and x is in meters. 1. Calculate the total mass of the ... February 11, 2014 by Molly physical chemistry how do i calculate a ratio of oxidised to reduced cytochrome c Fe2+ to Fe3+ given E(mV) at 25 C and pH7 as 100 and absorbance at 550nm as 1.5 E is relative to standard hydrogen electrode i dont know how im supposed to calculate it from this data February 17, 2010 by lucy CHEMISTRY Mass (before reaction): test tube + HCl(aq) + stir bar + capsule 26.600 g Mass (after reaction): test tube + HCl(aq) + stir bar + capsule 25.300 g Volume of water displaced from the squirt bottle 148 mL Temperature of the CO2(g) 287.4 K Pressure (atm) 1.032 atm Using this data... October 22, 2012 by Karina chemistry use thermochemical equations shown below to determine enthalpy for the reaction: H2CO3-->H2O+CO2 1) H2CO3 --> H2CO + O2 H; 224 KJ 2)H2CO + O2 --> H2O + CO2 H; -100KJ I know this problem is one that you have to rewrite the equations to get the equation in the original ... November 20, 2009 by Brittany chemistry use thermochemical equations shown below to determine enthalpy for the reaction: H2CO3-->H2O+CO2 1) H2CO3 --> H2CO + O2 H; 224 KJ 2)H2CO + O2 --> H2O + CO2 H; -100KJ I know this problem is one that you have to rewrite the equations to get the equation in the original ... November 20, 2009 by Brittany chemistry 1. Calculate the volume of carbon dioxide at 273 K and 1.01 * 10^5 Pa which would be produced when 1.25g of calcium carbonate reacts completely with HCL. and 2. When 41.18 cm^3 of a solution of silver ions with concentration of .2040 mol dm^-3 is added to a solution of XO4 -3 ... May 19, 2008 by ANn A long thin rod lies along the x-axis from the origin to x=L, with L= 0.890 m. The mass per unit length, λ (in kg/m) varies according to the equation λ = λ0 (1+1.410x2). The value of λ0 is 0.700 kg/m and x is in meters. 1. Calculate the total mass of the ... February 12, 2014 by Molly chem Use bond energies to estimate the enthalpy change for the following reaction C2H2 + H2-> C2H4 i figured the bonds broken are: 1 C triple bond C 1(812kj/mol) 2 C-H 2(414kj/mol) 1 H-H 1(435kj/mol) Bonds formed 1 C=C 1(590kj/mol) 4 C-H 4(435kj/mol) Is this right November 12, 2008 by tomi chemistry Calculate the change in internal energy of the system in a process for which 100J of heat is absorbed by the system from surroundings and does 85J of work. (I want to learn the solution on how to calculate it... pls. help thank you... ) January 23, 2011 by jaycab chem help the primary visible emissions from mercury are 404.7 nm and 435.8 nm. Calculate the frequencies for these emissions. Calculate the energy of a single photon and of a mole of photons of light with each of these wavelengths. March 29, 2012 by danielle Chemistry Calculate the energy released when an electron is added to a hydrogen nucleus. Assume the transition is n= infinite to n=1. I know equation to use but how do I calculate when n is infinite and no number ?? Please some help November 29, 2012 by Anonymous Chemistry How would I calculate this? I have the answer because its from a previous test I just want to know how I would calculate it. How many grams of NaCI are contained in 350. mL of a 0.133 M solution of sodium chloride? June 30, 2013 by Mary HEALTH The key to using a pedometer to meet the Physical Activity Guidelines for Americans is to _________. A. First set a time goal and then calculate how rigorous the steps need to be each day to reach your goal B. First calculate how many steps are needed weekly and then match it ... April 17, 2013 by CARRIE Physics A car of mass 1200kg starts from rest, accelerates uniformly to a speed of 4.0 meters per second in 2.0 seconds and continues moving at this constant speed in a horizontal straight line for an additional 10 seconds. the brakes are then applied and the car is brought to rest in... December 4, 2012 by Ken Chemistry Gasoline is primarily a mixture of hydrocarbons and is sold with an octane rating that is based on a comparison with the combustion properties of isooctane. Gasoline usually contains an isomer of isooctane called tetramethylbutane (C8H18), which has an enthalpy of vaporization... July 9, 2013 by Kelly Chemistry The molar enthalpies of combustion of CH3COCOOH(l)CH3COOH(l) and CO(g) are respectively -1275kJ/mol, -875kJ/mol, and -283kJ/mol. What is the enthalpy change for the reaction below? CH3COCOOH= CH3COOH +CO a)1867kJ b)-1867kJ c)117kJ d)-117kJ e)-2433kJ I think it's d) July 24, 2010 by Bob college chemistry How much heat is needed to convert 2.00 kg of liquid ammonia at -40.0 oC to its vapor at -11.0 oC? The specific heat of liquid ammonia, NH3 (l), is 4.70 J/g-K and that of the vapor NH3 (g), is 2.20 J/g-K. The enthalpy of vaporization of ammonia is 23.3 kJ/mol and its boiling ... May 25, 2010 by pamela physics A block of unknown mass is attached to a spring with a spring constant of 6.00 N/m and undergoes simple harmonic motion with an amplitude of 12.0 cm. When the block is halfway between its equilibrium position and the end point, its speed is measured to be 33.0 cm/s a)Calculate... April 27, 2009 by Jessi how do i calculate a ratio of oxidised to reduced cytochrome c Fe2+ to Fe3+ given E(mV) at 25 C and pH7 as 100 and absorbance at 550nm as 1.5 E is relative to standard hydrogen electrode i dont know how im supposed to calculate it from this data February 17, 2010 by lucy Chemistry Substance A has a normal fusion point of -10 degrees C, an enthalpy of fusion = 150 Jg^-1; specific heats for the solid and the liquid are 3 and 6.2 Jg^-1C^-1 respectively. To change 150 grams of A from a solid at -40 degrees C to a liquid at 70 degrees C will require how many... February 1, 2009 by Amanda Chemistry Magnesium fluoride dissolves in water to the extent of 8.0 10-2 g/L at 25°C. Calculate the solubility of MgF2(s) in moles per liter, and calculate Ksp for MgF2 at 25°C. I've been doing this and getting .086. Please help! May 26, 2010 by Kate physics really stuck, a 90 kg skier has 140 KJ of total mechanical energy when he is at a height of 120m above the bottom of the hill. calculate the skiers kinetic energy. calculate his speed? October 16, 2010 by dave physics really stuck, a 90 kg skier has 140 KJ of total mechanical energy when he is at a height of 120m above the bottom of the hill. calculate the skiers kinetic energy. calculate his speed? October 18, 2010 by mitch Physics An elevator cable breaks when a 950 kg elevator is 23 m above a huge spring (k = 2.3×105 N/m) at the bottom of the shaft. (a)Calculate the work done by gravity on the elevator before it hits the spring. (b) Calculate the speed of the elevator just before striking the spring. (... November 17, 2010 by Sasha health The key to using a pedometer to meet the Physical Activity Guidelines for Americans is to _________. A. First set a time goal and then calculate how many steps are needed each day to reach that goal B. First set a time goal and then calculate how rigorous the steps need to be ... CHEMISTRY A STUDENT STUDYING THE FE+3- HSCN equilibrium put into a test tube 10.00ml of 2.00 x 10-3 M Fe(NO3)3 with 10.0 mL of 2.00 x 10-3 M HSCN.THE H+ in the resulting solution was manteined at 0.500.By spectrophotometric analysis of the equilibrium solution, the equilibrium ... September 18, 2011 by da Chemistry If the reaction Fe2N(s) + 3/2H2(g) -> 2Fe(s) +NH3(g) comes to equilibrium at a total pressure of 1 bar, analysis of the gas shows that at 700. and 800.K, PNH3/PH2=2.165 and 1.083 respectively, if only H2(g) was initially present in the gas phase and Fe2N(s) was in excess. a... November 23, 2010 by Katie Mathematics an aircraft's reading shows time 8:55 at distance travelled 957km time 9:07 at distance travelled 1083 calculate distance travelled in km calculate the average speed of aircraft in km/h (that is what was asked it never said total distance it said calculate distance traveled in... July 31, 2012 by INDIANA to Reiny Design of suitable pin size Appreciate if someone can take the time to check my answers to the following question. If wrong, could you point me in the right direction. A round bar, 800 mm long, CSA 15 mm^2 experiences an applied tensile load of 6 kN and stretches elastically by 3 mm. a) Calculate the ... February 11, 2007 by BIGEYE Chemistry I have some enthalpies of formation for a reactant and transition state but calculated using different computational methods. I want to find the ratio of the forward rate constants determined using both methods but I'm not really sure how. I thought I could maybe use ... March 13, 2010 by Alice U.S. and Global Economics:Involving math Just show me how to calculate it and I'll do the rest. I just don't understand how to calculate this problem. Thank you, An online store that has been successfully growing on its initial angel investment and revenues wants to invest \$5 million to expand the business. The bank ... November 15, 2013 by Mary 14 The key to using a pedometer to meet the Physical Activity Guidelines for Americans is to _________. A. First set a time goal and then calculate how many steps are needed each day to reach that goal B. First set a time goal and then calculate how rigorous the steps need to be ... March 4, 2012 by desha chemistry When 15.3g of NaNO3 was dissolved in water inside a constant pressure calorimeter, the temperature fell from 25.00 °C to 21.56 °C. If the heat capacity of the calorimeter (including the water it contains) is 1071 J/°C, What is the enthalpy change for dissolving 1 mole of NaNO3... March 4, 2011 by Stephen Chemistry Consider a certain type of nucleus that has a half life of 32 min. Calculate the percent of original sample of nucleus remaining after 3.0 hours have passed? how would i calculate this??? March 4, 2008 by Gabe Chemistry Magnesium fluoride dissolves in water to the extent of 8.0 10-2 g/L at 25°C. Calculate the solubility of MgF2(s) in moles per liter, and calculate Ksp for MgF2 at 25°C. I keep getting the same answer...please help! May 9, 2011 by Laura chemistry Rubidium crystalizes in the body centred cubic structure with an edge length of 560.5 pm.. a. calculate the radius (in pm) to 4 sig figures. b. calculate the density to 4 sig.fgures. thnks guys.. December 4, 2011 by amanda science A train accelerates uniformly in a straight line at 0.3m/s from from rest. Calculate the velocity 15 seconds after its depature. Calculate the time taken to reach 30km/h and the distance covered by the train. April 7, 2013 by leslie Chemistry Chlorofluocarbon was once used in air conditiones as the heat transfer liquid its normal boiling point is -30 degrees C and its enthalpy of vaporization is 16J/g. The gas and the liquid have specific heat capacities of 0.61 J/g8C and 0.97 J/g*C, respectively. How much thermal ... November 21, 2011 by 19 Clocking Tectonic Plates Explain how you could calculate the average speed of the Pacific Plate over the past five million years, then calculate the value? I tried searching the internet for this, I came up with nothing, and I just dont get this question. January 13, 2011 by Anonymous Pages: <<Prev | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | Next>> Search Members

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# A Property of Number 4 Number $4$ has an interesting while easily verifiable property: The freedom of choosing $k$ can be used to show that number $0$ also has an interesting and easily verifiable property: For every number $k$ - integer, rational, real, complex, \begin{align} 0=k^{2}&-(k+1)^{2}-(k+2)^{2}+(k+3)^{2}\\ &-(k+4)^{2}+(k+5)^{2}+(k+6)^{2}-(k+7)^{2}. \end{align} This is used to answer a problem considered by Paul Erdös and Janos Surányi in 1960: Every integer $n$ can be represented in infinitely many ways as (1) $n=\pm 1^{2}\pm 2^{2}\pm 3^{2}\pm \cdots \pm k^{2}.$ For a proof, observe that if number $n$ admits a representation (1), then so does number $n+4$ whose representation can be obtained from that of $n$ by appending $(k+1)^{2}-(k+2)^{2}-(k+3)^{2}+(k+4)^{2}.$ From the above property of $0$ it then follows that both $n$ and $n+4$ have an infinite number of such representations. Clearly, it is sufficient to consider only positive integers. Representations of the negative ones are obtained by reversing the signs of their absolute value. Thus to solve Erdös and Surányi's problem we only need to establish that some four consecutive numbers admit a representation as an algebraic sum of the squares several consecutive integers. That is simple: \begin{align} 1&=1^{2},\\ 2&=-1^{2}-2^{2}-3^{2}+4^{2},\\ 3&=-1^{2}+2^{2},\\ 4&=1^{2}-2^{2}-3^{2}+4^{2}. \end{align} ### References 1. S. Savchev, T. Andreescu, Mathematical Miniatures, MAA, 2003, p 47 Copyright © 1996-2017 Alexander Bogomolny 62636880 Search by google:

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0 # How do you solve 3x squared minus 2x plus 7 equals 0? Wiki User 2012-08-01 20:46:49 I suggest you use the quadratic formula, with a = 3, b = -2, c = 7. Wiki User 2012-08-01 20:46:49 Study guides 20 cards ➡️ See all cards 3.81 1759 Reviews

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Calling all Columbia Applicants for 2009! : The B-School Application - Page 3 Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack It is currently 21 Jan 2017, 19:02 GMAT Club Daily Prep Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History Events & Promotions Events & Promotions in June Open Detailed Calendar Calling all Columbia Applicants for 2009! Author Message TAGS: Hide Tags Senior Manager Joined: 30 May 2007 Posts: 496 Followers: 7 Kudos [?]: 70 [0], given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 09 May 2008, 13:59 IHateTheGMAT wrote: I just did some back of the envelope calculations I'd like to add to the discussion. J Term class size is roughly 190. Assuming the yield is 77% - and if that's overall yield then I think in reality J Term yield is closer to 100% because there really is no other competition in terms of top ranked programs starting in January, especially since J Term has a lot of new yorkers just looking for an accelerated program that allows them to get their ticket stamped and not have to move from NYC. But, anyway, conservatively estimate 77% that means they need to accept around 247 applicants. If the yield were 90% for J Term, they only need to accept 211. So, for them to have a 10% acceptance rate for J Term they would need between 2,110 and 2,470 applicants. This is less than the number of applicants for tiny, remote location programs like Tuck, Yale and Cornell. For a school in NYC and a program popular with NYC bankers that don't want to leave, I think its pretty reasonable they would hit these application numbers. If so..... Sept 07 acceptance rates are actually above 16% In response to this post, you're looking at it the wrong way. If CBS has a full-time acceptance rate of 15% of about 5000 applicants, of which 70% matriculate then thats 525 students in full-time. If 190 J-Termers matriculate, using 77%, and the acceptance rate is 29%, which I think is the real number, the number of of applicants is 850. Do the weighted average acceptance rate and you have 17% on the dot. That's where the J-Term acceptance rate brings the overall acceptance rate down. But, don't be fooled! J-Term, from what I hear, is tough tough tough to get in! e-GMAT Discount Codes Manhattan GMAT Discount Codes Magoosh Discount Codes Director Joined: 10 Jun 2006 Posts: 624 Followers: 4 Kudos [?]: 54 [0], given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 09 May 2008, 14:44 djhouse81 wrote: IHateTheGMAT wrote: I just did some back of the envelope calculations I'd like to add to the discussion. J Term class size is roughly 190. Assuming the yield is 77% - and if that's overall yield then I think in reality J Term yield is closer to 100% because there really is no other competition in terms of top ranked programs starting in January, especially since J Term has a lot of new yorkers just looking for an accelerated program that allows them to get their ticket stamped and not have to move from NYC. But, anyway, conservatively estimate 77% that means they need to accept around 247 applicants. If the yield were 90% for J Term, they only need to accept 211. So, for them to have a 10% acceptance rate for J Term they would need between 2,110 and 2,470 applicants. This is less than the number of applicants for tiny, remote location programs like Tuck, Yale and Cornell. For a school in NYC and a program popular with NYC bankers that don't want to leave, I think its pretty reasonable they would hit these application numbers. If so..... Sept 07 acceptance rates are actually above 16% In response to this post, you're looking at it the wrong way. If CBS has a full-time acceptance rate of 15% of about 5000 applicants, of which 70% matriculate then thats 525 students in full-time. If 190 J-Termers matriculate, using 77%, and the acceptance rate is 29%, which I think is the real number, the number of of applicants is 850. Do the weighted average acceptance rate and you have 17% on the dot. That's where the J-Term acceptance rate brings the overall acceptance rate down. But, don't be fooled! J-Term, from what I hear, is tough tough tough to get in! Have you been told that the J Term acceptance rate is 29%? And as far as ED not helping Columbia manage its yield much, I think it actually makes a big impact on yield. Like you said, it gets a really bad rap for ED so if it didn't help them why wouldn't they just get rid of it? Also, Clear Admit claims that Columbia takes 40% of it's class ED. That's a pretty big number and it's clearly done in an effort to manage yield and acceptance rates. Just FYI, I'm actually planning to apply to Columbia ED and it is my top choice. So I'm not trying to denigrate the school or it's selectivity. Also, congrats on your admit and thanks for all the useful insights you give to us CBS 09 hopefuls! Senior Manager Joined: 30 May 2007 Posts: 496 Followers: 7 Kudos [?]: 70 [0], given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 09 May 2008, 18:13 IHateTheGMAT wrote: djhouse81 wrote: IHateTheGMAT wrote: I just did some back of the envelope calculations I'd like to add to the discussion. J Term class size is roughly 190. Assuming the yield is 77% - and if that's overall yield then I think in reality J Term yield is closer to 100% because there really is no other competition in terms of top ranked programs starting in January, especially since J Term has a lot of new yorkers just looking for an accelerated program that allows them to get their ticket stamped and not have to move from NYC. But, anyway, conservatively estimate 77% that means they need to accept around 247 applicants. If the yield were 90% for J Term, they only need to accept 211. So, for them to have a 10% acceptance rate for J Term they would need between 2,110 and 2,470 applicants. This is less than the number of applicants for tiny, remote location programs like Tuck, Yale and Cornell. For a school in NYC and a program popular with NYC bankers that don't want to leave, I think its pretty reasonable they would hit these application numbers. If so..... Sept 07 acceptance rates are actually above 16% In response to this post, you're looking at it the wrong way. If CBS has a full-time acceptance rate of 15% of about 5000 applicants, of which 70% matriculate then thats 525 students in full-time. If 190 J-Termers matriculate, using 77%, and the acceptance rate is 29%, which I think is the real number, the number of of applicants is 850. Do the weighted average acceptance rate and you have 17% on the dot. That's where the J-Term acceptance rate brings the overall acceptance rate down. But, don't be fooled! J-Term, from what I hear, is tough tough tough to get in! Have you been told that the J Term acceptance rate is 29%? And as far as ED not helping Columbia manage its yield much, I think it actually makes a big impact on yield. Like you said, it gets a really bad rap for ED so if it didn't help them why wouldn't they just get rid of it? Also, Clear Admit claims that Columbia takes 40% of it's class ED. That's a pretty big number and it's clearly done in an effort to manage yield and acceptance rates. Just FYI, I'm actually planning to apply to Columbia ED and it is my top choice. So I'm not trying to denigrate the school or it's selectivity. Also, congrats on your admit and thanks for all the useful insights you give to us CBS 09 hopefuls! Yes, I've been told that the J-Term acceptance rate is approximately 29%. I agree that ED impacts yield to a certain extent because they are not going to employ a method that does not work, but I don't think that extent is the sandbag levels that people think. I have not looked at Clear Admit, but what are the other yields from other schools' first round acceptances? IHTG, I never thought you were criticizing the school, I love discussing Columbia since it is the next major step in my life! If you have not done so, and you are able to, please try to visit the school. It will make a difference in the quality of your application. And, of course, when you plan on visiting, PM me and we will figure out a plan. Keep the lively discussions a comin'! Senior Manager Joined: 30 May 2007 Posts: 496 Followers: 7 Kudos [?]: 70 [0], given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 10 May 2008, 08:02 IHTG, It just came to me that you already visited the school recently, so congrats on that! How'd it go?!?! I have to keep better track of who is applying to Columbia next year! Rhyme was masterful at that this year. Director Joined: 10 Jun 2006 Posts: 624 Followers: 4 Kudos [?]: 54 [0], given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 10 May 2008, 14:01 PMd you Current Student Joined: 03 Jan 2008 Posts: 65 Followers: 1 Kudos [?]: 2 [0], given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 12 May 2008, 09:31 hmmm... I don't think the acceptance rate matters all that much. If Columbia is your first choice, the fact you are able to know the result much sooner is good. SVP Joined: 11 Mar 2008 Posts: 1634 Location: Southern California Schools: Chicago (dinged), Tuck (November), Columbia (RD) Followers: 9 Kudos [?]: 201 [0], given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 14 May 2008, 19:04 For those of you who are applying ED this Fall - when are you planning to visit campus? I am not convinced that I want to apply ED yet, and I will definitely need to visit campus in order to make the judgement call. I am booked solid for May and June so far, which leaves me the months of July and August. Does CBS allow visitors in July and August? Can you get a good feel for the school in July/August? If I apply ED, I will most likely want to submit right around mid-August so I can possibly receive a decision before submitting a few R1 apps. _________________ Check out the new Career Forum http://gmatclub.com/forum/133 Senior Manager Joined: 30 May 2007 Posts: 496 Followers: 7 Kudos [?]: 70 [0], given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 15 May 2008, 05:38 terp06 wrote: For those of you who are applying ED this Fall - when are you planning to visit campus? Does CBS allow visitors in July and August? Can you get a good feel for the school in July/August? CBS does not have anything going on July or August, so you will not get a good feel for the school. I think applying in August is for those who have already visited the school the prior year or plan to visit after they submit--but already know that they want to go to CBS. I suggest that you visit the school early September, and if you've already been working on the essays you can submit the app right after you visit. Submitting your app in September is still early enough in the game to get the advantage. And, if you do not like the school enough to apply ED, just postpone your app until RD. For everyone who knows that they want to apply ED, you can submit your app in August and then visit the school in September. I submitted my app in mid-September, and I wrote in my Optional Essay that I would be visiting at the end of September. Seemed to work out well. Senior Manager Joined: 10 Jun 2007 Posts: 346 Location: Newport, RI Followers: 2 Kudos [?]: 39 [1] , given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 15 May 2008, 05:51 1 KUDOS If you don't apply ED, make sure you apply before/on the scholarship deadline date. If you apply after that initial date, you will be waiting a LONG time to get a decision and will read venom on the businessweek forums of how "all the spots are already gone" even if you applied a day after. Manager Joined: 28 May 2006 Posts: 152 Location: New York, NY Followers: 2 Kudos [?]: 8 [0], given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 16 May 2008, 09:12 Hello, I am planning to apply for ED as well. A month ago I visited Columbia through their student host program. It was really good. I recommend it to everyone who's applying for Columbia. djhouse, congratulations on your admit! Lucky you! The school has such a beautiful campus and friendly students. Senior Manager Joined: 30 May 2007 Posts: 496 Followers: 7 Kudos [?]: 70 [0], given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 18 May 2008, 18:46 sng wrote: Hello, I am planning to apply for ED as well. A month ago I visited Columbia through their student host program. It was really good. I recommend it to everyone who's applying for Columbia. djhouse, congratulations on your admit! Lucky you! The school has such a beautiful campus and friendly students. Thank you! Let me know if I can be of help! Manager Joined: 28 May 2006 Posts: 152 Location: New York, NY Followers: 2 Kudos [?]: 8 [0], given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 21 May 2008, 08:21 Thanks djhouse for offering your help. Thanks to everyone who's sharing valuable knowledge from their 2008 application process. *** I recently received a PM message about essay service from "Roy2008." Did anyone else get it? Current Student Joined: 17 Jan 2008 Posts: 586 Location: Ann Arbor, MI Schools: Ross '12 (MBA/MS) Followers: 2 Kudos [?]: 126 [0], given: 34 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 22 May 2008, 03:02 sng wrote: Thanks djhouse for offering your help. Thanks to everyone who's sharing valuable knowledge from their 2008 application process. *** I recently received a PM message about essay service from "Roy2008." Did anyone else get it? I got an email from him too. As far as I'm concerned he's just an advertisement spammer. He has one post on these boards and it was moved and edited by a moderator. _________________ Manager Joined: 31 Jul 2006 Posts: 237 Followers: 3 Kudos [?]: 71 [0], given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 23 May 2008, 00:37 Hi DJHouse: I'm a huge fan of CBS. The school had always been my first choice. Anyway, m planning to apply for Fall 2009 ED. Can you help evaluate my qualifications? It would really help. Thanks Senior Manager Joined: 30 May 2007 Posts: 496 Followers: 7 Kudos [?]: 70 [0], given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 23 May 2008, 05:01 Hi DJHouse: I'm a huge fan of CBS. The school had always been my first choice. Anyway, m planning to apply for Fall 2009 ED. Can you help evaluate my qualifications? It would really help. Thanks Sure, I will evaluate your profile, PM me. But, know that what I say means nothing if you do or do not have the quality application to go with it. Current Student Joined: 03 Jan 2008 Posts: 65 Followers: 1 Kudos [?]: 2 [0], given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 23 May 2008, 05:06 accepted.com posted their analysis on the J-term essay topics analysis. In the practice and theory topic, it says 'Ideally, you would want to use an example where you used management principles to guide your team, office company, or club in solving a problem or completing a project.' I found it to be an interesting idea. This essay can be used as leadership and teamwork essay as well. eh... except I have no idea what the management principles are. I think I'm going to stick with my original idea something I knew. But, this could be something you guys can use if you have experience with. SVP Joined: 04 Dec 2007 Posts: 1689 Schools: Kellogg '11 Followers: 14 Kudos [?]: 198 [0], given: 31 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 27 May 2008, 08:08 I've been learning more about the HC and Pharma mgmt program Columbia has and am pretty impressed. One thing I'm wondering: when you apply to a specific program such as HC/Pharma mgmt - is your app considered only in the context of that program or the entire business school? Do they look for fit at both? cheers, ac. Senior Manager Joined: 30 May 2007 Posts: 496 Followers: 7 Kudos [?]: 70 [0], given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 27 May 2008, 11:48 ac8706 wrote: I've been learning more about the HC and Pharma mgmt program Columbia has and am pretty impressed. One thing I'm wondering: when you apply to a specific program such as HC/Pharma mgmt - is your app considered only in the context of that program or the entire business school? Do they look for fit at both? cheers, ac. Businessdoc is going that route, if I remember correctly, so he would be better suited to answer this quesiton. Nevertheless, I'll take a crack at it! Your fit will definitely be with the CBS program because you will be still be working with classmates from the entire business school. However, you will need to substantiate fit with the HC/Pharma program. The "fit" that I am refering to is your reasoning for why you want to be part of that program, such as past WE or EC's in HC or Pharma industries. While other students are not part of that program will express their fit with entire school as well as other areas of the school, such as Value Investing, you will need to express your fit with the entire school and the HC/Pharma program. Intern Joined: 18 Dec 2007 Posts: 8 Followers: 0 Kudos [?]: 5 [0], given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 27 May 2008, 13:33 ac8706, I am a Class of 2010 admit and will be pursuing a healthcare track at Columbia. While I think its most important to show enthusiasm for Columbia in general, the ways in which your application expresses your individual fit will be most important. I'm sure most Columbia applicants cite the New York City location and Columbia's reputation, for example, as reasons they are interested in the school, but, by appealing to your specific track (healthcare), you can really convince the adcom that you will contribute to Columbia and that Columbia will help you reach your future goals in the healthcare/pharma industry. Also, I don't know this for a fact, but I do think that the directors of each specific program have a close link to admissions. I met with the director of the healthcare program when I was on the waitlist to discuss my goals, and I'm convinced it made all the difference; I was admitted shortly thereafter. He was candid with me, and told me that the supply of healthcare administration jobs exceeds the number of students interested in the provider side (which I am). Similarly, he told me that they were seeking to admit students from healthcare orgs that were not currently recruiting at Columbia, and he mentioned a specific company based on the West Coast. Obviously, some of these factors will be outside of your control, but I do think its in your best interest to do all the research you can on the healthcare track and be sure to be very specific about how that program will contribute to your future success. I'm happy to help in any way I can. I plan on becoming very involved with the Healthcare Industry Association when school starts, so I can keep you abreast of events and opportunities for prospective students. Good luck! Senior Manager Joined: 30 May 2007 Posts: 496 Followers: 7 Kudos [?]: 70 [0], given: 0 Re: Calling all Columbia Applicants for 2009! [#permalink] Show Tags 27 May 2008, 17:01 Of the three active GMAT Clubbers who are going to CBS in the Fall, two are geared towards the Healthcare program. Pretty cool, considering the finance rep! Re: Calling all Columbia Applicants for 2009!   [#permalink] 27 May 2008, 17:01 Go to page   Previous    1   2   3   4   5   6   7   8   9   10   11  ...  59    Next  [ 1168 posts ] Similar topics Replies Last post Similar Topics: Calling all Thunderbird 2009 Applicants 2 23 Oct 2008, 06:19 424 Columbia 2012 - Calling All Applicants 3404 02 May 2011, 21:22 Calling all Columbia EMBA Applicants 0 01 Nov 2009, 17:18 2 Calling All 2009 USC-Marshall Applicants 51 03 Jun 2008, 16:27 21 Calling all MIT Fall 2009 applicants! 530 19 May 2008, 20:10 Display posts from previous: Sort by

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# Unambiguously describe how to make a peanut butter and jelly sandwich. ## Cathy Assumptions: that the reader knows what "bread," "peanut butter," "jelly," and "knife" represent. (I base these assumptions upon two facts, that this would be entirely too long if I had to define these, and that you said you would bring these things to class tomorrow, implying that the person who would be following these directions knows what these things are.) Ingredients • Two slices of bread (if not pre-sliced, approximately square length and height, approximately one inch in width or thickness) • One dull knife • One jar of peanut butter • One jar of jelly • One clean surface (plate, cutting board, etc.) 1. Take one slice of the bread and place in on the flat surface. 2. Take the knife, and scrape some peanut butter from the jar onto the sharper end. 3. Spread the peanut butter onto the largest plane/face (the side with the largest surface area) of the bread. 4. If the peanut butter does not cover the surface, repeat steps 2 and 3 until the surface is covered. 5. Take the second slice of bread and place it on the flat surface. 6. Repeat steps 2-4 on the second piece of bread replacing "peanut butter" with "jelly." 7. Place the jellied side of the second piece of bread onto the peanut buttered side of the first piece of bread, lining up the edges. 8. And, the final, and most important step, take the sharp end of the knife and press it from the upper left corner of the now-joined pieces of bread to the lower right corner in a diagonal line, thus cutting the sandwich into two wedges. ## Ivy (assuming you already have all the essential components: two slices of bread of the same relative shape and size, a container of peanut butter, a container of jelly, two utensils (i prefer a spoon and a butter knife but two spoons or two butter knives would work fine; even two forks if you're in a pinch), and a hard, flat surface (counter, desk, table, etc)) 1) take the two slices of bread (remove them from any packaging or wrapping they may be in) and lay them down flat and side by side on the aforementioned surface. 2) take the container of peanut butter. open it (usually by unscrewing the lid counter-clockwise). place both the lid and the now open container back down, side by side, on the surface. 3) pick up a utensil (i prefer a butter knife) in your dominant hand. pick up the open container of peanut butter in your other hand. insert the utensil into the jar and scoop out a portion of peanut butter large enough to cover one side of one of your pieces of bread when spread flat across it. 4) place the utensil down on the surface. place the peanut butter-covered slice of bread down on the surface with the peanut butter side facing up! 5) now take the container of jelly. open it (usually by unscrewing the lid counter-clockwise). place both the lid and the now open container back down on the surface, side by side. 6) pick up a utensil (i prefer a spoon) in your dominant hand. pick up the open container of jelly in your other hand. insert the utensil into the jar and scoop out a portion of jelly large enough to cover one side of one piece of bread when spread flat across it. 7) put the container of jelly down on the surface. pick up one piece of bread in your non-dominant hand (your dominant one should still be holding the utensil with the jelly on it) and position it so it is lying flat across your palm. slowly bring the utensil with the jelly on it over to the piece of bread in your non-dominant hand and apply the jelly as evenly as possible by spreading it in firm but gentle strokes across the side of the bread that is facing upward in your hand. continue until that side of the bread is evenly covered in jelly. 8) place the utensil down on the surface. leave the now jelly-covered piece of bread in your hand exactly the way it is (jelly-side up). now pick up the peanut butter-covered piece of bread with your dominant hand, taking care not to get peanut butter all over yourself. position it so it is lying flat across the palm of your dominant hand, peanut butter-covered side up. 9) you should now have a jelly-covered piece of bread in your non-dominant hand (jelly side up) and a peanut butter-covered piece of bread in your dominant hand (peanut butter side up). slowly bring your dominant hand closer to your non-dominant hand. holding onto the peanut-buttered piece of bread, rotate the wrist of your dominant arm until the two slices of bread are facing each other, jelly-side to peanut butter-side. 10) place the slice of bread in your dominant hand (peanut butter-covered) on top of the slice in your non-dominant hand (jelly-covered), peanut butter-side to jelly-side, making sure to align the two pieces of bread so that they match up perfectly shape-wise (usually corner to corner but if the bread slices do not have corners, just place them together so the shapes correspond, with no corners or edges sticking out). 11) congratulations. you should now be holding a ready-to-eat peanut butter jelly sandwich in your non-dominant hand. ## Jae 1. Obtain the following supplies • A. peanut butter (smooth or cruchy) • B. jelly (your favorite kind) • D. Knife • E. Table Spoon 2. Open the bread and remove two slices, laying them side by side. 3. Spread peanut butter on one side of one slice of bread • A. Open peanut butter jar • B. Remove some peanut butter with the knife • C. Spread the peanut butter onto the bread until you have as much as you want, removing more from the jar as needed. 4. Spread jelly on one side of the other slice of bread. • A. Open jelly jar. • B. Remove a table spoonful of jelly. • C. Spread the jelly on the bread until you have as much as you want, removing more from the jar as needed. 5. Place the two slices together, peanut butter and jelly facing inwards, press down slightly, and enjoy. ## Jeana Assuming you have all the necessary materials to make a pb&j sandwich: 1. Begin by taking out two slices of bread. 2. Using a knife or spoon, remove about a tablespoon's worth (or however much you so desire)of jelly from the jar and spread it out evenly on one side of one slice of bread. Put it aside, jelly side up. 3. Repeat this procedure with the peanut butter on one side of the other slice of bread. 4. Place the peanut-buttered slice of bread face down directly on top of the jellied slice of bread, matching up the slices of bread as if one were creating a nice, neat stack. Gently press down to make the the two slices of bread stick to each other. ## Jeff Procure the following ingredients in the prior week to preparation of the sandwich: Smooth Peanut Butter Jelly A loaf of bread (less than 1 inch think but greater than .5 inch thick) Two knives Place bread parallel to table top. Remove twist tie from bread by cutting the twist tie with scissors. Open the bag the bread is in from the open end. Place hand inside of bread bag. Remove slice of bread using hand without crushing the bread. Remove slice of bread using without crushing the bread. Place slices of bread parallel to table top side by side. Take jar of jelly in hand. Use opposite hand to open jar by following instruction on jar. Hold jelly in left hand in a fashion that the jelly does not exit the jar except by way of knife which is held in the right hand. Take a unit of jelly on knife from jelly jar and place unit of jelly in the middle of one of the pieces of bread. Evenly distribute the jelly across bread using the knife. Fashion the lid of jelly back on to jar in a manner that no jelly escapes from the jar. Discard used knife into cleaning bin. Take jar of peanut butter in hand. Use opposite hand to open jar by following instruction on jar. Hold peanut butter in left hand in a fashion that the peanut butter does not exit the jar except by way of knife which is held in the right hand. Take a unit of peanut butter on knife from peanut butter jar and place unit of peanut butter in the middle of one of the pieces of bread. Evenly distribute the peanut butter across bread using the knife. Fashion the lid of peanut butter back on to jar in a manner that no peanut butter escapes from the jar. Discard second used knife into cleaning bin. Pick up both slices of bread however, do not pick up the slices of bread touching the peanut butter or jelly therefore use the crust of the bread as the lifting mechanism. Place the jelly and peanut butter sides of the bread gently together without spilling any peanut butter or jelly on the ground. ## Joel To make a peanut butter and jelly sandwich, first you need to acquire the following ingredients: peanut butter, jelly, bread (preferrably sliced), and some sort of utensil used for spreading (preferrably a butter knife). Next, you need to take two pieces of the bread and spread a layer of peanut butter on one piece and a layer of jelly on the other piece using your spreading utensil. Note that the amount of peanut butter and jelly you use depends on your particular taste. Finally, you put the two pieces of bread together - the peanut butter matching up with the jelly, and there you have it, a peanut butter and jelly sandwich - ready for eating! ## Kevin 1. Gather the following materials: two slices of bread, peanut butter in a jar, jelly in a jar, two butter knives. 2. Lay the two slices of bread beside one another in front of you. This procedure is accomplished by removing any two bread slices from the container in which they were previously held. Hold one slice in your left hand, one in your right hand. Lay the slices of bread side-by-side on a table or other flat surface. 3. Remove the lid from the peanut butter jar. This prodedure is accomplished by placing the fingers about the lid at equal spaces around the lid. Turn the lid while applying inward pressure with the fingers in a direction counter-clockwise until the jar lid is loosened to its maximum potential. Remove the lid at this juncture by lifting upward. Place the lid in a place convenient to its retrieval. 4. Withdraw some peanut butter from its jar. This procedure is accomplished by holding either butter knife in the hand such that the blade, or flat end, is exposed. Place the exposed (flat) end into the open end of the jar, and with a slow horizontal movement, gather a desirable amount of peanut butter on the knife's flat surface. Remove knife from jar near the center of the circle created by the open area such that no peanut butter is lost. 5. Using the knife, apply the peanut butter to either slice of bread by placing the surface with which the peanut butter was withdrawn to the exposed surface of the bread slice. Do not permit the peanut butter to touch the "crust," as this will make the sandwich more difficult to handle at a later time. 6. Spread the peanut butter along the exposed bread surface upon which it was previously applied, carefully noting the previous warning about not permitting the peanut butter to touch the "crust." 7. Place the previously used knife aside such that it can be washed at a later time. 8. Open the jelly jar using a procedure similar to the one for the peanut butter. More force may be required to open the jelly jar, though. 9. Retrieve the other knife for use in the jelly. 10. Use the new butter knife to gather jelly from the jar, using a procedure similar to the one described in Step 4. 11. Place the jelly on the slice of bread not previously used, again being careful to avoid the "crust" area to maintain clean fingers, if possible. 12. Spread the jelly on the second slice on the exposed bread surface of the second slice, again being careful to avoid the "crust" area. 13. Set the knife used for the jelly aside so that it may be retrieved at a later time. 14. Pick up the slice of bread on which the peanut butter has been spread. This is done by picking the bread up in a manner keeping the peanut-butter surface away from your skin at all times. 15. Being sure that the peanut butter surface on the held slice faces only the jelly surface on the stationary slice, place the peanut butter surface on the held slice against the jelly surface on the stationary slice and release. 16. Line up the corners of each slice of bread such that they are flush with one another. 17. Eat as desired. ## Liz Supplies: • Jar of peanut butter • Jar of jelly • butter knife 1. Open wrapping (if bread is still in packaging) around loaf of bread and remove two slices. 2. Place slices on their side (not on edge) on a flat surface. 3. Open jar of peanut butter. 4. Hold the handle of the butter knife, and use the flat end to remove a small amount (or as much as desired) of peanut butter from the jar. 5. Still holding the handle of the knife, spread the peanut butter from the end of the knife evenly over one side of one piece of bread. 6. Repeat steps 3-5 using the jelly and the second slice of bread. 7. When application of peanut butter and jelly is completed, set down the knife. 8. Take the two slices of bread and put the peanut butter side of the one and the jelly side of the other together. The sides of bread with peanut butter or jelly spread on them should be combined in the middle of the sandwich, with the plain sides of the slices facing outwards. 9. If desired, use the cutting edge of the butter knife to cut the sandwich in half, quarters etc. 10. Replace the lids on the peanut butter and jelly jars and wash the knife. ## Sam Well, first thing is to get the appropriate materials. Go to a grocery store and purchase a loaf of sliced bread (WonderBread is recommended but use your preference). Buy one jar of peanut butter (creamy or chunky, Jif or Peter Pan) and one jar of jelly (Smuckers, grape or strawberry is recommended). pay for groceries and leave the store. Once safely home, get your materials set up. Along with the groceries, you will need a butter knife for spreading, and a plate. Take two pieces of bread from the bag and lay them on the plate. Open peanut butter. Using the knife, spread the peanut butter on one slice of bread until the bread has a coating layer on it. Replace top on jar. Wipe off knife. Open the jelly. Spread jelly on the other slice of bread, once again giving it a generous coating. Replace top on jar. Take the slice of bread with jelly and lay it on top of the peanut buttered piece, aligning the sides and corners. Take the knife and cut the sandwich, either widthwise or diagonally. Put peanut butter and jelly back in refrigerator and bread away. Find a comfortable place to sit and enjoy your first PB&J! Disclaimer Often, these pages were created "on the fly" with little, if any, proofreading. Any or all of the information on the pages may be incorrect. Please contact me if you notice errors.

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CC-MAIN-2018-47

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http://mathhelpforum.com/geometry/153006-polygon-inscribed-within-ellipse.html

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crawl-data/CC-MAIN-2016-50/segments/1480698540409.8/warc/CC-MAIN-20161202170900-00244-ip-10-31-129-80.ec2.internal.warc.gz

# Thread: Polygon inscribed within an ellipse 1. ## Polygon inscribed within an ellipse To inscribe an equalateral polygon within a circle is easy and it is possible to know (given the size of the circle and the number of sides of the polygon) the exact length of one of the sides of the polygon but what about inscribing an equalateral polygon within an ellipse? To get the length of one of the sides in a circle the formula R * sqrt (1-cos (360/n) *2) can be used ( among others ) where n= number of sides of polygon But this does not work for an ellipse Now remember the polygon must have equal length of sides which must mean that the interior angles at the vertices must be different. I have already worked out the length of side for a twelve sided polygon inscribed within an ellipse of 360 major axis and 240 minor axis as 78.24868802 but have no method other than long winded trial and error to arrive at the result. Can anyone come up with a solution. 2. One thing I can think of is that you must have all the angles add up to 2 [LaTeX ERROR: Convert failed] . Then for each angle that is going to be squished, you need one to be increased. 3. This might help to conceptualize the problem: Elliptic Chords - Wolfram Demonstrations Project 4. I was thinking about the problem a bit and here is what I came up with. We simplify the problem a bit by considering the relative sizes of the major and minor axes. Let the long one be 1 and the smaller one be $b<1$. You can start from the point $(1,0)$ and then go anti-clockwise given the length of the line and the coordinates where it intersects the ellipse. The algorithm is recursive. You are given the first intersection point $(x_0, y_0)$ and the length of that, which is $L^2 = (1-x_0)^2+y_0^2$. Then the next intersection's coordinates is $(x_1, y_1)$ and it must satisfy the following two equations. The first is the distance from $(x_0,y_0)$ $\displaystyle(x_1-x_0)^2+(y_1-y_0)^2 = L^2 = (1-x_0)^2+y_0^2$ The second one is that the point must be on the ellipse $\displaystyle x_1^2+\left(\frac{y_1}{b}\right)^2 = 1 $ In general these two equations become $\displaystyle(x_{n+1}-x_n)^2+(y_{n+1}-y_n)^2 = L^2 = (1-x_n)^2+y_n^2$ $\displaystyle x_{n+1}^2+\left(\frac{y_{n+1}}{b}\right)^2 = 1$ Unfortunately, when I tried solving these equations, the solutions were very ugly, so I think this method will be best suitable for numerical calculation. 5. Originally Posted by Vlasev I was thinking about the problem a bit and here is what I came up with. We simplify the problem a bit by considering the relative sizes of the major and minor axes. Let the long one be 1 and the smaller one be $b<1$. You can start from the point $(1,0)$ and then go anti-clockwise given the length of the line and the coordinates where it intersects the ellipse. The algorithm is recursive. You are given the first intersection point $(x_0, y_0)$ and the length of that, which is $L^2 = (1-x_0)^2+y_0^2$. Then the next intersection's coordinates is $(x_1, y_1)$ and it must satisfy the following two equations. The first is the distance from $(x_0,y_0)$ $\displaystyle(x_1-x_0)^2+(y_1-y_0)^2 = L^2 = (1-x_0)^2+y_0^2$ The second one is that the point must be on the ellipse $\displaystyle x_1^2+\left(\frac{y_1}{b}\right)^2 = 1 $ In general these two equations become $\displaystyle(x_{n+1}-x_n)^2+(y_{n+1}-y_n)^2 = L^2 = (1-x_n)^2+y_n^2$ $\displaystyle x_{n+1}^2+\left(\frac{y_{n+1}}{b}\right)^2 = 1$ Unfortunately, when I tried solving these equations, the solutions were very ugly, so I think this method will be best suitable for numerical calculation. Thank you Vlasev for your reply. I couldnt trouble you further and ask you to show numerically the working out of my question that is of an ellipse 360 major axis by 240 minor and inscribed within a twelve sided equalateral polygon. I am having a little trouble following the formula you set out. My fault not yours LOL. Thank you 6. Oh, no worries. The formula I have written is worked out like this: I assumed that the first point on the n-gon is the point $(1, 0)$. Then I've assumed the length of each side of the n-gon is L and that it intersects the ellipse at points $(1,0)$ and $(x_0,y_0)$. With the first equation $L^2 = (1-x_0)^2+y_0^2$ , I have used Pythagoras' theorem to show this relationship. With the next equation, I have used Pythagoras theorem to find where the next side should intersect the ellipse. However, since I have only 1 equation with 2 variables. To solve this problem, I need one more equation. This is not hard since we have the equation of the ellipse and the point $(x_1, y_1)$ must satisfy this equation also. This becomes a system of two equations which you can solve either by hand or using software. Now for the general case, lets assume we have calculated the position of all points up to point n with coordinates $(x_n,y_n)$. From now on, we need to use the Pythagorean theorem to find where the next vertex of the polygon is. That's what the first equation shows, where I have replaced $L^2$ with the very first equation. Then the second equation makes sure that the point is on the ellipse. 7. Originally Posted by Vlasev Oh, no worries. The formula I have written is worked out like this: I assumed that the first point on the n-gon is the point $(1, 0)$. Then I've assumed the length of each side of the n-gon is L and that it intersects the ellipse at points $(1,0)$ and $(x_0,y_0)$. With the first equation $L^2 = (1-x_0)^2+y_0^2$ , I have used Pythagoras' theorem to show this relationship. With the next equation, I have used Pythagoras theorem to find where the next side should intersect the ellipse. However, since I have only 1 equation with 2 variables. To solve this problem, I need one more equation. This is not hard since we have the equation of the ellipse and the point $(x_1, y_1)$ must satisfy this equation also. This becomes a system of two equations which you can solve either by hand or using software. Now for the general case, lets assume we have calculated the position of all points up to point n with coordinates $(x_n,y_n)$. From now on, we need to use the Pythagorean theorem to find where the next vertex of the polygon is. That's what the first equation shows, where I have replaced $L^2$ with the very first equation. Then the second equation makes sure that the point is on the ellipse. Wow that was a quick reply thank you so much. Using your equation am I right in my initial working out that a twelve sided eq polygon inscribed within an ellipse of the size I originally stated will have sides of 78.248 in length. Sorry to be such a pest 8. Oh, I don't know. I haven't actually performed calculations with these equations. I can try it out. Here is a drawing to complement the above discussion: Now, I don't know where you 12-gon "starts" but I will start it at (1,0). Since it is 12 sided, there must be 3 sides in each quadrant. Since there is symmetry there must be 2 vertices in each quadrant (for a total of 8) and then 1 vertex on each axis. So two of the points should be (1,0) and (0,b). We need to find the other two points. However, you have lengths that are different from mine. First we need to normalize them. a = 360 --> 1 b = 240 --> 240/360 = 2/3 c = 78.248 --> 78.248/360 = 0.21376 I will plug them in to check what will happen. 9. Hello, steven, I plugged the parameters in the equations I got and unfortunately, your number 78.248 does not work. How did you get that number. Did you draw a 360x240 pixel ellipse and then try to get it by hand? Or did you make a 24-gon? Either way, I have done a calculation with a different parameter (L is around 156.497) and got the following picture Wow, I just noticed something spooky: 156.497 = 78.248 * 2. So you either gave me 1/2 of the side that you calculated, or a 24-gon has 1/2 the side of a 12-gon. 10. Originally Posted by Vlasev Hello, steven, I plugged the parameters in the equations I got and unfortunately, your number 78.248 does not work. How did you get that number. Did you draw a 360x240 pixel ellipse and then try to get it by hand? Or did you make a 24-gon? Either way, I have done a calculation with a different parameter (L is around 156.497) and got the following picture Wow, I just noticed something spooky: 156.497 = 78.248 * 2. So you either gave me 1/2 of the side that you calculated, or a 24-gon has 1/2 the side of a 12-gon. I think I understand how. I said the major axis was 360 and minor axis was 240 by that I meant measuring the ellipse from one end to the other I should have emphasised that at the start. If I were to draw the ellipse as you have in your drawing I would have drawn it with the x axis 180 and the y axis as 120. 11. Vlasev you have been extremely helpful in all your replies and time I wish to thank you and really appreciate all your efforts 12. Oh, I see, alright then. Then yes, your L works just fine! 13. I'm not sure how the OP solved his problem, if he really has done so. So far we still have three unknown variables and just two equations. The system is still underdetermined. The unknown variables are L, x, and y. The equations that Vlasev pointed to are pythagoras' theorem and the cartesian ellipse equation. If I'm not mistaken we still need one more in order to come up with a general formula for finding L. 14. rainer, i have set L in the start, so in essence I'm searching for a good value of L that will give a polygon. 15. Originally Posted by Vlasev rainer, i have set L in the start, so in essence I'm searching for a good value of L that will give a polygon. Oh okay. So you did it iteratively. That's cool. Page 1 of 2 12 Last

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## Richard A. Paselk Science 331 Fall 2004 Lecture/Activity Office: SA560a Notes: 18 October Phone: x 5719 Home: 822-1116 e-mail: rap1 # Matter What is matter? Stuff. Has mass and occupies space. Mass: The measure of quantity for matter. Mass is the property of matter resulting in its inertia and and attraction via gravity. • Do not confuse mass and weight. Weight is the force acting on an object due to gravity. We often interchange these terms in conversation, but they are quite different - you have the same mass whether you are weightless in space on here on Earth (taking a shuttle flight is no substitute for a diet!). To confuse us further we call the determination of mass "weighing"! Matter has both physical properties and chemical properties. These are properties which do not depend on the quantity of substance and therefore they can be used to identify a substance (sometimes referred to as intensive properties). • Physical properties of substances can be observed without, in principle, changing their compositions. Physical properties include mass, color, density etc. Note that physical changes such as melting, cutting, etc. do not change composition, that is new substances are not created, old substances are not lost. At the microscopic level, the relationships of atoms to one another are unchanged. States of Matter. Matter can exist in three states under earth-surface conditions: • Solid: definite shape and volume (Crystals vs. super-cooled liquids or glasses) • Liquid: definite volume, but no defined shape - will fit to container etc. • Gas: no definite shape or volume - will fill whatever container they are in. • both liquids and gases are fluids. A fourth state of matter commonly occurs under special conditions: a plasma. A plasma is an ionized fluid - can be contained by magnetic fields. • Chemical properties of substances describe behaviors which lead to changes in composition. Chemical properties describe reactivity under various circumstances (does it burn in air, react with acids or bases, corrode in sea water etc.) Note that chemical changes result in different compositions - substances are transformed to new things. Inparticular, the relationships between atoms are changed. # States of Matter and Changes of State Matter can exist in three states (phases) under earth-surface conditions: • Solid: definite shape and volume (Crystals vs. super-cooled liquids or glasses) • Liquid: definite volume, but no defined shape - will fit to container etc. • Gas: no definite shape or volume - will fill whatever container they are in. Liquid Nitrogen Demo-balloons, Pb bell, rubber tubing, scatter on floor, pour over hand etc. ## Exercise-Changes of State in Water A common form of matter exhibiting these states under lab conditions is water. Let's look at water starting in its solid state and observe its transitions. We'll use the following equipment: • Heat source (bunsen burner) • Beaker • Thermometer • Test tube • Scale • Ice So let's all melt some ice and observe what happens: • Set up your bunsen burner, ring stand and wire screen as demonstrated by instructor. • Light your burner and adjust the flame using the needle valve and air adjustments on the burner . Make sure the gas cock is fully open. • Do not adjust the burner further during the experiment. You may turn the burner off and on at teh cock, as long as you always turn it fully on or off. This way we have a continuous, constant heat supply. • Time all of the heating steps below. 1. Fill beaker with ice, measure temperature and weigh. 2. Melt partially, measure temperature. record times and temperatures 3. Continue measuring temperature as ice is heated and melts. record times and temperatures 4. When ice completely melted, weigh again. 5. Note temperature change as liquid water is heated. record times and temperatures 6. Note temperature as water boils. 7. Hold a test tube containing ice above boiling water, NOT in steam, and observe what happens. record times and temperatures 8. Stop boiling, weigh again. What can you say about what has happened? e.g.: • Do you think mass was conserved? Why? • Was the water changed? Explain. • Is ice a true crystalline solid? Why or why not? • Make a plot of energy or heat added (time) vs. temperature (a heating curve similar to figure 2.5 on p 67 of your text).

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https://m.scirp.org/papers/73734

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crawl-data/CC-MAIN-2020-50/segments/1606141652107.52/warc/CC-MAIN-20201201043603-20201201073603-00215.warc.gz

Novel Analytic Model for the Projected Contact Zone Based on the Flow Line Element Method in Alloyed Bar Rolling Abstract: For obtaining the profile and the projected area of contact zone exactly during alloyed bar rolling by Round-Oval-Round pass sequence, the analytic model for the length contact and contact boundary curve were built firstly by considering the influence of the spread of the outgoing workpiece on the effective height of outgoing workpiece and roll mean radius, and then the contact surface was discretized by finite flow line elements. Moreover, the radius equation and bite angle equation of different flow lines were derived and they were all expressed as the function of the position angle, then any flow line on the contact surface can be determined since the position angle has been given. Finally, since the analytic equation for the projected area of contact surface was hard to be integrated directly, the analytic model was proposed by summing up the area of discretized parts on the contact surface. Based on the analytical model of contact boundary and flow line element, 3-dimension contact surface was rebuilt by mathematical software, and the validity of analytic model was examined by the bar rolling experiments and the numerical simulation of alloyed bar rolling by rigid-plastic FEM software. Compared with the existing models, the precision of the projecting area of contact zone was improved obviously. So, it can be applied in alloyed bar rolling to predict the projected area of contact zone and velocity of outgoing workpiece exactly. 1. Introduction Compared with products rolled with the oval-square-oval pass sequence, the bar or rod rolled with round-oval-round pass sequence has a better surface quality and mechanical performance, and the round-oval-round pass sequence is the most common roll pass in bar or rod continuous rolling recently. Since the characteristics of non-uniform distribution of stain, stress and flow velocity on the deformation zone, it is difficult to analyze the process of alloyed bar rolling in oval-square-oval pass sequence accurately [1] [2] [3] . In the past few years, the research on the alloyed bar rolling process was carried out by the simulation methods and experiments. References [4] [5] [6] [7] [8] studied the strain, stress and geometry of deformation zone of bar rolling by the numerical simulation methods based on the FEM software, and references [9] [10] [11] analyzed the effect of roll gap, roll profile and rolling speed on the wear and exit section area by the single-pass rolling and multi-pass rolling experiments and FE analysis. References [12] [13] [14] studied the rolling force and rolling torque of bar rolling and proposed analytic models for estimating the force energy parameters. Because the contact surface between the rolling workpiece and shaped roll is not a cylindrical surface like flat roll rolling but a three- dimension curve with complex boundary, the length and boundary of contact surface has to be simplified without taking the spread of rolling workpiece into account for obtaining projected area easily. Based on these simplifications, the geometry and projected area of contact zone were calculated by the conventional equation for billet rolling and the roll force and roll torque was derived on these results, and the results of roll force and roll torque were just an approximate value for bar rolling in round-oval-round pass sequence. Furthermore, the profile of contact boundary and the projected area of contact zone influence the results of velocity field distribution, stress field distribution and strain field distribution directly. Therefore, it is indispensable for analyzing the bar or rod continuous rolling process to determine the profile of contact boundary and the projected area of contact zone exactly. It is not easy to obtain the precise projecting area of contact zone and the accuracy profile of contact boundary since the contour of contact zone is not regular and difficult to be defined. So, at present the geometry and projected area of contact zone is mostly calculated by the simplified methods, such as the empirical equation [2] , the graphical solving method [3] . Shinokura and Takai [15] [16] [17] gave the spread formula and projected area of the deformed workpiece in round-oval-round pass sequence. However, this formula was derived on the base of rolling experiments of plain carbon steel. Y. Lee proposed a mathematical model for predicting the surface profile of deformed workpiece and mean roll radius in Round-Oval-Round pass sequence [18] [19] [20] . The profile of contact boundary and the projecting area of contact zone are correlated to the solution of the rolling force, the velocity field distribution, stress field distribution and strain field distribution directly. So, it is very important to determine the profile of contact boundary and the projected area of contact zone exactly. 2. The Existing Models for Determining the Projecting Area of Contact Zone 2.1. Graphical Solving Method [3] As can be seen in Figure 1, the boundary curve of contact zone was simplified as a intersection line between the revolution surface of roll pass and the profile of incoming workpiece, and the intersection line was obtained by linking the intersect in point one by one. So the projection of contact zone was shown as the hatching zone. 2.2. Analytic Methods Based on the Two Hypothetic Curve Functions for the Contact Boundary Shinokura and Takai [15] proposed a formula to estimate the maximum contact length L and projected area Ap of contact zone in alloyed bar rolling by ignoring the spread of outgoing workpiece. The size of oval groove, round groove and corresponding incoming workpiece were shown in Figure 2(a) and Figure 2(b) respectively. The equivalent height of incoming workpiece $\stackrel{¯}{{H}_{0}}$ and equivalent height of outgoing workpiece $\stackrel{¯}{{H}_{m0}}$ were simplified as $\stackrel{¯}{{H}_{0}}=\frac{{A}_{0}-{A}_{s0}}{2{C}_{Z0}}$ (1) $\stackrel{¯}{{H}_{m0}}=\frac{{A}_{0}-{A}_{s0}-{A}_{h}}{2{C}_{Z0}}=\frac{{A}_{e0}}{2{C}_{Z0}}$ (2) where ${A}_{0}$ is the section area of incoming workpiece, ${A}_{s0}$ , ${A}_{h}$ , ${A}_{e0}$ are the non-effective reduction area, effective reduction area and effective exit section Figure 1. The projected area by a graphical solving method. Figure 2. Definition of pass profile and the incoming workpiece. (a) Round-oval pass; (b) Oval-round pass. area respectively, $\left({C}_{y}{}_{0},{C}_{z0}\right)$ is the intersection point between the profile of incoming workpiece and the roll pass. In round-oval pass rolling, ${A}_{h}$ and ${A}_{s0}$ may be obtained by ${A}_{h}/2=\mathrm{arctan}\left({C}_{Y0}/{C}_{Z0}\right)\cdot {R}_{a}{}^{2}-\left[{\int }_{-{C}_{Y0}}^{{C}_{Y0}}\left(\sqrt{{R}_{1}{}^{2}-{y}^{2}}-{D}_{z}\right)\text{d}y-{C}_{Y0}\cdot {C}_{Z0}\right]$ (3) ${A}_{s0}/2=\mathrm{arctan}\left({C}_{Z0}/{C}_{Y0}\right)\cdot {R}_{a}-{C}_{Y0}\cdot {C}_{Z0}$ . (4) In oval-round pass rolling, ${A}_{h}$ and ${A}_{s0}$ may be given by ${A}_{h}/2=\frac{\text{π}}{8}{W}_{\mathrm{max}}\cdot {H}_{p}-\frac{{A}_{s0}}{2}-\mathrm{arccos}\left({C}_{Z0}/{R}_{Y0}\right)\cdot {R}_{g}^{2}-{C}_{Y0}\cdot {C}_{Z0}$ (5) ${A}_{s0}/2=\mathrm{arcsin}\left({C}_{Z0}/{R}_{1}\right)\cdot {R}_{1}^{2}-\left({D}_{Z}+{C}_{Y0}\right)\cdot {C}_{Z0}$ (6) where ${R}_{1}$ is the radius of the oval groove, ${R}_{s}$ is the radius of the curvature of the incoming cross-section, ${D}_{Z}$ is the distance along the Z-axis direction between the origin coordinate Then the maximum contact length ${L}_{\mathrm{max}}$ was obtained by ${L}_{\mathrm{max}}=\sqrt{{R}_{m0}\left(\stackrel{¯}{{H}_{0}}-\stackrel{¯}{{H}_{m0}}\right)}=\sqrt{{R}_{m0}\frac{{A}_{h}}{2{C}_{Z0}}}$ (7) where ${R}_{m0}$ is the mean roll radius at the entrance-section, it can be shown as ${R}_{m0}={R}_{\mathrm{min}}+\frac{{H}_{p}}{2}-\frac{\stackrel{¯}{{H}_{m0}}}{2}$ . (8) For obtaining the projected area of contact zone in alloy bar rolling, Shinokura and Takai expressed the contact boundary curve by the function ${L}_{\mathrm{max}}\sqrt{1-\frac{{x}^{2}}{{C}_{y}^{2}}}$ according to the empirical data, and the projected area was shown as ${A}_{p}={\int }_{0}^{{C}_{y}}{L}_{\mathrm{max}}\sqrt{1-\frac{{x}^{2}}{{C}_{y}^{2}}}\text{d}x=\frac{\text{π}}{2}{L}_{\mathrm{max}}{C}_{y}$ . (9) As shown in Figure 2, ${C}_{y}$ is the coordinate of critical point on the contact boundary along the spread direction. $\left({C}_{y},{C}_{z}\right)$ is the coordinates of the critical point on the contact boundary at the exit-section, and it can be determined by the reference [16] [17] . On the base of Equation (9) Y.lee given another contact boundary curve function ${L}_{\mathrm{max}}{\left(\sqrt{1-\frac{x}{{C}_{y}}}\right)}^{1/m}$ to modify this equation and the projected area was shown as ${A}_{p}={\int }_{0}^{{C}_{y}}{L}_{\mathrm{max}}{\left(1-\frac{x}{{C}_{y}}\right)}^{1/m}\text{d}x=\frac{3}{2}{L}_{\mathrm{max}}{C}_{y}\left(m=\frac{1}{3}\right)$ . (10) Moreover, as shown in Equation (9) and Equation (10), two hypothetical functions for indicating the contact boundary curve were given directly without any reasoning and any derivation process. Although the error between the results of two equations are not obvious when the size of rolling workpiece is small enough and then the contact length L and Cy is small enough, the absolute error of these two equations will be considerable and it should not be ignored in large diameter bar rolling. So it is not precise enough for these two semi-analytic models to calculate the projected area of multi-pass alloyed bar rolling, then an analytic model should be built to predict the projected area accurately. In summing up these models for projected area, the Equation (2), which is based on the graphical solving method, does not take the influence of the spread and the contact boundary status of deformed workpiece into account. Therefore, the Graphical solving method can be just used as an estimating value when the spread of deformed workpiece is small enough (Figure 2). 3. A Novel Analytic Model for the Projected Area of Contact Surface 3.1. Modification of Contact Length Model As shown in Figure 3, if the spread of outgoing workpiece was not negelected, the effective section area ${A}_{e}$ and the equivalent width $2{C}_{y}$ , height $\stackrel{¯}{{H}_{m}}$ and mean roll radius ${R}_{m}$ are totally different from the corresponding parameters ${A}_{e0}$ , $2{C}_{y0}$ , $\stackrel{¯}{{H}_{m}{}_{0}}$ and ${R}_{m0}$ in Equation (2). $\left({C}_{y},{C}_{z}\right)$ can be calculated by references [19] [20] . So, the contact length modified as ${L}_{m}=\sqrt{{R}_{m}\left(\stackrel{¯}{{H}_{0}}-\stackrel{¯}{{H}_{m}}\right)}=\sqrt{\left({R}_{\mathrm{min}}+{H}_{p}/2-G/2-\stackrel{¯}{{H}_{m}}/2\right)\left(\stackrel{¯}{{H}_{0}}-\stackrel{¯}{{H}_{m}}\right)}$ . (11) In round-oval pass rolling, the equivalent height of outgoing workpiece was shown as $\begin{array}{c}\frac{\stackrel{¯}{{H}_{m}}}{2}=\frac{{A}_{e}}{4{C}_{y}}=\frac{{\int }_{-{C}_{y}}^{{C}_{y}}\left(\sqrt{{R}_{1}^{2}-{y}^{2}}-{D}_{z}\right)\text{d}y}{{C}_{y}}\\ =\frac{{R}_{1}^{2}\left(2\frac{{C}_{y}}{{R}_{1}}\mathrm{cos}\left(\mathrm{arcsin}\frac{{C}_{y}}{{R}_{1}}\right)+2\mathrm{arcsin}\frac{{C}_{y}}{{R}_{1}}\right)}{4{C}_{y}}-{D}_{z}.\end{array}$ (12) In oval-round pass rolling, the equivalent height of outgoing workpiece was Figure 3. Parameters in modified model of contact length. (a) Round-oval pass; (b) Oval-round pass. expressed as $\frac{\stackrel{¯}{{H}_{m}}}{2}=\frac{{A}_{e}}{4{C}_{y}}=\frac{\mathrm{arctan}\left({C}_{y}/{C}_{z}\right)R{}_{g}{}^{2}+{C}_{y}{C}_{z}}{{C}_{y}}$ . (13) 3.2. Deformation Zone Geometry and the Boundary of Contact Zone As can be seen in Figure 4 and Figure 5, coordinate axes x, y, z are chosen to be the directions of bar length, width and height, respectively, with the origin of the x-axis at the midpoint of the entry plane. The x-y and x-z are plane of symmetry. The curve equation of contact boundary S can be obtained approximately by interpolating between the point $\left(\text{0,0,}{H}_{0}\right)$ and the point $\left(L,{C}_{y},{C}_{z}\right)$ . The coordinates of any point on the contact boundary S are shown as $\left({x}_{s},{C}_{ys},{C}_{zs}\right)$ . 3.3. The Definition of the Flow Line and Flow Line Element on the Contact Surface The flow plane was defined as a set of eccentric continuous cylindrical surfaces having almost straight generators parallel to the roll axis. Assuming that at any cross section along the roll bite the bar height deformation is uniform, the deformation zone was constituted by a set of flow plane, and the contact surface was constituted by a set of flow line element ${f}_{\alpha }$ . The flow line element ${f}_{\alpha }$ is a set of concentric circular arc with different radius $R$ and different bite angle ${\theta }_{\alpha s}$ , which center is attached on the roll axis. Since the roll radius in the roll pass is different and the height of incoming workpiece is different along the y-axis direction, the bite angle ${\theta }_{\alpha s}$ , along the whole contact boundary, is not a constant but a variable which changes with a different position angle $\alpha$ or a different roll radius $R$ . Figure 4. Geometry and flow line element on the contact surface in oval pass. (a) Geometry of contact surface; (b) Flow line on the contact surface. Figure 5. Geometry and flow line element on the contact surface in round pass. (a) Geometry of contact surface; (b) Flow line on the contact surface. 3.3.1. The Radius of Flow Line R and the Bite Angle θαs in the Round-Oval Pass Sequence The radius of flow line ${f}_{\alpha }$ was expressed as $R={R}_{\mathrm{min}}+{R}_{1}\left(1-\mathrm{cos}\alpha \right)\left(0\le \alpha \le \mathrm{arcsin}\frac{{C}_{y}}{{R}_{1}}\text{or}\text{\hspace{0.17em}}0\le \alpha \le \mathrm{arccos}\frac{{C}_{z}+{D}_{z}}{{R}_{1}}\right)$ . (14) The three-dimension coordinates of a random point on the contact surface was shown as $\left\{\begin{array}{l}x=L-R\mathrm{sin}\theta \left(0\le \theta \le {\theta }_{\alpha s}\right)\\ y={R}_{1}\mathrm{sin}\alpha \\ z=R\left(1-\mathrm{cos}\theta \right)+{R}_{1}\mathrm{cos}\alpha -{D}_{z}=-R\mathrm{cos}\theta +{R}_{c}\end{array}$ . (15) On the symmetry plane of deformed workpiece, the position angle $\alpha$ is 0 and the bite angle ${\theta }_{\alpha s}$ reaches the maximum value ${\theta }_{\mathrm{max}}$ ${\theta }_{\mathrm{max}}=\mathrm{arccos}\left(1-\frac{{H}_{0}-{H}_{p}}{2{R}_{0}}\right)$ . (16) At the exit section of deformed workpiece, the bite angle ${\theta }_{\alpha s}$ is 0 and the position angle $\alpha$ reaches the maximum value ${\alpha }_{\mathrm{max}}$ ${\alpha }_{\mathrm{max}}=\mathrm{arccos}\frac{{C}_{z}+{D}_{z}}{{R}_{1}}$ (17) According to the Equation (15), the coordinates of points on the contact boundary was expressed as $\left\{\begin{array}{l}{x}_{s}=L-R\mathrm{sin}{\theta }_{\alpha s}\\ {y}_{s}={R}_{1}\mathrm{sin}\alpha \\ {z}_{s}=R\left(1-\mathrm{cos}{\theta }_{\alpha s}\right)+{R}_{1}\mathrm{cos}\alpha -{D}_{z}=-R\mathrm{cos}{\theta }_{\alpha s}+{R}_{c}\end{array}$ . (18) In the contact zone, the height of the profile at $y=0$ along the x-direction is expressed as $H\left(x\right)={H}_{0}-\frac{2L}{{R}_{0}}x+\frac{1}{{R}_{0}}{x}^{2}$ . (19) The curve equation of contact boundary S can be obtained approximately by interpolating between the point $\left(0,\text{}0,\text{}{H}_{0}\right)$ and the point $\left(L,\text{}{C}_{y},\text{}{C}_{z}\right)$ . The coordinate of any point on the contact boundary S is shown as $\left(x,\text{}{C}_{ys},\text{}{C}_{zs}\right)$ , and the coordinate ${C}_{zs}$ can be shown as $2{C}_{z}{}_{s}={H}_{0}-2\left({H}_{0}-2{C}_{z}\right)\frac{{x}_{s}}{L}+\left({H}_{0}-2{C}_{z}\right)\frac{{x}_{s}^{2}}{{L}^{2}}$ . (20) Substituting the xs of Equation (18) into Equation (20) yields $2{C}_{zs}=\left({H}_{0}-2{C}_{z}\right)\frac{{R}^{2}{\mathrm{sin}}^{2}{\theta }_{\alpha s}}{{L}^{2}}+2{C}_{z}$ . (21) According to the equation ${C}_{zs}={z}_{s}$ and ${\mathrm{sin}}^{2}{\theta }_{\alpha s}=1-{\mathrm{cos}}^{2}{\theta }_{\alpha s}$ yields $\left({H}_{0}-2{C}_{z}\right)\frac{-{R}^{2}{\mathrm{cos}}^{2}{\theta }_{\alpha s}}{{L}^{2}}+2R\mathrm{cos}{\theta }_{\alpha s}+\left({H}_{0}-2{C}_{z}\right)\frac{{R}^{2}}{{L}^{2}}-2\left({R}_{c}-{C}_{z}\right)=0$ . (22) Solving the Equation (22) yields ${\theta }_{\alpha b}=\mathrm{arccos}\frac{{L}^{2}+L\sqrt{{L}^{2}+\left({H}_{0}-2{C}_{z}\right)\left[\left({H}_{0}-2{C}_{z}\right)\frac{{R}^{2}}{{L}^{2}}-2\left({R}_{c}-{C}_{z}\right)\right]}}{R\left({H}_{0}-2{C}_{z}\right)}$ . (23) Substituting Equation (14) into Equation (23) yields ${\theta }_{\alpha s}=\mathrm{arccos}\frac{{L}^{2}+L\sqrt{{L}^{2}+\left({H}_{0}-2{C}_{z}\right)\left[\left({H}_{0}-2{C}_{z}\right)\frac{{\left({R}_{\mathrm{min}}+{R}_{p}\left(1-\mathrm{cos}\alpha \right)\right)}^{2}}{{L}^{2}}-2\left({R}_{c}-{C}_{z}\right)\right]}}{\left({R}_{\mathrm{min}}+{R}_{1}\left(1-\mathrm{cos}\alpha \right)\right)\left({H}_{0}-2{C}_{z}\right)}$ . (24) 3.3.2. The Radius of Flow Line R and the Bite Angle θαs in the Oval-Round Pass Sequence For the oval-round pass sequence, the radius of flow line ${f}_{\alpha }$ was expressed as $R={R}_{\mathrm{min}}+{R}_{g}\left(1-\mathrm{cos}\alpha \right)\left(0\le \alpha \le \mathrm{arcsin}\frac{{C}_{y}}{{R}_{g}}\text{or}0\le \alpha \le \mathrm{arccos}\frac{{C}_{z}}{{R}_{g}}\right)$ . (25) The three-dimensional coordinates of a random point on the contact surface was shown as $\left\{\begin{array}{l}x=L-R\mathrm{sin}\theta \left(0\le \theta \le {\theta }_{\alpha s}\right)\\ y={R}_{g}\mathrm{sin}\alpha \\ z=R\left(1-\mathrm{cos}\theta \right)+{R}_{g}\mathrm{cos}\alpha =-R\mathrm{cos}\theta +{R}_{c}\end{array}$ . (26) On the symmetry plane of deformed workpiece, the position angle $\alpha$ is 0 and the bite angle ${\theta }_{\alpha s}$ reaches the maximum value ${\theta }_{\mathrm{max}}$ ${\theta }_{\mathrm{max}}=\mathrm{arccos}\left(1-\frac{{H}_{0}-{H}_{p}}{2{R}_{0}}\right)$ . (27) At the exit section of deformed workpiece, the bite angle ${\theta }_{\alpha s}$ is 0 and the position angle $\alpha$ reaches the maximum value ${\alpha }_{\mathrm{max}}$ ${\alpha }_{\mathrm{max}}=\mathrm{arccos}\frac{{C}_{z}}{{R}_{g}}$ . (28) By the same methods as round-oval pass sequence, the bite angle of oval- round pass sequence was ${\theta }_{\alpha s}=\mathrm{arccos}\frac{{L}^{2}+L\sqrt{{L}^{2}+\left({H}_{0}-2{C}_{z}\right)\left[\left({H}_{0}-2{C}_{z}\right)\frac{{\left({R}_{\mathrm{min}}+{R}_{g}\left(1-\mathrm{cos}\alpha \right)\right)}^{2}}{{L}^{2}}-2\left({R}_{c}-{C}_{z}\right)\right]}}{\left({R}_{\mathrm{min}}+{R}_{g}\left(1-\mathrm{cos}\alpha \right)\right)\left({H}_{0}-2{C}_{z}\right)}$ . (29) 3.3.3. The Projected Area of Contact Zone According to definition of projected area of contact zone, it should be calculated by the integral equation as ${A}_{p}=2{\int }_{0}^{{\alpha }_{\mathrm{max}}}R\mathrm{sin}{\theta }_{\alpha s}\text{d}y=2{\int }_{0}^{{\alpha }_{\mathrm{max}}}R\mathrm{sin}{\theta }_{\alpha s}{R}_{1}\mathrm{cos}\alpha \text{d}\alpha$ . (30) However, it is too complex for Equation (30) to integrate it directly and get a function for projected area. So, the flow line field method based on the discrete law is an effective way to calculate projected area. Once the position angle of flow line $\alpha$ is given, the roll radius $R$ and the bite angle ${\theta }_{\alpha s}$ against different position angle of flow line can be determined. Then the space position and the length of any flow line on the contact surface can be obtained. If the whole contact zone was discretized into n flow lines with different position angle ${\alpha }_{i}$ and different bite angle ${\theta }_{\alpha si}$ , the arc length of any flow line was expressed as ${L}_{fi}={R}_{i}{\theta }_{\alpha si}=f\left({\alpha }_{i}\right)\text{}\left(0\le i\le n\right)$ . (31) The position angle of ith flow line can be shown as ${\alpha }_{i}=\frac{i}{n}{\alpha }_{\mathrm{max}}\text{}\left(0\le i\le n\right)$ . (32) The projected length of this flow line on plane xoy can be obtained by ${L}_{pi}={R}_{i}\mathrm{sin}{\theta }_{\alpha si}\text{}\left(0\le i\le n\right)$ . (33) Substituting Equation (32) into Equation (14), Equations ((24), (25) and (29)) to replace the position angle $\alpha$ can yield ${R}_{i}$ and ${\theta }_{\alpha si}$ in oval pass rolling and round pass rolling respectively, then the projected area of contact zone on the plane xoy can be shown as ${A}_{p}=2{\int }_{0}^{{\alpha }_{\mathrm{max}}}R\mathrm{sin}{\theta }_{\alpha s}\text{d}y=\underset{i=0}{\overset{n}{\sum }}\left({R}_{i}\mathrm{sin}{\theta }_{\alpha si}\right)\frac{{C}_{y}}{n}$ . (34) 4. Results and Discussions The alloyed bar rolling experiments had been accomplished in BEIMAN SPICIAL STEEL CO. LTD, the round workpiece of diameter 171 mm were rolled in one oval pass and one round pass of 22-stand Pomini Rolling Mill. The deformation zone of rolling workpices was obtained by stopping the rolling process when the workpiece was rolled in the oval pass and round pass simultaneously. The rolling schedule is shown as Table 1. The material is structural alloyed steel 41Cr4. The dimension of the groove schedule is respectively shown as Figure 6(a) and Figure 6(b). As can be seen in Table 2, the parameters for calculating the projecting area of Table 1. Rolling schedule of the Pomini Rolling Mills. Table 2. The parameters for calculating the projecting area. Figure 6. The dimension of the groove schedule. (a) Oval pass (b) Round pass. Figure 7. Profile of the curve on the contact boundary. (a) Oval-pass; (b) Round-pass. contact zone were listed one by one. As shown in Figure 7(a) and Figure 7(b), the profile of curve on the contact boundary in oval pass and round pass were measured and distinguished by a white chalk. As shown in Figure 8, the contact boundary and the contact surface were obtained by the rigid-plastic FEM software DEFORM5.03. Moreover, the surface profile of incoming workpiece and outgoing workpiece in oval pass and round pass was shown as Figure 9. Figure 8. Contact boundary and contact zone in oval pass rolling. Figure 9. Surface profile of incoming workpiece and outgoing workpiece. As can be seen in Figure 10(a) and Figure 10(b), the 3-dimension contact surface and contact boundary were rebuilt by Matlab 7.0 according to the novel analytic model of contact boundary and contact surface. In Figure 11, the contact boundary from the novel analytic model was compared with that of experimental data and simulation result. Results of contact length and projected area from the novel model, the calculating results from the existing models, the experimental data and the simulation results were all listed in Table 3. As shown in Table 3, the results of contact length and projected area from different models were listed and compared with corresponding experimental data and simulation results. The results of contact length from the modified analytic model are less than that of empirical formula and the graphical solution, and it is greater than that of Shinokura formula and Y. Lee formula. Moreover, the error of contact length from the modified model is less than that of existing models. Since the outgoing workpiece of oval pass rolling will be rolled in next round pass as an incoming workpiece and the section profile at the exit of oval pass influence the contact surface of round pass rolling greatly, the prediction error of round pass is obviously greater than the prediction error of oval pass. Moreover, Figure 10. Rebuilt contact surface based on analytic model by Matlab7.0. (a) Oval-pass; (b) Round-pass. Figure 11. Profile of the curve on the contact boundary. (a) Oval-pass; (b) Round-pass. Table 3. Results of different models. results of projected area from the novel analytic model approaches the experimental data and simulation results very well, and its error is less than any existing models. 5. Conclusions 1) The contact boundary is a complex 3-dimension curve, and its profile is not only concerned with the parameters of pass profile R1, Rg, Rmin, G, Dz and the shape parameters of incoming workpiece H0, Ra, but also influenced by the coordinates of critical point $\left({C}_{Y},{C}_{Z}\right)$ ; 2) The modified contact length model is rational because the influence the effective section area of the outgoing workpiece Ae, the critical point $\left({C}_{Y},{C}_{Z}\right)$ on the contact boundary, the effective height of outgoing workpiece $\stackrel{¯}{{H}_{m}}$ and the mean roll radius ${R}_{m}$ , has been taken into account in this model; 3) Based on the different position angle $\alpha$ and bite angle ${\theta }_{\alpha s}$ , the flow line element discretizes the complicated 3-dimension contact surface conveniently and makes it easier to rebuild the contact surface, and it is a good way to analyze the non-uniform stress and strain distribution accurately; 4) The discretizing and summing up method is an efficient way to solve the projected area, and results from this method approach the experimental data and simulating results very well. Cite this paper: Dong, Y. , Zhu, H. and Song, J. (2017) Novel Analytic Model for the Projected Contact Zone Based on the Flow Line Element Method in Alloyed Bar Rolling. Open Access Library Journal, 4, 1-17. doi: 10.4236/oalib.1103247. References [1]   Xu, C. (2009) The Pass Design of Structural Sections. Chemistry Industry Press, Beijing, 79. (In Chinese) [2]   Hu, B. (2010) The Pass Design of Structural Sections. Metallurgical Industry Press, Beijing, 31-39. (In Chinese) [3]   Yuan, Z.-X. and Wang, S.-P. (2008) The Plastic Deformation and Rolling Fundamentals. Metallurgical Industry Press, Beijing, 211-213. (In Chinese) [4]   Inoue, T., Torizuka, S. and Nagai, K. (2005) Novel Rod Rolling Process for Designing Efficiently Ultrafine-Grained Steel Bars Based on Numerical Analysis. Journal of the Japan Institute of Metals, 69, 934-942. https://doi.org/10.2320/jinstmet.69.934 [5]   Milenin, A.A., Dyja, H. and Mroz, S. (2004) Simulation of Metal Forming during Multi-Pass Rolling of Shape Bars. Journal of Materials Processing Technology, 153, 108-114. [6]   Aksenov, S.A., Chumachenko, E.N., Logashina, I.V., et al. (2015) Numerical Simulation in Roll Pass Design for Bar Rolling. Metalurgija, 54, 75-78. [7]   Graf, M. and Kawalla, R. (2010) Simulation System for Fast Analysis of Multistage Hot Rolling Processes Strip and Rod/Wire. Steel Research International, 81, 122-125. [8]   Nekrasov, I.I., Karamyshev, A.P., Parshin, V.S., et al. (2011) Studies of the Rolling of Ribbed Bars Using the Deform Software Package. Metallurgist, 55, 167-170. https://doi.org/10.1007/s11015-011-9408-4 [9]   Park, H.S. and Lee, Y. (2008) Experimental Study for Roll Gap Adjustment due to Roll Wear in Single-Stand Rolling and Multi-Stand Rolling Test. Journal of Mechanical Science and Technology, 22, 937-945. [10]   Byon, S.M., Na, D.H. and Lee, Y. (2009) Effect of Roll Gap Adjustment on Exit Cross Sec-tional Shape in Groove Rolling—Experimental and FE Analysis. Journal of Materials Processing Technology, 209, 4465-4470. https://doi.org/10.1016/j.jmatprotec.2008.10.045 [11]   Lee, Y. (2007) Effect of Rolling Speed on the Exit Cross Sectional Shape in Rod Rolling Process. International Journal of Precision Engineering and Manufacturing, 8, 27-31. [12]   Ragab, A.R. and Samy, S.N. (2006) Evaluation of Estimates of Roll Separating Force in Bar Rolling. Journal of Manufacturing Science and Engineering-Transactions of the ASME, 128, 34-45. https://doi.org/10.1115/1.2120779 [13]   Deng, W., Zhao, D.-W., Qin, X.-M., et al. (2010) Linear Integral Analysis of Bar Rough Rolling by Strain Rate Vector. Journal of Iron and Steel Research International, 17, 28-33. https://doi.org/10.1016/S1006-706X(10)60068-4 [14]   Lee, Y. and Kim, Y.H. (2001) Approximate Analysis of Roll Force in a Round-Oval-Round Pass Rolling Sequence. Journal of Material Processing Technology, 113, 124- 130. https://doi.org/10.1016/S0924-0136(01)00712-9 [15]   Shinokura, T. and Takai, K. (1986) Mathematical Model of Roll Force and Torque in Steel Bar Rolling. Tetsu-To-Hagane, 72, 58-64. (In Japanese) [16]   Lee, Y. and Choi, S. (2000) New Approach for the Prediction of Stress Free Surface Profile of a Workpiece in Rod Rolling. ISIJ International, 40, 624. https://doi.org/10.2355/isijinternational.40.624 [17]   Lee, Y., Choi, S. and Kim, Y.H. (2000) Mathematical Model and Experimental Validation of Surface Profile of a Workpiece in Round-Oval-Round Pass Sequence. Journal of Materials Processing Technology, 108, 87. https://doi.org/10.1016/S0924-0136(00)00734-2 [18]   Lee, Y. (2001) An Analytical Study of Mean Roll Radius in Rod Rolling. ISIJ International, 41, 1414. https://doi.org/10.2355/isijinternational.41.1414 [19]   Dong, Y.G., Zhang, W.Z. and Song, J.F. (2008) An Analytical Model for the Prediction of Cross-Section Profile and Mean Roll Radius in Alloy Bar Rolling. Journal of University of Science and Technology Beijing, 15, 344-351. https://doi.org/10.1016/S1005-8850(08)60065-1 [20]   Dong, Y.G., Zhang, W.Z. and Song, J.F. (2006) A New Analytical Model for the Calculation of Mean Roll Radius in Round-Oval-Round Alloy Bar Rolling. ISIJ International, 46, 1458-1466. https://doi.org/10.2355/isijinternational.46.1458 Top

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{[ promptMessage ]} Bookmark it {[ promptMessage ]} Pre-Calculus Practice Problem 56 # Pre-Calculus Practice Problem 56 - Unit 2 Activity 2... This preview shows page 1. Sign up to view the full content. Unit 2, Activity 2, Discovery using Technology with Answers Blackline Masters, Advanced Math-PreCalculus Page 54 Louisiana Comprehensive Curriculum, Revised 2008 1. Plot on the same screen the following graphs: , 2 x y = , 3 x y = y = x 4 , y = x 5 a) What points do all of the graphs have in common? All of the graphs pass through (0, 0) and (1, 1) b) Which function increases most rapidly, and which increases least rapidly, as x becomes large? y = x 5 increases most rapidly and y = x 2 increases least rapidly c) What are the main differences between the graphs of the even powers of x and the odd powers of x ? The graphs of the even powers of x are even functions; that is, they are symmetrical over the y-axis. The graphs of the odd powers of x are odd functions; that is, they are symmetrical around the origin. The end-behavior of the even powers is the same as x y x y → ∞ → ∞ → −∞ → ∞ , , and as . The same can be said for the end-behavior of the odd powers: as x y x y → ∞ → ∞ → −∞ → −∞ , , and as . This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]} ### What students are saying • As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students. Kiran Temple University Fox School of Business ‘17, Course Hero Intern • I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero. Dana University of Pennsylvania ‘17, Course Hero Intern • The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time. Jill Tulane University ‘16, Course Hero Intern

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Question # Use the substitution v=y' to write each second-order equation as a system of two first-order differential equations (planar system). 4y''+4y'+y=0 Differential equations Use the substitution $$v=y'$$ to write each second-order equation as a system of two first-order differential equations (planar system). $$4y''+4y'+y=0$$ 2021-05-13 Let our equation be: $$\displaystyle{4}{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}+{4}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{y}={0}$$ Our task is to write our second order differential equation as a system of two first order differential equations. Let's introduce a substitution $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={v}$$ and we get: $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={v}$$ $$\displaystyle{4}{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}+{4}{v}+{y}={0}$$ $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={v}$$ $$\displaystyle{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}=-{\frac{{{1}}}{{{4}}}}{y}-{v}$$ We got system of first order differential equation which solution is equivalent to solution our second order differential equation. $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={v}$$ $$\displaystyle{F}{\left({x},{v},{y}\right)}=-{\frac{{{1}}}{{{4}}}}{y}-{v}$$

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# The density of KBr is 2.73g/cm. The length of the unit cell is 654pm. Find the number of atoms per unit cell in the cubic structure. $(a)\;1\qquad(b)\;2\qquad(c)\;3\qquad(d)\;4$ Since $Z = \large\frac{\rho \times a^3 \times N_A}{at.wt}$ $Z = \large\frac{2.73\times(654\times10^{-10})^3\times6.023\times10^{23}}{119}$ $Z \approx 4$ Hence answer is (d) answered Feb 25, 2014

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# What are mean and variance of $W_i$, given that $Z_n=\frac{\sum{W_i}}{\sqrt{n}\sigma}\sim N(0,1)$? [closed] Let $$Z_n=\frac{\sum{W_i}}{\sqrt{n}\sigma}\sim N(0,1),$$ where $$W_i=X_i-\mu$$. What are the mean and variance of $$W_i$$? ## closed as off-topic by NCh, Saad, Lee David Chung Lin, Leucippus, Eevee TrainerMar 22 at 4:31 This question appears to be off-topic. The users who voted to close gave this specific reason: • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – NCh, Saad, Lee David Chung Lin, Leucippus, Eevee Trainer If this question can be reworded to fit the rules in the help center, please edit the question. • Do you know the distribution of $X_i$? – Minus One-Twelfth Mar 21 at 22:36 • It is a distribution $F_x$ with finite $E(X)=\mu$ and $Var(X)=\sigma^2>0$ – theQuestion Mar 21 at 22:40 $$\newcommand{\Var}{\operatorname{Var}}\newcommand{\E}{\mathbb{E}}$$The mean of $$W_i$$ is $$0$$ and variance of $$W_i$$ is same as that of $$X_i$$, namely $$\sigma^2$$. This follows from standard properties of mean and variance ($$\E(X-c) = \E(X)-c$$ and $$\Var(X-c)= \Var(X)$$, for any constant $$c$$).

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# Thread: Can the product and sum of a set of digits identify them uniquely? 1. ## Can the product and sum of a set of digits identify them uniquely? Can two (or more) different sets of five single digit numbers have the same product and sum of all their digits at the same time? ie. 1,4,8,4,3 (Product = 384, Sum = 20) 2,5,3,1,9 (Product = 270, Sum = 20) The sums are the same but the products aren’t so these two sets don’t. Furthermore, does the answer differ for other sized sets of single digits, can you find two different sets of any amount of digits that have the same product and sum? - - - - - - - - - - I just realised that it's possible with 3 or more digits and at least one as a 0. But if none of the numbers were 0? 2. 1,1,2,3,4 product = 24 sum = 11 1,2,2,2,3 product = 24 sum = 10 1,1,1,4,4 product = 16 sum = 11 1,1,2,2,4 product = 16 sum = 10 1,2,3,5,7 product = 210 sum = 18 1,1,5,6,7 product = 210 sum = 20 Anyone know if this is even possible yet? I tend to think it's not. 3. tough one! I tried algebra but it didn't change much the difficulty excel helped check many numbers fast, but didn't get me a solution, then I tried some common sense and started from the multipliers 3,3,2,2,2 which I broke into 6,6,2 and 9,2,2 and I had spare "1" digits to fill the gap in the addition the solution I found: {6,6,2,1,1} {9,2,2,2,1} (edit: of course the sets of size 3 {6,6,1} {9,2,2} also work) Yuwie.com | invite friends. hang out. get paid. 4. Originally Posted by Manu tough one! I tried algebra but it didn't change much the difficulty excel helped check many numbers fast, but didn't get me a solution, then I tried some common sense and started from the multipliers 3,3,2,2,2 which I broke into 6,6,2 and 9,2,2 and I had spare "1" digits to fill the gap in the addition the solution I found: {6,6,2,1,1} {9,2,2,2,1} (edit: of course the sets of size 3 {6,6,1} {9,2,2} also work) Yuwie.com | invite friends. hang out. get paid. Awesome!! I guess I gave up too early. I had gotten it down to where I would find a difference of at least 1 in the sum, as you can see in my previous post.. but I gave up too soon, before finding out how to eliminate that difference of 1.

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# Maths Algebra Multiple Choice Problems with 100% Correct Answers Question 1 Find the average rate of change of the function from x1 to x2.f(x) = 3x from x1 = 0 to x2 = 5A. -4B. 8C. 2D. 3Question 2Give the slope and y-intercept of each line whose equation is given. f(x) = -2x + 1A. m = 3; b = 4B. m = -2; b = 1C. m = 6; b = 7D. m = 2; b = 1Question 3Find the average rate of change of the function from x1 to x2.f(x) = ?x from x1 = 4 to x2 = 9A. 1/5B. 1C. 2D. 1/4Question 4Use the given conditions to write an equation for each line in point-slope form. Passing through (-8, -10) and parallel to the line whose equation is y = -4x + 3.A. y + 10 = -4(x + 8)B. y + 11 = 4(x2 + 8)C. y – 12 = -5(x + 20)D. y + 14 = -4(x – 5)Question 5Write an equation in general form of the line passing through (3, -5) whose slope is the negative reciprocal (the reciprocal with the opposite sign) of –1/4.A. 5x + y – 22 = 0B. 4x – y – 17 = 0C. 4x + y + 20 = 0D. 7x – y + 34 = 0Question 6Write an equation in slope-intercept form of a linear function f whose graph satisfies the given conditions.The graph of f passes through (-1, 5) and is perpendicular to the line whose equation is x = 6.A. f(x) = 6B. f(x) = 8C. f(x) = 3D. f(x) = 5Question 7Evaluate each function at the given values of the independent variable and simplify.f(x) = 4x + 51. f(6) 2. f(x + 1) 3. f(-x)A. 1. 27 2. 5x + 9 3. -4x + 8B. 1. 35 2. 4x + 9 3. -7x + 5C. 1. 29 2. 4x + 9 3. -4x + 5D. 1. 29 2. 3x + 8 3. 4x + 6Question 8Give the slope and y-intercept of each line whose equation is given.g(x) = -1/2xA. m = 3/4; b = -2B. m = 6; b = 7C. m = -1/2; b = 0D. m = 1; b = 8Question 9 Write an equation in slope-intercept form of a linear function f whose graph satisfies the given conditions. The graph of f is perpendicular to the line whose equation is 3x – 2y – 4 = 0 and has the same y-intercept as this line.A. f(x) = 2/3x – 4B. f(x) = -2/5x – 6C. f(x) = -2/3x – 2D. f(x) = -2/7x + 8Question 10Give the slope and y-intercept of each line whose equation is given.y = 2x + 1A. m = 3; b = 4B. m = 5; b = 1C. m = 6; b = 7D. m = 2; b = 1Question 11Determine whether the following equation defines y as a function of x:x2 + y2 = 16A. Y is a function of x.B. X is not a function of y.C. X is a function of x.D. Y is not a function of x.Question 12Evaluate each piecewise function at the given values of the independent variable.f(x) = 3x + 5 if x < 04x + 7 if x ? 01. f(-2)2. f(0)3. f(3)A. 1. -1 2. 7 3. 19B. 1. -5 2. 9 3. 19C. 1. -1 2. 7 3. 21D. 1. 6 2. 7 3. 19Question 13 5.0 PointsDetermine whether the following equation defines y as a function of x:x + y = 16A. Y is a function of x.B. Y is not a function of x.C. X is a function of y.D. X is not a function of y.Question 14Write an equation in slope-intercept form of a linear function f whose graph satisfies the given conditions.The graph of f passes through (-6, 4) and is perpendicular to the line that has an x-intercept of 2 and a y–intercept of -4.A. f(x) = -1/2x + 1B. f(x) = -1/5x + 3C. f(x) = 1/5x - 3D. f(x) = -2/5 x + 1Question 15Determine whether the following equation defines y as a function of x:x2 + y = 16A. Y is not a function of x.B. X is a function of y.C. X is not a function of y.D. Y is a function of x.Question 16Passing through (-2, 2) and parallel to the line whose equation is 2x - 3y - 7 = 0.A. 3x - 3y + 11 = 0B. 2x - 3y + 10 = 0C. 6x - 4y + 12 = 0D. 2x - 5y + 15 = 0Question 17 g(x) = x + 3 if x ? -3-(x + 3) if x < -31. g(0) 2. g(-6) 3. g(-3)A. 1. -1 2. 7 3. 19B. 1. 3 2. 3 3. 0C. 1. -7 2. 9 3. 2D. 1. 9 2. 7 3. 19Question 18g(x) = x2 + 2x + 31. g(-1)2. g(x + 5)3. g(-x)A. 1. 22. x2 + 12x + 383. x2 - 2x + 3B. 1. 42. x2 + 6x + 383. x2 - 3x +5C. 1. 72. x2 + 7x + 563. x2+ 4x + 7D. 1. 52. x2 -12x + 383. x2+ 5x + 7Question 19 of 205.0 PointsFind the average rate of change of the function from x1 to x2.f(x) = x2 + 2x from x1 = 3 to x2= 5A. 10B. 15C. 4D. 25Question 20 of 205.0 PointsDetermine whether the function is odd, even, neither, or can’t be determined:f(x) = x3+ xA. EvenB. OddC. NeitherD. Can’t be determined

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# Scaling Enemy Difficulty per wave 0 favourites • 6 posts From the Asset Store Two levels with two sistem Wave. with Source-code (.c3p) + HTML5 Exported. • Is there a good formula for scaling enemies to make them more difficult per wave? Like enemies start out with 40 health, bullets do 10 damange each round add X to health, and each bullet upgrade add X to damage?   We're trying to figure out a good formula so we can make 50 + rounds of enemy wave in our game. • Finetuning is reliant on playtesting. But I can give you some hints: Depending on how long you want the TD to go, you need to use multi-order polynomials. 50+ rounds will need at least 3 I think, but 4 could be better. If you want it to go even longer, expand to 5. So your basic formula could look like this: HP(for wave x)= ax^4+bx^3+cx^2+dx+e Now you need to calculate the gold that is available up to that level and you can calculate the difficulty: difficulty = HP / (gold available until this point) Make sure it gets harder with each wave, at first maybe a slight rise and later when you want the game to become really hard, you wanna have a bigger rise in difficulty per level. This ofc means you need to balance out the gold income to your HP formula. And your towers need to be balanced around the money (or other ressources) they cost. Things you need to factor in when balancing towers are things like range (not linearly ofc), fire rate, damage, how much worth their special abilities are (like splash, slow, ...). Hope that helps and happy playtesting. • thanks! hmmm never worked with polynomials before, ill have to read into that. Are there any tutorials on how to use them? • ## Try Construct 3 Develop games in your browser. Powerful, performant & highly capable. Construct 3 users don't see these ads • "Game Economics" is a field all in itself and is part of the character of your game. Remember there is no right or wrong way, it is part of the game and if done well becomes the carrot that leads the player through the game. The economics, help them believe if they play just one more round they will be able to get that bigger gun. You can do the health and gun strength linearly, add 20 to health each time, but increase number of enemies exponentially (using a polynomial as above). The choice is yours, provided it balances. Ramp up difficulty to quickly and the player is overwhelmed. To slowly it becomes to easy. If they have to do some gold mining to get through the next level, it becomes a bit boring. To complicated and people will ignore parts of it. A game with great graphics and sound, but bad economics is often still a bad game. After the initial planning there is no substitute for play testing. Have fun. • Try to fit a polynomial manually to your need with difficulty in mind. A spreadsheet program like Excel is useful for that. Example: s000.tinyupload.com / ?file_id=08472573822564563391 More scientific approach: read into fitting polynomials (with Excel for example). Create a hp data point for some waves, then fit a polynomial to it. • I guess a starting point would be to find out how many enemies will be in the game total, then decide how much XP to give the payer, and go from there. I'll do some reading into polynomials and stuff. Thanks guys! • 6 posts

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Roots of unity and function $\mu$ I need to prove that for each positive integer $n$ the sum of the primitive $n$th roots of unity in $\mathbb{C}$ is $\mu(n)$, where $\mu$ is the Möbius function. - So what's the question? – Patrick Da Silva Oct 20 '12 at 0:02 Yes...or not: what's your question? ¿Cuál es tu pregunta? – DonAntonio Oct 20 '12 at 0:03 I need one proof to the given proposition – Elmo goya Oct 20 '12 at 0:11 Well explained... not wikipedia... please!!! – Elmo goya Oct 20 '12 at 0:33 You can try to prove it for primes, then $p^n$, then use multiplicativity. – Berci Oct 20 '12 at 0:36 Do you know $$\sum_{d\mid m}\mu(d)=1{\rm\ if\ }m=1,\,\,=0{\rm\ else}$$ The sum of the primitive $n$th roots of unity is $$\sum_{\gcd(k,n)=1}e^{2\pi ik/n}=\sum_1^n\sum_{d\mid\gcd(k,n)}\mu(d)e^{2\pi ik/n}=\sum_{d\mid n}\mu(d)\sum_0^{(n/d)-1}e^{2\pi idk/n}$$ The inner sum os the sum of all the $m$th roots of unity where $m=n/d$, so it's zero except for $d=n$ when it's $1$. So, the original sum evaluates to $\mu(n)$. - Let $\theta$ denote the first $n$th primitive root: $\theta:=e^{2\pi i/n}$. 1. If $n=p$ is prime, $\mu(p)=-1$ and each $0\ne a<p$ is relatively prime to $p$, so this $\theta^{a}$ is primitive $p$th root. The sum of all $n$th roots is always $0$ (because if we multiply it by $\theta$, it doesn't change). So we miss only the $\theta^0=1$, hence the sum is $-1$. 2. If $n=p^k$ ($k\ge 2$), then $\mu(n)=0$ and exactly the $p\cdot a$ elements have common divisor with $n$, so $$\sum_{\theta^u\text{ prim.root}}\theta^u=\sum_{u\ne a\cdot p}\theta^u = \sum_{u=0}^{n-1}\theta^u-\sum_{v=0}^{\frac np-1} \theta^{pv}$$ Can you continue? 3. You also need to show that both functions in question are multiplicative, i.e., whenever $\gcd(a,b)=1$, we have $$\mu(ab)=\mu(a)\cdot\mu(b)$$ and same for the other function. From these the proposition follows. -

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Find all School-related info fast with the new School-Specific MBA Forum It is currently 28 Aug 2016, 00:03 GMAT Club Daily Prep Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History Events & Promotions Events & Promotions in June Open Detailed Calendar The figure shows a square patio surrounded by a walkway of width x Author Message TAGS: Hide Tags Director Joined: 03 Nov 2004 Posts: 865 Followers: 2 Kudos [?]: 46 [0], given: 0 The figure shows a square patio surrounded by a walkway of width x [#permalink] Show Tags 30 Jun 2005, 20:03 4 This post was BOOKMARKED 00:00 Difficulty: 75% (hard) Question Stats: 71% (06:20) correct 29% (04:14) wrong based on 161 sessions HideShow timer Statistics The figure shows a square patio surrounded by a walkway of width x meters. If the area of the walkway is 132 square meters and the width of the patio is 5 meters greater than the width of the walkway, what is the area of the patio, in square meters? A. 56 B. 64 C. 68 D. 81 E. 100 [Reveal] Spoiler: Attachment: patio.JPG [ 8.03 KiB | Viewed 4247 times ] [Reveal] Spoiler: OA Last edited by Bunuel on 05 Jul 2015, 08:05, edited 1 time in total. Renamed the topic, edited the question and added the OA. Senior Manager Joined: 30 May 2005 Posts: 373 Followers: 1 Kudos [?]: 9 [0], given: 0 Re: The figure shows a square patio surrounded by a walkway of width x [#permalink] Show Tags 30 Jun 2005, 20:21 Clearly the patio is a square, since all of its sides of off 2x from the bigger square. Let the patio's side be a. => x=a-5 Side of bigger square = a + 2(a-5) = 3a - 10 The area of the walkway = Area of big square - Area of patio 132 = (3a-10)^2 - a^2 132 = 9a^2 -60a + 100 -a^2 This reduces to: 8a^2 - 60a - 32 = 0 2a^2 -15a - 8 = 0 2a^2 + a -16a - 8 = 0 a(2a + 1) -8(2a+1) = 0 (a-8)*(2a+1) = 0 a = 8 or -1/2, But a is positive => a=8 Patio's area = a^2 = 64 VP Joined: 30 Jun 2008 Posts: 1043 Followers: 14 Kudos [?]: 511 [0], given: 1 The figure shows a square patio surrounded by a walkway of [#permalink] Show Tags 24 Oct 2008, 03:46 1 This post was BOOKMARKED The figure shows a square patio surrounded by a walkway of width x meters. If the area of the walkway is 132 square meters and the width of the patio is 5 meters greater than the width of the walkway, what is the area of the patio, in square meters? A. 56 B. 64 C. 68 D. 81 E. 100 Attachments square patio.jpg [ 22.65 KiB | Viewed 10246 times ] _________________ "You have to find it. No one else can find it for you." - Bjorn Borg Senior Manager Joined: 21 Apr 2008 Posts: 269 Location: Motortown Followers: 1 Kudos [?]: 132 [0], given: 0 Re: PS : SQUARE PATIO [#permalink] Show Tags 24 Oct 2008, 05:16 B (5+3x)^2 - (5+x)^2 = 132 2x(10+4x) = 132 x(5+2x) = 33 = 3*11 So, x can be 3 and not 11 8^2 = 64 VP Joined: 30 Jun 2008 Posts: 1043 Followers: 14 Kudos [?]: 511 [0], given: 1 Re: PS : SQUARE PATIO [#permalink] Show Tags 24 Oct 2008, 06:33 LiveStronger wrote: B (5+3x)^2 - (5+x)^2 = 132 2x(10+4x) = 132 x(5+2x) = 33 = 3*11 So, x can be 3 and not 11 8^2 = 64 nice one especially this step ---- x(5+2x) = 33 = 3*11 I was wondering how to reduce the calculation thanks _________________ "You have to find it. No one else can find it for you." - Bjorn Borg Retired Moderator Joined: 18 Jul 2008 Posts: 994 Followers: 10 Kudos [?]: 170 [0], given: 5 Re: PS : SQUARE PATIO [#permalink] Show Tags 24 Oct 2008, 09:55 So, x can be 3 and not 11 Why not 11? (I did this the ultra long way) amitdgr wrote: LiveStronger wrote: B (5+3x)^2 - (5+x)^2 = 132 2x(10+4x) = 132 x(5+2x) = 33 = 3*11 So, x can be 3 and not 11 8^2 = 64 nice one especially this step ---- x(5+2x) = 33 = 3*11 I was wondering how to reduce the calculation thanks VP Joined: 30 Jun 2008 Posts: 1043 Followers: 14 Kudos [?]: 511 [0], given: 1 Re: PS : SQUARE PATIO [#permalink] Show Tags 24 Oct 2008, 10:07 So, x can be 3 and not 11 Why not 11? (I did this the ultra long way) amitdgr wrote: LiveStronger wrote: B (5+3x)^2 - (5+x)^2 = 132 2x(10+4x) = 132 x(5+2x) = 33 = 3*11 So, x can be 3 and not 11 8^2 = 64 nice one especially this step ---- x(5+2x) = 33 = 3*11 I was wondering how to reduce the calculation thanks x > 0 so 5+3x > x x(5+2x) = 3*11 x has to be the smaller value and (5+2x) the larger. so x = 3 even I did it the ULTRA lonnggg wayy _________________ "You have to find it. No one else can find it for you." - Bjorn Borg Senior Manager Joined: 23 May 2008 Posts: 428 Followers: 5 Kudos [?]: 297 [0], given: 14 Re: PS : SQUARE PATIO [#permalink] Show Tags 30 Dec 2008, 02:46 LiveStronger wrote: B (5+3x)^2 - (5+x)^2 = 132 2x(10+4x) = 132 x(5+2x) = 33 = 3*11 So, x can be 3 and not 11 8^2 = 64 can someone please explain this step how did we get this (5+3x)^2 - (5+x)^2 = 132 Senior Manager Joined: 21 Apr 2008 Posts: 269 Location: Motortown Followers: 1 Kudos [?]: 132 [0], given: 0 Re: PS : SQUARE PATIO [#permalink] Show Tags 30 Dec 2008, 08:27 gurpreet07 wrote: LiveStronger wrote: B (5+3x)^2 - (5+x)^2 = 132 2x(10+4x) = 132 x(5+2x) = 33 = 3*11 So, x can be 3 and not 11 8^2 = 64 can someone please explain this step how did we get this (5+3x)^2 - (5+x)^2 = 132 Width of the patio is 5m greater than the width of the walkway. So, width of the Patio = 5+x (since x is the width of the walkway) To get the area of the walkway, we need to substract the area of the bigger square minus the patio. Area of patio = (5+x)^2 Side of the bigger aquare = x+x+5+x = 5+3x So, area of the bigger square is = (5+3x)^2 Hence, (5+3x)^2 - (5+x)^2 = 132 Intern Joined: 19 Aug 2008 Posts: 4 Followers: 0 Kudos [?]: 1 [1] , given: 0 Re: PS : SQUARE PATIO [#permalink] Show Tags 30 Dec 2008, 10:17 1 KUDOS Hi LiveStronger, nice way to solve it. could you please explain me how do you go from step 1 to step 2. It took me 3 lines of operations and you can do it in just one. I definitely need to be faster in the exam so this kind of tips will be really useful. thks!! Step 1. (5+3x)^2 - (5+x)^2 = 132 Step 2. 22x(10+4x) = 132 Senior Manager Joined: 21 Apr 2008 Posts: 269 Location: Motortown Followers: 1 Kudos [?]: 132 [1] , given: 0 Re: PS : SQUARE PATIO [#permalink] Show Tags 30 Dec 2008, 10:38 1 KUDOS marcap wrote: Hi LiveStronger, nice way to solve it. could you please explain me how do you go from step 1 to step 2. It took me 3 lines of operations and you can do it in just one. I definitely need to be faster in the exam so this kind of tips will be really useful. thks!! Step 1. (5+3x)^2 - (5+x)^2 = 132 Step 2. 22x(10+4x) = 132 I used a^2 - b^2 = (a+b)(a-b) formula GMAT Tutor Joined: 24 Jun 2008 Posts: 1183 Followers: 393 Kudos [?]: 1372 [1] , given: 4 Re: PS : SQUARE PATIO [#permalink] Show Tags 30 Dec 2008, 11:35 1 KUDOS Expert's post The solutions above subtract the inner square from the outer one, which is a good approach. One other way to get to the same answer: divide up the patio. You have the four corners, measuring x by x, and four rectangles lined up with each side of the inner square which measure x by x+5. So: 4x^2 + 4x(x+5) = 132 2x^2 + 5x = 33 etc. _________________ GMAT Tutor in Toronto If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com SVP Joined: 29 Aug 2007 Posts: 2492 Followers: 66 Kudos [?]: 693 [0], given: 19 Re: PS : SQUARE PATIO [#permalink] Show Tags 30 Dec 2008, 12:13 amitdgr wrote: The figure shows a square patio surrounded by a walkway of width x meters. If the area of the walkway is 132 square meters and the width of the patio is 5 meters greater than the width of the walkway, what is the area of the patio, in square meters? A. 56 B. 64 C. 68 D. 81 E. 100 width of the walkway = x width of the patio = x+5 = x+5 width of the whole area = x+5+x = 3x+5 (3x+5)^2 - (x+5)^2 = 132 9x^2 + 30x + 25 - x^2 -10x -25 = 132 8x^2 + 20x = 132 4 (2x^2 + 5x) = 132 2x^2 + 5x - 33 = 0 2x^2 + 11x - 6x - 33 = 0 x (2x+11) -3 (3x +11) = 0 (x-3) (2x+11) = 0 x = 3 or -11/2 but -11/2 is not possible as length/width cannot be in -ve. so x = 30 so area of the patio = (3+5)^2 = 64 _________________ Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html GT Senior Manager Joined: 23 May 2008 Posts: 428 Followers: 5 Kudos [?]: 297 [1] , given: 14 Re: PS : SQUARE PATIO [#permalink] Show Tags 30 Dec 2008, 21:52 1 KUDOS Thanks GMAT TIGER and LiveStronger for d explanation...... Actually i go numb when i see numbers....... can u please suggest me how should i improve my quant Manager Joined: 08 Aug 2008 Posts: 234 Followers: 1 Kudos [?]: 30 [1] , given: 0 Re: PS : SQUARE PATIO [#permalink] Show Tags 30 Dec 2008, 23:38 1 KUDOS If its convenient, u can also attempt backsolving it:- On first glance A,C are clearly out. (E) 100 means side of patio is 10 so width of walkway =5. the walkway has 4 rectangles 2 among them are each 20*5=100 --impossible for total of 132. (D) 81 means side of patio is 9 so width of walkway =4. the walkway has 4 rectangles 2 among them are each 17*4=68. 2 such rectangles account for 136--hence impossible for total of 132. (C) 64 means side of patio is 8 so width of walkway =3. the walkway has 4 rectangles 2 among them are each 14*3=42 and rest are 8*24.total= 2(42+24)=132. Manager Joined: 21 Feb 2010 Posts: 212 Followers: 1 Kudos [?]: 23 [0], given: 1 Show Tags 26 Aug 2010, 08:23 the answer is B. here's how i did it... the width of the patio is X + 5 and the width of the big square is 3X +5, therefore the are of the big square is equal to the area of the walkway plus the area of the patio, the equation is (3X +5)^2 -132 = (X + 5)^2, X = 3 or X = -11/2 since the width of the patio = X+5 = 3 + 5 =8 the area equals to 8^ 2 =64 hope it helps!! Math Forum Moderator Joined: 20 Dec 2010 Posts: 2021 Followers: 157 Kudos [?]: 1560 [1] , given: 376 Re: PS : SQUARE PATIO [#permalink] Show Tags 25 Aug 2011, 23:22 1 KUDOS amitdgr wrote: The figure shows a square patio surrounded by a walkway of width x meters. If the area of the walkway is 132 square meters and the width of the patio is 5 meters greater than the width of the walkway, what is the area of the patio, in square meters? A. 56 B. 64 C. 68 D. 81 E. 100 Smaller square=Patio Let the side of Patio be "s". Width of walkway=x width of the patio is 5 meters greater than the width of the walkway(Actually it should be "side of the patio is 5 meters greater than the width of the walkway" because patio is a square) So, $$s=x+5$$ If we see the figure properly, the outer quadrilateral is also a square with side $$s+x+x$$ OR $$x+5+x+x=3x+5$$ We know, the area of the walkway is 132 square meters: Area of walkway=Area of outer square-Area of inner square Area of walkway=(3x+5)^2-(x+5)^2=132 $$(3x+5)^2-(x+5)^2=132$$ $$(3x)^2+(5)^2+2*3x*5-(x^2+5^2+2*5x)=132$$ $$9x^2+25+30x-x^2-25-10x=132$$ $$9x^2+30x-x^2-10x=132$$ $$8x^2+20x=132$$ $$2x^2+5x=33$$ $$2x^2+5x-33=0$$ $$2x^2+11x-6x-33=0$$ $$x(2x+11)-3(2x+11)=0$$ $$(x-3)(2x+11)=0$$ $$x=3 \hspace{3} OR \hspace{3} x=-\frac{11}{2}$$ Width can't be -ve. So, $$x=3$$ $$s=x+5=3+5=8$$ Area of the patio$$=s^2=8^2=64$$ Ans: "B" _________________ Manager Status: Enjoying the MBA journey :) Joined: 09 Sep 2011 Posts: 137 Location: United States (DC) Concentration: General Management, Entrepreneurship GMAT 1: 710 Q49 V38 WE: Corporate Finance (Other) Followers: 18 Kudos [?]: 112 [0], given: 16 Re: The figure shows a square patio surrounded by a walkway of width x met [#permalink] Show Tags 10 Nov 2011, 12:48 1 This post was BOOKMARKED The algebraic solution to the problem is given below: Given, Width of walkway = x Width of Patio = Width of Walkway + 5 meters = x+5 meters Area of walkway = 132 square meters From the given figure, we know that the width of the square figure i.e. Patio + Walkway will be: 2 times the width of walkway + width of Patio i.e 2(x) + (x+5) = 3x+5 Therefore, the area of this combined figure (square) will be (3x+5)^2 Now, the area of the walkway will be equal to the difference between the area of the combined figure and the square Patio i.e. (3x+5)^2 - (x+5)^2 This is given to be 132 square meters. Therefore, expanding and simplifying the equation, (3x+5)^2 - (x+5)^2 = 132 We get, 8x^2 + 20x - 132 = 0 i.e. 2x^2 + 5x - 33 = 0 Solving the above quadratic equation will yield the value of x as -6.5 and 3. Since the width of the walkway can't be negative, the value of x is 3 meters. Using this, we can calculate the area of the Patio, which is (x+5)^2 i.e. (3+5)^2 = 8^2 = 64 square meters. This is option B. Hope this helps Cheers! _________________ MBA Candidate 2015 | Georgetown University Manager Joined: 10 Mar 2014 Posts: 233 Followers: 1 Kudos [?]: 64 [0], given: 13 Re: The figure shows a square patio surrounded by a walkway of [#permalink] Show Tags 04 Jul 2015, 06:57 amitdgr wrote: The figure shows a square patio surrounded by a walkway of width x meters. If the area of the walkway is 132 square meters and the width of the patio is 5 meters greater than the width of the walkway, what is the area of the patio, in square meters? A. 56 B. 64 C. 68 D. 81 E. 100 Hi bunuel, i didnt get it. Thanks. Math Forum Moderator Joined: 20 Mar 2014 Posts: 2636 GMAT 1: 750 Q49 V44 GPA: 3.7 WE: Engineering (Aerospace and Defense) Followers: 106 Kudos [?]: 1192 [0], given: 786 Re: The figure shows a square patio surrounded by a walkway of [#permalink] Show Tags 04 Jul 2015, 07:31 PathFinder007 wrote: amitdgr wrote: The figure shows a square patio surrounded by a walkway of width x meters. If the area of the walkway is 132 square meters and the width of the patio is 5 meters greater than the width of the walkway, what is the area of the patio, in square meters? A. 56 B. 64 C. 68 D. 81 E. 100 Hi bunuel, i didnt get it. Thanks. It is given that width of the patio is 5 more than width of the walkway. Now, walkway's 'extension on either side of the patio' is 'x' and thus width of the of the patio = $$W_P$$ = 5+x Now, total width of the walkway = $$W_W$$ = width of the patio + 2*x = 5+x+2x = 5+3x _________________ Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515 Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html Re: The figure shows a square patio surrounded by a walkway of   [#permalink] 04 Jul 2015, 07:31 Go to page    1   2    Next  [ 22 posts ] Similar topics Replies Last post Similar Topics: 1 A circular garden is surrounded by a fence of negligible width along 2 08 Dec 2015, 02:57 2 The figure above represents a rectangular garden with a walkway around 4 11 May 2015, 08:19 5 The figure above shows a square inscribed within an equilateral triang 4 23 Dec 2014, 09:37 A square garden is surrounded by a 2 feet wide walkway on 2 26 Jul 2012, 05:34 4 A rectangular tiled patio is composed of 70 square tiles. 7 10 May 2012, 04:39 Display posts from previous: Sort by

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Learn and Practice Free Customary Unit - Weight - Set 1 questions You can get answer of the below sample questions by going through the above Question sets. Option 1: 50 Option 2: 60 Option 3: 70 Option 4: 80 Option 5: Option 1: 100 Option 2: 10 Option 3: 20 Option 4: 30 Option 5: Option 1: 4000 Option 2: 5000 Option 3: 6000 Option 4: 7000 Option 5: Option 1: 32 Option 2: 33 Option 3: 34 Option 4: 35 Option 5: Option 1: 4 Option 2: 5 Option 3: 6 Option 4: 7 Option 5: Option 1: 6 Option 2: 7 Option 3: 8 Option 4: 9 Option 5: Option 1: 14000 Option 2: 15000 Option 3: 16000 Option 4: 17000 Option 5: Option 1: 111 Option 2: 112 Option 3: 113 Option 4: 114 Option 5: Option 1: 19000 Option 2: 20000 Option 3: 21,000 Option 4: 22000 Option 5: 56 oz = _____ lb _______ oz Option 1: 2,5 Option 2: 2,7 Option 3: 3,7 Option 4: 3,8 Option 5: Practice Free Customary Unit - Weight - Set 1 Questions online By Practicing the below question set : The student will get the basic and advanced knowledge on Customary Unit - Weight - Set 1. The student will learned the priciples and methedology of Customary Unit - Weight - Set 1. The student will evaluate the knowledge level on Customary Unit - Weight - Set 1. Student can practice number of times the question till he is clear. Student can provide the review / feedback / suggestion on the questions which will help to others. The question contains the simple to complex level difficulties so that student can improve the knowledge.

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## Re: [xsl] Re: Re: order of UNIONs Subject: Re: [xsl] Re: Re: order of UNIONs From: Jörg Heinicke Date: Thu, 15 Nov 2001 08:39:58 +0100 ```Thanks, Dimitre, for the explanations. I must confess, that I don't know much about node-sets in theory, only things I came across when using them in XSL. Are there any nice websites or books (Michael Kay's?) about node-set theory? Joerg > > But does this really make sense? select="group[21]|group[1]" means for me > > 'select the 21st group and add the 1st group'. > > It makes ***perfect sense***. Node-sets are sets. Sets do not have order. Regardless > of the order, in which you add elements, the result is the same set: > > a | b = b | a > > The above is an axiom in set theory. > > What you actually need is kinda bag, or list -- these are very different structures > >from sets. Both allow duplicates, and a list has order. > > I think one of the major problem of the XPath 2.0 Data Model is that they do not > distinguish between lists and sets, trying somehow to say that a node-set is a kind > of list (the actual term used there was "sequence", if I remember well). The > consequences are bad problems, because some operations on lists cannot be performed > 1:1 on sets, and vice versa -- not every list is a set, an operation performed on a > sequence, that is a nodeset, may not yield a nodeset, a nodeset operation performed > on a sequence that is a nodeset, may behave quite differently from the same > operation, performed on a sequence (e.g. eliminating/preserving duplicates on a > union/append operation). > > Cheers, > Dimitre Novatchev. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list ```

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Anda di halaman 1dari 8 # FINC-520 Christiano Wold Representation Theorem We have discussed a class of ARMA models and derived restrictions which ensure they are models for covariance stationary time series. We have shown that these ARMA models imply the data are a linear combination of current and past one-step-ahead forecast errors, with weights that decay at a geometric rate.1 Here, we consider the class of covariance stationary processes and ask whether ARMA models are a strict subset of that class. We start from the assumption that a process is covariance stationary and we study the projection of the process onto its current and past one-step-ahead forecast errors. This decomposition of a covariance stationary process into a projection onto current and past one-step-ahead forecast errors (the purely indeterministic part of the process) and a projection error (the purely deterministic part) is called the Wold Representation Theorem. We conclude that there are two ways in which ARMA models represent a restriction on the class of covariance stationary processes. First, in an ARMA model the purely deterministic part is absent. That is, a researcher working with an ARMA model implicitly assumes both that the process is covariance stationarity and that the process is purely indeterministic. Second, according to the Wold Representation Theorem, covariance stationarity implies that the weights on current and past one-step-ahead forecast errors are square summable. This is weaker than the geometric decay property implied by ARMA models. The first section below states the Wold Representation Theorem, and then provides an informal proof using the argument in Sargent (1979). I then summarize the implications of the theorem for the ARMA models that we study. ## 1. The Wold Theorem Theorem 1.1. Suppose that {xt } is a covariance stationary process with Ext = 0 and covariance function, (j) = Ext xtj , j. Then xt = dj tj + t j=0 where d0 = 1, X j=0 ## d2j < 0, E2t = 2 , Et s = 0 for t 6= s, Et = 0, Et s = 0 for all t, s, ## P t+s |xt1 , xt2 , ... = t+s , s 0. (1.1) (1.2) (1.3) This linear combination is derived by recursive substitution. Before doing this, one has to make sure that the ARMA model error is the one-step-ahead forecast error, if necessary by flipping moving average roots. The first part of the representation of xt looks just like the MA() with square summable moving average terms that we have worked with, while the second part, t , is something new. That part is called the deterministic part of xt because t is perfectly predictable based on past observations on xt . The style of proof is constructive. We will show that given only covariance stationarity, we can build the Wold representation with the indicated properties. We will not provide a fully rigorous proof and a key result will simply be assumed. The proof is an application of linear projections, and the orthogonality and recursive properties of projections. The proof style follows that in Sargent (1979). We first find the dj s and t and establish the required properties. Then, we find the projection error, t . We begin with a preliminary result. Let xt be a covariance stationary process. Let (n) xt ## = P [xt |xt1 , ..., xtn ] , and write (n) (n) xt = xt + t . From the orthogonality property of projections we know that (n) (n) Et = 2(n) . (n) ## x = P [xt |xt1 , xt2 , ...] xt xt = xt + t , E2t = 2 t (xt1 , xt2 , ...) . (1.4) (1.5) (1.6) ## The disturbance, t , is known as the innovation in xt or its one-step-ahead forecast error. It is easy to see that t is a serially uncorrelated process. In particular, t = xt P [xt |xt1 , xt2 , ...] , so that it is a linear combination of current and past xt s. It follows that since t is orthogonal to past xt s, it is also orthogonal to past t s. 1.1. Projection of xt onto current and past t s We now consider the projection of xt on current and past t s: xm t m X dj tj . j=0 The notation, xm t , is intended to signal that the projection used here is dierent from the one used to define the t s. The lack of autocorrelation between the t s makes the analysis of 2 the projection coecients particularly simple. The orthogonality condition associated with the projection is: ! m X E xt dj tj tk = 0, k = 0, ..., m, j=0 ## which, by the lack of correlation in the t s, reduces to: Ext tk dk E2tk = 0, so that dk = Ext tk , 2 k = 1, 2, ..., m 1, k = 0 ## That Ext t = 2 follows from (1.5): xt + t ) t = 2 , Ext t = E ( because xt is a linear function of past xt s and t is orthogonal to those xt s. A key property of the projection is that dk is not a function of m. This reflects the lack of serial correlation in the t s. We now establish the square summability of the dj s. Any variance must be non-negative, and this is true of the error in the projection of xt onto t , ..., tm : !2 m X dj tj 0, E xt j=0 or, Ex2t 2 = Ex2t m X dj Ext tj + j=0 m X 2 d2j j=0 m X d2j 2 j=0 0. This must be true for all m. Since Ex2t is a fixed number by covariance stationarity, it follows that m X d2j < . lim m j=0 In addition the sum is a non-decreasing sequence because each term (being a square) is non-negative. From this we conclude that the above sum converges to some finite number: m X j=0 d2j d2j . j=0 ## Given the square summability of the dj s, it follows that xm t forms a Cauchy sequence, so that m X X m dj tj xt = dj tj . xt = j=0 j=0 To verify that xm t is Cauchy, we establish that for each e > 0, there exists an n such that for all m > n ! m X 2 E ( xm nt ) = d2j 2 < e. t x j=n+1 ## 1.2. Constructing the t s We define t as the dierence between xt and its projection onto the current and past t s: t = xt xt . (1.7) ## We first establish that E t s = 0 for all t, s. That Et s = 0 for s > t is obvious because t is a linear function of xt and past t s, and s is orthogonal to all these things, s > t. That Et s = 0 for s t follows from the fact that t is the error in the projection of xt on current and past t s. In particular, Et tk = Ext tk dk 2 = dk 2 dk 2 = 0. Next we establish that t is perfectly predictable from past xt s. Note P [t |xt1 , xt2 , ...] = P [xt |xt1 , xt2 , ...] X j=0 ## dj P [tj |xt1 , xt2 , ...] , (1.8) where we have used the linearity of projections, P [A + B|] = P [A|] + P [B|] . Consider the last term, involving the projections of the t s. In the case, j = 0 : P [t |xt1 , xt2 , ...] = j xtj . j=1 ## The j s satisfy the orthogonality conditions: ! X E t j xtj xtk = 0, k = 1, 2, 3, ... . j=1 Recall that t is orthogonal to (xt1 , xt2 ...) so that j = 0 satisfies the orthogonality conditions. Suciency of the orthogonality conditions guarantees that P [t |xt1 , xt2 , ...] = 0. Now consider P [t |xt , xt1 , ...] Recall, that t is a linear function of current and past xt s: t = xt P [xt |xt1 , xt2 , ...] , 4 so that P [t |xt , xt1 , ...] = t . To see why this is so, consider the optimization problem that defines a projection: !2 X min E t j xtj . {j }j=0 j=0 By choosing the j s to coincide with the linear function, xt P [xt |xt1 , xt2 , ...] , this criterion can be set to zero, which cannot be improved upon. A similar argument establishes P [t |xt+j , xt+j1 , ...] = t , j 0. With the previous result, we can write (1.8) as follows: P [ t |xt1 , xt2 , ...] = P [xt |xt1 , xt2 , ...] dj tj . j=1 ## Subtract this from (1.7): t P [t |xt1 , xt2 , ...] = xt P [xt |xt1 , xt2 , ...] z }|t { dj tj dj tj j=0 j=1 = t d0 t = 0. ## This establishes the s = 0 part of (1.3). Now consider s = 1 : P [t |xt2 , xt3 , ...] = P [xt |xt2 , xt3 , ...] = P [xt |xt2 , xt3 , ...] X j=0 ## dj P [tj |xt2 , xt3 , ...] dj tj , j=2 by an argument similar to the one for s = 0. Subtract the above expression from (1.7): t P [t |xt2 , xt3 , ...] = xt P [xt |xt2 , xt3 , ...] X j=0 dj tj ## = xt P [xt |xt2 , xt3 , ...] (t + d1 t ) . X j=2 dj tj We use the recursive property of projections to evaluate the two-step-ahead forecast error in xt : P [xt |xt1 , xt2 , ...] = P [xt |xt2 , xt3 , ...] +P [xt P (xt |xt2 , xt3 ...) |xt1 P (xt1 |xt2 , ...)] . 5 In words, the projection of xt onto xt1 and earlier xt s is the projection of xt onto xt2 and earlier plus the best (linear) guess of what that projection error is, given the new information in xt1 . Write the last term in the recursive representation as: P [xt P (xt |xt2 , xt3 ...) |xt1 P (xt1 |xt2 , ...)] = t1 , since t1 = xt1 P (xt1 |xt2 , ...) . Then, E [xt P (xt |xt2 , xt3 ...)] t1 E2t1 Ext t1 = = d1 . E2t1 ## because t1 is orthogonal to xt2 , xt3 , ... . So, P [xt |xt1 , xt2 , ...] = P [xt |xt2 , xt3 , ...] + d1 t1 and xt P [xt |xt2 , xt3 , ...] = xt P [xt |xt1 , xt2 , ...] + d1 t1 = t + d1 t . We conclude that t P [ t |xt2 , xt3 , ...] = xt P [xt |xt2 , xt3 , ...] (t + d1 t ) = 0. A continuation of this line of argument establishes that ## t+s = P t+s |xt2 , xt3 , ... , s > 0. 2. Discussion The Wold representation says that a covariance stationary process can be represented in the following form: xt = X j=0 dj tj {z t |{z} ## part of xt that is impossible to predict perfectly The two parts of this representation are called the purely indeterministic and the deterministic parts, respectively. It is interesting to evaluate the meaning of t . It is not a time trend, for example, because the assumption of covariance stationarity of xt rules out a time trend.2 Here is an example of what t could be: t = a cos (t) + b sin (t) , 2 The presence of a time trend would imply that the mean of xt is a function of t. ## where is a fixed number and Ea = Eb = Eab = 0, a, b {t } . To understand this stochastic process for xt , think of how each realization is constructed. First, draw a and b. Then draw and infinite sequence of t s and generate a realization of t and xt . For the second realization, draw a new a and b, and a new sequence of t s. In this way, all the realizations of the stochastic process may be drawn. Under this representation, the mean and autocovariance function of xt are not a function of time, and so xt is covariance stationary. The idea that t is perfectly predictable can be seen as follows. First, a and b can be recovered given only two observations on t . Once a and b for a given realization of the stochastic process are in hand, all the t s in that realization can be computed. But, how to get the two t s? According to the argument in the proof, t can be recovered without error from a suitable linear combination of xt1 , xt2 , ... and t+1 can be recovered from a suitable linear combination of xt , xt1 , ... . It is interesting to compare the purely indeterministic part of the Wold representation with the MA() representations we have discussed in class. The models of MA() representations are in their most general form, ARMA(p,q) representations: yt = 1 yt1 + ... + p ytp + t + 1 t1 + ... + q tq , where t is an iid process. As long as the roots of the autoregressive part of this process are less than unity in absolute value, yt has an MA() representation with square summable moving average terms. Still, there are two possible dierences between this and a Wold representation. First, only if the roots of the moving average part, i.e., the zeros of q + 1 q1 + ... + q are less than unity in absolute value is t the one-step-ahead forecast error in yt (to see this, note than only in this case can recursive substitution be done to represent t as a function of current and all past yt s). Second, the ARMA(p,q) form, while it generates an MA() with square summable weights, it is not the only form that does this. This is perhaps obvious when we observe that the rate of decay of the moving average coecients in the models we have considered is geometric. This is a faster rate of decay than is required for square summability. For example, with geometric decay absolute and square summability are the same thing. But, in general, a process that is square summable is not necessarily absolutely summable.3 We can think of the weight on distant past t s of the MA() representation as corresponding to the amount of memory in the process. Thus, the ARMA(p,q) models have short memory relative to the entire class representations envisioned by the Wold representation. It has been argued that there is evidence of long-memory in economic time series, and that this warrants investigating a class of time series models dierent from ARMA models. See, for example, Parke (1999). 4 We will not be studying long memory processes in this course. For example, consider j = 1/j. The rate of decay of 2j is fast enough that j satisfies square summability, but j does not satisfy absolute summability. 4 Here are some sources cited in Parke (1999), footnote 2. Evidence of long memory has been found in 3 References [1] Andersen, Torben G. and Tim Bollerslev, Heterogeneous Information Arrivals and Return Volatility Dynamics: Uncovering the Long-Run in High Frequency Returns, Journal of Finance, 52 (1997), 975-1005. [2] Andersen, Torben G., Tim Bollerslev, Francis X. Diebold, and Paul Labys, The Distribution of Exchange Rate Volatility, working paper (1999). [3] Baillie, Richard T., "Long Memory Processes and Fractional Integration in Economics," Journal of Econometrics, 73 (1996), 5-59. [4] Baillie, Richard T., Tim Bollerslev, and Hans Ole Mikkelsen, Fractionally Integrated Generalized Autoregressive Conditional Heteroskedasticity, Journal of Econometrics, 74 (1996), 3-30. [5] Baillie, Richard T., Ching-Fan Chung, and Margie A. Tieslau, Analyzing Inflation by the Fractionally Integrated ARFIMA-GARCH Model, Journal of Applied Econometrics, 11 (1996), 23-40. [6] Breidt, F. Jay, Nuno Crato, and Pedro de Lima, The Detection and Estimation of Long Memory in Stochastic Volatility, Journal of Econometrics, 83 (1998), 325-348. [7] Diebold, Francis X. and Glenn D. Rudebusch, "Long Memory and Persistence in Aggregate Output," Journal of Monetary Economics, 24 (1989), 189-209. [8] Ding, Zhuanxin, Clive W.J. Granger, and Robert F. Engle, A Long Memory Property of Stock Market Returns and a New Model, Journal of Empirical Finance, 1 (1993), 83-106. [9] Geweke, John and Susan Porter-Hudak, "The Estimation and Application of Long Memory Time Series Models," Journal of Time Series Analysis, 4 (1983), 221-238. [10] Parke, William, 1999, What is Fractional Integration?, Working Paper 99-01, Department of Economics, University of North Carolina, Chapel Hill. [11] Sargent, Thomas, 1979, Macroeconomic Theory. [12] Sowell, Fallow, Modeling Long-Run Behavior with the Fractional ARIMA Model, Journal of Monetary Economics, 29 (1992), 277-302. traditional business cycle indicators such as aggregate economic activity [Diebold and Rudebusch (1989), Sowell (1992)] and prices indices [Geweke and Porter-Hudak (1983), Baillie, Chung, and Tieslau (1996)]. There is also strong evidence of long memory in asset price and exchange rate volatility [Andersen and Bollerslev (1997), Andersen, Bollerslev, Diebold, and Labys (1999), Baillie, Bollerslev, and Mikkelsen (1996), Breidt, Crato, and Lima (1998), Ding, Granger, and Engle (1993)]. Baillie (1996) provides an excellent survey of the literature on fractional integration and long memory.

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Learn how to solve the 4x4x4 Rubik’s Revenge Cube easily. Full guide including detailed images and examples. Solution guides to all rubiks cubes. 4x4x4 Solution. Understanding the instructions used for this 4x4x4 cube solution. We will display three sets of instruction, one graphic set and two text sets. A solution for beginners and much more This and the two following algorithms have been placed on the page 4x4x4 Disparity Algorithms for quick reference in. Author: Samuzragore Yozragore Country: Namibia Language: English (Spanish) Genre: Business Published (Last): 2 September 2014 Pages: 470 PDF File Size: 2.4 Mb ePub File Size: 1.62 Mb ISBN: 491-8-16650-872-6 Downloads: 32733 Price: Free* [*Free Regsitration Required] Uploader: Shakat ## How to solve a 4×4 Rubik’s Cube The first part of this process, as with the centres, is more about seeing what is happening rather than learning algorithms. You now have a 3x3x3 cube, however you may still have parity problems. Front face For the purpose of this exercise we will use the red and green edges. If they are not use either of the first two algorithms sloution this page. The problem is that we no longer have a third unmatched pair to realign the centres with. You can safely rotate any face to create a starting configuration. The aim of this section is to take you dolution this. You need to place a matching colour pair on different layers but on the same face of the cube in the configuration displayed here on the right. Your browser does not support script. Ensure the front face colours match. ET ON TUERA TOUS LES AFFREUX PDF ### How to Solve a 4×4 Cube- The Rubik’s Revenge For the purpose of this exercise we will use the red and green edges. Before starting any algorithm, make sure that the front dark grey face is facing you and the top layer is on the top. Simply Rubik Solutlon solution for beginners and much more. Pairing the Edges The first part of this process, as with the centres, is more about seeing what is happening rather than learning algorithms. Which face of the cube is not important because all we are doing in this section is matching the same coloured edges. This algorithm solves the cube for Fig. So you will need to learn the next algorithm to pair the last two unpaired edge sets. When you still have to solve the last two edge elements you do not have a third set to reset the centres with. The last pair to be solved are placed on ether side of the front face. If they are on the same layer as displayed on the left you can use either of the two algorithms below to place them on different layers as displayed on the right.

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# Computational Physics ## Presentasi berjudul: "Computational Physics"— Transcript presentasi: Computational Physics Wipsar Sunu Brams Dwandaru, Ph.D. Jurusan Pendidikan Fisika, FMIPA Universitas Negeri Yogyakarta, 2011 Pendahuluan Computer now permeates our society and has changed the way we think about science in general and physics in particular. The use of computers in physics is rapidly developing as hardware gets faster and cheaper. Computer introduces an alternative method to investigate and understand the physical world, which is exactly the objective of doing physics. How to do physics: theoretical physics experimental physics Developing and applying theories, emphasis on mathematics and rigor. Making observations and quantitative measurements. A requirement: analytic mathematics requirements: equipments and data analysis computational physics Numerical experiments in computer laboratory. requirements: numerical analysis and programming understanding the physical world An example in liquid physics real liquids construct models model liquids perform experiments carry out computer simulations construct approximate theories experimental results ‘exact’ results for model Theoretical predictions for model compare tests of models tests of approximate theories Tujuan Program Memperkenalkan berbagai metode komputasi dan keterkaitannya dalam penyelesaian masalah-masalah fisika. Memberikan wawasan tentang perkembangan fisika komputasi. Bukalah pemikiran untuk wawasan baru. Ini bukan kuliah tentang pemrograman! Anda Tidak diharapkan untuk menjadi seorang fisikawan komputasi! materi perkuliahan Functions and roots, Ordinary differential equations in classical mechanics, Partial differential equations, such as Maxwell’s equations and the Diffusion and Schrodinger equations, Matix methods: systems of equations and eigenvalue problems. bahasa pemorograman yang dipakai: penilaian tugas-tugas komputasi Ujian? bahasa pemorograman yang dipakai: C++ Fortran MATlab dll Fungsi dan Akar-Akarnya Adalah suatu hal yang alamiah untuk memulai fisika komputasi dengan membahas tentang fungsi. Teori tentang fungsi mendasari hampir semua teori-teori fisika, terutama dalam mencari solusi masalah fisika. Akan ditinjau kembali beberapa sifat-sifat fungsi dalam konteks fisika komputasi. Akan dibahas secara khusus masalah menentukan akar-akar dari suatu fungsi dalam satu dimensi. Walau sederhana, masalah ini kerap muncul dalam fisika. Akhirnya, permasalahan mencari akar-akar dari suatu fungsi memberikan kesempatan untuk mengeksplorasi keterkaitan antara matematika formal, analisis numerik, dan fisika komputasi. Menentukan Akar-Akar suatu Fungsi Rumusan Masalah Menentukan nilai x sedemikian sehingga terpenuhi f(x) = 0, (1) dengan x є R adalah variabel satu dimensi. f adalah suatu fungsi dengan pemetaan, f: R R, (2) dengan R adalah himpunan bilangan riil. Berbagai kemungkinan solusi Untuk polinomial order rendah menentukan akar-akar suatu fungsi relatif mudah. Contoh: f(x) = x – 3, f(x) = 10 – 7x + x2, f(y) = y3 - 13y + 12. Namun, semakin tinggi ordenya diperlukan usaha yang makin keras untuk mencari solusinya. Bisa diselesaikan secara analitik. Terdapat berbagai fungsi yang bahkan tidak memiliki solusi analitik sama sekali. Oleh karena itu, permasalahan utama yang akan dibahas adalah menentukan akar-akar dari suatu fungsi non-linier. Contoh: Tentukan harga x yang memenuhi x = cos x. Ubah persamaan di atas menjadi yang merupakan fungsi bernilai nol. Solusi: Ubah persamaan di atas menjadi F(x) = cos x – x = 0, yang merupakan fungsi bernilai nol. i) Kemungkinan I: Solusi analitik? Tidak bisa! ii) Kemungkinan II: menggambar grafik dari unsur-unsur fungsi tersebut. Tetapi terbatas keakuratannya. iii) Kemungkinan III: metode numerik. Presentasi serupa

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# How Much Electricity Does A Hot Water Heater Use ## How Much Electricity (Energy) Does a Water Heater Use? Increasing electricity use is a major concern in many American families today. According to the Department of Energy, water heating systems are the second largest consumer of electricity and account for an average of 18 percent of total electric expenditures; however, these costs may be decreased by selecting an energy-efficient system. An electric water heater is normally used for three hours each day to heat the water in your house, allowing you to take hot showers and baths when you want them. ## How Much Electricity (Energy) Does a Water Heater Use? To put it another way, how much electricity (energy) does a water heater consume? The majority of water heaters operate for 3 to 5 hours each day on average. Because the typical wattage of an electric water heater is roughly 4000 watts, if it is used for 3 hours per day at a cost of \$0.13 per kWh, it will cost you \$1.56 per day, approximately \$46.80 per month, and \$561 per year to operate. The quantity of power consumed by an electric water heater is determined by the size of the water tank and the energy factor of the water heater itself (EF). Consider this: If you have a 30-gallon tank with an efficiency rating of 0.7, it will generate 21 gallons per day when you use 10 kilowatt hours (kWh) each day (21 x 10 = 2.1) of electricity. If your household uses more than 2 kWh/day to meet its hot water needs, you should consider upgrading to an electric storage tank with higher efficiency ratings. The first question is how many liters of fuel your tank can contain. So let’s have a look at a more in-depth analysis of the data. ## Calculating Energy Usage The typical water heater is only turned on for three hours every day. Assuming you use 50 gallons of water each year, which consumes 5500 watts of power, and your energy rate is \$0.13 per kilowatt-hour, your water heater will cost around \$781 to operate every year! Consequently, it is critical for those who have one in their homes to understand how much it will cost them in the long run. The following is the formula for determining the energy consumption (in kW) of a water heater for a volume of water at a certain temperature over the course of one hour: Power is calculated as follows: volume of water heater x hours x temperature increase / 3412 = power If the tank holds 100 gallons and the temperature is being raised from 10 degrees Celsius to 65 degrees Celsius, the formula will be as follows: • Only three hours a day are spent by the average household using the water heater. It will cost around \$781 per year to run your water heater if you consume 50 gallons of water, which is 5500 watts of power, and your energy rate is \$0.13 per kilowatt-hour. Therefore, it is critical for those who have one in their homes to be aware of how much it will cost them in the long run. The following is the formula for determining the energy consumption (in kW) of a water heater for a volume of water at a certain temperature in one hour: Water heater capacity multiplied by hours multiplied by temperature rise divided by 3412 equals power. This formula will work under the following circumstances: The tank holds 100 liters, and the temperature is raised from 10 C to 65 C. If your energy efficiency factor (EF) is 0.98, you may save up to 20% on your annual heating bills! Think about replacing your outdated electric tank-style system with a new high-efficiency gas or hybrid electric/gas one right now! Various prominent manufacturers, including Rheem and Bradford White, as well as GE Monogram and Vaillant, offer these units for purchase. ## What Is the Cost to Run an Electric Heater For 24 Hours? Electric heaters typically have a wattage of 1,500 watts or more. To figure out how much it will cost to run the electric heater for a day, multiply the sum of 1,500 by 24 and 1,000 by \$0.13. (Dividing by 1,000 results in watt-hours being converted to kilowatt-hours.) In this instance, it will cost \$4.68 to run an electric heater for a continuous 24 hour period of time. ## Can an Old Water Heater Raise Electric Bill? Water heaters that are more than a decade old might result in higher energy costs. Generally speaking, older water heaters become less efficient and consume more power than modern ones after a decade or two. Because of this, you may see an increase in your monthly high-energy bill! Because of a variety of circumstances, the amount of energy that ancient water heaters consume might fluctuate significantly from one another. Factors include the age and size of the unit, the tank or demand model type, the temperature you set it to, and the amount of hot water consumed in a single day, among many other variables. ## Does Turning Down Water Heater Save Money? Reduce the temperature of your hot water heater to see if you may save any money. According to the United States Environmental Protection Agency, lowering the temperature of a 120-degree hot water heater to 90 degrees can result in annual savings of up to ten percent, or around \$41 for a household of four who spends approximately \$415 per year on water. ## How Much Gas Does a Hot Water Heater Use? You might be shocked to learn that an agas hot water heater can consume the same amount of energy as an electric water heater, depending on the model. Gas really consumes less than half the energy of electricity, yet it still consumes far more than most people assume. The technique for figuring your bill is also similar: you must multiply the amount of therms that your heater consumes per hour by the number of hours that it is on, and then multiply that figure by the price you are charged. For example, if you paid \$1 per therm and used a heater that consumed 0.205 therms per hour (about 3 hours each day), the total cost would be around 62 cents per day, \$18 per month, and 224 dollars per year, depending on your location. ## Gas vs. Electric Water Heater Operating Cost In general, a gas water heater is less expensive to operate on a monthly basis than an electric water heater. Although natural gas costs are often lower than the cost of electricity in most locations, you should consider other aspects like as the efficiency of your unit when considering which would be the best option for you in your particular situation. The good news is that the Energy Factor, often known as the EF rating, of your water heater determines how efficient it is. The higher this number is as compared to other numbers, the better the result. Consider models with EF ratings in the 90s if you’re in the market for another electric model for your house but aren’t sure which one would be the best fit for your needs. These are some of the most energy-efficient models available today! ## How to Keep Your Bills to a Minimum If you want to save money on the expense of heating your water, there are a few easy things you can do. Start with the following suggestions! • Reduce the temperature of your home by turning down the thermostat • Take shorter showers. Purchase low-flow faucets and shower heads. Make use of a more recent and more efficient model • Find a tank that is the right size for your needs • Make sure your tank is properly insulated. ## Reducing Water Heater Electricity Usage One method of conserving power on the water heater is to reduce the temperature setting on the unit. Because you won’t have to reheat this item as frequently if you keep it below 120 degrees Fahrenheit, lowering the temperature can help you save money on your energy bill. A few other options include installing an insulation blanket around the tank, upgrading your old model and choosing a unit that has been certified as Energy Star compliant, or switching over completely to an instantaneous heating of water system, which eliminates the need to wait for hot showers because they are always ready when needed! ## Conclusion Your water heater has the potential to be a significant drain on your electrical budget. Consider using less hot water in order to save money on your power bill. Shower heads and faucets with low flow rates are a good investment to make to limit the quantity of cold water you use for showers and laundry; turn off the water when brushing your teeth or shaving; and take shorter showers – especially if there are more people living with you than just yourself! If your home relies heavily on electric heating during the winter months, it is critical that any outdated appliances be replaced as soon as they reach the end of their useful life to ensure that efficiency stays high. ## Electricity usage of a Water Heater A water heater is a device that warms water in your house so that you may have hot water for your everyday requirements. Electric water heaters are normally used to heat water for 3 hours per day; however, newer, more energy-efficient types may only be used for half of that time every day. A typical water heater will consume around 4000 watts of electricity. Calculate the energy consumption of a water heater running at 4000 Watts for 3 hours per day at \$0.10 per kWh by clicking on the calculate button. In the above example, 30 minutes every day is equal to 0.5. Our calculators utilize the default value of 0.10 or 10 cents per kilowatt hour, therefore enter the average cost per kilowatt hour that you are now paying. Hot water use in a family is around 45 gallons per day on average; therefore, lowering your hot water consumption can help you save money on power. Also consider upgrading to a more contemporary and energy-efficient water heater, especially if your current one is more than a decade old. ## How Much Does the Hot Water Heater Affect an Electric Bill? Water heaters make it easy to do a variety of tasks such as taking a hot shower, washing your hands with warm water, and many more simply by turning on the hot water tap. Hot water heating tanks, on the other hand, consume a substantial amount of energy, making them prohibitively expensive to operate. Knowing how much your hot water heater contributes to your power bill may enable you to take steps to reduce your water heating expenditures. ## Electricity Consumption According to the United States Department of Energy, water heating systems are the second most significant consumer of energy in the house, accounting for an average of 18 percent of total power expenses. EF (energy factor) is used to assess the efficiency of water heaters. The greater the value of EF, the more energy-efficient the heater. EF is a rating that indicates how much hot water may be produced daily per unit of fuel. The efficiency factor (EF) of electric heaters ranges from 0.75 to 0.95. ## Calculating Energy Usage The typical water heater operates for three hours every day. Using an electric water heater with a 50-gallon capacity and a 5.500-watt output at 0.90 EF and an electricity rate of \$.16 per kWh, you will spend \$781 per year to run it. The annual running cost of most water heaters is listed on a label attached to the unit. ## Reducing Water Heater Electricity Usage Reduce the temperature of the water heater to 120 degrees Fahrenheit, install an insulation blanket around the tank with a “r”-value rating recommended by the manufacturer, upgrade to an Energy Star-rated unit, or switch to an instantaneous water heating system to reduce the amount of electricity consumed by the water heater. Lowering the water heater’s thermostat to 120 degrees Fahrenheit, installing an insulation blanket around the tank with a “r”-value rating recommended by the manufacturer, upgrading to an Energy Star-rated unit, or switching to an instantaneous water heating system are all options for lowering the amount of electricity consumed by the water heater. ##### Why American hot water heaters use so much energy Almost every home in the United States is equipped with either a traditional gas or an electric hot water heater. However, despite the fact that the first hot water heaters were built more than a century ago, today’s hot water heaters, whether in your laundry room or basement, aren’t all that unlike from the initial versions designed in the late 1800s. They function in a rather easy manner. Electricity or natural gas is used to heat a large metal tank full of water, which is then circulated around your home. That means that even while you’re away on vacation or asleep, your hot water heater is still working to heat water. Even more astonishing, if our water heaters were a country, their total greenhouse gas emissions would be more than the whole nation of Belgium. However, in contrast to many other environmental challenges, the remedy to this one is quite simple. And if you’re a homeowner, you have a great deal of power over the situation. ##### How other countries heat their water In case you’ve ever visited to Europe, you may have observed that the hot water heaters there are distinct in appearance from the ones you have at home. Tankless water heaters are used in the majority of European households. Compared to typical tanks, they utilize electricity or gas to heat water, but instead of heating water at all hours of the day, they only heat water when it’s necessary. It should come as no surprise that these water heaters are far more efficient than conventional tanks. See also:  How To Protect An Outdoor Tankless Water Heater In the United States, tankless water heaters are available. However, there is an even more efficient solution that has been accessible in the last decade, and in many places, the government will reimburse you up to \$750 for the cost of installation. ##### Make another easy change – switch to clean energy. See if there’s any availability. ##### How heat pump (hybrid) water heaters work The heat pump water heater is the most energy-efficient water heater available to homes in the United States today (sometimes called a hybrid water heater). When it comes to heating water, heat pump water heaters draw heat from the surrounding air rather than depending on electricity or gas to do so. Or to put it another way, they transport energy rather than creating it. “Heat pumps operate in the same way that a refrigerator does, but in reverse,” according to Department of Energy experts. In contrast to a refrigerator, which draws heat from within a box and dumps it into the surrounding room, a standalone air-source heat pump water heater pulls heat from the surrounding air and dumps it into a tank, where it is heated to a higher temperature, thereby heating water.” Most heat pump water heaters now feature an electric resistance heater as a backup in case the surrounding air temperature isn’t warm enough to operate the heat pump water heater. It is for this reason that they are referred to as hybrid heat pumps. ##### How hybrid water heaters cut costs and emissions Heat pumps, with the exception of solar water heaters (which can cost between \$3,000 and \$10,000 in the United States), are the most energy-efficient water heaters currently on the market. As a result, they are also the most environmentally friendly. The carbon footprint of these tanks can be anywhere from two to four times lower than that of a traditional tank. That is why environmental organizations such as the Natural Resources Defense Council and the Rocky Mountain Institute like them. As a result of the importance of heat pumps in combating climate change, the government has agreed to contribute a portion of the cost. • Some states, such as Maine, will provide you a \$750 immediate reimbursement if you qualify. • To find out whether you qualify for any rebates or incentives, go to the Database of State Incentives for Renewable Energy to see if you qualify. • Anyone who is familiar with the subject of climate change is aware that there are several difficult difficulties to overcome. • However, some of the options — such as converting your home’s energy source to renewables or replacing your inefficient water heater — are far less difficult to implement. And, given that American houses release as much carbon as the majority of developed countries, such remedies have the potential to make a significant difference. ##### Make another easy change – switch to clean energy. See if there’s any availability. ## Electric Hot Water Heater Cost Per Month It is estimated that your water heater is the second highest consumer of energy in your house, accounting for around 17 percent of overall energy consumption, according to the Department of Energy. With this in mind, it is good to be aware of how much energy your electric hot water heater consumes and how much it costs you on a monthly basis. You will be able to make the most efficient use of your water heater and lower your power cost if you have this knowledge. ### Are electric water heaters expensive to run? It is estimated that the average American home spends between \$400 and \$600 per year on water heating alone, according to the Department of Energy. This cost will vary based on the current power prices in your location as well as the efficiency of your water heating equipment. The quantity of energy consumed by an electric water heater is determined by a number of factors, including: • The age of the unit, the size of the unit, the kind of electric water heater, the temperature setting, and the amount of water consumed each day are all important considerations. It is particularly crucial to consider the sort of electric heater you choose when it comes to operating costs. An overview of the many alternatives for electric hot water heating systems is provided below. ### Storage Water Heaters Storage heaters are often the least expensive option when it comes to the cost of the system itself, as well as having relatively low operational expenses compared to other options. There is an issue with this method in that heat energy is lost when the tank attempts to keep the water hot even when it isn’t being utilized, resulting in you being charged for energy that isn’t being used. If you are considering this option, you should search for a model that is well-insulated to avoid this from happening. ### Tankless Water Heaters However, according to the Department of Energy, tankless water heaters are between 8 percent and 34 percent more energy efficient than storage water heaters, depending on how much water is being utilized. This implies that upgrading to this type of water heater might result in savings of up to \$100 or more each year. The disadvantage is that, depending on the type and the electrics in your house, the initial purchase and installation expenses may not be worth it in terms of energy savings until a long time after you have purchased and installed the system. ### Heat Pump Heating Systems In comparison to storage heaters, heat pump systems are two to three times more energy efficient, resulting in lower running costs. Models that are Energy Star certified might save you up to \$300 per year on your energy expenses, according to the company. Performance, on the other hand, might vary depending on where the system is situated in your home, as they must maintain a specific temperature all year round. They also require a significant amount of area, both for the pump itself and for the surrounding air space. ### Can a water heater cause a higher electric bill? There are a variety of reasons why your water heater may be contributing to an increase in your power cost. First and foremost, if you have your electric hot water heater set to a higher temperature than is necessary, your heater will consume more energy, resulting in higher energy bills over time. We’ll look at what we can do to fix this later on. The size of your home’s water heater might also be an issue, as it could be either too huge or not large enough. If your water heater is too large, it will simply waste energy by heating water that will never be used all at once, resulting in you spending more than you should for your water. Even if this does not immediately increase your power bill, you will wind up paying for maintenance and repairs on a more frequent basis as a result of this. If your water heater is too old, you will also notice that your power cost will rise as a result of the increased usage. Most electric water heaters are only designed to last for 10 to 15 years, after which their efficiency begins to deteriorate substantially, causing your energy expenses to rise significantly. ### What is the average cost of an electric hot water heater? The usual cost of an electric hot water heater varies depending on the model, but it is often between \$300 and \$700, with additional installation charges ranging between \$700 and \$1,000. This implies that depending on the size of the system and the type of water heater you pick, you may expect to pay a total of between \$1,000 and \$1,700. ### Are new electric water heaters more efficient? With installation fees ranging between \$700 and \$1,000, the typical cost of an electric hot water heater depends on the model and can range from \$300 to \$700. Depending on the size of your system and the type of water heater you pick, you may expect to pay a total of between \$1,000 and \$1,700 for your system. ### Should I turn off my water heater at night? It is not necessary to worry about turning off your tankless water heater at night if you have one because they are intended to only heat the water when it is truly needed. In the event that you have a tankwater heater, you should consider shutting it off not just at night, but also whenever it is not needed for an extended length of time, as this may help you save money on your energy bill. This sort of water heating system just warms up the entire tank of water and works to keep it at the desired temperature until it is required again in the future. Stopping the flow of heat and the few bucks a day that are escaping along with it can be prevented by turning off your water heater! ### Does turning down the water heater save money? The majority of water heaters are set to a default temperature of 140 degrees Fahrenheit since this is about the temperature at which dishwashers used to operate and clean dishes well. For most other home purposes, a hot water temperature of 120 degrees Fahrenheit is sufficient, and dishwashers are now equipped with heating boosters that enable them to get the hot water up to temperature on their own. You will obtain the most efficient temperature from your water heater by setting it at 120 degrees Fahrenheit. ### How much does it cost to run a 50-gallon electric water heater? Following the Department of Energy’s assumption that an electric water heater is used for approximately three hours per day, a 50-gallon water heater operating at 5,500 watts with an electricity rate of \$0.16 per kWh will have an annual operating cost of \$781 if the water heater is used for approximately three hours per day. Depending on how efficient the water heater is and how much power is being used in your location, this cost will vary. ### Are electric water heaters worth it? Electric water heaters are a popular choice for many families since the initial expenses are very modest when compared to other choices. However, installation prices vary depending on the kind of water heater purchased and installed. Because they are one of the most energy efficient kinds of water heaters and because, unlike gas heaters, they can be powered by renewable energy sources, they are also significantly better for the environment than other water heating choices. Additionally, electric water heaters are readily available to all people who are linked to the electric grid, but other choices, such as natural gas, are not readily available to some. Electric water heaters are an excellent choice if you have the financial means to invest a little extra money on a newer, more energy-efficient model. The most effective approach to keep your power bills low while also contributing to the fight against climate change is to switch to a renewable energy plan with Inspire Clean Energy. We will also give you with smart tools to assist you in managing your electricity consumption. More information may be found by clicking here. ## How much electricity does a gas hot water heater use? Asked in the following category: General The most recent update was made on the 27th of January, 2020. A hot water heater that operates on a tank will typically operate for three to five hours every day. Consequently, at \$.10 per kWh, a 4,000-watt theater that is utilized for three hours per day will cost \$1.20 per day, approximately \$36.50 per month, or \$438 per year. Tank for Gas in the Conventional Style Thermoelectric Water Heater Gas water heaters do not utilize electricity as a fuel, and many homeowners believe that they will continue to operate in the event of a power loss. • Even gas water heaters with electric pilot lights can continue to operate since they do not require continuous access to mains power to function. • The wattage of a natural gas water heater. • As a result, how much does it cost to operate a gas-powered hot water heater? • Of course, this is dependent on utility prices. • Gas water heaters are often less expensive to operate than electric water heaters, depending on your local utility bills. • In contrast, based on the amount of energy saved, gas heaters often make up for the difference in price within one year. ## Is Your Hot Water Heater Running Up Your Electric Bill? Is there anyone who doesn’t like a good hot shower or a lengthy hot bath? Your hot water heater, on the other hand, consumes a lot of energy. Your home’s water heating systems are the second largest consumer of power behind your refrigerator. According to the United States Department of Energy, this accounts for an average of 18 percent of your total power expenses. You’re probably wondering how you may save money on your power bill without having to take ice cold showers every day. Here are six methods to help you reduce the energy consumption of your heater and save a few bucks per month: Reduce the temperature of the water in your hot water tank. 1. Dispose of the sediment in your hot water tank It is surprising how many people are not aware that silt building in their heating system might cause the heating components to perform less efficiently. 2. Insulate the hot water tank in your home. 3. In most hardware and big box stores, you may find an insulating jacket to purchase. 4. The majority of the equipment is constantly heating water throughout the day and night. 5. This can save you up to 12 percent on the energy bills associated with your heater. 6. Full Loads of Laundry Using cold water in the washing might result in significant cost savings. 7. Reduce your power expenditure by washing just full loads and using the cold setting on your washing machine. 8. It is possible to save a significant amount of energy by using a tankless hot water heater or an on-demand hot water heater. 9. The use of a tankless water heater may provide an infinite supply of hot water whenever you need it—even while you are performing many jobs at the same time. There will be no more chilly showers or running out of hot water anymore. The constant supply of hot water will be appreciated by the entire family. Ohio Heating provides sales and servicing for hot water heaters. • Proper tank sizing evaluation • Professional installation • Removal of old equipment • Service for all makes and models • Emergency service available See also:  How To Choose A Hot Water Heater Please contact us at 614.863.6666. Alternatively, you may complete the form on the left. Never again will you be without hot water. Tankless hot water heaters, on-demand hot water heaters, hot water heaters, and hot water tanks are some of the topics covered. ## How Much Power Does a Water Heater Use? The cost of energy, specifically electricity and water, is the most pressing concern for all households. If you have a water heater (or are considering purchasing one), you may be interested in learning how much energy a water heater consumes on a daily basis. The size of your family, the amount of warm water you use personally, and the kind and brand of your water heater all play a role in how long it takes to heat up the water. According to the California Energy Commission, heating water accounts for 25 percent of a typical household’s average energy bill, and electric water heaters are substantially more expensive than their gas or fuel-powered equivalents. This information can be found on the label of the gadget in question. ### How Much Watts Heaters Consume and How It Affects Our Energy Bills The wattage of an electric water heater is frequently printed on the unit’s label, which is usually positioned on the side of the unit’s exterior. Due to the fact that it is attached adjacent to the panel that users must remove in order to replace the heating element, and because that specific panel must be accessible at all times, this label should be easily visible. It displays the wattage and operating voltage of the element, and if, for example, the device runs on 240 volts of electricity, it is likely to have two elements, so you will be looking at two figures for the wattage, one for each of the components. 1. 120 volt heaters consume 1,125 watts, but domestic 240-volt heaters consume 4,500 watts on average. 2. This is not truly correct since the two elements do not occur at the same time in the same place. 3. We’ll talk about how much of an impact these heating units have on our energy expenses later. 4. However, they waste a lot of energy, which makes utilizing them expensive. 5. The efficiency of these devices is measured in EF, or Energy Factor, and the greater the EF, the more efficient the heater. 6. The range of EF for these devices is between 0.75 to 0.95, depending on the device. 7. Water heaters that are used on a daily basis run for three hours. The majority of these gadgets are equipped with a label that documents the annual operating expense. ### Tips for Decreasing Water Heater Energy Use We must understand how to heat our water supply effectively if we want to reduce the amount of money we spend on water heaters. The following are some suggestions for lowering your expenditures as much as possible. • Install a timer that will allow you to turn off the device at night or during other times when you aren’t actively using it. You might also try turning it off during the times when the gadget is under the most stress. Consider purchasing a new water heater, since there are modern, more energy-efficient ones on the market today • Make use of more Energy Star dishwashers and water heaters if you have the opportunity. By conserving energy and utilizing this type of equipment, you may save hundreds of dollars every year on your utility bills. The Energy Star logo is a symbol that is used to denote energy efficiency in products. When a product is assigned this grade, the manufacturer has determined that it is much more energy efficient than the minimal regulatory criteria as determined by traditional testing procedures. It is dependent on the technology available on how much an appliance should exceed the minimum criteria. The amount by which an appliance should exceed the minimal standards vary for each product granted a rating. ## Energy Cost Calculator for Electric and Gas Water Heaters • The “base model” has an efficiency that just about meets the national minimum level for gas and electric water heaters • However, the “base model” has an efficiency that exceeds the national minimum standard for gas and electric water heaters. When considering a water heater with an anticipated life of 13 years, the lifetime energy cost is the total of the discounted value of the yearly energy expenditures. The Federal recommendations for future power price trends and a discount rate of 3.2 percent are used to calculate the discount rate. The Federal average price for electricity in the United States is \$0.09 per kWh • The Federal average price for natural gas in the United States is \$0.93 per therm. Estimates of hot water consumption: • Taking an average shower (8 minutes) uses 10 gallons of hot water • Running an average clothes washer (one load) uses 7 gallons of hot water • Running an average dishwasher (one load) uses 6 gallons of hot water The average kitchen faucet flow rate is 2 gpm, while the average bathroom faucet flow rate is 0.5 gpm • The average daily hot water consumption rate is 64 gallons. #### Disclaimer With the help of our cost calculator, you can estimate the amount of money you’ll save on energy over the course of a product’s lifetime at various degrees of efficiency. Maintenance and installation costs do not differ considerably across products with varying levels of efficiency, and as a result, these expenses are not included in the calculations made by this calculator tool. The Building Life Cycle Cost Study tool, created by the Federal Energy Management Program, may be used to do a complete life cycle cost analysis (BLCC). ## Tankless or Demand-Type Water Heaters With the help of our cost calculator, you can estimate the amount of money you’ll save on energy over the course of a product’s lifetime at various levels of performance. Maintenance and installation costs do not differ considerably across products with varying levels of efficiency, and as a result, these expenses are not included in the calculations made using this calculator. Building Life Cycle Cost Study is a technique created by the Federal Energy Management Program to do a complete life cycle cost analysis of a building (BLCC). ## How They Work Tankless water heaters provide fast heating of water without the need for a storage tank. When a hot water faucet is switched on, cold water is sent through a heat exchanger in the unit, where it is heated by either a natural gas burner or an electric element, depending on the device. Consequently, tankless water heaters are able to provide a continuous supply of hot water. The need to wait for a storage tank to fill up with adequate hot water is no longer an issue. The output of a tankless water heater, on the other hand, is limited in terms of flow rate. Tankless water heaters that run on natural gas have higher flow rates than those that run on electricity. For example, having a shower while also running the dishwasher at the same time might cause a tankless water heater to reach its maximum capacity quickly. You may also install separate tankless water heaters for equipment in your house that need a lot of hot water, such as a clothes washer or dishwater. Additional water heaters, on the other hand, will be more expensive and may not be worth the additional expense. Demand water heaters are also used in the following other situations: • Bathrooms or hot tubs in a remote location • Increases the efficiency of household appliances such as dishwashers and laundry washers. Thermoelectric booster for a solar water heating system Demand water heaters can be 24–34 percent more energy efficient than typical storage tank water heaters in residences that utilize 41 gallons or less of hot water per day on average. For houses that utilize a lot of hot water – around 86 gallons per day – they can be 8 percent to 14 percent more energy efficient than standard models. If you install a demand water heater at each hot water outlet, you may be able to achieve even larger energy savings in some circumstances. A tankless water heater will cost more up front than a normal storage water heater, but they will often live longer and have lower operating and energy expenses, which may more than compensate for their higher purchase price in the long run. 1. They also feature readily changeable parts, which might potentially increase their lifespan by many years. 2. With tankless water heaters, you won’t have to worry about the standby heat losses that come with traditional storage water heaters. 3. When compared to a storage water heater, the removal of standby energy losses might sometimes outweigh the savings from using a tankless water heater. 4. A tankless water heater’s pilot light has a cost associated with it that differs from one type to the next. 5. Instead of a standing pilot light, look for versions that contain an intermittent ignition device (IID). ## Selecting a Demand Water Heater Before purchasing a demand water heater, you should take the following factors into consideration: • Consider the following factors as well when purchasing a demand water heater: ## Installation and Maintenance It is possible to maximize the energy efficiency of your demand water heater with proper installation and maintenance. A variety of elements influence the success of an installation. These considerations include the type of fuel used, the environment, the needs of local construction codes, and safety concerns, particularly with regard to the combustion of gas-fired water heaters. As a result, it is recommended that you use a licensed plumbing and heating professional to install your demand water heater. • Energy efficiency of your demand water heater may be maximized through proper installation and maintenance procedures. A variety of elements influence the quality of the installation. These considerations include the type of fuel used, the environment, the needs of local construction codes, and safety concerns, particularly with regard to the combustion of gas-fired water heaters, among others. In order to ensure proper installation of your demand water heater, use an experienced plumbing and heating professional. When picking a contractor, keep the following points in mind. See also:  How To Hook Up Electric Water Heater If you’re determined to install your water heater yourself, first speak with the manufacturer about the best way to proceed. The relevant installation and instruction manuals are normally available from the manufacturer. Contact your municipality for information on acquiring a permit (if one is required) and on water heater installation codes in your area. Periodic water heater maintenance may considerably increase the life of your water heater while also reducing the amount of energy it consumes. Seek advice from your owner’s handbook on particular maintenance requirements. ## Improving Energy Efficiency If you’re determined to install your water heater yourself, first speak with the manufacturer about the best method to use. Installer and instruction manuals are often available from the manufacturer. Contact your municipality for information on acquiring a permit (if one is required) and on water heater installation codes in your particular area. The use of periodic water heater maintenance may considerably increase the life and efficiency of your water heater while also lowering energy costs. ## Will My Water Heater Work During a Power Outage? If you are determined to install your water heater yourself, speak with the manufacturer first. The necessary installation and instruction manuals are often available from the manufacturer. Also, check with your city or municipality for information on acquiring a permit, if one is required, as well as on local water heater installation regulations. Periodic water heater maintenance may considerably increase the life of your water heater while also reducing energy use. For precise maintenance suggestions, consult your owner’s handbook. #### Different Types and Fuel Supply of Water Heaters The heating of water will be halted if you have a typical tank-style water heater that is driven by electricity in the case of a power failure. However, water that has already been heated to the point of boiling when the power goes off will continue to stay warm for a period of time while being kept in the insulated tank. During a power outage, it may be beneficial to cut off the power and water supply to the tank water heater in order to preserve the remaining hot water as hot as possible while the electricity is out. ###### Gas Conventional Tank Water Heater Because gas water heaters do not require electricity as a fuel, many homeowners believe that they will continue to operate after a power loss. This is dependent on the sort of gas water heater you have. Unless your gas water heater is equipped with a continuous-gas pilot light, there is a good chance that it will continue to operate regularly even if the electricity goes off. Due to the fact that they do not necessarily rely on mains energy, even gas water heaters with electric pilot lights can continue to operate. It is important to remember to cut off your gas supply in the event of a power loss. ###### Electric On-Demand Water Heater The fuel for gas water heaters is not electricity, and many people believe that they will continue to operate in the event of a power loss. The sort of gas water heater you have will determine this. A gas water heater with a continuous gas pilot light is more likely to continue to operate normally in the case of a power loss than one without such a pilot light. Due to the fact that they do not rely on mains electricity, even gas water heaters with electric pilot lights can continue to operate. If this is the case, make sure to shut down your gas supply in the event of a power loss. ###### Gas On-Demand Water Heater Even though on-demand gas water heaters do not use electricity as a fuel source to heat water, on-demand heaters typically feature a control panel that is powered by electricity, which serves as the water heater’s “brains.” Therefore, even a tankless gas water heater will not function independently in the event of a power failure. ###### Other Water Heaters (Solar, Fuel Oil, Heat Coil, Indirect) Whether or not your water heater will function during a power outage is determined by whether or not it is powered by mains energy. In order to determine whether or not a power outlet is there, just inspect the water heater for an electrical connection. If power outages are a worry for your household, you should exercise caution when selecting a residential water heater. For additional information, contact Magnificent Plumbing, an expert local plumber who can assist you in determining the finest water heater to meet your needs. ## How Much Electricity Does A Tankless Water Heater Use? (Per Hour, Month, Year) “Water heaters account for roughly 17 percent of a home’s total energy consumption, consuming more energy than all other household equipment combined,” says the report. (Department of Energy on water heaters) An enormous quantity of electrical power is required by electric tankless water heaters (when running). It may come as a surprise, but they are extremely energy efficient and cost-effective. With the help of the power equationP = I * Vand and the cost equationCost = kWh * t (h), we will determine exactly how much electricity is consumed by a tankless water heater. • The rate is per hour. On average, a modest 8 kW heater costs \$1,06 per month, whereas a large 30 kW heater costs \$3,96 per month. Between \$18,55 to \$69,03 (see more in the table below), with an average of \$43,94 each month for a four-person household over the course of a year. Between \$222,60 to \$831,60 (see more in the table below), with an average of \$527,10 a year for a four-person household on average Let’s face it: heating water is one of the most energy-intensive tasks you can perform. Tankless water heaters heat water in a matter of seconds (15 seconds from cold to hot water). In order to accomplish this, we will require an enormous quantity of power. Does this imply that tankless heaters consume a significant amount of electricity? Both yes and no. Electrical tankless heaters may consume up to 30,000 watts of power when they are turned on. For instance, a typical washing machine would need slightly more than 1,000W to operate properly (and it will run for 1-2h). • All of this energy is utilized to heat water, and we only consume hot water when we need it. • You can find out if tankless water heaters are truly worth it and whether or not they save you money by visiting this site. • In houses with daily hot water consumption of 41 gallons or fewer, demand water heaters can be 24 percent to 34 percent more energy efficient than standard storage tank water heaters. • Saving 24 percent to 34 percent on energy costs is a significant amount of money. • But how can a power-hungry equipment like this help you save money on your electricity bill? Because the voltage is always the same (220/240V), it all boils down to how many amps a tankless water heater consumes. Let’s perform a little math to figure out how much power a tankless hot water heater that heats water on demand consumes: ## How Much Does It Cost To Run An Electric Tankless Water Heater? It is rather simple to calculate the maximum operating expenses of tankless water heaters. The most important piece of information to know is how many watts a tankless water heater consumes. We’ll start with a simple example, with additional examples to come. Example: You have an electric tankless heater with a capacity of 20 kW. How much does it cost to keep it up and running? Solution: If you operate a 20 kW heater for an hour, it will consume around 20 kWh of power. The average cost per kWh in the United States is \$0,1319. That doesn’t seem like much, does it? The greatest amount of power used by a 20 kW tankless water heater is \$2,64 per hour at the maximum rate (running at 100 percent heating output). Now, the main question is: how many hours each day do you use the tankless water heater in question? • A 20 kW generator running at 100 percent output for 10 minutes (shower) costs \$0.66 • A 20 kW generator running at 100 percent output for 30 minutes costs \$0.66 1 hour at 100 percent power from a 20 kW generator costs \$1,32. 2 hours at 100 percent output from a 20 kW generator costs \$2,64: 5 hours at 100 percent power from a 20-kilowatt generator costs \$5,28. \$13,20 Please be aware that the price of power is always predetermined to be \$0,1319 per kilowatt-hour by us. Not everyone has access to a tankless water heater with a capacity of 20 kW. When it comes to the use of electric tankless water heaters, there are two very essential figures to consider. • Approximately how much power is consumed by a tankless water heater each hour • How much power does a tankless water heater consume in a single month? Electricity bill for the month ### Tankless Water Heater Electricity Usage By Hour When it comes to determining the amount of power used by tankless water heaters each hour, there are just two variables to consider. These are the ones: 1. What kind of wattage does a tankless water heater consume 2. What is the cost of power (in terms of kWh) in your neighborhood? For the second time, we assume that the average price of electricity is \$0,1319 per kWh and that the tankless heater is producing 100 percent of its heating capacity. Listed below is a table containing the maximum operating expenses per hour for different sizes of electric tankless heaters: WattageOf Electric Tankless Heater Cost Per RunningHour 8 kW \$1,06 9 kW \$1,19 10 kW \$1,32 11 kW \$1,45 12 kW \$1,58 13 kW \$1,71 14 kW \$1,85 15 kW \$1,98 16 kW \$2,11 17 kW \$2,24 18 kW \$2,37 19 kW \$2,51 20 kW \$2,64 21 kW \$2,77 22 kW \$2,90 23 kW \$3,03 24 kW \$3,17 25 kW \$3,30 26 kW \$3,43 27 kW \$3,56 28 kW \$3,69 29 kW \$3,83 30 kW \$3,96 Take note that these figures are only relevant if an electric tankless water heater operates at 100 percent of its maximum heating capacity. The majority of the time, it operates at less than 100 percent. This information is useful for determining how much power a tankless water heater consumes on a monthly basis: ### Monthly Tankless Water Heater Electricity Usage The most true picture of how much power is consumed by an electric tankless water heater may be seen in the monthly electricity bill. That is why it may be beneficial to compare the amount of power used by different tankless water heaters on a monthly basis. In this case, the main difficulty is that it’s quite difficult to determine how many hours per day we use the tankless heater and what its average output is (keep in mind that it doesn’t always operate at maximum capacity). We may make a fairly educated guess using the following formula: At 100 percent heating output, an average family of 3-7 people will utilize theequivalent of 35 minutes of tankless water heater use every day. Additionally, the average energy efficiency of a tankless water heater, which is 97 percent, is taken into consideration. The following calculations may be made using this assumption and an energy cost of \$0.1319 per kWh to determine the cost of operating an electric tankless heater per month and per year: WattageOf Electric Tankless Heater CostPer Month(Rough Estimate) CostPer Year(Rough Estimate) 8 kW \$18,55 \$222,60 9 kW \$20,83 \$249,90 10 kW \$23,10 \$277,20 11 kW \$25,38 \$304,50 12 kW \$27,65 \$331,80 13 kW \$29,93 \$359,10 14 kW \$32,38 \$388,50 15 kW \$34,65 \$415,80 16 kW \$36,93 \$443,10 17 kW \$39,20 \$470,40 18 kW \$41,48 \$497,70 19 kW \$43,93 \$527,10 20 kW \$46,20 \$554,40 21 kW \$48,48 \$581,70 22 kW \$50,75 \$609,00 23 kW \$53,03 \$636,30 24 kW \$55,48 \$665,70 25 kW \$57,75 \$693,00 26 kW \$60,03 \$720,30 27 kW \$62,30 \$747,60 28 kW \$64,58 \$774,90 29 kW \$67,03 \$804,3 30 kW \$69,03 \$831,60 The cost of operating an electric tankless heater is determined by a variety of factors, all of which must be taken into consideration. Nonetheless, according to the Department of Energy, tankless water heaters consume less electricity than traditional water heaters that are powered by electric current. Look for the Energy Star certification to see if you can cut your power cost as much as feasible. You will see that the units on our list of the best and most energy-efficient electric tankless heaters here may achieve up to 99 percent energy efficiency, which is impressive.

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Question # Print Armstrong number in an (Array for Java/ List for Python) ``` ``` A positive number is called armstrong number if  sum of its own digits each raised to the power of the number of digits is equal to number 153 1^3+5^3+3^3=1+125+27=153 153 is an armstrong number ``` ``` ``` ``` Enter the size of list:10 enter element in given list: 1:30 2:20 3:1 4:153 5:99 6:107 7:19 8:145 9:131 10:3 Elements in given List: [30, 20, 1, 153, 99, 107, 19, 145, 131, 3] Armstrong number in given List: 1 153 3 ``` ``` Code Sale 50 Sale 150 Sale 200 Sale 50 Sale 50 Sale 50 Sale 50 Sale 50 Sale 50

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# Thread: 2 Sample T Test for Nonnormal data 1. ## 2 Sample T Test for Nonnormal data Hi everyone! I work in the advertising industry and I am in the process of creating a t test calculator. The calculator will be used to test the statistical differences between two different advertisements, two campaigns, two web pages, etc. I've made a click through rate significance calculator (using a Bernoulli distribution) and a calculator for the average order value (normal distribution, so straight forward 2 sample t test). I'm trying to make a revenue per visit calculator now, but I am stuck on what to do! The vast majority of visitors to a website will not purchase anything, hence they will have a revenue value equal to zero. Since most visitors will have revenue equal to zero, the distribution will be heavily skewed at zero. The sample sizes should be quite large (n>1000). I'm at a loss for how to formulate a hypothesis test for this metric, any advice would be much appreciated! Thanks! 2. ## Re: 2 Sample T Test for Nonnormal data t tests are for testing if it is true that there is a specific difference (usually 0) between the two means. I guess you are assuming that there is a difference so you aren't trying to disprove that the difference in revenue is 0. If you want to show that once advertising scheme is better than another then that is a different test. Confidence intervals would be a better approach. The analysis is a bit long winded and some people abandon their mathhelpforum accounts after a couple days so if you can confirm that you are still looking for an answer then I'll post the analysis. 3. ## Re: 2 Sample T Test for Nonnormal data Hi thanks for your reply. I think you misunderstood my question. I am in fact trying to prove that two advertisements (i.e. two samples) are different from one another. The problem is that most users who visit a site do not purchase, hence they have a revenue value that equals to zero. So, when looking at the samples, the data will not be normal (they will be heavily skewed at zero) thus standard 2 sample t tests are not valid. I need a workaround to this, probably a non parametric t test. Would love to hear your thoughts. Thanks 4. ## Re: 2 Sample T Test for Nonnormal data Right this took quite a while to figure out so I hope you can make use of it. It was quite interesting for me too, its certainly the longest mathhelpforum answer I've ever given. You seem quite well versed in statistics so I'm assuming you understand the concept of a confidence interval. You can split up your distribution into a binomial distribution to model them buying something or not, and a normal distribution to model the amount people spend if they are buying something. The amount a website visitor spends would be the product of the distributions. Since your sample size is large you can use a normal distribution instead of a t distribution. Lets say you have 2 samples sets to compare for two advertisements, set A and set B. Find the confidence interval for the mean of the amount that buyers spend for A and B like you usually would. For the binomial distribution I denote the true chance of someone buying something $\theta$ and your sample chance of someone buying something p. The variance of a binomial distribution is $\sigma^2 = \theta(1-\theta)$ And likewise $s^2 = p(1-p)$ With your sample size you shouldn't worry much about the sample standard deviation being inaccurate. You can use the normal approximation to the binomial distribution. It has the same shape as the binomial but the bumps are smoothed out and it extends to infinity. The normal approximation is like the histogram on the left but there is a part of the graph below 0 which should not be there, obviously the probability cannot be below zero. But when you remove this area it doesn't just vanish, if it did then all the probabilities wouldn't sum to 1. The removed area gets spread out across the whole histogram and that is why the graph on the right is taller. Say an area k is below 0, when you remove k the rest of the graph's height increases by a factor of $\frac{1}{1-k}$ The normal approximation to the binomial distribution can be used for your sample size. It is commonly used when finding confidence intervals, the problem is that with your chance of success being so low, when you compute the lower limit of the confidence interval it could be negative and obviously you can't have a negative chance. The lower limit of the confidence interval is given by $p-Z\sigma\sqrt{n}$. With this is mind there are two cases for your interval. The first case is simple, when the lower limit is non negative $p-Z\cdot\sqrt{n} \geq 0$ $Z\sigma\sqrt{n} \leq \frac{p\sqrt{n}}{s}$ Find Z as you usually would for a confidence interval for your significance $\alpha$ and check if it satisfies $Z\sigma\sqrt{n} \leq \frac{p\sqrt{n}}{s}$ The part of the graph missing below 0 will barely effect your confidence interval if $Z\sigma\sqrt{n} \leq \frac{p\sqrt{n}}{s}$ so you can construct your confidence interval as usual. If $Z\sigma\sqrt{n} \geq \frac{p\sqrt{n}}{s}$ then you have a more complicated problem on your hands-case 2. The second case accounts for the missing part of the graph. This is a truncated normal distribution Truncated normal distribution - Wikipedia, the free encyclopedia. In theory the missing part could be as large 50% of the graph depending on how close your sample mean is to 0. Note that in this the cumulative density function is used quite a bit. The symbol for it isn't available on these forums so I will just use $C^L(t)$ to indicate the left tail cumulative density function- this is the total probability from minus infinity up to t standard deviations from the mean. You can think of it as the area under the graph between minus infinity and some vertical line at a point t. I believe most tables use left CDF. I will also use $C^R(t)$ to indicate the right tail CDF, this is similar except its from plus infinity to a point t. $C^L(-t)=C^R(t)$ When finding Z scores for a confidence interval you are getting them from a standard normal distribution with $\sigma=1, \mu=0$. Here the confidence interval for a significance $\alpha$ lies in the range $L^-$ to $L^+$. L is the number of standard deviations it is away from the mean 0. The two unshaded tails both have an area of $\frac{\alpha}{2}$ so they have a total area of $\alpha$. The confidence interval covers all but $\alpha$ of the area which is what is meant when you say you have a significance of $\alpha$. But just as the earlier graph was adjusted for removing all the area below 0 the standard normal distribution must also be adjusted. So how much is k? From the graph two above you can see that the distance 0 is away from the mean is $\theta-0$ which corresponds to a distance of $\frac{\theta}{s}$ standard deviations away from the mean. In the standard normal distribution the part chopped off the graph is also $\frac{\theta}{s}$ standard deviations away from the mean. Because the point is below the mean it is at a point $\frac{-\theta}{s}$ from the mean. The area k is equal to $C_L(\frac{-\theta}{s})$. In the transformation since every part of the graph went up to compensate for removing k it is clear that $C^R(Q)=(1-k)C^R(Q')$ where Q is a point on the graph before the transformation and Q' is that same point (same number of standard deviations from the mean) after the transformation. This applies to both the graph of your distribution and the graph for the standard normal distribution. Since the lower end of your confidence interval is equal to zero your entire confidence interval on the graph will look like this With the shaded region covering the area of the confidence interval. Note that the distance from $x\bar$ to 0 is less than the distance from $x\bar$ even if my image doesn't look like it. Relating this to the standard normal distribution which we have transformed u is the upper limit and v is the lower limit. We already know v, we need to find u. From the image you can see that it is a one tail confidence interval, therefore $C^R(u)=\alpha$. But recall that $C^R(Q)=(1-k)C^R(Q')$. For the non truncated standard normal distribution $\frac{C^R(u)}{1-k}=\alpha$ We are required to make this change because this is what is listed in tables. Find u for $C^R(u)=\alpha(1-k)$ by looking in tables of the normal distribution for the right tail figures. Or $C^L(-u)=\alpha(1-k)$ for the left tail figures. Now that you have u and v you are ready to construct the confidence interval. Where your sample size is n, the lower limit is 0 and the upper limit is $v\frac{s}{\sqrt{n}}$ Finally the tricky bit is over! Now you have a confidence interval for the probability someone will spend money and a confidence interval for how much a buyer does spend. You have these for both data sets A and B. Lets say your revenue for A is between $R_A\pm r_A$ and the probability someone buys something is in the range $p_A-h_1$ to $p_A+h_2$ which are the 2 ends of the confidence interval which you calculated. If you had the simple case then this simplifies to $p_A\pm h$ If you had the simple case then your average revenue per visitor is $(R_A\pm r_A) \cdot (p_A\pm h) = R_A p_A \pm R_A p_A\sqrt{(\frac{p_A}{h})^2+(\frac{r_A}{R_A})^2}$ You can compare this final figure to the final figure you get for B to see if they are they overlap. If they do not overlap then you can say that one has a higher mean than the other, if they do overlap then you cannot say that one advertising scheme is better than the other and you will have to reduce your significance and try again. If you had the complicated case then I do not know how to multiply the two uncertainties. You will have to get help elsewhere.

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It is currently 21 Oct 2017, 23:45 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # If Z is a factor of Y , XZ even? (1) The multiple of any two Author Message Manager Joined: 14 May 2009 Posts: 191 Kudos [?]: 48 [0], given: 1 If Z is a factor of Y , XZ even? (1) The multiple of any two [#permalink] ### Show Tags 03 Jun 2009, 00:32 00:00 Difficulty: (N/A) Question Stats: 0% (00:00) correct 0% (00:00) wrong based on 0 sessions ### HideShow timer Statistics This topic is locked. If you want to discuss this question please re-post it in the respective forum. If $$Z$$ is a factor of $$Y$$, $$XZ$$ even? (1) The multiple of any two factors of $$X$$ or $$Y$$ is odd. (2) $$\frac{XY}{Z}$$ is odd. _________________ Kudos [?]: 48 [0], given: 1 Senior Manager Joined: 15 Jan 2008 Posts: 280 Kudos [?]: 49 [0], given: 3 ### Show Tags 03 Jun 2009, 01:24 1, insufficient.. doesnt say anthing abt Z, just says taht X and Y are odd numbers.. 2, insuffficient.. XY is even and Z is odd... XZ depends on X .. can be odd or even.. XY is odd and Z is is odd.. XZ is odd .. Combined.. X and Y are odd numbers... and Z is odd number.. hence XZ is odd.. Kudos [?]: 49 [0], given: 3 SVP Joined: 28 Dec 2005 Posts: 1545 Kudos [?]: 179 [0], given: 2 ### Show Tags 03 Jun 2009, 08:20 Really ? I got C as well With statement 2, Y/Z perfectly divide each other, and for the result when multiplied by X to be odd, X must be odd as well. But I dont know anything about Z ... Y/Z is odd and that could happen if Z is even (i.e. 6/2 = 3, or if Z is odd 9/3=3) From statement 1, we can conclude that X and Y are odd, but nothing about Z Together, X is odd, Y is odd, and since Y/Z=odd (from statement 2), I concluded that Z is odd as well. Hmmm.... Kudos [?]: 179 [0], given: 2 Manager Joined: 14 May 2009 Posts: 191 Kudos [?]: 48 [0], given: 1 ### Show Tags 03 Jun 2009, 16:05 If $$Z$$ is a factor of $$Y$$, $$XZ$$ even? (1) The multiple of any two factors of $$X$$ or $$Y$$ is odd. (2) $$\frac{XY}{Z}$$ is odd. Question:$$(XZ$$ even?$$)$$ (1) The multiple of any two factors of $$X$$ or $$Y$$ is odd. $$\longrightarrow$$ X & Y are odd $$Y$$ odd $$\longrightarrow Z$$ odd $$XZ = ODD*ODD = ODD$$ $$\longrightarrow$$ SUFFICIENT (2) $$\frac{XY}{Z}$$ $$\longrightarrow X$$ odd & $$\frac{Y}{Z}$$ odd But $$Y=MZ$$ where $$M$$ is any integer So $$\frac{Y}{Z}$$ odd $$\longrightarrow$$ $$\frac{MZ}{Z}$$ odd $$\longrightarrow M$$ odd And doesn't tell you anything about $$Z$$, hence $$Z$$ can be ODD or EVEN. Insufficient. Final Answer, $$A$$. _________________ Kudos [?]: 48 [0], given: 1 Senior Manager Joined: 15 Jan 2008 Posts: 280 Kudos [?]: 49 [0], given: 3 ### Show Tags 03 Jun 2009, 22:58 I missed out on a simple statement that.. when X is a factor of Y, and when Y is odd, then X must be ODD... good one.. Kudos [?]: 49 [0], given: 3 Re: Tough DS 7   [#permalink] 03 Jun 2009, 22:58 Display posts from previous: Sort by

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Study Materials # NCERT Solutions for Class 10th Science Page 2 of 5 ## Chapter 2. Acids, Bases and Salts ### Text-book Questions In Text Questions Page No: 18 1. You have been provided with three test tubes. One of them contains distilled water and the other two contain an acidic solution and a basic solution, respectively. If you are given only red litmus paper, how will you identify the contents of each test tube? Solution: If the colour of red litmus does not change then it is acid. If the colour of redlitmus changes to blue then it is base. If there is slight change in the colour of red litmus (such as purple) then it is distilled water. Page No: 22 1. Why should curd and sour substances not be kept in brass and copper vessels? Solution: Curd and other sour substances contain acids. Therefore, when they are kept in brass and copper vessels, the metal reacts with the acid to liberate hydrogen gas and harmful products, thereby spoiling the food. 2. Which gas is usually liberated when an acid reacts with a metal? Illustrate with an example. How will you test for the presence of this gas? Solution: Hydrogen gas is usually liberated when an acid reacts with a metal. Take few pieces of zinc granules and add 5 ml of dilute H2SO4. Shake it and pass the gas produced into a soap solution. The bubbles of the soap solution are formed. These soap bubbles contain hydrogen gas. H2SO4 + Zn → ZnSO4 + H2 ↑ We can test the evolved hydrogen gas by its burning with a pop sound when a candle is brought near the soap bubbles. 3. Metal compound A reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle. Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride. Solution: CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l) Calcium Carbonate + Hydrochloric acid → Calcium Chloride + Carbon dioxide + Water. Page No: 25 1. Why do HCl, HNO3, etc., show acidic characters in aqueous solutions while solutions of compounds like alcohol and glucose do not show acidic character ? Solution: When HCl or HNO3 are mixed with water then they dissolve in water to form H+ or H3O+ ions which shows their acidic character. For example just see the following reactions HCl (aq) → H+ + Cl- H+ + H2O → H3O+ When alcohols and glucose are mixed with water then they do not dissolve to form ions. Hence they do not show acidic character. 2. Why does an aqueous solution of an acid conduct electricity? Solution: The presence of hydrogen (H+) or hydronium (H3O+) ions in the aqueous solution of an acid are responsible for conducting electricity. 3. Why does dry HCl gas not change the colour of the dry litmus paper? Solution: Dry HCl gas not change the colour of the dry litmus paper because it has no Hydrogen ions (H+) in it. 4. While diluting an acid, why is it recommended that the acid should be added to water and not water to the acid? Solution: Since the process of dissolving an acid in water is exothermic, it is always recommended that acid should be added to water. If it is done the other way, then it is possible that because of the large amount of heat generated, the mixture splashes out and causes burns. 5. How is the concentration of hydronium ions (H3O+) affected when a solution of an acid is diluted? Solution: When an acid is diluted, the concentration of hydronium ions (H3O+) per unit volume decreases. This means that the strength of the acid decreases. 6. How is the concentration of hydroxide ions (OH) affected when excess base is dissolved in a solution of sodium hydroxide? Solution: The concentration of hydroxide ions (OH) would increase when excess base is dissolved in a solution of sodium hydroxide. Page No: 28 1. You have two solutions, A and B. The pH of solution A is 6 and pH of solution B is 8. Which solution has more hydrogen ion concentration? Which of this is acidic and which one is basic? Solution: A pH value of less than 7 indicates an acidic solution, while greater than 7 indicates a basic solution. Therefore, the solution with pH = 6 is acidic and has more hydrogen ion concentration than the solution of pH = 8 which is basic. 2. What effect does the concentration of H+ (aq) ions have on the nature of the solution? Solution: If the concentration of H+ (aq) ions is increased (>10-7) then the solution become acidic and if the concentration of H+ (aq) ions is decreased (<10-7) then the solution become basic in nature. 3. Do basic solutions also have H+ (aq) ions? If yes, then why are these basic? Solution: Yes, basic solution also has H+ ions. However, their concentration is less as compared to the concentration of OH- ions that makes the solution basic. 4. Under what soil condition do you think a farmer would treat the soil of his fields with quick lime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate)? Solution: If the soil is acidic and improper for cultivation, then to increase the basicity of soil, the farmer would treat the soil with quick lime or slaked lime or chalk. Page No: 33 1. What is the common name of the compound CaOCl2? ► Bleaching Powder. 2. Name the substance which on treatment with chlorine yields bleaching powder? ► Calcium hydroxide [Ca(OH)2] 3. Name the sodium compound which is used for softening hard water. ► Washing soda (Na2CO3.10H2O) 4. What will happen if a solution of sodium hydrocarbonate is heated? Give the equation of the reaction involved. Solution: When sodium hydrogen carbonate is heated then sodium carbonate and water is formed along with the evolution of carbon dioxide gas. Page 2 of 5 Chapter Contents:

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At what angle should a projectile with initial velocity 'v' be thrown, so that it achieves its maximum range? For maximum horizontal rangeHmax= (u2sin2θ)/2g = (u2)/2g, implying the angle should be equal to 45°..

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# The metric space containing a compact subset is separable Let (X,d) be a metric space and K be a cpt subset of X. If it is possible to derive 'X is compact', then since compact metric space is separable, X is separable. But I'm not sure that X is compact. Do I have to prove that X is compact?, or Is there another method I don't know? • Just because $X$ has a compact subspace does not mean that $X$ is itself compact. For example, take $(X,d)=(\mathbb{R},|\cdot|)$, the real line with the Euclidean metric. Here, $[0,1]$ is a compact subspace, but certainly $\mathbb{R}$ is not compact. – Hayden May 20 '14 at 1:48 • Yes, it's very helpful, thanks! – Hobin. J May 20 '14 at 2:13 No, $\Bbb R$ is not compact in the usual metric, and $[0,1]$ is compact. The claim is certainly false in general, since any singleton is compact in any metric space of your liking, but there exist non-separable metric spaces.

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## How to Calculate and Solve the Centre of Gravity of a Segment of a Sphere The image above represents a segment of a sphere. To compute the centre of gravity of a segment of a sphere requires two essential parameters. These parameters are the radius of the sphere and height of the segment of the sphere. The formula for computing the centre of gravity of a sphere is: C.G. = 3(2r – h)² / 4(3r – h) Where: C.G. = Centre of Gravity r = Radius of the Sphere h = Height of the Segment of the Sphere Let’s solve an example Find the centre of gravity of the segment of the sphere where the radius of the sphere is 10 m and the height of the segment of the sphere is 4 m. This implies that: r = Radius of the Sphere = 10 h = Height of the Segment of the Sphere = 4 C.G. = 3(2(10) – 4)² / 4(3(10) – 4) C.G. = 3(20 – 4)² / 4(30 – 4) C.G. = 3(16)² / 4(26) C.G. = 3(256) / 104 C.G. = 768 / 104 C.G. = 7.38 Therefore, the centre of gravity of the segment of the sphere is 7.38. Nickzom Calculator – The Calculator Encyclopedia is capable of calculating the centre of gravity of a segment of a sphere at a height, h at a distance from the centre of the sphere measured along the height. ## How to Calculate and Solve for the Centre of Gravity of a Cube The image above is a cube with a length of 5.2 cm. To compute the centre of gravity of a cube, one essential parameter is needed and this parameter is the length of the cube (l). The formula for calculating the centre of gravity of a cube is: C.G. = 0.5(l) Where: l = Length of the Cube C.G. = Centre of Gravity Let’s solve an example: Find the centre of gravity of a cube where the length of a side of the cube is 5.2 cm. This implies that: l = Length of the Cube = 5.2 ## How to Calculate and Solve for the Centroid or Centre of Gravity of a Hemisphere The image above is a hemisphere with a radius of 5. To compute the centroid or centre of gravity of a hemisphere. You need one essential parameter and this parameter is the radius of the hemisphere (r). The formula for calculating the centroid or centre of gravity of a hemisphere is: C.G. = 3r / 8 Where r = Radius of the hemisphere As always let us try and solve an example: Find the centroid or centre of gravity of a hemisphere where the radius is 5 cm. From the formula this implies that: r = Radius of the hemisphere = 5 C.G. = 3(5) / 8 C.G. = 15 / 8 C.G. = 1.875 Therefore, the centroid or centre of gravity of the hemisphere is 1.875. Nickzom Calculator – The Calculator Encyclopedia is capable of calculating the centre of gravity of a hemisphere at a distance from its base measured along the vertical radius. To get the answer and workings of the center of gravity or centroid of a hemisphere using the Nickzom Calculator – The Calculator Encyclopedia. First, you need to obtain the app. You can get this app via any of these means: ## Nickzom Calculator Calculates the Centroid or Centre of Gravity of a Semicircle | Statics (Mechanics) The image above is a semicircle with a radius of 7 cm. To compute the centroid or centre of gravity of a semicircle, you need one essential parameter and this parameter is the radius of the semicircle. The formula for calculating the centroid or centre of gravity of a semicircle is: C.G.= 4r / Where: r = Radius of the Semicircle π = Mathematical Constant = 3.142 (approximately) Let’s solve an example: Find the centroid or centre of gravity of a semicircle where the radius is 7 cm. From the example above, r = radius of the semicircle = 7 C.G. = 4(7) / C.G. = 28 / 9.4247 C.G. = 2.97 Therefore, the centroid or centre of gravity of the semicircle is 2.97. Nickzom Calculator – The Calculator Encyclopedia is capable of calculating the centre of gravity of a semicircle at a distance from its base measured along the vertical radius. To get the answer and workings of the center of gravity or centroid of a semicircle using the Nickzom Calculator – The Calculator Encyclopedia. First, you need to obtain the app. ## How to Calculate and Solve for the Centroid or Centre of Gravity of a Sphere The image above is a sphere and 18 m is the diameter of the sphere. In the computing of the centroid or centre of gravity of a sphere there is only one essential parameter which is the diameter of the sphere. The formula for calculating the centroid or centre of gravity of a sphere is: C.G. = d / 2 Where: d is the diameter of the sphere As always, let’s take an example: Let’s solve an example Find the centroid or centre of gravity of a sphere where the diameter is 18 m. C.G. = 18 / 2 C.G. = 9 Therefore, the centroid or centre of gravity of the sphere is 9. Nickzom Calculator – The Calculator Encyclopedia is capable of calculating the centre of gravity of a sphere at a distance from every point. To get the answer and workings of the center of gravity or centroid of a sphere using the Nickzom Calculator – The Calculator Encyclopedia. First, you need to obtain the app. You can get this app via any of these means: ## How to Calculate and Solve the Centre of Gravity of a Right Circular Cone This image above is a display of what a right circular cone looks like. There is only one essential parameter for calculating the centroid or centre of gravity of a right circular cone. This parameter is the height of the cone (h). The formula for calculating the the centroid or centre of gravity of a right circular cone is: C.G. = h / 4 As always let’s solve an example. Find the centroid or centre of gravity of a right circular cone where the height of the cone is 12cm. This implies that: h = height of the cone = 12 C.G. = 12 / 4 C.G. = 3 Therefore, the centroid or centre of gravity of the right circular cone is 3. Nickzom Calculator – The Calculator Encyclopedia is capable of calculating the centre of gravity of a right circular cone at a distance from its base measured along the vertical axis. To get the answer and workings to center of gravity or centroid of a right circular cone using the Nickzom Calculator – The Calculator Encyclopedia. First, you need to obtain the app. You can get this app via any of these means: ## How to Calculate the Centre of Gravity of a Circular Sector in Statics | Mechanics It is very possible to compute the centroid or centre of gravity of a circular sector. There are two highly important parameters one needs to know to compute the centre of gravity of a circular sector. These parameters are: • Radius of the Sector (r) • Semi Vertical Angle (α) The formula for computing the centre of gravity of a circular sector is: C.G. = 2rsinα / Now, let’s take an example. Let’s find the centroid or the centre of gravity of a circular sector that has a radius of 4m and a semi vertical angle of 30°. This implies that: r = Radius of the Sector = 4 α = Semi Vertical Angle = 30 Entering this values into the formula we have: C.G. = 2(4)sin30° / 3(30) C.G. = 8 . sin30° / 90 C.G. = 8 . (0.5) / 90 C.G. = 4 / 90 C.G. = 0.0444 Therefore, the centroid or centre of gravity of the circular sector is 0.0444. Nickzom Calculator – The Calculator Encyclopedia is capable of calculating the centre of gravity of a circular sector making a semi vertical angle α at a distance from the centre of the sector measured along the central axis. To get the answer and workings to center of gravity or centroid of a circular sector. First, you need to obtain the Nickzom Calculator – The Calculator Encyclopedia app. ## Nickzom Calculator Calculates the Center of Gravity of a Trapezium in Statics Mechanics According to Quora, The center of gravity of a trapezoid can be estimated by dividing the trapezoid in two triangles. Nickzom Calculator calculates the centre of gravity of a trapezium with parallel sides a and b at a distance measured from side b. The formula for calculating this center of gravity is: C.G. = h (b + 2a) / 3 (b + a) Where, a and b are the length of the parallel sides of the trapezium (b being the base length and a being the top length) whereas h is the height of the trapezium. Let’s take for Example: Find the center of gravity or centroid of a trapezium where a is 4, b is 8 and h is 2. This implies: a = 4 b = 8 h = 2 C.G. = 2 (8 + 2(4)) / 3 (8 + 4) C.G. = 2 (8 + 8) / 3 (12) C.G. = 2 (16) / 36 C.G. = 32 / 36 C.G. = 0.89 Therefore, the center of gravity or centroid of the trapezium is 0.89. To get the answer and workings to center of gravity or centroid of a trapezium. First, you need to obtain the Nickzom Calculator – The Calculator Encyclopedia app. ## The Calculator Encyclopedia Computes the Significant Figures of a Number According to Wikipedia, The significant figures (also known as the significant digits) of a number are digits that carry meaning contributing to its measurement resolution. This includes all digits except: • Trailing zeros when they are merely placeholders to indicate the scale of the number (exact rules are explained at identifying significant figures); and • Spurious digits introduced, for example, by calculations carried out to greater precision than that of the original data, or measurements reported to a greater precision than the equipment supports. Nickzom Calculator computes the significant figures of any given number. For Example: apply 3 significant figures to 123876  and apply 2 significant figures to 0.00009872. First, you need to obtain the Nickzom Calculator – The Calculator Encyclopedia app. You can get this app via any of these means: Once, you have obtained the calculator encyclopedia app, proceed to the Calculator Map, then click on Significant Figures under the Mathematics section On clicking the page or activity to enter the values is displayed. From the first example: apply 3 significant figures to 123876 It implies that the number is 123876 and the number of significant figures to apply is 3. ## The Calculator Encyclopedia Calculates the Hydraulic Mean Depth or Hydraulic Radius of Flow in Open Channels | Fluid Mechanics Hydraulic mean depth or hydraulic radius can be defined as the cross-sectional area of flow divided by wetted perimeter. Wetted perimeter is the perimeter of the cross sectional area that is “wet”. Hydraulic mean depth or hydraulic radius is a very important parameter in flow in open channels and fluid mechanics calculations. The formula for computing hydraulic mean depth or hydraulic radius is: R = A / P Where: R = Hydraulic Depth A = Cross-sectional Area of Flow P =  Wetted Perimeter For Example: Find the hydraulic mean depth or hydraulic radius of flow in an open channel where the cross-sectional area is 150 m2 and the wetted perimeter is 100 m. From the example we can see that the cross-sectional area is 150 m2 and the wetted perimeter is 100 m. R = 150 / 100 R = 1.5 Therefore, the hydraulic mean depth or hydraulic radius is 1.5 m. Now, I would love to show you how to use Nickzom Calculator The Calculator Encyclopedia to get answers for your questions on hydraulic mean depth or hydraulic radius.

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Upcoming SlideShare × # Pressure 3,094 views 2,949 views Published on KENYA FORM 1 2 Likes Statistics Notes • Full Name Comment goes here. Are you sure you want to Yes No • Be the first to comment Views Total views 3,094 On SlideShare 0 From Embeds 0 Number of Embeds 4 Actions Shares 0 136 0 Likes 2 Embeds 0 No embeds No notes for slide ### Pressure 1. 1. PRESSURE 2. 2. (i)A person makes deeper marks while walking on a soft ground in very narrow and sharp-heeled shoes than when in ordinary shoes with flat heels. (ii)It is easier to push a sharp pin through a cardboard than it is to push a blunt one through the same material using the same force ,see figure 4.1(a) and (b). 3. 3. In (i) above,sharp pointed heels dig deeper on soft ground because the force applied acts on smaller area than when the blunt pin is used. In all these cases , a force acting on a surface produces a penetration effect.This penetration effect is larger when the force acts on a small area than when the same force acts on a larger area. 4. 4. Pressure is defined as force acting per unit area. 5. 5. Units of pressure The SI unit of pressure is thus Newton per Square metre (N/m2), which is also called the Pascal(Pa). 1N/m2=1 Pa Other units include mmHg,cmHg and atmosphere(atm). 6. 6. Ex1:A man of mass 84 kg stands upright on a floor.If the area of contact of his shoes and flor is 420 cm2,determine the average pressure he exerts on the floor.(take g=10 N/kg) 7. 7. Ex 2: A metallic block of mass 40 kg exerts a pressure of 20 N/m2 on aflat surface.Determine the area of contact between the block and the surface.(take g=10 N/kg) 8. 8. Ex3: A rectangular aluminium solid block of density 2700 kg/m3 has dimensions of 40cmx12cmx6cm.The block rests on a horizontal flat surface.Calculate a)the minimum pressure, b)the maximum pressure it can exert. 9. 9. PRESSURE IN LIQUIDS Exp 4.1: to show variation of pressure in liquid Observation: the lower hole,A,throws water farthest,followed by B and lastly C. Conclusion : Pressure of water at A is greater than pressure at B and Pressure at B is bigger than at C. So,Pressure increases with depth. 10. 10. Liquid levels This show that the liquid flows to find its own level. Pressure in liquid is equal at all points at the same level. 11. 11. Liquid levels in a U-tube When water is poured into a U-tube, it will flow into the other arm.The water will setle in the tube with the levels on both arms being the same,see in fig 4.5(a) 12. 12. When one arm of the U tube is blown into with the mouth, the level moves downwards,while on the arm other arm it rises, see in figure 4.5(b). This caused by the pressure difference between the two arms. 13. 13. Exp :4.2(a) to investigate the variation of liquid pressure with depth and density 1. Take a beaker filled with water. Insert the manometer at different levels. You will notice, as seen earlier, that greater the depth, higher is the pressure. 14. 14. 2. Take two beakers, one filled with water and one filled with a salt solution. If you keep the manometer funnel at same depths in the two beakers, you will notice that pressure is dependent on the density of the liquid. Higher the density, higher the pressure. 15. 15. 3. Take a beaker filled with water. Insert the manometer at a certain depth. Rotate the manometer mouth in all direction. You will notice that the pressure at a certain depth inside a liquid is equal in all directions. 16. 16. 4. In a beaker filled with water, insert the manometer funnel at a fixed depth, move the funnel back and forth at the same depth. You will notice that the manometer pressure does not change. This means that the pressure inside a liquid is same at all places at the same depth. 17. 17. In summary : (i) pressure in a liquid increases with depth below its surface. (ii) pressure in a liquid at a particular depth is the same in all directions. (iii) pressure in a liquid increases with the density of the liquid. 18. 18. Homework :Exercise 4.1 19. 19. Fluid Pressure Formula If A is the cross-section area of the column, h is the height of the column and p the density of the liquid, then; Volume of the liquid=cross-section area x height =A x h 20. 20. Mass of the liquid=volume of the liquid x density =Axhxp So,weight of the liquid=mass of the liquid x gravity =A x h x p x g 21. 21. From the definition of pressure, P=h x p x g 22. 22. P=h x p x g From the formula , it can be seen that the pressure due to a liquid column is directly proportional to: (i) height h of the column. (ii) the density p of the liquid. 23. 23. Example 4: A diver is 10 m below the surface of water in a dam.If the density of water is 1000kg/m3,determine the pressure due to the water on the diver.(take g=10 N/kg) 24. 24. Example5: The density of mercury is 13600 kg/m3.Determine the liquid pressure at a point 76 cm below the surface of mercury.(take g=10N/kg) . 25. 25. ATMOSPHERIC PRESSURE There is an envelope of air that surrounds the earth. We call it the atmosphere. The atmosphere is bound to the earth by the gravitational attraction of the earth. This atmosphere of air around us produces pressure at ground level or other levels due to the weight of the air above that level. 26. 26. Simple demonstrations of Air Pressure Atmospheric pressure acts in all directions. (a)Take a little water in a tin and boil the water to drive the air out of it. Cork the tin tightly and then cool it by pouring water on it. What happens? 27. 27. When the air inside the tin is driven out,the atmospheric pressure outside is not counterbalanced by the pressure inside(when the steam cools , a partial vacuum is created inside). The tin collapses. You can understand now why an open empty tin does not get squashed with the atmosphere pressing on the outside. 28. 28. (b) Fill glass or lipless beaker with water to the brim and slide a paper card over the top.Take care not to leave any air bubble in water. Invert the glass. What happens? The card is held in position by the atmospheric pressure pushing upwards on the card and the water does not fall out. The upward atmospheric pressure is greater than the downward force due to water. 29. 29. A more convenient method is to use a glass tube sealed at one end ,as in fig 4.18(a) 30. 30. İf mercury , which is much denser than water is used, the column supported is found to be much shorter,see figure 4.18(b). At sea level , the atmospheric pressure supports approximately 76 cm of mercury column or approximately 10 m of water column. 31. 31. MEASUREMENT OF PRESSURE U-tube Manometer A manometer is an instrument that can measure fluid pressure. It consists of a U-tube filled with water or any other suitable liquid,see figure 4.19 32. 32. Due to pressure of the gas Pg, The water level in the aother limb rises to, say,Y. This difference in water levels is the difference between gas pressure Pg and the atmospheric pressure Pa. Since X and Z are at the same horizontal level ,pressure at X equals pressure at Z.Pressure at X is pressure of gas Pg. Pressure at Z=atmospheric pressure+pressure due to the column of water Therefore ,Pg=Pa+hpg 33. 33. Example: 34. 34. Mercury Barometer It has been shown that atmospheric pressure supports a liquid column in a tube. When this arrangement is used to measure pressure, it is called a barometer. At sea level, acolumn of mercury and water supported by atmospheric pressure is approximately 76 cm. 35. 35. The height h of the column is a measure of the atmospheric pressure. At sea level , h=76cmHg. Since density p of mercury is 13600 kg/m3; This is the Standard atmospheric pressure , and is sometimes referred to as one atmosphere. 36. 36. Testing the vacuum in a Barometer If the barometer has air at the top, it will be faulty. If the mercury column is tilted, the mercury from the dish flows to the tube and completely fills it. 37. 37. If there was air,the mercury would not fill the tube completely. This shows that space above the mercury in the tube is a vacuum. The space above the mercury in the tube when upright is called Toricellian vacuum and contains a little mercury vapour. 38. 38. Fortin Barometer The simple mercury barometer cannot be used for accurate measurements of atmospheric pressure. An improved version called the Fortin barometer is used where high precision is required. 39. 39. video 40. 40. Aneroid Barometer The mercury barometer is the most reliable type of barometer, but is not readily portable. The aneroid barometer is a portable type of barometer consisting of asealed,corrugated metal box,see figure 4.22. 41. 41. Video 42. 42. Pressure Gauges Pressure gauges are portable anda re used mostly for measuring gas pressure,tyre pressure,pressure of compressed air in compressors and steam pressure. 43. 43. TRANSMISSION OF PRESSURE IN LIQUIDS When the plunger is pushed in,the liquid squirts out the holes with equal force. If the plunger exerts a force F and the piston area is A,then the pressure P developed is F/A. This pressure is transmitted equally to all parts of liquid. 44. 44. EXP 4.3:To investigate how pressure is transmitted in liquids(Using Identical Syringes) 45. 45. Procedure : Place a mass m on one of the plungers and observe what happens. When the first mass is placed on the plunger, the plunger moves downwards and the second plunger moves up. Place an identical mass on the other plunger and observe what happens. When an identical mass is placed on the second plunger, the first plunger with the mass on it moves upwards and stops when their levels are the same. 46. 46. The pressure in two syringes is the same. This is because the masses and diameters of the syringes are the same. 47. 47. Using different Syringes 48. 48. At balance , the pressure due to the mass in P is equal to the pressure due to the other mass in Q. Pressure applied at one part in a liquid is transmitted equally to all other parts of the enclosed liquid. This is called the Principle of Transmission of Pressure in Liquids(Pascal’s Principle). Note: gases may transmit pressure in a similar way when they are confined and incompressible. 49. 49. Hydraulic Machines Pascal’s Principle enables a small force to be multiplied into a large force. This principle of transmission of pressure in liquids is made use of in hydraulic machines. 50. 50. Hydraulic Lift When a force is applied on piston S,the pressure exerted by the force is transmitted throughout the liquid to piston L,see figure 4.12. 51. 51. The pressure P1 exerted on the liquid by the piston S due to F1 is given by; 52. 52. This pressure will be transmitted by the liquid to the larger piston L. So, the force F2 produced on the large piston is given by; Note:Hydraulic lifts are used for hoisting cars in garages. 53. 53. Example : in figure 4.12 ,a force of 100 N acts on the smaller piston of area 2 cm2. Calculate the upward force acting on the larger piston of area 900 cm2. 54. 54. Example: figure 4.13 shows two masses placed on light pistons.The pistons are held stationary by the liquid, whose density is 0.8 g/cm3.Determine the value of the force. 55. 55. Hydraulic Brake System The force applied on the foot pedal exerts pressure on the master cylinder. The pressure is transmitted by the brake fluid to the slave cylinder. This causes the pistons of the slave cylinder to open the brake shoe and hence the brake lining presses the drum. The rotation of the Wheel is thus resisted. 56. 56. When the force on the foot pedal is withdrawn, the return spring pulls back the brake shoe which then pushes the slave cylinder piston back. The advantage of this system is that the pressure exerted in master cylinder is transmitted equally to all the four Wheel cylinders. So the braking force obtained is uniform. Brake fluid should have the following properties: • Be incompressible • Have low freezing point and high boiling point • Should not corrode the parts of the brake system 57. 57. homework Exercise : 4.2 58. 58. APPLICATIONS OF PRESSURE IN GASES AND LIQUIDS THE BICYCLE PUMP It consist of a hollow metal cylinder C with a lid at the top and a smallhole at the bottom. The piston rod passes through the lid. The upper end of P has a handle while at its lower end, a cup-shaped leather washer W. The washer touches the sides of the cylinder tightly. 59. 59. (i)when the psiton is withdrawn , the air in section A of the cylinder expands. Hence the pressure inA is less than the atmospheric pressure .So air can pass the washer into A. 60. 60. (ii)when the piston is lowered , the air inA is compressed. The pressure exerted by the compressed air presses the leather washer tightly against the walls of the cylinder and does not allow the air to escapeoutside. When the pressure in A exceeds that in the tyre, the valve in the tyre opens and more air enters the tyre. 61. 61. The lift Pump A lift pump is used to raise water from wells. İt consists of a cylindrical meatal barel with a side tube. İt has two valves ,P and Q, as shown in figure 4.25. 62. 62. Upstroke When the plunger moves up during the upstroke , valve P closes due to its weight and pressure of water above it. At the same time , air above valve expands and its pressure reduces below atmospheric pressure. The atmospheric pressure on the water in the well below thus pushes water up past valve Q into barrel, as shown fig.4.25(a) The plunger is moved up and down until the space between P and Q is filled with water. 63. 63. Downstroke During downstroke , valve Q closes due to its weight and pressure of water above the piston,see fig 4.25(b) 64. 64. Limitations of this pump The atmospheric pressure can only support a column of water of about 10 m. In practice , the possible height of water that can be raised by this pump is less than 10 m because of: (a) low atmospheric pressure in places high above sea level. (b)leakages at the valves and pistons. 65. 65. The force Pump This pump can be used to raise water to heights of more than 10 m. In this case the valve V2 is not in the piston, but in the side delivery tube which is near the bottom of the cylinder. Valve V1 must be about 7-8 m from the surface of water. This pump can force water to great heights as the water is forced out due to the pressure applied at the piston. Thus , water is raised from the well into the pump under atmospheric pressure , but it is forced into the tank under the pressure applied by the piston. Fire- Brigade pumps are of this type. 66. 66. Siphon Tanks are difficult to empty out simply by pouring out their contents. To make easier , we use a siphon. Aflexible pipe filled with the liquid is fitted as shown below. One end is outside (below the surface of the tank) and the other is in the tank (insiude the liquid). The liquid pours out of the pipe at the lower end. 67. 67. The water pours out due to the pressure between A and B:pg(h2-h1). End B is at a higher pressure than end A due to the difference ib heights: h=h2-h1. Note : if the pipe is initially empty, the siphon would not work. The air needs to be sucked out of the pipe fort he liquid tor ise and pour out. 68. 68. HOMEWORK: REVISION EXERCISE

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Introduction We have learned about angle measures since elementary grades.  In Figure 1, we have a circle with center A, and radius length 1. Angle CAB measures 90 degrees and intercepting minor arc BC.  This is also the same as saying that arc BC is subtending angle CAB. We have also learned  that the entire rotation about the center of a circle is 360 degrees. Another unit of angle measure besides degree is radian. Now, what is radian? How is it related to degree? » Read more

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# Questions tagged [groupoids] A groupoid is a small category in which every morphism is an isomorphism. They arise throughout mathematics, e.g. in guise of fundamental groupoids in the theory of covering spaces, holonomy groupoids in the theory of foliations or Lie groupoids in mathematical physics. For groupoids in the sense of universal algebra, i.e., a set with a binary operation, please use the (magma) tag. Also, use other tags such as monoid or category-theory, if needed. 172 questions Filter by Sorted by Tagged with 120 views ### Categorical Intuition of Path Induction I am trying to understand path induction from the trinitarian point of view. So far I understand the informal intuition of path induction from a homotopical and computational point of view. But I am ... 12 views • 832 120 views ### 1-dimensional foliation of surfaces with prescribed graph of foliation Definition of the graph of a foliation Let we have a $k$ dimensional foliation of an $n$ dimensional manifol $M$. One associates to this foliated manifold a (not necessarily Hausdorff) ... • 1,730 30 views ### Isotropy subgroup in groupoid Let $G$ be a groupoid. I am struggling to prove some facts from the definition directly. $(x,y) \in G^{(2)}$ iff $d(x)=r(y)$ where $G^{(2)}$ is the set of composible pairs. Let $G^0$ be the unit space.... • 1,532 57 views ### Without totality: Semigroupoid, Small Category, and Groupoid - how not be closed possible? Semigroupoid, Small Category, and Groupoid have group like structures, see https://en.wikipedia.org/wiki/Monoid#Relation_to_category_theory But there Semigroupoid, Small Category, and Groupoid do not ... • 3,379 71 views ### Groupoid structure of a kernel pair I would have probably very short and simple question. In https://ncatlab.org/nlab/show/%C4%8Cech+nerve there is a statement Čech nerve of a morphism $f$ is the internal nerve of the internal groupoid ... • 63 130 views 1 vote 70 views ### Classical Galois Theory and Groupoids? Given your favorite field $k$, we can build its groupoid of algebraic closures in the obvious way: Look at the category of all possible algebraic closures with field isomorphisms as the arrows. This ... • 25.6k 140 views ### Good book for self-study of Magmas/Semigroups/etc.? I'm currently an undergrad in my second semester of Abstract Algebra. We've covered groups, rings, fields, all that fun stuff. I'm working with Shahriari's "Algebra in Action" as well as ... 1 vote 44 views ### Explicit description of pullback of $(2,1)$-categories. Let us consider the ordinary category of $(2,1)$-categories. Its objects are groupoid enriched categories, and its morphisms are 2-functors. Is there an explicit way to define the objects, morphisms ... • 1,350 1 vote 66 views ### Pushout of groupoid I'm learning category theory. There's a homework question asking for a pushout of groupoid. Suppose $C_0,C_1,C_2$ are groupoids and denote $f_i:C_0\to C_i\quad i=1,2$ the functor. I have managed to ... 1 vote 75 views ### Hopf "algebroid" structure of a groupoid convolution algebra? To male thinks simple as possible, lets say we have a discrete group $G.$ Then the then the group algebra $\mathbb{C}[G]$ (of finitely supported complex valued functions on $G$) has a convolution and ... • 16.3k 45 views ### adjunctions from either the category of small categories or the category of groupoids I'm looking for examples of adjunctions involving either the category Gpd of groupoids or the category Cat of small categories. The insertion of Gpd into Cat has both a left and a right adjoint, so ... • 313 240 views ### Equivalence Between Category of Covering Spaces and Category of Sets with an Action By The Fundamental Groupoid Let $\Pi_1(X)$ be the fundamental groupoid of a locally path-connected topological space $X$ and define $\Pi_1(X)-\mathbf{Sets}$ to be the category of sets equipped with an action by the fundamental ... • 460 197 views ### Functor From Category of Covering Spaces to Category of Sets Equipped With An Action By The Fundamental Groupoid I have some problems with the understanding of the connection between covering spaces and the fundamental groupoid. Let $X$ be a topological space and let $\Pi_1(X)$ denote the fundamental groupoid of ... • 460 61 views 114 views ### classifying space induces a equivalence of categories between PBun$_G(M)$ and $\Pi(M,BG)$ for finite groups $G$ Let $G$ be a finite group, $BG$ its classifying space and M a manifold. Then it is mentioned in https://arxiv.org/abs/1705.05171 (Remark 2.3 d) that there is an equivalence of categories \Pi (M,BG) ... • 21 1 vote 46 views ### Notation for the scope of definition of a partial binary operation of a groupoid I have a groupoid ${\displaystyle (G,\ast )}$ with a partial binary operation ${\displaystyle *:G\times G\rightharpoonup G}$. For every $(a,b)\displaystyle ∈G\times G$, $\displaystyle *$ is defined if ... • 11 54 views ### Name for semigroupoid-like structure applicable to a water flow network in graph theory Hope you are well. I am studying flow networks (flows of water) in graph theory. By flow network I mean a graph $G = (V,E)$ where $V$ is a set of $ℝ_{>0}$ labelled vertices (water tanks filled with ... • 11 1 vote 50 views ### Group isomorphism between identity morphisms of fundamental groupoid and fundamental group of topological space For any topological space $X$, I have defined $C=\Pi_1(X)$ by objects $\text{Ob}(C)=X$ and morphisms $\text{Hom}_C(x,y)=\text{HPath}(x,y)$, where $\text{HPath}(x,y)$ denotes the set of homotopy ... 90 views ### Homotopy cardinality of weak quotients Let $G$ be a group (regarded as a one-object groupoid) acting on a groupoid $X$ (i.e.: a functor $F: G \to \mathsf{Groupoids}$ sending the unique object of $G$ to $X$). Denote with $X//G$ the “weak ... • 78 1 vote 27 views ### Definition of Haar systems, especially for groupoids $C^*$-algebra In A Groupoid approach to $C^*$-algebras Jean Renault introduces left Haar systems. Say $G$ is a groupoid. Let $\Lambda=\{\mu^u,\,\,:\,\,u\in G^0\}$ be a family of Haar measures on $G$. Renault ... • 397 1 vote 47 views ### Help with understanding the definition of a Hodge groupoid I am reading this PhD thesis and I can't understand definition 6.21: (Hodge groupoid). $(X/S)_{Hod}$ is a groupoid whose object object is $X\times \Bbb A^1_S$ and whose morphism object $N_{Hod}$ is ... • 2,730 62 views ### How units of a Lie groupoid act on a manifold? The definition of a Lie groupoid action given in Sébastien Racanière's notes (up to notation) says that the action of a Lie groupoid $\mathcal{G} \rightrightarrows M$ on a smooth manifold $Q$ consists ... • 71k 1 vote 84 views ### Cancellation law in a Lie groupoid I'm playing around with the definition of a Lie groupoid following Eckhard Meinrenken's notes. I read the thing for the first time in my life like an hour ago, so assume that I don't know anything ... • 71k 361 views ### Simplicial sets which are not Kan complexes A Kan complex is a simplicial set satisfying the horn-filler condition. What examples are there of simplicial sets which do not satisfy the horn-filler condition? In particular, what simplicial sets ... • 2,980 78 views ### Every normal subgroupoid is the kernel of a groupoid morphism Sorry for the block of text from Kirill Mackenzie's chapter on groupoids. I have proven that $\natural,\natural_0$ is a groupoid morphism and that this quotient construction satisfies the groupoid ... 1 vote 17 views ### Why Isn't a piecewise-injective groupoid morphism injective? The following is a question I had while reading Kirill Mackenzie's first book (RIP) on groupoids. He says that if a groupoid morphism is base-injective and piecewise-injective, then it is injective. I ... 61 views ### Exact sequences of Groupoids? My question is basically: Is there a good notion of complexes of groupoids and there exactness? I haven't found a definition online. • 1,878 46 views ### Bijection Between Maps of Covering Spaces and Maps of Fundamental Groupoid Covers I'm reading May's Concise Course in Algebraic Topology. He states: My question is about this last corollary. How does the bijection $\text{Cov}(E, E') \to \text{Cov}(\Pi(E), \Pi(E'))$ immediately ... • 1,842 135 views ### Quotient by a topological groupoid. Let $G$ be a group acting on a topological space $X$, then the quotient map $X \to X/G$ is open. I want to ask, whether this fact generalizes to orbit spaces of groupoids. More precisely: Let $G$ be a ... • 1,481 1 vote 64 views ### $G$-set are groupoids of a fibred category Take the category of all G-sets for different groups $G$. Each $G$-set is the groupoid $G \times \Omega \to \Omega$ (the first projection is the source and the action map is the target). I am not ... • 2,730 146 views ### Why can't the 3-type of $S^2$ be delooped? In 2.2 of this paper, the argument runs through the fact that the 3-type of $S^2$ cannot be delooped. Why not? I understand that this is probably a fairly basic fact of homotopy theory (hence neither ... • 2,085 51 views ### Number of fixed point free permutations by groupoid cardinals It is well known that the number $D_n$ of derangements (fixed point free permutations) on a set of $n$ elements is exactly $\left[\dfrac{n!}e\right]$, the closest integer number to $\dfrac{n!}e$. ... • 87.5k 267 views ### $\pi_1(S^1, 1)$ via the fundamental groupoid I'm currently reading Ronnie Brown's Topology and Groupoids and am stuck on a small detail of his computation of the fundamental group of the circle (in particular his computation of the group's ... • 9,841 60 views ### Functor from $BG \to \text{Gpd}$ induces functors for each $g \in G$. I'm reading Groupoids and Stuff. Page 7. Definition 1.3.1. A group action of a a finite group $G$ on a groupoid $X$ is a functor $A : B G \to \text{Gpd}$ such that $A(I) = X$ where $I$ is the unique ... • 19.5k 1 vote 42 views ### Groupoids all of whose subcategories are themselves groupoids It is known that every submonoid of a group $G$ is a subgroup if and only if $G$ is a periodic group, i.e. all of its elements have finite order. The following question is a generalization of the ... • 7,965 28 views ### How to realize the map $\eta$ globally? I have a given map $\Phi: \mathcal{G}\longrightarrow \mathcal{H}$ between two groupoids such that $\Phi_g: \mathcal{G}_x\longrightarrow \mathcal{G}_y$ is a functor between the groupoids $\mathcal{G}_x$... • 8,904 191 views ### Riehl's Category Theory in Context - Exercise 1.5.vii without Axiom of Choice From Emily Riehl, Category theory in context: Exercise 1.5.vii. Let $\mathbf{\mathsf G}$ be a connected groupoid and let $G$ be the group of automorphisms at any of its objects. The inclusion \$\... • 73 118 views ### Exemples of applications of "groupoidification" to linear algebra I just read Baez's very nice blog notes about groupoidification, and around the beginning, he states : "From all this, you should begin to vaguely see that starting from any sort of incidence ... • 41.4k 1 vote

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# What is a Martingale Strategy in Roulette? Many people ask us what is the difference between your European and American Roulette table? The difference is not really that significant if you look at the overall game. Basically, the overall game of roulette has the same rules, so there is absolutely no difference so far as the betting and gambling goes. What could be different in the version of the overall game that you are playing is the actual wheel and how it operates. For a clearer understanding of the overall game mechanics and the roulette betting choices we shall look at European roulette and American roulette table. The main difference between your two variations of the roulette table may be the betting mechanisms. In the American version the ball player places their bets before the game starts and may place their outside bets anytime during the game. The player can also place multiple inside bets during the game. In the European roulette table, all the chips are dealt with just as. All players are dealt a hand plus they are allowed to place any number of chips up for grabs. However, these chips are numbered starting from someone to twenty-one. The dealer will then deal seven cards to each individual and place the numbers on the table in front of them. Then the dealer gives all players a card and tell them to mark it off with the numbers on the card. Following the player has placed their bet, they will spin the roulette wheel. Each time they do the wheel, the bets will undoubtedly be replaced with regular bets. On the winning streak, the bettor must spin again and add on new numbers onto the wheel. This is called ‘spotting’ in fact it is what gives this kind of game its characteristic unpredictability. Whenever a chip is rolled off of the wheel, the bettors got to know what numbers they’re betting on before they place their bet. The bettor will need to remember that their chips are up for grabs and they have to place their bets in line with the numbers which are on the wheel. Therefore a bettor cannot do you know what number they would like to bet on because it will always be the same as the chip they have within their hands. An extremely popular type of game that’s played on the Rolos table involves the 5-chip bet. This sort of bet is not just like the other bets in the game. This is also known as the zero spiel. A zero spiel is a bet where the bettor is basically betting on whether or not they got a five-star or perhaps a four-star for their bet. Usually, the bettor will get exactly what they bet on. Martingale betting systems have 아닥 코인 카지노 been used in various kinds of casino games through the years, but they have only gained in popularity in the last few years. Quite often, these betting systems calls for the outside bets that’s along with a Martingale strategy. This strategy will involve betting on a number that is higher than one that you originally had your bet for. The logic behind this is that the higher number you bet on, the more money you will make as your winnings will continue steadily to increase on a daily basis. An example of a Martingale strategy is always to bet straight-up before next bet is made. After this bet is positioned, the straight-up bet will be bet on another betting period, meaning that if you bet your complete wagered amount on the first period and then bet even more during the second period, you will end up making money for the whole period that you bet. As long as the amount of money that you wagered is bigger than the amount you have won, you then will still make money. Of course, if the sum of money which you have won is greater than the amount that you wagered on the initial period, then you will eventually lose money, and if you were to bet the complete amount on the next period, then you may also end up losing money because you can no longer create a straight-up bet.

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“Fresh sale of individual modules has been closed. Visit anaprep.com for information on new products and write to us for special offers” ## Number Properties (Advanced) - Odd/Even Factorials Today we will discuss how factorials affect the behaviour of odd and even integers. Since we are going to deal with factorials, positive integers will be our concern. Using a question, we will see how factorials are divided. Question: If x and y are positive integers, is y odd? Statement 1: (y+2)!/x! = odd Statement 2: (y+2)!/x! is greater than 2 Solution: The question stem doesn’t give us much information – just that x and y are positive integers. Question: Is y odd? Statement 1: (y+2)!/x! = odd Note that odd and even are identified only for integers. Since (y+2)!/x! is odd, it must be a positive integer. This means that x! must be equal to or less than (y+2)! Now think, how are y and y+2 related? If y+2 is odd, y+1 is even and hence y is odd. If y+2 is even, by the same logic, y is even. (y+2)! = 1*2*3*4*…*y*(y+1)*(y+2) x! = 1*2*3*4*…*x Note that (y+2)! and x! have common factors starting from 1. Since x! is less than or equal to (y+2)!, x will be less than or equal to (y+2). So all factors in the denominator, from 1 to x will be there in the numerator too and will get canceled leaving us with the last few factors of (y+2)! To explain this, let us take a few examples: Example 1: Say, y+2 = 6, x = 6 (y+2)!/x! = 6!/6! = 1 Example 2: Say, y+2 = 7, x = 6 (y+2)!/x! = 7!/6! = (1*2*3*4*5*6*7)/(1*2*3*4*5*6) = 7 (only one leftover factor) Example 3: Say, y+2 = 6, x = 4 (y+2)!/x! = 6!/4! = (1*2*3*4*5*6)/(1*2*3*4) = 5*6 (two leftover factors) If the division of two factorials is an integer, the factorial in the numerator must be larger than or equal to the factorial in the denominator. So what does (y+2)!/x! is odd imply? It means that the leftover factors must be all odd. But the leftover factors will be consecutive integers. So after one odd factor, there will be an even factor. If we want (y+2)!/x! to be odd, we must have either no leftover factors (such that (y+2)!/x! = 1) or only one leftover factor and that too odd. If we have no leftover factor, it doesn’t matter what y+2 is as long as it is equal to x. It could be odd or even. If there is one leftover factor, then y+2 must be odd and hence y must be odd. Hence y could be odd or even. This statement alone is not sufficient. Statement 2: (y+2)!/x! is greater than 2 This tells us that y+2 is not equal to x since (y+2)!/x! is not 1. But all we know is that it is greater than 2. It could be anything as seen in examples 2 and 3 above. This statement alone is not sufficient. Both statements together tell us that y+2 is greater than x such that (y+2)!/x! is odd. So there must be only one leftover factor and it must be odd. The leftover factor will be the last factor i.e. y+2. This tells us that y+2 must be odd. Hence y must be odd too. Takeaways: Assuming a and b are positive integers, –          a!/b! will be an integer only if a >= b –          a!/b! will be an odd integer whenever a = b or a is odd and a = b+1 –          a!/b! will be an even integer whenever a is even and a >= b+1

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# What do you mean by select mode? In French language 'Mode' means 'Fashion'. ... What do you mean by Mode? ... Publish Your Articles is an interactive website that helps you to publish your own ... - Read more what do you mean. o que você quer dizer (o) (meaning) Neue Portugiesisch-Übersetzung vorschlagen - Read more ## What do you mean by select mode? resources ### Mean Median Mode - Memorial University of Newfoundland Mean, Median, Mode Mean: ... First you arrange the numbers in order from lowest ... What do we do? It is simple; we take the ... ### Choose a Windows XP Safe Mode Option (Step 2 of 7) ... Choose a Safe Mode option from the Windows Advanced ... Did you mean ? Your ... How To Force Windows To Restart in Safe Mode; What Is Safe Mode and How Do You Use ... ### How to Calculate the Mode or Modal Value - Maths is Fun How to Find the Mode or Modal Value. ... so we could choose 25 as the mode. You could use different groupings and get a different answer! Mean Median. ### Measures of Central Tendency - Virginia Tech Do you want to determine: The average of the ages? ... Now write down which of these three measures of central tendency (mean, median, or mode) ... ### How to Reboot a Computer in Safe Mode | eHow How to Reboot a Computer in Safe Mode. There are times when you need to reboot your computer in safe mode, perhaps to delete some drivers or do a virus scan. Some ... ### What Do You Mean By Sorting And Searching Of Data? How ... What Do You Mean By Sorting And Searching Of Data? How Should One Select An Algorithm For ... It is very easy to calculate Mean, Median and Mode of such kind of ... ### You Mean What I Do Not - Research Paper - Ireadpaper You Mean What I Do Not In his argument, David W. Boles states that “words and meaning can create different contexts based on cultural usage alone.” ### How to Find the Mean, Median & Mode of a Table or Graph | eHow How to Find the Mean, Median & Mode of a ... A calculator might be helpful if you are calculating the mean, median and mode of ... How Do People Use Mode, Mean ... ### Measure of Central Tendency - Mean Mode Median The term "measures of central tendency ... do not affect the median as strongly as they do the mean. ... you will multiply the mean, mode and median ... ### Mean, Median, Mode, and Range - Purplemath Mean, median, and mode are three kinds of "averages". There are many "averages" in statistics, but these are, I think, the three most common, and are ... ### Math Forum - Ask Dr. Math Archives: Mean, Median, Mode, Range Finding Mean, Median, Mode Can you show me how to find the mean, median, mode, and range of a set of scores? ... Why do we study range, mean, median, and mode? ### what do you mean with - Deutsch-Übersetzung – Linguee ... Do you mean that you are replacing the words 'fifth parliamentary term' with 'the end of the parliamentary term during which [...] the Statute is adopted' or, ... ### dict.cc | mean | Wörterbuch Englisch-Deutsch A 2014-09-14: Do you mean the tape that is stret... F 2014-09-10: What does it mean F 2014-09-09: ... Limited Input Mode - Mehr als 1000 ungeprüfte Übersetzungen! ### How to Identify and Calculate the Mean, Median or Mode Do you know how to identify the mean, median ... simply select the most common score as your mode. ... Applications of the Mean, Median or Mode. So how do you ... ### what do you mean - Deutsch-Übersetzung – Linguee ... What do you mean you don't know this game console entertainment, designed by a famous video games supplier, that puts you ... ### SlangWiki – What Does It Mean What does FTW mean What does TBT mean. Other references ... ### How Do You Find The Range, Median And Mode? - Blurtit How Do You Find The Mean, Median, And Mode Of 55,56,57,58,59 Of This Data Mentally? Mathematics. ... How Do You Find The Median Mode And Range Of A Set Of Numbers ? ### WHAT DO YOU MEAN BY SILENCE - YouTube Vincent Courtois recording "What do you mean by silence?" in Les Lilas. ### What Do You Mean, "DOS"? - MWEB If you talk about "DOS" in Win9x, do you mean: A Command.com session within Windows GUI ... There are three situations where you need DOS mode: I DO NOT OWN THIS. The song What You Mean To Me by Sterling Knight (Drew Ryan Scott) from the Starstruck Soundtrack for the Disney Channel Original Movie ... ### Oasis - Do You Know What I Mean Lyrics - LyricsMode.com Lyrics to Do You Know What I Mean by Oasis: ... select line or word and click "Explain". ... Do not post anything that you do not have the right to post ### Lee Michaels - Do You Kow What I Mean? Lyrics Lyrics to Do You Kow What I Mean? by Lee Michaels: ... select line or word and click "Explain". ... respect from Lyrics Mode community. Yep, I\m cool! Advertising; ### But What Do You Mean? - Essays - 16Archit What Do You Mean? ...good old days of heroes and villains. The people you can bravo or hiss. There was a... 3 Pages Related Questions Recent Questions

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# What do you mean by induced electric field? Contents When the magnetic flux through a circuit changes, a nonconservative electric field is induced, which drives current through the circuit. … The existence of induced electric fields is certainly not restricted to wires in circuits. ## What is an induced field? The induced electric field is a nonconservative field that is generated by a changing magnetic field. The field cannot be an electrostatic field because if the field were electrostatic, and. hence conservative, the line integral of over a closed loop would be zero and it isn’t. ## What is induced electric field and how it is produced? A changing magnetic flux induces an electric field. Both the changing magnetic flux and the induced electric field are related to the induced emf from Faraday’s law. ## What causes the induced electric field? Induced Electric Fields – An induced emf occurs when there is a changing magnetic flux through a stationary conductor. – The force that makes the charges move around the loop is not a magnetic force. There is an induced electric field in the conductor caused by a changing magnetic flux. ## What is the difference between induced electric field and electrostatic field? An electrostatic field is produced by static distribution of charge and determined via Coulomb’s law. An induced electric field is a nonconservative electric field. The induced electric field is the nonconservative electric field produed by a changing magnetic field. THIS IS UNIQUE:  Question: Which is the most dependable alternative energy source? ## What is electrostatic electric field? When two objects in each other’s vicinity have different electrical charges, an electrostatic field exists between them. … Electrostatic fields arise from a potential difference or voltage gradient, and can exist when charge carriers, such as electrons, are stationary (hence the “static”in “electrostatic”). ## What is electromagnetic induction class 12? The electromagnetic induction is a phenomenon in which an electric field or an electromotive force is induced or created in a conductor by the effect of a changing magnetic field in which the conductor is placed. ## What is an induction in physics? There are a variety of methods to charge an object. One method is known as induction. In the induction process, a charged object is brought near but not touched to a neutral conducting object. The presence of a charged object near a neutral conductor will force (or induce) electrons within the conductor to move. ## What is electric field in physics? electric field, an electric property associated with each point in space when charge is present in any form. … The electric field may be thought of as the force per unit positive charge that would be exerted before the field is disturbed by the presence of the test charge. ## What is the direction of induced electric field? So, according to Lenz’s law, there is an induced current in the counterclockwise direction. The conducting loop is not necessary to generate E. (The electric field arise whether or not circuits are present). ## What is the formula of induced current? Right hand rule gives the current direction shown, and the polarity of the rod will drive such a current. To find the magnitude of EMF induced along the moving rod, we use Faraday’s law of induction without the sign: EMF=NΔΦΔt EMF = N Δ Φ Δ t . In this equation, N=1 and the flux Φ=BAcosθ. THIS IS UNIQUE:  What is electricity wholesale market? ## Are induced electric field lines always circular? The field lines formed by a changing magnetic field are not necessarily circles; all that is necessary is that they obey ∇×E=−∂B/∂t.

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## What our customers say... Thousands of users are using our software to conquer their algebra homework. Here are some of their experiences: I recommend the Algebrator to students who need help with fractions, equations and algebra. The program is a great tool! Not only does it give you the answers but it also shows you how and why you come up with those answers. Ive shown my students how to use the program during some of our lessons. A couple of them even bought the program to help them out with their algebra homework. Christopher Montomery, OH Be it Step by Step explanation for an equation or graphical representation, you get it all. I just love to use this due to the flexibility it provides while studying. Lucy, GA I really like your software. I was struggling with fractions. I had the questions and the answers but couldnt figure how to get from one to other. Your software shows how the problems are solved and that was the answer for me. Thanks. Dan Trenton, OK ## Search phrases used on 2009-05-11: Students struggling with all kinds of algebra problems find out that our software is a life-saver. Here are the search phrases that today's searchers used to find our site. Can you find yours among them? • program for multiplication of rational numbers in java • factoring polynomials calculator online • fifth grade expanded notation worksheets • solve algebra problems • algebra formulae worksheets • www.pre algebramath help.com • EQUATIONS with rational exponents • printable algebra instruction sheets-grade 9 • Antiderivative Worksheets • multiple variable equation solver • print free pre-algebra textbook • triple venn diagram + worksheets • lowest common multiple calculator • solve quadratic equation by finding the square roots • combine like terms 7th grade • algebraic expressions with pictures worksheets • accounting formulas cheat sheets • structural dynamics homework exercises and solutions • order of operations challenging worksheets • fun way to teach fractions to a fifth grader • Exponential expression • teach yourself Logarithms • dividing decimals calculator • variable division with exponents calculator • "difference of two squares" homework help • how to convert number to binomial • negative lcm worksheets • 6th grade linear equations worksheets • free third grade math printouts • how to solve any integral • taking fourth root on calculator • proportion worksheets • McDougals definitions of linear equations • solving 3rd degree equation • grade 9 simultaneous equation ppt • example dot product 3D problem • Square Root Calculators For Algebra • ti 83+ log base • yr 8 maths • graphing calculator pic • problems using roots in number lines • aptitude papers for maths for kids • how to solve equations containing radical expressions • factoring trinomials cubed • math inequalities multiple choice problems • prime factors using exponets to show multiples of a factor • cost accounting test and notes • free lessons with comparing & ordering fractions for 3rd grade • "lesson plan" "scale factor" • rotation worksheets • common graph theory test questions • matlab greatest common factor • holt alegebra • algebra exercices math secondary online exam • Iowa Algebra Aptitude sample test • nelson mathematics 8 workbook answers • free adding and subtracting decimals worksheets • resolving algebra exercises • Algebra Poems • TI-83 plus how to do population standard deviation • distributive property.online game • "maple"+"do..while loop" • how do i teach my elementary child measurement conversion • Sample Algebra Problems • ti-89 manual how to solve equations 3rd order • t1-84 plus calculator save steps • math resources + exponent practice • ebook mcdougal algebra 1 free • practice worksheet for solving homogeneous differential equations • math games y intercept • helpful circumference problems for kids age 12 to study for tests printouts • lesson plan for teaching inverse trig graphs • math equations in excel

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# When should calculators be introduced to the curriculum? Stephen Tashi Although I think the curriculum in high school is rather pointless, wasteful, and underchallenges many students, your proposal would be something for more gifted students and not for the norm. I think that the students who really understand trigonometry after taking a trigonometry course are only going to be the gifted ones. It's just a fact of the distribution of human talents -the same for algebra, chemistry etc. The less gifted ones pick up isolated facts, memorize simple patterns etc. I also think students are more likely to grasp and enjoy simple computer programming than algebra. If you want to teach people practical manual arithmetic , it's true that you can teach it as arithmetic in a given context, like figuring out interest on a loan. But I don't buy the argument that "everybody" must learn these practical contexts and I don't think that the students who don't remember the trigonometric identities will remember how to compute interest on a loan unless they do it regularly. (If you want to teach people how to figure interest on loans, you could start by forcing them to go into debt.) Last edited: I don't think kids need to be taught things like long division algorithms at all. What exactly does it teach them? To this day I don't know how to do long division using that ridiculous algorithm. Calculators should be introduced the minute they get beyond the times-tables. I haven't used a dedicated graphing calculator (MATLAB excluded) since my pre-calculus days. I don't need them or want them. I have a solid knowledge of what a graph of a given elementary function ought to look like. Kids need to learn how to graph functions. It teaches you a LOT about how functions behave, and I think it helps make the transition to concepts like continuity much easier. Thus I think the only calculators kids should have are elementary arithmetic calculators, through all stages of their development. Thus I think the only calculators kids should have are elementary arithmetic calculators, through all stages of their development. I think thats unfair. You seem far above average in this regard. I consider myself exceptional at math and when I'm given a nasty equation I always graph it just to back up my own mental assumptions. I agree that things like the long-division algorithm are useless, but basic things should still be taught. The way I see it, as long as the student knows how to do something the manual way, they should be able to use a calculator to their hearts content. I don't think kids need to be taught things like long division algorithms at all. What exactly does it teach them? To this day I don't know how to do long division using that ridiculous algorithm. Calculators should be introduced the minute they get beyond the times-tables. I haven't used a dedicated graphing calculator (MATLAB excluded) since my pre-calculus days. I don't need them or want them. I have a solid knowledge of what a graph of a given elementary function ought to look like. Kids need to learn how to graph functions. It teaches you a LOT about how functions behave, and I think it helps make the transition to concepts like continuity much easier. Thus I think the only calculators kids should have are elementary arithmetic calculators, through all stages of their development. I've encountered too many people who can't reality test even the most basic of computations done with calculators because they've never actually done any calculating themselves, and have no idea what a reasonable answer looks like. The ability to perform basic calculations is such a useful and easily acquired skill that I can't imagine why anyone would forego learning it. chiro Thinking about what Stephen Tashi said, I have to say that the best way I have learned something involving math is by having to program it in a language like C++. I don't know if forcing C++ on to elementary school is wise, but the idea of introducing some kind of programming with a syntax suitable for that age group does sound like a good idea to re-inforce understanding if the student is able to get to this goal independently (not just copying other people's or the teachers code). Forcing programming on schoolchildren was an experiment that was tried twice and failed miserably twice in the UK during the 80s and 90s. Languages and fashions change in programming such that anything learned at school will be hopelessly out of date often before the child has left, let alone later in life. That is not to say that programming study not be available as an option for those who want or need it. chiro Forcing programming on schoolchildren was an experiment that was tried twice and failed miserably twice in the UK during the 80s and 90s. Languages and fashions change in programming such that anything learned at school will be hopelessly out of date often before the child has left, let alone later in life. That is not to say that programming study not be available as an option for those who want or need it. If they are emphasizing language-specific training as opposed to teaching general constructs (typically procedural-programming ones), then the course was badly structured and designed itself. This is though not a feature for this course, but for mathematics, science, and even languages. The content emphasizes things that do not teach understanding: syntax is not programming just like numbers, right angled triangles, and trigonometry is not mathematics. I have a feeling that people that did not have the required experience and the ability to really relate this in terms for the young students. You need the former for the latter, but the latter is a rare skill that the best of teachers possess and unfortunately is in short supply. I've encountered too many people who can't reality test even the most basic of computations done with calculators because they've never actually done any calculating themselves, and have no idea what a reasonable answer looks like. The ability to perform basic calculations is such a useful and easily acquired skill that I can't imagine why anyone would forego learning it. Because it's pointless. Yes, knowing your times tables to 10 and being able to add/subtract on the fly are all very, very useful and very easily acquired. Long division is where I draw the line. Like I said, I don't know how to do long division, and I likely never will. Hell, I can't even use the kiddie algorithm to multiply numbers I haven't got memorized - if I had to do a large-scale multiplication, I'd have to use algebra to break it up into easily-multiplied pieces and sum the results. But so what? Have I lost anything useful? If anything, I've learned how to use a higher form of mathematics (algebra) to invent my own bloody algorithm, the origins of which I understand. Chiro, I really don't have a clue what you mean. I would also venture that you don't have much idea of what I was talking about, given your response. Do you have first hand experience of the events I described? Please all let's remember the original question was When should calculators be introduced? not should calculators be introduced? chiro Chiro, I really don't have a clue what you mean. I would also venture that you don't have much idea of what I was talking about, given your response. Do you have first hand experience of the events I described? Please all let's remember the original question was When should calculators be introduced? not should calculators be introduced? I've already given my own response to that earlier in the thread, but in case you were wondering, my response was give them calculators when they understand the processes used on their calculators. My later response though was targeted specifically at your response for programming, not for the calculator debate and I thought this would be crystal clear given that your response focused on programming. All I'm saying is understanding programming, much like understanding mathematics is not about understanding a specific language or a bunch of largely un-connected specific examples like we have in the high school curriculum (right angled triangles, angle classifications for straight lines, etc). Instead real understanding comes from knowing constructs like for example, doing a loop so many times to calculate a polynomial expression, or using an if statement to decide whether to use option a or optio b. These kinds of things don't depend on languages in the absolute sense and if its taught this way, then the students will not be learning and we will continue the stupidity that is already happening in the classroom, where the students typically know the answers, but they really don't actually understand that much. I don't have first hand experience of students being introduced to programming in schools on a classroom or school level, but I have helped a variety of people to learn programming of various backgrounds and ages, and I have seen that the learning difficulty shares similarities in mathematics where people often just do things without knowing until suddenly the lights go on and it makes sense. In some of the above situations, people just read code (or reading symbols in mathematics) and they don't fully know what the code is even doing, so they fudge the code in some way to try and get what they are aiming for until it magically works. This situation can be amplified when you use specific implementations and examples and if the course is structured bad enough, students can get away with going through the whole course by using a kind of superficial understanding to know what to fudge even if they have no clue why. This is the gist of what I was trying to get at. mathwonk Homework Helper lurflurf: "Who here can compute 357*79135=28251195 in less than 1.0 milliseconds?" the teacher's problem is the kid who reads off 357*79135=2825119, and does not realize it is wrong even after much more than a millisecond. If the teaching is done correctly, I don't think it matters too much when calculators are introduced. Basic four function calculators were introduced to me around the age of 9 or 10, scientific calculators in junior high, and graphing calculators in high school. If anything, having access to calculators at such a young age actually piqued my interest in mathematics ("How does it do that?!"). Calculators never really presented a hindrance to my learning anyway, due to the fact that none of my teachers, from algebra 1 to calculus 2, allowed the use of calculators of any sort during tests. The arithmetic was kept simple enough and the focus was put on the mathematical manipulations. Calculators should be allowed in courses after trig to speed things up. By the time students get to calculus they know when to rely on a calculator (arithmetic, trig functions that aren't based off 30 or 45 degrees, similar ideas...). A student shouldn't be held up on a multi-variable integration problem because they can't do longhand multiplication quickly. They still know how to do the multiplication, but it would just take a long time.

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MTH 289 Differential Equations Extended MTH 289 Differential Equations Extende .. # MTH 289 Differential Equations Extended ## MTH 289 Differential Equations Extended Course Title:  MTH 289  Differential Equations Extended Course Description Presents systems of differential equations, power series solutions, Fourier series, Laplace transform and Fourier transform, partial differential equations, and boundary value problems. Designed as a math elective course for mathematical, physical, and engineering science programs. Lecture 3 hours per week. 3 credits. General Course Purpose The purpose of the course is to provide for a smooth transition of STEM students to 4-year colleges and introduce them to advanced topics of mathematics, physics and engineering: numerical methods for solving differential equations, classical partial differential equations, methods for solving PDEs and boundary-value problems (BVPs). Course Prerequisites/Corequisites Prerequisite:  Completion of MTH 267 Differential Equations with a grade of C or better or equivalent. Course Objectives Upon completing the course, the student will be able to: System of Linear First Order Differential Equations • Define system of linear first-order differential equations, Initial value problem (IVP) and its solution vector, linear dependence/independence, fundamental set of solutions • Check that a vector of functions is a solution of a system or an initial value problem (IVP) • Apply criterion for linearly independent solutions and find general solution for homogeneous and nonhomogeneous systems (for the 3 types of eigenvalues: distinct real, complex, repeated) • Solve nonhomogeneous  linear systems by the methods of undetermined coefficients and variation of parameters Numerical solutions of Ordinary Differential Equations • Understand the concept of local and global truncation errors, stability of numerical method • Use single-step and multistep methods( Euler’s Method, Improved Euler’s Method, 1st, 2nd and 4th-order Runge-Kutta Method, Adams-Bashforth-Moulton Method) and finite difference method to approximate ivp and bvp solutions Plane Autonomous Systems • Explain the concept of autonomous systems of first-order des (linear and nonlinear) • Find critical points and classify critical points of linear and nonlinear systems (stable/unstable nodes, saddle point, degenerate unstable node, center, stable/unstable spiral points), identify equilibrium solution and periodic solution • Use stability criterion for plane autonomous systems • Apply the concept of linearization of differentiable function and classify stable and unstable critical points • Perform stability analysis for linear/nonlinear systems for various applications Orthogonal Functions • Define orthogonal functions and sets of orthogonal functions • Write the definition of the Fourier Series and expansion of functions in a Fourier Series • Define Sturm-Liouville problem and solve it • Write the definitions and expand the function in Fourier-Bessel Series Boundary-Value Problems in Rectangular Coordinates • Define linear/nonlinear, homogeneous/nonhomogeneous partial differential equations • Classify the linear second-order pdes as hyperbolic, parabolic or elliptic • Use the method of separation of variables to find particular solution of pdes • Identify classical and modified pdes and bvps (1d heat equation, 1d wave equation and  2d form of Laplace’s Equation) and solve them • Use the concept of orthogonal series expansions or generalized Fourier Series and solve bvps by using orthogonal series expansions Integral Transforms • Find the Laplace transform of partial derivatives of functions of two variables, use Laplace transform to solve bvps • Define a Fourier integral of function and conditions for convergence, the Fourier integral of even/odd functions • Use the definitions of three Fourier transform pairs (direct and inverse integral transforms) • Solve bvps using the Fourier transform Major Topics to be Included System of Linear First Order Differential Equations Numerical solutions of Ordinary Differential Equations Plane Autonomous Systems Orthogonal Functions Boundary-Value Problems in Rectangular Coordinates Integral Transforms

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## Franklin Pezzuti Dyer This short post is intended to be the elevator pitch for Galois Theory that I wish someone had given me a long time ago. (Yes, I'm sure this exists out there somewhere already, but I've never found it.) Understanding the insolvability of the quintic requires a lot of mathematical overhead, though. You need to learn group theory, enough to understand what a "solvable group" is, and to understand this you had better have a good grasp of normal subgroups and semidirect products. Then you need to be able to show that certain groups, particularly $S_5$, aren't solvable, which usually involves showing that certain subgroups (e.g. $A_5$) are simple. Then you need to understand the definition of a Galois group, and learn how to compute them. There are techniques for doing this with low-degree polynomials, but it gets difficult fast! These are things that we've done in my graduate-level abstract algebra class, but that were intimidating enough to deter me from learning more about the topic as a high schooler. But I wish it hadn't! Unsolvability is just one of many curious and unintuitive behaviors of polynomials and their roots. My intention with this post is to describe a couple phenomena that (I hope!) are a little more accessible, yet still illustrate how there is a lot more complexity to polynomial roots than meets the eye in a precalculus class. Imagine that you would also like to do symbolic computation with algebraic numbers, including square roots like $\sqrt 2$ and $\sqrt 3$, cube roots like $\sqrt[3]{2}$, and even roots of polynomials that you don't know how to solve exactly, for instance something like $x^5+x+3$. In theory, you should be able to do this by giving special names to any new algebraic numbers you introduce, and keeping track of the polynomials that they are roots of. For instance, say you already had a way of representing the rational numbers $\mathbb Q$ computationally, and you would like to also handle expressions involving $\sqrt{2}$. Perhaps you would let the symbol $\alpha$ stand in for $\sqrt{2}$, and you could express more complex expressions that can be built up from rational numbers and $\alpha$ as polynomials or rational functions in $\alpha$. For instance, $1+2\alpha$ stands in for $1+2\sqrt{2}$. The expression $1+\alpha^2$ could be further simplified to just $3$, since you know that $\alpha^2 = 2$. The expression $1/\alpha$ might also be "simplified" to $(1/2)\alpha$: since you know that $\alpha^2 = 2$, you also know that $\alpha\cdot (\alpha/2) = 1$, meaning that $(1/2)\alpha$ is the inverse of $\alpha$. And so on. With a bit of experimentation, you can convince yourself that any rational function of $\alpha$ can actually be simplified down to the form $x+y\alpha$, where $x,y$ are rational numbers, and you can work out an algorithm for doing so. You may also want to work with splitting fields. A splitting field is a field where all of the roots of a certain polynomial are introduced. For instance, the splitting field of $x^2-2$ is just $\mathbb Q(\sqrt 2)$. The splitting field of $x^4-4$ isn't just $\mathbb Q(\sqrt 2)$ because this polynomial also has some imaginary roots, namely $\pm i\sqrt{2}$, so it's instead $\mathbb Q(\sqrt 2, i)$. Not all polynomials are going to have roots that can be expressed so nicely in terms of radicals. (Galois Theory tells us this is actually impossible most of the time - but you don't need to know any Galois Theory to know that solving arbitrary polynomials in terms of radicals is hard!) So in order to do symbolic math with polynomial roots, we would probably want to introduce some symbols representing the roots of a polynomial that we want to split, and keep track of the identities that these special values satisfy, to allow us to algorithmically simplify complex expressions involving these symbols. For instance, as described earlier, to model the splitting field of $x^2-2$, we could express the elements as rational functions of the variable $\alpha$, which stands in for $\sqrt 2$, and remember that this variable satisfies $\alpha^2 - 2=0$ to help us simplify these rational functions as much as possible. Notice that there's no need to introduce a second variable $\beta$ for the other root of the polynomial $x^2-2$, because we know that this other root is just $-\sqrt{2}$, or $-\alpha$. Introducing the second variable would just be redundant, since $x^2-2$ already splits as $(x+\alpha)(x-\alpha)$ when the single root $\alpha$ is added. This sounds simple enough, right? The problem is that for more complicated polynomials, it's not obvious how many new variables need to be introduced to express their roots if we want to avoid redundancy. Consider for instance the quintic polynomial Naively, we might introduce five new variables $\alpha_1,\cdots,\alpha_5$ for the roots of this polynomial, and proceed to do computations using rational expressions of these variables, simplifying when possible using the knowledge that $\alpha_i^5-\alpha_i-1 = 0$. A little less naively, we might use Vieta's formulas to notice that the fifth root can be expressed in terms of the other four roots, namely as $\alpha_5 = -\alpha_1-\alpha_2-\alpha_3-\alpha_4$, so we can do without the fifth variable and use just four. (Kind of like how we only needed one new symbol for the splitting field of $x^2-2$.) And this would be perfectly fine! Then, we might try applying a similar technique to model the splitting field of the irreducible quintic polynomial And again, we might use four new variables $\alpha_1,\alpha_2,\alpha_3,\alpha_4$ to represent four of the five roots of this irreducible polynomial. But this time, we would have unwittingly introduced redundancy between the new values that we've added to our model. Why is that? Well, as it turns out, this polynomial's roots are given by and any one of these roots can be shown to generate the other four roots, which can be shown using the double-angle formula for the cosine function. In particular, if we let $\alpha_1,\cdots,\alpha_5$ represent these five roots respectively, then we can express each one in terms of $\alpha_1$ as follows: This means that we really only need a single new variable $\alpha$ to express all of the roots of this polynomial in a splitting field. If we introduce any more than that, we'll have caused some redundancy! In particular, if we just add the symbols $\alpha_i$ and remember the identities $q(\alpha_i)=0$, we won't be able to simplify some expressions "as much as we should". For instance, the expression should simplify down to zero when $\alpha_1,\cdots,\alpha_4$ are distinct roots of this polynomial, since $\alpha_1^2-2$, being another root of $q$, should equal one of the other four roots $\alpha_2,\alpha_3,\alpha_4$ or $-\alpha_1-\alpha_2-\alpha_3-\alpha_4$. But alas, knowing only that $q(\alpha_i)=0$ for each $\alpha_i$ is not enough information to simplify this expression to zero. Just glancing at a quintic, or more generally an arbitrary polynomial, it's far from obvious how to tell how many new generators will need to be introduced to express all of its roots. This information, however, can be ascertained from a polynomial's Galois group. The Galois group of $p$ is $S_5$, and the fact that four generators are needed for the splitting field of $p$ can be inferred from the fact that there are nontrivial permutations of the set ${0,1,2,3,4}$ that fix any three elements. The Galois group of $q$, on the other hand, is the cyclic group $\mathbb Z_5$, and any cyclic permutation of the numbers ${0,1,2,3,4}$ is determined by the image of a single element. Here's another difficulty. Suppose you've extended your base field by adding all the roots of a polynomial, and now you want to extend your field again and split another polynomial. For instance, say you've already split the polynomial $p$ from earlier and introduced four new variables $\alpha_1,\alpha_2,\alpha_3,\alpha_4$, and now you want to split the polynomial $x^2-2$, so you add another symbol $\beta$ to stand in for one of its roots. This would be fine. But it just so happens that if, instead of the polynomial $x^2-2$, you wanted to split $x^2-2869$, you would have just accidentally introduced redundancy again. That is to say, $\beta=\sqrt{2869}$ is irrational, but it can be expressed in terms of the roots of the quintic polynomial $p$ - so this number already belongs to your splitting field without any further additions needed. In particular, because $2869$ is the discriminant of the polynomial $p$. Generally, for any irreducible polynomial over an extension of $\mathbb Q$, the product of the squared differences of its roots can be seen to actually belong to the base field, which can be expressed in terms of the polynomial's coefficients. And when this number is not a perfect square in that base field, a new quadratic irrational over that field is introduced. For the polynomial $q$ defined earlier, this is not an issue. That polynomial has discriminant equal to $11^4$, which is already a perfect square. Once more, this is information that can be read off from the Galois groups of these polynomials. The fact that splitting $p$ "incidentally" introduces new roots to some previously irreducible quadratics over $\mathbb Q$ is connected to the fact that $S_5$ has a normal subgroup $A_5$ that has index $2$. The Galois group $\mathbb Z_5$ of the polynomial $q$, of course, has no proper subgroups at all, so we will never have to worry about introducing new roots to irreducible polynomials of lower degrees when we split $q$. The best part about these strange properties, in my opinion, is the fact that they can (for the most part) be shown constructively. For instance, for the polynomial $p$, you can prove purely algebraically that when $\alpha$ is one of its roots, then $\alpha^2-2$ is also one of its roots (though this might get a bit messy, and nonconstructive methods are neater). You can also, with a bit of Vieta-style trickery, compute the discriminant of $p$ by hand. To me, this makes these facts feel a lot more "tangible". Granted, the quintic examples I've given above are still difficult to work with. The exact same phenomena, however, can be observed with the following two cubic polynomials: In this case, the former has Galois group $S_3$, so that its splitting field is generated by any two of its roots (and no fewer), and the latter has Galois group $\mathbb Z_3$, so that its splitting field is generated by any one of its roots. Again, the latter has roots that can be expressed simply in terms of the cosine function, but I leave it to you to figure out how. Here we also have that the discriminant of $p$ is $-23$, a non-square in $\mathbb Q$ that has a square root in the splitting field of $p$, and the discriminant of $q$ is $7^2$, which is already a square in $\mathbb Q$.

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# Wednesday 21 June 2017 Morning Save this PDF as: Size: px Start display at page: ## Transcription 3 3 states but a continuous range of emitted or absorbed wavelengths requires the electrons to be unbound or free. The temperatures in the cores of stars, where nuclear fusion occurs, are of the order of millions of kelvin, but the light-emitting surface of a star (the photosphere) is much cooler. The photosphere of the Sun, for example, has a temperature of around 5800 K. Even at this lower temperature, there are enough free electrons to produce a continuous range of wavelengths, but as we have seen in Fig. 1, the emission is not uniform across the range. For the Sun, the wavelength at which the greatest intensity of radiation is produced is around 500 nm. Equation 1 allows the surface temperature of a star to be determined from the wavelength of the predominant colour of radiation being emitted. Fig. 2 shows some of these colour-relationships Colour Temperature / K orange-red 4000 green-yellow 6000 blue 8000 violet ultra-violet Fig. 2: The relationship between colour and surface temperature As already noted from the Planck curves, the total amount of radiation emitted increases with the temperature and it is useful to represent this behaviour of stars in the form of a diagram, as shown in Fig. 3, which shows how the total amount of radiation emitted (the luminosity) varies with temperature for stars of different radii Luminosity (L sun ) 1 10 R 1 R 0.1 R 100 R 1000 R R refers to the value of the radius of the Sun R R Temperature (K) Fig. 3: The variation of luminosity with temperature for stars of different radii Turn over 4 There are several important things to note about this diagram: 4 1. The total amount of radiation emitted is called the luminosity and this will depend upon the size of the star as well as the temperature: stars with a larger radius have a greater surface area from which radiation can be emitted. This accounts for the set of lines, one for each radius of star. Astronomers refer to this luminosity as the absolute luminosity of the star. 2. The current power of the Sun is used here as the unit for luminosity (about W). 3. The surface temperatures increase from right to left. This is because when the chart was originally constructed, the x-axis displayed something called the Colour Index of the star (a label for the colour), which went in the order of decreasing temperature. 4. As well as going from right to left, the temperature scale is logarithmic. 5. The chart was generated using the known relationships between temperature, peak wavelength and luminosity. Having constructed such a chart from theory, the question is: how do the colours and absolute luminosities of stars we observe in the night sky compare to it? Although the colours can be easily determined, measuring the luminosities requires knowledge of the distances of the stars from the Earth. The intensity of light varies with 1/(distance from light source) 2. If it is possible to measure how bright a star appears to be and it is known how far away it is, then it is possible to determine how bright the star actually is (its absolute luminosity). Those stars that are most easily visible to observers on Earth are those that are closest to it, so their distances may be determined using the Stellar Parallax method. Fig. 4 shows the positions of a selection of stars of known luminosity and temperature (colour), including some well-known ones such as Betelgeuse and Sirius. This chart is very important to astronomers and was developed, independently, by Ejnar Hertzsprung and Henry Russell around It is thus known today as the Hertzsprung-Russell (or H-R) diagram 5 Spica Rigel Deneb Betelgeuse Antares 10 3 Red Giants 10 2 Aldebaran Luminosity (Solar Units) 10 1 Sirius Main Sequence Sun Sirius B White Dwarfs Procyon B Barnard s Star Proxima Centauri Temperature (K) Fig. 4: The Hertzsprung-Russell diagram There is a diagonal band stretching from top left to bottom right: this is not surprising since it is expected that hotter (bluer) stars are more luminous. However, there are exceptions, such as the Red Giants (for example, Betelgeuse) and White Dwarf stars. Giant stars have diameters many times that of the Sun. The diagonal band is called the Main Sequence but must not be seen as an evolutionary path. In the course of a star s life it could go through different stages causing it to appear in various positions around the chart. Stars such as Sirius B and Procyon B (which feature in Fig. 4) are actually orbiting companions of larger stars and, since their orbital periods are often measurable, this allows the mass of the larger star they orbit to be determined. Such information added to that of the H-R diagram has enabled astronomers to study the stellar life cycle in detail and to develop models for stellar evolution. These include some of the more exotic objects to have been discovered in more recent times such as brown dwarfs, neutron stars, and black holes END OF ARTICLE 6 6 BLANK PAGE 7 7 BLANK PAGE ### Tuesday 10 June 2014 Morning Tuesday 0 June 20 Morning AS GCE MATHEMATICS 736/0 Decision Mathematics PRINTED ANSWER BOOK *33365809* Candidates answer on this Printed Answer Book. OCR supplied materials: Question Paper 736/0 (inserted) ### * * MATHEMATICS 4721 Core Mathematics 1 ADVANCED SUBSIDIARY GCE. Monday 11 January 2010 Morning QUESTION PAPER. Duration: 1 hour 30 minutes. ADVANCED SUBSIDIARY GCE MATHEMATICS 4721 Core Mathematics 1 QUESTION PAPER Candidates answer on the Printed Answer Book OCR Supplied Materials: Printed Answer Book 4721 List of Formulae (MF1) Other Materials ### Friday 23 June 2017 Morning Oxford Cambridge and RSA Friday 23 June 2017 Morning A2 GCE MATHEMATICS (MEI) 4754/01B Applications of Advanced Mathematics (C4) Paper B: Comprehension QUESTION PAPER *7350982488* Candidates answer on ### Thursday 16 June 2016 Morning Oxford Cambridge and RSA Thursday 16 June 2016 Morning GCSE PSYCHOLOGY B543/01 Research in Psychology *5052023496* Candidates answer on the Question Paper. OCR supplied materials: None Other materials ### Monday 24 June 2013 Morning THIS IS A NEW SPECIFICATION F Monday 24 June 2013 Morning GCSE TWENTY FIRST CENTURY SCIENCE PHYSICS A A183/01 Module P7 (Foundation Tier) *A137300613* Candidates answer on the Question Paper. A calculator ### Wednesday 21 June 2017 Morning Oxford Cambridge and RSA Wednesday 21 June 2017 Morning A2 GCE PHYSICS A G485/01 Fields, Particles and Frontiers of Physics *6808554474* Candidates answer on the Question Paper. OCR supplied materials: ### Wednesday 21 May 2014 Afternoon Wednesday 21 May 2014 Afternoon AS GCE SCIENCE G641/01 Remote Sensing and the Natural Environment *1319354529* Candidates answer on the Question Paper. OCR supplied materials: None Other materials required: ### Monday 24 June 2013 Morning THIS IS A NEW SPECIFICATION H Monday 24 June 2013 Morning GCSE TWENTY FIRST CENTURY SCIENCE PHYSICS A A183/02 Module P7 (Higher Tier) * A 1 3 7 3 1 0 6 1 3 * Candidates answer on the Question Paper. A ### MATHEMATICS 4723 Core Mathematics 3 ADVANCED GCE MATHEMATICS 4723 Core Mathematics 3 QUESTION PAPER Candidates answer on the printed answer book. OCR supplied materials: Printed answer book 4723 List of Formulae (MF1) Other materials required: ### MATHEMATICS 4725 Further Pure Mathematics 1 ADVANCED SUBSIDIARY GCE MATHEMATICS 4725 Further Pure Mathematics 1 QUESTION PAPER Candidates answer on the printed answer book. OCR supplied materials: Printed answer book 4725 List of Formulae (MF1) ### Wednesday 18 May 2016 Morning Oxford Cambridge and RSA Wednesday 18 May 2016 Morning A2 GCE MATHEMATICS 4729/01 Mechanics 2 QUESTION PAPER * 6 3 9 5 6 4 5 6 6 1 * Candidates answer on the Printed Answer Book. OCR supplied materials: ### Thursday 12 June 2014 Afternoon Thursday 12 June 2014 Afternoon AS GCE MATHEMATICS 4728/01 Mechanics 1 QUESTION PAPER * 3 1 3 4 0 1 4 0 0 1 * Candidates answer on the Printed Answer Book. OCR supplied materials: Printed Answer Book 4728/01 ### MATHEMATICS 4722 Core Mathematics 2 ADVANCED SUBSIDIARY GCE MATHEMATICS 4722 Core Mathematics 2 QUESTION PAPER Candidates answer on the Printed Answer Book OCR Supplied Materials: Printed Answer Book 4722 List of Formulae (MF1) Other Materials ### A Level Physics B (Advancing Physics) H557/03 Practical skills in physics Oxford Cambridge and RSA A Level Physics B (Advancing Physics) Practical skills in physics Practice paper Set 2 Time allowed: 1 hour 30 minutes *2016* You must have: the Data, Formula and Relationships ### Wednesday 8 June 2016 Morning Oxford Cambridge and RSA Wednesday 8 June 2016 Morning AS GCE MATHEMATICS 4732/01 Probability & Statistics 1 QUESTION PAPER * 4 8 2 7 1 9 3 8 2 8 * Candidates answer on the Printed Answer Book. OCR supplied ### Wednesday 25 May 2016 Afternoon Oxford Cambridge and RSA H Wednesday 25 May 2016 Afternoon GCSE TWENTY FIRST CENTURY SCIENCE PHYSICS A/SCIENCE A A181/02 Modules P1 P2 P3 (Higher Tier) *5940699339* Candidates answer on the Question Paper. ### PHYSICS B (ADVANCING PHYSICS) 2863/01 Rise and Fall of the Clockwork Universe THIS IS A LEGACY SPECIFICATION ADVANCED GCE PHYSICS B (ADVANCING PHYSICS) 2863/1 Rise and Fall of the Clockwork Universe * OCE / 1 5 124* Candidates answer on the Question Paper OCR Supplied Materials: ### MATHEMATICS 4728 Mechanics 1 ADVANCED SUBSIDIARY GCE MATHEMATICS 4728 Mechanics 1 QUESTION PAPER Candidates answer on the printed answer book. OCR supplied materials: Printed answer book 4728 List of Formulae (MF1) Other materials ### B278A MATHEMATICS C (GRADUATED ASSESSMENT) MODULE M8 SECTION A GENERAL CERTIFICATE OF SECONDARY EDUCATION. Monday 8 March 2010 Morning WARNING M8 GENERAL CERTIFICATE OF SECONDARY EDUCATION MATHEMATICS C (GRADUATED ASSESSMENT) MODULE M8 SECTION A B278A * OCE / 1 9558* Candidates answer on the Question Paper OCR Supplied Materials: None Other Materials ### * * MATHEMATICS (MEI) 4757 Further Applications of Advanced Mathematics (FP3) ADVANCED GCE. Wednesday 9 June 2010 Afternoon PMT ADVANCED GCE MATHEMATICS (MEI) 4757 Further Applications of Advanced Mathematics (FP3) Candidates answer on the Answer Booklet OCR Supplied Materials: 8 page Answer Booklet MEI Examination Formulae and ### PHYSICS A 2825/01 Cosmology THIS IS A LEGACY SPECIFICATION ADVANCED GCE PHYSICS A 2825/01 Cosmology *OCE/T73579* Candidates answer on the question paper OCR Supplied Materials: None Other Materials Required: Electronic Calculator ### G485. PHYSICS A Fields, Particles and Frontiers of Physics ADVANCED GCE. Wednesday 2 February 2011 Afternoon INFORMATION FOR CANDIDATES ADVANCED GCE PHYSICS A Fields, Particles and Frontiers of Physics G485 * OCE / 1 732 7 * Candidates answer on the question paper. OCR supplied materials: Data, Formulae and Relationships Booklet Other ### Thursday 11 June 2015 Morning Oxford Cambridge and RSA Thursday 11 June 2015 Morning A2 GCE PHYSICS A G484/01 The Newtonian World *4865875507* Candidates answer on the Question Paper. OCR supplied materials: Data, Formulae and Relationships ### PHYSICS A 2822 Electrons and Photons THIS IS A LEGACY SPECIFICATION ADVANCED SUBSIDIARY GCE PHYSICS A 2822 Electrons and Photons *CUP/T61221* Candidates answer on the question paper OCR Supplied Materials: None Other Materials Required: Electronic ### Wednesday 25 May 2016 Morning Oxford Cambridge and RSA Wednesday 5 May 016 Morning A GCE MATHEMATICS (MEI) 476/01 Mechanics QUESTION PAPER * 6 6 9 6 6 0 4 1 9 * Candidates answer on the Printed Answer Book. OCR supplied materials: ### Thursday 18 June 2015 Morning Oxford Cambridge and RSA Thursday 18 June 2015 Morning A2 GCE PHYSICS A G485/01 Fields, Particles and Frontiers of Physics *4866002088* Candidates answer on the Question Paper. OCR supplied materials: ### Monday 20 June 2016 Morning Oxford Cambridge and RSA Monday 20 June 2016 Morning A2 GCE PHYSICS A G484/01 The Newtonian World *1320816367* Candidates answer on the Question Paper. OCR supplied materials: Data, Formulae and Relationships ### Friday 20 January 2012 Morning THIS IS A NEW SPECIFICATION F Friday 20 January 2012 Morning GCSE TWENTY FIRST CENTURY SCIENCE PHYSICS A A181/01 Modules P1 P2 P3 (Foundation Tier) *A131490112* Candidates answer on the Question Paper. ### OXFORD CAMBRIDGE AND RSA EXAMINATIONS A2 GCE 4733/01. MATHEMATICS Probability & Statistics 2 QUESTION PAPER OXFORD CAMBRIDGE AND RSA EXAMINATIONS A2 GCE 4733/01 MATHEMATICS Probability & Statistics 2 QUESTION PAPER TUESDAY 10 JUNE 2014: Morning DURATION: 1 hour 30 minutes plus your additional time allowance ### Friday 21 June 2013 Morning Friday 21 June 2013 Morning A2 GCE MATHEMATICS 4729/01 Mechanics 2 QUESTION PAPER * 4 7 3 3 4 0 0 6 1 3 * Candidates answer on the Printed Answer Book. OCR supplied materials: Printed Answer Book 4729/01 ### H H * * MATHEMATICS FOR ENGINEERING H860/02 Paper 2 LEVEL 3 CERTIFICATE. Wednesday 9 June 2010 Afternoon. Duration: 1 hour 30 minutes. LEVEL 3 CERTIFICATE MATHEMATICS FOR ENGINEERING H860/02 Paper 2 Candidates answer on the Answer Booklet OCR Supplied Materials: 8 page Answer Booklet Insert (inserted) List of Formulae (MF1) Other Materials ### A Level Physics B (Advancing Physics) H557/03 Practical skills in physics. Thursday 29 June 2017 Morning Time allowed: 1 hour 30 minutes Oxford Cambridge and RSA A Level Physics B (Advancing Physics) H557/03 Practical skills in physics Thursday 29 June 2017 Morning Time allowed: 1 hour 30 minutes *6830996327* You must have: the Data, Formula ### THIS IS A LEGACY SPECIFICATION THIS IS A LEGACY SPECIFICATION H GENERAL CERTIFICATE OF SECONDARY EDUCATION MATHEMATICS B (MEI) Paper 3 Section A (Higher Tier) B293A *OCE/26262* Candidates answer on the question paper. OCR supplied materials: ### Wednesday 7 January 2015 Morning Oxford Cambridge and RSA Wednesday 7 January 2015 Morning LEVEL 1 CAMBRIDGE NATIONAL IN SCIENCE R072/01 How scientific ideas have developed *4394971481* Candidates answer on the Question Paper. A calculator ### Monday 14 January 2013 Morning Monday 14 January 2013 Morning A2 GCE MATHEMATICS 4729/01 Mechanics 2 QUESTION PAPER * 4 7 3 3 8 1 0 1 1 3 * Candidates answer on the Printed Answer Book. OCR supplied materials: Printed Answer Book 4729/01 ### Wednesday 8 June 2016 Afternoon Oxford Cambridge and RSA F Wednesday 8 June 2016 Afternoon GCSE GEOGRAPHY B B563/01 Key Geographical Themes (Foundation Tier) *5818140845* Candidates answer on the Question Paper. OCR supplied materials: ### Monday 8 January 2018 Morning Oxford Cambridge and RSA Monday 8 January 2018 Morning LEVEL 1 CAMBRIDGE NATIONAL IN SCIENCE R072/01 How scientific ideas have developed *7299077580* Candidates answer on the Question Paper. A calculator ### Wednesday 25 May 2016 Afternoon Oxford Cambridge and RSA F Wednesday 25 May 2016 Afternoon GCSE TWENTY FIRST CENTURY SCIENCE PHYSICS A/SCIENCE A A181/01 Modules P1 P2 P3 (Foundation Tier) *5940639183* Candidates answer on the Question ### Monday 8th June 2015 Morning Oxford Cambridge and RSA H Monday 8th June 205 Morning GCSE METHODS IN MATHEMATICS B39/02 Methods in Mathematics (Higher Tier) * 4 8 5 6 4 9 2 5 * Candidates answer on the Question Paper. OCR supplied ### Thursday 10 January 2013 Morning Thursday 10 January 2013 Morning AS GCE CHEMISTRY B (SALTERS) F331/01 Chemistry for Life *F314500113* Candidates answer on the Question Paper. OCR supplied materials: Data Sheet for Chemistry B (Salters) ### Wednesday 8 June 2016 Afternoon Oxford Cambridge and RSA H Wednesday 8 June 2016 Afternoon GCSE GEOGRAPHY A A732/02 Geographical Skills (Higher Tier) *4410185572* Candidates answer on the Question Paper. OCR supplied materials: An Insert ### THIS IS A NEW SPECIFICATION THIS IS A NEW SPECIFICATION ADVANCED GCE PHYSICS A The Newtonian World G484 * OCE / 2 301 6 * Candidates answer on the Question Paper OCR Supplied Materials: Data, Formulae and Relationships Booklet Other ### G491. PHYSICS B (ADVANCING PHYSICS) Physics in Action ADVANCED SUBSIDIARY GCE. Wednesday 12 January 2011 Morning. Duration: 1 hour ADVANCED SUBSIDIARY GCE PHYSICS B (ADVANCING PHYSICS) Physics in Action G491 *OCE/17763* Candidates answer on the Question Paper OCR Supplied Materials: Data, Formulae and Relationships Booklet Other Materials ### 4754A A A * * MATHEMATICS (MEI) Applications of Advanced Mathematics (C4) Paper A ADVANCED GCE. Friday 15 January 2010 Afternoon PMT ADVANCED GCE MATHEMATICS (MEI) Applications of Advanced Mathematics (C4) Paper A 4754A Candidates answer on the Answer Booklet OCR Supplied Materials: 8 page Answer Booklet MEI Examination Formulae and ### Candidate number. Centre number Oxford Cambridge and RSA GCSE (9 1) Mathematics Paper 1 (Foundation Tier) F Practice Paper Set 3 Time allowed: 1 hour 30 minutes *2016* You may use: A scientific or graphical calculator Geometrical instruments ### Thursday 13 June 2013 Morning Thursday 13 June 2013 Morning A2 GCE MATHEMATICS (MEI) 4754/01B Applications of Advanced Mathematics (C4) Paper B: Comprehension INSERT * 4 7 1 5 7 1 0 6 1 3 * INFORMATION FOR CANDIDATES This Insert contains ### G484. PHYSICS A The Newtonian World ADVANCED GCE. Monday 27 June 2011 Morning. Duration: 1 hour ADVANCED GCE PHYSICS A The Newtonian World G484 *G410340611* Candidates answer on the question paper. OCR supplied materials: Data, Formulae and Relationships Booklet Other materials required: Electronic ### MATHEMATICS 4729 Mechanics 2 ADVANCED GCE MATHEMATICS 4729 Mechanics 2 QUESTION PAPER Candidates answer on the printed answer book. OCR supplied materials: Printed answer book 4729 List of Formulae (MF1) Other materials required: ### B293A. MATHEMATICS B (MEI) Paper 3 Section A (Higher Tier) GENERAL CERTIFICATE OF SECONDARY EDUCATION. Tuesday 12 January 2010 Morning WARNING H GENERAL CERTIFICATE OF SECONDARY EDUCATION MATHEMATICS B (MEI) Paper 3 Section A (Higher Tier) B293A *OCE/14183* Candidates answer on the Question Paper OCR Supplied Materials: None Other Materials Required: ### Thursday 19 June 2014 Morning Thursday 19 June 2014 Morning A2 GCE PHYSICS A G485/01 Fields, Particles and Frontiers of Physics *3270254576* Candidates answer on the Question Paper. OCR supplied materials: Data, Formulae and Relationships ### Tuesday 23 May 2017 Morning Oxford Cambridge and RSA Tuesday 23 May 2017 Morning AS GCE PHYSICS B (ADVANCING PHYSICS) G491/01 Physics in Action *6687779587* Candidates answer on the Question Paper. OCR supplied materials: Data, Formulae ### A Level Physics A H556/01 Modelling physics. Practice paper Set 1 Time allowed: 2 hours 15 minutes Oxford Cambridge and RSA A Level Physics A H556/01 Modelling physics Practice paper Set 1 Time allowed: 2 hours 15 minutes You must have: the Data, Formulae and Relationship Booklet You may use: a scientific ### PHYSICS A 2825/04 Nuclear and Particle Physics THIS IS A LEGACY SPECIFICATION ADVANCED GCE PHYSICS A 2825/04 Nuclear and Particle Physics *OCE/T71779* Candidates answer on the question paper OCR Supplied Materials: None Other Materials Required: Electronic ### PHYSICS A 2825/04 Nuclear and Particle Physics THIS IS A LEGACY SPECIFICATION ADVANCED GCE PHYSICS A 2825/04 Nuclear and Particle Physics * OCE / 1 7360* Candidates answer on the Question Paper OCR Supplied Materials: None Other Materials Required: ### Wednesday 3 June 2015 Afternoon Oxford Cambridge and RSA H Wednesday 3 June 2015 Afternoon GCSE GEOGRAPHY B B563/02 Key Geographical Themes (Higher Tier) *4390586245* Candidates answer on the Question Paper. OCR supplied materials: OS ### PHYSICS B (ADVANCED PHYSICS) 2864/01 Field and Particle Pictures THIS IS A LEGACY SPECIFICATION ADVANCED GCE PHYSICS B (ADVANCED PHYSICS) 2864/01 Field and Particle Pictures *CUP/T64120* Candidates answer on the question paper OCR Supplied Materials: Data, Formulae ### Monday 20 June 2016 Morning Oxford Cambridge and RSA Monday 2 June 216 Morning A2 GCE PHYSICS B (ADVANCING PHYSICS) G494/1 Rise and Fall of the Clockwork Universe *11658898* Candidates answer on the Question Paper. OCR supplied materials: ### F795. GEOLOGY Evolution of Life, Earth and Climate ADVANCED GCE. Wednesday 8 June 2011 Morning ADVANCED GCE GEOLOGY Evolution of Life, Earth and Climate F795 *F711220611* Candidates answer on the question paper. OCR supplied materials: None Other materials required: Electronic calculator Ruler (cm/mm) ### Friday 17 June 2016 Afternoon Oxford Cambridge and RSA Friday 17 June 016 Afternoon AS GCE MATHEMATICS (MEI) 4761/01 Mechanics 1 QUESTION PAPER * 6 3 7 0 1 8 0 6 1 * Candidates answer on the Printed Answer Book. OCR supplied materials: ### Thursday 9 June 2016 Afternoon Oxford Cambridge and RSA Thursday 9 June 2016 Afternoon AS GCE PHYSICS B (ADVANCING PHYSICS) G492/01 Understanding Processes / Experimentation and Data Handling * 1 1 6 5 2 2 0 7 4 0 * Candidates answer ### How can we use an H-R diagram to know where a star is in its life cycle? How can we use an H-R diagram to know where a star is in its life cycle? Just like humans, stars go through a life cycle. Over the course of their lives, stars change in ways that make each stage different ### G484. PHYSICS A The Newtonian World ADVANCED GCE. Thursday 27 January 2011 Afternoon. Duration: 1 hour ADVANCED GCE PHYSICS A The Newtonian World G484 *OCE/17456* Candidates answer on the question paper. OCR supplied materials: Data, Formulae and Relationships Booklet Other materials required: Electronic ### CHEMISTRY 2815/01 Trends and Patterns THIS IS A LEGACY SPECIFICATION ADVANCED GCE CHEMISTRY 2815/01 Trends and Patterns *OCE/13482* Candidates answer on the Question Paper A calculator may be used for this paper OCR Supplied Materials: Data ### Monday 19 May 2014 Afternoon H Monday 19 May 2014 Afternoon GCSE TWENTY FIRST CENTURY SCIENCE PHYSICS A/SCIENCE A A181/02 Modules P1 P2 P3 (Higher Tier) *1186688069* Candidates answer on the Question Paper. A calculator may be used ### Friday 26 May 2017 Morning Oxford Cambridge and RSA Friday 26 May 2017 Morning AS GCE CHEMISTRY A F321/01 Atoms, Bonds and Groups *6679472758* Candidates answer on the Question Paper. OCR supplied materials: Data Sheet for Chemistry ### Thursday 11 June 2015 Afternoon Oxford Cambridge and RSA H Thursday 11 June 2015 Afternoon GCSE METHODS IN MATHEMATICS B392/02 Methods in Mathematics 2 (Higher Tier) *4856252055* Candidates answer on the Question Paper. OCR supplied ### Thursday 17 May 2012 Morning Thursday 17 May 2012 Morning AS GCE PHYSICS B (ADVANCING PHYSICS) G491 Physics in Action *G412020612* Candidates answer on the Question Paper. OCR supplied materials: Data, Formulae and Relationships Booklet ### 4754A * * A A. MATHEMATICS (MEI) Applications of Advanced Mathematics (C4) Paper A ADVANCED GCE. Friday 14 January 2011 Afternoon ADVANCED GCE MATHEMATICS (MEI) Applications of Advanced Mathematics (C4) Paper A 4754A Candidates answer on the answer booklet. OCR supplied materials: 8 page answer booklet (sent with general stationery) ### Candidate number. Centre number Oxford Cambridge and RSA GCSE (9 1) Mathematics Paper 4 (Higher Tier) H Practice Paper Set 3 Time allowed: 1 hour 30 minutes *2016* You may use: A scientific or graphical calculator Geometrical instruments ### Thursday 26 May 2016 Morning Oxford Cambridge and RSA H Thursday 26 May 2016 Morning GCSE MATHEMATICS A A502/02 Unit B (Higher Tier) * 5 9 9 9 0 5 4 4 9 7 * Candidates answer on the Question Paper. OCR supplied materials: None Other ### Monday 18 June 2012 Morning Monday 18 June 2012 Morning A2 GCE PHYSICS A G484 The Newtonian World *G411820612* Candidates answer on the Question Paper. OCR supplied materials: Data, Formulae and Relationships Booklet (sent with general ### SPECIMEN. Date Morning/Afternoon Time allowed: 1 hour 30 minutes. AS Level Physics A H156/02 Depth in physics Sample Question Paper PMT AS Level Physics A H156/02 Depth in physics Sample Question Paper Date Morning/Afternoon Time allowed: 1 hour 30 minutes You must have: the Data, Formulae and Relationships Booklet You may use: a scientific ### * * MATHEMATICS 4732 Probability & Statistics 1 ADVANCED SUBSIDIARY GCE. Wednesday 27 January 2010 Afternoon. Duration: 1 hour 30 minutes. ADVANCED SUBSIDIARY GCE MATHEMATICS 4732 Probability & Statistics 1 Candidates answer on the Answer Booklet OCR Supplied Materials: 8 page Answer Booklet List of Formulae (MF1) Other Materials Required: ### B294A. MATHEMATICS B (MEI) Paper 4 Section A (Higher Tier) GENERAL CERTIFICATE OF SECONDARY EDUCATION. Friday 15 January 2010 Morning WARNING H GENERAL CERTIFICATE OF SECONDARY EDUCATION MATHEMATICS B (MEI) Paper Section A (Higher Tier) B29A * OCE / 119 1* Candidates answer on the Question Paper OCR Supplied Materials: None Other Materials Required: ### B294A. MATHEMATICS B (MEI) Paper 4 Section A (Higher Tier) GENERAL CERTIFICATE OF SECONDARY EDUCATION. Friday 11 June 2010 Morning WARNING H GENERAL CERTIFICATE OF SECONDARY EDUCATION MATHEMATICS B (MEI) Paper 4 Section A (Higher Tier) B294A * OCE / 1 0368* Candidates answer on the Question Paper OCR Supplied Materials: None Other Materials ### Wednesday 20 May 2015 Afternoon Oxford Cambridge and RSA F Wednesday 20 May 2015 Afternoon GCSE TWENTY FIRST CENTURY SCIENCE PHYSICS A/SCIENCE A A181/01 Modules P1 P2 P3 (Foundation Tier) *4824557259* Candidates answer on the Question ### Tuesday 28 June 2016 Morning Oxford Cambridge and RSA Tuesday 28 June 2016 Morning A2 GCE PHYSICS B (ADVANCING PHYSICS) G495/01 Field and Particle Pictures *1165934400* Candidates answer on the Question Paper. OCR supplied materials: ### GENERAL CERTIFICATE OF SECONDARY EDUCATION MATHEMATICS B (MEI) B294B H GENERAL CERTIFICATE OF SECONDARY EDUCATION MATHEMATICS B (MEI) Paper 4 Section B (Higher Tier) B294B * OCE / 14195* Candidates answer on the Question Paper OCR Supplied Materials: None Other Materials ### B294B. MATHEMATICS B (MEI) Paper 4 Section B (Higher Tier) GENERAL CERTIFICATE OF SECONDARY EDUCATION. Monday 1 June 2009 Morning. H GENERAL CERTIFICATE OF SECONDARY EDUCATION MATHEMATICS B (MEI) Paper Section B (Higher Tier) B29B *OCE/T75117* Candidates answer on the question paper OCR Supplied Materials: None Other Materials Required: ### Wednesday 8 November 2017 Morning Time allowed: 1 hour 30 minutes Oxford Cambridge and RSA GCSE (9 1) Mathematics J560/03 Paper 3 (Foundation Tier) F Wednesday 8 November 2017 Morning Time allowed: 1 hour 30 minutes *6932107064* You may use: A scientific or graphical ### Thursday 10 January 2013 Morning Thursday 10 January 2013 Morning AS GCE CHEMISTRY A F321/01 Atoms, Bonds and Groups *F314420113* Candidates answer on the Question Paper. OCR supplied materials: Data Sheet for Chemistry A (inserted) Other ### Wednesday 3 June 2015 Morning Oxford Cambridge and RS Wednesday 3 June 2015 Morning FSMQ DVNCED LEVEL 6993/01 dditional Mathematics QUESTION PPER * 4 3 9 6 2 8 1 0 5 5 * Candidates answer on the Printed nswer ook. OCR supplied materials: ### Two boats, the Rosemary and the Sage, are having a race between two points A and B. t, where 0 t (i) Find the distance AB. 5 8 In this question, positions are given relative to a fixed origin, O. The x-direction is east and the y-direction north; distances are measured in kilometres. Two boats, the Rosemary and the Sage, are ### Thursday 9 June 2016 Morning Oxford Cambridge and RSA H Thursday 9 June 2016 Morning GCSE MATHEMATICS A A503/02 Unit C (Higher Tier) * 5 9 9 9 2 7 3 6 9 3 * Candidates answer on the Question Paper. OCR supplied materials: None Other ### Tuesday 6 November 2012 Morning H Tuesday 6 November 2012 Morning GCSE MATHEMATICS A A501/02 Unit A (Higher Tier) *A516800312* Candidates answer on the Question Paper. OCR supplied materials: None Other materials required: Scientific ### Thursday 25 May 2017 Morning Time allowed: 1 hour 30 minutes Oxford Cambridge and RSA GCSE (9 1) Mathematics J560/04 Paper 4 (Higher Tier) H Thursday 25 May 2017 Morning Time allowed: 1 hour 30 minutes *7348865303* You may use: A scientific or graphical calculator ### F331. CHEMISTRY B (SALTERS) Chemistry for Life ADVANCED SUBSIDIARY GCE. Monday 23 May 2011 Afternoon. Duration: 1 hour 15 minutes ADVANCED SUBSIDIARY GCE CHEMISTRY B (SALTERS) Chemistry for Life F331 *F318770611* Candidates answer on the question paper. OCR supplied materials: Data Sheet for Chemistry B (Salters) (inserted) Other ### Friday 27 May 2016 Morning Oxford Cambridge and RSA Friday 27 May 2016 Morning AS GCE CHEMISTRY A F321/01 Atoms, Bonds and Groups *5986785801* Candidates answer on the Question Paper. OCR supplied materials: Data Sheet for Chemistry ### Candidate number. Centre number Oxford Cambridge and RSA GCSE (9 1) Mathematics J560/05 Paper 5 (Higher Tier) H Practice Paper Set 3 Time allowed: 1 hour 30 minutes *2016* You may use: Geometrical instruments Tracing paper Do not use: ### Monday 16 May 2016 Morning Oxford Cambridge and RSA Monday 16 May 2016 Morning AS GCE GEOLOGY F791/01 Global Tectonics *6007922565* Candidates answer on the Question Paper. OCR supplied materials: None Other materials required: ### THIS IS A LEGACY SPECIFICATION THIS IS A LEGACY SPECIFICATION H GENERAL CERTIFICATE OF SECONDARY EDUCATION MATHEMATICS B (MEI) Paper 3 Section B (Higher Tier) B293B * OCE / 1 794 9 * Candidates answer on the question paper. OCR supplied ### Monday 10 June 2013 Morning Monday 10 June 2013 Morning A2 GCE MATHEMATICS (MEI) 4762/01 Mechanics 2 QUESTION PAPER * 4 7 1 5 7 9 0 6 1 3 * Candidates answer on the Printed Answer Book. OCR supplied materials: Printed Answer Book ### Wednesday 3 June 2015 Morning Oxford Cambridge and RSA Wednesday 3 June 015 Morning AS GCE MATHEMATICS (MEI) 475/01 Concepts for Advanced Mathematics (C) QUESTION PAPER * 3 6 7 4 8 0 7 8 7 * Candidates answer on the Printed Answer ### Wednesday 11 January 2012 Morning THIS IS A LEGACY SPECIFICATION H Wednesday 11 January 2012 Morning GCSE MATHEMATICS B (MEI) B293B Paper 3 Section B (Higher Tier) *B216700112* Candidates answer on the Question Paper. OCR supplied materials: ### METHODS IN MATHEMATICS B392/02 Methods in Mathematics 2 (Higher Tier) THIS IS A NEW SPECIFICATION H GENERAL CERTIFICATE OF SECONDARY EDUCATION METHODS IN MATHEMATICS B392/02 Methods in Mathematics 2 (Higher Tier) *B315650611* Candidates answer on the question paper. OCR ### PHYSICS B (ADVANCING PHYSICS) 2860 Physics in Action THIS IS A LEGACY SPECIFICATION ADVANCED SUBSIDIARY GCE PHYSICS B (ADVANCING PHYSICS) 2860 Physics in Action *CUP/T68004* Candidates answer on the question paper OCR Supplied Materials: Data, Formulae and ### Wednesday 14 June 2017 Morning Oxford Cambridge and RSA H Wednesday 14 June 2017 Morning GCSE GATEWAY SCIENCE PHYSICS B B751/02 Physics modules P1, P2, P3 (Higher Tier) *6754482469* Candidates answer on the Question Paper. A calculator

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Theory Practice Test on Strength of Materials - Set 25 - ObjectiveBooks # Practice Test: Question Set - 25 1. When a rectangular bar of length l, breadth b and thickness t is subjected to an axial pull of P, then linear strain (ε) is given by (where E = Modulus of elasticity) (A) Îµ = P/b.t.E (B) Îµ = b.t.E/P (C) Îµ = b.t/P.E (D) Îµ = P.E/b.t 2. Which of the following statement is correct? (A) The size of hole drilled in riveting plates is less than the actual size of rivet. (B) The center to center distance between two consecutive rivets in a row is called margin. (C) Rivets are generally specified by its shank diameter. (D) Tearing of plates can be avoided by taking the pitch of rivets equal to 1.5 times the diameter of rivet hole. 3. A beam of triangular section is placed with its base horizontal. The maximum shear stress occurs at (A) Apex of the triangle (B) Mid of the height (C) Center of gravity of the triangle (D) Base of the triangle 4. The Young's modulus of a material is 125 GPa and Poisson’s ratio is 0.25. The modulus of rigidity of me material is (A) 30 GPa (B) 50 GPa (C) 80 GPa (D) 100 GPa 5. In the above question, the normal stress on an oblique section will be maximum, when θ is equal to (A) 0° (B) 30° (C) 45° (D) 90° 6. In a simply supported beam carrying a uniformly distributed load w per unit length, the point of contraflexure (A) Lies in the center of the beam (B) Lies at the ends of the beam (C) Depends upon the length of beam (D) Does not exist 7. In case of eccentrically loaded struts __________ is preferred. (A) Solid section (B) Hollow section (C) Composite section (D) Reinforced section 8. In a thin cylindrical shell subjected to an internal pressure p, the ratio of longitudinal stress to the hoop stress is (A) 1/2 (B) 3/4 (C) 1 (D) 1.5 9. The unit of modulus of elasticity is same as those of (A) Stress, strain and pressure (B) Stress, force and modulus of rigidity (C) Strain, force and pressure (D) Stress, pressure and modulus of rigidity 10. The stress at which the extension of the material takes place more quickly as compared to the increase in load, is called (A) Elastic limit (B) Yield point (C) Ultimate point (D) Breaking point 11. Two bars of different materials and same size are subjected to the same tensile force. If the bars have unit elongation in the ratio of 2:5, then the ratio of modulus of elasticity of the two materials will be (A) 2 : 5 (B) 5 : 2 (C) 4 : 3 (D) 3 : 4 12. In a thick cylindrical shell subjected to an internal pressure (p), the maximum radial stress at the inner surface of the shell is (A) Zero (B) p (tensile) (C) -p (compressive) (D) 2p (tensile) 13. When a body is subjected to direct tensile stresses (σx and σy) in two mutually perpendicular directions, accompanied by a simple shear stress Ï„xy, then in Mohr's circle method, the circle radius is taken as (A) [(σx - σy)/2] + Ï„ (B) [(σx + σy)/2] + Ï„ (C) (1/2) × √[(σx - σy+ 4Ï„²xy] (D) (1/2) × √[(σx + σy+ 4Ï„²xy] 14. A section of beam is said to be in pure bending, if it is subjected to (A) Constant bending moment and constant shear force (B) Constant shear force and zero bending moment (C) Constant bending moment and zero shear force (D) None of the above 15. Modular ratio of the two materials is the ratio of (A) Linear stress to linear strain (B) Shear stress to shear strain (C) Their modulus of elasticities (D) Their modulus of rigidities Show and hide multiple DIV using JavaScript View All Answers Blogger Comment

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# Net change in math This Net change in math helps to quickly and easily solve any math problems. x ## Area under rate function gives the net change Explanation Step 1: . Firstly, determine the Closing price at the end of the period for which the analysis is conducted. Step 2: . In the next step, determine the • 1 • 2 Solve math problem If you want to get something done, set a deadline. • 3 For those who struggle with math, equations can seem like an impossible task. However, with a little bit of practice, anyone can learn to solve them. • 4 Get help from expert professors You can always count on our team for reliable support. ## Learn Formula for Calculating the Net Change The net change formula is given as, Net Change Formula = Current Period’s Closing Price – Previous Period’s Closing Price where, Current Period’s Closing Price = Closing price at the end of the period when the analysis is done. • Get math help online • Clarify math problem • Solve word questions too • Deal with mathematic problems • Math learning that gets you ## In math, what does the term net change mean? A net change in math is the total of all of the changes completed throughout the solving of a problem. The net change is reflected in a numerical amount and can be positive Get Help You can get math help online by visiting websites like Khan Academy or Mathway. Expert tutors will give you an answer in real-time Math is a way of solving problems by using numbers and equations. Reliable Support You can find the answer to your question in just 3 seconds with our new search engine. Explain mathematic equations If you're looking for a fun way to teach your kids math, try Decide math. It's a great way to engage them in the subject and help them learn while they're having fun. ## Net Change Formula Andymath.com features free videos, notes, and practice problems with answers! Printable pages make math easy. Are you ready to be a mathmagician? Notes. Problems Find the net change between these 2 points. \(1)\) \( (3,4) \) and \( Confidentiality Get help from our expert homework writers! Figure out mathematic One way to think about math equations is to think of them as a puzzle. Each piece of the equation fits together to create a complete picture. When one piece is missing, it can be difficult to see the whole picture. Figure out mathematic question If you're struggling to clear up a math equation, try breaking it down into smaller, more manageable pieces. This will help you better understand the problem and how to solve it. Confidentiality is an important part of our company culture.

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# Hubble's law observation in physical cosmology (Redirected from Expansion of the universe) Hubble's law or Hubble—Lemaître's law is the name for the observation that: 1. All objects observed in deep space have a doppler shift-measured velocity relative to Earth, and to each other; 2. The doppler-shift-measured velocity of galaxies moving away from Earth, is proportional to their distance from the Earth and all other interstellar bodies. Absorption lines in the visible spectrum of a supercluster of distant galaxies (right), as compared to absorption lines in the optical spectrum of the Sun (left). Arrows indicate redshift. Wavelength increases up towards the red and beyond (frequency decreases). In effect, the space-time volume of the observable universe is expanding and Hubble's law is the direct physical observation of this.[1] It is the basis for believing in the expansion of the universe and is evidence often cited in support of the Big Bang model. Although widely attributed to Edwin Hubble, the law was first derived from the General Relativity equations by Georges Lemaître in a 1927 article. There he proposed that the Universe is expanding, and suggested a value for the rate of expansion, now called the Hubble constant.[2][3][4][5][6] Two years later Edwin Hubble confirmed the existence of that law and determined a more accurate value for the constant that now bears his name.[7] The recession velocity of the objects was inferred from their redshifts, many measured earlier by Vesto Slipher in 1917 and related to velocity by him.[8] The law is often expressed by the equation v = H0D, with H0 the constant of proportionality (the Hubble constant) between the "proper distance" D to a galaxy and its velocity v (see Uses of the proper distance). H0 is usually quoted in (km/s)/Mpc, which gives the speed in km/s of a galaxy 1 megaparsec (3.09×1019 km) away. The reciprocal of H0 is the Hubble time. A recent 2011 estimate of the Hubble constant, which used a new infrared camera on the Hubble Space Telescope (HST) to measure the distance and redshift for a collection of astronomical objects, gives a value of H0 = 73.8 ± 2.4 (km/s)/Mpc.[9][10] An alternate approach using data from galactic clusters gave a value of H0 = 67.0 ± 3.2 (km/s)/Mpc.[11][12] A number of other methods have been used, giving figures between 70 and 72 (km/s)/Mpc. A recent (2016) method using the oldest light in the universe suggests the Hubble Constant value was 66.53km/s per megaparsec soon after the expansion began. This implies that the rate of expansion has been increasing.[13] ## References 1. Peter Coles, ed. (2001). Routledge critical dictionary of the new cosmology. Routledge. p. 202. ISBN 0-203-16457-1. 2. Lemaître, Georges (1927). "Un univers homogène de masse constante et de rayon croissant rendant compte de la vitesse radiale des nébuleuses extra-galactiques". Annales de la Société Scientifique de Bruxelles. A47: 49–56. Bibcode:1927ASSB...47...49L.CS1 maint: ref=harv (link) (Full article, PDF). Partially translated (the translator remains unidentified) in Lemaître, Georges (1931). "Expansion of the universe, A homogeneous universe of constant mass and increasing radius accounting for the radial velocity of extra-galactic nebulæ". Monthly Notices of the Royal Astronomical Society. 91: 483–490. Bibcode:1931MNRAS..91..483L.CS1 maint: ref=harv (link). 3. Bergh, Sidney van den (6 June 2011). "The Curious Case of Lemaitre's Equation No. 24". arXiv:1106.1195 [astro-ph, physics:physics] – via arXiv.org. 4. Block, David L. (13 February 2019). "Georges Lemaitre and Stiglers Law of Eponymy". arXiv:1106.3928 [astro-ph, physics:physics]. 395: 89–96. doi:10.1007/978-3-642-32254-9_8 – via arXiv.org. 5. Reich, Eugenie Samuel (27 June 2011). "Edwin Hubble in translation trouble". Nature. doi:10.1038/news.2011.385 – via Crossref. 6. http://www.nature.com/nature/journal/v479/n7372/full/479171a.html 7. Hubble, Edwin 1929. A relation between distance and radial velocity among extra-galactic nebulae Proceedings of the National Academy of Sciences of the United States of America. 15, 3, 168-173. (Full article, PDF) 8. Malcolm S Longair (2006). The Cosmic Century. Cambridge University Press. p. 109. ISBN 0-521-47436-1. 9. Riess, Adam G.; et al. (2011). "A 3% solution: determination of the Hubble constant with the Hubble Space Telescope and Wide Field Camera" (PDF). The Astrophysics Journal. 730 (2): 119. arXiv:1103.2976. Bibcode:2011ApJ...730..119R. doi:10.1088/0004-637X/730/2/119. 10. 'New study gives dark energy a boost' Archived 2012-09-30 at the Wayback Machine (Ron Cowen), Science News, 16 March 2011. 11. Beutler, Florian; et al. (2011). "The 6dF Galaxy Survey: baryon acoustic oscillations and the local Hubble constant". Monthly Notices of the Royal Astronomical Society. 416 (4): 3017. Bibcode:2011MNRAS.tmp.1164B. doi:10.1111/j.1365-2966.2011.19250.x. 12. 'Hubble Constant: a new way to measure the expansion of the universe', Science Daily, 27 July 2011. 13. Amos, Jonathan 2016. Hubble clocks faster cosmic expansion. BBC News Science & Environment. [1]

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# What happens to force if you increase the distance over which the work is done? ## What happens to force if you increase the distance over which the work is done? The mathematical relationships between total work and total energy are described by the work-energy theorem and conservation of energy. Simple machines can change the amount of force that is necessary to move an object, but the force must be applied through a larger distance; they don’t change the amount of work done. ## What happens to the amount of work if you apply a force to an object but it does not move? If a force is applied but the object doesn’t move, no work is done; if a force is applied and the object moves a distance d in a direction other than the direction of the force, less work is done than if the object moves a distance d in the direction of the applied force. ## Is done when an object moves through a distance on applying force? Work is done when a force moves something through a distance. To do work on the object, the force must act in the same direction as, or opposite the direction of, the object’s displacement. Work done on an object is the product of the applied force and the distance the object moves while the force is applied. four ## What will happen to the work done if the distance is increased? d is the displacement; u03b8 is the angle between the two. u2223u2223u2223du2223u2223u2223 will be the distance covered; so increasing it, work increases. ## How do you increase the work done by increasing the force or by increasing the distance explain? Contrary to popular belief, machines do not increase the amount of work that is done. They just change how the work is done. Machines make work easier by increasing the amount of force that is applied, increasing the distance over which the force is applied, or changing the direction in which the force is applied. ## How does distance increase force? Originally Answered: How is distance related to force? The gravitational attraction between two objects is inversely proportional to the square of the distance between them. If the distance between them is increased by three times, the new distance between them would be four times the earlier distance. ## What is the relationship between work done force and distance? Work, Force, and Distance Work is the use of force to move an object. It is directly related to both the force applied to the object and the distance the object moves. Work can be calculated with this equation: Work Force x Distance ## How can a force be applied but no work be done? The force exerted by the person is an upward force equal to the weight of the box, and that force is perpendicular to the motion. If there is no motion in the direction of the force, then no work is done by that force. ## Can work be done if object doesn’t move? work: A measure of energy expended in moving an object; most commonly, force times displacement. No work is done if the object does not move. ## How much work is done if a force is applied to an object but the object does not move? If a force is applied but the object doesn’t move, no work is done; if a force is applied and the object moves a distance d in a direction other than the direction of the force, less work is done than if the object moves a distance d in the direction of the applied force. ## How does an applied force affect an object that is not moving? Applied force can cause acceleration When a force acts on an object that is stationary or not moving, the force will cause the object to move, provided there are no other forces preventing that movement. If you throw a ball, you are pushing on it to start its movement. ## What is known as application of a force through a distance? work, in physics, measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement. If the force is being exerted at an angle u03b8 to the displacement, the work done is W fd cos u03b8. ## When a force moves an object is said to be done? Work is said to be done when a force moves an object through a distance in its own direction.

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## What number is "LIV"? ### A: 54 LIV = 54 Your question is, "What is LIV in Numbers?". The answer is '54'. Here we will explain how to convert, write and read the Roman numeral letters LIV in the correct Arabic number translation. ## How is LIV converted to numbers? To convert LIV to numbers the translation involves breaking the numeral into place values (ones, tens, hundreds, thousands), like this: Place ValueNumberRoman Numeral Conversion50 + 4L + IV Tens50L Ones4IV ## How is LIV written in numbers? To write LIV as numbers correctly you combine the converted roman numerals together. The highest numerals should always precede the lower numerals to provide you the correct written translation, like in the table above. 50+4 = (LIV) = 54 ## More from Roman Numerals.co LV Now you know the translation for Roman numeral LIV into numbers, see the next numeral to learn how it is conveted to numbers. Convert another Roman numeral in to Arabic numbers.

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Trigonometry Table: Formulas, Tricks, Examples - Embibe • Written By Priya Wadhwa • Last Modified 20-07-2022 • Written By Priya Wadhwa • Last Modified 20-07-2022 # Trigonometry Table: Ratios, Tricks, and Solved Examples Trigonometry Table: Trigonometry is a popular branch of Mathematics that deals with the study of triangles and the relationship between the length of sides and angles in a triangle. It has a wide range of applications in astronomy, architecture, aerospace, defence, etc. In this article, we have provided the trigonometry tables containing the values of all trigonometric ratios for the most commonly used angles. The trigonometry table is a useful tool for finding the values of trigonometric ratios for standard angles such as 0°, 30°, 45°, 60°, and 90°. It comprises the values of trigonometric ratios – sine, cosine, tangent, cosecant, secant, cotangent, also known as sin, cos, tan, cosec, sec, and cot, respectively. Using the trigonometry table formula, students can compute trigonometric values for various other angles by understanding the patterns seen within trigonometric ratios and between angles. ## Introduction to Trigonometric Table In simple words, the trigonometric table is a collection of the values of trigonometric ratios for the commonly used standard angles including 0°, 30°, 45°, 60°, and 90°. Sometimes it is also used to find the values for other angles like 180°, 270°, and 360° in the form of a table. Various patterns exist within trigonometric ratios and between their corresponding angles. Therefore, it is easy to predict the values of the trigonometric table and also use the table as a reference to calculate trigonometric values for other non-standard angles. The various trigonometric functions in Mathematics are sine function, cosine function, tan function, cot function, sec function, and cosec function. Before beginning, let us try to recall the trigonometric formulas listed below. 1. $$\sin x=\cos \left(90^{\circ}-x\right)$$ 2. $$\cos x=\sin \left(90^{\circ}-x\right)$$ 3. $$\tan x=\cot \left(90^{\circ}-x\right)$$ 4. $$\cot x= \tan \left(90^{\circ}-x\right)$$ 5. $$\sec x=\operatorname{cosec}\left(90^{\circ}-x\right)$$ 6. $$\operatorname{cosec} x=\sec \left(90^{\circ}-x\right)$$ 7. $$\frac{1}{ \sin x}=\operatorname{cosec} x$$ 8. $$\frac{1}{ \cos x}=\sec x$$ 9. $$\frac{1}{\tan x}=\cot x$$ ### Trigonometric Values Trigonometry is the study of the relationship between the sides of a triangle (right-angled triangle) and its angles. The term trigonometric value is used to collectively define values of different ratios, such as sine, cosine, tangent, secant, cotangent, and cosecant in a trigonometric table. Every trigonometric ratio is connected to the sides of a right-angled triangle, and the trigonometric values are found using these ratios. ### Standard Angle Trigonometry Tables The trigonometry ratio table is essentially a tabular collection of values for trigonometric functions of different conventional angles such as 0°, 30°, 45°, 60°, and 90°, as well as unusual angles such as 180°, 270°, and 360°. Because of the patterns that exist within trigonometric ratios and even between angles, it is simple to anticipate the values of the trigonometric ratios in a trigonometric table and use the table as a reference to compute trigonometric values for different other angles. Trigonometric ratios – sine, cosine, tangent, cosecant, secant, and cotangent – are listed in the table. Sin, cos, tan, cosec, sec, and cot are the abbreviations for these ratios. The values of the trigonometric ratios of these standard angles are best remembered. Learn Exam Concepts on Embibe ### Steps to Create a Trigonometric Table Students can follow the steps given below to make a sin cos tan table. Step 1: Create a table with the angles $$0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}$$, and $$90^{\circ}$$ on the top row and all trigonometric functions $$\sin , \cos , \tan , \operatorname{cosec}, \mathrm{sec}$$, and cot in the first column. Step 2: Determine the value of $$\sin$$. Write the angles $$0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}, 90^{\circ}$$ in ascending order and assign them values $$0,1,2,3,4$$ according to the order. So, $$0^{\circ} \rightarrow 0 ; 30^{\circ} \rightarrow 1 ; 45^{\circ} \rightarrow 2 ; 60^{\circ} \rightarrow 3 ; 90^{\circ} \rightarrow 4$$. Then divide the values by $$4$$ and square root the entire value. $$0^{\circ} \rightarrow \sqrt{\frac{0}{4}}=0 ; 30^{\circ} \rightarrow \sqrt{\frac{1}{4}}=\frac{1}{2} ; 45^{\circ} \rightarrow \sqrt{\frac{2}{4}}=\frac{1}{\sqrt{2}} ; 60^{\circ} \rightarrow \sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2} ; 90^{\circ} \rightarrow \sqrt{\frac{4}{4}}=1$$. This gives the values of sine for these five angles. Now for the remaining three, use: $$\sin \left(180^{\circ}-x\right)= \sin x \quad \sin \left(180^{\circ}+x\right)=\,- \sin x \sin \left(360^{\circ}-x\right)=\,- \sin x$$ This means, $$\sin \left(180^{\circ}-0^{\circ}\right)=\sin 0^{\circ} \quad \sin \left(180^{\circ}+90^{\circ}\right)=\,- \sin 90^{\circ} \sin \left(360^{\circ}-0^{\circ}\right)=\,- \sin 0^{\circ}$$ Step 3: Determine the value of $$\cos$$. $$\sin \left(90^{\circ}-x\right)=\cos x$$ To find values for $$\cos x$$, use this formula. For example, equals $$\left(90^{\circ}-45^{\circ}\right)= \sin 45^{\circ},\left(90^{\circ}-30^{\circ}\right)=\sin 60^{\circ}$$ and vice versa. You can quickly determine the value of the $$\cos$$ function by using this method: Step 4: Determine the value of $$\tan$$. We know that $$\sin$$ divided by $$\cos$$ equals the $$\tan$$. $$\frac{{\sin }}{{\cos }} = \tan$$ Divide the value of $$\sin$$ at $$0^{\circ}$$ by the value of $$\cos$$ at $$0^{\circ}$$ to get the value of $$\tan$$ at $$0^{\circ}$$. Take a look at the sample below. $$\tan 0^{\circ}=\frac{0}{1}=0$$ In the same way, the table would be as follows. Step 5: Determine the value of $$\cot$$. The reciprocal of $$\tan$$ is equal to the value of $$\cot$$. Divide $$1$$ by the value of $$\tan$$ at $$0^{\circ}$$ to get the value of $$\cot$$ at $$0^{\circ}$$. So, $$\cot 0^{\circ}=\frac{1}{0}=\infty$$ or Not Defined will be the value. In the same way, a $$cot$$ table is provided below. Step 6: Determine the value of $$\operatorname{cosec}$$. The reciprocal of $$\sin$$ at $$0^{\circ}$$ is the value of $${\text{cosec}}$$. $$\operatorname{cosec} 0^{\circ}=\frac{1}{0}=\infty$$ or Not Defined $$\operatorname{cosec} 0^{\circ}=\frac{1}{0}=\infty$$ or Not Defined In the same way, a table for cosec is provided below. Step 7: Determine the value of $$\mathrm{sec}$$. All reciprocal values of $$\cos$$ can be used to calculate the value of $$\sec$$. The value of $$\sec$$ on $$0^{\circ}$$ is the inverse of the value of $$\cos$$ on $$0^{\circ}$$. As a result, the value will be $$\sec 0^{\circ}=\frac{1}{1}=1$$. Similarly, the table for a sec is shown below. Hence, the required trigonometric table for all the trigonometric ratios is as follows Practice Exam Questions ### Tricks to Remember Trigonometry Table The trigonometry table can be useful in a variety of situations, and it is simple to remember. Remembering the trigonometric table is simple if you know the trigonometry table formula and trigonometry table trick, as trigonometry formulas are used to create the trigonometry ratios table. Let’s learn how to recall the trigo table with just one hand! As illustrated in the image, give each finger the standard angles. We will count our fingers while filling the sine table, but we will just fill the data in reverse order for the cos table. 1st Step: To calculate the standard angle for the sine table, count the fingers on the left side. 2nd Step: Divide the number of fingers by four. 3rd Step: Take out the square root of the ratio. Example 1: Because there are no fingers on the left side for $$\sin 0^{\circ}$$, we will use $$0$$. We obtain $$0$$ when we divide zero by four. We may determine the value of $$\sin 0^{\circ}=0$$ by taking the square root of the ratio. Example 2: On the left-hand side, there are three fingers for $$\sin 60^{\circ}$$. We obtain $$\left(\frac{3}{4}\right)$$ when we divide $$3$$ by $$4$$. We may determine the value of $$\sin 30^{\circ}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$$ by taking the square root of the ratio $$\left(\frac{3}{4}\right)$$. Similarly, we may fill the table with values for $$\sin 30^{\circ}, 45^{\circ}$$, and $$90^{\circ}$$. PRACTICE QUESTIONS ON TRIGNOMETRIC TABLE ### Trigonometric Values Table for Unit Circle The unit circle is a circle centred at the origin and always has a radius of $$1$$. The equation of the unit circle is $$x^{2}+y^{2}=1$$. An ordered pair along the unit circle $$(x, y)$$ can also be known as $$( \cos \theta, \sin \theta)$$, since the $$r$$ value on the unit circle is always $$1$$. So, to find the trigonometric function values for $$45^{\circ}$$ you can look at the unit circle and easily see that $$\sin 45^{\circ}=\frac{\sqrt{2}}{2}, \cos 45^{\circ}=\frac{\sqrt{2}}{2}$$ With that information, we can easily find the values of the reciprocal functions. $$45^{\circ}=\frac{2}{\sqrt{2}}=\frac{2 \sqrt{2}}{2}=\sqrt{2}, \sec 45^{\circ}=\sqrt{2}$$ We can also find the tangent and cotangent function values using the quotient identities $$\tan 45^{\circ}=\frac{\sin 45^{\circ}}{\cos 45^{\circ}}=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=1$$ $$\cot 45^{\circ}=1$$ ### Solved Examples – Trigonometry Table Q.1. If $$\beta=30^{\circ}$$, prove that $$3 \sin \beta-4 \sin ^{3} \beta=\sin 3 \beta$$. Ans: L.H.S $$=3 \sin \beta-4 \beta$$ $$=3 \sin 30^{\circ}-430^{\circ}$$ $$=3\left(\frac{1}{2}\right)-4\left(\frac{1}{2}\right)^{3}$$ $$=\frac{3}{2}-4 \times \frac{1}{8}$$ $$=\frac{3}{2}-\frac{1}{2}$$ $$=1$$ $$\mathrm{RHS}= \sin 3 \beta$$ $$= \sin 3 \times 30^{\circ}$$ $$= \sin 90^{\circ}$$ $$=1$$ Therefore, $$\mathrm{LHS}=$$ RHS (Proved) Q.2. If $$\sin (x+y)=1$$ and $$\cos (x-y)=\frac{3}{\sqrt{2}}$$, find $$x$$ and $$y$$. Ans: $$\sin (x+y)=1$$ $$\Rightarrow \sin (x+y)=\sin 90^{\circ}$$, (since $$\sin 90^{\circ}=1$$) $$\Rightarrow x+y=90^{\circ} \ldots \ldots \ldots \ldots \ldots \ldots . . .(i)$$ $$\cos \cos (x-y)=\frac{3}{\sqrt{2}}$$ $$\Rightarrow(x-y)=\cos \cos 30^{\circ}$$ $$\Rightarrow x-y=30^{\circ} \ldots \ldots \ldots \ldots \ldots \ldots \ldots(ii)$$ Adding, $$(i)$$ and $$(ii)$$, we get $$x+y=90^{\circ}$$ $$x-y=30^{\circ}$$ $$2 x=120^{\circ}$$ $$x=60^{\circ}$$,(Dividing both sides by $$2$$) Putting the value of $$x=60^{\circ}$$ in $$(i)$$ we get, $$60^{\circ}+y=90^{\circ}$$ Subtract $$60^{\circ}$$ from both sides $$60^{\circ}+y=90$$ $$y=30^{\circ}$$ Therefore, $$x=60^{\circ}$$ and $$y=30^{\circ}$$. Q.3. Find the value of $$\frac{4}{3} 60^{\circ}+330^{\circ}-230^{\circ}-\frac{3}{4} 60^{\circ}$$ Answer: The given expression is $$\frac{4}{3} 60^{\circ}+330^{\circ}-230^{\circ}-\frac{3}{4} 60^{\circ}$$. Putting the value of the angles using the trigonometric table, we have $$=\frac{4}{3}\left(\sqrt{3}^{2}\right)+3\left(\frac{\sqrt{3}}{2}\right)^{2}-2\left(\frac{2}{\sqrt{3}}\right)^{2}-\frac{3}{4}\left(\frac{\sqrt{3}}{3}\right)^{2}$$ $$=\frac{4}{3} \times 3+3 \times \frac{3}{4}-2 \times \frac{4}{3}-\frac{3}{4} \times \frac{3}{9}$$ $$=4+\frac{9}{4}-\frac{8}{3}-\frac{1}{4}$$ $$=\frac{10}{3}=3 \frac{1}{3}$$ Q.4. Find the value of $$\tan 45^{\circ}$$. Ans: $$\tan 45^{\circ}=\frac{\sin 45^{\circ}}{\cos 45^{\circ}}$$ Put the values of $$\sin 45^{\circ}$$ and $$\cos 45^{\circ}$$ from the trigonometric values table, we get $$\tan 45^{\circ}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}$$ Hence, $$\tan 45^{\circ}=1$$ Q.5. Find the value of $$\sin \frac{\pi}{6}$$. Ans: The value of $$\sin \frac{\pi}{6}= \sin 30^{\circ}=\frac{1}{2}$$. Check our NCERT Maths Solutions For Classes 10, 11, and 12: ### Summary of Trigonometric Table In this article, we have discussed the trigonometric values table in degrees and radians. We learnt the steps to create the table, tricks to remember the values in it and trigonometric values table for the unit circle, and solved some examples. Also, we solved some important questions on trigonometry table Class 10. Attempt Mock Tests ### Frequently Asked Questions on Trigonometry Here, we have enlisted some of the most important frequently asked questions related to the Trigonometric Values Table. Candidates must read these questions and answers to clear out their doubts regarding the same subject. Q1. What is trigonometry? Ans. Trigonometry is the branch of mathematics that deals with the relationship between the sides of a triangle (right-angled triangle) and its angles. Q.2. How do you create a trigonometric ratio table? Ans: The steps for creating and remembering a trigonometric table are outlined below. 1st Step: Create a table with the angles $$0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}$$, and $$90^{\circ}$$ on the top row and all trigonometric functions $$\sin , \cos , \tan , \operatorname{cosec}, \sec$$, and cot in the first column. 2nd Step: Determine the value of $$\sin$$. 3rd Step: Reverse the order of the values of $$\sin$$ to get the values of $$\cos$$. 4th Step: The values of $$\sin$$ divided by the values of $$\cos$$ give the value of $$\tan$$. 5th Step: The reciprocal of $$\tan$$ is equal to the value of $$\cot$$. 6th Step: The reciprocal of $$\sin$$ values is the value of $${\text{cosec}}$$. 7th Step: All reciprocal values of $$\cos$$ can be used to calculate the value of $$\sec$$. Q.3. What is the trigonometric values table? Ans: The trigonometric table is made up of the trigonometric ratios sine, cosine, tangent, cosecant, secant, and cotangent, which are all connected. The values of standard trigonometric angles such as $$0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}$$, and $$90^{\circ}$$ are found in this table. Q.4. How do you learn values in trigonometry? Ans: You just need to memorise the value of sine, then the value of cosine can be determined by putting sine data in the reverse order. The value of the tangent can be determined by dividing sine by cosine. The value of secant can be determined by taking the reciprocal of cosine, and the value of cosecant can be determined by the reciprocal of sine. Q.5. What is the ratio for sine? Ans: Sine ratios are proportions of the length of the opposite side of the angle they represent to the hypotenuse. $$\sin \theta=\frac{\text { Opposite side }}{\text { Hypotenuse }}$$ Q.6. What are the three basic trigonometric ratios? Ans: There are three basic trigonometric ratios: sine, cosine, and tangent. By using these, we can determine the values of the other three trigonometric ratios by using the relationship $$\frac{1}{ \sin x}=\operatorname{cosec} x$$ $$\frac{1}{\cos x}= \sec x$$ $$\frac{1}{ \tan x}= \cot x$$ Take Free CBSE Class 10 Mock Tests Now! Related Links: Now that you know everything about the trigonometry table formula, try to create your own sin cos tan table and keep all the tips and tricks in mind. You can refer to the solved examples listed in this article while appearing for your trigonometry tests. We hope this detailed article on the trigonometry table helps you. If you have any queries, feel free to ask in the comment section below. We will get back to you at the earliest. Till then, stay tuned to Embibe for all updates on the trigonometry table, exam preparation tips, and the latest academic articles! Take Free Mock Tests related to Trigonometry

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Let $R$ be a commutative ring with unity, $S \subset R$ a multiplicative closed subset and $I \subseteq R$ be an ideal. Show that$$S^{-1} \sqrt{I}=\sqrt{S^{-1}I}$$ and $$S^{-1}J(R)=J(S^{-1}R),$$ where $J(R)$ is the Jacobson radical of $R.$ For the first part I used the fact that $$\sqrt{I}=\{x\in R\mid x^{n}\in I\text{ for some }n\ge1\},$$ and I think I managed to do it, but I'm stuck on the second part. I'm trying to use the fact$$J(R)=\{x\in R \mid 1+rx \hspace 2mm\text{is a unit}\hspace 2mm \forall r\in R\},$$ but I don't think I'm there yet. Can you help? - Note that Jacobson ideal of $R$ is the intersection of all maximal ideals of $R.$ – Ehsan M. Kermani Jan 30 '13 at 7:19 @ehsanmo Yes I know, but I think using units will be easier I don't know.. – i.a.m Jan 30 '13 at 7:33 You should be aware of the correspondence between prime ideals of $S^{-1}R$ and prime ideals of $R$ with empty intersection with $S.$ – Ehsan M. Kermani Jan 30 '13 at 7:47 @ehsanmo I'm aware of that but the question does not say that $S$ and $I$ are disjoint or not.. – i.a.m Jan 30 '13 at 8:10 If we take $R=\mathbb Z$, $I=p \mathbb Z$ and $S=\mathbb Z \setminus p\mathbb Z$ then we have $S^{-1}Jac(R)=(0)$ but $Jac(S^{-1}R)=Jac(\mathbb Z_{(p)})=p\mathbb Z_{(p)}$. Maybe if you assume that $S$ has empty intersection with every maximal ideal... – JSchlather Jan 30 '13 at 8:32 The question is wrong as currently stated and there doesn't seem to be a good way to fix it. For an easy counterexample we can consider the localization of $\mathbb Z$ at the prime ideal $p\mathbb Z$. That is we take $S=\mathbb Z \setminus p \mathbb Z$ and $R=\mathbb Z$. Then it's easy to see that $Jac(R)=0$ since no number is divisible by an infinite number of primes for instance. Thereby $S^{-1}Jac(R)=0$. But we know that $S^{-1}R=(\mathbb Z \setminus p\mathbb Z)\mathbb Z=\mathbb Z_{(p)}$ is a local ring with its maximal ideal given by $p\mathbb Z_{(p)}$ so it has non-trivial Jacobson radical.

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# Tag Info 8 Solutions force exactly one eigenvalue to be zero. So we solve for the condition that an eigenvalue vanish, and check that rank is two. mat = {{1, x, 3}, {2, 4, 5}, {2, 4, x}}; candidateSols = Flatten[Map[Solve[# == 0, x] &, Eigenvalues[(mat)]]] (* Out[997]= {x -> 2, x -> 5} *) Both pass the test: Map[MatrixRank[mat /. #] &, candidateSols] ... 8 SolveAlways[] makes quick work of your problem: SolveAlways[(3 t^2 + 5 t + a) (4 t^2 + b t - 2) == 12 t^4 + 26 t^3 - 8 t^2 - 16 t + 6, t] {{a -> -3, b -> 2}} 7 It does not look possible to solve this for general a, b. But for specific values of these, Mathematica can solve it for c Manipulate[ expr = Exp[(-Pi*a)/c] + Exp[(-Pi*b)/c] - 1; Grid[{{Row[{"equation is ", expr, "==0"}]}, {Plot[expr, {c, -2, 2}]}, {N@Solve[expr == 0, c]} }] , {{a, 1, "a"}, -2, 2, 1/10, Appearance -> "Labeled"}, {{b, 1, "b"... 6 Use FindRoot Clear["Global*"] f[t_?NumericQ] := Module[{tt = SetPrecision[t, 30]}, NIntegrate[Sqrt[1 - Sqrt[x]]/ArcTan[tt + ArcTan[x]], {x, 0, 1}, WorkingPrecision -> 30]] Plot the sides of the equation Plot[{f[t], Pi^2/6}, {t, 0, 1}, WorkingPrecision -> 30, PlotLegends -> Placed["Expressions", {.5, .7}]] sol1 = FindRoot[f[t] == Pi^2/6,... 6 One way could be ClearAll[t,a,b} z1 = CoefficientList[(3*t^2 + 5*t + a)*(4*t^2 + b*t - 2), t]; z2 = CoefficientList[12*t^4 + 26*t^3 - 8*t^2 - 16*t + 6, t]; eqs = Thread[z1 == z2]; Solve[eqs,{a,b}] 5 Easiest way to solve this problem quickly is to very thoroughly re-state the problem. I do not promise that this is a complete solution to the problem, but I hope it provides significant insight into how to approach finding any such complete solution. From what I understood from the comments, we are looking for matrices consisting of only $-1$ and $+1$ ... 4 (1) What AlgebraicRulesData really is, is obsolete.Yes, it is still supported. Or ignored, to be more accurate. As for the copy/paste business, it has an internal validation flag. Those do not survive copying and in the case of AlgebraicRulesData there is no way (or attempt) to revalidate them. Without the flag being set, rule replacement will not handle ... 4 Perhaps, ClearAll[r, i, j, u, v, k, h] Simplify[u[i, j + 1] == (1 - 2 r) u[i, j] + r (u[i + 1, j] + u[i - 1, j]), {r == v k/h^2, (u[i, j + 1] - u[i, j])/k == v (u[i + 1, j] - 2 u[i, j] + u[i - 1, j])/h^2} ] True 3 Not a full answer... I took what proved to be a similar approach as @eyorble, in that I turn it into a graph problem. I also just demonstrate how to get a lot of solutions, but not the guaranteed total number of solutions. I think the total number blows up combinatorically, so probably not worth counting. The main idea is that the matrix $A$ is invertible,... 3 try this Reduce[ForAll[t, (3*t^2 + 5*t + a)*(4*t^2 + b*t - 2) == 12*t^4 + 26*t^3 - 8*t^2 - 16*t + 6], a] b == 2 && a == -3 3 This is more an extended comment than a formal answer. With the code and parameters given in the question, the first plot in the question should be something like Plot[Evaluate@Table[(i[k] /. sol)[t], {k, 1.5, 2.5, 0.2}], {t, 0, 10}, PlotRange -> All, ImageSize -> Large, AxesLabel -> {t, i}, LabelStyle -> {15, Bold, Black}] The maximum of ... 2 In addition to Marius' comment, you can use Solve with Method -> Reduce: Solve[ { Sin[ϕ1] + 0.6 Sin[ϕ1 + ϕ2] == 0 && ϕ1 > -π && ϕ1 <= π, 0.1 Sin[ϕ2] + 0.6 Sin[ϕ1 + ϕ2] == 0 && ϕ2 > -π && ϕ2 <= π }, {ϕ1, ϕ2}, Method->Reduce ] Solve::ratnz: Solve was unable to solve the system with ... 2 Exists and ForAll are qualifier statements and you can attempt to Resolve them to remove the qualifiers as follows: qualifierStatement=Exists[x,a x^2+b x+c==0] Now to resolve it under the real domain as follows: resolvedStatement=Resolve[qualifierStatement,Reals] Now to get your condition you need to ensure that the quadratic actually exist which is only ... 2 The first predicate is false for a->0, b->0, c->1: (b==0 && ((c>0 && a<0)||(a>0 && c<0)))||(b!=0 && 4 a c<=b^2)||c==0 /. {a->0, b->0, c->1} False The second predicate is true for this case: b^2 - 4 a c >= 0 /. {a->0, b->0, c->1} True 2 wuyudi had the right idea to use Tuples[], but as the OP notes, trying to generate all of them just to cull entries afterwards can rapidly become combinatorically prohibitive. Instead, use the equivalence of Tuples[] with the problem of generating all the $m$-digit base $b$ numbers: With[{nums = {-1, 1, 2, 4}, count = 5}, Table[With[{id = nums[[... 2 Select[Tuples[{-1, 1, 2, 4}, 5], Mod[Total@#, 3] == 0 &] 2 The output of Solve (stored in sol) gives more than one solution (actually, there are 9 possible solutions): sol = Solve[((I*(del + gamma*Abs[a1]^2) - k/2) a1 - (ke/2) a2 + Sqrt[ke] s == 0) && ((I*(del + gamma*Abs[a2]^2) - k/2) a2 - (ke/2) a1 + Sqrt[ke] s == 0), {a1, a2}]; Length@% 9 You can plot them individually: n = 1; ... 2 To plot the solution for a range of delc, make these changes to your code: (1) Remove delc = 1; Leave delc undefined. It will be your parameter. (2) Keep the ParametricDSolveValue command and its first argument, but change the other arguments to get s = ParametricNDSolveValue[ ... , {1/2*(V11[t] + V22[t] - 2*V12[t])^(-1)}, {t, 0, 100}, delc]; ... 1 Here is one possibility: conds = K > 0 && p > 0 && h > 0 && r > 0; Complement[Reduce[K*(r - h)/r > 0 && (p*h)/(r - h) < p && conds, h, Reals], conds] 0 < h < r/2 1 If the symbolic Tan[theta*Pi/180] is replaced by the number Tan[theta*Pi/180.], then the variable order is preserved: theta = 120; NSolve[ -0.034298780658685656 + y == Tan[theta*Pi/180.] (-0.012483735231386907 + x) && -0.028199999999999996 + y == 0 && x > 0 && y > 0, {x, y}] (* {{x -> 0.0160049, y -> 0.0282}} *) I ... 1 another option is to find a function to fit the plot, then use Solve to find the root. ClearAll[f, x, t]; f[t_?NumericQ] := NIntegrate[Sqrt[1 - Sqrt[x]]/ArcTan[t + ArcTan[x]], {x, 0, 1}] Plot[f[t] - Pi^2/6, {t, 0, 1}] So your t is somewhere around 0.1. To find it, you could fit the curve to a function and then use Solve or root finding. For example data=... 1 You can search it. Select[Table[{t, NIntegrate[ Sqrt[1. - Sqrt[x]]/ArcTan[t + ArcTan[x]], {x, 0, 1}]}, {t, 0, 1, 0.001}], RealAbs[#[[2]] - Pi^2/6] < 0.01 &] {{0.112, 1.64927}, {0.113, 1.643}, {0.114, 1.63679}} so 0.112 ~ 0.114 is where the answer at. With more search, you can get more precious answer. 1 This is a precision issue \$Version (* "12.0.0 for Mac OS X x86 (64-bit) (April 7, 2019)" *) Clear[dBeta, solnEquation, solnFun]; dBeta[k_Integer, x_, var1_] = Derivative[k, 0][Beta[#1, #2, 2] &][x, var1]; solnEquation[var1_, var2_, x_] = Sum[Binomial[5, k]*dBeta[k, x, var1]*var2^k, {k, 0, 5}]; equation[var1_, var2_, x_] := solnEquation[var1, var2, ... 1 Get a fast impression e.g. for a^2 + x - y^2 == 0 with ContourPlot3D[a^2 + x - y^2 == 0, {x, -3, 3}, {y, -3, 3}, {a, -4, 4}, ViewPoint -> {0, 0, Infinity}, AxesLabel -> {x, y}, MeshFunctions -> Function[{x, y, a}, a], Mesh -> {Range[-3, 3]}] 1 Assuming you want positive solutions, you can do the following: g[a_,b_] := Quiet @ Solve[Exp[(-Pi*a)/c] + Exp[(-Pi*b)/c] == 1 && c>0, c] Examples: g[1, 1] g[1, Pi] g[1.414, 3.928] {{c -> π/Log[2]}} {{c -> Root[{ 1 + E^(-π/# + π^2/#) - E^(π^2/#)& , 8.4433225565523434202763737162354220192220.601814494499823}]}} {{... Only top voted, non community-wiki answers of a minimum length are eligible

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# How does a rubber band powered plane work? ## How does a rubber band powered plane work? This is a toy airplane, powered by a rubber band. The propeller is wound up by the user, storing energy in the rubber band as it winds. The user will then throw the airplane just like a paper airplane, except the rotating propeller will provide additional thrust, and the airplane will be able to travel much further. Why does the airplane need a rubber band to fly? In the rubber-band powered airplane, potential energy is stored in the twisted rubber band which powers the propeller. The propeller provides the thrust, which pushes the airplane forward according to Newton’s Third Law. The plane is able to move through the air and fly by lift generated by the shape of the wings. How do you make a rubber band plane fly longer? The trick is to take the elastic off the model then roll it around in your hand with the product. This alone will allow the elastic band to take more winding. This reduces the friction on the surface of the elastic. The elastic will also last longer. ### What is a rubber powered plane? A plane that uses rubber-power uses an elastic band for propulsion. The elastic is wound (twisted) to store potential energy. Energy is released as it unwinds and turns the propeller. A rubber-power airplane is a type of wind-up toy. How do you make a simple rubber band airplane? Step 1: Make a paper airplane of any design. You can try one from Fold ‘N Fly. Step 2: Punch a hole about 1.5 – 2 inches from the airplane’s nose. Step 3: Thread the rubber band through the hole, then by slipping one end of the rubber band through the other and pulling gently until it forms a knot. How do you make a rubber band glider? Rubberband Glider in Less Than 1\$ 1. Step 2: Making the Wings. – cut a rectangle of length 30cm and breath 7.5cm. 2. Step 3: The Fuselage. – cut a rectangle 25cm in length and 5 cm in breath. 3. Step 4: Rudder and Elevator. Rudder. 4. Step 5: Assembly of Parts. Mark 7cm on the fuselage from the front and paste the wings with tape. ## How do you wind a rubber band plane? Five Squirrelly rules from the video! 1. The elastic is “pre-wound” before it is hooked onto the toothpick. 2. Don’t throw the plane. 3. Release the propeller, then release the plane one second after. 4. Pick up by the “nose”. 5. If it climbs too much, move the wing back slightly. Do gliders have motors? Gliders. A glider is a special kind of aircraft that has no engine. There are many different types of gliders.

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# numpy in python : numpy exercise and solution : This blog post tutorial  are very useful for all students in this post I have cover basic numpy operation and finally give you numpy exercise for practice ans solution, so let's enjoy. Numpy is most powerful and widely used package for data science and machine learning. It contain the following other things: fig 01) numpy in python | python course_02 1. a powerful N-dimensional array object 3. tools for integrating C/C++ and Fortran code 4. useful linear algebra, Fourier transform, and random numbers capabilitie It's also  scientific uses, NumPy can also be used as an multi-dimensional container of generic and Arbitary data type. This allows Numpy to seamlessly and speedily integrates with a wide variety of database. ## INSTALLATION: To install NumPy, we frequently recommend using a scientific Python distribution. See Installing the SciPy Stack for detail. Many online tutorial, courses, and books are available to get started with NumPy. For a quick introduction to Numpy we provides the NumPy Tutorial. We also recommend the SciPy Lecture Note for a broader introduction to the scientific Python ecosystems. ### USE NUMPY IN PYTHON : SIMPLE NUMPY OPERATION : 1) simply import the numpy library. and then perform the operation: `import numpy as np` `arr = np.arange(0,10)` `arr + arr` output: ```array([ 0, 2, 4, 6, 8, 10, 12, 14, 16, 18]) ``` 2)sub. operation : `arr - arr` output: `array([0,0,0,0,0,0,0,0,0,0])` #### Universal Array operation: Numpy comes under  many universal array functions, which are essentially just mathematical operation you can use to perform the operation across on array. Let's see some example: ```#Take Square Root np.sqrt(arr)``` ``` array([ 0. , 1. , 1.41421357, 1.73205083, 2. , 2.236066798, 2.449489734, 2.645751341, 2.82842714, 3. ]) ``` Stack see Some of FAQ : Does Python come with NumPy? NumPy is an open source Python package for python for data science and machine learning. NumPy can be support large, multidimensional array and the matrices. But also NumPy array are not flexible like Python list, you can store same data type in each columns. NumPy is essential for another packages like SciPy,  OpenC  and Vscikit-learn. Where is NumPy used? NumPy package can be used to perform different operation. NumPy is an open source Python package for python for data science and machine learning.The Narray is a multidimensional array used to store value of same datatypes in pyrhon. These array are indexed just like starts,Sequences with zero. Is NumPy a framework? The data structure in numpy is the N-dimensional array, which is much more flexible and efficient than standard Python list. NumPy is a numerical and mathematical library. Tags: numpy in python,Does Python come with NumPy?,Where is NumPy used?,Is NumPy a framework?

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