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# How to move an object along a circumference of another object? I am so out of math that it hurts, but for some of you this should be a piece of cake. I want to move an object around another along its ages or circumference on a simple circular path. At the moment my game algorithm knows how to move and position a sprite just at the edge of an obstacle and now it waits for the next point to move depending on various conditions. So the mathematical problem here is how to get (aX, aY) and (bX, bY) positions, when I know the Centre (cX, cY), the object position (oX, oY) and the distance required to move (d) • Is d a linear distance or is it an arc? Jun 26, 2012 at 22:37 • It is a linear distance in pixels Jun 26, 2012 at 22:44 • Are you familiar at all with what vectors are and basic operations on them? Jun 26, 2012 at 23:00 • @Patrick No, I guess I will have to do a course on Vectors. Since this is a frame by frame animation the code should be fast and optimised. Jun 26, 2012 at 23:24 (CAVEAT: I'm using two approximations here: the first takes d as an arc length, and the second takes it as an orthogonal length. Both of these approximations should be good for relatively small values of d, but they don't fulfill the precise question as clarified in comments.) The math on this, fortunately, is relatively straightforward. First of all, we can find the relative vector from our center position to our current position: deltaX = oX-cX; deltaY = oY-cY; And once we have this relative vector, then we can know the radius of the circle we're working on by finding the length of it: radius = sqrt(deltaX*deltaX+deltaY*deltaY); What's more, from our relative vector we can find the precise angle that the line from cX to oX is at: curTheta = atan2(deltaX, deltaY); Now things get a little bit trickier. First of all, understand that the circumference of a circle — that is, the 'arc length' of an arc with an angular measure of 2π — is 2πr. In general, the arc length of an arc with an angular measure of θ along a circle of radius r is just θr. If we were to use the d in your diagram as the arc length, and since we know the radius, we can find the change in theta to get us to the new position by just dividing out: deltaTheta = d/radius; // treats d as a distance along the arc For the case where d needs to be a linear distance, things are a little more complicated, but fortunately not much. There, d is one side of an isoceles triangle whose other two sides are the radius of the circle (from cX/cY to oX/oY and aX/aY respectively), and bisecting this isoceles triangle gives us two right triangles, each of which has d/2 as one side and radius as the hypotenuse; this means that the sine of half our angle is (d/2)/radius, and so the full angle is just twice this: deltaTheta = 2*asin(d/(2*radius)); // treats d as a linear distance Notice how if you took the asin out of this formula, and cancelled the 2s, this would be the same as the last formula; this is the same as saying that sin(x) is approximately x for small values of x, which is a useful approximation to know. Now we can find the new angle by just adding or subtracting: newTheta = curTheta+deltaTheta; // This will take you to aX, aY. For bX/bY, use curTheta-deltaTheta Once we have the new angle, then we can use some basic trig to find our updated relative vector: newDeltaX = radius*cos(newTheta); and from our center position and our relative vector we can (finally) find the target point: aX = cX+newDeltaX; aY = cY+newDeltaY; Now, with all this, there are some big caveats to be aware of. For one, you'll notice that this math is mostly floating-point, and in fact it almost has to be; trying to use this method to update in a loop and rounding back to integer values at every step can do everything from making your circle not close (either spiralling inward or outward every time you go around the loop) to not getting it started in the first place! (If your d is too small, then you might discover that the rounded versions of aX/aY or bX/bY are exactly where your start position oX/oY was.) For another, this is very expensive, especially for what it's trying to do; in general, if you know your character is going to be moving in a circular arc, you should plan out the whole arc in advance and not tick it from frame to frame like this, since many of the most expensive calculations here can be front-loaded to cut down on costs. Another good way to trim back the costs, if you really want to update incrementally like this, is to not use trig in the first place; if d is small and you don't need it to be exact but just very close, then you can do a 'trick' by adding a vector of length d to oX/oY, orthogonal to the vector towards your center (note that a vector orthogonal to (dX, dY) is given by (-dY, dX) ), and then shrink it down to the right length. I won't explain this code quite so step-by-step, but hopefully it'll make sense given what you've seen so far. Note that we 'shrink down' the new delta vector implicitly in the last step, where we add it to our center to get the updated point: deltaX = oX-cX; deltaY = oY-cY; newDeltaX = deltaX+orthoX; newDeltaY = deltaY+orthoY; newLength = sqrt(newDeltaX*newDeltaX+newDeltaY*newDeltaY); • Steven I think I am going to try the approximation first as this is just a game where is more important to feel natural and interesting than precise. Also speed matters. Thank you for this long and good tutorial! Jun 26, 2012 at 23:39 • Wow, Steven your approximation is working like a dream! Can you tell me how to change your code to get bX, bY. I am not clear yet on your orthogonal concept... Jun 27, 2012 at 0:07 • Sure! You'll really want to understand vector math at some point, and once you do I suspect this will be second nature; to get bX/bY you just have to go 'the other way around' - in other words, rather than adding the (particular) orthogonal vector, just subtract it. In terms of the code above, this would be 'newDeltaX = deltaX-orthoX; newDeltaY = deltaY-orthoY;', followed by the same computation of newLength, and then 'bX=cX+newDeltaXradius/newLength; bY=cY+newDeltaYradius/newLength;'. Jun 27, 2012 at 0:16 • Basically, that code would point newDeltaX/newDeltaY in the direction of bX/bY (instead of in the direction of aX/aY), then trim to fit and add to the center just as you would to get aX/aY. Jun 27, 2012 at 0:17 Form a triangle using the two sides you already have (one side is from 'c' to 'o', the other from 'o' to 'a'), and the third side goes from 'a' to 'c'. You don't know where 'a' is just yet, just imagine there is a point there for now. You'll need trigonometry to calculate the angle of the angle that is opposite to the side 'd'. You have the length of the sides c<->o and c<->a, because they're both the radius of the circle. Now that you have the length of the three sides of this triangle you cannot yet see, you can determine the angle that is opposite to the 'd' side of the triangle. Here's the SSS (side-side-side) formula if you need: http://www.teacherschoice.com.au/maths_library/trigonometry/solve_trig_sss.htm Using the SSS formula you have the angle (which we'll call 'j') that is opposite to side 'd'. So, now we can calculate (aX, aY). // This is the angle from 'c' to 'o' float angle = Math.atan2(oY - cY, oX - cX) // Add the angle we calculated earlier. angle += j; Vector2 a = new Vector2( radius * Math.cos(angle), radius * Math.sin(angle) ); Make sure the angles you're calculating are always in radians. If you need to calculate the radius of the circle, you can use vector subtraction, subtract point 'c' from point 'o', then get the length of the resulting vector. float lengthSquared = ( inVect.x * inVect.x + inVect.y * inVect.y + inVect.z * inVect.z ); Something like this should do, I believe. I do not know Java, so I guessed on the exact syntax. Here's the image given by user Byte56 to illustrate what this triangle might look like: • I was making an answer, but this is it. You're welcome to use the image I made :) i.imgur.com/UUBgM.png Jun 26, 2012 at 23:06 • @Byte56: Thanks, I didn't have any image editor handle to illustrate. Jun 26, 2012 at 23:11 • Note that the radius has to be computed too; there should be more straightforward ways of getting j than the full SSS calculation, since we have an isoceles triangle.) Jun 26, 2012 at 23:16 • Yes, that seems simple, even to me! Android does not have Vector2 so I guess I can just use the values separately. Intrestingly I found Vector2 class manually created for Android here: code.google.com/p/beginning-android-games-2/source/browse/trunk/… Jun 26, 2012 at 23:19 • (I've tweaked my own answer to find the correct linear distance - the second calculation of deltaTheta there, as 2*asin(d/(2*radius)), is how you would find j here.) Jun 26, 2012 at 23:29 To make obj2 rotate around obj1 maybe try: float angle = 0; //init angle //call in an update obj2.x = (obj1.x -= r*cos(angle)); obj2.y = (obj1.y += r*sin(angle)); angle-=0.5; • This doesn't show how to get the angle in the first place, and you appear to be showing how to orbit, instead of finding the coordinates like the question asks. Jun 26, 2012 at 22:43 • Lewis, thanks for showing how to orbit an object around a point. It may come useful... Jun 26, 2012 at 23:49

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# If y=xe^7 is a solution of a linear homogeneous 2nd order DE, then another solution might be If $y=x{e}^{7x}$ is a solution of a linear homogeneous 2nd order DE, then another solution might be? You can still ask an expert for help • Live experts 24/7 • Questions are typically answered in as fast as 30 minutes • Personalized clear answers Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Damarion Pierce For a second order linear homogenous equations, If roots of auxiliary equation are 1. real and distinct then the solution is $y={C}_{1}{e}^{{m}_{1}x}+{C}_{2}{e}^{{m}_{2}x}$ 2.real and equal then the solution is $y={C}_{1}{e}^{mx}+{C}_{2}x{e}^{mx}$ 3.complex then the solution is $y={e}^{\alpha x}\left({C}_{1}\mathrm{cos}\left(\beta x\right)+{C}_{2}\mathrm{sin}\left(\beta x\right)\right)$ Given $y=x{e}^{7x}$ is the solution the solution is similar for the case of real and equal roots therefor the other solution is $y={e}^{7x}$

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0 # What is the circumference of a circle with a diameter of 28 inches? Wiki User 2012-08-20 15:34:46 Best Answer Circumference = pi*diameter = 87.96 inches. Wiki User 2012-08-20 15:34:46 This answer is: Study guides 20 cards ➡️ See all cards 3.74 818 Reviews ## Add your answer: Earn +20 pts Q: What is the circumference of a circle with a diameter of 28 inches? Write your answer... Submit Still have questions? People also asked

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# Linear Regression And Correlation Coefficient Worksheet Answers Saturday, May 7th 2022. | Worksheet Linear Regression And Correlation Coefficient Worksheet Answers - There are a lot of affordable templates out there, but it can be easy to feel like a lot of the best cost a amount of money, require best special design template. Making the best template format choice is way to your template success. And if at this time you are looking for information and ideas regarding the Linear Regression And Correlation Coefficient Worksheet Answers then, you are in the perfect place. Get this Linear Regression And Correlation Coefficient Worksheet Answers for free here. We hope this post Linear Regression And Correlation Coefficient Worksheet Answers inspired you and help you what you are looking for. Linear Regression And Correlation Coefficient Worksheet Answers. A give an example of two variables that you believe would be positively correlated. What is difference between correlation and correlation coefficient? The plot to the right shows 5 data points and the least squares line. Try the free mathway calculator and problem solver below to practice various math topics. B) the value of the pearson correlation coefficient is 1.0. ### What Is Difference Between Correlation And Correlation Coefficient? Worksheets are work on correlation and regression, linear regression work, algebra 2 honors linear and quadratic regression work, name period date practice work linear regression, regression line work, ap statistics review linear. Linear regression and correlation coefficient worksheet with worksheet 05 more linear regression. Recall that the least squares line minimizes the squares of the residuals. ### Scuba Divers Have Maximum Dive Times They Cannot Exceed When Going To Different Depths. Worksheet by kuta software llc algebra 1 hw 1.3 linear regression & correlation name_____ id: Determining if there is a relationship A give an example of two variables that you believe would be positively correlated. ### Once You Find Your Worksheet (S), You Can Either. The first category establishes a causal relationship between two variables, where the dependent variable is continuous and the predictors are either categorical (dummy coded), dichotomous,. It denoted by the letter \( r \) (which is why it is also. The plot to the right shows 5 data points and the least squares line. ### Questions The Linear Regression Answers. 1 date_____ period____ ©z l2e0n1x6m ikiuwt_av bsqosfytpwmapr[ev iluljcm.x i za_lilm crdisgahntfsc rryeys[edrlv[eudg. Labor force at various times throughout history years after 1900 50 percentage 70 80 100 296 33.4 38.1 42.5 45.3 520 90 enter the. Correlation just shows that richer countries have both more the correlation coefficient, r, is a measure of the strength of the relationship between or among variables. ### Guided Notes And Worksheet (With Answer Keys) Which Cover Finding An. Compute the linear regression equation by the least square method. However, a transformation of variable is necessary since the sampling distribution is skewed when there is a correlation. To predict the values, use options and then type in the x value of your variable there. Customer Feedback Forms: A Simple Guide To Collecting Valuable Data In 2023 was posted in December 26, 2022 at 1:34 pm. If you wanna have it as yours, please click the Pictures and you will go to click right mouse then Save Image As and Click Save and download the Customer Feedback Forms: A Simple Guide To Collecting Valuable Data In 2023 Picture.. Don’t forget to share this picture with others via Facebook, Twitter, Pinterest or other social medias! we do hope you'll get inspired by ExcelKayra... Thanks again! If you have any DMCA issues on this post, please contact us! tags: , ,

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' # Search results Found 802 matches Evaporation - Penman Equation (Shuttleworth modification) The Penman equation describes evaporation (E) from an open water surface, and was developed by Howard Penman in 1948. Penman’s equation requires ... more Auger electron spectroscopy - Energetics of Auger transitions (more rigorous model) Auger electron spectroscopy is a common analytical technique used specifically in the study of surfaces and, more generally, in the area of materials ... more Knudsen number ( related to mean free path) The Knudsen number (Kn) is a dimensionless number defined as the ratio of the molecular mean free path length to a representative physical length scale. ... more Auger electron spectroscopy - Energetics of Auger transitions Auger electron spectroscopy is a common analytical technique used specifically in the study of surfaces and, more generally, in the area of materials ... more Conic constant In geometry, the conic constant (or Schwarzschild constant, after Karl Schwarzschild) is a quantity describing conic sections, and is represented by the ... more Wavelength - Sinusoidal Wave In physics, the wavelength of a sinusoidal wave is the spatial period of the wave—the distance over which the wave’s shape repeats, and the inverse ... more Spatial resolution The angular resolution may be converted into a spatial resolution,by multiplication of the angle (in radians) with the distance to the object. For a ... more Apsis - Periapsis minimum distance An apsis, plural apsidesis a point of greatest or least distance of a body in an elliptic orbit about a larger body. For a body orbiting the Sun the ... more Apsis - Apoapsis maximum distance An apsis, plural apsidesis a point of greatest or least distance of a body in an elliptic orbit about a larger body. For a body orbiting the Sun the ... more Worksheet 288 ...can't find what you're looking for? Create a new formula ### Search criteria: Similar to formula Category

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# 9306894 ## 9,306,894 is an even composite number composed of four prime numbers multiplied together. What does the number 9306894 look like? This visualization shows the relationship between its 4 prime factors (large circles) and 16 divisors. 9306894 is an even composite number. It is composed of four distinct prime numbers multiplied together. It has a total of sixteen divisors. ## Prime factorization of 9306894: ### 2 × 3 × 1087 × 1427 See below for interesting mathematical facts about the number 9306894 from the Numbermatics database. ### Names of 9306894 • Cardinal: 9306894 can be written as Nine million, three hundred six thousand, eight hundred ninety-four. ### Scientific notation • Scientific notation: 9.306894 × 106 ### Factors of 9306894 • Number of distinct prime factors ω(n): 4 • Total number of prime factors Ω(n): 4 • Sum of prime factors: 2519 ### Divisors of 9306894 • Number of divisors d(n): 16 • Complete list of divisors: • Sum of all divisors σ(n): 18643968 • Sum of proper divisors (its aliquot sum) s(n): 9337074 • 9306894 is an abundant number, because the sum of its proper divisors (9337074) is greater than itself. Its abundance is 30180 ### Bases of 9306894 • Binary: 1000111000000011000011102 • Base-36: 5JH8U ### Squares and roots of 9306894 • 9306894 squared (93068942) is 86618275927236 • 9306894 cubed (93068943) is 806147112517537164984 • The square root of 9306894 is 3050.7202428277 • The cube root of 9306894 is 210.3463219723 ### Scales and comparisons How big is 9306894? • 9,306,894 seconds is equal to 15 weeks, 2 days, 17 hours, 14 minutes, 54 seconds. • To count from 1 to 9,306,894 would take you about twenty-three weeks! This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!) • A cube with a volume of 9306894 cubic inches would be around 17.5 feet tall. ### Recreational maths with 9306894 • 9306894 backwards is 4986039 • The number of decimal digits it has is: 7 • The sum of 9306894's digits is 39 • More coming soon! MLA style: "Number 9306894 - Facts about the integer". Numbermatics.com. 2022. Web. 19 August 2022. APA style: Numbermatics. (2022). Number 9306894 - Facts about the integer. Retrieved 19 August 2022, from https://numbermatics.com/n/9306894/ Chicago style: Numbermatics. 2022. "Number 9306894 - Facts about the integer". https://numbermatics.com/n/9306894/ The information we have on file for 9306894 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun! Keywords: Divisors of 9306894, math, Factors of 9306894, curriculum, school, college, exams, university, Prime factorization of 9306894, STEM, science, technology, engineering, physics, economics, calculator, nine million, three hundred six thousand, eight hundred ninety-four. Oh no. Javascript is switched off in your browser. Some bits of this website may not work unless you switch it on.

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## Algebra 1 $w=13,1$ $w^2-14w+13=0$ or, $w^2-14w=-13$ Compare it with the standard form of quadratic equation $ax^2+bx+c$, we have $a=1, b=-14$ Therefore, $b^2=4ac$ $\implies$ $c=\dfrac{b^2}{4a}$ Thus, $c=\dfrac{b^2}{4a}=\dfrac{(-14)^2}{4}=49$ To complete the square, add $4$ on both sides. $w^2-14w+49=-13+49$ $\implies (w-7)^2=36$ $\implies w-7=6$ and $\implies w-7=-6$ or, $w=13,1$

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## Saturday, October 12, 2019 ### On the entanglement between superfluidity, superconductivity, and entanglement. On the entanglement between superfluidity, superconductivity, and entanglement. A lot of misconceptions and mistakes is being done simply because people do not know what they are talking about; meaning, they do not know the exact meaning of the words they use, so they use those words very loosely, without employing the specific meaning of them – because they don’t know that, because they don’t know the definition of that, because they did not have a good education on the foundations of science, and they are too confident, or lazy to do it on their own. When I talk science, I know what I’m talking about, and I want everyone to know that as well, so we would use the same words assuming the same meaning for all of them. That is why I start this post from some important definitions (that I borrowed from another post: On a Definition Of Science, but sine it is also mine, that’s not plagiarism). An object is something that represents the focus of our attention. This is what we are talking about. An object can be physical (e.g. we can touch it; often, we also call it a “system”) or abstract (e.g. a symbol). A small physical object localized at a specific place in space is called a particle (usually, we say a “system” when we focus on many particles or on a large object). For each particle we can assign a set of parameters, and a specific set of values of those parameters we call “a state of a particle”. That state may evolve, i.e. change in time; when that happens, we call it "a process", or a “behavior”. Now we can talk about the meaning of another term – correlation. In general, correlation is interdependence between two objects – abstract (i.e. states) or physical (i.e. particles). I will illustrate what “interdependence” means in a couple of paragraphs. I believe that this definition is better than saying that correlation is a relationship, and it goes well beyond statistics. There are different types of correlations, for example: 1) A temporal correlation – when two states of the same object correlate. 2) A processual, or behavioral, or spatial, or inter-systemic correlation – when states of at least two different objects correlate (most often not at the same time). We concentrate on the latter correlation. For a behavioral correlation interdependence means that a change in a state of one object affects a change in a state of another one, and can be illustrated using a simple model. Let's say, we observe a behavior (or an evolution) of a state of object #1. Then we repeat the observation; we observe the evolution of the same object keeping all but one circumstances to be exactly the same as in the previous observation, but this time we either remove, or add another object or change a state of one of the previously present objects. If that action does NOT affect the behavior of object #1, we say, that objects #1 and #2 do not correlate, there is no correlation between the states of those two object, correlation is absent. Otherwise, a correlation exists, it is present, evolutions of the objects correlate. Just to make things absolutely clear, “otherwise” means, that the action of adding or removing an object or changing the state of an object that is not object #1 affects the evolution of object #1, as compared to the previous observation; and hence a correlation exists, it is present, evolutions of the objects correlate. There are plenty of well-known examples of correlations. In an outer space, in the absence of any stars a comet travels straight ahead without speeding up or slowing down. But if a star is present on its path, the path is getting bent, hence the star affects the comet, there is correlation between those two objects, and that specific correlation has specific name – interaction, and described by a specific physical quantity – a force; specifically, the force of gravity, that is described by the Newton’s law of gravity. If you take two plastic cups and connect them with a string in a certain manner, stretch the string and start making sounds on one end, speaking into one cup, a second cup also starts vibrate and making a sound. In physics there are important laws that describe correlations between different states of different objects; those laws have a name of conservation laws. For example, if in outer space a grenade blows up and separates in exactly two identical halves; those two objects – halves – will fly away from each other. But if we catch one half of this grenade and measure its velocity, we will be able to predict the exact velocity of another half; and that prediction is governed by the law of consideration of a linear momentum. Quantum mechanics also describes many correlations between quantum objects, quantum particles. Some of those correlations are very similar to correlations in the macroscopic world; for example, the law of conservation of linear momentum works for quantum particles as well. However, the vast majority of physicists believe that quantum objects also exhibit a different type of a correlation - a correlation that is specific only for the quantum world, a correlation that does not exists for macroscopic objects, a correlation that has a name “entanglement”. Many people have been writing about quantum entanglement, including yours truly. The discussion about entanglement goes deep to the foundations of quantum mechanics. Albert Einstein didn't like it; he thought that entanglement as it was understood in his time, and how it is still understood, demonstrates that quantum mechanics is either not local theory, or includes “spooky” interactions that happen faster than the speed of light (and naturally, Einstein could not accept it). And idea of entanglement can be described by a thought experiment proposed by Albert Einstein (and that has different variations). In order to conduct this is experiment, first one needs to prepare two entangled particles, so those two particles would be a part of one quantum system. Then they fly away from each other; and when an observer affects a state of one particle, the second particle should immediately change it state in a specific and predictable way. Some physicists argue that the term is wrong; that this not some kind of new quantum type correlation called “entanglement”, this is just another manifestation of some law of conservation, that works on the quantum level in the same way it works in the classical world. The majority of physicists believe that entanglement is a uniquely quantum phenomenon and cannot be explained by application of laws of conservation. I’ve been back-and-force on this, but eventually I settled on the idea of entanglement, but not because of to-be-fashionable experiments with a couple of entangled particles (even if they are produced in bulk) – those experiments are ideologically trivial, although technologically challenging and advanced. There are much more interesting demonstrations of quantum entanglement, and those are superfluidity and superconductivity. Let's start from superfluidity. It was discovered in 1937 in liquid Helium below about 2 K (at the normal pressure). In its superfluid state, quote: “the liquid … behaves as if it consists of two components: a normal component, which behaves like a normal fluid, and a superfluid component with zero viscosity The two-fluid model is clear and intuitive. At the absolute zero, T = 0 K, all Helium is superfluid, i.e. does not experience friction and can flow without experiencing resistance. Above a certain temperate called critical (at a given pressure), Helium behaves like a regular normal fluid with internal and external viscosity. Above absolute zero, but below a critical temperate Helium is “composed” of two components: one components is “normal” and behaves like a regular liquid as it was above the critical temperature, but the second component is superfluid, like it would be at zero temperature. The mystery here is what happens to atoms below the critical temperature? If you would look at the Helium atoms above or below the critical temperature, you would not see any difference. And yet, they behave in the very different way! Why? There must be some difference on the atomic level, otherwise what would be the source/reason for a superfluid component to exist? The answer is – quantum entanglement! When temperature drops below the critical one, quantum correlations between many atoms make those atoms to be entangled and to form one highly/strongly correlated quantum state, also called a coherent state. Atoms in that coherent state form the superfluid component. At T = 0 K, all atoms belong to this coherent state. Above zero, some atoms belong to the coherent state, and others don’t, and those that do not form the normal component of Helium. Above critical temperature coherent state does not exist. The difference between atoms in a normal state and atoms in a superfluid coherent state is very significant (although, atoms constantly “jump” from one state into another and back, on average, the number of atoms in each state remains constant). That difference is described by how those atoms react to an attempt to be excited, for example by a simple hit from another atom. If we hit (a fancy term is “collide”) an atom in a normal component (state, phase) we can transfer to that atom practical any amount of energy, from very little to very large. When we make this atom, that is a part of a normal component, collide with a different atom, for example with an atom in a wall of a vessel or a tube holding Helium, this collision results in a transfer of energy from an atom in a wall to the atom in the normal component, or back. This interaction is responsible for internal friction and for the friction between the normal component and the walls of a pipe, and when the fluid travels through a pipe this interaction is responsible for the existence of resistance/friction/viscosity. However, if we try to hit/knock/collide an atom in the coherent state, that atom is entangled with all other atoms in that state, with all other atoms that compose a superfluid component. So, when we hit a single atom in that state, we hit all of them in that state. And in order to excite this whole system of correlated entangled atoms, a small energy is not enough anymore to excite it. The energy has to be above a certain threshold; if the energy is below that threshold, the atoms in the entangled stated do not “feel” it, “ignore” the interaction. And that is why this component of a fluid, that is composed of correlated atoms in a coherent entangled state, can travel through pipes without feeling any resistance. That makes this component superfluid. When the system goes through the critical temperature in one direction or another, atoms do not disappear, or new atoms do not appear; but when the temperature drops below the critical one, some atoms become entangled, they become strongly correlated in a specific quantum way, and they form a new phase, a superfluid phase. What is happening when the temperature rises above the critical? Note, when we said that above critical temperature coherent state does not exist, that statement was not exactly accurate. According to the general theory of critical transitions, a critical transition does not happen instantly. Even above a critical temperature in some parts of a material and for a short periods of time coherent state may exist. But it is very short-living, or metastable, and the regions with the coherent state are small and do not overlap so they cannot cover the whole sample of a material. When temperature is getting closer to the critical value, those short-living short-range pockets of a new phase getting larger and live longer and below the critical temperature finally occupy the whole material space. So, in general, the traces of a new phase may be observed even above the critical temperature. A superfluid transition is not related to existence or absence of new particles, but related to the absence or presence of strong quantum correlations between large numbers of particles. Those correlations may exist even above the critical temperature in a form of metastable, short-range and short-living pockets. Superconductivity was discovered in 1911 but explained only in 1957. To that time, a well-developed model could explain many electrical phenomena based on the idea that electrons in conductors could be treated as a carrying electric charge fluid. Electric current could be seen as a river, as a stream of electric charges traveling in the same direction. Atoms or ions would present islands that could slow that stream down. If you would give this electric fluid a push, those islands would eventually make it stop - and this is a nature of electric resistance in conductors. Turns out, however, that under certain circumstances those islands could also help fluid to travel. We could imagine that when electron fluid flows around an island, sometimes some of those electrons may start traveling together in a common whirl. Of course, this is just a visual representation of a fact that some electrons can move in a correlated manner - not an actual motion of the electrons. So, the islands (atoms, ions) could not just slow down the motion of the electrons, but also could make them moving in a correlated manner. And then those correlated electrons can correlate with other correlated electrons all across a conductor, forming one strongly correlated entangled state. When that happens, a macroscopic number of electrons get entangled in a “superfluid electric component”, that can travel with no resistance, and is called a “supercurrent”. When electrons form one highly correlated fully entangled macroscopic state similar to a superfluid component in Helium, and when electrons in that state interact with atoms, in the event when that interaction is not strong enough, if it doesn't go over a certain threshold of energy, those entangled electrons just ignore that interaction and keep exist in the unchanged entangled state. Those electrons that get correlated in “a whirl”, due their motion around the same island (atom, ion) has a name a Cooper pair. Below a critical temperature electrons in those pairs form an entangled superconductive state. However, like Helium atoms do not disappear above critical temperature, Cooper pairs also should not completely disappear immediately above critical temperature. Although, the number of Cooper pairs should significantly decrease with even the slightest increase in the temperature, I bet, accurate experiments would demonstrate their existence even in the normal state. And now it is a natural time to talk about high-temperature superconductivity. It was discovered in 1986, and to this day there is no commonly accepted explanation of this phenomenon. In high-temperature superconductors electrons cannot be treated any more as a fluid. A better model would treat electrons rather as a solid, or as a crystal where electrons spend most of the time at certain locations, and from time to time hop from one location to another one. However, since electrons form a superconductive state, it means that under certain circumstances, at least some of the electrons are getting entangled in a coherent strongly correlated state that allows them to travel through the material without feeling the existence of other electrons or atoms or ions. We can try to understand the nature of this phenomenon building on another very important and common feature that all those highly entangled states have in common. In a superfluid or in a superconductor, that highly correlated state with entangled particles (atoms of electrons), due to strong correlations represents a state with a very high order, or a very law chaos, as compared with the regular normal state or a component of a material. Hence, the entropy of this state is equal to zero. Entropy is a measure of order v. chaos; more chaos - higher entropy; and when a macroscopic state is strongly correlated and highly entangled, the chaos is so low that the entropy is zero. In all superfluid or superconductive systems, below a certain temperature macroscopic parts of a system simply stop “generating” an entropy. At the absolute zero temperature, the entropy of all systems is zero, but in certain systems under certain circumstances when temperature is rising above zero, some parts of this system still remain “feeling” themselves like they still at zero temperature, forming a highly ordered, strongly correlated entangled state. With the rise in temperature, the portion of the entangled part of a system gradually decreases to zero and becomes zero at a critical temperature (with some short-living short-range pockets above the critical temperature). This idea may guide us toward better understanding of the nature of high-temperature superconductivity. It is assumed that in materials exhibiting high-temperature superconductivity, it happens when an original order of electrons is destroyed or disturbed by doping. In an original material (under ideal circumstances, at zero temperature) electrons from a “crystal”. The simplest model would look like this: A plus or a minus in the picture represents that fact that electrons have a feature called a “spin”; we can imagine a spin as an arrow attached to an electron that can have only two directions, in or our (or up v. down, or left v. right – but always only TWO). And in an ideal case, two neighboring electrons would prefer having opposite spins (because that makes their “life” easier, or energy lower). In this “crystal”, electrons prefer stay at their places, because they “hate” – meaning, electrically repel – each other. Hence, jumping to a place already occupied by another electrons is restricted, almost forbidden. Hence, in this state the material a is not a good conductor, but an insulator. However, doping can change that. Doping is procedure that can change the number of electrons, for example, take some electrons out form their locations. This is done by inserting some other atoms that can attract electrons and keep those electrons on those atoms. In that case, the structure of the material is not so ordered anymore, it has some random holes – empty spaces at the locations from where electrons were removed. For a better visibility, in the picture those holes are noted by black circles (instead of just keeping those places empty). Now we have to use the most powerful human ability – imagination. Other animals do not have it. Albert Einstein said that imagination is the true sign of intelligence. No single AI professional has any idea how to model it (e.g. The new stage of the race for AI domination AI). Imagine that ALL black holes would move on step diagonally down and the electron moved from that place one step diagonally up (to take the place where the hole was). This action is shown by arrows in the next picture. As the result, you get this: Now, compare the two pictures below, the first one is an excerpt from the ordinal picture (before the jump), and the second on is an excerpt from the final picture (after the jump): ---------------------------------------------------------------------------------------------- These two pictures are identical! They show two DIFFERENT but IDENTICAL states of the system. Of course, because our pictures have a finite size, we could see some differences at the edges, but for a very large system (or with periodic boundary conditions) the deference would be negligible. This simple illustration makes us to make a conclusion that when all electrons would jump simultaneously in the same direction – that is diagonal – this actions does NOT change the state of the system. In a certain sense – this action does not make the state more chaotic, less ordered – as long as all electrons are entangled. It is naturally to assume, that this is the state we are looking for, the state with zero entropy. Based on this idea, we could assume that in high-temperature superconductors electrons jump: (a) in the same direction; (b) diagonally. I do not follow specialized literature, but I know that I am not the first who suggested that in high-temperature superconductors electrons may travel in the same direction (even I wrote on this and published it on arXiv; although in a very primitive way – sorry for the visual picture used to support my model, I like visual pictures, as you may have noticed already). I also heard of a suggestion that electrons jump diagonally (honestly, do not remember where). But to my best knowledge, I am the first one who makes both statements as a model for high-temperature superconductivity. Due to very specific geometry of the model, it could be tested using a simple mechanical motion of a sample of a high-temperature superconductor – the movement along diagonals or along the edges would affect supercurrent differently. However, for this experiment one would need a thin, ideally a single, sheet of a high-temperature superconductor. This still maybe a challenging technological obstacle. Dr. Valentin Voroshilov P.S. On this page (if you click on this link) you will find some piece on the foundation of quantum mechanics. Recently, it has become fashionable to run actual or thought experiments on quantum entanglement. However, ALL of them are based on ONE specific interpretation of quantum mechanics and completely ignore and do not mention the fact that so far there are several different interpretations of it. In my publications I attacked different aspects of different publications on the matter. Appendix Everyone who claims he/she knows what quantum mechanics is about must read the original EPR paper (so, ask the guy - have you read EPR? that is a litmus test for you should you even listen to the guy). It has many layers, more than just the thought experiment they use to claim that quantum mechanics is not a complete theory (e.g. click on this link and scroll down to Appendix III). The fact of the matter is that this experiment does show that quantum mechanics is different from classical mechanics. When this matter is accepted, one has a choice: (a) follow the strategy "shut up and calculate" and do not spend any time on trying to make the theory "complete" or (b) spend some time on trying to make the theory "complete". In the latter case, one can be inventing different approaches - some are mentioned in the four pieces about a cat: But the simplest (thank you Occame!) way to resolve all the mysteries of quantum mechanics would be to assume that - yes, faster than light interactions do exist! Tachyons are responsible for that "spooky action at distance". There is a whole world of particles that cannot travel slower than the speed of light! And that world interacts with our world, where particles cannot travel faster than the speed of light. Simple! Imagine a sea of tachyons. Every known particle can have its counterpart in that sea - tachyo-electron, tachyo-proton, etc. Due to fluctuations, for a teeny-tiny instant of time, those tachyons may enter our world, become a so called virtual particle, and interact with our particles. But even more interesting process happens when our particles can disappear from our world and enter the world of tachyons, spend there a teeny-tiny instant of time and come back again - but at a different location, or with a different speed. Of course, until tachyons are found, they are just a theory, a mathematical abstract. But so was the Higgs boson. BTW: tachyons, or in general the world of faster than light particles, can explain such phenomenon as tunneling. A classical particle cannot escape a potential well - when it has not enough energy. But a quantum particle can "tunnel" through. Why? Because due to interactions with tachyons it may "accidentally" (a scientific name - fluctuations) gain energy enough to get "over the well".

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FutureStarr 27 Is What Percent of 30 OR ## 27 Is What Percent of 30 via GIPHY A number 27 is what percent of 30. ### Percent via GIPHY CGPA Calculator X is What Percent of Y Calculator Y is P Percent of What Calculator What Percent of X is Y Calculator P Percent of What is Y Calculator P Percent of X is What Calculator Y out of What is P Percent Calculator What out of X is P Percent Calculator Y out of X is What Percent Calculator X plus P Percent is What Calculator X plus What Percent is Y Calculator What plus P Percent is Y Calculator X minus P Percent is What Calculator X minus What Percent is Y Calculator What minus P Percent is Y Calculator What is the percentage increase/decrease from x to y Percentage Change Calculator Percent to Decimal Calculator Decimal to Percent Calculator Percentage to Fraction Calculator X Plus What Percent is Y Calculator Winning Percentage Calculator Degree to Percent Grade Calculator Practice the questions given in the worksheet on percentage of a number. We know, to find the percent of a number we obtain the given number and then multiply the number by the required percent i.e., x % of a = x/100 × a 1. Find the following: (i) 22 % of 140 (Source: www.math-only-math.com) ### Number via GIPHY In calculating 27% of a number, sales tax, credit cards cash back bonus, interest, discounts, interest per annum, dollars, pounds, coupons,27% off, 27% of price or something, we use the formula above to find the answer. The equation for the calculation is very simple and direct. You can also compute other number values by using the calculator above and enter any value you want to compute.percent dollar to pound = 0 pound To calculate percentages, start by writing the number you want to turn into a percentage over the total value so you end up with a fraction. Then, turn the fraction into a decimal by dividing the top number by the bottom number. Finally, multiply the decimal by 100 to find the percentage. (Source: percentagecalculator.guru) ### Percentage CGPA Calculator X is What Percent of Y Calculator Y is P Percent of What Calculator What Percent of X is Y Calculator P Percent of What is Y Calculator P Percent of X is What Calculator Y out of What is P Percent Calculator What out of X is P Percent Calculator Y out of X is What Percent Calculator X plus P Percent is What Calculator X plus What Percent is Y Calculator What plus P Percent is Y Calculator X minus P Percent is What Calculator X minus What Percent is Y Calculator What minus P Percent is Y Calculator What is the percentage increase/decrease from x to y Percentage Change Calculator Percent to Decimal Calculator Decimal to Percent Calculator Percentage to Fraction Calculator X Plus What Percent is Y Calculator Winning Percentage Calculator Degree to Percent Grade Calculator (Source: percentagecalculator.guru) ## Related Articles • #### Supply Chain Resume Summary August 07, 2022     |     Javeria Ijaz • #### Fut20 Future Stars in new york 2022 August 07, 2022     |     Muhammad Arsalan • #### Eros Miami OR August 07, 2022     |     Abid Ali • #### mva online services August 07, 2022     |     Abid Ali • #### A Lab Tech Resume August 07, 2022     |     Muhammad Waseem • #### Ashley OR August 07, 2022     |     Muhammad Arsalan • #### Top Homes For Sale in Sherwood, Oregon August 07, 2022     |     Ayaz Hussain • #### Hot Air Balloon Miami August 07, 2022     |     sheraz naseer • #### craigslist okc apartments for rent August 07, 2022     |     Shaveez Haider • #### Walmart Resume Example August 07, 2022     |     sheraz naseer • #### What Percent Is 20 Out Of 22 August 07, 2022     |     Javeria Ijaz • #### Solve For X Calculator With Steps Free August 07, 2022     |     Fahad Nawaz • #### A Craigslist oc free stuff August 07, 2022     |     m aqib • #### A Resume for 16 Year Old With No Experience August 07, 2022     |     Muhammad Waseem • #### Computer Hardware Engineer Resume Format OR August 07, 2022     |     Asif Niazi

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# Factorise x² - 25, thank you :) Answer: (x-5) (x+5) because if you times the two xs you'll get x squared and if you times negative 5 and positive 5, you'll get negative 25!! Hope this helps!!! ## Related Questions if you werevto draw three different parallelograms each with a base of 6 units and a height of 4 units how would the areas compare? If you created 3 different parallelograms with these dimensions, the areas would still be the same. They would look different in the tilt that is created, but if you were to take the slanted side to the right/left of the height and move it over beside the other side of the parallelogram, you would create the same size rectangle in all examples, giving you the exact same area. Please write the word version of this algebra expression of 3x-7 Three times x then minus 7 You would 3 times x minus 7 Answer: Advantages: It is quick and easy, when done repetitively it allows you to do mental math very fast Step-by-step explanation: Disadvantages: Does not allow you to multiply fractions, which come up a lot in everyday situations or jobs, takes a long time to do exponents 3p^4(4p+7p+4p+1) ???????? HEY DEAR 45p^5 + 3p^4 step by step explanation:- 3p^4( 4p+7p+4p+1) 3p^4 (11p + 4p) +1 3p^4 (15 p +1) multiplying 3p^4 to both 15p and 1 we get:- 3p^4× 15p +3p^4× 1 =45p^5 +3p^4

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# Readers ask: How To Make A Right Turn In A School Bus? ## How many seconds does it take a school bus to make a left turn? To make a right turn, we assign a delay of 10 seconds. To turn left, we assign a delay of 30 seconds. The net result of this is that BusBoss will prefer to go straight. If not straight, then a right turn. ## How do you make the right turns? Right turns–To make a right turn, drive close to the right edge of the road. If there is a bike lane, drive into the bike lane no more than 200 feet before the turn. Watch for pedestrians, bicyclists, or motorcyclists who may get between your vehicle and the curb. Begin signaling about 100 feet before the turn. ## How do left turns work? Left Turn: 1. Turn on the left turn signal before you make the turn and slow down. 2. Look both ways and make sure that the oncoming lanes are clear. 3. Make the turn from the designated lane (use left lane). 4. Do not enter into the right lane. In some states, it is illegal to enter the right lane after the turn is completed. You might be interested:  Readers ask: How Much Weight On The Top School Bus? ## Who has the right away when turning left on a solid green light? If you’re turning left at a green light, pull out into the intersection but wait to turn left until all oncoming traffic has passed. If you’re turning left at a four-way stop or uncontrolled intersection, you should give the right-of-way to any oncoming drivers going straight, even if you got there first. ## How do school bus stop signs work? When a school bus has the stop sign out and the red lights flashing, drivers are required to stop: on two-lane, undivided roads, in both directions. on roads that are divided to separate the directions of travel, in the direction that the bus is traveling. ## When you come to a four way stop who has the right away? Yield to right. When two vehicles arrive at a 4-way stop at the same time, and are located side-by-side, the vehicle furthest to the right has the right of way. If three vehicles arrive at the same time, the car furthest left should continue to yield until both of the other cars to the right of them have passed. ## How much room does a school bus need to turn around? STANDARD 40′ BUS A typical inner turning radius of a standard 40-foot bus is 21.5 feet, which is required to clear the curb. At its tightest turning angle, the rear overhang of the back bumper extends out to 43.3 feet. ## Is bus driving stressful? According to Dr Tage Kristensen, of the National Institute of Occupational Health in Copenhagen, bus drivers are under more stress than executives. Bus drivers are practically public property, open to abuse and accusation. After a while it gets to you, and makes you mean. You might be interested:  Quick Answer: The Magic School Bus Rides Again What Happened To Phoebe? ## Is driving bus hard? Driving a bus is not that much more complicated than driving a car. You do have to be more careful, given the length and weight of a bus, as well as the responsibility you have to your passengers. However, the basics are the same, though you may need to learn to shift if you’ve never driven a standard before. ## Is driving a school bus difficult? School districts across the country are struggling to fill vital school bus driver positions. There’s a reason that great school bus drivers are difficult to find and keep: driving a school bus requires more than just driving skill and patience. The job requires a unique skill set, and it’s not as easy as you’d think. ## Are wide right turns illegal? The only vehicles that can make wide right turns legitimately or legally are tractor-trailers. Tractor-trailers are legally allowed to make wide right turns because they are larger vehicles that have a more difficult time turning a corner and landing in the far right lane. ## Do you press the gas when turning? As you begin turning the wheel, release the pressure on the brake. Through the apex of the turn, there shouldn’t be any pressure on the brake. Instead, apply light pressure to gas pedal as you come out of the turn. ## Do you accelerate when turning? Common driving wisdom holds that you get more traction during a turn when you are accelerating. For many drivers, it’s standard practice to decelerate before entering a turn, then accelerate once they are half way through (past the apex).

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Solved # Print a Title message in this conversion list Posted on 2003-02-23 Medium Priority 185 Views How can I include a printed title at the top of this conversion list such as "Currency Conversion," ??  I keep getting compile errors for wrong number of args. /* Prints a message on the screen */ #include <studio.h> main() { float Euro, Peso, Yen, Pound, Shekel; /*Declare how much \$1 USD is when converted to these currencies */ /*www.xe.com -- Live mid-market rates as of 2003.02.24 01:31:21 GMT */ Euro = 0.932543; Peso = 10.9554; Yen = 118.411; Pound = 0.634517; Shekel = 4.87100; /*Print the conversion for how much 1 of each currency is in USD*/ printf("1 Euro = %f USD  \n", (float)1/Euro); printf("1 Peso = %f USD  \n", (float)1/Peso); printf("1 Yen = %f USD  \n", (float)1/Yen); printf("1 Pound = %f USD  \n", (float)1/Pound); printf("1 Shekel = %f USD  \n", (float)1/Shekel); return 0; } 0 Question by:gmahler5th [X] ###### Welcome to Experts Exchange Add your voice to the tech community where 5M+ people just like you are talking about what matters. • Help others & share knowledge • Earn cash & points • 3 • 3 LVL 9 Expert Comment ID: 8005741 what is it you are trying to print dind fully understand 0 LVL 9 Expert Comment ID: 8005746 maybe /* Prints a message on the screen */ #include <studio.h> main() { float Euro, Peso, Yen, Pound, Shekel; /*Declare how much \$1 USD is when converted to these currencies */ /*www.xe.com -- Live mid-market rates as of 2003.02.24 01:31:21 GMT */ Euro = 0.932543; Peso = 10.9554; Yen = 118.411; Pound = 0.634517; Shekel = 4.87100; printf("Currency Conversion\n"); /*Print the conversion for how much 1 of each currency is in USD*/ printf("1 Euro = %f USD  \n", (float)1/Euro); printf("1 Peso = %f USD  \n", (float)1/Peso); printf("1 Yen = %f USD  \n", (float)1/Yen); printf("1 Pound = %f USD  \n", (float)1/Pound); printf("1 Shekel = %f USD  \n", (float)1/Shekel); return 0; } 0 Author Comment ID: 8005784 I keep getting the following error when I used the print command that you added. C:\Program Files\Miracle C\examples\test.c: line 18: wrong # args in function call 'printf("1 Euro = %f USD  \n", (float)1/Euro)' aborting compile I am trying to print the list that I posted, but the the title "Currency Conversion" as the title that appears first, before the list is printed. 0 Author Comment ID: 8005799 I keep getting the following error when I used the print command that you added. C:\Program Files\Miracle C\examples\test.c: line 18: wrong # args in function call 'printf("1 Euro = %f USD  \n", (float)1/Euro)' aborting compile I am trying to print the list that I posted, but the the title "Currency Conversion" as the title that appears first, before the list is printed. 0 LVL 9 Accepted Solution tinchos earned 80 total points ID: 8005801 /* Prints a message on the screen */ #include <stdio.h> main() { float Euro, Peso, Yen, Pound, Shekel; /*Declare how much \$1 USD is when converted to these currencies */ /*www.xe.com -- Live mid-market rates as of 2003.02.24 01:31:21 GMT */ Euro = 0.932543; Peso = 10.9554; Yen = 118.411; Pound = 0.634517; Shekel = 4.87100; printf("Currency Conversion\n"); /*Print the conversion for how much 1 of each currency is in USD*/ printf("1 Euro = %f USD  \n", (float)1/Euro); printf("1 Peso = %f USD  \n", (float)1/Peso); printf("1 Yen = %f USD  \n", (float)1/Yen); printf("1 Pound = %f USD  \n", (float)1/Pound); printf("1 Shekel = %f USD  \n", (float)1/Shekel); return 0; } this code is compiling 0 Author Comment ID: 8005886 Sweeet!  Thanks! I dont' know what I was doing wrong for this not to compile at first, but it seems to be working now. 0 ## Featured Post Question has a verified solution. If you are experiencing a similar issue, please ask a related question This tutorial is posted by Aaron Wojnowski, administrator at SDKExpert.net.  To view more iPhone tutorials, visit www.sdkexpert.net. This is a very simple tutorial on finding the user's current location easily. In this tutorial, you will learn ho… This is a short and sweet, but (hopefully) to the point article. There seems to be some fundamental misunderstanding about the function prototype for the "main" function in C and C++, more specifically what type this function should return. I see so… The goal of this video is to provide viewers with basic examples to understand opening and writing to files in the C programming language. The goal of this video is to provide viewers with basic examples to understand opening and reading files in the C programming language. ###### Suggested Courses Course of the Month14 days, 20 hours left to enroll

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Is there a way to retrieve points of a room/space (corners)? Tried the element.location node, but all it retrieves is the center reference point of a room/space. Was wondering how I can get the vertices of a room? try to use Room.Boundary it will give you all the curves , then use start or end points to get the corner Here is another way to find the vertice of the room. Room vertice.dyn (8.4 KB) Hi, yes it’s possible to extract any kind of information as long as you have the room “Solid”. I am not sure of what you mean by “Room vertices”. Here’s what i suggest : • Use the node “Explode” to split your single room solid • From the surface list you get exctract the local axes using the node PrincipalDirectionsAtParameter. It gives you a list of 2 vectors for each surface of your original solid. • Look at the X,Y or Z component of those vectors to determine wether you save or delete the associated surface. I suggest to use the FilterByBoolMask node to achieve this point. I enclose here a little DesignScript code I did few weeks ago. Coments and variable names are in french… tell me if you want me to translate it to make it clearer. The aim of this script is to retreive only the floor and ceilling of the input rooms CODEBLOCK // Surfaces dĂ©limitant chaque pièce (volume murs/sol/plafond) RoomSurf=RoomList.Explode(); RoomSurf=List.Flatten(RoomSurf,1); // Couples de vecteurs directeurs de ces surfaces VectList=RoomSurf.PrincipalDirectionsAtParameter(0,0); // Composante Z de chaque vecteur directeur ZComp=VectList.Z; ZComp=Flatten(ZComp); // CrĂ©ation de deux listes : index pairs et impairs NbSurf=Count(Flatten(RoomSurf)); indexpairs=0…2NbSurf-1…2; indeximpairs=1…2 NbSurf…2; // Comparer les composantes Z des 2 vecteurs directeur de chaque surface Z1=List.GetItemAtIndex(ZComp,indexpairs); Z2=List.GetItemAtIndex(ZComp,indeximpairs); // Z identiques -> surface plane -> on la sauvegarde ListeATrier=Flatten(RoomSurf); SurfPlan=List.GetItemAtIndex(CompareList,0); // On conserve une seule des deux surfaces (sol ou plafond) c=Count(SurfPlan); index=0…c…2; FinalSurf=List.RemoveItemAtIndex(SurfPlan,index); Surface.PerimeterCurves(FinalSurf); 1 Like Obviously what i suggest is just an example, you’ll probably need to adapt and change it. So in a rectangular room, I’m getting 6 vertex points instead of just 4, got any ideas on how to clear the two ambiguous points? Select the node Element.Geometry and switch from Graph preview to 3D Preview Navigation. You will see the geometry of your room. Your room is probably not a parallelepiped if it has more than 8 vertices. Yes, you can look at the UV couple of vector for each of your 6 surfaces. As does the script I just post above

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in # How To Make Paper Snowflakes Easy Cut some jute rope in different lengths and tie them around your paper bag snowflakes. Fold the triangle in half again to make a smaller triangle. How to Make a Six Sided Paper Snowflake by Craftiments ### Because real life snowflakes are symmetrical, we’ll need to do some careful measuring to create a symmetrical paper snowflake of. How to make paper snowflakes easy. First, s tart with a square sheet of paper and fold it into a triangle (image 1). To make a paper snowflake, cut a piece of paper into a square shape. Trim off excess.) fold paper in half diagonally to make a triangle. Get these 30 easy diy paper snowflake patterns to make your own diy snowflakes to feel the pure spirits of winters and snow. You can print them or create your own. You can make a square from a rectangular piece of paper simply by folding one corner down to form an isosceles triangle and trimming off the excess paper. The best way with this is to make up your own design. First, you’ll need to cut your paper into a square: How to make paper snowflakes: Create attractive and easy paper snowflakes and learn how to use them in your home decors and other use as well. Cut off the unfolded part of the paper so that you have just the folded square remaining. All you need to do is subscribe to our newsletter at the bottom of this post (make sure it’s the form that has the picture of the snowflake template) and the free snowflake templates will be sent straight to your inbox. Now fold the large triangle again in half. Take a square piece of paper and fold it in half diagonally to make a triangle. Start by bringing the right point towards the left by one third. You can learn to make paper snowflakes either to decorate your home or use it as a theme for […] The width of the strip determines the size of the final snowflakes. Here is a sample of the template. You can use a protractor, or just estimate the size of the sections, which is what we did here. Fold it in half again to make a smaller triangle. Then, fold it in half again (image 2). Take one corner of the paper and fold it diagonally so that it touches the opposite edge of the paper. 1.prepare a piece of a4 paper. Fold the paper in a simple way. You can also embellish your paper snowflakes with glitter glue to make them even. All you need is a pair of scissors, a square piece of paper and a crafty mind. Top tip if you’re starting with an a4 piece of paper or a rectangular shape, make a square by folding one corner down to form a triangle and cutting off the excess paper. Cut diagonally across the top of the folded triangles. Divide and conquer your triangle! Open up the cut triangle and you will have a square paper snowflake. Using scissors, make cuts into the sides of the paper, such as triangles, rectangles, or round shapes. How to make paper snowflakes with kids. Fold paper triangle in half so that the pointed corners meet. How to make paper snowflakes, the easy way read all of the steps before starting. Show kids where to cut the paper by using a simple snowflake cutting template. Use any colors of tissue paper you like. Since we are going to make four kinds of snowflakes, we cut it into four small pieces of square paper. Fold it in half to create a large triangle. Begin with a square piece of paper. Divide this triangle into three (see diagram below). Winter brings the loveliest of all the fantasies, and as long as frozen has entered our lives, snowflakes are quite the charmers, now! Some like them for christmas, but you may like them any time! Keep your paper folded in a triangle shape. Creativity with paper and scissor. Fold this smaller triangle into thirds. First you’ll need to print off the snowflake templates. Your folded paper should look like this (image 5). Three dimensional paper snowflakes look magnificent hanging in a window or on a wall. Fold a 15cm square of white paper in half diagonally to form a triangle. Snowflakes just really make everyone get in the christmas mood, and they are really easy to make out of many different materials. Finally, fold the right side over the left (image 4). Young kids can make beautiful snowflakes with a piece of paper folded into fourths. Fold the paper in half diagonally to form a triangle, then fold it in half twice more. How to make paper snowflakes: You only need paper, scissors and a sprinkle of festive spirit for this super easy activity! We also use a craft stick to make folds crisper. How to make paper snowflakes. There are so many ways to enjoy the presence of diy snowflakes in our lives. Display your super cute craft proudly! See second picture.) i usually make two snowflakes for every 8.5×11 piece of paper, so i first cut the paper in half, and then make a square from each half. Start with a square piece of paper. In nature, no two snowflakes are exactly the same. I have two tips to help kids learn to cut out paper snowflakes. Fun for kids or adults, they are easy to make. Next, fold the paper in half at the arrow but pull the tip of the triangle down like so (image 3). Cut a variety of shapes into the sides of the triangle. This will make a smaller triangle. Use a template or pattern: They can also be the huge frozen charmer, which by the way. Try your best to make clean and straight creases! Check out more paper snowflake instructions for: Fold one outside side edge in over the central triangle, then fold the other side in. Learn how to make paper snowflakes the easy way. Our printable paper snowflake templates (you can grab it at the end of this tutorial) regular paper (the lighter the better as heavier paper is tricky to cut) scissors; Pin on Papirne pahuljice Pin on All Time Favorite Printables How to make paper Snowflakes Holiday crafts for kids Paper snowflakes are SO SIMPLE and super inexpensive to How to make paper Snowflakes Winter crafts for kids Paper Snowflake Easy Tutorial Paper snowflakes, Paper origami snowflake Christmas Pinterest Origami Pin on Paper Snowflakes How to make Paper Snowflake Method step by step DIY Super Easy Paper Snowflake Craft in 2020 Winter crafts Origami Snowflake Instructions Free Printable Papercraft How to Make Giant Paper Snowflakes Step by Step Photo How To Make A Paper Snowflake3 Pop up christmas cards How to Make Paper Snowflakes & Ornaments Using Doilies How to make five pointed paper snowflake, origami My Snowflake Obsession Continues in 3D! Crafts, Holiday Book Page Decorating Book page crafts, Christmas paper Snowflakes out of coffee filters ) Christmas diy kids 3D Paper Snowflake Tutorial Designed by ArtsnCraft4u Easy

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# AP Statistics Section 15 C ## Presentation on theme: "AP Statistics Section 15 C"— Presentation transcript: AP Statistics Section 15 C The most common hypothesis about the slope is _________ The most common hypothesis about the slope is _________. A regression line with slope 0 is _________. That is, the mean of y (does/does not) change when x changes. So this says that there is no true linear relationship between x and y. Put another way, says there is ___________ between x and y in the population from which we drew our data. You can use the test for zero slope to test the hypothesis of zero correlation between any two quantitative variables. Note that testing correlation makes sense only if the observations are ______. The test statistic is just the standardized version of the least-squares slope b.   To test the hypothesis compute the t statistic     Once again, we use the t-distribution with n - 2 degrees of freedom and as always our p-value is the area under the tail(s).     Regressions output from statistical software usually gives t and its two-sided P-value. For a one-sided test, divide the P-value in the output by 2. Example 15.6: The hypothesis says that crying has no straight-line relationship with IQ. The scatterplot we constructed shows that there is a relationship so it is not surprising that the computer output given on the previous page of notes give t = ______ with a two-sided P-value of _____. There is (strong/weak) evidence that IQ is correlated with crying. Example 15. 7: A previous example (3 Example 15.7: A previous example (3.5) looked at how well the number of beers a student drinks predicts his or her blood alcohol content (BAC). Sixteen student volunteers at Ohio State University drank a randomly assigned number of cans of beer. Thirty minutes later, a police officer measured their BAC. Here are the data. Student: 1 2 3 4 5 6 7 8 Beers: 9 BAC: 0.10 0.03 0.19 0.12 0.04 0.095 0.07 0.06 10 11 12 13 14 15 16 0.02 0.05 0.085 0.09 Here is the Minitab output for the blood alcohol content data:   The regression equation is BAC = Beers S = R-Sq = 80.0% Note: the actual calculated values are slightly different from these Predictor Coef StDev T P Constant -1.00 0.332 Beers 7.48 0.000 Test the hypothesis that the number of beers has no effect on BAC Test the hypothesis that the number of beers has no effect on BAC.   Hypotheses: The population of interest is __________________ H0: _______ In words, ____________________________ H1: _______ In words, _________________________________ where is ___________________________ Conditions: Calculations: 7.48 Interpretation:

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Metamath Proof Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  MPE Home  >  Th. List  >  mplascl Structured version   Visualization version   GIF version Theorem mplascl 19317 Description: Value of the scalar injection into the polynomial algebra. (Contributed by Stefan O'Rear, 9-Mar-2015.) Hypotheses Ref Expression mplascl.p 𝑃 = (𝐼 mPoly 𝑅) mplascl.d 𝐷 = {𝑓 ∈ (ℕ0𝑚 𝐼) ∣ (𝑓 “ ℕ) ∈ Fin} mplascl.z 0 = (0g𝑅) mplascl.b 𝐵 = (Base‘𝑅) mplascl.a 𝐴 = (algSc‘𝑃) mplascl.i (𝜑𝐼𝑊) mplascl.r (𝜑𝑅 ∈ Ring) mplascl.x (𝜑𝑋𝐵) Assertion Ref Expression mplascl (𝜑 → (𝐴𝑋) = (𝑦𝐷 ↦ if(𝑦 = (𝐼 × {0}), 𝑋, 0 ))) Distinct variable groups:   𝜑,𝑦   𝑦,𝐵   𝑦,𝐷   𝑓,𝐼,𝑦   𝑅,𝑓,𝑦   𝑦,𝑊   𝑦,𝑋   0 ,𝑓,𝑦 Allowed substitution hints:   𝜑(𝑓)   𝐴(𝑦,𝑓)   𝐵(𝑓)   𝐷(𝑓)   𝑃(𝑦,𝑓)   𝑊(𝑓)   𝑋(𝑓) Proof of Theorem mplascl StepHypRef Expression 1 mplascl.x . . . 4 (𝜑𝑋𝐵) 2 mplascl.b . . . . 5 𝐵 = (Base‘𝑅) 3 mplascl.p . . . . . . 7 𝑃 = (𝐼 mPoly 𝑅) 4 mplascl.i . . . . . . 7 (𝜑𝐼𝑊) 5 mplascl.r . . . . . . 7 (𝜑𝑅 ∈ Ring) 63, 4, 5mplsca 19266 . . . . . 6 (𝜑𝑅 = (Scalar‘𝑃)) 76fveq2d 6107 . . . . 5 (𝜑 → (Base‘𝑅) = (Base‘(Scalar‘𝑃))) 82, 7syl5eq 2656 . . . 4 (𝜑𝐵 = (Base‘(Scalar‘𝑃))) 91, 8eleqtrd 2690 . . 3 (𝜑𝑋 ∈ (Base‘(Scalar‘𝑃))) 10 mplascl.a . . . 4 𝐴 = (algSc‘𝑃) 11 eqid 2610 . . . 4 (Scalar‘𝑃) = (Scalar‘𝑃) 12 eqid 2610 . . . 4 (Base‘(Scalar‘𝑃)) = (Base‘(Scalar‘𝑃)) 13 eqid 2610 . . . 4 ( ·𝑠𝑃) = ( ·𝑠𝑃) 14 eqid 2610 . . . 4 (1r𝑃) = (1r𝑃) 1510, 11, 12, 13, 14asclval 19156 . . 3 (𝑋 ∈ (Base‘(Scalar‘𝑃)) → (𝐴𝑋) = (𝑋( ·𝑠𝑃)(1r𝑃))) 169, 15syl 17 . 2 (𝜑 → (𝐴𝑋) = (𝑋( ·𝑠𝑃)(1r𝑃))) 17 mplascl.d . . . 4 𝐷 = {𝑓 ∈ (ℕ0𝑚 𝐼) ∣ (𝑓 “ ℕ) ∈ Fin} 18 mplascl.z . . . 4 0 = (0g𝑅) 19 eqid 2610 . . . 4 (1r𝑅) = (1r𝑅) 203, 17, 18, 19, 14, 4, 5mpl1 19265 . . 3 (𝜑 → (1r𝑃) = (𝑦𝐷 ↦ if(𝑦 = (𝐼 × {0}), (1r𝑅), 0 ))) 2120oveq2d 6565 . 2 (𝜑 → (𝑋( ·𝑠𝑃)(1r𝑃)) = (𝑋( ·𝑠𝑃)(𝑦𝐷 ↦ if(𝑦 = (𝐼 × {0}), (1r𝑅), 0 )))) 2217psrbag0 19315 . . . 4 (𝐼𝑊 → (𝐼 × {0}) ∈ 𝐷) 234, 22syl 17 . . 3 (𝜑 → (𝐼 × {0}) ∈ 𝐷) 243, 13, 17, 19, 18, 2, 4, 5, 23, 1mplmon2 19314 . 2 (𝜑 → (𝑋( ·𝑠𝑃)(𝑦𝐷 ↦ if(𝑦 = (𝐼 × {0}), (1r𝑅), 0 ))) = (𝑦𝐷 ↦ if(𝑦 = (𝐼 × {0}), 𝑋, 0 ))) 2516, 21, 243eqtrd 2648 1 (𝜑 → (𝐴𝑋) = (𝑦𝐷 ↦ if(𝑦 = (𝐼 × {0}), 𝑋, 0 ))) Colors of variables: wff setvar class Syntax hints:   → wi 4   = wceq 1475   ∈ wcel 1977  {crab 2900  ifcif 4036  {csn 4125   ↦ cmpt 4643   × cxp 5036  ◡ccnv 5037   “ cima 5041  ‘cfv 5804  (class class class)co 6549   ↑𝑚 cmap 7744  Fincfn 7841  0cc0 9815  ℕcn 10897  ℕ0cn0 11169  Basecbs 15695  Scalarcsca 15771   ·𝑠 cvsca 15772  0gc0g 15923  1rcur 18324  Ringcrg 18370  algSccascl 19132   mPoly cmpl 19174 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-8 1979  ax-9 1986  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590  ax-rep 4699  ax-sep 4709  ax-nul 4717  ax-pow 4769  ax-pr 4833  ax-un 6847  ax-inf2 8421  ax-cnex 9871  ax-resscn 9872  ax-1cn 9873  ax-icn 9874  ax-addcl 9875  ax-addrcl 9876  ax-mulcl 9877  ax-mulrcl 9878  ax-mulcom 9879  ax-addass 9880  ax-mulass 9881  ax-distr 9882  ax-i2m1 9883  ax-1ne0 9884  ax-1rid 9885  ax-rnegex 9886  ax-rrecex 9887  ax-cnre 9888  ax-pre-lttri 9889  ax-pre-lttrn 9890  ax-pre-ltadd 9891  ax-pre-mulgt0 9892 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-3or 1032  df-3an 1033  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-eu 2462  df-mo 2463  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-ne 2782  df-nel 2783  df-ral 2901  df-rex 2902  df-reu 2903  df-rmo 2904  df-rab 2905  df-v 3175  df-sbc 3403  df-csb 3500  df-dif 3543  df-un 3545  df-in 3547  df-ss 3554  df-pss 3556  df-nul 3875  df-if 4037  df-pw 4110  df-sn 4126  df-pr 4128  df-tp 4130  df-op 4132  df-uni 4373  df-int 4411  df-iun 4457  df-iin 4458  df-br 4584  df-opab 4644  df-mpt 4645  df-tr 4681  df-eprel 4949  df-id 4953  df-po 4959  df-so 4960  df-fr 4997  df-se 4998  df-we 4999  df-xp 5044  df-rel 5045  df-cnv 5046  df-co 5047  df-dm 5048  df-rn 5049  df-res 5050  df-ima 5051  df-pred 5597  df-ord 5643  df-on 5644  df-lim 5645  df-suc 5646  df-iota 5768  df-fun 5806  df-fn 5807  df-f 5808  df-f1 5809  df-fo 5810  df-f1o 5811  df-fv 5812  df-isom 5813  df-riota 6511  df-ov 6552  df-oprab 6553  df-mpt2 6554  df-of 6795  df-ofr 6796  df-om 6958  df-1st 7059  df-2nd 7060  df-supp 7183  df-wrecs 7294  df-recs 7355  df-rdg 7393  df-1o 7447  df-2o 7448  df-oadd 7451  df-er 7629  df-map 7746  df-pm 7747  df-ixp 7795  df-en 7842  df-dom 7843  df-sdom 7844  df-fin 7845  df-fsupp 8159  df-oi 8298  df-card 8648  df-pnf 9955  df-mnf 9956  df-xr 9957  df-ltxr 9958  df-le 9959  df-sub 10147  df-neg 10148  df-nn 10898  df-2 10956  df-3 10957  df-4 10958  df-5 10959  df-6 10960  df-7 10961  df-8 10962  df-9 10963  df-n0 11170  df-z 11255  df-uz 11564  df-fz 12198  df-fzo 12335  df-seq 12664  df-hash 12980  df-struct 15697  df-ndx 15698  df-slot 15699  df-base 15700  df-sets 15701  df-ress 15702  df-plusg 15781  df-mulr 15782  df-sca 15784  df-vsca 15785  df-tset 15787  df-0g 15925  df-gsum 15926  df-mre 16069  df-mrc 16070  df-acs 16072  df-mgm 17065  df-sgrp 17107  df-mnd 17118  df-mhm 17158  df-submnd 17159  df-grp 17248  df-minusg 17249  df-mulg 17364  df-subg 17414  df-ghm 17481  df-cntz 17573  df-cmn 18018  df-abl 18019  df-mgp 18313  df-ur 18325  df-ring 18372  df-subrg 18601  df-ascl 19135  df-psr 19177  df-mpl 19179 This theorem is referenced by:  subrgascl  19319  subrgasclcl  19320  evlslem1  19336  mdegle0  23641 Copyright terms: Public domain W3C validator

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# 451 Kilometer/Hour/Second in Millimeter/Hour Squared Acceleration Kilometer/Hour/Second Millimeter/Hour Squared 451 Kilometer/Hour/Second = 1623600000000 Millimeter/Hour Squared ## How many Millimeter/Hour Squared are in 451 Kilometer/Hour/Second? The answer is 451 Kilometer/Hour/Second is equal to 1623600000000 Millimeter/Hour Squared and that means we can also write it as 451 Kilometer/Hour/Second = 1623600000000 Millimeter/Hour Squared. Feel free to use our online unit conversion calculator to convert the unit from Kilometer/Hour/Second to Millimeter/Hour Squared. Just simply enter value 451 in Kilometer/Hour/Second and see the result in Millimeter/Hour Squared. ## How to Convert 451 Kilometer/Hour/Second to Millimeter/Hour Squared (451 km/h/s to mm/h2) By using our Kilometer/Hour/Second to Millimeter/Hour Squared conversion tool, you know that one Kilometer/Hour/Second is equivalent to 3600000000 Millimeter/Hour Squared. Hence, to convert Kilometer/Hour/Second to Millimeter/Hour Squared, we just need to multiply the number by 3600000000. We are going to use very simple Kilometer/Hour/Second to Millimeter/Hour Squared conversion formula for that. Pleas see the calculation example given below. $$\text{1 Kilometer/Hour/Second} = \text{3600000000 Millimeter/Hour Squared}$$ $$\text{451 Kilometer/Hour/Second} = 451 \times 3600000000 = \text{1623600000000 Millimeter/Hour Squared}$$ ## What is Kilometer/Hour/Second Unit of Measure? Kilometer/Hour/Second or Kilometer per Hour per Second is a unit of measurement for acceleration. If an object accelerates at the rate of 1 kilometer/hour/second, that means its speed is increased by 1 kilometer per hour every second. ## What is the symbol of Kilometer/Hour/Second? The symbol of Kilometer/Hour/Second is km/h/s. This means you can also write one Kilometer/Hour/Second as 1 km/h/s. ## What is Millimeter/Hour Squared Unit of Measure? Millimeter/Hour Squared or Millimeter per Hour Squared is a unit of measurement for acceleration. If an object accelerates at the rate of 1 millimeter/hour squared, that means its speed is increased by 1 millimeter per hour every hour. ## What is the symbol of Millimeter/Hour Squared? The symbol of Millimeter/Hour Squared is mm/h2. This means you can also write one Millimeter/Hour Squared as 1 mm/h2. ## Kilometer/Hour/Second to Millimeter/Hour Squared Conversion Table (451-460) Kilometer/Hour/Second [km/h/s]Millimeter/Hour Squared [mm/h2] 4511623600000000 4521627200000000 4531630800000000 4541634400000000 4551638000000000 4561641600000000 4571645200000000 4581648800000000 4591652400000000 4601656000000000 ## Kilometer/Hour/Second to Other Units Conversion Table Kilometer/Hour/Second [km/h/s]Output 451 kilometer/hour/second in meter/second squared is equal to125.28 451 kilometer/hour/second in attometer/second squared is equal to125277777777780000000 451 kilometer/hour/second in centimeter/second squared is equal to12527.78 451 kilometer/hour/second in decimeter/second squared is equal to1252.78 451 kilometer/hour/second in dekameter/second squared is equal to12.53 451 kilometer/hour/second in femtometer/second squared is equal to125277777777780000 451 kilometer/hour/second in hectometer/second squared is equal to1.25 451 kilometer/hour/second in kilometer/second squared is equal to0.12527777777778 451 kilometer/hour/second in micrometer/second squared is equal to125277777.78 451 kilometer/hour/second in millimeter/second squared is equal to125277.78 451 kilometer/hour/second in nanometer/second squared is equal to125277777777.78 451 kilometer/hour/second in picometer/second squared is equal to125277777777780 451 kilometer/hour/second in meter/hour squared is equal to1623600000 451 kilometer/hour/second in millimeter/hour squared is equal to1623600000000 451 kilometer/hour/second in centimeter/hour squared is equal to162360000000 451 kilometer/hour/second in kilometer/hour squared is equal to1623600 451 kilometer/hour/second in meter/minute squared is equal to451000 451 kilometer/hour/second in millimeter/minute squared is equal to451000000 451 kilometer/hour/second in centimeter/minute squared is equal to45100000 451 kilometer/hour/second in kilometer/minute squared is equal to451 451 kilometer/hour/second in inch/hour/minute is equal to1065354330.71 451 kilometer/hour/second in inch/hour/second is equal to17755905.51 451 kilometer/hour/second in inch/minute/second is equal to295931.76 451 kilometer/hour/second in inch/hour squared is equal to63921259842.52 451 kilometer/hour/second in inch/minute squared is equal to17755905.51 451 kilometer/hour/second in inch/second squared is equal to4932.2 451 kilometer/hour/second in feet/hour/minute is equal to88779527.56 451 kilometer/hour/second in feet/hour/second is equal to1479658.79 451 kilometer/hour/second in feet/minute/second is equal to24660.98 451 kilometer/hour/second in feet/hour squared is equal to5326771653.54 451 kilometer/hour/second in feet/minute squared is equal to1479658.79 451 kilometer/hour/second in feet/second squared is equal to411.02 451 kilometer/hour/second in knot/hour is equal to876673.87 451 kilometer/hour/second in knot/minute is equal to14611.23 451 kilometer/hour/second in knot/second is equal to243.52 451 kilometer/hour/second in knot/millisecond is equal to0.24352051930556 451 kilometer/hour/second in mile/hour/minute is equal to16814.3 451 kilometer/hour/second in mile/hour/second is equal to280.24 451 kilometer/hour/second in mile/hour squared is equal to1008858.27 451 kilometer/hour/second in mile/minute squared is equal to280.24 451 kilometer/hour/second in mile/second squared is equal to0.077844002138622 451 kilometer/hour/second in yard/second squared is equal to137.01 451 kilometer/hour/second in gal is equal to12527.78 451 kilometer/hour/second in galileo is equal to12527.78 451 kilometer/hour/second in centigal is equal to1252777.78 451 kilometer/hour/second in decigal is equal to125277.78 451 kilometer/hour/second in g-unit is equal to12.77 451 kilometer/hour/second in gn is equal to12.77 451 kilometer/hour/second in gravity is equal to12.77 451 kilometer/hour/second in milligal is equal to12527777.78 451 kilometer/hour/second in kilogal is equal to12.53 Disclaimer:We make a great effort in making sure that conversion is as accurate as possible, but we cannot guarantee that. Before using any of the conversion tools or data, you must validate its correctness with an authority.

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Our users: As a teacher, much of my time was taken up by creating effective lesson plans. Algebra Helper allows me to create each lesson in about half the time. My kids love it because I can spend more time with them! Once they are old enough, I hope they will find this program useful as well. I originally bought Algebra Helper for my wife because she was struggling with her algebra homework. Now only did it help with each problem, it also explained the steps for each. Now my wife uses the program to check her answers. David Aguilar, CA Algebra Helper is far less expensive then my old math tutor, and much more effective. John Tusack, MI A friend recommended this software and it really helped me get all my homework done faster and right. I strongly recommend it. Candice Rosenberger, VT 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? 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# What formula markup is this? Here's a web page from 2013 using some variant of TeX-style formula markup. Modern Chrome and Firefox no longer render the formula syntax on this page. Excerpt: $$P(text{CTR}_A,text{CTR}_B|text{data})$$ is a two-dimensional function of $$text{CTR}_A$$ and $$text{CTR}_B.$$ So to find $$P(text{CTR}_A>text{CTR}_B|text{data})$$ we have to add up all the probabilities in the region where $$text{CTR}_A>text{CTR}_B$$: $$!P(text{CTR}_A > text{CTR}_B|text{data}) = iintlimits_{text{CTR}_A > text{CTR}_B} P(text{CTR}_A,text{CTR}_B|text{data}) dtext{CTR}_A dtext{CTR}_B.$$ To actually calculate this integral will require a few insights. The first is that for many standard $$A/B$$ tests, $$A$$ and $$B$$ are independent because they are observed by non-overlapping populations. Keeping this in mind, we have: $$!P(text{CTR}_A,text{CTR}_B|text{data}) = P(text{CTR}_A|text{data}) P(text{CTR}_B|text{data}).$$ Questions: 1. What library / technology was meant to interpret this markup? 2. Is there an easy way to render these formulas today? Maybe I could save the HTML from these pages and run it through some renderer? Or maybe there's an online interpreter? • Chrome and Firefox have never been able to natively render these. The code quoted is also very wrong as all \ are missing. Web pages will need javascript libraries like mathjax or katex to be loaded on the page in order for them to render this in a browser. Both of which are off topic here. Commented Oct 23, 2020 at 6:09 • I would guess that it was mathjax (or jsmath, its predecessor) but an update to the blog software has stripped all the \ making it unusable Commented Oct 23, 2020 at 9:41 • Both comments are right. I will gladly accept either one as an answer. Commented Oct 27, 2020 at 21:51

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Cody # Problem 8. Add two numbers Solution 830693 Submitted on 17 Feb 2016 by sjoerd This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. ### Test Suite Test Status Code Input and Output 1   Pass %% a = 1; b = 2; c_correct = 3; assert(isequal(add_two_numbers(a,b),c_correct)) c = 3 2   Pass %% a = 17; b = 2; c_correct = 19; assert(isequal(add_two_numbers(a,b),c_correct)) c = 19 3   Pass %% a = -5; b = 2; c_correct = -3; assert(isequal(add_two_numbers(a,b),c_correct)) c = -3 ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting!

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New Online Book! Handbook of Mathematical Functions (AMS55) Conversion & Calculation Home >> Measurement Conversion Measurement Converter Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC. Conversion Result: ```pace = 0.762 meter (length) ``` Related Measurements: Try converting from "pace" to agate (typography agate), cable length, en (typography en), fermi, finger, French, Greek fathom, hand, Israeli cubit, li (Chinese li), m (meter), micron, mile, nail (cloth nail), nautical mile, palm, parasang, pica (typography pica), point (typography point), UK mile (British mile), or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: pace = .02147766 actus (Roman actus), 420 agate (typography agate), 1.67 Biblical cubit, 1,200 bottom measure, .00347222 cable length, .00833333 city block (informal), 1.98E-09 earth to moon (mean distance earth to moon), .66666667 ell, 34.29 finger, .00378788 furlong (surveyors furlong), 1.65 Greek cubit, 9.88 Greek palm, 30 inch, 8.05E-17 light yr (light year), 3.79 link (surveyors link), .00013715 nautical league, .00019405 ri (Japanese ri), 2.57 Roman foot, .00047348 UK mile (British mile), .90936647 vara (Mexican vara). Feedback, suggestions, or additional measurement definitions? Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks!

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# Show that this PDE can be reduced to the heat equation [closed] Consider the partial differential equation: $$\frac{\partial u}{\partial t}=\frac{\partial^2u}{\partial x^2} + a \frac{\partial u}{\partial x} + bu$$ for the function $u (x; t)$ where $a$ and $b$ are constants. By using substitution of the form $u(x,t) = \exp(\alpha x+\beta t)v(x,t)$; And suitable choice of constants alpha and beta, show that the PDE can be reduced to the heat equation $$\frac{\partial v}{\partial t}=\frac{\partial^2v}{\partial x^2}.$$ - ## closed as off-topic by This is much healthier., Claude Leibovici, RecklessReckoner, Hanul Jeon, Carl MummertJul 15 '14 at 10:03 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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Claude Leibovici, RecklessReckoner, Hanul Jeon, Carl Mummert If this question can be reworded to fit the rules in the help center, please edit the question. Just differentiate $v$ with respect to $x,x^2$ and $t$. Add the terms and you should see your reslut. Note that $v$ is determined by the substitution. Usually, you should show a bit more effort and tell us what you have tried so far. – Quickbeam2k1 Feb 4 '14 at 6:36 you could compare the equations $$\beta v+ \frac{\partial v}{\partial t} \\=\alpha^2v+\frac{\partial^2 v}{\partial x^2}+2\alpha\frac{\partial v}{\partial x}+a\alpha v+a\frac{\partial v}{\partial x}+bv$$ and the old one and choose the constant that fits your request.

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# Q2: The Discriminant, D We continue from the previous lessons and summarize what we had achieved therein when we engineered a standard quadratic expression to break down into two linear expressions A+B and A – B where B contains a guy called D, in short and in long form is called DISCRIMINANT. This is an important guy and we need to get introduced to him. This is another way of looking at the quadratic expression. When we equate it to zero, we can arrive at the two values of x as shown. The +- sign comes in front of a root, as you know, since the square of both -ve & +ve numbers become +ve after squaring. Equating an expression to zero means that y=0 since y is the name of the expression i.e. y is f(x), y is a function of x and that function is the expression, in this case, the quadratic expression. So, when the expression = 0, what it means is that y=0 and if you remember, what is represented in coordinate geometry as y=0? Yes you are right if you said it is the X AXIS. The technical name of the x-axis is y=0 because anywhere on the x-axis, the value of x keeps changing but the value of y always remains ZERO. Similarly remember that the technical name of y-axis is x=0. So, finally, when we put an expression equal to zero i.e. y=0, graphically what happens is we find those values of x where y=0 And that is precisely those points where the graph cuts the x axis. This understanding is very very important: ​that to put any algebraic expression equal to 0 precisely means where the graph of that function cuts the x-axis. The role of D: DISCRIMINANT: see where it cuts the x-axis We will deal with the formula shown and D is the guy we need to get familiar with as it influences the geometry. The U shaped diagram of the quadratic can cut the x axis at two places, can touch it at one point or be above the x axis. This is shown in this diagram. Now we can see in the equation that we have two values of x and inside the formula there is a guy called square root of D. And we have identified three cases. When D is zero, x has only one value i.e. – b/2a. Since x has only one value, it is the diagram II where the U touches x axis at only one point When D is positive i.e. 4, 40, 54, whatever, it will have a root and it will have two roots +& -, as you know. So, x will have two values and it is our third diagram where the U cuts the x axis at two points. And what if D is less than zero i.e. -1, -33 -2.65 or whatever -ve value. What happens then. Remember that a negative number has no Real roots since the square of both + ve and – ve numbers become positive upon squaring. Thus there are no values of square root of D and thus when D is negative, x does not have a Real value and thus, it will not cut the x axis. Vital Math Skill: Algebra <-> Geometry Look at the diagram once again. Now look at the algebra. The skill that you must develop is to see how the diagram comes from the algebra and also how the algebra comes from the diagram. You may think I am telling you the same thing twice by inverting the sentence. But, we will show you both by experimenting with the elements in the graph as well as by experimenting with the elements in the algebra and see how a tweak in algebra affects the geometry and vice versa. Once we can do both of these things, we will have a 360 degree view of any concept in mathematics. In fact, this is the strength which distinguishes mathematicians from others.

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## What does DCT mean? What is the full form of DCT? 1, The full form of DCT is Discrete Cosine Transform. It’s used on Academic & Science ,Mathematics in Worldwide Discrete Cosine Transform (DCT) is a mathematical transform related to the Fourier transform and widely used in digital data compression. 2, The full form of DCT is Dual Clutch Transmission. It’s used on Technology ,Automotive in Worldwide Dual Clutch Transmission (DCT) or Double-Clutch Transmission, is a type of automatic transmission that automates clutch and shift operation while retaining the essential structure and direct acceleration feel of a manual transmission. As the name suggests, Dual Clutch Transmission features two clutches (dual), instead of one, for odd and even gear sets, which operate independently of each other. Just like any other automatic transmission system, it also doesn’t have a clutch pedal. DCT offers acceleration and shifting while minimizing the power gaps that occur during shifting with a manual transmission. 3, The full form of DCT is Distal Convoluted Tubule. It’s used on Medical ,Anatomy & Physiology in Worldwide Distal Convoluted Tubule (DCT) is a part of the functional unit of the kidney. 4, The full form of DCT is Dominated Convergence Theorem. It’s used on Academic & Science ,Mathematics in Worldwide Dominated Convergence Theorem (DCT) is a mathematical theorem in the theory of integration. 5, The full form of DCT is DopaChrome Tautomerase. It’s used on Medical ,Genetics in Worldwide Dopachrome Tautomerase (DCT) is a human gene. 6, The full form of DCT is Direct Coombs Test. It’s used on Medical ,Tests in Worldwide Direct Coombs Test (DCT) is a blood test for autoimmune hemolytic anemia. 7, The full form of DCT is Discourse-Completion Task. It’s used on Academic & Science ,Language & Linguistics in Worldwide Discourse-Completion Task (DCT) is a pragmatics research method in linguistics. 8, The full form of DCT is Discovery Channel Telescope. It’s used on Academic & Science ,Astronomy & Space Science in United States Discovery Channel Telescope (DCT) is a aperture telescope in Coconino National Forest, Arizona, United States. 9, The full form of DCT is Dreams Come True. It’s used on Arts ,Musical groups in Japan Dreams Come True (DCT) is a Japanese band. 10, The full form of DCT is Danescourt. It’s used on Transport & Travel ,Railway Station Codes in United Kingdom Danescourt railway station (DCT) is a railway station in Danescourt, Cardiff, Wales, United Kingdom. 11, The full form of DCT is Dreams Come True. It’s used on Associations & Organizations ,Regional Organizations in United States Dreams Come True (DCT) is a non-profit organization in Jacksonville, Florida, United States. 12, The full form of DCT is Director Control Tower. It’s used on Governmental ,Military in Worldwide Director Control Tower (DCT) was a facility equipped to a navy ship incorporating the gun-laying sights and often a rangefinder.

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Spectral BIST # Spectral BIST ## Spectral BIST - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. Spectral BIST Alok Doshi Anand Mudlapur 2. Overview • Introduction to spectral testing • Previous work • Application of RADEMACHER – WALSH spectrum in testing and design of digital circuits • Spectral techniques for sequential ATPG • Spectral methods for BIST in SOC • Spectral Analysis for statistical response compaction during BIST • Modifying spectral TPG using selfish gene algorithm • Proposed improvements • Results 3. Introduction Projection of time varying vectors in the frequency domain. PI1 : 11001100 PI2 : 11110000 PI3 : 00111100 PI4 : 01010101 4. Basic idea • Meaningful inputs (e.g., test vectors) of a circuit are not random. • Input signals must have spectral characteristics that are different from white noise (random vectors). • Sequential circuit tests are not random: • Some PIs are correlated. • Some PIs are periodic. • Correlation and periodicity can be represented by spectral components, e.g., Hadamard coefficients. 5. Statistics of Test Vectors 100% coverage test Test vectors are not random: • Correlation: a = b, frequently. • Weighting: c has more 1s than a or b. a a 00011 b 01100 c 10101 b c 6. Primary Motivation • We want to extract the information embedded in the input signals and output responses • Hence, we apply signal processing techniques to extract this information • In order to meet the above objective, we make use of frequency decomposition techniques. i.e. A signal can be projected to a set of independent, periodic waveforms that have different frequencies. • This set of waveforms, forms the basis matrix 7. Primary Motivation (cont.) • The projection operation reveals the quantity that each basis vector contributes to the original signal • This quantity is called decomposition coefficient • With the aid the decomposed information, one can easily enhance the important frequencies and suppress the unimportant ones • This process leads us to a new and better quality signal, easing the complexity (in our case i.e. of test generation) 8. Hadamard Transform • The projection matrix choosen is the Hadamard transform as it is a well-known non-sinusoidal orthogonal transform used in signal processing. • A Hadamard matrix consists of only 1’s and -1’s, which makes it a good choice for the signals in VLSI testing (1 = logic 1, -1 = logic 0). Each basis (row/column) in the Hadamard matrix is a distinct sequence that characterizes the switching frequency between 1s and -1s. 9. Hadamard Transform (cont.) • Hadamard matrices are square matrices containing only 1s and –1s. They can be generated recursively using the formula, where, H(0) = 1 and n = log2N 11. Hadamard transforms over FFT • The Walsh functions can be interpreted as binary (sampled) versions of the sin and cos, which are the basic functions of the Discrete Fourier Transform. • This interpretation led to the name BInary FOurier REpresentation (BIFORE). • The inverse Hadamard transform is given by the transpose of the Hadamard matrix scaled by the factor 1/N. 12. Hadamard transforms over FFT (cont.) • The flow graph of the Hadamard transform is shown below: x[0] X[0] x[1] X[1] The above flow graph resembles the radix-2 decimation-in-frequency FFT But it must be observed that Hadamard transform has no twiddle factors since its basis functions are square waves of either 1 or -1. Hence the Hadamard transform requires no multiplications and only N log (N) additions 13. Applying spectral techniques to generate vectors for a sequential circuit [ Agrawal et al. `01] Replace with compacted vectors Test vectors (initially random vectors) Fault simulation-based vector compaction Compacted vectors Append new vectors Stopping criteria (coverage, CPU time, vectors) satisfied? Yes Stop Extract spectral characteristics (e.g., Hadamard coefficients) and generate vectors No 14. Applying spectral techniques to generate vectors for a sequential circuit (cont.) • In seq. circuits, faults may need a biased input value • Static compaction is used to aid test generation. The resulting vector set has to retain the fault coverage • Vectors are appended before every iteration and static compaction is performed to remove the unwanted vectors that were appended 15. Applying spectral techniques to generate vectors for a sequential circuit (cont.) • Initially the test set consists of random vectors • Static compaction filters all unnecessary vectors • Now the predominant patterns are identified at the inputs using the spectral information using the Hadamard transform. • New vectors are generated based on the information obtained • Using this process, one can traverse the vector space only using the basis vectors 16. Applying spectral techniques to generate vectors for a sequential circuit (cont.) • This is of particular interest, since it drives the circuit to hard-to-reach states that require specific vectors at the PIs, making it easier to detect hard-to-detect faults • Each row/column in a Hadamard matrix is a basis vector, carrying a distinct frequency component. • Consider H(2) for example; the four basis vectors are [1 1 1 1], [1 0 1 0], [1 1 0 0], and [1 0 0 1] • Any bit sequence of length 4 can be represented as a linear combination of these basis vectors 17. Applying spectral techniques to generate vectors for a sequential circuit (cont.) • Ex: [1 0 0 0] -1 x [ 1 1 1 1] + 1 x [ 1 -1 1 -1] + 1 x [1 1 -1 -1] + 1 x [1 -1 -1 1] • Ex: [1 1 1 0] +1 x [ 1 1 1 1] + 1 x [ 1 -1 1 -1] + 1 x [1 1 -1 -1] - 1 x [1 -1 -1 1] 18. Applying spectral techniques to generate vectors for a sequential circuit (cont.) Let a, be the input bit sequence forprimary input i. for (each primary input i in test set) coefficient vector ci= H x ai for (each value in the coefficient matrix [c0, ..., cn]) if (absolute value of coefficient < cutoff) Set the coefficient to 0. else Set the coefficient to 1 or -1, based on its abs value. for (each primary input i) extension vector ei= modified ci x H if (weight > 0) Extend the vector set with value 1 to PI i. else if (weight < 0) Extend the vector set with value -1 to PI i. else if (weight == 0) Randomly extend the vector set with either 1 or -1 19. Applying spectral techniques to generate vectors for a sequential circuit (cont.) Cut-off = 4 [0 1 0 0 0 0 0 0] x H(3) = [1 -1 1 -1 1 -1 1 -1] 20. Applying spectral techniques to generate vectors for a sequential circuit (cont.) Cut-off = 4 21. Applying spectral techniques to generate vectors for a sequential circuit (cont.) Faults detected per iteration for b12 benchmark circuit 22. To be continued on Tuesday 11/16 ….. 23. Spectral Methods for BIST in a SOC environment [ A. Giani et al. `01] • This method of built-in-self-test (BIST) for sequential cores on a system-on-a-chip (SOC) generates test patterns using a real-time program that runs on an embedded processor. • This method resulted in higher fault coverage compared to the LFSR based random pattern generation techniques, without incurring the overhead of additional test hardware. 24. Spectral Methods for BIST in a SOC environment (cont.) 25. Random pattern generation with Selfish Gene algorithm for testing digital sequential circuits [ J. Zhang et al. ITC `04] • A selfish gene (SG) algorithm differs from the genetic algorithm (GA) because it evolves genes (characteristics) that provide higher fitness rather than evolving individuals with higher fitness. • The objects of evolution are the Hadamard spectral matrix, non-linear digital signal processing (DSP) filtering cutoff values, vector holding time, and relative input phase shifts, which are all modeled as genes. 26. Random pattern generation with Selfish Gene algorithm for testing digital sequential circuits (cont.) Holding theorem 27. Random pattern generation with Selfish Gene algorithm for testing digital sequential circuits (cont.) Phase Shifting 28. Random pattern generation with Selfish Gene algorithm for testing digital sequential circuits (cont.) Original Sequence Phase shift of 1 I1 is the fault propagation sequence 29. Random pattern generation with Selfish Gene algorithm for testing digital sequential circuits (cont.) • Step 1: On the first iteration, generate random vectors and compact them, but on subsequent iterations, use the compacted test sequence of the last iteration. • Step 2: In all algorithms for each vector in the compacted sequence, randomly generate a vector holding time between 0 and 64 and hold the vector accordingly to extend the sequence. If the holding time is 0, discard the vector. Holding a vector for some clock cycles is very important for high fault coverage. Further expand the test sequence. • Step 3: Evolve the genotype using the expanded sequence to get better fault coverage. • Step 4: Fault simulate and compact the best sequence. If the fault coverage is satisfactory or 125 iterations are finished, stop. Otherwise, iterate. Basic Algorithm 30. Random pattern generation with Selfish Gene algorithm for testing digital sequential circuits (cont.) • Step 2: Hold vectors to form a new sequence Snew. After that, the following procedure is performed 50 times: { Generate a random perturbation ε in the range (-0.05; +0.05) for each bit i in the original sequence. Flip each bit with pi + ε > 0.5. Do this twice, to generate two new sequences, S1 and S2, which are fault-simulated. The winning sequence has the highest fault coverage. Change the bit-flipping probabilities in the gene. If the bit was flipped (not flipped) in both the winner and the loser, there is no change. If it flipped in the winner but not in the loser, pi = pi+0.01, otherwise pi = pi – 0.01}. Finally, only the bit with highest pi in each 8-bit chunk of each PI bit stream is flipped, provided that its pi > 0.5 (only one flip/chunk). This sequence becomes the extended sequence Snew. • Step 3: The final probabilities pi as evolved in Step 2 are discarded and reset to 0.5 for the next round of holding and perturbation. Bit – perturbation and selfish gene algorithm 31. Spectral analysis for Statistical Response Compaction during BIST [ O. Khan et al. ITC `04] • Five new spectral response compactors SRC1-5 are presented. • The Hadamard matrix H(1) is used to perform spectral analysis. • Each of the SRC’s calculate either the auto-correlation of testing responses at primary outputs (POs) with the two spectral tones, each a row in H(1), or the cross-correlation between different POs. • The response compactors store the correlation coefficients, i.e., the spectral content in terms of the tones in H(1), in two counters, which represent the BIST signature. 32. Spectral analysis for Statistical Response Compaction during BIST (cont.) Hadamard matrix H(1) used for spectral analysis 33. Spectral analysis for Statistical Response Compaction during BIST (cont.) PO1 Add: 4 (1+1+0+1+1+0) Sub: 0 (-1+1+0-1+1+0) 34. Spectral analysis for Statistical Response Compaction during BIST (cont.) 0 1 0 1 1 1 SRC1 101011 0 0 1 0 1 0 0 1 1 1 1 0 0 0 1 1 0 1 1 0 35. Spectral analysis for Statistical Response Compaction during BIST (cont.) • SRC1 has higher area overhead compared to a MISR. • It was found that SRC1 was completely free of aliasing. • Holding test vectors for a number of clock cycles can improve the fault coverage in sequential circuits, but can potentially cause aliasing in a MISR. 36. Spectral analysis for Statistical Response Compaction during BIST (cont.) SRC2 37. Spectral analysis for Statistical Response Compaction during BIST (cont.) • This technique has a much lower area overhead than SRC1, since there is only one counter for the entire circuit, rather than one counter for each PO. • SRC2 has a slightly higher aliasing rate than the MISR. 38. Spectral analysis for Statistical Response Compaction during BIST (cont.) • To reduce hardware overhead, the subtract counter and the hardware associated with it was eliminated from SRC1, and only the first spectral tone was used in SRC3. It was observed that the aliasing probability for SRC3 was higher than for SRC1. The hardware overhead for SRC3 is slightly less than that for SRC1. • SRC4 is identical to SRC3 except that the second spectral tone from SRC1 was implemented instead of the first. The overhead is slightly higher than for SRC3 because a subtracter requires an extra inverter. 39. Spectral analysis for Statistical Response Compaction during BIST (cont.) SRC5 40. Spectral analysis for Statistical Response Compaction during BIST (cont.) Hadamard Transform (HT) block 41. Spectral analysis for Statistical Response Compaction during BIST (cont.) Causes of aliasing in SRC’s :- • Counter overflow; n = [log2(Length of test set)] • Bit flipping; 42. Spectral analysis for Statistical Response Compaction during BIST (cont.) • They have used Upadhyayula’s method to design a spectral TPG, the basic idea of which is to hold vectors at PI’s of circuits for multiple clocks to increase fault coverage. • This new spectral BIST system has a 91.26% shorter test sequence than for a conventional LFSR pattern generator and MISR system, with at least 8.24% higher fault coverage. 43. Proposed BIST scheme • Design of TPG • Response Compactor • MISR • Spectral compaction [Bushnell et al.] T P G CUT M I S R 44. Proposed BIST scheme (cont.) x [ 1 1 -1 1] 1 1 1 1 [ 2 -2 2 2] 1 -1 1 -1 = 1 1 -1 -1 1 -1 -1 1 45. Proposed BIST scheme (cont.) 0 X[0] 1 x[0] 2 -2 2 X[1] 1 x[1] -1 X[2] 2 -1 x[2] 2 -1 X[3] 2 1 x[3] 0 -1 -1 46. Proposed BIST scheme (cont.) x[0] x[1] x[2] x[3] CUT 47. Thank You

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# 04 - Lecture 04 Todays class: Reminders about summation... This preview shows pages 1–5. Sign up to view the full content. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: Lecture 04 Todays class: Reminders about summation notation Summation using for and while loops for and while loops in Matlab Branching with if blocks in Matlab Absolute and relative errors [We will look in more detail at invoking functions using functions as arguments (function functions in Matlab ) later. Dont worry, I wont forget this!] Reminder 1 X k =1 a k = a 1 2 X =1 a = a 1 + a 2 3 X p =1 a p = a 1 + a 2 + a 3 4 X q =1 a q = a 1 + a 2 + a 3 + a 4 n X k =1 a k = a 1 + a 2 + a 3 + ... + a n = n X =1 a Summing with for loop Write a Matlab function sum_with_for that accepts a vector as input and returns the sum of its elements. Have your function use a for loop to compute the sum. function total = sum_with_for( a ) %SUM_WITH_FOR computes the sum of a vector's elements with for loop % Input: % a = vector of input numbers % Output: % total = computed sum total = 0; % Initialise running total n = length(a); % Determine # elements in sum for k=1:n % k plays role of summation index total = total + a(k); % Increment total end % end of for loop; k incremented end Summing with while loop Write a Matlab function sum_with_while that accepts a vector as input and returns the sum of its elements. Have your function use a... View Full Document ## This note was uploaded on 06/20/2011 for the course MATH 2070 taught by Professor Aruliahdhavidhe during the Winter '10 term at UOIT. ### Page1 / 11 04 - Lecture 04 Todays class: Reminders about summation... This preview shows document pages 1 - 5. Sign up to view the full document. View Full Document Ask a homework question - tutors are online

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The OEIS is supported by the many generous donors to the OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A156740 Triangle T(n, k, m) = round( Product_{j=0..m} binomial(2*(n+j), 2*(k+j))/binomial( 2*(n-k+j), 2*j) ), where m = 7, read by rows. 5 1, 1, 1, 1, 153, 1, 1, 4845, 4845, 1, 1, 74613, 2362745, 74613, 1, 1, 735471, 358664691, 358664691, 735471, 1, 1, 5311735, 25533510145, 393216056233, 25533510145, 5311735, 1, 1, 30421755, 1056158828725, 160324910200455, 160324910200455, 1056158828725, 30421755, 1 (list; table; graph; refs; listen; history; text; internal format) OFFSET 0,5 LINKS G. C. Greubel, Rows n = 0..30 of the triangle, flattened FORMULA T(n, k, m) = round( Product_{j=0..m} b(n+j, k+j)/b(n-k+j, j) ), where b(n, k) = binomial(2*n, 2*k) and m = 7. Sum_{k=0..n} T(n, k, 7) = A151614(n). EXAMPLE Triangle begins as: 1; 1, 1; 1, 153, 1; 1, 4845, 4845, 1; 1, 74613, 2362745, 74613, 1; 1, 735471, 358664691, 358664691, 735471, 1; 1, 5311735, 25533510145, 393216056233, 25533510145, 5311735, 1; MATHEMATICA b[n_, k_]:= Binomial[2*n, 2*k]; T[n_, k_, m_]:= Round[Product[b[n+j, k+j]/b[n-k+j, j], {j, 0, m}]]; Table[T[n, k, 7], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Jun 19 2021 *) PROG (Magma) A156740:= func< n, k | Round( (&*[Binomial(2*(n+j), 2*(k+j))/Binomial(2*(n-k+j), 2*j): j in [0..7]]) ) >; [A156740(n, k): k in [0..n], n in [0..12]]; # G. C. Greubel, Jun 19 2021 (Sage) def A156740(n, k): return round( product( binomial(2*(n+j), 2*(k+j))/binomial(2*(n-k+j), 2*j) for j in (0..7)) ) flatten([[A156740(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 19 2021 CROSSREFS Cf. A086645 (m=0), A156739 (m=6), this sequence (m=7), A156741 (m=8), A156742 (m=9). Cf. A151614 (row sums). Sequence in context: A157881 A099117 A109778 * A095226 A346630 A165340 Adjacent sequences: A156737 A156738 A156739 * A156741 A156742 A156743 KEYWORD nonn,tabl AUTHOR Roger L. Bagula, Feb 14 2009 EXTENSIONS Definition corrected to give integral terms and edited by G. C. Greubel, Jun 19 2021 STATUS approved Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified May 30 14:11 EDT 2023. Contains 363055 sequences. (Running on oeis4.)

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pathfinding and chasing This topic is 4879 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic. Recommended Posts Hi. I was reading about pathfinding to try to implement it to my pacman game, but i still have lot of doubts. First, all the articles of the A* algorithm says the was pathfinding is with it, but never mentioned how to get the minimun number of squares(tiles, or whatever it is)between one object and another. I tried to find a formula but i was unsuccesful. My other doubt is about the enemy chasing pacman. I think pathfinding will work in some cases but i need something a little bit different, for example...... I need the enemy just moving and detecting the new ways to go around i mean, if the enemy encounters a cross where there are 3 paths to choose, choose one like in random, and then if the pacman is seen, then start chasing him, until it loses it. What do you think, what algorithm do you recomend me?, What can i do to do this? Share on other sites Quote: Original post by DosproFirst, all the articles of the A* algorithm says the was pathfinding is with it, but never mentioned how to get the minimun number of squares(tiles, or whatever it is)between one object and another. I tried to find a formula but i was unsuccesful. The minimum might be manhattan distance (difference in X plus difference in Y), or pythagorean distance (the square root of the difference in X plus the difference in Y squared), depending on whether you can make diagonal moves or not. In most Pacman games you can't, so the former makes sense. You can count it and verify it for yourself. Quote: for example...... I need the enemy just moving and detecting the new ways to go around i mean, if the enemy encounters a cross where there are 3 paths to choose, choose one like in random, and then if the pacman is seen, then start chasing him, until it loses it.What do you think, what algorithm do you recomend me?, What can i do to do this? Well, the algorithm is right there, in what you said: if at junction: if can see pacman: follow pacman else: pick random direction Just think about it in broad terms, make some notes, and most of the answers will come to you. Obviously you then have to decide exactly what 'follow pacman' and 'pick random direction' mean, but such is the nature of problem-solving - start with the outline and drill down. Share on other sites . Edited by GDHalFas Share on other sites I dont care having to enlarge my credits list.... But thanks for the help, i will try to implement it myself and also see whats the code about..... Share on other sites WHOA. what languaje did you use? I need jut to know. I thought it was made in C/C++ Share on other sites Actually, there's a very easy way to go about chasing, probably even less intense than using A* or checking visibility. Do what ants do with Pac-man. So, you have two different entities, ghosts and pac-man. As pac-man moves along, he leaves a pheromone trail that gradually evaporates. The ghosts do the same, but they have a tendency to follow the path with the stronger pheromone as they come to cross roads. This will cause the ghosts to chase after pac-man when they pick up his scent. If done right, this will also create clustering among the ghosts as the chase after pac-man. Of course, pheromone trails aren't deterministic. Its a probabilitic process. So, at a crossroad, a ghost may still chose a different path probabilistically. Its just that they will have the tendency to move in the direction (to the next square or tile) with a strong pheromone scent than the others around it. The slight randomness will inject some randomness into the ghosts' movement, not to mention possibly create a swarming like behavior. This is completely based on Ant Colony Optimization and Ant System research. You should be able to find the stuff on google, along with the probabilistic function used to determine the next space to move to. Share on other sites When you expand a node in a*, give each node a counter, witch = the counter of the node it expanded from + 1. When you hit the pathfinding goal, that node's counter should be the number of steps your shortest path would need. With this you could add a treshold, so that if the number of steps to reach pack-man is > X, you lost him. This wil not take corners int acount, so seeing packman should use another algo. Since there are no open spaces in pack-man, a simple _is_he_on_my_horisontal_row_before_any_walls_ and _is_he_on_my_vertical_colum_before_any_walls_ test should do it! this way you have two states: CHASE: find shortest path. if path it to long, your LOST. LOST: move in one direction, and see if you could move in another. if you SEE pack_man, CHASE him. find shortest path to CENTER, and go that way. -Anders Share on other sites Really make me think. No i have this problem. Remember that pacman is always moving, and the a* works perfect with static units. So just guessing, i think that the a* will try to find the best path but with each movement, this will make the enemy move to one side and the opposite if you the same, its not bad, but its no what i want. I want that for example: ------------------------------------------- N --------- P ------------------------------------------- N is enemy and P pacman. So with a* the algorithm would go left, then down and finally right. Well, if in that movements the pacman decide to change some tiles to the right, then the enemy should just cancel the left moving and now start moving right which is now the best path. I want to take an actio and do not cancel it until its really necesary, i mean: if pacman moves to right, continue te the left path and instead of going to the previos pacman tile, then go to the new one. Its kinda wreid. but i need IA not REALLY intelligent, i will try to implement some algorithms to see what happens... THanks for the help Share on other sites Think of the movement as static tile-steps, and then think about the smooth movement as interpolation of positions for nicer graphics / eye-candy (big AAA titles do this :-) ) So: When pac-man or goast are moving from TileA, to TileB, give them the _tile_ position at TileB, because they cant change direction until they reach B, and its nice that the AI plans a little ahead! :-) This way, the gosts wil chase the tile pac-man are going into, and not pac-mans real position! (I hope you have tiles in your game, or else, getting A* nodes would be a problem.) How you implement this is not so important, as long as you have some _tile_ positions for all units. your AI dont realy need to know that they have smooth movement from A to B (some care should be taken in the gost-vision system :-) ), and taking this information away from your AI should only make it simpler and better. You also only need to recalc pathfinding and stuff when a unit reaches its _tile_ position, since this is the only time it can change direction! Not so big speedup for pac-man, but think about 1000 units in an epic rts! -Anders Share on other sites Rule #1 in developing anything: "When you have a hammer, everything looks like a nail." So, the first question here is, do we really need A*? If you're using this pac-man game as a demonstration of how A* works, then I guess the rest of my arguments are pointless and you can skip this post. If, this is not an application specific for demonstrating A*, then we can pose the question of do you really need it? Yes, its nice and all when it comes to finding shortest paths in tile based games, but then it implies a global view? Do ghosts have an omnipotent global view? Do you want them to have a global view? Personally, I don't think ghosts should have global views because then it will make escaping from them that much tougher, consider that they move just that slightly faster than pac-man anyways. So, then what may happen if you use A* is that the game turns into how long can you survive with every ghost chasing you constantly? Traditionally, I probably don't have anything to back this up, but I feel ghosts in pac-man just randomly move around until they actually "see" pac-man. Then they simply follow pac-man. So, if you do A* during the "chase" process, then that might make sense, but will still make escaping near impossible, since ghosts will literally come at you from all sides. Since ghosts move slightly faster than pac-man anyways, I think that for game balance purposes, something as complex as A* may not be needed to get the effect you want. Just random movement, short range visibility checks, and simple chasing should suffice. Of course, this is all personal opinion. So, its not that I'm criticizing you for using A*, just that I'm presenting a different prespective. 1. 1 Rutin 44 2. 2 3. 3 4. 4 5. 5 • 11 • 9 • 12 • 10 • 13 • Forum Statistics • Total Topics 632984 • Total Posts 3009712 • Who's Online (See full list) There are no registered users currently online ×

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https://quantumweird.wordpress.com/category/arrow-of-time/

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# Five Major Problems with Wormholes Wormholes are supposed to be shortcuts from one time and place to another time and place.   For example, drive your spaceship into one end and exit near some other star, perhaps 1000 light years away.   Drive back through and return to earth.  Simple enough. If a wormhole is ever created for passage of man or machine by some future civilization, then there will be some major problems to overcome other than the biggie… creating the wormhole in the first place.  I believe this is the first time most, if not all, of these problems have been identified. Although the wormhole supposedly bends/warps time and space, there is a fundamental limit to how fast you can get from here to there, no matter how much time and space are warped.   That limit is c and it applies to the Long Way Around (LWA) path length.  First let me tell you why I think so as it is key to the some of the rest of my list of problems. A common wormhole is created by every photon that exists.   For example, a photon does a space/time warp from Proxima Centauri (the nearest star to our sun) to our eye.  The distance and time the photon experiences is zero.  It does not age during the trip and the total distance is zero at the moment of creation.   However, it still takes 4.22 years to get here, the time light takes to travel the total distance from that star to ours. Einstein’s equations say that the photon traveling at c has a total path length of zero and travel time of zero duration.  I believe that applies to every photon.   However, we know that the photon takes 4.22 light years and travels about 28 trillion miles from that star to our eye as we measure or calculate it.   Even though the path the photon sees is zero length and the time it ages is zero time during the trip, it still does not arrive until the entire 4.22 light years elapses. It is my theory that this is because the space/time warp of our photon wormhole connects the emission point on Proxima Centauri and the landing point in our eye only in a virtual sense and only in the first instant of its creation. After that first instance, the photon moves away from the emitter at light speed and the path behind it expands as the photon travels along it at c.   The photon’s path to our eye always remains zero length, but it traverses the path at c, leaving an expanded path behind until the entire path is traversed.   The photon never transfers its energy until the entire path is completed at the maximum velocity of c. My first wormhole problem is that the time required is no less than the long way around travel time at c.   Anything entering the wormhole is imposibly close to the other end (as for our photon example), but cannot actually get there until the path from the entry point expands behind the object moving at c throughout the entire trip, the LWA, just as it does for the photon wormhole. Even if the wormhole spans a time/space warp of 1000 light years, it will still take no less than 1000 years to get from here to there even if the wormhole appears to be of zero length.   The crew of the space ship that manages to get into a wormhole would not age during the trip, a distinct advantage for the crew and the ship’s lifetime.  It would seem to be instantaneous and if it were indeed reversible, then the return trip would be just as fast.  Drive into one end and return immediately and likely not be but a few hours older.   However any companions that were left behind on earth would be dead nearly 2000 years.   All this assumes the problems that follow can be solved. The second problem is that a wormhole cannot be established before it is created at each end.  If  one end is created today and the other is somehow created on a distant star, the wormhole would not be operable until the second wormhole is created, presumably at least the normal space ship travel time from one construction site to the other, even if the construction crew travels at c.    Unless the wormhole acts like a reversible time machine, a much more difficult arrangement, it will take the same amount of time each way through the wormhole with the arrow of time aging both ways and it cannot begin to be used as a shortcut until both ends are finished.    It would take a very patient civilization to plan for such a feat. My third problem involves getting into any wormhole that moves you along at light speed.  The nose of the ship would presumably be accelerated to light speed even before the crew compartment made it into the opening.   The result would be powdered spaceship and crew with photons leading the way, larger particles and atoms dragging behind, but no survivors or anything recognizable. The fourth problem is getting out of the wormhole.  Let’s say somehow you can get your space ship in and up to speed.   Everything going out the other end arrives there at light speed.   A huge blast of various rays and light burping out the other end, frying anything loitering near the exit.  A great light show, but hardly useful for the crew wanting to get from here to there in a hurry, or their greeting party for that matter.  The wormhole turns out to be a great ray gun! My fifth problem involves reversibility.  We assume that entering the wormhole at either end establishes the direction of travel.  However, it appears to me that it is very likely that the arrow of time exists only in the direction of the creation of the wormholes.  That is, from the first wormhole to the second.  Items entering the first one created would be moving in an arrow of time from the earliest time to the latest.   Items trying to enter the second wormhole to come back would be rejected in a smoldering heap or blast of rays.   If that logic is reversed, the problem still exists:  One way only! ### Arrow of Time Established? I believe this applies to photons and particles in general.  The equations for physics always seem to allow collisions to be reversable and there are no laws that would not allow any set of particle interactions to be reversible.   However, it is my opinion that photons are not reversible for the reasons listed above.  They are zipping through non reversible wormholes.   Energy is transferred from point of creation to some other point where it is absorbed or transferred to another particle and can’t go back though the wormhole as it is a one way street, from first end created to the second end and never the other way around.  That means the arrow of time always moves forward and is never reverseable.  It can be stopped but never reversed. ### SuperLumal Transmission? As a side note, for the reasons listed in the problems listed above, there will be no speedup of communications through a wormhole.  No superlumal transmissions, no advantage over sending it across space the normal way, and very likely, no two way communications.   I hope these revelations do not stop any projects in progress as science will advance no matter what.  8>)  Photon wormholes are the best anyone will be able to do. Oldtimer Copyright 2007, James A. Tabb  (may be reproduced with full credits) ## Does Time Exist? There is no question that we experience what we call time.   There is a precision with which we can measure the progression of events over time that is phenomenally accurate.   Things age and particles decay over “time” and it is consistent.    However, physical laws that use time as a reference work equally well for time reversal – going backward – a particle hitting another particle, generating other particles and emitting photons will work just as well running backward according to physics.   We just have never experienced time reversal and this disconnect with the laws of physics seems to be a mystery.  This disconnect is used by many to express the opinion that time exists.   However the fact remains that equations of space and time break down at certain points and time falls out of some of them as an unnecessary factor. Think of this: photons live in “null” time.  They live and die in the same instant because they travel at the speed of light and therefore if time exists for them, they do not experience it.   They experience zero flight time over zero distance no matter how far apart the start and finish line are.  They live in a go-splat world.    A photon leaving a star a billion light years away destroys itself in our eye the instant it is emitted, having not aged even a fraction of a nanosecond in its long trip.   Space and time are that warped! The space and the time have been warped because of the speed of the photon.  It travels at the speed of light.   Our very definition of speed involves time so when we say the speed of light we assume that time exists, but for the photon time does not exist. A photon experiences zero distance and zero time due to its incredible speed.   Every photon that lights our office or illuminates our book arrives the instant it is emitted.  It has not aged even though we can calculate that it moved from the bulb to our book and then to our eye at about one nanosecond per foot of travel.  The photon did not experience the “time” that we measure or calculate.  It aged not at all.  Time does not exist for any particle moving at c.  It only exists for us as calculated or measured in a laboratory.  But does it exist as a real dimension?  Does it have a physical basis? A photon in flight between point a and point b is invisible to any and all observers.  It does not exist in flight and can only be detected at b when it actually arrives.   The photon in flight experiences null time – time zero – no time – non-existent time, and travels a null path – or no path at all, regardless of the length of its travel.   Time for the photon does not exist, nor does distance.  Those measurements of time and distance for the photon are for our domain only – the human one. Now consider an extension of that thought – most of the particles that make up our world vibrate and exchange energy with each other.  That occurs even at temperatures close to zero.   There is also a froth of virtual particles that pop into and out of existence continually at all times even in a so-called perfect vacuum.     All the energy exchanged through photons is timeless because all photons are moving at c.   Even gravity moves at c.  Gravity is also timeless within its self.   The exception is for atoms that bump into each other and exchange energy through vibration and bumping.   Or do they?   Do they actually touch or isn’t there an exchange of particles  moving at c that keep them apart? If the energy transfer by photons is timeless, the photons are timeless, gravity is timeless all due to the speed of light as experienced by the particles that carry them, then does time exist or are we merely measuring external events by counting uniform progressions that we experience and can see? I know and acknowledge that we can measure the speed of a photon to a very high precision.  I know that we can measure the speed of gravity as other planets tug on ours and on each other.    The measurement is based on the progression of the components of our clocks.   We do live in a dimension that experiences progression of events in one direction which we call time. However, we can measure but we cannot see.  We can observe the effects but not the event.   The truth is that whenever something is traveling at c, simultaneous observations are impossible.  Every observer of the same event sees something different.    Have you ever seen time?  Maybe the change in a clock, which is actually only a measure of repetitive events, whether a wind up (measuring escapement events) or a NBS clock counting cycles of an atomic nature, but not timeWe can’t see time, only experience it.  We can’t measure time, only define it. Time for us may be just a projection of ourselves on a line defined by a progression of events that occur in a uniform manner, but it may not really exist.    We are bundles of energy made up of atoms and particles in extraordinarily rapid motion.  Take us down to the quantum world and we are made up of many quadrillions of particles exchanging energy among themselves in mostly empty space.    In such huge numbers there is an average motion and an average progression of events that may make up our concept of time.  Certainly our most accurate “clocks” are merely counting cycles of an atomic nature.   Even the National Bureau of Standards admit they are “not measuring time, but only defining it“. Does time exist just for us because we experience this progression in a uniform manner? Perhaps it is not actually an extra dimension as we have been so often told. Do you think time exists as a dimension in the same manner as x, y, z?  Is time real?  If you have been following my last two posts, you will understand it is the lack of time, at least on the photon level, that explains quantum weirdness.   And explains it well. What do you think? Oldtimer Enjoy!

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https://www.wur.nl/en/article/msc-thesis-topic-uncertainty-propagation-in-global-soil-organic-carbon-stock-estimation.htm

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Student information # MSc thesis topic: Uncertainty propagation in global soil organic carbon stock estimation There is much worldwide interest in accurate estimation of the global soil organic carbon stock. This is because soil carbon storage can help mitigate climate change. For instance, the 4per1000 initiative (https://www.4p1000.org/) aims to increase the soil organic carbon stock with 0.4 per cent each year. If this can be achieved, it would compensate for all emissions from fossil fuels. But evaluation of the effectiveness of 4per1000 measures requires that we can accurately estimate the (changes) in the global organic carbon stock. The aim of this thesis research is to contribute to this requirement by using the SoilGrids global soil map to compute the global carbon stock and analyse how uncertainties in the SoilGrids product propagate to the global carbon stock estimate. You will first get acquainted with the SoilGrids product by studying which statistical and geo-informatics techniques were used to derive the product. Next you will compute the global carbon stock based on 3D global soil property maps of soil organic carbon concentration, bulk density and coarse fragments. You will compare the result with other estimates of the global soil organic carbon stock as published in the scientific literature. Once this is done you will model the uncertainty by building a geostatistical model of the discrepancy between SoilGrids carbon stock predictions and point values of the carbon stock. This will be a fairly simple model but regional variations will need to be taken into account, while another difficulty lies in handling big data issues. It may be useful to tackle the big data issues by first working for a smaller part of the world (e.g. France or USA) and next scale up to the globe. Once the geostatistical model is derived it will be used to create ‘possible realities’ using spatial stochastic simulation techniques. These possible realities are next used as input to a Monte Carlo uncertainty propagation analysis, which ultimately yields an estimate of the global carbon stock while also quantifying the estimation error. If time permits you will also analyse the sensitivity of the results to assumptions and parameter settings. Objectives • Understand the scientific literature on global soil organic carbon stock estimation • Model the uncertainty in SoilGrids products by developing a geostatistical model of the errors in the SoilGrids carbon stock map and by calibrating this model using point data of observed differences • Be able to handle big data issues by familiarising with appropriate software and hardware environments • Run a Monte Carlo uncertainty propagation analysis and interpret results Literature Requirements • Solid background in geostatistical modelling, such as obtained through the Spatial Modelling and Statistics course • Experience with programming in R • Geo-informatics skills, in particular for handling big-data situations Theme(s): Modelling & visualisation

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Margin Charting A top investment site online said few investors understand margin as well as they should. That includes me. Drop this code with one paste into an existing algorithm at the end of initialize() to chart any overnight margin where fees can run high. schedule_function(margin_overnight, date_rules.every_day(), time_rules.market_open()) context.margin_ttl = 0 context.mdays = 0.0 # a count of days for margin average def margin_overnight(context, data): ''' Chart overnight margin, current, total and average per day. ''' if context.mdays: margin_one_night = abs(min(0, context.portfolio.cash)) # Margin held overnight context.margin_ttl += margin_one_night margin_average = context.margin_ttl / context.mdays record(Margin = margin_one_night) # Last night's carried margin record(Mrgn_Ttl = context.margin_ttl) # Total overnight margin record(Mrgn_Avg = margin_average) # Average overnight margin record(PnL = context.portfolio.pnl) # Actual profit context.mdays += 1.0 Why does this matter? Realistic evaluation of algorithms and comparison can only be done with knowledge of their amount of margin. If you start with $1 and buy a share of SPY every day, your returns will appear astronomical, the returns curve in that case is unfortunately an illusion, most of the apparent profit would be stock value owed eventually in cash back to the broker, even without considering the fees. About 600% returns below. And what are the fees? Margin | Interactive Brokers https://gdcdyn.interactivebrokers.com/en/index.php?f=marginnew&p=overview1 Overnight Margin Calculations. Stocks have additional margin requirements when held overnight. For overnight margin requirements for stocks, click the Stocks ... Adjusted Gross Margin - Investopedia www.investopedia.com/terms/a/adjusted-gross-margin.asp Adjusted gross margin goes one step further than gross margin because it includes these inventory carrying costs, which greatly affect the bottom line of a product's profitability. Can anyone add some broker fees to the code? 61 Loading... Total Returns -- Alpha -- Beta -- Sharpe -- Sortino -- Max Drawdown -- Benchmark Returns -- Volatility -- Returns 1 Month 3 Month 6 Month 12 Month Alpha 1 Month 3 Month 6 Month 12 Month Beta 1 Month 3 Month 6 Month 12 Month Sharpe 1 Month 3 Month 6 Month 12 Month Sortino 1 Month 3 Month 6 Month 12 Month Volatility 1 Month 3 Month 6 Month 12 Month Max Drawdown 1 Month 3 Month 6 Month 12 Month # https://www.quantopian.com/posts/margin def initialize(context): schedule_function(trade, date_rules.every_day(), time_rules.market_open(minutes=1)) schedule_function(margin_overnight, date_rules.every_day(), time_rules.market_open()) context.margin_ttl = 0 context.mdays = 0.0 # a count of days for margin average def margin_overnight(context, data): ''' Chart overnight margin, current, total and average per day. Intraday margin is ignored. ''' if context.mdays: margin_one_night = abs(min(0, context.portfolio.cash)) # Margin held overnight context.margin_ttl += margin_one_night margin_average = context.margin_ttl / context.mdays record(Margin = margin_one_night) # Last night's carried margin record(Mrgn_Ttl = context.margin_ttl) # Total overnight margin record(Mrgn_Avg = margin_average) # Average overnight margin context.mdays += 1.0 def trade(context, data): order(sid(8554), 1) record(PnL = context.portfolio.pnl) There was a runtime error. 6 responses Can someone who knows from margin vet this, added a fee and logging if margin call. Maybe can lead to a correct version schedule_function(margin_overnight, date_rules.every_day(), time_rules.market_open()) context.margin_ttl = 0 context.margin_cost = 0.0 # Fees for overnight margin, 3% ? Not sure. context.mdays = 0.0 # a count of days for margin average def margin_overnight(context, data): ''' Chart overnight margin, current, total, average per day and an assigned cost. Intraday margin is ignored. ''' c = context if c.mdays: margin_one_night = abs(min(0, c.portfolio.cash)) # Margin held overnight c.margin_cost += margin_one_night * .03 c.margin_ttl += margin_one_night margin_average = c.margin_ttl / c.mdays record(Night_Margin = margin_one_night) # Last night's carried margin record(Ttl_Cost_Mrgn = c.margin_cost) # Fee on overnight margin record(Avg_Mrgn = margin_average) # Average overnight margin record(PnL = c.portfolio.pnl) # Profit without margin # Profit if overnight margin were 3%. right? wrong? record(PnL_w_Mrgn = c.portfolio.pnl - c.margin_cost) #record(Ttl_Mrgn = c.margin_ttl) # Total overnight margin if margin_one_night > .5 * context.portfolio.portfolio_value: # anywhere close? log.info('margin call, margin {} vs 50% of portfolio = {}'.format( int(margin_one_night), int(.5 * context.portfolio.portfolio_value))) c.mdays += 1.0 I feel like this method is creating false margin calls on days when there are large gains while holding a long position. (EDIT: Though.. Maybe not.. I can't read the equations very well because I don't understand all the code yet. I just know I'm getting tons of calls on an algo that doesn't drawdown more than 30%. I don't understand how it could be generating calls if I haven't lost more than 50% of my original purchase price?) When you purchase a long position on margin you only put up 50% of the value at purchase. As the market value increases your DEBIT stays the same. What changes as the Market value changes is your equity. So Equity = Long Market Value - Debit Balance. So, if I buy 1000 shares of Widget Corp at$20.00 a share for a total of $20,000 I will have a DEBIT balance of$10,000 (Reg T requires 50% for between days. You are also maintenance margin called if LMV hits 25% of your purchase price during intraday trading.) If the stock goes up say 20% I have a new LMV of $24,000. So my account now looks like this. ( LMV of$24,000 Debit of $10,000$14,000 of equity. ($4,000 of EXCESS equity) I could use the excess equity to purchase more securities. When you do that the debit created is 100% of the purchase price. So say I did that. My account would now look like this. (I'll buy 166 more shares @$24) LMV of $27,984 ($24,000 + $3,984 of the new purchase using excess equity.) Debit of$13,984 ($10,000 +$3,984 of new debit from the purchase using excess equity.) Equity of $14,000 (No net change) Excess equity of$6. (Equity - Debit) Also, margin interest is ANNUAL so to get the true cost of margin you need to calculate the daily rate (Annual Interest Rate/365)=Daily interest rate. I think it may be hard to code this in a proper way without it getting entirely too complex. I feel like we would need to create order types. Three types. Long (Cash) - No extra requirements Long (Margin) - 50% due at purchase and 50% as a debit for Reg T. 25% Maintenance margin. Short (Margin) - 50% due at purchase and 50% as a credit. Keep in mind that maintenance margin is 30% on shorts instead of 25% I'm trying to think of ways to do this, but I'm still very new to this whole coding thing. Here's a good reference for LONG margin. http://www.dummies.com/test-prep/series-7/how-to-calculate-the-numbers-in-long-margin-accounts-on-the-series-7-exam/ Great vetting so with that added info now I hope someone will write some code for modeling margin costs. Thanks WF. I can help with equations and work flow for addressing this, however until I learn more about code I'm not very good at building something out myself. I currently work at a trade desk and am studying for my series 66 when I'm not at the desk. Hopefully soon I'll take that exam and then I can devote more time to learning more about python. If you want to prod me for any info let me know. I've been building strategies in excel for awhile now cause at work we have plugins that pull in our market feeds, but now I have to learn how to build in an "actual" language. ALSO.. THANK YOU so much for your PvR tools. I use them on the regular now when I'm looking at algos. Your PVR tool has actually helped me catch a few double purchasing bugs in some of the code I was trying to build. Hopefully someone into margin and python will roll up their sleeves and dig in. Good to hear PvR was helpful and thanks for mentioning it. I think part of the problem might lay in the way quantopian places the orders. No matter how I align my orders quontopian seems to put them in alphabetical order and process them. So, this results in buys before sells a LOT, which would cause the cash in my account to drop considerably for a moment. Does anyone know how to override this behavior?

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## Fraction Worksheets Conversion :: Addition :: Subtraction :: Multiplication :: Division ## Conversions Fractions - addition, fractions - subtraction, fractions - multiplication, fractions - division. Confused by basic fractions? Scared of fractions ? Want to learn how to add, subtract, multiply, or divide really complicated fractions? Let's start with a basic introduction to fractions. You're used to working with whole numbers such as 1, 2, 3 etc., but fractions allow you to go in between. Here are some good lessons that will introduce the concepts of fractions: Introduction to Fractions (from the BBC) Varnelle's Primary Math (mathforum.org) NRICH Introduction (Maths.org) Once you've got a handle on what a fraction is, you will want to do things with fractions. It's easy enough to add 1/4 to 1/4, but how do you add 1/3 to 3/7? How do you divide fractions? Check out these lessons on performing arithmetic with fractions: Dividing Fractions (FreeMathHelp.com) Multiplying Fractions (FreeMathHelp.com) Multiplying Fractions (math.com) Multiply and Divide Fractions (themathpage.com) Multiplying Simple Fractions (sosmath.com) Looking for even more help with fractions? Guaranteach , a major math video site, has provided over 500 free videos in their fractions category! ## Fractions Worksheets Welcome to the fractions worksheets page at Math-Drills.com where the cup is half full! This is one of our more popular pages most likely because learning fractions is incredibly important in a person's life and it is a math topic that many approach with trepidation due to its bad rap over the years. Fractions really aren't that difficult to master especially with the support of our wide selection of worksheets. This page includes Fractions worksheets for understanding fractions including modeling, comparing, ordering, simplifying and converting fractions and operations with fractions. We start you off with the obvious: modeling fractions. It is a great idea if students can actually understand what a fraction is, so please do spend some time with the modeling aspect. Relating modeling to real life helps a great deal too as it is much easier to relate to half a cookie than to half a square. Ask most students what you get if you add half a cookie and another half a cookie, and they'll probably let you know that it makes one delicious snack. The other fractions worksheets on this page are devoted to helping students understand the concept of fractions. From comparing and ordering to simplifying and converting... by the time students master the material on this page, operations of fractions will be a walk in the park. ## General Use Fractions Printables Fraction circles. Fraction circle manipulatives are mainly used for comparing fractions, but they can be used for a variety of other purposes such as representing and identifying fractions, adding and subtracting fractions, and as probability spinners. There are a variety of options depending on your purpose. Fraction circles come in small and large versions, labeled and unlabeled versions and in three different color schemes: black and white, color, and light gray. The color scheme matches the fraction strips and use colors that are meant to show good contrast among themselves. Do note that there is a significant prevalence of color-blindness in the population, so don't rely on all students being able to differentiate the colors. Suggested activity for comparing fractions: Photocopy the black and white version onto an overhead projection slide and another copy onto a piece of paper. Alternatively, you can use two pieces of paper and hold them up to the light for this activity. Use a pencil to represent the first fraction on the paper copy. Use a non-permanent overhead pen to represent the second fraction. Lay the slide over the paper and compare the two circles. You should easily be able to tell which is greater or lesser or if the two fractions are equal. Re-use both sheets by erasing the pencil and washing off the marker. Adding fractions with fraction circles will involve two copies on paper. Cut out the fraction circles and segments of one copy and leave the other copy intact. To add 1/3 + 1/2, for example, place a 1/3 segment and a 1/2 segment into a circle and hold it over various fractions on the intact copy to see what 1/2 + 1/3 is equivalent to. 5/6 or 10/12 should work. ## Fraction Strips Fractions strips are often used for comparing fractions. Students are able to see quite easily the relationships and equivalence between fractions with different denominators. It can be quite useful for students to have two copies: one copy cut into strips and the other copy kept intact. They can then use the cut-out strips on the intact page to individually compare fractions. For example, they can use the halves strip to see what other fractions are equivalent to one-half. This can also be accomplished with a straight edge such as a ruler without cutting out any strips. Pairs or groups of strips can also be compared side-by-side if they are cut out. Addition and subtraction (etc.) are also possibilities; for example, adding a one-quarter and one-third can be accomplished by shifting the thirds strip so that it starts at the end of one-quarter then finding a strip that matches the end of the one-third mark (7/12 should do it). Teachers might consider copying the fraction strips onto overhead projection acetates for whole class or group activities. Acetate versions are also useful as a hands-on manipulative for students in conjunction with an uncut page. The "Smart" Fraction Strips include strips in a more useful order, eliminate the 7ths and 11ths strips as they don't have any equivalents and include 15ths and 16ths. The colors are consistent with the classic versions, so the two sets can be combined. ## Modeling Fractions Worksheets Besides using the worksheets below, you can also try some more interesting ways of modeling fractions. Healthy snacks can make great models for fractions. Can you cut a cucumber into thirds? A tomato into quarters? Can you make two-thirds of the grapes red and one-third green? ## Modeling fractions with groups of shapes Fractions can represent parts of a group or parts of a whole. In these worksheets, fractions are modeled as parts of a group. ## Ratio and Proportion Worksheets Picture ratios. More picture ratios can be found on holiday and seasonal pages. Try searching for picture ratios to find more. ## Equivalent fractions The equivalent fractions models worksheets include only the "baking fractions" in the A versions. To see more difficult and varied fractions, please choose the B to J versions after loading the A version. ## Comparing & Ordering Fractions Worksheets Comparing proper fractions. Comparing fractions involves deciding which of two fractions is greater in value or if the two fractions are equal in value. There are generally four methods that can be used for comparing fractions. First is to use common denominators . If both fractions have the same denominator, comparing the fractions simply involves comparing the numerators. Equivalent fractions can be used to convert one or both fractions, so they have common denominators. A second method is to convert both fractions to a decimal and compare the decimal numbers. Visualization is the third method. Using something like fraction strips , two fractions can be compared with a visual tool. The fourth method is to use a cross-multiplication strategy where the numerator of the first fraction is multiplied by the denominator of the second fraction; then the numerator of the second fraction is multiplied by the denominator of the first fraction. The resulting products can be compared to decide which fraction is greater (or if they are equal). ## Comparing Proper and Improper Fractions The worksheets in this section also include improper fractions. This might make the task of comparing even easier for some questions that involve both a proper and an improper fraction. If students recognize one fraction is greater than one and the other fraction is less than one, the greater fraction will be obvious. ## Comparing Proper, Improper and Mixed Fractions This section additionally includes mixed fractions. When comparing mixed and improper fractions, it is useful to convert one of the fractions to the other's form either in writing or mentally. Converting to a mixed fraction is probably the better route since the first step is to compare the whole number portions, and if one is greater than the other, the proper fraction portion can be ignored. If the whole number portions are equal, the proper fractions must be compared to see which number is greater. ## Ordering fractions on a Number Line Many of the same strategies that work for comparing fractions also work for ordering fractions. Using manipulatives such as fraction strips, using number lines, or finding decimal equivalents will all have your student(s) putting fractions in the correct order in no time. We've probably said this before, but make sure that you emphasize that when comparing or ordering fractions, students understand that the whole needs to be the same. Comparing half the population of Canada with a third of the population of the United States won't cut it. Try using some visuals to reinforce this important concept. Even though we've included number lines below, feel free to use your own strategies. ## Ordering fractions The ordering fractions worksheets in this section do not include a number line, to allow for students to use various sorting strategies. ## Simplifying & Converting Fractions Worksheets Rounding fractions. Rounding fractions helps students to understand fractions a little better and can be applied to estimating answers to fractions questions. For example, if one had to estimate 1 4/7 × 6, they could probably say the answer was about 9 since 1 4/7 is about 1 1/2 and 1 1/2 × 6 is 9. ## Simplifying fractions Learning how to simplify fractions makes a student's life much easier later on when learning operations with fractions. It also helps them to learn that different-looking fractions can be equivalent. One way of demonstrating this is to divide out two equivalent fractions. For example 3/2 and 6/4 both result in a quotient of 1.5 when divided. By practicing simplifying fractions, students will hopefully recognize unsimplified fractions when they start adding, subtracting, multiplying and dividing with fractions. ## Operations with Fractions Worksheets Multiplying fractions. Multiplying fractions is usually less confusing operationally than any other operation and can be less confusing conceptually if approached in the right way. The algorithm for multiplying is simply multiply the numerators then multiply the denominators. The magic word in understanding the multiplication of fractions is, "of." For example what is two-thirds OF six? What is a third OF a half? When you use the word, "of," it gets much easier to visualize fractions multiplication. Example: cut a loaf of bread in half, then cut the half into thirds. One third OF a half loaf of bread is the same as 1/3 x 1/2 and tastes delicious with butter. ## Fillable, Savable, Layout Conceptually, dividing fractions is probably the most difficult of all the operations, but we're going to help you out. The algorithm for dividing fractions is just like multiplying fractions, but you find the inverse of the second fraction or you cross-multiply. This gets you the right answer which is extremely important especially if you're building a bridge. We told you how to conceptualize fraction multiplication, but how does it work with division? Easy! You just need to learn the magic phrase: "How many ____'s are there in ______? For example, in the question 6 ÷ 1/2, you would ask, "How many halves are there in 6?" It becomes a little more difficult when both numbers are fractions, but it isn't a giant leap to figure it out. 1/2 ÷ 1/4 is a fairly easy example, especially if you think in terms of U.S. or Canadian coins. How many quarters are there in a half dollar? Adding fractions requires the annoying common denominator. Make it easy on your students by first teaching the concepts of equivalent fractions and least common multiples. Once students are familiar with those two concepts, the idea of finding fractions with common denominators for adding becomes that much easier. Spending time on modeling fractions will also help students to understand fractions addition. Relating fractions to familiar examples will certainly help. For example, if you add a 1/2 banana and a 1/2 banana, you get a whole banana. What happens if you add a 1/2 banana and 3/4 of another banana? A common strategy to use when adding mixed fractions is to convert the mixed fractions to improper fractions, complete the addition, then switch back. Another strategy which requires a little less brainpower is to look at the whole numbers and fractions separately. Add the whole numbers first. Add the fractions second. If the resulting fraction is improper, then it needs to be converted to a mixed number. The whole number portion can be added to the original whole number portion. ## Subtracting Fractions There isn't a lot of difference between adding and subtracting fractions. Both require a common denominator which requires some prerequisite knowledge. The only difference is the second and subsequent numerators are subtracted from the first one. There is a danger that you might end up with a negative number when subtracting fractions, so students might need to learn what it means in that case. When it comes to any concept in fractions, it is always a good idea to relate it to a familiar or easy-to-understand situation. For example, 7/8 - 3/4 = 1/8 could be given meaning in the context of a race. The first runner was 7/8 around the track when the second runner was 3/4 around the track. How far ahead was the first runner? (1/8 of the track). ## Various Operations Fractions Worksheets Mixing up the signs on operations with fractions worksheets makes students pay more attention to what they are doing and allows for a good test of their skills in more than one operation. ## Multiplying and dividing fractions This section includes worksheets with both multiplication and division mixed on each worksheet. Students will have to pay attention to the signs. ## Operations with Negative fractions Although some of these worksheets are single operations, it should be helpful to have all of these in the same location. There are some special considerations when completing operations with negative fractions. It is usually very helpful to change any mixed numbers to an improper fraction before proceeding. It is important to pay attention to the signs and know the rules for multiplying positives and negatives (++ = +, +- = -, -+ = - and -- = +). ## Order of Operations with Fractions Worksheets Order of operations with fractions. The order of operations worksheets in this section actually reside on the Order of Operations page, but they are included here for your convenience.. ## Order of operations with decimals & fractions mixed • Kindergarten • Learning numbers • Comparing numbers • Place Value • Roman numerals • Subtraction • Multiplication • Order of operations • Drills & practice • Measurement • Factoring & prime factors • Proportions • Shape & geometry • Data & graphing • Word problems • Children's stories • Leveled Stories • Context clues • Cause & effect • Compare & contrast • Fact vs. fiction • Fact vs. opinion • Main idea & details • Story elements • Conclusions & inferences • Sounds & phonics • Words & vocabulary • Early writing • Numbers & counting • Simple math • Social skills • Other activities • Dolch sight words • Fry sight words • Multiple meaning words • Prefixes & suffixes • Vocabulary cards • Other parts of speech • Punctuation • Capitalization • Cursive alphabet • Cursive letters • Cursive letter joins • Cursive words • Cursive sentences • Cursive passages • Grammar & Writing • Math by topic ## Fractions Worksheets Our fraction worksheets start with the introduction of the concepts of " equal parts ", "parts of a whole" and "fractions of a group or set"; and proceed to operations on fractions and mixed numbers. Fraction worksheets Fractions to decimals Fraction multiplication and division Converting fractions, equivalent fractions, simplifying fractions Fraction to / from decimals Fraction multiplication and division worksheets Fraction to / from decimals Topics include: • Identifying "equal parts" • Dividing shapes into "equal parts" • Parts of a whole • Fractions in words • Coloring shapes to make fractions • Writing fractions • Fractions of a group or set • Word problems: write the fraction from the story • Equal parts • Numerators and denominators of a fraction • Writing fractions from a numerator and denominator • Reading fractions and matching to their words • Writing fractions in words • Identifying common fractions (matching, coloring, etc) • Fractions as part of a set or group (identifying, writing, coloring, etc) • Using fractions to describe a set • Comparing fractions with pie charts (parts of whole, same denominator) • Comparing fractions with pie charts (same numerator, different denominators) • Comparing fractions with pictures (parts of sets) • Comparing fractions with block diagrams • Understanding fractions word problems • Writing and comparing fractions word problems • Identifying fractions • Fractional part of a set • Identifying equivalent fractions • Equivalent fractions - missing numerators, denominators • 3 Equivalent fractions • Comparing fractions with pie charts (same denominator) • Comparing proper fractions with pie charts • Comparing proper or improper fractions with pie charts • Compare mixed numbers with pie charts • Comparing fractions (like, unlike denominators) • Compare improper fractions, mixed numbers • Simplifying fractions (proper, improper) • Completing whole numbers • Subtracting like fractions • Subtracting a fraction from a whole number or mixed number • Subtracting mixed numbers • Converting fractions to / from mixed numbers • Converting mixed numbers and fractions to / from decimals • Fractions word problems • Adding like fractions (denominators 2-12) • Adding like fractions (all denominators) • Adding fractions and mixed numbers (like denominators) • Subtracting like fractions (denominators 2-12) • Subtracting fractions from whole numbers, mixed numbers • Subtracting mixed numbers from mixed numbers or whole numbers • Comparing improper fractions and mixed numbers with pie charts • Comparing proper and improper fractions • Ordering 3 fractions • Identifying equivalent fractions (pie charts) • Writing equivalent fractions (pie charts) • Equivalent fractions with missing numerators or denominators ## Grade 4 fractions to decimals worksheets • Convert decimals to fractions (tenths, hundredths) • Convert decimals to mixed numbers (tenths, hundredths) • Convert fractions to decimals (denominator of 10 or 100) • Convert mixed numbers to decimals (denominator of 10 or 100) • Adding like fractions (denominators 2-25) • Adding mixed numbers and / or fractions (like denominators) • Adding unlike fractions & mixed numbers • Subtracting fractions from whole numbers and mixed numbers (same denominators) • Subtracting mixed numbers with missing subtrahend or minuend) • Subtracting unlike fractions • Subtracting mixed numbers (unlike denominators) • Word problems on adding and subtracting fractions ## Grade 5 fraction multiplication and division worksheets • Multiply fractions by whole numbers • Multiply fractions by fractions • Multiply improper fractions • Multiply fractions by mixed numbers • Multiply mixed numbers by mixed numbers • Missing factor questions • Divide whole numbers by fractions (answers are whole numbers) • Divide a fraction by a whole number and vice versa • Divide mixed numbers by fractions • Divide fractions by fractions • Mixed numbers divided by mixed numbers • Word problems on multiplying and dividing fractions • Mixed operations with fractions word problems ## Grade 5 converting, simplifying & equivalent fractions • Converting improper fractions to / from mixed numbers • Simplifying proper fractions • Simplifying proper and improper fractions • Equivalent fractions (2 fractions) • Equivalent fractions (3 fractions) ## Grade 5 fraction to / from decimals worksheets • Convert decimals to fractions (tenths, hundredths), no simplification • Convert decimals to fractions (tenths, hundredths), with simplification • Convert decimals to mixed numbers • Convert fractions to decimals (denominators of 10 or 100) • Convert mixed numbers to decimals (denominators of 10 or 100) • Convert mixed numbers to decimals (denominators of 10, 100 or 1000) • Convert fractions to decimals (common denominators of 2, 4, 5, ...) • Convert mixed numbers to decimals (common denominators of 2, 4, 5, ...) • Convert fractions to decimals, some with repeating decimals • Adding  fractions and mixed numbers • Adding mixed numbers (unlike denominators) • Subtract unlike fractions • Subtract mixed numbers (unlike denominators) ## Grade 6 fraction multiplication and division worksheets • Fractions multiplied by whole numbers • Fractions multiplied by fractions • Mixed numbers multiplied by fractions • Mixed numbers multiplied by mixed numbers • Whole numbers divided by fractions • Fractions divided by fractions • Mixed numbers divided by mixed nuymbers • Mixed multiplication or division practice ## Grade 6 converting, simplifying and equivalent fractions worksheets • Convert improper fractions to / from mixed numbers • Simplify proper fractions • Simplify proper and improper fractions • Equivalent fractions (4 fractions) ## Grade 6 fraction to / from decimals worksheets • Convert decimals to fractions, with simplification • Convert decimals to mixed numbers, with simplification • Convert fractions to decimals (denominators are 10 or 100) • Convert fractions to decimals (various denominators) • Convert mixed numbers to decimals (various denominators) ## Related topics Decimals worksheets Word problem worksheets Sample Fractions Worksheet What is K5? K5 Learning offers free worksheets , flashcards  and inexpensive  workbooks  for kids in kindergarten to grade 5. Become a member  to access additional content and skip ads. Our members helped us give away millions of worksheets last year. We provide free educational materials to parents and teachers in over 100 countries. If you can, please consider purchasing a membership (\$24/year) to support our efforts. Members skip ads and access exclusive features. This content is available to members only. • For a new problem, you will need to begin a new live expert session. • You can contact support with any questions regarding your current subscription. • You will be able to enter math problems once our session is over. • I am only able to help with one math problem per session. Which problem would you like to work on? • Does that make sense? • I am currently working on this problem. • Are you still there? • It appears we may have a connection issue. 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If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. ## Unit 1: Decimal place value Unit 2: add decimals, unit 3: subtract decimals, unit 4: add and subtract fractions, unit 5: multi-digit multiplication and division, unit 6: multiply fractions, unit 7: divide fractions, unit 8: multiply decimals, unit 9: divide decimals, unit 10: powers of ten, unit 11: volume, unit 12: coordinate plane, unit 13: algebraic thinking, unit 14: converting units of measure, unit 15: line plots, unit 16: properties of shapes. ## Teach Fractions Through Word Problems • Math Tutorials • Pre Algebra & Algebra • Exponential Decay Teaching fractions can often seem like a daunting task. You may hear many the groan or sigh when you open a book to the section on fractions. This does not have to be the case. In fact, most students will not dread a topic once they feel confident working with the concept. The concept of a “fraction” is abstract. Visualizing apart versus a whole is a developmental skill not fully grasped by some students until middle or high school. There are a few ways to get your class embracing fractions, and there are a number of worksheets you can print out to nail the concept home for your students. ## Make Fractions Relatable Children, in fact, students of all ages prefer a hands-on demonstration or an interactive experience to pencil-and-paper math equations. You can get felt circles to make pie graphs, you can play with fraction dice, or even use a set of dominoes to help explain the concept of fractions. If you can, order an actual pizza. Or, if you happen to celebrate a class birthday, well perhaps make it a "fraction" birthday cake. When you engage the senses, you have a higher engagement of the audience. Also, the lesson has a great chance of permanence, too. You can print fraction circles so your students can illustrate fractions as they learn. Have them touch the felt circles, let them watch you create a felt circle pie representing a fraction, ask your class to color in the corresponding fraction circle. Then, ask your class to write the fraction out. ## Have Fun with Math As we all know, not every student learns the same way. Some children are better at visual processing than auditory processing. Others prefer tactile learning with hand-held manipulatives or may prefer games. Games make what could be a dry and boring topic more fun and interesting. They provide that visual component that might make all the difference. There are plenty of online teaching tools with challenges for your students to use. Let them practice digitally. Online resources can help solidify concepts. ## Fraction Word Problems A problem is, by definition, a situation that causes perplexity. A primary tenet of teaching through problem-solving is that students confronted with real-life problems are forced into a state of needing to connect what they know with the problem at hand. Learning through problem-solving develops understanding. A student's mental capacity grows more complex with time. Solving problems can force them to think deeply and to connect, extend, and elaborate on their prior knowledge. ## Common Pitfall Sometimes you can spend too much time teaching fraction concepts, like "simplify," "find the common denominators," "use the four operations," that we often forget the value of word problems. Encourage students to apply their knowledge of fraction concepts through problem-solving and word problems. • Free Christmas Math Worksheets • Fraction Worksheets and Printables • Thanksgiving Math Worksheets and Activities for Kids • A Delicious Way to Teach Fractions • Halloween Worksheets, Printables, and Activities • Subtraction of Fractions With Common Denominators • Halloween Math Worksheets & Printable Activities • Christmas Word Problem Worksheets • Free Christmas Worksheets for the Holidays • Fraction Tests and Worksheets • Why Learning Fractions Is Important • Free Math Word Problem Worksheets for Fifth-Graders • Free Easter Worksheets Over Reading, Math, and More • 2nd Grade Math Word Problems • Quiz 8th-Graders With These Math Word Problems By clicking “Accept All Cookies”, you agree to the storing of cookies on your device to enhance site navigation, analyze site usage, and assist in our marketing efforts. • Kindergarten • Number charts • Skip Counting • Place Value • Number Lines • Subtraction • Multiplication • Word Problems • Comparing Numbers • Ordering Numbers • Odd and Even • Prime and Composite • Roman Numerals • Ordinal Numbers • In and Out Boxes • Number System Conversions • More Number Sense Worksheets • Size Comparison • Measuring Length • Metric Unit Conversion • Customary Unit Conversion • Temperature • More Measurement Worksheets • Writing Checks • Profit and Loss • Simple Interest • Compound Interest • Tally Marks • Mean, Median, Mode, Range • Mean Absolute Deviation • Stem-and-leaf Plot • Box-and-whisker Plot • Permutation and Combination • Probability • Venn Diagram • More Statistics Worksheets • Shapes - 2D • Shapes - 3D • Lines, Rays and Line Segments • Points, Lines and Planes • Transformation • Ordered Pairs • Midpoint Formula • Distance Formula • Parallel, Perpendicular and Intersecting Lines • Scale Factor • Surface Area • Pythagorean Theorem • More Geometry Worksheets • Converting between Fractions and Decimals • Significant Figures • Convert between Fractions, Decimals, and Percents • Proportions • Direct and Inverse Variation • Order of Operations • Squaring Numbers • Square Roots • Scientific Notations • Speed, Distance, and Time • Absolute Value • More Pre-Algebra Worksheets • Translating Algebraic Phrases • Evaluating Algebraic Expressions • Simplifying Algebraic Expressions • Algebraic Identities • Systems of Equations • Polynomials • Inequalities • Sequence and Series • Complex Numbers • More Algebra Worksheets • Trigonometry • Math Workbooks • English Language Arts • Summer Review Packets • Social Studies • Holidays and Events • Worksheets > • Pre-Algebra > • Fractions > ## Fraction Word Problem Worksheets Featured here is a vast collection of fraction word problems, which require learners to simplify fractions, add like and unlike fractions; subtract like and unlike fractions; multiply and divide fractions. The fraction word problems include proper fraction, improper fraction, and mixed numbers. Solve each word problem and scroll down each printable worksheet to verify your solutions using the answer key provided. Thumb through some of these word problem worksheets for free! Represent and Simplify the Fractions: Type 1 Presented here are the fraction pdf worksheets based on real-life scenarios. Read the basic fraction word problems, write the correct fraction and reduce your answer to the simplest form. Represent and Simplify the Fractions: Type 2 Before representing in fraction, children should perform addition or subtraction to solve these fraction word problems. Write your answer in the simplest form. Conjure up a picture of how adding fractions plays a significant role in our day-to-day lives with the help of the real-life scenarios and circumstances presented as word problems here. (15 Worksheets) Subtracting Fractions Word Problems Worksheets Crank up your skills with this set of printable worksheets on subtracting fractions word problems presenting real-world situations that involve fraction subtraction! Multiplying Fractions Word Problems Worksheets This set of printables is for the ardently active children! Explore the application of fraction multiplication and mixed-number multiplication in the real world with this exhilarating practice set. Fraction Division Word Problems Worksheets Gift children a broad view of the real-life application of dividing fractions! Let them divide fractions by whole numbers, divide 2 fractions, divide mixed numbers, and solve the word problems here. Related Worksheets » Decimal Word Problems » Ratio Word Problems » Division Word Problems » Math Word Problems » Fractions Become a Member Membership Information What's New? Printing Help Testimonial Members have exclusive facilities to download an individual worksheet, or an entire level. • Math Worksheets • ELA Worksheets • Science Worksheets • Online Worksheets • Become a Member • Kindergarten • Skip Counting • Place Value • Number Lines • Subtraction • Multiplication • Word Problems • Comparing Numbers • Ordering Numbers • Odd and Even Numbers • Prime and Composite Numbers • Roman Numerals • Ordinal Numbers • Big vs Small • Long vs Short • Tall vs Short • Heavy vs Light • Full and Empty • Metric Unit Conversion • Customary Unit Conversion • Temperature • Tally Marks • Mean, Median, Mode, Range • Mean Absolute Deviation • Stem and Leaf Plot • Box and Whisker Plot • Permutations • Combinations • Lines, Rays, and Line Segments • Points, Lines, and Planes • Transformation • Ordered Pairs • Midpoint Formula • Distance Formula • Parallel and Perpendicular Lines • Surface Area • Pythagorean Theorem • Significant Figures • Proportions • Direct and Inverse Variation • Order of Operations • Scientific Notation • Absolute Value • Translating Algebraic Phrases • Simplifying Algebraic Expressions • Evaluating Algebraic Expressions • Systems of Equations • Slope of a Line • Equation of a Line • Polynomials • Inequalities • Determinants • Arithmetic Sequence • Arithmetic Series • Geometric Sequence • Complex Numbers • Trigonometry • Social Studies Worksheets • Educational Games • Interactive Lessons ## Fraction Word Problems Worksheets • Pre-Algebra > • Fractions > Traverse through our printable fraction word problems worksheets for a wide range of real-world scenarios that involve identifying fractions and equivalent fractions and performing four operations with proper fractions, improper fractions, and mixed numbers. With exclusive theme-based word problems on adding, subtracting, multiplying, and dividing fractions, our free worksheets will see the young beginners and skilled learners practice with relish! These pdf worksheets are most recommended for students in grade 3 through grade 7. CCSS: 3.NF, 4.NF, 5.NF, 6.NS Identifying Fractions | Worksheet #1 Savor a potpourri of everyday circumstances that have fractions in place. Read the scenarios carefully and pinpoint the fractions that take part in them to solve this free identifying fractions worksheet. Identifying Fractions | Worksheet #2 Children in 3rd grade and 4th grade need to pore over the fraction word problems, identify the numerators as parts and denominators as whole, and solve them. Equivalent Fractions Word Problems Here's an exclusive set that hands out well-researched word problems on equivalent fractions, so compelling that you would instantly want to add it to your repertoire! Grab these printable practice worksheets now! Casting around for story problems on fraction addition? Watch kitchen ingredients, handicrafts, and many other familiar objects contribute their bits toward the cause of fractions for addition. Subtracting Fractions Word Problems Plug into this bunch of pdf worksheets with interesting word problems on fraction subtraction for grade 3, grade 4, grade 5, and grade 6, and steal the limelight! Multiplying Fractions Word Problems Unleash your hidden skills at fraction multiplication through our fraction word problems worksheets, where it's raining relatable situations. Children will discover a whole new way of embracing fractions in their daily lives. Dividing Fractions Word Problems A free, must-have resource for 5th grade, 6th grade, and 7th grade students. Tap into the student’s inner genius and watch them perform fraction division with renewed passion! ## Exclusive Online Worksheets • Dividing Fractions Word Problems | Customary • Dividing Fractions Word Problems | Metric • Dividing Fractions Related Printable Worksheets ▶ Subtracting Fractions ▶ Multiplying Fractions ▶ Dividing Fractions Members can share the worksheet with students instantly via WhatsApp, Email, or Google Classroom. Members can add worksheets to “My Collections”, save them as folders, and download each folder as a workbook or a .zip file. Members can learn to solve the problems step by step with an example. Members can filter the worksheets by topic. Become a member for \$2.50/month (billed annually), and gain instant access to 20,000+ printable and digitally fillable worksheets. • Solve equations and inequalities • Simplify expressions • Factor polynomials • Graph equations and inequalities • All solvers • Arithmetics • Determinant • Percentages • Scientific Notation • Inequalities ## What can QuickMath do? QuickMath will automatically answer the most common problems in algebra, equations and calculus faced by high-school and college students. • The algebra section allows you to expand, factor or simplify virtually any expression you choose. It also has commands for splitting fractions into partial fractions, combining several fractions into one and cancelling common factors within a fraction. • The equations section lets you solve an equation or system of equations. You can usually find the exact answer or, if necessary, a numerical answer to almost any accuracy you require. • The inequalities section lets you solve an inequality or a system of inequalities for a single variable. You can also plot inequalities in two variables. • The calculus section will carry out differentiation as well as definite and indefinite integration. • The matrices section contains commands for the arithmetic manipulation of matrices. • The graphs section contains commands for plotting equations and inequalities. • The numbers section has a percentages command for explaining the most common types of percentage problems and a section for dealing with scientific notation. ## Math Topics More solvers. • Simplify Fractions Numbers Place value ... Decimal numbers ... Estimating and rounding ... Adding/subtracting decimals ... Multiplying decimals ... Dividing decimals ... Percent ... Exponents ... Square roots ... Signed integers ... Adding and subtracting integers ... Multiplying and dividing integers ... Properties of integers ... Numbers Unit Quiz Ratios and Proportions Ratios ... Proportions ... Distance, rates, and time ... Similar figures Ratios and proportions Unit Quiz Factoring Factors and multiples ... Greatest common factor (GCF) ... Least common multiple (LCM) ... Factoring Unit Quiz Fractions Fraction definitions ... Reducing fractions ... Adding and subtracting fractions ... Multiplying fractions ... Dividing fractions ... Adding and subtracting mixed numbers ... Multiplying mixed numbers ... Dividing mixed numbers ... Fractions Unit Quiz The Language of Algebra Definitions ... Order of operations ... Writing equations ... Writing inequalities ... The Language of Algebra Unit Quiz The Basics of Algebra Useful properties ... Exponents ... Evaluating expressions ... Like terms ... Simplifying ... The Basics of Algebra Unit Quiz Equations and Inequalities Solving addition and subtraction equations ... Solving multiplication equations ... Solving division equations ... Solving inequalities ... Formulas .... Two-step equations and inequalities ... Equations and Inequalities Unit Quiz Graphing Equations and Inequalities The coordinate plane ... Slope and y-intercept ... Graphing linear equations ... Graphing Equations and Inequalities Unit Quiz Geometry Building Blocks Geometry words ... Coordinate geometry ... Pairs of lines ... Classifying angles ... Angles and intersecting lines ... Circles ... Geometry Building Blocks Unit Quiz Polygons Polygon basics ... Triangles ... Quadrilaterals ... Area of polygons and circles ... Polygons Unit Quiz Relations and Sizes Congruent figures ... Similar figures ... Squares and square roots ... The Pythagorean Theorem and right triangle facts ... Relations and Sizes Unit Quiz Three-dimensional Figures Space figures ... Prisms ... Pyramids ... Cylinders, cones, and spheres ... Three-dimensional Figures Unit Quiz ## Homework help The downsides of free homework help.. First, and the main downside is the inability to check the result. People helping you with your task might not know the answers themselves. And what assistance would you get in this case? Wrong answers! Second, it's the proficiency of the strangers on the other side of the screen. Who are they? College professors who came to provide homework help for free? Or just students like you who might not be able to cope with a simple task. They might not now anything about the subject, or might be way too self-confident and give you wrong answers. Think twice before using the help from these people. In the next part of the article we describe the most popular sources of homework assistance and give our honest opinion about them. ## The best websites for homework help. When a student is looking for help he always (no exaggeration!) expects it to be free. Of course, being a college student always means being tight on budget and looking for the ways to save money even more. But it should be related to college homework help. Sadly, there are thousands of examples when a bad homework resulted in bad grades, and a poor student had to learn more to pass the exam. That's why we don't recommend to look for answers on the websites like Reddit or Chegg. Why? We explain below. Chegg homework help is a paid service. You need to buy a monthly subscription to use it. Is it worth it? Depends on your professor. If the tasks you get are strictly from a textbook, then it might be good. To cope with a creative teachers this website has nothing to offer. The same goes for Reddit homework help. It might be useful when you're looking for solutions on a standard task, the one that dozens of people are struggling with as well. There's nothing these services can offer if you're homework is unique and created by a teacher himself. What to do in this case? Read our recommendations below. ## We recommend to try it! There's an easy and cheap solution that will help you succeed in your studies. A personal assistance with homework created just for your tasks. No need to scroll pages looking for similar tasks and subjects, no need to copy from the screen and guess, if the results are correct. Professionals will perform the task for you! All you have to do is to provide it and enjoy a personalized approach and high quality service. After that you'll never come back to Reddit in search of answers! ## Making Sense of Fractions: This Tactic Helped Students Grasp a Key Math Topic • Share article Fractions are an important building block in students’ mathematical foundations. Understanding how they work and why is crucial for success in Algebra and more advanced math courses. But fractions are also notoriously difficult to master. Studies have shown that a significant portion of students— about one third —don’t make much progress in their understanding of the topic between 4th grade, when operations with fractions are typically introduced, and 6th grade, when students are expected to be fluent in fractional arithmetic. Fractions are so challenging in part because they don’t operate in the same way that whole numbers do, said Nancy C. Jordan, a professor of learning sciences at the University of Delaware. Numbers of the same magnitude can look very different: Take 2/4 and 8/16, for example. And sometimes, when the numbers in a fraction grow bigger, the magnitude actually gets smaller—1/4 is bigger than 1/8, for instance. On Tuesday, Jordan presented her work on a fraction sense intervention for struggling 6th graders at an Institute of Education Sciences Math Summit, an online conference hosted by the U.S. Department of Education’s research wing. Jordan’s work, which is funded by an IES grant , will scale up a program that she and her colleagues found improved students’ understanding of fraction concepts and measurement, as well as their ability to apply that understanding to solve problems. The intervention “aims to make explicit mathematical connections,” said Jordan, demonstrating how fraction magnitudes are represented across different contexts. ## Parts of a whole vs. values on a number line Traditional fraction instruction emphasizes fractions as part of a whole, said Jordan. Think about an eight-slice pizza with two slices missing to represent 6/8, for example, or a group of four circles with three colored in to represent 3/4. But teaching fractions this way, rather than representing them as numbers with their own magnitudes, can lead to misunderstandings, Jordan said. She shared examples of student work from pre-tests in her research. In one question, students were asked to shade in 3/4 of eight circles. To get the question correct, students would need to shade in 6 circles. But when students got the question wrong, many shaded in three circles, because they thought of 3/4 as three parts—rather than a value between the numerals 0 and 1. In Jordan’s intervention, teachers use a number line to represent fractions. This allows teachers to show fraction equivalence on the number line—to demonstrate, for example, that 3/4 is the same distance between 0 to 1 as 6/8. Teachers also link the number line to other fraction representations: fraction bars, a collection of items, or liquid in a measurement cup. Teachers then help students connect these representations to numbers and equations, and students get regular practice distinguishing between and performing different operations. Teaching fractions with a number line isn’t a new practice. It was emphasized in the Common Core State Standards introduced in 2010, which at the time represented a major shift in how fractions were taught in schools. The underlying idea behind this change is that number lines help students put fractions into context—demonstrating their relationship to integers. But Jordan said that presenting fractions as part of a whole is still a common teaching method—as are procedural shortcuts that can leave students with little conceptual understanding of why operations work the way they do. She gave the example of the “butterfly” method of adding and subtracting fractions, which relies a multiplication trick to find a common denominator. Jordan said her intervention demonstrates that even if students have misconceptions, it’s possible to help them develop deeper understanding in older elementary grades: “Even students who are struggling mightily with basic fractions after three years of instruction can learn to make sense of fractions.” Edweek top school jobs. ## 24/7 Homework Help: Get Help Online The downsides of getting free homework help. When we aren’t able to deal with the pile of hometasks ourselves, we think about getting any help. If our parents had to ask their parents or other seniors for help, then we have a wonderful opportunity to check, for example, Chegg: homework help of any kind can be found there. Also there are dozens of other platforms that offer homework help of different types. For example, we all love to scroll through Reddit threads sometimes. However, apart from gossiping and meaningless talks, there might be something important and serious there as well. Reddit homework help is also quite popular, as most tasks in schools and colleges are taken from the same books. But have your ever thought that free homework help might be dangerous for you? First of all, it’s a wrong answer to the task. Just imagine that you copy something without thinking, and bring it to class. Oops, apparently, the answer is wrong, but you can’t even explain how you were working on this task, because you wasn’t… One of the most tricky aspects here is math homework help. Sometimes, the right answer is not enough. A student should follow a certain algorithm when solving the task. And professional services will help him with that. No need to worry about any type math: algebra homework help, geometry and other subjects are available. As a result, you will save your time, avoid mistakes and errors, pay a reasonable price and, mot importantly, will be confident in the correct answer. How all these became possible? We are very lucky to live in the data-driven era, when anything we want is available with one click of the mouse (in fact, you don’t even need a mouse). College homework help is available online at any time you need it. Forget about time zones, don’t think about living in a small town, when the power of the whole world is in front of you! Get help with tasks of any complexity, regardless of the subject. Not need to be a pro in accounting: homework help will change the way you think about studies! Lorem ipsum dolor sit amet, consectetur adipisicing elit. Perferendis qui quam iure natus rerum minima ut aut minus nam fugiat odit, id perspiciatis, at laudantium autem? Incidunt, ab dolorum fugit, obcaecati sapiente aliquid Lorem ipsum dolor sit amet, consectetur adipisicing elit. Perferendis qui quam iure natus rerum minima ut aut minus nam fugiat odit, id perspiciatis, at laudantium autem? Incidunt, ab dolorum fugit, obcaecati Customer Reviews Bennie Hawra #### IMAGES 1. Fractions Homework Help 2. Pin by Twisty Noodle on Fractions in 2021 3. Fraction homework help sheets year 7 4. 5th Grade Fractions Practice Worksheet 5. Fractions Homework Help 6. Equivalent Fractions Worksheet #### VIDEO 1. Understanding Fractions 2. CONVERT THE MIXED FRACTIONS TO IMPROPER FRACTIONS 4. Fraction|| भिन्‍न || in hindi Part-2 5. Learn to DIVIDE FRACTIONS in just 3 minutes! 6. Question 6: Equivalent expressions of a fraction 1. Fraction Worksheets Fraction Worksheets Worksheets » Fractions Conversion :: Addition :: Subtraction :: Multiplication :: Division Worksheet Black/White Circle to Fraction B/W Fraction to Circle B/W Line to Fraction B/W Fraction to Line B/W Conversions Worksheet Simplify Fractions Mixed to Improper Fractions Improper to Mixed Fractions Fractions - Addition Worksheet 2. Fractions NRICH Introduction (Maths.org) Once you've got a handle on what a fraction is, you will want to do things with fractions. It's easy enough to add 1/4 to 1/4, but how do you add 1/3 to 3/7? How do you divide fractions? Check out these lessons on performing arithmetic with fractions: Adding Fractions (FreeMathHelp.com) 3. Understand fractions Unit 1 Intro to multiplication Unit 2 1-digit multiplication Unit 3 Intro to division Unit 4 Understand fractions Unit 5 Place value through 1,000,000 Unit 6 Add and subtract through 1,000,000 Unit 7 Multiply 1- and 2-digit numbers Unit 8 Divide with remainders Unit 9 Add and subtract fraction (like denominators) Unit 10 Multiply fractions 4. Fractions Worksheets News Most Popular Fractions Worksheets this Week General Use Fractions Printables Fraction Circles Fraction Strips Modeling Fractions Worksheets Modeling fractions with groups of shapes Modeling fractions with rectangles Modeling fractions with circles Ratio and Proportion Worksheets Picture ratios Equivalent fractions Equivalent ratios 5th grade 16 units · 130 skills. Unit 1 Decimal place value. Unit 2 Add decimals. Unit 3 Subtract decimals. Unit 4 Add and subtract fractions. Unit 5 Multi-digit multiplication and division. Unit 6 Multiply fractions. Unit 7 Divide fractions. Unit 8 Multiply decimals. 6. Fraction Calculator Fraction Calculator. Step 1: Enter the fraction you want to simplify. The Fraction Calculator will reduce a fraction to its simplest form. You can also add, subtract, multiply, and divide fractions, as well as, convert to a decimal and work with mixed numbers and reciprocals. We also offer step by step solutions. 7. Browse Printable Fraction Worksheets Search Printable Fraction Worksheets. Perhaps your child is just beginning to learn the differences between halves, thirds, and fourths. Or maybe she's a fraction master who can simplify fractions and multiply mixed numbers with whole numbers. Either way, we have fractions worksheets designed to assist students at all learning levels. 8. Math.com Homework Help Hot Subject: Fractions · Study Tips · Wonders of Math Search Fraction Definitions Reducing fractions Adding and subtracting Multiplying Dividing Adding and subtracting Multiplying Dividing Unit Quiz Factoring Factors and multiples Greatest common factor (GCF) Least common multiple (LCM) Unit Quiz Tables, Formulas & Calculators · Fraction-Decimal Conversion Table 9. Dividing fractions (practice) Course: 6th grade > Unit 2. Lesson 5: Dividing fractions by fractions. Understanding division of fractions. Dividing fractions: 2/5 ÷ 7/3. Dividing fractions: 3/5 ÷ 1/2. Dividing fractions. Dividing mixed numbers. Divide mixed numbers. Writing fraction division story problems. 10. Fractions worksheets for grades 1-6 Free Book! Fraction worksheets for grades 1-6, starting with the introduction of the concepts of "equal parts", "parts of a whole" and "fractions of a group or set"; and proceeding to reading and writing fractions, adding, subtracting, multiplying and dividing proper and improper fractions and mixed numbers. Equivalent fractions. 11. Mathway Algebra Free math problem solver answers your algebra homework questions with step-by-step explanations. Unit 1: Decimal place value 0/1300 Mastery points Decimal place value intro Decimals on the number line Decimals in expanded form Decimals in written form Decimals in different forms Comparing decimals Rounding decimals Unit 2: Add decimals 0/1000 Mastery points 13. Fraction Worksheets and Ratio Homework Make Fractions Relatable. Children, in fact, students of all ages prefer a hands-on demonstration or an interactive experience to pencil-and-paper math equations. You can get felt circles to make pie graphs, you can play with fraction dice, or even use a set of dominoes to help explain the concept of fractions. If you can, order an actual pizza. 14. Printable Fractions Worksheets for Teachers These fractions worksheets are a great resource for children in Kindergarten, 1st Grade, 2nd Grade, 3rd Grade, 4th Grade, and 5th Grade. Click here for a Detailed Description of all the Fractions Worksheets. If you're looking for a great tool for adding, subtracting, multiplying or dividing mixed fractions check out this online Fraction Calculator. 15. Fractions Learn about fractions using our free math solver with step-by-step solutions. 16. Fraction Word Problems Worksheets Fraction Word Problem Worksheets. Featured here is a vast collection of fraction word problems, which require learners to simplify fractions, add like and unlike fractions; subtract like and unlike fractions; multiply and divide fractions. The fraction word problems include proper fraction, improper fraction, and mixed numbers. 17. Browse Printable 5th Grade Fraction Worksheets Your students won't have to fear fractions with the help of our fifth grade fractions worksheets and printables. Designed to challenge fifth graders and prepare them for middle school math, these fifth grade fractions worksheets give students practice in adding, subtracting, multiplying, dividing, and simplifying fractions, as well converting proper and improper fractions, and even applying ... 18. Fraction Word Problems Worksheets Fraction Word Problems Worksheets. Traverse through our printable fraction word problems worksheets for a wide range of real-world scenarios that involve identifying fractions and equivalent fractions and performing four operations with proper fractions, improper fractions, and mixed numbers. With exclusive theme-based word problems on adding ... 19. Step-by-Step Math Problem Solver Example: 2x-1=y,2y+3=x What can QuickMath do? QuickMath will automatically answer the most common problems in algebra, equations and calculus faced by high-school and college students. The algebra section allows you to expand, factor or simplify virtually any expression you choose. 20. Math.com Homework help Free math lessons and math homework help from basic math to algebra, geometry and beyond. Students, teachers, parents, and everyone can find solutions to their math problems instantly. 21. Math Homework Help Online There's an easy and cheap solution that will help you succeed in your studies. A personal assistance with homework created just for your tasks. No need to scroll pages looking for similar tasks and subjects, no need to copy from the screen and guess, if the results are correct. Professionals will perform the task for you! 22. Making Sense of Fractions: This Tactic Helped Students Grasp a Key Math On Tuesday, Jordan presented her work on a fraction sense intervention for struggling 6th graders at an Institute of Education Sciences Math Summit, an online conference hosted by the U.S ... 23. Homework Help Service Get help with tasks of any complexity, regardless of the subject. Not need to be a pro in accounting: homework help will change the way you think about studies! Choose you subject and get the best homework help: math, physics, biology, chemistry, literature, etc. Lorem ipsum dolor sit amet, consectetur adipisicing elit. 24. Fraction Homework Help Be it anything, our writers are here to assist you with the best essay writing service. With our service, you will save a lot of time and get recognition for the academic assignments you are given to write. This will give you ample time to relax as well. Let our experts write for you. With their years of experience in this domain and the ...

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# Question: A book publisher monitors the size of shipments of its A book publisher monitors the size of shipments of its textbooks to university bookstores. For a sample of texts used at various schools, the 95% confidence interval for the size of the shipment was 250 ± 45 books. Which, if any, of the following interpretations of this interval is/are correct? (a) All shipments are between 205 and 295 books. (b) 95% of shipments are between 160 and 340 books. (c) The procedure that produced this interval generates ranges that hold the population mean for 95% of samples. (d) If we get another sample, then we can be 95% sure that the mean of this second sample is between 160 and 340. (e) We can be 95% confident that the range 160 to 340 holds the population mean. Sales2 Views127

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Click to Chat 1800-2000-838 +91-120-4616500 CART 0 • 0 MY CART (5) Use Coupon: CART20 and get 20% off on all online Study Material ITEM DETAILS MRP DISCOUNT FINAL PRICE Total Price: Rs. There are no items in this cart. Continue Shopping ` a mixture of co and co2 is found to have a density of 1.5 gm/lit at 30  degree celsius and 730 mm .what is the composition of the mixture.` one year ago Sakshi 646 Points ``` Hi..For a mixture of CO and CO2, d = 1.50 g/litre P = 730 / 760 atm, T = 303K PV =w / m RT; PV w / Vm RT 730 / 760 = 1.5 / m × 0.0821 × 303 =∴ 38.85 i.e. molecular weight of mixture of CO and CO2= 38.85 Let % of mole of CO be a in mixture then Average molecular weight = a × 28 + (100 – a) 44 / 100 38.85 = 28a + 4400 – 44a / 100 a = 32.19 Mole % of CO = 32.19 Mole % of CO2= 67.81 Regards ``` one year ago Think You Can Provide A Better Answer ? ## Other Related Questions on Most Scoring Topics in IIT JEE View all Questions » • Complete JEE Main/Advanced Course and Test Series • OFFERED PRICE: Rs. 15,000 • View Details • Complete JEE Main/Advanced Course and Test Series • OFFERED PRICE: Rs. 15,000 • View Details Post Question

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## Reflection: Diverse Entry Points Border Tiles: Seeing Structure in Algebraic Expressions - Section 2: Opening This year I am working with many special education students in my Algebra 1 class.  In addition, all of my students have had significant interruptions to their education.  Several have been out of school for over one year.  When I introduced today's mathematical task, some students struggled to get started. I had to tell them again and again NOT to count the tiles! Other students found ways to find the number of tiles, but were at a loss for how to describe (or diagram) how they found them.  What I did with those students was to ask them (one-on-one) how they found the number of tiles and I scribed what they said. From there, we tried to translate that writing into a diagram or a mark up of the original paper.  Students were using really interesting and different ways to figure out the number of tiles, they just weren't always comfortable describing those ways. Differentiating the Lesson Opening Diverse Entry Points: Differentiating the Lesson Opening # Border Tiles: Seeing Structure in Algebraic Expressions Unit 2: Multiple Representations: Situations, Tables, Graphs, and Equations Lesson 1 of 17 ## Big Idea: Don't count! Students find different approaches for counting tiles around a garden and then write rules that represent their methods. Print Lesson 6 teachers like this lesson Standards: Subject(s): Math, algebraic expression, Algebra, Graphing (Algebra), combining like terms, distributive property, Algebraic expressions, equation 60 minutes ### Amanda Hathaway ##### Similar Lessons ###### The Cell Phone Problem, Day 1 Algebra II » Rational Functions Big Idea: Real world modeling of rational functions. Cell phone signal strength, can you hear me now? Favorites(6) Resources(20) Fort Collins, CO Environment: Suburban ###### Maximizing Volume - Day 1 of 2 12th Grade Math » Functioning with Functions Big Idea: A classic maximization problem is used to investigate relative extrema. Favorites(4) Resources(12) Troy, MI Environment: Suburban ###### Graphing Linear Functions in Standard Form (Day 1 of 2) Algebra I » Graphing Linear Functions Big Idea: Students will analyze the importance of intercepts in linear function, and use them to graph lines that are in an unfamiliar format. Favorites(43) Resources(16) Washington, DC Environment: Urban

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2010 Zero Admits... now what? : The B-School Application Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack It is currently 25 Feb 2017, 11:34 ### 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 # 2010 Zero Admits... now what? Author Message TAGS: ### Hide Tags Intern Joined: 27 Mar 2011 Posts: 36 GMAT 1: 710 Q47 V40 Followers: 1 Kudos [?]: 9 [1] , given: 1 ### Show Tags 27 Mar 2011, 07:33 1 KUDOS Good Afternoon, Although I have frequented this forum sporadically over the last year or so, I finally decided to create an account to leave a few posts. As the subject indicates, my year-long application misadventure has brought me to this rather unpleasant of situations. Inspired by the 2008 admits thread, I thought this might be a good place to brush off the dust, commiserate with others, and prepare for the next leg of the journey. Without further ado, here goes a rundown of my profile along with a brief postmortem. Though I hope there are not too many others in my situation, if you are, please feel free to leave your information as well as a little contribution for posterity. Who knows, after reading over the 2008 zero admits - where are they now follow up thread, there may be hope yet! GMAT Score: 710 GPA: 3.4 Work Experience : 5 years Location: United States #Applications: 3 #Dings: 3 #Interviews: 1 #WL: 0 Which ding hurt the most and why? Chicago definitely hurt the most. Despite a dismal interview, I still harbored a tiny shred of hope. What next? To be honest, I am not sure. I am fairly determined (as the name indicates) to re-apply next year. In order to avoid repeating my present situation, I plan to apply during Round 1 to give me sufficient time to apply to additional schools during Round 2 if needed. Unfortunately, this gives me very little time to do much to improve my candidacy beyond where it is currently at. I am currently a Peace Corps volunteer in West Africa and my service should conclude in July. I will probably attempt to take 1-2 graduate business courses in the interim (My GMAT quant percentile was only 77). In my attempt to remain positive, I see this as a rather unique opportunity. Basically, after I apply in October, I will be accepted, and will matriculate the following August/September. In effect, this gives me a year to play with. Given that I am still fairly uncertain of what I'd like to do long term, this gives me a year to try things out. Ideally, I think I'll try to work with either a start-up or with a local government organization back home. While this does not give me much time to improve my application for next admissions cycle, I think it may help me to figure out whether I would like to do either of these things later in life. Why the lack of success? Where to begin? Here, I had minimal access to electricity, which did not allow me to research programs very effectively or to put in as much effort to my essays as I could have. I only applied to three schools (Kellogg, Wharton, and Chicago - What was I thinking?), and only during Round 2. Some of my essays were certainly not as well (or as strategically) written as they could have been. Also, my career goals were definitely a little flimsy. If I am able to firm those up over the next few months, I will probably re-apply to at least one of the above schools, though I will also need to be a bit more realistic about my chances and will need include a bit of a safety net. Congratulations to everyone who was admitted this application cycle! I hope to be in your shoes next time around. Intern Joined: 01 Mar 2011 Posts: 7 Followers: 0 Kudos [?]: 5 [3] , given: 0 ### Show Tags 28 Mar 2011, 08:47 3 KUDOS More of a lurker and don't usually comment, but was right there with you last year. Similar profile, US applicant, 720 GMAT, 3.3 GPA, 6 years work exp. I applied to Kellogg, Chicago & Yale. Had my heart set on Kellogg (like 110% set), thought we (me and significant other) were moving to Chicago for his work, so applied to Booth as well. Threw in Yale b/c I've worked in non-profs, and everyone kept telling me to apply there. Which is probably not a great reason to apply to any school, but I digress. Was totally bummed to go zero for 3, had given up hope entirely....And then I randomly had a conversation with a friend's spouse who recently graduated from MIT Sloan who told me that the admission's process is a game, and that you've just got to play it right, and was a game worth playing to achieve my personal/career goals. She offered to look over my resume, essays, etc., and gave me some excellent advice about applying again this year. I'd heard all this before, but she put it all together in a persuasive and eloquent manner and just made me reconsider abandoning hope so easily. (Maybe that's why the Sloanie's get the big bucks!) I revised my list of schools to better match my background and future career goals. Everything from my resume to my essays to my Rec letters (I did NOT write these, but upon my friend's suggestions, wrote detailed primers with examples my recommenders had witnessed - but may have needed a refresher on - that highlighted key traits and characteristics important to that particular school) were a gazillion times better. In fact, I got a hand-written note on a mailed acceptance letter with a note commenting on my Rec letters! Definitely felt good! In retrospect, my previous apps were just crappy. Not my A-game. Or even B-game. Did not present a cohesive story or put me in the most favorable light. I think I was having a hard time expressing exaclty why I wanted to go to bschool and what I wanted to do afterwards, and this definitely came across in the application process. In fact, my previous apps were SOOO bad, I did not feel comfortable re-applying to Kellogg - it was just a tragic app, awkward interview, and felt that I had killed my chance. This may be my only regret from this cycle, was not giving Kellogg another go, but everything works out for a reason. Fast forwarding to now. This year, I got flat out rejected from MIT, but I'm cool with that b/c it's a stellar school and I gave it my best try. But I also got into Ross, Kenan-Flagler, Goizueta, Georgia Tech and am waiting on McCombs. Not quite sure where I'll end up yet (waiting on \$), but this is an infinitely better place to be in than last year! Now, I realize this isn't H/S/W, or Booth/Haas/Columbia/etc., But I took the time to visit each of the schools I applied to, sat in on classes, researched their websites, talked to professors/students/alumni, walked around town, and felt extremely happy with my list - I could honestly see myself being happy at each of them. I didn't apply to any schools that I had any reservations about, or wouldn't attend. And with the exception of MIT, I didn't apply to a plethora of schools I liked, but had hardly any chance to get into, which for me would be have been MIT, Stanford & Kellogg. Others may take a different philosophy on this, but it worked for me. And saved me a bit of money on increasingly crazy application fees! All this to say - it get's better! And this is just my personal experience with the process. Maybe for you, it's just a subtle tweak here and there and you can re-apply to the same schools and be golden the next time around. I'm a believer that everything happens for a reason and you'll discover that if your heart is still set on b-school, that this minor setback will make you a stronger person and a better candidate next time around. Good luck! Founder Affiliations: AS - Gold, HH-Diamond Joined: 04 Dec 2002 Posts: 14548 Location: United States (WA) GMAT 1: 750 Q49 V42 GPA: 3.5 Followers: 3785 Kudos [?]: 23622 [1] , given: 4554 ### Show Tags 27 Mar 2011, 16:08 1 KUDOS Expert's post _________________ Founder of GMAT Club US News Rankings progression - last 10 years in a snapshot - New! Just starting out with GMAT? Start here... Need GMAT Book Recommendations? Best GMAT Books Co-author of the GMAT Club tests GMAT Club Premium Membership - big benefits and savings Manager Joined: 19 Aug 2010 Posts: 62 Followers: 3 Kudos [?]: 8 [1] , given: 0 ### Show Tags 03 Jun 2011, 21:18 1 KUDOS Replying to this thread to update my status. I was able to get initial support to enroll in CCMBA at Fuqua from my immediate supervisor, but our senior management rejected the plan. So now, I will not be attending Fuqua or anywhere else. I'm really just providing this information as a resource to future applicants. In my case, what I've learned after a year of living on this forum and P&Q, I would be much happier if I had a real option that I could enroll in at this point. I don't have such an option, because: 1. I underestimated the competitiveness of the applicant pool to the top schools and overestimated my appeal. 2. I failed to consider the broader range of schools and plans for my life that might have made me happy. I wouldn't suggest that anyone drop a hundred or two hundred K on a degree from a crap school. But I've personally change my mind a lot about what I want/ed from an MBA program and how I could make that work for me (and my family). It's been an interesting year, but now I think I'm signing off. Manager Status: Berkeley Haas 2013 Joined: 23 Jul 2009 Posts: 191 Followers: 1 Kudos [?]: 38 [0], given: 16 ### Show Tags 27 Mar 2011, 08:53 This process is so long and so tough, that many people do lose hope after failing. I for one was starting to doubt my own capabilities for a while. But one needs to understand that after one stage in the application, luck plays some role.... From your post, you sound very positive and I am sure with minor tweaks to the essays and with a focused story, you will get into more than one school this time around. Good luck Manager Joined: 14 Nov 2010 Posts: 239 Followers: 8 Kudos [?]: 12 [0], given: 7 ### Show Tags 27 Mar 2011, 09:02 You have a non-traditional profile with solid stats and possibly a good story in you. Have you reached out to current MBA students at your target BSchools who were previously Peace Corps? They are normally a goldmine of intelligence on essays and selling yourself to adcoms. 3 of the most competitive BSchools worldwide, it was top or bust. Have you considered top 15? Manager Status: CBS Class of 2014 Joined: 27 Mar 2011 Posts: 53 Concentration: Strategy, Technology Schools: Columbia - Class of 2014 GMAT 1: 720 Q50 V38 GPA: 3.7 WE: Consulting (Energy and Utilities) Followers: 0 Kudos [?]: 11 [0], given: 6 ### Show Tags 27 Mar 2011, 09:11 I am on the same board as you are. Applied to 6 out of the 7 M7 schools and rejected at all without an interview. Time for some careful planning for 2012. Which school hurt most: HBS (dream school) and MIT What to do next: Try to evaluate myself and solidify my goals and make my ambitions specific and precise. Its a long process and kudos to everyone who got admitted. But I will have to work hard to be in all their shoes next year. Manager Joined: 19 Aug 2010 Posts: 62 Followers: 3 Kudos [?]: 8 [0], given: 0 ### Show Tags 27 Mar 2011, 16:38 Well, I'm not there yet, but I feel like I'm just inches away. Wharton: Ding no Int I actually loved this school, which is just so unfortunate. Kellogg: Ding w/Int I didn't like this school at all: not my interviewer, not the students, not the staff. So no loss here. Stanford: (Soon to be) Ding w/ no Int This place is clearly incredible, but obviously so out of my league. Still not really clear on why I applied. Fuqua: Waitlist Wasn't sure that this school was for me, and eventually got totally convinced that it was 100% the place for me. Then they stuck me in limbo. Haas: No Int No news yet, but in this case, that doesn't mean good news. So now what? Good question. Maybe quit my job. Maybe leave the country. Maybe both. I'd be thrilled to get off the waitlist at Fuqua, but that's certainly nothing I can count on. I'll certainly try to keep this ball in the air through the May R3 date, but after that I imagine that I'll either feel dinged or be literally dinged. Either way, I'm moving on. Manager Joined: 19 Aug 2010 Posts: 62 Followers: 3 Kudos [?]: 8 [0], given: 0 ### Show Tags 21 Apr 2011, 19:43 Just updating this post with my last ding, Haas. Nominally still waiting around for the 5/9 R3 Fuqua WL decisions, considering doing P/T via Cross Continent. In truth, Cross Continent maybe is the program I should have been looking at in the first place. But I'd still like to have to approached this process differently, had I known how little traction I was going to get with the schools I selected. Manager Status: Go Duke! Affiliations: ACFE Member, CalCPA Member Joined: 16 Jul 2010 Posts: 172 Location: Northern California Schools: Duke - Fuqua WE 1: CPA & CFE with 7 years of experience Followers: 6 Kudos [?]: 24 [0], given: 0 ### Show Tags 29 Apr 2011, 08:35 jsmith54 wrote: Just updating this post with my last ding, Haas. Nominally still waiting around for the 5/9 R3 Fuqua WL decisions, considering doing P/T via Cross Continent. In truth, Cross Continent maybe is the program I should have been looking at in the first place. But I'd still like to have to approached this process differently, had I known how little traction I was going to get with the schools I selected. Hi there, I read that you asked for some additional info on the CCMBA on another thread and thought I'd reply here. Have you looked into it any further? Have you talked with their admissions? For me, I was contacted very recently out of nowhere by their Assistant Director of Admissions and asked if I was interested in the Cross-Continent Program. I had also applied for the full time but wasn't accepted way back in round 1, so you can imagine my surprise many months later. She said that she had my application passed to her for consideration, which was nice to hear. Before this contact I had already been accepted into 3 other programs and just have to make my final decision. When I told her this she said that they would work with me to expidite the process, which they certainly have! Since just over a week ago when I was initially contacted I have exchanged emails not only with admissions, but I've talked with the same assistant director, financial aid, and they set up conversations with two current students. The two students were amazing to talk to and had very few negative things to say. One conversation went so well that he invited me to a mixer for current and incoming students with alumni attending as well. I have to say that the program has blown me away with how nice and professional everyone is. Also, the international component is just impressive to me. I've been to quite a few countries and the idea of studying and learning within a different culture is quite attractive to me. Let me know what you decide. Who knows, we may end up meeting at the China residency. Manager Joined: 19 Aug 2010 Posts: 62 Followers: 3 Kudos [?]: 8 [0], given: 0 ### Show Tags 01 May 2011, 19:29 I replied to VinceCPA via PM, but I thought I'd throw this general comment out to the thread. It's great to have an 11th hour option to be considering. I know some people will go back to the drawing board for the upcoming R1 apps at this point, but that wouldn't have worked for me. That said, U of Phoenix wouldn't have worked for me either. If I'm going to do what I want to do, I need a good MBA, and I need to start now. Not being totally in the zero admits group yet is really great for my remaining bits of sanity. Intern Joined: 08 Dec 2009 Posts: 43 Followers: 0 Kudos [?]: 8 [0], given: 2 ### Show Tags 03 May 2011, 09:16 Hey Guys in the same boat. Spent most of 2010 strategising and writing the essays. Had a pretty balanced set of schools too but alas 5 applications 2 Interviews and 1 waitlist. I am considering applying for 2012 session. European MBAs from Oxford, IESE, Bocconi may come into play now as they will certainly help me save one year. Lets all work together like the 2008 zero admits did and may be within 7 8 months we might be able to start a new thread on where the 2011 zero admits land up in 2012. Let the games begin! _________________ GMAT 720(94%) (Q 49(87%) V 39(87%)) Similar topics Replies Last post Similar Topics: 6 Admitted students...now what? 82 21 Dec 2007, 04:21 4 The 2010 Chat NOW thread! 85 04 Aug 2009, 08:36 45 2008 Zero admits - Where are they now ?? 32 14 Apr 2009, 06:48 82 The 2008 ZERO admits club 80 03 Apr 2008, 19:51 Applying for 2009 or 2010.. What to do now? 4 23 Sep 2007, 08:52 Display posts from previous: Sort by

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Re: Rebuilding polygon from CoefficientList? • To: mathgroup at smc.vnet.net • Subject: [mg20193] Re: [mg20164] Rebuilding polygon from CoefficientList? • From: BobHanlon at aol.com • Date: Tue, 5 Oct 1999 04:04:23 -0400 • Sender: owner-wri-mathgroup at wolfram.com ```Holger, reconstructPolynomial[coefMatrix_List, vars_List] := Module[{maxPwrs = Dimensions[coefMatrix] - 1, factors, k}, factors = Table[#[[1]]^k, {k, 0, #[[2]]}] & /@ Transpose[{vars, maxPwrs}]; Flatten[coefMatrix].Flatten[Outer[Times, Apply[Sequence, factors]]]]; poly = a*x^3*y + b*x^2*y^2*z + c*x*y^3*z^2 + d*z^2 + e*t*x*y*z; var = {x, y, z}; coef = CoefficientList[poly, var]; reconstructPolynomial[coef, var] == poly True var = {t, x, y, z}; coef = CoefficientList[poly, var]; reconstructPolynomial[coef, var] == poly True Bob Hanlon In a message dated 10/4/1999 3:10:53 AM, strauss at ika.ruhr-uni-bochum.de writes: >I have a mixed polynomial poly in several variables vars. > >cl = CoefficientList[poly, vars] > >gives a multi-dimensional matrix of coefficients. > >Can anyone help with an algorithm/expression that >re-constructs the original poly given cl and vars? >(In practice, I'd like to manipulate the coefficients before >reconstructing the polynomial; otherwise this wouldn't >make sense). >The algorithm must be able to handle any number of vars. >I've found a solutions for a small and fixed number of vars >using some ugly nested For loops. However, I suppose >that there must be a more efficient solution using some cute >matrix operations. > ``` • Prev by Date: Re: Greek alphabet • Next by Date: RE: Rebuilding polygon from CoefficientList? • Previous by thread: Re: Rebuilding polygon from CoefficientList? • Next by thread: RE: Rebuilding polygon from CoefficientList?

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Variables 1 / 21 # Variables - PowerPoint PPT Presentation Variables. Members: Fine 083-2 Mill 108-2 Pooppup 189-6. What is Variable?. A variable is any factor, trait or condition that you are testing in the experiment. It can be varied or changed according to the experimenter . There are many types of variables existed. Independent variable. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. ## PowerPoint Slideshow about 'Variables' - tessa Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - Presentation Transcript ### Variables Members: Fine 083-2 Mill 108-2 Pooppup 189-6 What is Variable? • A variable is any factor, trait or condition that you are testing in the experiment. It can be varied or changed according to the experimenter. • There are many types of variables existed. A variable that leads, influences or predicts the outcome of the study. It lies at the heart of any experimental design. EXAMPLE You are interested in how stress affects heart rate in humans. independent variable : the stress dependent variable : the heart rate. You can directly manipulate stress levels in your human subjects and measure how those stress levels change heart rate. A WORD OF CAUTION In identifying independent variables students often get confused with distinguishing between the independent variable and the levels of the independent variable. You may understand that there are three independent variables, when there is only one independent variable with three levels LEVELS OF INDEPENDENT VARIABLES A teacher is doing a study of Instructional Methods. One class was taught only with lecture and no visuals Another class was taught using the book and worksheets A third class was taught only using PowerPoint In this example there is only one independent variable – Instructional Methods; but there are three levels of that variable. Dependent Variable • A dependent variable is what you measure in the experiment and what is affected during the experiment. • It is called dependent because it "depends" on the independent variable. • In a scientific experiment, you cannot have a dependent variable without an independent variable Example: Dependent Variable • For example, if we want to measure therate of growth of the plants. • Therefore, the rate of growth of the plants is the dependent variable.(the thing that we are measuring) • Therefore, we can see that the Independent Variable is water for the plants. • Thedependentvariable isthe rate of plant growth. Mr. Wuu is driving at a constant rate of 65 miles per hour on the expressway. The function d=65h represents the no. of miles d he has traveled after h hours. Which quantity is the dependent variable, independent variable? Which is neither? 1.Miles traveled 2.Speed 3.65 miles 4.Hours traveled Extraneous variable Undesired (add error to the experiment) May influence the relation between the dependent and in dependent variable Often research studies do not find evidence to support the hypotheses because of unnoticed extraneous variables that influenced the results. Extraneous variable (Cont.) A major goal in research design is to decrease or control the influence of extraneous variables as much as possible. If, however, a variable cannot be controlled for, it becomes what is known as a confounding variable. Types of Extraneous Variable Participant Variables -related to individual characteristics of each subject that may impact how the subject responds -example: background differences, mood, anxiety, intelligence 2. Situational Variables -relate to the environment that may impact the individual subject - example: taking a test in a chilly room, the temperature would be considered an extraneous variable Example A researcher is doing research on the effectiveness of a computer mathematic lesson on long division compared to a typical mathematic lesson by measuring with the same standard test. The sample group the researcher choose was a class of fourth grade students form SatitKhonKaen. The researcher divided the class randomly into two equal groups. The first group used the computer lesson and the second group learnt with a typical lecturer for the exact same amount of time. Then took the exact same standard test. The sample group the researcher choose was a class of fourth grade students form SatitKhonKaen. The researcher divided the class randomly into two equal groups. -The first group used the computer lesson -The second group learnt with a typical lecturer for the exact same amount of time. Then took the exact same standard test. Example(cont.) Independent variable: the fourth grade class form SatitKhonKaen, the number of students in each group, the time of learning, the standard test, lesson on long division Dependent variable: The score on the standard test Extraneous Variable: Gender, age, each student’s background, mood, anxiety, intelligence, the enviornment of the room used for testing Reference

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Testing with models [From Rick Marken (960617.2045)] Me re: Peter Burke's experiment The Test for the Controlled variable is done by seeing how well the model fits the data. Jeff Vancouver 960617.15:30] Excuse me! What is the controlled variable? Fit? No. Of course not. The _hypothesized_ controlled variables are the variables controlled by the model. In Burke's model there were three controlled variables that I described in my post: 1) participation in the exchanges 2) the actual number of points received and 3) the time taken to accept the offer. A good example of using modelling to Test for controlled variables is described in Powers "A Feedback Model of Behavior" (Chapter 3 in LCS). Here Bill applied a control model to determine the variable controlled by rats in a shock avoidance experiment. Bill tested for the controlled variable by trying two different definitions of the variable controlled by the model: rate of shock and probability of shock. The model that fit best was the one controlling rate of shock -- suggesting that the variable controlled by the rats is rate rather than probability of shock. I suggest that you what Peter tried to do in his study, which was to determine what an organism is controlling by matching a control model to the organism's behavior. The best guess at the controlled variable is the perceptual function that let's the model perform most like the organism. Testing for controlled variables with a model is usually _ not_ the best way to determine what an organism is controlling; it was the only approach Bill could use with the rats because the data had already been collected. A far more efficient and accurate procedure would have been to do the Test for the controlled variable in the usual way; by introducing disturbances that should have a pronounced effect on a variable if it is _not_ controlled. This allows you to hone in on better and better definitions of the controlled variable. Once you have very precise data -- clear-cut control -- you can compare the data to a control model; since you already know the variable that should be controlled by the model, what you will learn from the model is something about the organism's gain and dynamics. Where is the word (or concept of) model in the Test? In the concept of a controlled _perceptual_ variable. The control model controls a representation (perception) of some state of affairs in the world; when you find evidence that some function of environmental events that is being protected from disturbance you have, in fact, tested a model that says organisms are controlling their perceptions. It seems that Burke a) guessed at some controlled variables, b) set up a situation that disturbed those variables, and c) measured the state of the variables from an external perspective. Peter didn't do the test this way, probably becuase there were several possible controlled variables involved. Also, it may have been difficult to calculate the expected value of the hypothesized controlled variables under the assumption of no control; so there was no frame of reference against which to judge whether the variables were controlled or not. The model actually provides a frame of reference for evaluating whether control is happening; in this case the model shows how the hypothetical controlled variables would behave if the WERE (rather than were not) unded control. The [ecological validity] criticism is not that the findings are worthless (note, I would suggest it be a discussion point), but that one be sure to fully appreciate the limitations in the data. OK. I'll accept that. Best Rick [from Jeff Vancouver 960620.12:35] [From Rick Marken (960617.2045)] Me re: Peter Burke's experiment >The Test for the Controlled variable is done by seeing how well the model >fits the data. Jeff Vancouver 960617.15:30] >Excuse me! What is the controlled variable? Fit? No. Of course not. Phew, my world almost colapsed there. A good example of using modelling to Test for controlled variables is described in Powers "A Feedback Model of Behavior" (Chapter 3 in LCS). Here Bill applied a control model to determine the variable controlled by rats in a shock avoidance experiment. Bill tested for the controlled variable by trying two different definitions of the variable controlled by the model: rate of shock and probability of shock. The model that fit best was the one controlling rate of shock -- suggesting that the variable controlled by the rats is rate rather than probability of shock. I suggest that you what Peter tried to do in his study, which was to determine what an organism is controlling by matching a control model to the organism's behavior. The best guess at the controlled variable is the perceptual function that let's the model perform most like the organism. Testing for controlled variables with a model is usually _ not_ the best way to determine what an organism is controlling; it was the only approach Bill could use with the rats because the data had already been collected. A far more efficient and accurate procedure would have been to do the Test for the controlled variable in the usual way; by introducing disturbances Now I understand. There is the TEST the "usual way" and than there are other ways. The other way that is demonstrated by Burke and Powers is that of comparing a model's behavior with subjects' behavior, where the model is a control system model. Fine, welcome to cognitive psychology. Only the nature of the model is not always a control system model in cognitive psychology. Perhaps we should not confuse the method with the content of the theory being tested. But I just want to make clear what has been established (so that Rick has something to argue against). Sometimes, when exigencies (like archival data) or ethics (like not being able to disturb certain variables) prevent us from using the TEST, we may then use the test. Given that Rick's version of the test (comparing results to a control theory model) assumes the model, neither the TEST nor the test can be used to test the theory. So, there must be more methods available as Rick wishes to test the theory. Stay tuned? ME: >The [ecological validity] criticism is not that the findings are >worthless (note, I would suggest it be a discussion point), but that one >be sure to fully appreciate the limitations in the data. OK. I'll accept that. You are teasing me - I can change Rick's mind? Later JEff

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# Integration of a PDE solution over a custom-defined subdomain Preamble: In my previous question I raised a problem of a custom-chosen integration of the solution of a PDE. In that case it was an integration along a line lying within the PDE domain. The solution given there is correct, but oversimplified, since I have oversimplified my original question. The problem, however, seems to me of general interest, since (1) it seems to push the limits of the present Mma capability and (2) it will be of a general use. Let me first formulate the question in a general form, and then go down to examples with the corresponding codes. The question formulation in a general form: Provided one has a PDE defined in a complex (possibly multiply connected) domain, let us understand how one can integrate the result over a subdomain (that may or may not include holes). Discussion of the question: Indeed, if one finds the PDE solution, it is not to admire it. One often needs to operate on it. Operation often means integration, evaluating derivatives, search for extrema etc. Even more, often (at least in my case) such an operation is necessary within a subdomain of the domain, or along a line lying within the domain, or so. Furthermore, all these operations can be worked around, if the domain is simple. A) If one only needs to integrate the solution of the whole multiply connected domain, he can comfortably use the construct Integrate[expression[x, y], Element[{x, y}, mesh]] that takes care not to integrate over the holes. I assume, for instance, that the PDE in question is defined in a domain within the (x, y) plane. B) If the domain is multiply connected, but simple enough, it may be divided into marked subdomains from the very beginning (see here: "Quad Meshes" and "Element Meshes with Subregions" and few others), and functions may be introduced that are nonzero only in the necessary subdomains and equal to the PDE solution there. This requires tedious work on the level of the mesh building, but might be a workaround. C) If, however, the domain is complex, the task to build the mesh with the subdomains may become close to impossible. The example I give below. So my question is related to the following problem, whether or not there are any built-in mechanisms enabling one to do something like Integrate[expression[x, y], Element[{x, y}, meshPart]], but to integrate over a part of the mesh, rather than over the whole, such that the possible holes are automatically taken care of. Or, alternatively, if there is a workaround for such a task. This, of course, can be viewed in a more general perspective, not only in regard to integration. My focus on it is simply related to my personal tasks. I realize, of course, that such a mechanism/workaround may simply not exist. Example: This is a trapezoid: coord = {{0, 0}, {1/3, 1}, {2/3, 1}, {1, 0}}; trap = Polygon[coord]; This gives a list of $n$ disks with the radius $r$, such that they do not intersect the top and the bottom of the trapezoid: distr[r_] := ProbabilityDistribution[(1 + y)/NIntegrate[(1 + y), {y, 2 r, 1 - 2 r}], {y, 2 r, 1 - 2 r}] lst[r_, n_Integer] := Table[{RandomReal[{#/3, 1 - #/3}], #} &[RandomVariate[distr[r]]], {n}] They are distributed linearly along y, but this is not important for our further discussion. This gives the trapezoid with $n$ holes of the radius $r$: fd[r_, n_Integer] := Fold[RegionDifference, trap, Disk[#, r] & /@ lst[r, n]]; This makes the mesh with $n$ holes of the radius $r$: Needs["NDSolveFEM"]; mesh[r_, n_Integer] := ToElementMesh[fd[r, n], "MaxCellMeasure" -> r/20, "MeshQualityGoal" -> Automatic, "ImproveBoundaryPosition" -> False, "MeshOrder" -> 1]; Like the mesh m1, for example: m1 = mesh[0.05, 10] m1["Wireframe"] This solves the equation (for electric potential u, the Laplace equation with u equal to $1$ at the top and $0$ at the bottom): nds1 = NDSolveValue[{Inactive[Div][Inactive[Grad][u[x, y], {x, y}], {x, y}] == 0, DirichletCondition[u[x, y] == 0, y == 0], DirichletCondition[u[x, y] == 1, y == 1]}, u[x, y], {x, y} \[Element] m1]; and this shows the solution for the potential: Show[{ ContourPlot[Evaluate[nds1], {x, y} \[Element] m1, PlotRange -> {0, 1}, ColorFunction -> "TemperatureMap", PlotLegends -> Automatic, ImageSize -> 300], m1["Wireframe"] }] Now we come to the point: look at the two integrals below. The first integrates the solution over the mesh, the second over the whole trapezoid: NIntegrate[Evaluate[nds1], {x, y} \[Element] m1] NIntegrate[Evaluate[nds1], {y, 0, 1}, {x, y/3, 1 - y/3}, Method -> "LocalAdaptive"] (* 0.197248 0.222601 *) The first is smaller, since it takes care not to integrate over the holes, while the second does the integration there as well. Indeed, if we add the area of 10 holes to the first integral, we get: 0.197248 + Pi * 0.05^2 * 10 (* 0.275788 *) which is pretty close to the value of the second one (this works, since u varies within the domain rather slowly). It is quite clear that to do this correctly we should have explicitly excluded the holes from integration in the second integral. This is, however, difficult to automate, if the holes are generated randomly. Nevertheless, the task of the integration over the whole domain is feasible. It is solved by the former approach: NIntegrate[Evaluate[nds1], {x, y} \[Element] m1]. Let us now divide the trapezoid into several parts by horizontal lines, and integrate u[x, y] within each part, such that the holes are excluded from the integration. ?? Reverting to the old strategy of using Boole seems efficient on the test case: SeedRandom[0]; (* to give a reproducible result *) m1 = mesh[0.05, 10]; m1["Wireframe"] Clear[x, y, u]; nds1 = NDSolveValue[{Inactive[Div][Inactive[Grad][u[x, y], {x, y}], {x, y}] == 0, DirichletCondition[u[x, y] == 0, y == 0], DirichletCondition[u[x, y] == 1, y == 1]}, u[x, y], {x, y} ∈ m1]; Table[ (* slices of height 0.2 *) With[{x1 = x1}, NIntegrate[nds1*Boole[x1 <= x <= x1 + 0.2], {x, y} ∈ m1]], {x1, 0, 0.8, 0.2}] Total@% (* checks *) NIntegrate[nds1, {x, y} ∈ m1] (* {0.00496725, 0.0451698, 0.0640033, 0.0425635, 0.00529073} 0.161995 0.161995 *) • Could you please kindly comment, what does With[{x1 = x1},... construct do? Aug 10, 2015 at 9:06 • @AlexeiBoulbitch It replaces every occurrence of x1 in the body with the value of x1. Since NIntegrate is HoldAll, the x1 would not automatically evaluate; I forget if I tried it without the With; it's possible (maybe likely) that NIntegrate evaluates the integrand before Table is done. See tutorial/IntroductionToDynamic#2125133640. Aug 10, 2015 at 10:01 One way to do it is to specify an ExtralopationHandler (see section on extrapolation) and have it return 0. for queries outside the domain. For example: nds1 = NDSolveValue[{Inactive[Div][ Inactive[Grad][u[x, y], {x, y}], {x, y}] == 0, DirichletCondition[u[x, y] == 0, y == 0], DirichletCondition[u[x, y] == 1, y == 1]}, u[x, y], {x, y} \[Element] m1, "ExtrapolationHandler" -> {(0. &), "WarningMessage" -> False}]; NIntegrate[Evaluate[nds1], {x, y} \[Element] m1] NIntegrate[Evaluate[nds1], {y, 0, 1}, {x, y/3, 1 - y/3}, What this does is that for every query point that is outside the domain, 0. is returned in stead of the extrapolated value, which is the default. It's a bit unfortunate that NIntegrate rejects Indeterminate like Overflow or Infinity. I would found "ExtrapolationHandler" -> {(Indeterminate &),... cleaner but the behavior of NIntegrate is not going to change; it's been that way too long. The default behavior of Interpolation is to return an InterpolatingFunction that does extrapolation if queried outside of it's domain. For many interpolation scenarios this is a reasonable choice. For the FEM, however, this will give almost certainly a wrong solution outside the domain. The reason is we do not know anything about the region beyond it's boundary conditions. So the "ExtrapolationHandler" was introduced to allow to change the behavior of how InterpolatingFunction behave when queried outside the domain. My feeling is that Indeterminate is a good choice since it reflects the state of affairs: it is indeterminate what value should be given on extrapolation. Numerical 0. is a good choice in this specific case because 0 is invariant for integration - it does not alter the result. But arguing from a engineering perspective the value outside the region is not 0. We do not know what it is. The first argument to the ExtrapolationHandler is a function that returns the value to be returned by the InterpolatingFunction in cases of extrapolation and it can be (if I recall correctly) any expression. The "WarningMessage" -> ... option takes True or False and controls if a message is issued on extrapolation or not.

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## Introduction: Mindo Activity Scene - Tubbing- Instructable Hi! My name is Paula Sanchez and I am in tenth grade, in the subject of Technology, our teacher assigned us a very cool task, which was to do a project of making a design in Tinkercad representing something significant from somewhere in Ecuador. I chose Mindo, a beautiful and touristy place, there is a great variety of flora and fauna. There are many activities to do and one of those is tubbing. This activity is very entertaining since you go on top of some balls by a river that goes very fast and full of rocks. You go with a guide who handles the balls and you soak up everything, it is very entertaining and that's why I chose this activity and designed it. Here I show you step by step how I did it. ## Step 1: Choose the Shape for the Balls Go to the right where the shapes are and choose 'toroide' ## Step 2: Duplicate and Align When you have the shape, you have to duplicate it, by selecting it and choosing a button on the top left, do that 6 times and then align them to form the shape of all of them together. ## Step 3: Coloring the Balls Then you need to color all the balls, the color is black, for this you select the object, click where it says solid, and choose the color black. Do the same with the other ones. ## Step 4: Adding the Rocks Then you go to the right, you click where it says 'generadores de formas' and choose 'todos'. Then you go down and you will see some numbers, like pages, go to the page 7 and select the 'Hi-res Parabolo..' shape. Duplicate those shapes by 8 and change their shape with the arrows around the shape so they are different shapes. ## Step 5: Coloring the Rocks Then, as you did with the balls, color the rocks but with the color gray and you place them around the balls and.... you are done!!!!

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# Thread: Problem using Round - any ideas 1. ## Problem using Round - any ideas Hi I'm having trouble with using round to round a vat calculation. I have done separated out the lines that I believe show the problem. I can't work out why the first sprintf produces an incorrect rounded value, yet the second sprintf produces the required results. Driving me mad - can anyone put my mind at rest and stop my hair from turning grey? I'm sure it's something simple. The code is: <?php \$number = 691.30; \$withvat = \$number * 1.15; \$total = \$number * 1.15; \$vat = \$total - \$number; ?> <?=sprintf("%.2F", round(\$vat, 2));?><br> <?=sprintf("%.2F", round(103.695, 2));?> 2. Originally Posted by mambo-no-6 Hi I'm having trouble with using round to round a vat calculation. I have done separated out the lines that I believe show the problem. I can't work out why the first sprintf produces an incorrect rounded value, yet the second sprintf produces the required results. Driving me mad - can anyone put my mind at rest and stop my hair from turning grey? I'm sure it's something simple. The code is: <?php \$number = 691.30; \$withvat = \$number * 1.15; \$total = \$number * 1.15; \$vat = \$total - \$number; ?> <?=sprintf("%.2F", round(\$vat, 2));?><br> <?=sprintf("%.2F", round(103.695, 2));?> http://en.php.net/round check out the manual. More people have encountered this problem. As a result some of them wrote their own function. Maybe you could look into that. 3. It's always better to do math with money using integer cents/pence values rather than float dollars/pounds But why round to 2dp when you're using sprintf to format to 2dp anyway 4. By the way, you would be better off using: PHP Code: ``` <?php\$number = 691.3;\$vat = \$number * 0.15;\$total = \$number + \$vat;?>....<?php printf('£%.2f', \$vat); ?> ``` 5. Keep in mind, printf() truncates PHP Code: ``` printf('%.2f', 1.555); // 1.55 echo round(1.555, 2); // 1.6  ```

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The Physics Hypertextbook Opus in profectus # Orbital Mechanics I ## Problems ### practice 1. Geosynchronous/geostationary Earth orbiting satellites There is a special class of satellites that orbit the Earth above the equator with a period of one day. 1. How will such a satellite appear to move when viewed from the surface of the Earth? 2. What type of satellites use this orbit and why is it important for them to be located in this orbit? (Keep in mind that this is a relatively high orbit. Satellites not occupying this band are normally kept in much lower orbits.) 3. Determine the orbital radius at which the period of a satellite's orbit will equal one day. State your answer in… 1. kilometers 2. multiples of the Earth's radius 3. fractions of the moon's orbital radius 2. Locate the L1, L2, and L3 Lagrange points for the Earth-Sun system using dynamical principles. State your answers as distances… 1. from the Sun and Earth in meters 2. from the Earth as multiples of the Moon's orbital radius 3. from the Sun as multiples of the Earth's orbital radius 3. TRAPPIST-1 is a planetary system 39 light years from our solar system in the constellation of Aquarius. The star at the center is much less massive than the Sun and only slightly larger than Jupiter. The table below provides the period and semimajor axis for six of the planets in this system that were reasonably well known in 2016. Convert these values to their SI equivalents and use them to determine the mass of the TRAPPIST-1 star. (These data are also available as a tab delimited text file.) Six exoplanets of the star TRAPPIST-1 Source: Gillon, Triaud, Demory, et al; 2017. planet period (days) semimajor axis (au) b 01.51087081 0.01111 c 02.42182330 0.01521 d 04.04961000 0.02144 e 06.09961500 0.02817 f 09.20669000 0.03710 g 12.35294000 0.04510 4. Some sort of binary star problem would be good here. Write it! ### conceptual 1. Two related questions 1. If objects in Earth orbit are weightless, why can't astronauts throw objects like baseballs or screwdrivers into outer space? Since they're weightless, it should be possible to heave them to the moon, planets, or distant stars. What's wrong with this thinking? 2. Why would casually discarding junk overboard from a space station or space shuttle be a bad idea? 2. One way to send a spaceship to the planet Mars would be to point it in the general direction of the Red Planet, ignite the rocket engines, and let it go. This method won't work, however. Give two reasons why this procedure would never result in a successful mission, no matter how precisely the spacecraft was aimed. 3. trajectories-satellite.pdf The accompanying pdf file shows a satellite in a circular orbit about the Earth. Sketch the new path that the satellite would take if its speed were changed abruptly in the ways described. 4. Spacecraft in extreme near Earth orbit are subject to small but (in the long run) non-negligible amounts of aerodynamic drag from the upper regions of the Earth's atmosphere. 1. What happens to the altitude and speed of such a satellite over time? 2. Sketch the path of a satellite in such an orbit. 5. Pluto was discovered in 1930, but it's mass wasn't known with any accuracy until 1978 when Pluto's moon Charon was discovered. What was it about Charon's discovery that enabled astronomers to finally determine the mass of Pluto? ### numerical 1. Satellite motion 1. Calculate the speed needed for the space shuttle to travel around the Earth in a circular orbit at an altitude of 350 km above the Earth's surface. 2. Calculate the period of the space shuttle at this same orbit. ### statistical 1. The table below gives the orbital period in days and orbital radius in millions of meters for Jupiter's four largest satellites named the Galilean moons in honor of their discoverer, the Italian scientist Galileo Galilei (1564–1642). Use this data to determine the mass of Jupiter. The Galilean moons of Jupiter moon period (days) distance (106 m) Io 1.769137786 422 Europa 3.551181041 671 Ganymede 7.154552960 1070 Callisto 16.68901840 1883

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 Sample Problems > Usage > 3D_domains > 3d_lenses # 3d_lenses Navigation:  Sample Problems > Usage > 3D_domains > # 3d_lenses { 3D_LENSES.PDE This problem considers the flow of heat in a lens-shaped body of square outline.  It demonstrates the use of FlexPDE in problems with non-planar extrusion surfaces. Layer 1 consists of a flat bottom with a paraboloidal top. Layer 2 is a paraboloidal sheet of uniform thickness. Plots on various cut planes show the ability of FlexPDE to detect intersection surfaces. }     title '3D Test - Lenses'     coordinates       cartesian3     Variables       u     definitions       k = 0.1       heat = 4     equations      U: div(K*grad(u)) + heat   = 0     extrusion     surface z = 0     surface z = 0.8-0.3*(x^2+y^2)     surface z = 1.0-0.3*(x^2+y^2) boundaries { implicit natural(u) = 0 on top and bottom faces } Region 1 layer 2 k = 1   { layer specializations must follow regional defaults } start(-1,-1) value(u) = 0     { Fixed value on sides } line to (1,-1) to (1,1) to (-1,1) to close select painted plots contour(u) on x=0.51       as "YZ plane" contour(u) on y=0.51       as "XZ plane" contour(u) on z=0.51       as "XY plane cuts both layers and part of outline" contour(u) on z=0.75       as "XY plane cuts both layers, but not the outline" contour(u) on z=0.8       as "XY plane cuts only layer 2" contour(u) on z=0.95       as "XY plane cuts small patch of layer 2" contour(u) on z=0.95 zoom as "small cut patch, zoomed to fill frame" contour(u) on surface 1   as "on bottom surface" contour(u) on surface 2   as "on paraboloidal layer interface" contour(u) on x=y         as "oblique plot plane" contour(u) on x+y=0       as "another oblique plot plane" end

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Number of terms in Balanced boolean functions? I know that balanced boolean functions are required for creating S-boxes. I want to create four balanced Boolean functions for 4X4 Sbox and as far as I’ve understood, the maximum order of the balanced Boolean function should be <= (number of var)/2. But… how many terms should there be in my boolean functions? • is the answer satisfactory? Nov 2, 2016 at 22:28 You can represent such an Sbox as a mapping from $GF(2^4)$ to itself as well as a univariate polynomial mapping from $GF(2)^4$ to itself and what's sparse or simple in one representation can be complex in the other.

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1. ## Distrubition help needed! I'm sorry for asking a maths question again. XD Q8. The smallest 84% of stainless steel nails are less than 49.7mm long, and the largest 97.5% are longer than 41.6mm long. a. Caculate the likely mean length of the nails and an appropriate value for the standard deviation. What do I do? >.< 2. Originally Posted by Hikaru I'm sorry for asking a maths question again. XD Q8. The smallest 84% of stainless steel nails are less than 49.7mm long, and the largest 97.5% are longer than 41.6mm long. a. Caculate the likely mean length of the nails and an appropriate value for the standard deviation. What do I do? >.< This is a normal distribution. We can say that the chances that a nail is shorter than 41.6mm is 2.5% And the chances that a nail is longer than 49.7mm is 16% $P(X < 41.6) = 0.025$ $P \left( \frac{X - \mu}{\sigma} < \frac{41.6 - \mu}{\sigma} \right) = 0.025$ $P \left( Z < \frac{41.6 - \mu}{\sigma} \right) = 0.025$ $\Phi \left( \frac{41.6 - \mu}{\sigma} \right) = 0.025$ $\frac{41.6 - \mu}{\sigma} = -1.96$ ===== $P(X > 49.7) = 0.16$ $1 - P(X \leq 49.7) = 0.16$ $P \left( \frac{X - \mu}{\sigma} \leq \frac{49.7 - \mu}{\sigma} \right) =$ $P \left( Z \leq \frac{49.7 - \mu}{\sigma} \right) = 0.84$ $\Phi \left( \frac{49.7 - \mu}{\sigma} \right) = 0.84$ $\frac{49.7 - \mu}{\sigma} = 0.996$ ==== From here on it's simply a matter of setting those two equations equal to each other and solving. 3. Originally Posted by Hikaru I'm sorry for asking a maths question again. XD Q8. The smallest 84% of stainless steel nails are less than 49.7mm long, and the largest 97.5% are longer than 41.6mm long. a. Caculate the likely mean length of the nails and an appropriate value for the standard deviation. What do I do? >.< Let X be the random variable length of nail and assume X ~ Normal( $\mu, ~ \sigma)$. $\Pr(X < 49.7) = 0.84 \Rightarrow \Pr\left( Z < \frac{49.7-\mu}{\sigma}\right) = 0.84$ But $\Pr(Z < a) = 0.84 \Rightarrow a \approx 0.99446$. Therefore: $\frac{49.7-\mu}{\sigma} = 0.99446$ .... (1) $\Pr(X > 41.6) = 0.975 \Rightarrow \Pr\left( Z > \frac{41.6-\mu}{\sigma}\right) = 0.975$ But $\Pr(Z > a) = 0.975 \Rightarrow a \approx -1.95996$. Therefore: $\frac{41.6 -\mu}{\sigma} = -1.95996$ .... (2) Now solve equations (1) and (2) simultaneously. For checking purposes: I get $\mu = 46.97$, correct to two decimal places. 4. Originally Posted by mr fantastic Let X be the random variable length of nail and assume X ~ Normal( $\mu, ~ \sigma)$. $\Pr(X < 49.7) = 0.84 \Rightarrow \Pr\left( Z < \frac{49.7-\mu}{\sigma}\right) = 0.84$ But $\Pr(Z < a) = 0.84 \Rightarrow a \approx 0.99446$. Therefore: $\frac{49.7-\mu}{\sigma} = 0.99446$ .... (1) $\Pr(X > 41.6) = 0.975 \Rightarrow \Pr\left( Z > \frac{41.6-\mu}{\sigma}\right) = 0.975$ But $\Pr(Z > a) = 0.975 \Rightarrow a \approx -1.95996$. Therefore: $\frac{41.6 -\mu}{\sigma} = -1.95996$ .... (2) Now solve equations (1) and (2) simultaneously. Mr F, your distribution table seems to be more accurate than mine... 5. Originally Posted by janvdl This is a normal distribution. We can say that the chances that a nail is shorter than 41.6mm is 2.5% And the chances that a nail is longer than 49.7mm is 16% $P(X < 41.6) = 0.025$ $P \left( \frac{X - \mu}{\sigma} < \frac{41.6 - \mu}{\sigma} \right) = 0.025$ $P \left( Z < \frac{41.6 - \mu}{\sigma} \right) = 0.025$ $\Phi \left( \frac{41.6 - \mu}{\sigma} \right) = 0.025$ $\frac{41.6 - \mu}{\sigma} = -1.96$ ===== $P(X > 49.7) = 0.16$ $1 - P(X \leq 49.7) = 0.16$ $P \left( \frac{X - \mu}{\sigma} \leq \frac{49.7 - \mu}{\sigma} \right) =$ $P \left( Z \leq \frac{49.7 - \mu}{\sigma} \right) = 0.84$ $\Phi \left( \frac{49.7 - \mu}{\sigma} \right) = 0.84$ $\frac{49.7 - \mu}{\sigma} = 0.996$ ==== From here on it's simply a matter of setting those two equations equal to each other and solving. That's a first. Getting beaten to the punch on a statistics question 6. Originally Posted by janvdl Mr F, your distribution table seems to be more accurate than mine... And my TI-89 is even more accurate ..... 7. Originally Posted by mr fantastic And my TI-89 is even more accurate ..... Ah okay I get it My calculator isn't capable of doing that... 8. Originally Posted by janvdl Ah okay I get it My calculator isn't capable of doing that... There are many on-line calculators that can do it for you. For example: z score Calculator

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### Building Tetrahedra Can you make a tetrahedron whose faces all have the same perimeter? A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground? ### Areas and Ratios What is the area of the quadrilateral APOQ? Working on the building blocks will give you some insights that may help you to work it out. ##### Stage: 4 Short Challenge Level: See all short problems arranged by curriculum topic in the short problems collection The diagram shows a shaded shape bounded by circular arcs with the same radius. The centres of three arcs are the vertices of an equilateral triangle; the other three centres are the midpoints of the sides of the triangle. The sides of the triangle have length 2. What is the difference between the area of the shaded shape and the area of the triangle? If you liked this problem, here is an NRICH task that challenges you to use similar mathematical ideas. This problem is taken from the UKMT Mathematical Challenges.

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View more editions Engineering Mechanics Statics # TEXTBOOK SOLUTIONS FOR Engineering Mechanics Statics 12th Edition • 1392 step-by-step solutions • Solved by publishers, professors & experts • iOS, Android, & web Over 90% of students who use Chegg Study report better grades. May 2015 Chegg Study Survey PROBLEM Chapter: Problem: SAMPLE SOLUTION Chapter: Problem: • Step 1 of 7 Draw the free body diagram of the concrete panel and pulley. ` • Step 2 of 7 Consider the entire system is in equilibrium. Find the force exerted on the pulley from the vertical equilibrium equations: …… (1) Here, force exerted on the pulley is , and the weight of the concrete panel is w. • Step 3 of 7 Consider free body diagram of joint A, • Step 4 of 7 Equate the sum of forces in direction. Equate the sum of forces indirection. Substitute for and for. …… (2) • Step 5 of 7 Consider the triangle ABC Take length of the cable’s AB and AC as ‘L’ and the distance between OB and OC as‘d’ • Step 6 of 7 Calculate the angle made at B and C from the triangle ABC. …… (3) Substitute for in equation (2). …… (4) • Step 7 of 7 The equation (4) tells that, longer the cables the less the force in each cable. To illustrate that consider: Trail – 1: Take, and fix and substitute in equation (4). Trail – 2: Take, and fix and substitute in equation (4). Hence, it is shown that longer the cables the less the force in each cable. Corresponding Textbook Engineering Mechanics Statics | 12th Edition 9780136077909ISBN-13: 0136077900ISBN: R. C. HibbelerAuthors:

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Last visit was: 11 Sep 2024, 23:26 It is currently 11 Sep 2024, 23:26 Toolkit 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 Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How SORT BY: Tags: Show Tags Hide Tags Manager Joined: 13 Mar 2009 Posts: 116 Own Kudos [?]: 1681 [482] Given Kudos: 60 GMAT 2: 680  Q48  V35 GMAT 3: 710  Q47  V41 Math Expert Joined: 02 Sep 2009 Posts: 95451 Own Kudos [?]: 657790 [78] Given Kudos: 87242 Senior Manager Joined: 31 Aug 2009 Posts: 288 Own Kudos [?]: 1057 [59] Given Kudos: 20 Location: Sydney, Australia Q49  V41 General Discussion Manager Joined: 11 Aug 2008 Posts: 78 Own Kudos [?]: 98 [14] Given Kudos: 8 Re: Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How [#permalink] 14 Kudos From your ans I think the stem means 2/3 of 60 animals are both pigs and cows, not either pigs or cows. I think there is some problem with stem here? Math Expert Joined: 02 Sep 2009 Posts: 95451 Own Kudos [?]: 657790 [6] Given Kudos: 87242 Re: Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How [#permalink] 4 Kudos 2 Bookmarks ngoctraiden1905 From your ans I think the stem means 2/3 of 60 animals are both pigs and cows, not either pigs or cows. I think there is some problem with stem here? How it's possible for an animal to be BOTH a pig and a cow? 2/3 of 60 animals are either pigs or cows. So there are total of 40 cows and pigs. Yansta8's solution is correct. C it is. Intern Joined: 23 Feb 2010 Posts: 10 Own Kudos [?]: 22 [17] Given Kudos: 0 Q49  V50 Re: Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How [#permalink] 15 Kudos 1 Bookmarks The key word in statement 1 is "more" than twice as many. If "more" wasn't there, you would be right. From the stem we know that #pigs + #cows = 40 Statement 1: If you had 1 pig and 39 cows, you have more than twice as many cows as pigs (a lot more than twice the number actually). If you had 2 pigs and 38 cows, the statement still holds true. Therefore, insufficient by itself. Statement 2: If you had 13 pigs and 27 cows, the statement holds true. If you had 14 pigs and 26 cows, the statement still holds true. Therefore, insufficient by itself. Together: Start with the first posible option according to statement 2: 13 pigs and 27 cows works for both statements. The next option (14 pigs and 26 cows) violates statement 1 because you don't have more than twice as many cows as pigs. Every option after this will have the same problem. Therefore you only have one option - sufficient to determine the answer. C. Tutor Joined: 16 Oct 2010 Posts: 15297 Own Kudos [?]: 67986 [6] Given Kudos: 442 Location: Pune, India Re: Pigs and cows [#permalink] 3 Kudos 3 Bookmarks enigma123 Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How many of the animals are cows? (1) The farm has more than twice as many cows at it has pigs. (2) The farm has more than 12 pigs. The wording of the question is a little ambiguous but this is what I infer from it: Of the 60 animals, 2/3 i.e. 40 are either pigs or cows. So the total number of pigs and cows on the farm = 40. Stmnt 1: The farm has more than twice as many cows at it has pigs. Of the 40, number of cows (c) is more than twice of number of pigs (p) Many possible solutions: c = 30, p = 10 c = 31, p = 9 etc Stmnt 2: The farm has more than 12 pigs. Many possible solutions: c = 27, p = 13 c = 26, p = 14 etc Both together: p must be at least 13 so c can be at most 27 which satisfies stmnt 1. If p = 14, c must be 26 but this doesn't satisfy stmnt 1. As p increases, c decreases and c will never be more than twice of p. So the only possible value is p = 13, c = 27 Director Joined: 21 Sep 2012 Status:Final Lap Up!!! Affiliations: NYK Line Posts: 726 Own Kudos [?]: 1920 [0] Given Kudos: 70 Location: India GMAT 1: 410 Q35 V11 GMAT 2: 530 Q44 V20 GMAT 3: 630 Q45 V31 GPA: 3.84 WE:Engineering (Transportation) Re: Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How [#permalink] HI all I am facing little difficulty in interpretation of the question, My specifc doubts are:- 1. Here either or means C+p = 40 , since a pig cannot be a cow at the same time. What if the question had a scenario, where a case of both was possible.....should we consider c+p- both = 40 in that case. I am haveing a doubt with either or statement 2. In case of question stating "What was the number of cows", what should we infer that it requires number of only cow ie cow - both or only cow + both, i am facing difficulty...... 3."I. The farm has more than twice as many cows as it has pigs " Can we interpret this as 'For every pig there were more than twice cow" ie c/p>2/1 Pls help me in clearing my doubts... Regards Archit Tutor Joined: 16 Oct 2010 Posts: 15297 Own Kudos [?]: 67986 [4] Given Kudos: 442 Location: Pune, India Re: Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How [#permalink] 3 Kudos 1 Bookmarks Archit143 HI all I am facing little difficulty in interpretation of the question, My specifc doubts are:- 1. Here either or means C+p = 40 , since a pig cannot be a cow at the same time. What if the question had a scenario, where a case of both was possible.....should we consider c+p- both = 40 in that case. I am haveing a doubt with either or statement 2. In case of question stating "What was the number of cows", what should we infer that it requires number of only cow ie cow - both or only cow + both, i am facing difficulty...... 3."I. The farm has more than twice as many cows as it has pigs " Can we interpret this as 'For every pig there were more than twice cow" ie c/p>2/1 Pls help me in clearing my doubts... Regards Archit Responding to a pm: Either A or B means either A or B or both. If they mean to say that both should not be included then they will say 'Either A or B but not both' What is the number of A? implies all A (including those who can be B too) What is the number of only A ? implies those A who are B too are not to be counted. 3."I. The farm has more than twice as many cows as it has pigs " Can we interpret this as 'For every pig there were more than twice cow" ie c/p>2/1 Yes, that's correct. You can write it as c/p > 2 or as c > 2p (same thing) Manager Joined: 10 Mar 2013 Posts: 135 Own Kudos [?]: 500 [2] Given Kudos: 2412 GMAT 1: 620 Q44 V31 GMAT 2: 610 Q47 V28 GMAT 3: 700 Q49 V36 GMAT 4: 690 Q48 V35 GMAT 5: 750 Q49 V42 GMAT 6: 730 Q50 V39 GPA: 3 Re: Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How [#permalink] 1 Kudos 1 Bookmarks Could we add across inequalities as I did below? If not, when is it allowed to add across inequalities? (1) c > 2p NS (2) p > 12 NS (1) + (2) c + p > 2p + 12 40 > 2p + 12 2p < 28 p < 14 Because (2) gives p > 12, with the above statement, p = 13 and we can solve for c. S Tutor Joined: 16 Oct 2010 Posts: 15297 Own Kudos [?]: 67986 [4] Given Kudos: 442 Location: Pune, India Re: Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How [#permalink] 2 Kudos 2 Bookmarks TooLong150 Could we add across inequalities as I did below? If not, when is it allowed to add across inequalities? (1) c > 2p NS (2) p > 12 NS (1) + (2) c + p > 2p + 12 40 > 2p + 12 2p < 28 p < 14 Because (2) gives p > 12, with the above statement, p = 13 and we can solve for c. S You can add inequalities as long as both the inequalities have the same sign. a > b c > d gives a+c > b + d Think logically: a is greater than b and c is greater than d so a+c will be greater than b+d because you are adding the larger numbers together. So what you have done above is correct. If the inequality signs are different, you cannot add them. Also, you can subtract inequalities when they have opposite signs but prefer not to do that to avoid confusion. Just flip the sign of one inequality by multiplying it by -1 and then add them up. Manager Joined: 10 Mar 2014 Posts: 135 Own Kudos [?]: 685 [0] Given Kudos: 13 Re: Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How [#permalink] Bunuel Syed yangsta8 Question stem says that 2/3 of 60 are pigs or cows. That means 40 animals are pigs or cows. So all we need to have sufficiency is either number of pigs or number of cows. 1) Farm has more than 2 cows for 1 pig (at least 26.6 animals of the 40 are cows). This just tells us that the farm has at least 27 pigs and that the max number of cows is 13. For example the farm could have 30 pigs and 10 cows. Not a definitive number. Insufficient. 2) Farm has more than 12 pigs. Again not enough info. Insuff 1+2) Statements together tell us: 12 < number of pigs is <=13 Which means number pigs = 13 Number of cows = 27. I do agree with 'yangsta8' that the answer is 'C' but (if I am NOT wrong), Stmnt#1 says the number of Cows are more than twice than the pigs. Thus, the cows could be 27 and Pigs could be 13; and For example, the farm could have 30 cows and 10 pigs. ------------------------------- Hi Bunuel and Yangsta8 - Please correct me if I am wrong (which is very much possible)! c+p=40 (1) c>2p --> min # of cows is 27 and max # pigs is 13, so there can be any combination not violating this and totaling 40. Not sufficient (2) p>12 Not sufficient (1)+(2) p>12 but max of p is 13, hence p=13 --> c=27 You are right there can be 27 cows (min) and 13 pigs (max) or 30 cows and 10 pigs. Think there was simple typo from yangsta8. Hi bunnel, I need clarification here 2/3 are either pigs or cows i am understanding (2/3)*60 = 40 (cows or pigs) how it can be c+p = 40 question is saying either cow or pig so this can be c or p but not C+P Thanks. Tutor Joined: 16 Oct 2010 Posts: 15297 Own Kudos [?]: 67986 [3] Given Kudos: 442 Location: Pune, India Re: Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How [#permalink] 2 Kudos 1 Bookmarks PathFinder007 2/3 are either pigs or cows i am understanding (2/3)*60 = 40 (cows or pigs) how it can be c+p = 40 question is saying either cow or pig so this can be c or p but not C+P Thanks. If I may add, 40 are either pigs or cows does not mean that either all 40 are pigs or all 40 are cows. It means of all the 40, some are pigs and the rest are cows. Say, if you say that 90% of the students are either from Michigan or Ohio, it means that the rest of the 10% are from other states but 90% belong to these two states. How many of the 90% are from Michigan and how many are from Ohio, we don't know but we know that together they account for 90% of the class. Intern Joined: 06 Jul 2014 Posts: 2 Own Kudos [?]: [0] Given Kudos: 20 Re: Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How [#permalink] What about the 20 extra animals ? Why can't they be cows or pigs as well within that set ? Nothing is mentioned about these 20 animals. We just know that 40 animals are either pigs or cows. Does anyone get my point ? Having 13 pigs, 27 cows, and 20 other animals is a possiblity. Having 14 pigs, 30 cows and 16 other animals is another possibility. Following that logic, E is the right answer. Math Expert Joined: 02 Sep 2009 Posts: 95451 Own Kudos [?]: 657790 [1] Given Kudos: 87242 Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How [#permalink] 1 Kudos rbaudoin10 What about the 20 extra animals ? Why can't they be cows or pigs as well within that set ? Nothing is mentioned about these 20 animals. We just know that 40 animals are either pigs or cows. Does anyone get my point ? Having 13 pigs, 27 cows, and 20 other animals is a possiblity. Having 14 pigs, 30 cows and 16 other animals is another possibility. Following that logic, E is the right answer. The phrase "Of the 60 animals on a certain farm, 40 are either pigs or cows" means: 1. The total number of pigs and cows combined is 40. Thus, c + p = 40. 2. The remaining 20 animals are neither pigs nor cows. Therefore, these 20 animals could be another species, such as sheep. Tutor Joined: 16 Oct 2010 Posts: 15297 Own Kudos [?]: 67986 [1] Given Kudos: 442 Location: Pune, India Re: Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How [#permalink] 1 Kudos rbaudoin10 What about the 20 extra animals ? Why can't they be cows or pigs as well within that set ? Nothing is mentioned about these 20 animals. We just know that 40 animals are either pigs or cows. Does anyone get my point ? Having 13 pigs, 27 cows, and 20 other animals is a possiblity. Having 14 pigs, 30 cows and 16 other animals is another possibility. Following that logic, E is the right answer. To add to what Bunuel said: It is similar to: Of the 100 people in a room, 40 are men. What does this mean to you? It means that 60 are not men, right? Similarly, Of the 60 animals on a certain farm, 40 are either pigs or cows. This means that rest of the 20 are neither pigs nor cows! When you are given concrete numbers, it implies that the group has been considered. If only some people were considered, you would have been given "Of the 100 people in a room, at least 40 are men." Manager Joined: 15 Aug 2013 Posts: 176 Own Kudos [?]: 348 [1] Given Kudos: 23 Re: Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How [#permalink] 1 Bookmarks Bunuel shrivastavarohit Guys I don't know how long this post has been here however I thought I would add my 2 cents and see if what I feel should be the answer. Question says something either a cow or a pig form the ratio of 2/3. I don't feel 60 number is of much significance since this is DS. 1) says the mix is 1 to 2. So if by simple math I find what's remaining out of 2/3 the ratio comes out to be 1/3. At this moment for me the information in the statement 1 becomes helpful. Since 2wice of 1/3 is 2/3 and we know that 2wice as many cows hence this information is sufficient. Posted from my mobile device Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How many of the animals are cows? (1) The farm has more than twice as many cows as it has pigs --> so we have is $$c>2p$$ and not $$c=2p$$ --> as $$c+p=40$$ (p=40-c and c=40-p) --> $$40-p>2p$$, $$13.3>p$$, $$p_{max}=13$$ and $$c_{min}=27$$. Many combinations are possible: (27,13), (28, 12), ... Not sufficient. (2) $$p>12$$. Not sufficient (1)+(2) $$p>12$$ but $$p_{max}=13$$, hence $$p=13$$ --> $$c=27$$. Sufficient. One more thing: if the ratio indeed were $$c=2p$$, then the question would be flawed as solving $$c=2p$$ and $$c+p=40$$ gives $$p=13.3$$, but # of pigs can not be a fraction it MUST be an integer. Hi Bunuel, Two questions: -How do you know that Cmin is 27? I can se how Pmax is 13, but I don't really see how that translates onto Cmin? -If I were to add Stm 1 and Stm 2, I get c-p>12. That, algebraically, doesn't help solve the problem Am I correct? Math Expert Joined: 02 Sep 2009 Posts: 95451 Own Kudos [?]: 657790 [1] Given Kudos: 87242 Re: Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How [#permalink] 1 Kudos russ9 Bunuel shrivastavarohit Guys I don't know how long this post has been here however I thought I would add my 2 cents and see if what I feel should be the answer. Question says something either a cow or a pig form the ratio of 2/3. I don't feel 60 number is of much significance since this is DS. 1) says the mix is 1 to 2. So if by simple math I find what's remaining out of 2/3 the ratio comes out to be 1/3. At this moment for me the information in the statement 1 becomes helpful. Since 2wice of 1/3 is 2/3 and we know that 2wice as many cows hence this information is sufficient. Posted from my mobile device Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How many of the animals are cows? (1) The farm has more than twice as many cows as it has pigs --> so we have is $$c>2p$$ and not $$c=2p$$ --> as $$c+p=40$$ (p=40-c and c=40-p) --> $$40-p>2p$$, $$13.3>p$$, $$p_{max}=13$$ and $$c_{min}=27$$. Many combinations are possible: (27,13), (28, 12), ... Not sufficient. (2) $$p>12$$. Not sufficient (1)+(2) $$p>12$$ but $$p_{max}=13$$, hence $$p=13$$ --> $$c=27$$. Sufficient. One more thing: if the ratio indeed were $$c=2p$$, then the question would be flawed as solving $$c=2p$$ and $$c+p=40$$ gives $$p=13.3$$, but # of pigs can not be a fraction it MUST be an integer. Hi Bunuel, Two questions: -How do you know that Cmin is 27? I can se how Pmax is 13, but I don't really see how that translates onto Cmin? -If I were to add Stm 1 and Stm 2, I get c-p>12. That, algebraically, doesn't help solve the problem Am I correct? There are total of 40 pigs and cows. We know that there are at most 13 pigs (13 or less). Thus there are at least 27 cows. Or: p + c = 40 and p <= 13 --> (40 - c) <= 13 --> c >= 27. Hope it's clear. Intern Joined: 22 May 2016 Posts: 2 Own Kudos [?]: 1 [0] Given Kudos: 8 Re: Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How [#permalink] I believe answer should be E. As question demands number of cows among animals, we can form many other combinations from given 1st and 2nd statement. For ex. Pigs = 30, Cows = 10. S1: More than twice as many cows as pigs. 30 > 10*2 S2: Number of pigs more than 12. 30>12 Similarly many other combinations are possible like, (28,12), (32,8) etc. Kindly help. Tutor Joined: 16 Oct 2010 Posts: 15297 Own Kudos [?]: 67986 [1] Given Kudos: 442 Location: Pune, India Re: Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How [#permalink] 1 Kudos ngoyal2 I believe answer should be E. As question demands number of cows among animals, we can form many other combinations from given 1st and 2nd statement. For ex. Pigs = 30, Cows = 10. S1: More than twice as many cows as pigs. 30 > 10*2 S2: Number of pigs more than 12. 30>12 Similarly many other combinations are possible like, (28,12), (32,8) etc. Kindly help. S1: The farm has more than twice as many cows as pigs implies Number of cows is more than twice the number of pigs. C > 2P Hence Pigs = 30, Cows = 10 doesn't work. Neither do the other examples you have given. Re: Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How [#permalink] 1   2 Moderator: Math Expert 95451 posts

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3(p-q) when p=9 and q=2 Question Updated 9/11/2017 12:20:45 PM This conversation has been flagged as incorrect. Flagged by alfred123 [9/11/2017 12:20:42 PM] f Original conversation User: 3(p-q) when p=9 and q=2 Weegy: 18 Expert answered|charlencharg|Points 1592|Rating ***** (+318 -0) Question Updated 9/11/2017 12:20:45 PM This conversation has been flagged as incorrect. Flagged by alfred123 [9/11/2017 12:20:42 PM] Rating 8 3(p - q) when p = 9 and q = 2; 3(9 - 2); = 3(7); = 21

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# Bonus Calculations [closed] According to our company (UAE based company) policy yearly bonus given on below mentioned formula 60% of employees performance + 40% company growth My basic salary is 9540 and our company have yearly growth is 4% My rating is 7 out of 10 Anybody have idea ,how to calculate yearly bonus depending above mentioned formula. Thanks ## closed as off-topic by Jan Doggen, Dan Pichelman, keshlam, gnat, Masked ManJun 21 '16 at 0:26 This question appears to be off-topic. The users who voted to close gave this specific reason: • "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address only a specific company or position are of limited use to future visitors. Questions seeking legal advice should be directed to legal professionals. For more information, click here." – Jan Doggen, Dan Pichelman, keshlam, gnat, Masked Man If this question can be reworded to fit the rules in the help center, please edit the question. • That's not enough information. There's nothing in your company policy that says how much of that "growth" goes into employee bonuses; and of course without knowing company valuation we don't know how much money that 4% represents anyway. – Ernest Friedman-Hill Jun 20 '16 at 19:27 As a comment says, there's not really enough information here to provide a sure way to project your bonus. That may be an intentional decision from your employer. That said, one possible method of calculating your bonus would be to insert your rating and the growth amount into the formula for the computation. However, it seems likely that we'd also have to assume some maximum bonus, which may be based on your salary, experience, level, etc. For argument's sake (and easy computation), let's say that your maximum bonus is 10% of your salary. Then, putting all this together, the formula would be: Bonus = Max bonus * (performance + growth) which expands to: Bonus = 10% * Salary * (.6 * rating + .4 * growth) Plugging in the numbers, you get: Bonus = 0.1 * 9540 * (0.6 * 0.7 + 0.4 * 0.04) = 415.94 However, I stress that this is speculative. To really find out, you'll have to bring it up with your employer; however, don't be surprised if they won't tell you any more. • Why the downvote? – GreenMatt Jun 20 '16 at 19:56 That's not enough information. There's nothing in your company policy that says how much of that "growth" goes into employee bonuses; and of course without knowing company valuation we don't know how much money that 4% represents anyway. The bottom line is that a "formula" like that can be used to explain how much one employee gets relative to another, but not how much any one employee gets in absolute terms.

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# Functions and Graphs Multiple Choice Questions Test 2 Tests pdf Download Practice math test 2 on functions and graphs MCQs, grade 6 cartesian plane multiple choice questions and answers. Cartesian plane revision test has math worksheets, answer key with choices as coordinate axes, dimensional axes, vertical axis and horizontal axis of multiple choice questions (MCQ) with cartesian plane quiz as on cartesian plane, ox and oy are classified as for competitive exam prep. Free math study guide to learn cartesian plane quiz to attempt multiple choice questions based test. MCQ. On Cartesian plane, Ox and Oy are classified as 1. dimensional axes 2. coordinate axes 3. vertical axis 4. horizontal axis D MCQ. On Cartesian plane, point 'O' is known as 1. function 2. ordinate 3. origin 4. coordinate C MCQ. In ordered pair (-2, -3), abscissa is 1. 3 2. −2 3. 2 4. −3 B MCQ. In ordered pair (-3, -4), ordinate or y-ordinate is 1. −4 2. −7 3. −3 4. 3 A MCQ. Point P in Cartesian plane is located by an ordered pair called 1. (c, b, a) 2. (a, b, c) 3. (a, b) 4. (b, a) C

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# 10.4. Array of Strings# In previous sections of this chapter, we show strings as a special array of characters that is null-terminated. If we were to have an array of strings, it can be created in two ways: a 2D array of characters each row is terminated by a '\0', where each row is a string, or a 1D array of char*, where each element is pointing towards a 1D array that is a string. The first ## 10.4.1. 2D array of characters# We can declare a 2D array, where each row has a string holding a month name. For example, in the following code, in line 4, we declare a 2D array named months, with $$10$$ columns. Recall, we must define the column size, and the row size is not necessary to define if we are initializing, as it will be known from the number of elements. Each row stores a month name. For rows that are not completely filled if the string size is smaller than 10, similar to str in char str[10] = "Hi";, the remaining elements are filled with null-characters. This $$12$$ by $$10$$ array is shown in Fig. 10.14. The elements of the 2D array are all stored in the Code #include <stdio.h> int main(void) { char months[][10] = {"January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"}; for (int month_ind = 0; month_ind < 12; month_ind++) { months[month_ind][0] = months[month_ind][0] - 'A' + 'a'; printf("%s, ", months[month_ind]); } return 0; } In Fig. 10.14, the elements of the 2D array are all stored in the stack of the main memory. This means we can change any of the characters in the 2D array. For example, in line $$9$$, we can change the first character of each month to the lower case. We can do so by subtracting the ASCII code of the first upper case letter in the alphabet 'A' then adding the ASCII code of the first lower case letter in the alphabet 'a'. This shifts the ASCII to lower case letters. In line $$10$$, we can print each string/row using %s. Note Since months is a 2D array of characters, you cannot change an entire row in one line, e.g. months[0] = "january";. This is because months[0] is an array identifier, not a pointer variable. Refer to notes in Section 10.1.1.3 for an explanation of what can and cannot be done, while dealing with strings. ## 10.4.2. 1D array of char*# The other way to declare an array of strings, we can declare an array of char* as we do in line $$4$$ of the following code. To initialize element to point to a string, we can do it as in lines $$6$$$$17$$. Code int main(void) { char* months[12]; months[0] = "January"; months[1] = "February"; months[2] = "March"; months[3] = "April"; months[4] = "May"; months[5] = "June"; months[6] = "July"; months[7] = "August"; months[8] = "September"; months[9] = "October"; months[10] = "November"; months[11] = "December"; return 0; } We can declare and initialize the array of character pointers as in line $$4$$$$6$$ of the following code. Code (Declare and Initialize) #include <stdio.h> int main(void) { char* months[12] = {"January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"}; for (int month_ind = 0; month_ind < 12; month_ind++) { // CANNOT do // months[month_ind][0] = months[month_ind][0] + 'A' + 'a'; // as each string is in the constants segment of the main memory printf("%s, ", months[month_ind]); } return 0; } Here, months is a 1D array of character pointers and each pointer is pointing to a constant string in the constants segment of the main memory. This is illustrated in the following figure. Note Since months is a 1D array of character pointers and each pointer is pointing to a constant string in the constants segment of the main memory, you cannot change an individual element of the string, e.g. months[0][0] = "j";. This is because months[0][0] is constant and cannot be changed. Instead, you can change what the pointer points to, e.g. months[0] = "january";. Refer to notes in Section 10.1.1.3 for an explanation of what can and cannot be done, while dealing with strings. 0 Questions X

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1999 E&amp;M C1 Solution # 1999 E&amp;M C1 Solution - E& M SOLUTIONS 1999... This preview shows pages 1–3. Sign up to view the full content. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: E & M SOLUTIONS #' 1999 Physics C Solutions Distribution of points E & M 1 (15 points) (a) 4 points For [Ring the relationship between potzmial and charge 1 pain: _ 1 2. '— 47t£° 7' Solving for Q: Q = 41reoVr For correct substitutions for the potential and radius 1 poim 90 = Mum—2000 V)(0.20 m) or (-2000 vx0.20 m)/( 9 x109 N-nf/c1 ) Q1 2 -—16001r€°C or — 4.4 x10" C Forth: correct magnitude of go 1 point For the negative sign 1 point (b) Spoims LFoa'indicatingthatthcdecu'icfieldisw-o Ipoim ii. Theehargeonfinsphqccanbctrcamdasapoimchargeatitscm 1 Q0 E = —— 47560 ,r1 a. E = (9 x 10" N-nf/c2 {ii-\$43] -3965 599.5 -- E— r2 c or r1 C whuensmmetus Formyoftheaboveaqn-essionsme lpoim fii.ForindicaLingthatthe¢-lecu'icfiddism lpoint iv.Fmindicatingthatthedearicfiddisza-o lpoim ForhavinganfananswasomrectORfcrsmnemmfionofusingthcmclosed chargeORfusomemmfionofGauss’law Ipoim E 8: H SOLUTIONS #1 1999 Physics C Solutions Distribution of points E & M I (continued) (c) 3 points 5 AV=D;—V‘=-j.Edr For recognition of the need to take the difference of the potentials at radii a and b, or for writing the definite integral (with limits) 1 point b 90 dr [Ar/I ' 41:60 I r1 o 47:60 :- a [AV] = Q“ _ b 421-50 17 a For correct substitution of variables or numerical values for Q, a, and b 1 point For the correct answer 1 point 59. [AV] — 81mg 01' 1000V (Alternate solution) (Alternate points) For recognition of the need to take the difi‘erenee of the potentials at radii a and b 1 point AV = V‘ - V. 99 1 Qo 1 M - [a] - (a) For correct substitution of Q, , a, and b I point 90 I 1 AV " 47:60 (E — E] For the correct answer 1 point 590 (Alternate solution) (Alternate points) V = g C For using the above relationship I point For substituting QL1 from part (a) and C from part (d) alternate solution I point For the correct answer 1 point lAV| = SQ” or 1000 V 81m, E I M SOLUTIONS #‘I 1999 Physics C Solutions Distribution of points E & M I (continued) (d) 2 points 0 C=='—°— V For using the above relationship 1 point For substituting Q,0 from part (a) and AV from part (c) 1 point C = 4.4 x104 C 1000 V C = 4.4x10‘" F {Alternate solution) (Alternate points} For writing the equation for the capacitance of the spherical capaCitor 1 point h 4369a!) C ‘ b - a (0.02 mX0.04 m) (9 x 109 N-mI/c2 )(o.04 m — 0.02 m) For the correct answer I point C = 4.4x10‘“ F C: For correct units on two answers and no incorrect units 1 point ... View Full 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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Expert Reviewed # How to Calculate Payroll It's the responsibility of the employer to accurately calculate payroll for his or her employees. Errors can result in employees having too much or too little of their pay withheld for taxes, Social Security, Medicare and other deductions, which causes an inconvenience for them at tax time. Miscalculations can also result in the company coming under review and possibly facing penalties by government agencies, such as the Internal Revenue Service. ### Part 1 Filling out the Forms 1. 1 Have employees complete federal and state employee withholding forms. When starting a new job, all employees must complete a federal Employee Withholding Allowance Certificate, also known as a W-4 form.[1] The name and structure of the state form will vary from state to state. • Information provided on Employee Withholding Allowance Certificates lets a company know how much federal and state income tax to withhold from each employee's pay based on filing status and the number of exemptions he or she claims. Keep in mind, the more exemptions claimed, the less that the employee has withheld from his or her paycheck. However, that employee might owe money when it comes time to pay taxes. • Keep in mind that you might be doing business in a state with no state income tax, in which case there is no form for state withholding. Since the state isn't asking people to pay taxes, there's no reason that your employees will need to fill out a withholding form.[2] 2. 2 Verify that the forms are signed. Remember, those federal and state documents are not valid if they're not signed. 3. 3 Double-check the math on the forms. Although the math on those forms can be very easy (it's typically as simple as adding a few 1's), it's best to double-check it just to be certain that the employee added the numbers correctly. ### Part 2 Calculating Net Pay 1. 1 Determine the employee's gross pay. Before you can begin to calculate payroll, you must know what the employee's gross income is. This is determined by multiplying the number of hours worked in a pay period by the hourly rate. For example, if an employee works 40 hours in a pay period and earns \$15 an hour, you would multiply 40 times \$15 to get a gross pay of \$600. 2. 2 Obtain federal and state income tax tables. Federal Income Tax tables break down the amount of federal income tax an individual owes based on pay, exemptions, and filing status. The Internal Revenue Service posts current tax tables online.[3] You may find state income tax tables online by visiting your state comptroller’s office website. 3. 3 Apply federal and state income taxes. You'll use the tax tables that you've retrieved to apply the correct amount of federal and state income tax to withhold. • For federal taxes, you'll look up the withholding amount based on the employee's gross pay, filing status, and the number of exemptions claimed. Then, you'll deduct that amount from the gross pay. • For states taxes, consult your state's department of revenue website for instructions about how much to withhold. 4. 4 Apply Social Security tax rates. Calculating the amount of Social Security tax to pay is easy as it is a fixed percentage of an employee’s earnings. Employers must bear in mind that they too are responsible for paying social security taxes. The current tax rate for Social Security is 6.2% for the employee.[4] 5. 5 Deduct Medicare taxes. Like the Social Security tax, Medicare taxes are also a fixed percentage of a person’s earnings. In addition, employers are also responsible for paying Medicare taxes. Employees are currently taxed at 1.45% for Medicare.[5] 6. 6 Subtract other deductions. Employees may have voluntary contributions or mandatory deductions that need to be reduced from their gross pay. • Examples of voluntary contributions include 401(k) contributions, deferred compensation programs, long-term disability, and flexible spending accounts. • Examples of mandatory deductions include child support and alimony. 7. 7 Finalize net pay. The amount remaining after these deductions are subtracted will be net pay. Go back over your calculations and make sure that you haven't make any mistakes. ## Community Q&A • How does split calculator work in processing payroll? • What is the labor and industries tax deduction is Washington state? • How do I calculate federal and state taxes? ## Tips • Purchase payroll software to reduce the possibility of human error. Payroll software programs can also increase efficiency with features that include the ability to print reports and end of year W-2 forms. ## Article Info Categories: Accounting and Regulations In other languages: Español: calcular una nómina, Português: Calcular Folha de Pagamentos, 中文: 计算每月薪资, Русский: сделать расчет заработной платы, Italiano: Calcolare i Salari, Français: calculer le salaire de ses employés (USA), Deutsch: Eine Gehaltsabrechnung durchführen, Bahasa Indonesia: Menghitung Penggajian Karyawan di Amerika Serikat Thanks to all authors for creating a page that has been read 56,540 times.

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HOME TOPICS DEFINITIONS TABLES LAWS INVENTIONS EXPERIMENTS QUIZ VIDEOS Wien's Displacement Law The Wien's Displacement Law state that the wavelength carrying the maximum energy is inversely proportional to the absolute temperature of a black body. i.e λmax x T = b Where, λmax  =  Wavelength of maximum intensity ( meters ) T  =  Temperature of the blackbody ( kelvins ) b  =  Wien's displacement constant  =  2.8977685 ± 51 × 10-3 meters·kelvins For optical wavelengths, it is often more convenient to use the nanometer in place of the meter as the unit of measure. In this case, b  =  Wien's displacement constant  =  2.8977685 ± 51 × 106 nm·K The law is named after German physicist Wilhelm Wien.

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1. ### Graphing and Applying Coordinate Dilations • Resource ID: M8M2L3* • Subject: Math Given a coordinate plane or coordinate representations of a dilation, the student will graph dilations and use those graphs to solve problems. 2. ### Connecting Multiple Representations of Functions • Resource ID: A1M1L8 • Subject: Math The student will consider multiple representations of linear functions, including tables, mapping diagrams, graphs, and verbal descriptions. 3. ### Determining Reasonable Domains and Ranges (Verbal/Graph) • Resource ID: A1M2L2 • Subject: Math Given a graph and/or verbal description of a situation (both continuous and discrete), the student will identify mathematical domains and ranges and determine reasonable domain and range values for the given situations. 4. ### Developing the Concept of Slope • Resource ID: M8M2L4* • Subject: Math Given multiple representations of linear functions, the student will develop the concept of slope as a rate of change. 5. ### Using Multiplication by a Constant Factor • Resource ID: M7M2L6 • Subject: Math Given problems involving proportional relationships, the student will use multiplication by a constant factor to solve the problems. 6. ### Generating Different Representations of Relationships • Resource ID: M8M2L8* • Subject: Math Given problems that include data, the student will generate different representations, such as a table, graph, equation, or verbal description. 7. ### Predicting, Finding, and Justifying Data from a Table • Resource ID: M7M2L2 • Subject: Math Given data in table form, the student will use the data table to interpret solutions to problems. 8. ### Interpreting Scatterplots • Resource ID: A1M2L4 • Subject: Math Given scatterplots that represent problem situations, the student will determine if the data has strong vs weak correlation as well as positive, negative, or no correlation. 9. ### Making Predictions and Critical Judgments (Table/Verbal) • Resource ID: A1M2L5 • Subject: Math Given verbal descriptions and tables that represent problem situations, the student will make predictions for real-world problems. 10. ### Collecting Data and Making Predictions • Resource ID: A1M2L6 • Subject: Math Given an experimental situation, the student will write linear functions that provide a reasonable fit to data to estimate the solutions and make predictions. 11. ### Writing Expressions to Model Patterns (Table/Pictorial → Symbolic) • Resource ID: A1M3L2 • Subject: Math Given a pictorial or tabular representation of a pattern and the value of several of their terms, the student will write a formula for the nth term of a sequences. 12. ### Analyzing the Effects of the Changes in m and b on the Graph of y = mx + b • Resource ID: A1M4L8 • Subject: Math Given algebraic, graphical, or verbal representations of linear functions, the student will determine the effects on the graph of the parent function f(x) = x. 13. ### Writing Equations of Lines • Resource ID: A1M4L9 • Subject: Math Given two points, the slope and a point, or the slope and the y-intercept, the student will write linear equations in two variables. 14. ### Predicting, Finding, and Justifying Data from a Graph • Resource ID: M8M2L10 • Subject: Math Given data in the form of a graph, the student will use the graph to interpret solutions to problems. 15. ### Predicting, Finding, and Justifying Data from an Equation • Resource ID: A1M2L7* • Subject: Math Given data in the form of an equation, the student will use the equation to interpret solutions to problems. 16. ### Determining the Domain and Range for Linear Functions • Resource ID: A1M4L3 • Subject: Math Given a real-world situation that can be modeled by a linear function or a graph of a linear function, the student will determine and represent the reasonable domain and range of the linear function using inequalities. 17. ### Investigating Methods for Solving Linear Equations and Inequalities • Resource ID: A1M5L3 • Subject: Math Given linear equations and inequalities, the student will investigate methods for solving the equations or inequalities. 18. ### Selecting a Method to Solve Equations or Inequalities • Resource ID: A1M5L3b • Subject: Math Given an equation or inequality, the student will select a method (algebraically, graphically, or calculator) to solve the equation or inequality. 19. ### Determining Intercepts and Zeros of Linear Functions • Resource ID: A1M4L10

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## Saturday, November 20, 2010 ### Introduction: Don't be confused with the title, this article has nothing to do with "how to calculate an exponent of a given matrix", rather it will discuss on how to use this technique to solve a specific class of problems. Sometimes we face some problems, where, we can easily derive a recursive relation (mostly suitable for dynamic programming approach), but the given constraints make us about to cry, there comes the matrix exponentiation idea. The situation can be made more clear with the following example: Let, a problem says: find f(n) : n'th Fibonacci number. When n is sufficiently small, we can solve the problem with a simple recursion, f(n) = f(n-1) + f(n-2), or, we can follow dynamic programming approach to avoid the calculation of same function over and over again. But, what will you do if the problem says, given 0 < n < 1000000000, find f(n) % 999983 ? No doubt dynamic programming will fail! We'll develop the idea on how and why these types of problems could be solved by matrix exponentiation later, first lets see how matrix exponentiation can help is to represent recurrence relations. ### Prerequisite: Before continuing, you need to know: • Given two matrices, how to find their product, or given the product matrix of two matrices, and one of them, how to find the other matrix. • Given a matrix of size d x d, how to find its n'th power in O( d3log(n) ). ### Patterns: What we need first, is a recursive relation, and what we want to do, is to find a matrix M which can lead us to the desired state from a set of already known states. Let, we know k states of a given recurrence relation, and want to find the (k+1)th state. Let M be a k x k matrix, and we build a matrix A:[k x 1] matrix from the known states of the recurrence relation, now we want to get a matrix B:[k x 1] which will represent the set of next states, i.e. M x A = B, as shown below: ` | f(n) | | f(n+1) | | f(n-1) | | f(n) |M x | f(n-2) | = | f(n-1) | | ...... | | ...... | | f(n-k) | |f(n-k+1)|` So, if we can design M accordingly, job's done!, the matrix will then be used to represent the recurrence relation. • #### Type 1: Lets start by the simplest one, f(n) = f(n-1) + f(n-2). So, f(n+1) = f(n) + f(n-1) Let, we know, f(n) and f(n-1); we want to get f(n+1) From the above situation, matrix A and B can be formed as shown below: Matrix A Matrix B `| f(n) || f(n-1) |` `| f(n+1) || f(n) |` [Note: matrix A will be always designed in such a way that, every state on which f(n+1) depends, will be present] So, now, we need to design a 2x2 matrix M such that, it satisfies M x A = B as stated above. The first element of B is f(n+1) which is actually f(n) + f(n-1). To get this, from matrix A, we need, 1 f(n) and 1 f(n-1). So, the 1st row of M will be [1 1]. `| 1 1 | x | f(n) | = | f(n+1) || ----- | | f(n-1) | | ------ |` Similarly, 2nd item of B is f(n) which we can get by simply taking 1 f(n) from A. So, the 2nd row of M is [1 0]. `| ----- | x | f(n) | = | ------ || 1 0 | | f(n-1) | | f(n) |` [I hope you know how a matrix multiplication is done and how the values ar assigned!] Thus we get the desired 2 x 2 matrix M: `| 1 1 | x | f(n) | = | f(n+1) || 1 0 | | f(n-1) | | f(n) |` If you are confused about how the above matrix is calculated, you might try doing it this way: We know, the multiplication of an n x n matrix M with an n x 1 matrix A will generate an n x 1 matrix B, i.e. M x A = B. The k'th element in the product matrix B is the product of k'th row of the n x n matrix M with the n x 1 matrix A in the left side. In the above example, the 1st element in B is f(n+1) = f(n) + f(n-1). So, it's the product of 1st row of matrix M and matrix B. Let, the first row of M is [x y]. So, according to matrix multiplication, ` x * f(n) + y * f(n-1) = f(n+1) = f(n) + f(n-1)⇒ x = 1, y = 1` Thus we can find the first row of matrix M is [1 1]. Similarly, let, the 2nd row of matrix M is [x y], and according to matrix multiplication: ` x * f(n) + y * f(n-1) = f(n)⇒ x = 1, y = 0` Thus we get the second row of M is [1 0]. • #### Type 2: Now, we make it a bit complex: find f(n) = a * f(n-1) + b * f(n-2), where a, b are some constants. This tells us, f(n+1) = a * f(n) + b * f(n-1). By this far, this should be clear that the dimension of the matrices will be equal to the number of dependencies, i.e. in this particular example, again 2. So, for A and B, we can build two matrices of size 2 x 1: Matrix A Matrix B `| f(n) || f(n-1) |` `| f(n+1) || f(n) |` Now for f(n+1) = a * f(n) + b * f(n-1), we need [a b] in the first row of objective matrix M instead of [1 1] from the previous example. Because, now we need a of f(n)'s and b of f(n-1)'s. `| a b | x | f(n) | = | f(n+1) || ----- | | f(n-1) | | ------ |` And, for the 2nd item in B i.e. f(n), we already have that in matrix A, so we just take that, which leads, the 2nd row of the matrix M will be [1 0] as the previous one. `| ----- | x | f(n) | = | ------ || 1 0 | | f(n-1) | | f(n) |` So, this time we get: `| a b | x | f(n) | = | f(n+1) || 1 0 | | f(n-1) | | f(n) |` Pretty simple as the previous one... • #### Type 3: We've already grown much older, now lets face a bit complex relation: find f(n) = a * f(n-1) + c * f(n-3). Ooops! a few minutes ago, all we saw were contiguous states, but here, the state f(n-2) is missing. Now? what to do? Actually, this is not a problem anymore, we can convert the relation as follows: f(n) = a * f(n-1) + 0 * f(n-2) + c * f(n-3), deducing f(n+1) = a * f(n) + 0 * f(n-1) + c * f(n-2). Now, we see that, this is actually a form described in Type 2. So, here, the objective matrix M will be 3 x 3, and the elements are: `| a 0 c | | f(n) | | f(n+1) || 1 0 0 | x | f(n-1) | = | f(n) || 0 1 0 | | f(n-2) | | f(n-1) |` These are calculated in the same way as Type 2. [Note, if you find it difficult, try on pen and paper!] • #### Type 4: Life is getting complex as hell, and Mr. problem now asks you to find f(n) = f(n-1) + f(n-2) + c where c is any constant. Now, this is a new one and all we have seen in past, after the multiplication, each state in A transforms to its next state in B. f(n) = f(n-1) + f(n-2) + c f(n+1) = f(n) + f(n-1) + c f(n+2) = f(n+1) + f(n) + c ................................. so on So, normally we can't get it through the previous fashions. But, how about now we add c as a state? ` | f(n) | | f(n+1) |M x | f(n-1) | = | f(n) | | c | | c |` Now, its not much hard to design M according to the previous fashion. Here it is done, but don't forget to verify on yours: `| 1 1 1 | | f(n) | | f(n+1) || 1 0 0 | x | f(n-1) | = | f(n) || 0 0 1 | | c | | c |` • #### Type 5: Lets put it altogether: find matrix suitable for f(n) = a * f(n-1) + c * f(n-3) + d * f(n-4) + e. I would leave it as an exercise to reader. The final matrix is given here, check if it matches with your solution. Also find matrix A and B. `| a 0 c d 1 || 1 0 0 0 0 || 0 1 0 0 0 || 0 0 1 0 0 || 0 0 0 0 1 |` [Note: you may take a look back to Type 3 and 4] • #### Type 6: Sometimes, a recurrence is given like this: `f(n) = if n is odd, f(n-1) else, f(n-2)In short:f(n) = (n&1) * f(n-1) + (!(n&1)) * f(n-2)` Here, we can just split the functions in the basis of odd even and keep 2 different matrix for both of them and calculate separately. Actually, there might appear many different patterns, but these are the basic patterns. • #### Type 7: Sometimes we may need to maintain more than one recurrence, where they are interrelated. For example, let a recurrence relation be: g(n) = 2g(n-1) + 2g(n-2) + f(n), where, f(n) = 2f(n-1) + 2f(n-2). Here, recurrence g(n) is dependent upon f(n) and the can be calculated in the same matrix but of increased dimensions. Lets design the matrices A, B then we'll try to find matrix M. Matrix A Matrix B `| g(n) || g(n-1) || f(n+1) || f(n) |` `| g(n+1) || g(n) || f(n+2) || f(n+1) |` Here, g(n+1) = 2g(n) + 2g(n-1) + f(n+1) and f(n+2) = 2f(n+1) + 2f(n). Now, using the above process, we can generate the objective matrix M as follows: `| 2 2 1 0 || 1 0 0 0 || 0 0 2 2 || 0 0 1 0 |` So, these are the basic categories of recurrence relations which are used to be solved by this simple technique. ### Analysis: Now that we have seen how matrix multiplication can be used to maintain recurrence relations, we are back to out first question, how this helps us in solving recurrences on a huge range. Recall the recurrence f(n) = f(n-1) + f(n-2). `M x | f(n) | = | f(n+1) | | f(n-1) | | f(n) |` ............(1) How about we multiply M multiple times? Like this: `M x M x | f(n) | = | f(n+1) | | f(n-1) | | f(n) |` Replacing from (1): `M x M x | f(n) | = M x | f(n+1) | = | f(n+2) | | f(n-1) | | f(n) | | f(n+1) |` So, we finally get: `M^2 x | f(n) | = | f(n+2) | | f(n-1) | | f(n+1) |` Similarly we can show: `M^3 x | f(n) | = | f(n+3) | | f(n-1) | | f(n+2) |` `M^4 x | f(n) | = | f(n+4) | | f(n-1) | | f(n+3) |` `.............................................................................................` `M^k x | f(n) | = | f(n+k) | | f(n-1) | |f(n+k-1)|` Thus we can get any state f(n) by simply raising the power of objective matrix M to n-1 in O( d3log(n) ), where d is the dimension of square matrix M. So, even if n = 1000000000, still this can be calculated pretty easily as long as d3 is sufficiently small. ### Related problems: UVa 10229 : Modular Fibonacci UVa 10870 : Recurrences UVa 11651 : Krypton Number System UVa 10754 : Fantastic Sequence UVa 11551 : Experienced Endeavour Regards, Zobayer Hasan. 1. Thank you, Zobayer. 2. Hey Zobayer. Do have any idea to solve that: we consider F(n) function determined by following term if n is quadratic value then F(n)=0 otherwise f(n) = [1 / {sqrt(n)}] So what is S = F(1)+F(2)+...+F( N^2 ). I think there is no formula exists to solve it. But How do i solve it easier?? 3. I don't think I can solve this type using matrix exponentiation, I can go only for linear ones. I don't think quadratic equations can be solved. But, still this could be solved, like, sometimes approximation formulas can be found, or, you can also try divide and conquer / dp techniques. Might help. 4. Really nice tutorial! Helped me to solve this problem on latest Codechef - http://www.codechef.com/COOK05/problems/SEEDS/ Keep it up! 5. Hey Zobayer! I have a problem where the recursive relation is like this: f(n,h,k) = f(n-1,h-1,k+1) + f(n-1,h,k-1) + f(n-1,h+1,k)+1 do you have any idea how to solve this by matrix exponentiation..? 6. Is it not possible with dp? I don't think it's possible with matrix exponentiation as all of the parameters are not shifting at a same order. I am sorry if I am wrong, actually I don't know the answer of your question :( 7. Hi, Zobayar. Can you explain type-7 matrix A and B. According to me it should be | g(n) | A = | g(n-1) | | f(n) | | f(n-1) | B = | g(n+1) | | g(n) | | f(n+1) | | f(n) | 8. @Aakash, if you want to find out g(n+1) then you will be needing: g(n+1) = 2g(n) + 2g(n-1) + f(n+1) So, you must have f(n+1) on the left side. Hope you get it. 9. This comment has been removed by the author. 10. nice question. look carefully, you can always do something like this, say for n = 3 1 + A + A^2 + A^3 = (1 + A) + A^2(1 + A) so divide and conquer is possible. I have a post related to this, which u may wish to see. http://zobayer.blogspot.com/2009/12/power-series-evaluation.html thanks for commenting. 11. Thanks for the informative post! It would be more helpful, if you could point out/give links to resources about finding n'th power of matrix in O(d^3 * log(n)). 12. I think, you can find b^n in log(n) right? a recursive algorithm might look like this: f(b, n): ----If n == 0 return 1 ----If n is odd return f(b, n-1) * b; ----else return square(f(b, n/2)) See, you are multiplying b here which is definitely O(1), however, for matrix, b will be a dxd matrix, and to multiply two dxd matrix, you need O(d^3). So the actual complexity is O(d^3 lg(n)) 13. hey can you tell about uva 11651 I am not able to figure it out 15. I tried the HDU 2802 problem using matrix exponentiation. I got TLE, so tried storing the matrices for A, A^2, A^4,... to quicken the calculation of A^n. But still got TLE! Can this be solved using matrix exponentiation for sure? Because there is this Chinese blog, which tells there is some cycle form - http://hi.baidu.com/xiinho/blog/item/1307e6156a9be10d4b90a741.html 16. I could not do it with matrix expo either, but I have listed it here because I have found this problem in the list of matrix expo problems somewhere, can't remember now where, many days passed. Sorry! 17. f(n) = n*f(n-1) + f(n-2) Can this be solved using matrix exponentiation ? 18. Hi Zobayer, Can you show how to find the result for f(n) = a* f(n-1) + b? Thanks :) 1. This is quite simple, I think you can try the following relation { { a, 1 }, { 0, 1 } } * { { f(n), b } } = { f(n+1), b } 19. Hmm...the new theme is cool ;-), nice article btw. Although I knew about matrix exponentiation before but learned a couple of new things here. Very well explained too. 20. Can we do it for recurrence having constant as a function of n. Like for example f(n) = f(n-1) + f(n-2) + n ? 1. I don't think so, or to be precise, I can't. 2. Yes, we can do it. Just define g(n) = n => g(n) = g(n-1) + 1, g(1) = 1. Then, f(n+1) = f(n) + f(n-1) + g(n+1) = f(n) + f(n-1) + (n+1). This if of type 7 as given in the article. We can solve it as: | f(n+1) | | 1 1 1 0 | | f(n) | | f(n) | = | 1 0 0 0 | | f(n-1) | | g(n+2) | | 0 0 1 1 | | g(n+1) | | 1 | | 0 0 0 1 | | 1 | We need to find out { { f(n+1) }, { f(n) }, { g(n+2) }, { 1 } }. We know { { f(n) }, { f(n-1) }, { g(n+1) }, { 1 } }. The matrix is 1 1 1 0 1 0 0 0 0 0 1 1 0 0 0 1 And thank you Zobayer for this wonderful article. It helped me a lot. 21. Muhammed HedayetJune 7, 2012 at 1:28 AM Cool. I know Matrix exponentiation technique and have solved some problems using this technique, but nice organized article. This article let me think in a new fashion. Thanks. Want more. . . :-) 22. Very nice post, keep up the great work. 23. I didn't know about the topic before read the article. Many many thanks for this. 24. Many many thanks to your for this article. Because the topic is new for me.... ans thanks again. 25. Many many thanks to you for this article. Because the topic is new for me........ thanks again. 26. good work dude..really nice 27. Thanks Vaia.It's really helpful.Thanks a lot :) 28. Thanks for sharing. Keep up the good work !!! 29. Can we do it if we have a n on the right side? For example: f(n) = a*f(n-1) - n? 1. Yes we can. Just define g(n) = n => g(n) = g(n-1) + 1, g(1) = 1. Then f(n) = a*f(n-1) - g(n). This is of type 7 as given in the article. We need to find out the 1 column matrix: f(n+1) g(n+2) 1 We know: f(n) g(n+1) 1 The matrix is a -1 0 0 1 1 0 0 1 2. And can we solve f(n) = a*n*f(n-1)+b*n*f(n-2)? I tried treating n like a constant but it just doesn't work. And I couldn't transform it to any type mentioned by Zobayer. 3. @Piotr Kąkol, I do not think it can be solved using matrix exponentiation methods, or at least not that I know of. 4. Yeah, I figured that if it could be, you'd make more than 7 types. Anyway, thanks for the article. Very useful! 5. Matrix exponentiation is applicable only for solving linear recurrences (and recurrences convertible to linear ones). 6. Yash you are right. 30. Superb explanation. Very clear. Thanx a ton 31. good job.....well written 32. suberb... post.. and so simple to understand.. thanx... 33. Nicely written. There aren't many good tutorials on this topic. 34. Great post!! 35. Great post. I managed to solve the first 2 :) 1. Good luck with the rest too :) Thanks for visiting :) 36. Thanks a lot bro. I wanted to learn about this and at the right time found this blog. 1. Thanks for visiting, I'm glad that it helped :) 37. while(1) printf("ধন্যবাদ,ভাইয়া :)"); 38. Hi , Thanks for writing this beautiful note :) Can you please explain, how can we build the M for the following type of function? Thanks in advance. g(n)=g(n-1)+g(n-2)+f(n-3) where f(n)=f(n-1)+f(n-2)+g(n-3) 1. you can do it using a 6x6 matrix. all you need to do is derive this matrix f(n+1) f(n) f(n-1) g(n+1) g(n) g(n-1) and what you know is this matrix f(n) f(n-1) f(n-2) g(n) g(n-1) g(n-2) so the matrix will be 1 1 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 0 0 1 0 0 0 0 0 0 1 0 39. I'm having trouble understanding the Formation of this matrix ... f(n+1) f(n) f(n-1) g(n+1) g(n) g(n-1) I'm writing my reasoning , please correct it ...... I need to find , f(n+1) ... f(n+1)=f(n)+f(n-1)+g(n-2) Here , f(n+1) function's dependencies are f(n) , f(n-1) and g(n-2) ... now , I need to find out the the dependencies of g(n-2) to , get the value of g(n-2) , g(n-2)=g(n-3)+g(n-4)+f(n-5) //g() again is related to f() , I'm finding this confusing// then , f(n+1) total dependencies are f(n),f(n-1),g(n-3),g(n-4),f(n-5) .......so , A becomes a matrix will all the up stated values , and B will be a matrix with same functions but '1' added to the arguments . Please , Explain the fault in the reasoning a give a bit detail on what is really going on there . Thanks . 1. Yes, there are some mistakes in your reasoning, I mean, you are correct about the dependencies, but the fact is, if you already know what g(n-2) is, you wont need to expand it further. In matrix exponentiation algorithm, with M*A = B form, the 1 column matrix you showed is B here, and A is what you already have. No need to follow recursions any more. Just try to test if recurrences hold. As I have written above, just multiply it and see for yourself you u can get the right side or not. first row is 1 1 0 0 0 1, and the 1 column A matrix is f(n) f(n-1) f(n-2) g(n) g(n-1) g(n-2) so, multiplying, we get the first row of B which is f(n)*1 + f(n-1)*1 + f(n-2)*0 + g(n)*0 + g(n-1)*0 + g(n-2)*1 => f(n) + f(n-1) + g(n-2) => f(n+1). similarly, second row is 1 0 0 0 0 0, this is f(n) from third row, 0 1 0 0 0 0, we get f(n-1) from fourth row 0 0 1 1 1 0, we get f(n-2) + g(n) + g(n-1) => g(n+1) from fifth row 0 0 0 1 0 0, we get g(n) from sixth row, 0 0 0 0 1 0, we get g(n-1) so, multiplying the matrix M, we get f(n+1) f(n) f(n-1) g(n+1) g(n) g(n-1) Now, imagine, if we already know f(n) f(n-1) f(n-2) g(n) g(n-1) g(n-2) we can then find M^n to get f(n+1) and g(n+1) at the same time, don't follow typical recurrence reasoning here. I think the later part of the post describes how this is done actually. I don't know how can I make it more clear. 2. Correction, I wanted to write this in the last: Now, imagine, if we already know f(2) f(1) f(0) g(2) g(1) g(0), then we can find M^n to get f(2+n), g(2+n) at the same time. the other values are just carried with the process. 3. Thanks , I think I got it finally . :) I was taking the whole thing as the formal recursion and was trying to trace back and ...from f() I was trying to trace back through g() .....again f() ...all that made me confused ... sorry but I will nag you one more time :) I'm writing another recurrence function , please say whether it's okay or not :) Let , f(n)=f(n-1)+g(n-2)+h(n-3) g(n)=g(n-1)+h(n-2)+f(n-3) h(n)=h(n-1)+f(n-2)+g(n-3) then , B will be |f(n+1)| |f(n) | |f(n-1)| |g(n+1)| |g(n) | |g(n-1)| |h(n+1)| |h(n) | |h(n-1)| A will be : |f(n) | |f(n-1)| |f(n-2)| |g(n) | |g(n-1)| |g(n-2)| |h(n) | |h(n-1)| |h(n-2)| So , M will be | 1 0 0 0 1 0 0 0 1 | | 1 0 0 0 0 0 0 0 0 | | 0 1 0 0 0 0 0 0 0 | | 0 0 1 1 0 0 0 1 0 | | 0 0 0 1 0 0 0 0 0 | | 0 0 0 0 1 0 0 0 0 | | 0 1 0 0 0 1 1 0 0 | | 0 0 0 0 0 0 1 0 0 | | 0 0 0 0 0 0 0 1 0 | :) Thanks 4. Yeah, its correct! 40. it realy helped me to get thrgh the concept so easily..thank u... 41. Thanks Sir for this post. 42. দারুণ আইডিয়া... 44. That was an awesome article. However i failed to understand a certain part in type 7. I was hoping if you could explain how to arrive at this equation g(n+1) = 2g(n) + 2g(n-1) + f(n+1). from the main equation :g(n) = 2g(n-1) + 2g(n-2) + f(n), where, f(n) = 2f(n-1) + 2f(n-2). I could form the equations for all the other cases/types ,but I don't know how to work with equations with more than one recurrence. It would be very helpful if you could explain a bit briefly. 1. Hello there, about the recursion, isn't that putting n+1 instead of n ? 45. Sorry my previous question was rather silly. I don't understand why there are 4 rows in matrix A and B. I will try to explain what i have understood. Please rectify me if I'm wrong anywhere. Given: g(n) = 2g(n-1) + 2g(n-2) + f(n) --->(1) f(n) = 2f(n-1) + 2f(n-2)---->(2) replacing n by (n+1) in equation (1). g(n+1) = 2g(n) + 2g(n-1) + f(n+1) RHS of the equation is the present set of states. so they should be in matrix A. hence matrix A should be: | g(n) | | g(n-1) | | f(n+1) | and matrix B should be containing the next set of states: | g(n+1) | | g(n) | | f(n+2) | All the types from 1 to 6 were solved this way,but here in case 7 the fourth row was introduced. Can you tell me the reason behind that or why its necessary? 1. I was hoping if you could clear my doubt mentioned above. :) 2. Sorry for late response, got carried away by the chores of daily life. Anyway, its nothing different from the other types. Look closely, to know f(n+1), you need to know both f(n) and f(n-2). So, if you only use three rows, how are you going to get the value of f(n-2)? So, you will need two g()s and two f()s in the matrix. Hope it is clear now. The thing is, you need to have all the dependencies on the LHS matrix. 3. Oops! sorry about the type, its not f(n-2) its f(n-1). I think you got that as well, It can be done using DP, but if x and y are as large as 10^6 then one has to look around! Matrix Expo possible? It can be done using DP, but if x and y are as large as 10^6 then one has to look around! Matrix Expo possible? 1. No, I don't think it's possible, or at least not in my knowledge. 48. Any hope with Matrix Expo? Another property of the function f(x,y) was f(x,y)=f(y,x) 1. I don't think that can be done using matrix exponentiation. If you can express one variable in terms of the other one, then some thoughts can be made, but I have no clue how to handle function on more than one variable recurrences using matrix exponentiation. 49. Please explain Type 7 . It's given that g(n) = 2g(n-1) + 2g(n-2) + f(n) and f(n) = 2f(n-1) + 2f(n-2). So g(n+1)=2g(n)+2g(n-1)+f(n+1) and f(n+1)=2f(n)+2f(n-1) so shouldn't matrix A be [g(n) ,g(n-1) , f(n) , f(n-1)] . Please explain how did you got | g(n) | | g(n-1) | | f(n+1) | | f(n) | Someone also has asked this doubt but I could not understand your explanation for that.Please explain asap. 50. Nice Article! Can u please tell how to make approach for 11651? Thanks 52. zobayer why are we considering f(n)&f(n-1)in matrix exponentation after all we want to find f(n+1) 1. do you mean why we are keeping them on right hand side? they are generated as partial result, but we are not worried about them, as long as we find f(n+1). but on the left side matrix, we need them both because f(n+1) depends on both. 53. T(n)= max ( a+ T(n-b), c+ T(n-d) ). how can we solve this with matrix exponentiation ? 1. Seems more like dynamic programming to me. I have never tried anything of this sort, lets see if someone else can help you. 54. Man read this topic for the first time and you nailed it. Keep it up. And thanks for writing such a brilliant post. Are there more of such useful kinds... looking forward for reading them too. Thanks 55. Hi Zobayer Hasan, I am new to competitive programming but good at DS and Algo concepts. Can u please tell me where to start. Please share your experiences how have u mastered all tricks. How many years does it take. How much time do we need to spend coding. 56. Can you give one tutorial like this for dynamic programming. Greedy and for graph algorithms. It would be of great help 1. Well, I could find some for you searching google but then I wouldn't know if the material is good or not. To be honest, learning something deeply cannot be achieved by reading random tutorials, casually learnt knowledge does not persist much long in our brain, but the knowledge you gain when you get stuck with a problem goes a long way. My suggestion would be, do not waste time following tutorials, follow problems, and if you cant solve them, then find those solutions, you'll learn lot faster, in my opinion. :) That being said, there are good tutorials at http://topcoder.com/tc 57. Hey Can you Please explain the matrix for odd even case. I could not figure it out :( 58. hi very nice article - what about f(2n) = f(n) + f(n-1) + n for n > 1 and f(2n+1) = f(n-1) + f(n) + 1 for n >= 1 ?? any thoughts appreciated 59. oh i forgot to mention f(0) = 1 , f(1) = 1, and f(2) = 2 - thank you 60. hey ! Zobayer, thanks a ton man ! a 61. Just wanted to say that this was a really helpful, clearly-explained post! Great work! 62. Great Notes ,thankx. 63. It was a great Help . Thanks Man :) 64. ভাইয়া, UVa 11651 : Krypton Number System প্রব্লেমটাতে ম্যাক্সিমাম স্কোর 1e9 না হয়ে আরো কম হলে dp দিয়ে সল্ভ করতে পারতাম, কিন্তু ওটাতে লিনিয়ার রিকারেন্স কিভাবে হচ্ছে সেটা ধরতে পারতেছি না বলে matrix exponentiation এপ্লাই করতে পারছি না... কাইন্ডলি যদি হেল্প করতেন... :) 65. f(n)=f(n-1)+f(n-2) if(n is odd) f(n)=f(n-1)-f(n-2) if (n is even) how can we solve this with matrix exponentiation ? 66. ম্যাট্রিক্স এক্সপোনেনশসিয়েশন বোঝার জন্য এটা একটা সবথেকে BEST reference.. thank you! 67. Step 6 টা নিয়া যদি একটু বিস্তারিত লিখতেন তাহলে ভালো হইত (আলাদা আলাদা ব্যাপারটা বুঝিনাই) ... 68. http://codeforces.com/problemset/problem/821/E can you explain formation of matrix for this question

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Question # Consider a 1 liter solution in which silver (Ag+) has a concentration of 10^-2.94 moles/L. Then... Consider a 1 liter solution in which silver (Ag+) has a concentration of 10^-2.94 moles/L. Then if 0.53g of NaCl is added to the solution and it comes to equillibrium, show that AgCl will precipitate. Determine the weight in grams of AgCl that will precipitate from the 1L solution. The solution is ideal. The Ksp of AgCl is 10^-9.75. Assume the NaCl completely dissolves. The molar mass of NaCl is 58.4 g/mol. The Ksp of AgCl is Hence, precipitation of AgCl will occur. After precipitation, Before precipitation During precipitation, the change in Since total volume is 1 L, the molar concentration is equal to the number of moles. Change in number of moles of silver ions during precipitation is 0.001148 moles. It is equal to the number of moles of AgCl formed. The molar mass of AgCl 143.3 g/mol. The mass of AgCl precipitated

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Home > Error Bars > How To Represent Standard Deviation As Error Bars In Excel How To Represent Standard Deviation As Error Bars In Excel Contents On the Layout tab, in the Analysis group, click Error Bars, and then click More Error Bar Options. Reply sana says: June 21, 2016 at 7:46 am I am trying to add Error Bars in data based on standard deviation in Excel 2007-2010. You can remove either of these error bars by selecting them, and then pressing DELETE. In the dialog box you can enter an absolute value or a formula to treat all data points equally. navigate here Error bars express potential error amounts that are graphically relative to each data point or data marker in a data series. Her projects include everything from course development and webinars for business training clients such as Fred Pryor Seminars to email, website, and content marketing strategy for small businesses in the Kansas City area. Any other feedback? Learn more You're viewing YouTube in German. look at this web-site How To Add Error Bars In Excel 2013 Melde dich an, um dieses Video zur Playlist "Später ansehen" hinzuzufügen. Anzeige Autoplay Wenn Autoplay aktiviert ist, wird die Wiedergabe automatisch mit einem der aktuellen Videovorschläge fortgesetzt. If your missing error bar shows as just a black line at the top of the column, this might be the case. You will need to store your individual std deviation data somewhere in the workbook in its own range. (see image below) Follow the directions in the paragraph beginning with "If you need to specify your own error formula" to assign your custom deviations to the data points in the series. (If your data has different positive and negative values, you'll also need a column for that data.) Since you will be working with two series - male and female - you'll have to perform the process twice, once for each series. Graphically you can represent this in error bars. In this case, 5 measurements were made (N = 5) so the standard deviation is divided by the square root of 5. Notice the range of energy values recorded at each of the temperatures. How To Add Error Bars In Excel 2010 I would add it at the end (last column) and give it a different color in the chart. In order to post comments, please make sure JavaScript and Cookies are enabled, and reload the page. Blast from the Past: Error Bars function similarly in Excel 2007-2010, but their location in the user interface changed in 2013. Click on the chart, then click the Chart Elements Button to open the fly-out list of checkboxes. Wird verarbeitet... In this over-simplified example, we selected B4:E4 cell range for both positive and negative error values. Error Bars Excel 2016 Sprache: Deutsch Herkunft der Inhalte: Deutschland Eingeschränkter Modus: Aus Verlauf Hilfe Wird geladen... Wird verarbeitet... Resulting X &Y error bars will be the same size and won't vary with each value. Custom Error Bars Excel More precisely, the part of the error bar above each point represents plus one standard error and the part of the bar below represents minus one standard error. Melde dich an, um unangemessene Inhalte zu melden. How To Add Error Bars In Excel 2013 Or, you can enter a cell range that contains an Error result that you need for each individual data point. How To Add Error Bars In Excel Mac Die Bewertungsfunktion ist nach Ausleihen des Videos verfügbar. Wird verarbeitet... http://sysreview.com/error-bars/how-to-insert-standard-deviation-error-bars-in-excel.html Schließen Ja, ich möchte sie behalten Rückgängig machen Schließen Dieses Video ist nicht verfügbar. On the Layout tab, in the Analysis group, click Error Bars. Reply Rob Douglass says: September 14, 2016 at 8:47 pm HI, I have calculated the means and my std dev of my data points. How To Add Individual Error Bars In Excel Wähle deine Sprache aus. Reply Excel Tips and Tricks from Pryor.com says: June 23, 2016 at 1:43 pm Sorry to hear that adding error bars to your data has been frustrating! No, but you can include additional information to indicate how closely the means are likely to reflect the true values. his comment is here If your column represents 100,000,000 and your error is only 10, then the error bar would be very small in comparison and could look like it's either missing or the same as the other bars. Click More Error Bar Options, and then under Vertical Error Bars or Horizontal Error Bars, click the display and error amount options that you want to use. How To Add Error Bars In Excel 2016 I manage to specify all my standard deviation values. But when I select the range for both positive and negative values. If you look back at the line graph above, we can now say that the mean impact energy at 20 degrees is indeed higher than the mean impact energy at 0 degrees. Line chart showing error bars with Standard deviation(s) of 1.3   If you need to specify your own error formula, select Custom and then click the Specify Value button to open the Custom Error Bars dialog box. I followed your example where I selected a range B4:E4 cell range for both positive and negative error values. You can remove either of these error bars by selecting them, and then pressing Delete. What Are Error Bars You can do this with error bars.

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Build a 15,000 rpm Tesla Turbine using hard drive platters 9 Steps ## Step 9: Complete turbine and movie Please post (or email me) any questions or comments and I'll do my best to answer them. Steven Remove these ads by Signing Up maikel2 says: May 4, 2013. 3:46 AM a conventional rotary compressor runs more than 100k rpm to develop vacuum to sucks outside air continuously thus compressed it.if this device run at that speed can it become a compressor??? FFX says: Sep 17, 2012. 6:07 PM Make a good gearbox and put it to a generator which will charge the compressor ;) 4lifenerdfighter says: Nov 2, 2012. 11:05 AM Law of conservation of energy prevents this. Gear backlash, for example? Air speed being lost to exhaust? bart.p says: Jun 3, 2012. 11:21 PM great project! have you ever done any test on how much vacuum this thing can generate? im currently designing a turbine that would pump air and have the shaft spun with an electric motor, one of my constrains is the intake size so i need to suck in as much air as possible and preferably compressed it as well bart.p says: Jun 3, 2012. 11:22 PM den316a says: May 8, 2012. 8:38 AM my Question is how much torck does this have can u run a generator off of it? chairchild says: Oct 24, 2010. 4:25 PM if you added some exhaust port, in-line with the cutouts on the disc, efficiency would increase massively. The way it's working in this setup, is pretty much choking the output down to pitiful levels - easily one of the nicest-looking ones I've seen in a while though and the perpetual motion comments? LMAO!! KimberlyP says: Jan 10, 2012. 1:32 AM Remember that Lord Kelvin was LMAO about the idea of powered flight! If perpetual in the sense of forever, well highly improbable. As to coupled occilator's with irreversible flows it happens in biology. Thermodynamics While the 1st law seems on the sturdiest ground experimentally. The 2nd law as applied to thermodynamics is about energy density. Objects at a particular temperature if unrestricted by activation energies move spontaneously from high to low, and from dense to less dense. 3rd law that you can never break even except at 0 Kelvin is the add on law that may be cracked and or found exceptions too. Zeroth Law is just common sense as the commutative law of addition. It sets T. 3rd law is valid when considering reversible flows, but if one looks at coupled occilator s using irreversible flows out of phase one can see immediately that this is not necessarily so. As to Boltzman you can argue that with irreversible flows that the state is not exactly returned to the same probabilistic state as before, but the averages can be! Think about that! That fits right in with Poincare thinking which informed Boltzmans work. So there is times arrow locally and as we extend the domain we get back to symmetry. Something exhibited in the real world, in chaos theory, and many other physical phenomena. Geometrically we can think of fractals. So while it is true that in a simple expression of reversible processes the 3rd law is true, that we can only break even at 0 K and also can never reach 0 K because of Zeroth. We can always couple irreversible flows, which from a probabilistic standpoint cannot return to original state, but from a physical energy standpoint can return to the same average energy. Jar Sqwuid says: Nov 16, 2011. 3:53 PM Perpetual energy? That's all fine and dandy if you can find metal that will never rust or corrode, or other materials that are otherwise impervious to the various elements. As well as trying to complete Tesla's failed project to transmit energy wirelessly, because otherwise you're going to need a load of veeeeerrrrrryyyy long extension cables. Awesome project, really :) I wanna try it! T_T_ says: Nov 21, 2011. 10:39 AM tesla's wireless energy plan didn't fail, it was canceled. raybent says: Apr 11, 2011. 11:17 AM Actually, perpetual "motion" is entirely possible - you just have to choose the correct speed. Once adjusted to the correct speed, the friction reduces to zero. Hence it will run at that speed forever. The correct speed? Zero. Remember - zero is a valid number. beehard44 says: Nov 16, 2011. 4:31 AM I agree. Perpetual motion IS possible. It's just that it'll need regular maintenance and i doubt if it'll be able to produce excess energy. Still, nothing is impossible. c.doyle says: Sep 5, 2011. 10:03 AM Even if you reduce standard friction to 0, which is virtually impossible as it is, there is still 'quantum friction' that occurs even in a vacuum: http://www.newscientist.com/article/mg20927994.100-vacuum-has-friction-after-all.html. And perpetual motion means that the machine generates more energy than it uses, which at 0 for 0 it wouldn't. luig says: Nov 7, 2011. 9:03 PM you are right but it would be a Flywheel energy storage device instead. http://en.wikipedia.org/wiki/Flywheel_energy_storage theanthropicalthane says: Aug 6, 2011. 11:22 PM Correct me if I'm wrong, but my understanding of the number 'zero' is that zero is not a number, but a representation of the absence of a value. So it's not a number, but the absence of a number. KimberlyP says: Apr 26, 2011. 6:44 AM Superfluidity, Superconductivity.... Are other ways to reducing to 0. Get them working at Standard Temps and you will have yourself a Nobel Prize. jpetrill reese says: Apr 11, 2011. 3:51 PM Well Yes, I guess if a disk spinning at "zero" speed counts as being in perpetual motion then it is possible. Sadly it would be able to do exactly zero work. ilpug says: May 17, 2011. 7:44 PM it would be able to do that amount of work forever though! Jun 5, 2011. 12:19 AM we need to go to the perpetual energy subculture with this. it's elegant and true! the kind of argument i expect to see on xkcd.com kanhai says: Jul 8, 2011. 6:32 AM can i know how much pressure we will need for this working at 10000 rpm seansarvela says: Apr 7, 2011. 3:56 PM your design looks good. Im making one that is similar, although mine looks terrible. CodfishCatfish says: Jun 24, 2010. 3:44 PM In my opinion it is unwise to say never. As far as perpetual motion is concerned I don't think it is far off in being sussed. We have the science of hear and now so are limited by discovery. They say the day we invent the superconductor that runs at room temperatiure we can build a generator in a matchbox. As yet we have not discovered other planets minerals or structures to say that one element combined with another may not unlock this potential. I think the experiment touches on what NASA should be working towards generating electricity in space from sources other than chemical or solar or even nuclear and where the experiment would significantly reduce any friction in outerspace where gravity does not act on the structure. You spin a satellite in space and it would spin forever (we would like to think) and in future we may find a metal that does exactly what a superconductor does and defies gravity. Perpetual motion is not impossible just not with todays knowledge. Think if you could defy gravity to raise a weight and then reapply gravity to let it fall and gather the energy and repeat the process. Use a superconductor as the weight and then you have perpetual motion. Awesome insight and a very well thought out project 10/10.... tanmanknex says: Mar 31, 2011. 6:28 PM The problem with perpetual motion is as soon as you apply any load like to make a generator for example, it increases the energy needed to keep it perpetual, thus removing the perpetuality. It needs to be better than perpetual motion, it needs to generate more energy as it goes, thus speeding up. Think of it like this system below: E-5+5=E E is energy -5 is the amount of energy required to keep it going +5 is the amount of energy produced Now imagine there's a load of 1 on it E-6+5=E-1 It will slow. switch62 says: Jun 25, 2010. 3:36 PM True perpetual motion is probably not possible but you can get close even with todays technology. We already have some very large perpetual motion happening. That is anything in orbit in space. Like the moon orbiting earth, or the planets orbiting the sun. It is still not "true" perpetual motion but since it will take untold billions of years to all stop, it may as well be perpetual. In space there is almost nothing except the thin gasses and solar winds to cause friction, and gravity keeps it all going. We also have magnetic bearings that in a vacuum can allow things to spin forever, almost. Even the interaction of magnetic fields causes minute amounts of friction. Even if you had a perpetual motion machine, all it can do is keep the same amount of energy that started it moving in the first place. As soon as you try to take some energy out, like generate electricity or get mechanical work, it will slow down or stop. I won't say never to perpetual motion but our current laws of physics don't seem to allow for it to exist, or be useful. KimberlyP says: Apr 25, 2011. 4:58 AM Yes and then you have the Clarendon labs Clock running off a Zamboni pile ringing that bell for about 100 years. The reason the battery has not worn down yet is its a very good regenerator. Also you could have thermal power as in a coupled chemical occilator which would pass its thermal energy back and forth for a very long time and its resulting fluid power from the phase change would be greater than the losses of the thermal energy. It's not perpetual, in that in time you will lose mass and you will lose energy, but if most of the energy is internal and insulated it can continue for a long time as long as it gets the small kick. legless says: Mar 6, 2011. 12:32 AM Everywhere here you say perpetual motion does happen but also keep saying "almost" or not "true" PM. Which is it - perpetual motion or almost? The motion of satellites in orbit is not perpetual. It decays. The distance of the moon from Earth is increasing every year. The speed of rotation of the Earth is slowing every year. All due to other forces and losses which cause the system to decay. Gravity powers the universe but there are still "losses". KimberlyP says: Apr 26, 2011. 6:49 AM In the end resistance is futile! That or resistance makes the effort futile, I forget which... :P The best that can be done is highly regenerative systems which are coupled occilators. In the end you only delay the inevitable. such a cruel world. switch62 says: Mar 6, 2011. 3:01 AM BTW your example of the moon moving away from earth and the earth's spin slowing down is very interesting.  It's actually a transfer of energy between the earth and moon.  Gravity, the spin of the earth, and the tidal effect is actually speeding up the moon (thereby increasing orbit distance) and slowing the spin of the earth. In 16 billion years it will stabilise, the moon's orbit will be 1.6 times that of today and an earth day will be 55 current days long.  You might then see the moon's orbit start to decay.  But in 7 billon years the sun will be a red giant and destroy the earth and moon, so it's all academic. switch62 says: Mar 6, 2011. 2:29 AM Depends on your definition of perpetual :) strict definition: continuing or enduring forever. practical definition: continuing an extremely long time. Satellites are still in contact with a very thin atmosphere and so the orbit will decay.  They aren't really in space. Since the orbit of planets and moons will last billions of years, for all practical purposes it may as well be perpetual.  It is of course not "true" perpetual motion as it will not last forever. Yes, there will always be losses in any system (current physics) but it is the scale of the losses compared to the energy in the system that will determine how long the system will be in motion. For me billions of years is practically perpetual motion. runical1991 says: Dec 23, 2010. 11:15 AM I think it could be useful. think about is. If it is possible and the machine is not too big, it could serve as a new kind of battery, since you can get back all of the energy you put in. So there we would have our perfect energy storage ;-) correct me if I'm wrong, but I think it would be very useful, if it were possible... switch62 says: Dec 23, 2010. 3:13 PM Yes. I know in the 80's engineers were looking at energy storage using flywheels on magnetic bearings. The flywheels were designed to act as a motor to convert electical energy into rotational energy, and then act as a generator to get electrical energy out again. It would need to spin at very high rpm or be very heavy. The problem is the bigger the flywheel the stronger the magnetic bearings need to be, I think there were/are limits to this and you could not make a large enough device to be practical. Also a large flywheel would need to be stationary as any large movments might cause the magnetic bearings to make physical contact. With the energies involved that could literally rip the device apart (explode). The size of a flywheel to hold the same amount as a car battery would probably be much bigger and heavier than a car battery. Even if you could make it small enough you would never get 100% efficiency due to electrical resistance, magnetic "friction", etc.  A room tempreture super conductor would solve some of these problems, and make it very efficient. It seems chemical storage at the moment is a more practical solution.  Chemical batteries are fairly simple devices and easy to make.  Batteries are becoming smaller with greater capacities every year.  New technologies will make them even more so. KimberlyP says: Apr 26, 2011. 6:54 AM http://techportal.eere.energy.gov/technology.do/techID=25 They have a higher energy density storage than conventional batteries. Also Ultracaps can't store the energy density but the can charge and discharge large amounts of power quickly with low ESR. Low ESR = essentially resistance to discharging their power. Conventional batteries if you try to discharge to quick will heat up and resist discharging. hyratel says: Feb 10, 2011. 6:46 PM Porsche has built a race-class car with a flywheel dynamic storage. another automaker has done similar, but iirc it's not on mass-production yet. notanemo says: Sep 14, 2010. 2:31 AM as the speed of something increases, newtons law of motion becomes more wrong. try out something that 'could' be perpetual motion, then speed it up, and you could take energy out of it. KimberlyP says: Apr 26, 2011. 6:15 AM I'm assuming you are referencing MV^2? Yes but losses tend to go up too. However if you had room temperature superconductors and you spun in an absolute vacuum. (Absolute vacuum being impossible) You could accelerate to the speed of light, except what would you make the material from? But if you could and couple the two of them together, but then the problem would be switching which would be limited to the speed of light. Information problem. argh Nature gets you again. Now if you can find away around the speed limit of the Universe.... ? Happy hunting. zaney says: Jun 3, 2010. 1:52 PM as a thought, would it be possible to connect 2 of these together in such a way that with a motor attached to one you could pump enough air to run the second, the second being set up as a small generator to run the motor? Dinuc says: Feb 13, 2011. 10:15 AM Unfortunately not. Motors and generators are not 100% efficient. The best motors are about 95% efficient and the best generators about 80%. The rest of the energy is lost to heat and noise. The Tesla turbine is also not 100% efficient, let's assume it is 90 %. So if you put 100 units of energy into the motor, you get 95 units out to turn the turbine. The turbine then outputs 85.5 units of energy (95 x 90%) into the other turbine, which outputs 76.95 units of energy (85.5 x x90%). This turns the generator which outputs 61.56 units of energy (76.95 x 80%) which is fed back into the motor ad the cycle begins again. So your system runs down very quickly. MichelMoermans says: Jun 5, 2010. 9:20 AM I think your trying to make a perpetuum mobile out of it (or however you say that in English) a device that needs no extra power once it is running. Unfortunatly that is impossible due to friction. Everything that connects in there is using friction and that uses up some power. I can't give you the high tech science explenation but I do know that if you should try it it would work fine in the beginning after which it would slow down and stop. TarScrap says: Jun 8, 2010. 12:57 PM "Perpetual motion machine" is the phrase you're looking for. Friction and waste heat both mean that the total amount of energy in the closed system would bleed out over time. One would have to attain > 100% efficiency in order to keep it running, much less do any actual work with it. KimberlyP says: Apr 25, 2011. 5:16 AM Query me this? If hypothetically you had a box where you could lose no heat or mass. (Even Silicon Aerogels lose heat) How would you lose energy? Also if this system was a coupled occilator where the natural reaction is pressure out of phase with temperature such that they could never reach equilibrium unless they lost mas or heat, what would you have? I should say this is academic as there is no perfect insulator or seal. But there are chemical systems which act in this manner and they require only small energy inputs from the outside to make up for losses. Their energy flows are much higher, how does one explain this. Resonance. And internal energy is conserved and these small energy inputs feed one another. Take a look at the science of chaos or google chemical occilator. So you can get more energy out of a device than you put in but you cant get more energy out that the entire energy flow of the system! Think of it as a flow issue. Some of the flow can be converted to work but you cant convert 100% to work and you most certainly cant generate more than the flow. Thing is in dynamic systems you can have a dual flow system. Say heat and a fluid flow. A chemical occilator passes heat an fluid back and forth but they do this slightly out of phase with one another. In facts its required or they would be at equilibrium. The heat is conserved as is the mass. The fluid flow can drive a turbine and not lose energy. The reason they don't reach equilibrium is that when one evolves a fluid its endothermic and the other is exothermic. In the real world this is temperature sensitive and they always leak mass in enough time. MichelMoermans says: Jun 8, 2010. 1:23 PM Yep that's what I was looking for :) Thanks. Although no-one will ever succeed in making it it would be the perfect solution for all the problems we are facing today. Imagine a world where your car powers itself while driving :D One gallon of gass and the thing could run forever :D That would be pretty sweet :D tdawg309 says: Nov 24, 2010. 7:27 PM I know this wont remove all friction, but on something like this a magnetic bearing system would be of great help. pocketspy says: Jun 24, 2010. 10:05 AM Well, with today's electric motors and generators/alternators you could build a car that uses these to run the the car and produce it's own electricity and never need to be plugged into an outlet. iEdd says: Jun 24, 2010. 3:45 PM No you couldn't. It's impossible due to primary school physics. notanemo says: Sep 14, 2010. 2:32 AM newtons law? if so, that becomes more wrong the faster something gets. figuratively, this alows for perpetual motion iEdd says: Sep 14, 2010. 4:11 AM I was actually referring to conservation of energy. Newton's laws don't become "more wrong the faster something gets", they remain correct in all inertial reference frames. Relativity and non-inertial reference is a different story. In any case, while perpetual motion may be theoretically possible (but always practically impossible) if there were a lack of losses in a system, there is NO SUCH THING as overunity. That is, no system can produce more energy than it consumes. Anyone who suggests otherwise needs to have a physics book thrown at them. KimberlyP says: Apr 26, 2011. 6:22 AM True! You can have local phenomena which conserves the energy and passes it back and forth minimizing losses. IN such a case it may look like you are getting more energy than you put in. However, you are only inputing energy from the outside for losses. If you map the internal energy flow you are always getting less energy than the internal flow. The Famous Carnot engine as it nears optimum has a poor work ratio to the energy flow, even if you had a perfect regenerator. Ericsson Engine has the greatest work output for a Carnot equivalent. pocketspy says: Sep 14, 2010. 7:47 AM Here's what I was thinking. A car has four tires. I've seen electric cars with very powerful electric motors used on all four wheels. I think the tesla car does that. Couldn't you design a car that uses these and a couple of Alternators or generators to produce electricity and store it. For example you have the 4 generators and four electric motors. While you drive in the city you could be using only two of the motors while all four generators are producing electricity. Any extra power could be sent to the batteries. A computer could monitor use and automatically engage motors or generators as needed or use the batteries as needed. Throw some solar panels on topand you could be charing the batteries all day long. I was also thinking that if you used a hydraulic accumulator, you could store energy in that system for acceleration and braking, then you wouldn't have to use as much electricity to get the car up to speed. (I saw a video of a vehicle that used one, several years ago.)And you'd have less wear and tear on the motors. I figure some one out there could figure it out. Hell, if we can make 4g phones that can run your life, we should be able to produce a fully self-powered electric car. What do you think? robcull says: Oct 6, 2010. 12:30 PM Look, energy in always equals energy out. Period. If the gas you put in supplies X amount of energy, then the energy out (be it kinetic, electrical, heat, whatever) must also equal X. Adding the alternators, as you mentioned, or adding peltier/seebeck chips to the engine (converts heat to electricity), would increase the efficiency (some of the energy that would have been wasted is now being put back into making the car do what you want it to do- move). This is not perpetual motion. Adding the solar panels is another story... they're just another source of energy (just like you put gas in the tank, you'd be putting sunlight in the panel). This is also not perpetual motion. . A perpetual motion device could be an electric motor attached to an alternator (assuming both are 100% efficient). Once you give it a turn and start the motion initially, that kinetic energy is turned into electricity by the alternator, which powers the motor, which turns the alternator again. It is a 100% efficient setup. If, however, you had the alternator ALSO power even a small LED, then the device would not move perpetually, because some of the energy is "leaking" out through the LED (each time the electricity gets back to the motor, there would be slightly less). If, however, the alternator was SOMEHOW able to convert & output more electrical energy than the kinetic energy put in, then you would have higher than 100% efficiency and would be able to generate power perpetually. That's impossible, though. legless says: Mar 6, 2011. 12:24 AM You're right. Energy in does equal energy out. However energy out will include many things like the kind of output you want but will have components of loss. This doesn't mean that the energy is "lost" but that it is in a form you don't want. Incandescent bulbs for instance produce light (the thing you want) but they also produce a lot of heat (probably not the thing you want). Even if you use that heat for something else there are still further losses so the system will require more energy in that you can ever get back in any form. Theory is fine for theorists. Obviously one could build perfect machines if they could be made 100% efficient but they can't. Even if one adds devices to retrieve the unwanted output and convert it back to the wanted output there will still be losses. Certain physical laws just must be obeyed. KimberlyP says: Apr 26, 2011. 6:32 AM Heat and mass are fundamental. Deign a device which alternates those in opposite phase. The electric generator does not consume heat, it consumes the mechanical force generated by the heat acting on the mass flow. If you had a perfect insulator that would allow no heat to escape and had two chemical occilators occilating between endothermic and exothermic reactions driving the mass flow at a set temperature for evolving a gas... Ah but no such thing as a perfect insulator, and its only instructive as a thought experiment. My point is you can with aerogels get a very good insulator, and you can design with regeneration in mind, and therefore have very compact energy efficient machines. Where the only energy input is to make up for the losses. Another problem with the above example is not only would you need the perfect insulator but permeability would also have to be 0. Gases like Hydrogen and Helium diffuse through anything, its just a matter of time. H3 III a neet example. iEdd says: Sep 14, 2010. 12:55 PM You're trolling, right? A generator will create a force in the direction that opposes its rotation. The higher the current draw from the generator, the more pronounced the effect. If you put a generator on any part of the car, there is a braking force. With an alternator, it's not too big, but with anything that generates a large amount of power, it's huge. Think of electromagnetic braking - it converts the car's kinetic energy into electrical energy. That's where it comes from. You can't make energy from nothing. A 4G phone obeys the laws of physics. Overunity does not. zoltzerino says: Jun 9, 2010. 4:57 AM Cars already power themselves...it's called the internal combustion engine...UGH, honestly! *Joke sarcasm* I know exactly what you're on about ;-D navein333stein says: Jan 7, 2011. 8:16 AM what is the inlet pressure used for this tesla turbine. and what is the power produced by it moritata says: Feb 23, 2010. 10:55 PM steve very nice djr6789 says: Mar 12, 2009. 9:30 AM what are the uses for this? wildanddangerous says: Aug 1, 2009. 11:37 AM A 15,000 rpm fan lol! khanguy says: Mar 21, 2009. 3:36 PM Tesla dreamed of this as a cheap, efficient, and easy to create engine. He didn't like that the piston engine was so complicated. This can be used as an engine that can be made at home, or a pump( gas stations use it). Tesla wished that for this to be the engine of cars and possibly "aeroplanes." (he says this in an interview). pocketspy says: Jun 24, 2010. 10:11 AM It you use a hydraulic accumulator, you could run the generator with the fluid in a vehicle. When brake pressure is applied it could be designed to recharge the accumulator. Accumulators are used to start aircraft turbine engines all the time. Also if the tesla is connected to a electrical generator you could be producing electricity to run lights, charge batteries, etc.. REA says: Apr 10, 2009. 7:44 PM so it can be used as a way to convert energy? khanguy says: Apr 12, 2009. 3:04 PM It converts kenetic energy of a newtonian fluid( it also includes air) into rotation. So, yes it is a way to convert energy. But remember, if you attach a motor to the axle it can become a pump as well. ElectricUmbrella says: Jun 23, 2009. 6:00 AM Or a generator. djr6789 says: Mar 22, 2009. 10:09 AM ok. thanks this helped alot ill have to try and build one soon rednecknathan09 says: May 14, 2009. 4:16 PM how do u connect a electric motor to it cause i got an idea and i wouldnt mind ur opinion but i would need ur help ElectricUmbrella says: Jun 23, 2009. 5:59 AM Place the body of the motor on a stationary frame, and attach the shaft of the motor to the shaft of the turbine. tater_3001 says: Jul 16, 2008. 11:36 AM if you were to add a connecting piece to connect the platters like a water wheel except in a spiral fashion down the spindle, would the platters move faster and create more energy? countable says: Feb 25, 2009. 10:39 AM Thats basically the same design as a car turbocharger, so yes it would move faster but i'm not too sure about the energy created. Also, adding fins inside it would make it cease to be a tesla turbine because it works by creating adhesion between gas and platter. Steam Head says: Oct 27, 2008. 9:10 PM How would steam be as a motive power? That would give the turbine some extra "grunt" I would think. Fantastic Instructable. mrfixitrick says: Nov 7, 2007. 11:46 PM Great instructable! I, too, have an Instructable on how to build a Basic Tesla Turbine...except from CD discs, and CD Spindle...Wish I knew about your instructable when I recently did mine! For a really different take on the Tesla Turbine check out my new Steampunk Tesla CD Turbine with Pumpkincutter Attachment in action...http://www.youtube.com/watch?v=aoRh8T_VX-Y classicrocker12 says: Jul 11, 2006. 11:18 PM excellent instrustions! a trip to tap plastics (for the acrylic), my neighbor's junk box, and my garage later i made this thing with no trouble! i used this to prove a point to a friend who had negative thoughts on using steam to genterate electricity; low-pressure steam made in an old pipe cement can drove the turbine which drove a small electric motor, generating enough power to power a small light and a fan. i redrew mine 3d in autocad, ill try to post my plans here later... maybe prank says: May 26, 2006. 4:38 AM Cool project, and a beautiful machining job. I wonder if you could make one that turns at high torque with a low-speed flow (wind turbine...) sbtroy (author) says: May 27, 2006. 8:25 PM Thanks Prank. One big perk to my job is that I have access to a fully equiped instrument shop. I think decreasing the space between the platters would increase the torque. The boundary layer is very thin and any air flow between the platters but outside of this region isn't contributing to the torque. Also, air is compressible while water and oil are not. Using a more viscous fluid at a lower velocity might increase the torque as well.

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Time Left - 30:00 mins # Mini Mock Test Math/ Reasoning (Week 3)- 19.06.2022 Attempt now to get your rank among 777 students! Question 1 Select the set of classes, the relationship among which is best illustrated by the following Venn diagram. Question 2 Select the Venn diagram the best illustrates the relationship among the following classes. Neptune, Planet, Moon Question 3 In the following Venn Diagram, shows information about the person speak languages. The person who speaks the Malayalam language? Question 4 Identify the diagram that represents the relationship among the given classes. Cat, Animal, Pet Question 5 Identify the Venn diagram that best represents the relationship between the given classes. Loudspeaker, Sound, Stereo Question 6 Which two signs should be interchanged to make the given equation correct? 156 – 13 + 9 x 18 ÷ 5 = 169 Question 7 Which two signs should be interchanged to make the given equation correct? 51 + 18 – 5 × 12 ÷ 3 = 137 Question 8 If 5 # 9 @ 7 = 52 and 3 @ 9 # 2 = – 89, then 7 # 6 @ 9 = ? Question 9 Choose the correct signs from the options that will make the equation true. 28  7   4 = 8 Question 10 Which option gives the two signs that need to be interchanged to make the given equation correct? 6 – 20 ÷ 12 × 7 + 1 =70 Question 11 Study the given pattern carefully and select the number that can replace the question mark (?) in it. 3 13 4 6 67 31 5 ? 20 Question 12 Study the given pattern carefully and select the number that can replace the question mark (?) in it. Question 13 In the following question, select the missing number from the given alternatives. Question 14 In the following questions, select the number which can be placed at the sign of question mark (?) from the given alternatives. Question 15 Select the number that can replace the question mark (?). Question 16 The difference of simple interest on sum of money for 8 years and 10 years is ₹200. If the rate of interest is 10% p.a, then what is the sum of money? Question 17 A certain sum becomes 5 times in 3 years, at simple interest, then in how many years will it become 15 times at the same rate? Question 18 Find the difference between the compound interest and the simple interest on 32,000 at 10% p.a. for 4 years. Question 19 If the compound interest on a certain sum of money for 2 years at 5% p.a. is ₹328, then the sum is equal to: Question 20 In 4 years, ₹25000 at the rate of 20% per annum compounded yearly will amount to Question 21 A men invested Rs.2,400 at 5% and Rs.5,400 at 9% simple interest. What amount (in Rs.) will he get from his investments after 6 years? Question 22 A person deposits ₹8000 in a bank which pays 8% p.a. simple interest. The amount after 8 years will be: Question 23 In how many years, shall Rs. 3500 invested at the rate of 10% simple interest per annum, amount to Rs. 4500? Question 24 Find the simple interest on Rs. 74,000 at  per annum for a period of 8 months? Question 25 In how many years, will a sum of Rs. 5,000 yield a simple interest of Rs. 2,000 at an interest rate of 10% p.a.? Question 26 Two times of A’s salary is five times of B’s salary and four times of B’s salary is twice C’s salary. What is A’s salary if C’s salary is Rs 1,600? Question 27 The ratio of two numbers is 3 : 5. If eight is added to the first, and seven to the second, then the ratio becomes 2 : 3. What will be the ratio become if six is added to each? Question 28 If 7A = 4B = 14C, then what is A : B : C? Question 29 What is the ratio of the mean proportional between 1.2 and 10.8 to the third proportional to 0.2 and 1.2? Question 30 Two numbers A and B are, respectively, 80% and 20% more than a third number C. The ratio of the numbers A to B is: • 777 attempts

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# how to know that the point is intersect with the mesh? Show 40 post(s) from this thread on one page Page 1 of 2 12 Last • Aug 21st 2010, 08:29 AM siteq how to know that the point is intersect with the mesh? hello friends...i'm new here... i have a problem in determining the mesh that intersect with the point... i have 1 point and 2 mesh... i have tried many times searching in the google, but it doesn't work... i hope anyone here can help me to solve my problem..?? thank you...(Happy) • Aug 21st 2010, 07:10 PM ojones Give more mathematical details. • Aug 21st 2010, 10:45 PM siteq opss..sory2x.. i give an example.. if i have 2 points P0 = (2,-3,1) and P1=(-5,4,-3) and 2 planes Plane1 => 4x-2y+2z=5 and plane 2 => 5x-3y-6z=10 my question is, what is the condition that i can used to select which planes is intersect with P0? thank you... • Aug 21st 2010, 11:34 PM yeKciM Quote: Originally Posted by siteq opss..sory2x.. i give an example.. if i have 2 points P0 = (2,-3,1) and P1=(-5,4,-3) and 2 planes Plane1 => 4x-2y+2z=5 and plane 2 => 5x-3y-6z=10 my question is, what is the condition that i can used to select which planes is intersect with P0? thank you... did you mean that you need to know how does you can see which plane have (any) points (P_1 pr P_0) ? $\alpha : 4x-2y+2z=5$ $\beta : 5x-3y-6z=10$ and you need to know is it $P_0(2,-3,1) \in \alpha$ or is it $P_0(2,-3,1) \in \beta$ ? ? ? • Aug 21st 2010, 11:51 PM siteq nope.. actually i have 2 types of planes, and these planes are formed from a triangular mesh.. type 1 : original planes with 18 triangular meshes type 2 : simplified planes with 16 triangular mesh what i want to do is, to find the normal distance between the original planes to the simplified planes . and my lecturer give me a clue. i need to find distance from the points (from original planes) to the simplified plane , and the simplifies plane must intersect with normal line(from original planes).. i hope my explanation can help u...thank you very much.. • Aug 21st 2010, 11:56 PM ojones Siteq: I'm not familiar with the notion of a mesh. What subject is this coming from and can you define it? • Aug 22nd 2010, 12:06 AM siteq calculus vector... its okat ojones..actually i just want to know... if we have 3 points P0, P1 and P2, and we have 2 planes PLANE_1 and PLANE_2. is there any condition that i can used to determine that the normal line from P0 is intersect to PLANE_1 OR PLANE_2 ? • Aug 22nd 2010, 12:23 AM Vlasev Can you post the actual question that you have been given? • Aug 22nd 2010, 12:23 AM yeKciM line can't be "normal" to the point P_0 .... but to be normal to the any plane there is condition... if you have (make) line : $\displaystyle \frac {x-x_1}{m}= \frac {y-y_1}{n}=\frac {z-z_1}{p}$ and plane $Ax+By+Cz+D=0$ than you have vectors $\vec{q}=(m,n,p)$ $\vec{n} = (A,B,C)$ and from basic of vectors you have 2 conditions :D (to be parallel or orthogonal) $Am+Bn+Cp=0$ $\displaystyle \frac {A}{m}=\frac {B}{n}=\frac {C}{p}$ • Aug 22nd 2010, 12:34 AM siteq ahaa..ok2x..i know how to get m,n and p...but then, u stated there Am + Bn + Cp =0 is this the condition for these 2 vectors become parallel and A / m = B / n = C /p for these vectors become orthogonal ?? • Aug 22nd 2010, 12:43 AM ojones Siteq: There's no such thing as a normal line to a point. You're not forming your question clearly I'm afraid. • Aug 22nd 2010, 12:50 AM siteq aha..actually that one i just copy from my lecturer..maybe 'normal' thanat she mean in her email is the normal from the plane..i'm sorry for that...i miss understood...my mistake... maybe the correct is, the normal plane to a point..... • Aug 22nd 2010, 12:55 AM yeKciM Quote: Originally Posted by siteq ahaa..ok2x..i know how to get m,n and p...but then, u stated there Am + Bn + Cp =0 is this the condition for these 2 vectors become parallel and A / m = B / n = C /p for these vectors become orthogonal ?? there is one vector attitude that goes like this : "two vectors are orthogonal if their scalar product is zero". .. meaning that $(\vec{x}, \vec{y}) =0$ ... Quote: Originally Posted by siteq aha..actually that one i just copy from my lecturer..maybe 'normal' thanat she mean in her email is the normal from the plane..i'm sorry for that...i miss understood...my mistake... maybe the correct is, the normal plane to a point..... again normal to point... :D:D:D look, nothing line or plane can be normal to point... try imagine that in your head or draw it for yourself ... you have point in space and how do you think you should position line (or plane) so that it will be normal to point :D:D:D:D P.S. and please give exact question which you got ... so there will not be any more misunderstanding :D • Aug 22nd 2010, 05:50 PM siteq the exact question is not in english...and that's why i cannot put it here and i need to translate into english and then it become missunderstanding...heehee.. • Aug 22nd 2010, 06:23 PM yeKciM Quote: Originally Posted by siteq the exact question is not in english...and that's why i cannot put it here and i need to translate into english and then it become missunderstanding...heehee.. try giving it to someone who will be able to translate it for you :D Show 40 post(s) from this thread on one page Page 1 of 2 12 Last

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``` SUBROUTINE DLAZQ4( I0, N0, Z, PP, N0IN, DMIN, DMIN1, DMIN2, DN, \$ DN1, DN2, TAU, TTYPE, G ) * * -- LAPACK auxiliary routine (version 3.1) -- * Univ. of Tennessee, Univ. of California Berkeley and NAG Ltd.. * November 2006 * * .. Scalar Arguments .. INTEGER I0, N0, N0IN, PP, TTYPE DOUBLE PRECISION DMIN, DMIN1, DMIN2, DN, DN1, DN2, G, TAU * .. * .. Array Arguments .. DOUBLE PRECISION Z( * ) * .. * * Purpose * ======= * * DLAZQ4 computes an approximation TAU to the smallest eigenvalue * using values of d from the previous transform. * * I0 (input) INTEGER * First index. * * N0 (input) INTEGER * Last index. * * Z (input) DOUBLE PRECISION array, dimension ( 4*N ) * Z holds the qd array. * * PP (input) INTEGER * PP=0 for ping, PP=1 for pong. * * N0IN (input) INTEGER * The value of N0 at start of EIGTEST. * * DMIN (input) DOUBLE PRECISION * Minimum value of d. * * DMIN1 (input) DOUBLE PRECISION * Minimum value of d, excluding D( N0 ). * * DMIN2 (input) DOUBLE PRECISION * Minimum value of d, excluding D( N0 ) and D( N0-1 ). * * DN (input) DOUBLE PRECISION * d(N) * * DN1 (input) DOUBLE PRECISION * d(N-1) * * DN2 (input) DOUBLE PRECISION * d(N-2) * * TAU (output) DOUBLE PRECISION * This is the shift. * * TTYPE (output) INTEGER * Shift type. * * G (input/output) DOUBLE PRECISION * G is passed as an argument in order to save its value between * calls to DLAZQ4 * * Further Details * =============== * CNST1 = 9/16 * * This is a thread safe version of DLASQ4, which passes G through the * argument list in place of declaring G in a SAVE statment. * * ===================================================================== * * .. Parameters .. DOUBLE PRECISION CNST1, CNST2, CNST3 PARAMETER ( CNST1 = 0.5630D0, CNST2 = 1.010D0, \$ CNST3 = 1.050D0 ) DOUBLE PRECISION QURTR, THIRD, HALF, ZERO, ONE, TWO, HUNDRD PARAMETER ( QURTR = 0.250D0, THIRD = 0.3330D0, \$ HALF = 0.50D0, ZERO = 0.0D0, ONE = 1.0D0, \$ TWO = 2.0D0, HUNDRD = 100.0D0 ) * .. * .. Local Scalars .. INTEGER I4, NN, NP DOUBLE PRECISION A2, B1, B2, GAM, GAP1, GAP2, S * .. * .. Intrinsic Functions .. INTRINSIC MAX, MIN, SQRT * .. * .. Executable Statements .. * * A negative DMIN forces the shift to take that absolute value * TTYPE records the type of shift. * IF( DMIN.LE.ZERO ) THEN TAU = -DMIN TTYPE = -1 RETURN END IF * NN = 4*N0 + PP IF( N0IN.EQ.N0 ) THEN * * No eigenvalues deflated. * IF( DMIN.EQ.DN .OR. DMIN.EQ.DN1 ) THEN * B1 = SQRT( Z( NN-3 ) )*SQRT( Z( NN-5 ) ) B2 = SQRT( Z( NN-7 ) )*SQRT( Z( NN-9 ) ) A2 = Z( NN-7 ) + Z( NN-5 ) * * Cases 2 and 3. * IF( DMIN.EQ.DN .AND. DMIN1.EQ.DN1 ) THEN GAP2 = DMIN2 - A2 - DMIN2*QURTR IF( GAP2.GT.ZERO .AND. GAP2.GT.B2 ) THEN GAP1 = A2 - DN - ( B2 / GAP2 )*B2 ELSE GAP1 = A2 - DN - ( B1+B2 ) END IF IF( GAP1.GT.ZERO .AND. GAP1.GT.B1 ) THEN S = MAX( DN-( B1 / GAP1 )*B1, HALF*DMIN ) TTYPE = -2 ELSE S = ZERO IF( DN.GT.B1 ) \$ S = DN - B1 IF( A2.GT.( B1+B2 ) ) \$ S = MIN( S, A2-( B1+B2 ) ) S = MAX( S, THIRD*DMIN ) TTYPE = -3 END IF ELSE * * Case 4. * TTYPE = -4 S = QURTR*DMIN IF( DMIN.EQ.DN ) THEN GAM = DN A2 = ZERO IF( Z( NN-5 ) .GT. Z( NN-7 ) ) \$ RETURN B2 = Z( NN-5 ) / Z( NN-7 ) NP = NN - 9 ELSE NP = NN - 2*PP B2 = Z( NP-2 ) GAM = DN1 IF( Z( NP-4 ) .GT. Z( NP-2 ) ) \$ RETURN A2 = Z( NP-4 ) / Z( NP-2 ) IF( Z( NN-9 ) .GT. Z( NN-11 ) ) \$ RETURN B2 = Z( NN-9 ) / Z( NN-11 ) NP = NN - 13 END IF * * Approximate contribution to norm squared from I < NN-1. * A2 = A2 + B2 DO 10 I4 = NP, 4*I0 - 1 + PP, -4 IF( B2.EQ.ZERO ) \$ GO TO 20 B1 = B2 IF( Z( I4 ) .GT. Z( I4-2 ) ) \$ RETURN B2 = B2*( Z( I4 ) / Z( I4-2 ) ) A2 = A2 + B2 IF( HUNDRD*MAX( B2, B1 ).LT.A2 .OR. CNST1.LT.A2 ) \$ GO TO 20 10 CONTINUE 20 CONTINUE A2 = CNST3*A2 * * Rayleigh quotient residual bound. * IF( A2.LT.CNST1 ) \$ S = GAM*( ONE-SQRT( A2 ) ) / ( ONE+A2 ) END IF ELSE IF( DMIN.EQ.DN2 ) THEN * * Case 5. * TTYPE = -5 S = QURTR*DMIN * * Compute contribution to norm squared from I > NN-2. * NP = NN - 2*PP B1 = Z( NP-2 ) B2 = Z( NP-6 ) GAM = DN2 IF( Z( NP-8 ).GT.B2 .OR. Z( NP-4 ).GT.B1 ) \$ RETURN A2 = ( Z( NP-8 ) / B2 )*( ONE+Z( NP-4 ) / B1 ) * * Approximate contribution to norm squared from I < NN-2. * IF( N0-I0.GT.2 ) THEN B2 = Z( NN-13 ) / Z( NN-15 ) A2 = A2 + B2 DO 30 I4 = NN - 17, 4*I0 - 1 + PP, -4 IF( B2.EQ.ZERO ) \$ GO TO 40 B1 = B2 IF( Z( I4 ) .GT. Z( I4-2 ) ) \$ RETURN B2 = B2*( Z( I4 ) / Z( I4-2 ) ) A2 = A2 + B2 IF( HUNDRD*MAX( B2, B1 ).LT.A2 .OR. CNST1.LT.A2 ) \$ GO TO 40 30 CONTINUE 40 CONTINUE A2 = CNST3*A2 END IF * IF( A2.LT.CNST1 ) \$ S = GAM*( ONE-SQRT( A2 ) ) / ( ONE+A2 ) ELSE * * Case 6, no information to guide us. * IF( TTYPE.EQ.-6 ) THEN G = G + THIRD*( ONE-G ) ELSE IF( TTYPE.EQ.-18 ) THEN G = QURTR*THIRD ELSE G = QURTR END IF S = G*DMIN TTYPE = -6 END IF * ELSE IF( N0IN.EQ.( N0+1 ) ) THEN * * One eigenvalue just deflated. Use DMIN1, DN1 for DMIN and DN. * IF( DMIN1.EQ.DN1 .AND. DMIN2.EQ.DN2 ) THEN * * Cases 7 and 8. * TTYPE = -7 S = THIRD*DMIN1 IF( Z( NN-5 ).GT.Z( NN-7 ) ) \$ RETURN B1 = Z( NN-5 ) / Z( NN-7 ) B2 = B1 IF( B2.EQ.ZERO ) \$ GO TO 60 DO 50 I4 = 4*N0 - 9 + PP, 4*I0 - 1 + PP, -4 A2 = B1 IF( Z( I4 ).GT.Z( I4-2 ) ) \$ RETURN B1 = B1*( Z( I4 ) / Z( I4-2 ) ) B2 = B2 + B1 IF( HUNDRD*MAX( B1, A2 ).LT.B2 ) \$ GO TO 60 50 CONTINUE 60 CONTINUE B2 = SQRT( CNST3*B2 ) A2 = DMIN1 / ( ONE+B2**2 ) GAP2 = HALF*DMIN2 - A2 IF( GAP2.GT.ZERO .AND. GAP2.GT.B2*A2 ) THEN S = MAX( S, A2*( ONE-CNST2*A2*( B2 / GAP2 )*B2 ) ) ELSE S = MAX( S, A2*( ONE-CNST2*B2 ) ) TTYPE = -8 END IF ELSE * * Case 9. * S = QURTR*DMIN1 IF( DMIN1.EQ.DN1 ) \$ S = HALF*DMIN1 TTYPE = -9 END IF * ELSE IF( N0IN.EQ.( N0+2 ) ) THEN * * Two eigenvalues deflated. Use DMIN2, DN2 for DMIN and DN. * * Cases 10 and 11. * IF( DMIN2.EQ.DN2 .AND. TWO*Z( NN-5 ).LT.Z( NN-7 ) ) THEN TTYPE = -10 S = THIRD*DMIN2 IF( Z( NN-5 ).GT.Z( NN-7 ) ) \$ RETURN B1 = Z( NN-5 ) / Z( NN-7 ) B2 = B1 IF( B2.EQ.ZERO ) \$ GO TO 80 DO 70 I4 = 4*N0 - 9 + PP, 4*I0 - 1 + PP, -4 IF( Z( I4 ).GT.Z( I4-2 ) ) \$ RETURN B1 = B1*( Z( I4 ) / Z( I4-2 ) ) B2 = B2 + B1 IF( HUNDRD*B1.LT.B2 ) \$ GO TO 80 70 CONTINUE 80 CONTINUE B2 = SQRT( CNST3*B2 ) A2 = DMIN2 / ( ONE+B2**2 ) GAP2 = Z( NN-7 ) + Z( NN-9 ) - \$ SQRT( Z( NN-11 ) )*SQRT( Z( NN-9 ) ) - A2 IF( GAP2.GT.ZERO .AND. GAP2.GT.B2*A2 ) THEN S = MAX( S, A2*( ONE-CNST2*A2*( B2 / GAP2 )*B2 ) ) ELSE S = MAX( S, A2*( ONE-CNST2*B2 ) ) END IF ELSE S = QURTR*DMIN2 TTYPE = -11 END IF ELSE IF( N0IN.GT.( N0+2 ) ) THEN * * Case 12, more than two eigenvalues deflated. No information. * S = ZERO TTYPE = -12 END IF * TAU = S RETURN * * End of DLAZQ4 * END ```

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MullOverThings Useful tips for everyday # What is a parameter gradient? ## What is a parameter gradient? Gradient descent is used to minimize a cost function J(W) parameterized by a model parameters W. The gradient (or derivative) tells us the incline or slope of the cost function. The value of the gradient G depends on the inputs, the current values of the model parameters, and the cost function. ## What is a gradient in neural networks? An error gradient is the direction and magnitude calculated during the training of a neural network that is used to update the network weights in the right direction and by the right amount. ## What are gradients in machine learning? In machine learning, a gradient is a derivative of a function that has more than one input variable. Known as the slope of a function in mathematical terms, the gradient simply measures the change in all weights with regard to the change in error. ## How do neural networks use gradient descent? Gradient descent is an optimization algorithm which is commonly-used to train machine learning models and neural networks. Training data helps these models learn over time, and the cost function within gradient descent specifically acts as a barometer, gauging its accuracy with each iteration of parameter updates. ## Is gradient descent an activation function? Desirable features of an activation function Vanishing Gradient problem: Neural Networks are trained using the process gradient descent. The gradient descent consists of the backward propagation step which is basically chain rule to get the change in weights in order to reduce the loss after every epoch. ## How to update network parameters with gradient vector? Once the gradient vector is obtained, we’ll update the network parameters by subtracting the corresponding gradient value from their current values, multiplied by a learning rate that allows us to adjust the magnitude of our steps. ## When to zero out gradients in a training loop? So, the default action is to accumulate (i.e. sum) the gradients on every loss.backward () call. Because of this, when you start your training loop, ideally you should zero out the gradients so that you do the parameter update correctly. ## Why are gradients important in a neural network? Thus gradients are key to the power of a neural network. However, gradients often get smaller as the algorithm progresses down to the lower layers. So lower layer weights are unchanged, which leads the training to never converge to a good solution. ## How are error gradients used in backpropagation? Then it propagates the error gradient back to the network from the output layer to the input layer ( backpropagation ). Once the algorithm has computed the gradient of the cost function with regards to each parameter in the network, it uses the gradients to update each parameter with a Gradient descent step.

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#### Howdy, Stranger! It looks like you're new here. If you want to get involved, click one of these buttons! #### Categories Member Posts: 51 Hi, Iv'e just started with Pascal and I'm doing some easy simple programs. Iv'e got some C++ experience, so for this problem I decided to use if/then/else in a way that seems logical to me. Only thing is, when I type a number larger than 32767, the program doesn't produce the correct output. Can somebody plz help me out? [code] begin Write('Enter an integer: '); WriteLn; clrscr; begin if InputInteger <= 32767 then WriteLn('Integer is: ', InputInteger); if InputInteger > 32767 then WriteLn('Invalid integer!, try again'); end end. [/code] • Member Posts: 6,349 : Hi, : : Iv'e just started with Pascal and I'm doing some easy simple programs. Iv'e got some C++ experience, so for this problem I decided to use if/then/else in a way that seems logical to me. : Only thing is, when I type a number larger than 32767, the program doesn't produce the correct output. : Can somebody plz help me out? : : [code] : begin : : Write('Enter an integer: '); : : WriteLn; : clrscr; : : begin : if InputInteger <= 32767 then WriteLn('Integer is: ', InputInteger); : : if InputInteger > 32767 then WriteLn('Invalid integer!, try again'); : : end : end. : [/code] : Did you type-cast the InputInteger as integer or as a word. In TP the standard integer is a 16bit signed integer. This would give it a maximum of 32767 and give you probably an overflow into the sign bit. • Member Posts: 51 : : Hi, : : : : Iv'e just started with Pascal and I'm doing some easy simple programs. Iv'e got some C++ experience, so for this problem I decided to use if/then/else in a way that seems logical to me. : : Only thing is, when I type a number larger than 32767, the program doesn't produce the correct output. : : Can somebody plz help me out? : : : : [code] : : begin : : : : Write('Enter an integer: '); : : : : WriteLn; : : clrscr; : : : : begin : : if InputInteger <= 32767 then WriteLn('Integer is: ', InputInteger); : : : : if InputInteger > 32767 then WriteLn('Invalid integer!, try again'); : : : : end : : end. : : [/code] : : : Did you type-cast the InputInteger as integer or as a word. In TP the standard integer is a 16bit signed integer. This would give it a maximum of 32767 and give you probably an overflow into the sign bit. : Hi, thanks for your help. Iv'e used InputInteger as type integer. When I enter 123456 say for input, it doesn't print the invalid value message, but it gives me a value like -8765 or something. Maybe you tried to tell me this, so should I change InputInteger to another type or something? Thanks • Member Posts: 6,349 : : : Hi, : : : : : : Iv'e just started with Pascal and I'm doing some easy simple programs. Iv'e got some C++ experience, so for this problem I decided to use if/then/else in a way that seems logical to me. : : : Only thing is, when I type a number larger than 32767, the program doesn't produce the correct output. : : : Can somebody plz help me out? : : : : : : [code] : : : begin : : : : : : Write('Enter an integer: '); : : : : : : WriteLn; : : : clrscr; : : : : : : begin : : : if InputInteger <= 32767 then WriteLn('Integer is: ', InputInteger); : : : : : : if InputInteger > 32767 then WriteLn('Invalid integer!, try again'); : : : : : : end : : : end. : : : [/code] : : : : : Did you type-cast the InputInteger as integer or as a word. In TP the standard integer is a 16bit signed integer. This would give it a maximum of 32767 and give you probably an overflow into the sign bit. : : : : Hi, : : thanks for your help. Iv'e used InputInteger as type integer. When I enter 123456 say for input, it doesn't print the invalid value message, but it gives me a value like -8765 or something. Maybe you tried to tell me this, so should I change InputInteger to another type or something? : : Thanks : I tried to tell you that. In Pascal every number variable type (integer or real) has a minimum and a maximum value, based on the memory it occupies. I memory serves me well, you should use the LongInt type (32bits signed integer) for those kinds of numbers. I don't have TP here, but the maximum ranges for integers are in the help files. • Member Posts: 46 [b][red]This message was edited by adibichea at 2004-3-11 12:49:0[/red][/b][hr] if you realy want to see if number is betwin -32768..32767 you could try this: [code] var s: string; r: real; c: integer; begin if s <> '' then begin val(s,r,c); if c = 0 then begin if (r>=-32768)and(r<=32767) then writeln('Number Integer') else writeln('Number Not Integer'); end else writeln('Is not a number! Please input a number '); end; end. [code] good luck!

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Login # In parallelogram ABCD, E is mid-point of CD and through D, a line is drawr parallel to EB to meet CB produced at point G and to cut AB at point F. Prove that: <br> 2 xx AD = GC Updated On: 17-04-2022 Get Answer to any question, just click a photo and upload the photo and get the answer completely free, UPLOAD PHOTO AND GET THE ANSWER NOW! Click here to get PDF DOWNLOAD for all questions and answers of this Book - ICSE Class 9 MATHS Answer Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams. Transcript pahle Deewana the question is in ABC D is the midpoint of CT and fruity a line is drawn parallel to ab to meet C be produced at point to cut a bi at point P prove that 28 is equals to Jise so pasta for what were given a given that ABCD ready will be pushed to BC and CD AB de Villiers to DC because ABCD is a parallelogram are equal to BC and every case study solution now and the two lines of the triangle are having to determine if there is a triangle into the midpoint of the two lines then that I will be parallel to the third line and will be if A and B are the midpoints of this DC and ac line then will be parallel to each and every so what is the midpoint theorem let's crack it by mid point theorem the line passing through to midpoint of two lines of a triangle of a triangle then that line will be perpendicular to parallel to the third line and will be half of its which means we can see in triangle DP that means and e is the midpoint of BC which means be has to be because according to that we can say in this line is parallel then it will be the midpoint so here will use the concept of reverse so there are four rivers which means the terms and conditions will be reversed that means if the line is parallel the points to the midpoints ok ab is parallel to even so is the midpoint given so we can say that b is the midpoint of this so I can say is the midpoint of CD that means therefore b is the midpoint of gcb gcb how far DP will be to in 2GB RBC is equal to as this is the midpoint can say that this part will be half of this part that made this part is twice of this path and one more thing we know that BC is equal to 8 so that implies that is equal to ok so this is too difficult to thank you

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+ # TCS National Qualifier 2 Coding Question. You are given a string, and your task is — print the frequency of each character. ###### Direction for solving the problem 1. Take a line from STDIN. ` aaaabbBcddee ` 2 . Get all the different characters in a given string using set (). ` set = {a, b, B, c, d, e} # unordered set ` 3. Iterate for different characters (len (set)), because we only need to print a character once and it counts in the input line ` range 0 to 5 ie total 6 element ` 4. In each iteration, take the first character and output it and its count. ` now for 0 input_string [0] is `a` and its count is 4 ` 5 Remove all occurrences of the first character, this is will make the next character first. ` remove `a` by replacing all` a` in string by "" new input string will be bbBcddee ` 6. Repeat the same process, go to step 4. 7. Either print the value to STDOUT on each iteration (python3) or print in one go (python2), your output will be the same as ` a4b2B1c1d2e2 ` < p> Examples: ` Input: aaaabbBcddee Output: a4b2B1c1d2e2 Input: aazzZ Output: a2z2Z1 ` ` ` ` # Python2 code here input_string = raw_input () temp_string = "" for _ in range ( len ( set (input_string))):   temp_string + = input_string [ 0 ] + str (input_string.count (input_string [ 0 ]))  input_string = input_string.replace (input_string [ 0 ]," ") print temp_string ` ` ` ` # Python3 code here ` ` input_string ` ` = ` ` input ` ` () ` ` for ` ` _ ` ` in range ( len ( set (input_string))): ``   print (input_string [ 0 ] + str (input_string.count (input_string [ 0 ])), end = "") input_string = input_string.replace (input_string [ 0 ]," ") ` Write your interview experience or email it to [email protected] {{{ error}}}

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# NATURAL NUMBERS – INTRODUCTION, UTILITY IN PROGRAMMING LANGUAGE ## INTRODUCTION: We come across this term while doing maths. Mathematics is an integral part of our education system. Mathematics helps us to find ease in every prospect of our life. Starting from buying a product in the market to programming, maths is always the base of all. Here we can find the logic using basic and simple maths and then with that we can implement it using different languages. So today let us find out the different ways in which we can actually find out the logic behind performing a task using natural numbers in the field of programming. ## NATURAL NUMBER IN ENGLISH: Well as we can see from the name, Natural numbers are those numbers that are already present in the number system and we use it to do basic calculations such addition, subtraction, multiplication or division. We get to know more about these numbers as we go deep down the maths and then slowly, we also use maths in order to perform the example works. Suppose when we want to give examples regarding any derivations or any other further examples, we can use these natural numbers in order to find the proper idea regarding the number. The same goes for the case of the computer where we get to execute different ways of finding the different programs or execute various types of performance. ## NATURAL NUMBERS: Natural numbers are those numbers that start from1 and is extended till infinity. READ  FOR LOOP JAVA - JAVA STATEMENTS Now, let us see how we can either print or implement the natural numbers using java: For (i=1; i<=n;i++) { System.out.println(“the natural numbers are”,n); } This is the program snippet in java to print the natural numbers where n is the limit till where we would want to print the numbers. ## MATHEMATICAL UNDERSTANDING: Mathematics is essential in a number of fields such as natural sciences, engineering, medicine, finances, and social sciences. Applied mathematics has led to entirely new mathematical disciplines, such as statistics and some parts of game theory as well. The 1st step to Indian maths was kept by Aristotle who defined it as a science of quantity. Slowly with the increase in modernization and development, we are getting more and more focused on the technical parts that are also being a way to respond to artificial intelligence and many others. ## UTILITY IN PROGRAMMING LANGUAGE: Any programming language be it java or python or c or c++, all of them basically uses the same snippet just with a different way of broadcasting them. just like the way the different television models have different attire to look at yet they all show the same shows at the same time. Mathematicians have been using probability in a huge way in order to find than the maximum utility of maths in different fields and then from there now the computers have taken up a new way of understanding these theories mathematically so that it becomes easier for us to find the answer and code and decode it respectively. There have been a number of terms such a data structure, DBMS and many others that have used the insight of mathematics but some of the simple logical applications still have the same amount of importance just as mathematics has in our lives. READ  The Swift Programming Language - Introduction, Features ## Algebraic Properties Satisfied by the Natural Numbers The addition (+) and multiplication (×) operations on natural numbers as defined above have several algebraic properties: Closure under addition and multiplication: for all natural numbers a and b, both a + b and a × b are natural numbers. Associativity: for all natural numbers a, b, and c, a + (b + c) = (a + b) + c and a × (b × c) = (a × b) × c. Commutativity: for all natural numbers a and b, a + b = b + a and a × b = b × a. Existence of identity elements: for each number a, a + 0 = a and a × 1 = a. Distributivity of multiplication over addition for all natural numbers a, b, and c, a × (b + c) = (a × b) + (a × c). No nonzero zero divisors: if a and b are natural numbers such a × b = 0, then a = 0 or b = 0 (or both).

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# Math 5050–1, Spring 2009 Homework 6 (Final) ```Math 5050–1, Spring 2009 Homework 6 (Final) Announcements. There will be no lecture on Monday April 27th, and Wednesday April 29th. This assignment is due on Wed. April 29th, and is to be turned in to the Math Department (JWB, ground level) before the end of Wednesday April 29th. I will have the assignments mailed to me in France the following Thursday, early in the morning. Therefore, late assignments cannot possibly be accepted. Please do not put the assignments in my mailbox, or under my office door; if you do, then you can expect your grade to come in after I return, which will be late in May. 1. (Theory question) Let {Xt }t≥0 denote 1-dimensional Brownian motion throughout. You will need the following form of Itô’s formula, discussed in the lectures: Z t 1 f (t , Xt ) = f (0 , X0 ) + Mt + f˙(s , Xs ) + f 00 (s , Xs ) ds, 2 0 where {Mt }t≥0 is a continuous, mean-zero martingale. (a) Let λ &gt; 0 be fixed. Use Itô’s formula to show that λ2 t f (t , Xt ) = exp λXt − 2 defines a martingale, where f (t , x) := exp(λx − 12 λ2 t). (b) Let X0 := 0, and T := min{t &gt; 0 : |Xt | = 1} denote the first time Brownian motion leaves the interval [−1 , 1]. Show that 1 2 E eλXT − 2 λ T = 1. (c) Show also that 1 2 E e−λXT − 2 λ T = 1. (Hint: Redo (b), but start with f (t , x) := exp(−λx − 21 λ2 t).) (d) Conclude that 1 2 E cosh (λXT ) e− 2 λ T = 1. (Hint: 2 cosh θ := eθ + e−θ .) (e) Deduce “Lévy’s formula”: 1 2 E e− 2 λ T = 1 cosh λ for all λ &gt; 0. Equivalently [θ := 12 λ2 ], E e−θT = 1 √ cosh 2θ for all θ &gt; 0. (f) We can compute the moments of T from the preceding; for instance, the first two moments are obtained by setting θ := 0 in the following: d E e−θT = −E T e−θT dθ and d2 E e−θT = E T 2 e−θT . 2 dθ Use this and your answer to (e) to find E(T ) and Var(T ). [You know already that E(T ) = E(XT2 ) = 1, using a different martingale computation.] 2. (Simulation question) Let {Xt }t≥0 denote 1-dimensional Brownian motion. Consider the two curves defined by f (t) := 1 + 2t1/3 and g(t) := −1 + t2/3 . Let T denote the first time that Brownian motion hits either one of the two [it will hit them eventually, so this T is defined fine]. Simulate the probability that Brownian motion hits f before g. More precisely: What is P {XT = f (T )}? 2 ```

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•      Sign In • Create Account # tom_mai78101 Member Since 09 Jan 2007 Offline Last Active Today, 12:17 AM ### In Topic: What is the Quaternion equivalent for multiplying two 3x3 rotation matrices t... 10 September 2016 - 01:28 AM Ah, it seems I was missing a conjugate to achieve rotation multiplication with quaternions. No wonder I was confused about this. I will heed your warning. Is there anything else I am missing in regards to quaternion alternatives to rotation matrix multiplications? ### In Topic: What is the Quaternion equivalent for multiplying two 3x3 rotation matrices t... 09 September 2016 - 11:38 PM The result of composing two rotations depends on the order in which they are applied, so you may have to be a bit careful with the order of operands. In particular, it could be the case that your first operation should be `total = local_rotation * new_rotation', but it's hard to know without being familiar with the exact conventions you are using. The order I'm using is the order going from the left to the right. Same with the example I used in the GIF at the very top. I thought I was consistent enough and therefore there's no need to explain or define the multiplication order for both the Mat3x3 and the quaternions. Sorry. As for the code snippet I mentioned, I was referring to your quote saying: The equivalent of matrix multiplication when using quaternions is... quaternion multiplication! (Surprise!) To me, it sounds like you were referring that, assuming the order of operations is exactly the same (going from left to right, and the placement is result = old * new The pattern of the following operation: Mat3x3 result = Mat3x3 old * Mat3x3 new, where operator* denotes the multiplication, and Mat3x3 is the type. Is exactly the same, literally and logically, as the following operation (just the pattern): Quaternion result = Quaternion old * Quaternion new, where operator* denotes the multiplication, and Quaternion is the type. I would like to seek clarification on this. Thanks again. ### In Topic: What is the Quaternion equivalent for multiplying two 3x3 rotation matrices t... 09 September 2016 - 08:58 PM The equivalent of matrix multiplication when using quaternions is... quaternion multiplication! (Surprise!) So, this is correct? // Generate new local_rotation, matrix-wise // local_rotation = rotation matrix representing the current orientation // new_rotation = rotation matrix representing the difference. (Rotation difference from new_rotation to local_rotation). // total = rotation matrix representing (the final new orientation) total = local_rotation * new_rotation // Generate new local_rotation, quaternion-wise // local_rotation = quaternion representing the current orientiation. // total = quaternion representing (the local_rotation + the difference from the old orientation to the the final new orientation.) total = local_rotation * total ### In Topic: Porting a physics engine: What do these variables stand for? 07 September 2016 - 08:59 PM Were comments added after the thread was started? Because there are comments explaining them right in those areas: // in stands for "incoming" // out stands for "outgoing" // I stands for "incident" // R stands for "reference" // extent, as in the extent of each OBB axis But yeah, those variable names... huh... Yep, just added them in. Unfortunately I didn't come up with the names for a lot of these things. Extent. Incident. Bleh. Also it's much easier to just derive things on paper and copy down into the code the equations, so a lot of abstract symbols come up, like e and whatnot. Apologies if any of that is confusing. Hopefully, some of the issues/pull requests are handled. I noticed there is one which resolved some other confusing parts in your code, but I have no idea if it is better. Thanks again. ### In Topic: Porting a physics engine: What do these variables stand for? 07 September 2016 - 07:53 PM Not sure. Nonetheless, yay! Thanks. PARTNERS

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Econ2_Practice_exam_2 # Econ2_Practice_exam_2 - page 1 of 6 Economics 2 Winter 2009... This preview shows pages 1–3. Sign up to view the full content. page 1 of 6 Economics 2 your name _______________________________ Winter 2009 your TA’s name __________________________ day and time of your discussion section ________ PRACTICE SECOND EXAM DIRECTIONS: No calculators, books, or notes of any kind are allowed. All papers and notebooks must remain closed and on the floor at all times throughout the exam, and students are not allowed to leave the examination room until finished. Answer all questions in the space provided with the exam. HINTS: Feel free to use either of the following formulas if you find them useful. Area of a triangle = (1/2) (base) (height) Area of a trapezoid = (1/2) (base1 + base2) (height) PART I: MULTIPLE CHOICE—circle the correct answer (4 points each, 76 points total) Questions 1 through 3 refer to an advertising game between Alice’s company and Bonnie’s company. Alice advertises Alice doesn’t advertise Bonnie advertises Alice gets 30 Alice gets 40 Bonnie gets 30 Bonnie gets 60 Alice gets 40 Alice gets 50 Bonnie doesn’t advertise Bonnie gets 20 Bonnie gets 50 1.) Which of the following statements is correct? a.) Alice has a dominant strategy and Bonnie has a dominant strategy b.) Alice has a dominant strategy but Bonnie does not have a dominant strategy c.) Alice does not have a dominant strategy but Bonnie does have a dominant strategy d.) Alice does not have a dominant strategy and Bonnie does not have a dominant strategy 2.) How would you characterize the Nash equilibrium for this game? a.) In the Nash equilibrium, Alice and Bonnie will both get the same amount b.) In the Nash equilibrium, Alice will get more than Bonnie c.) In the Nash equilibrium, Alice will get less than Bonnie d.) There is no Nash equilibrium for this game 3.) How would you characterize this game? a.) prisoner’s dilemma b.) ultimatum bargaining game c.) tit-for-tat d.) none of the above This preview has intentionally blurred sections. Sign up to view the full version. View Full Document page 2 of 6 4.) Which of the following is the best characterization of tit-for-tat? a.) if the other person defected last round, then you cooperate this round b.) if the other person defected last round, then you defect this round c.) if the other person defected last round, then you refuse to play this round d.) if the other person cooperated last round, then you defect this round 5.) In game theory, a threat by Player A is said to be credible if a.) the threatened action is within the power of Player A to carry out b.) the threatened action would be in the interests of Player A to carry out c.) the threatened action is exactly what Player B would want Player A to do d.) the threatened action by A is exactly the same action that Player B was planning to do himself 6.) Consider the following version of the Ultimatum Bargaining Game. The researcher gives ten one- dollar bills to Carlos. Carlos is allowed to choose an integer X between 1 and 9 such that Carlos gets \$X and Miguel gets the remaining \$(10 - X). Miguel is then allowed to say either yes, he accepts the offer, in which case Carlos gets to take home X and Miguel gets to take home 10 - X, or no, he This is the end of the preview. Sign up to access the rest of the document. ## This note was uploaded on 09/17/2009 for the course ECON ECON 2 taught by Professor Hamilton during the Spring '09 term at UCSD. ### Page1 / 6 Econ2_Practice_exam_2 - page 1 of 6 Economics 2 Winter 2009... This preview shows document pages 1 - 3. Sign up to view the full document. View Full Document Ask a homework question - tutors are online

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Scale Properties This feature is in the alpha tier. For more information on feature tiers, see API Tiers. 1. Introduction The Scale Properties algorithm is a utility algorithm that is used to pre-process node properties for model training or post-process algorithm results such as PageRank scores. It scales the node properties based on the specified scaler. Multiple properties can be scaled at once and are returned in a list property. The input properties must be numbers or lists of numbers. The lists must all have the same size. The output property will always be a list. The size of the output list is equal to the sum of length of the input properties. That is, if the input properties are two scalar numeric properties and one list property of length three, the output list will have a total length of five. There are a number of supported scalers for the Scale Properties algorithm. These can be configured using the `scaler` configuration parameter. List properties are scaled index-by-index. See the list example for more details. In the following equations, `p` denotes the vector containing all property values for a single property across all nodes in the graph. 1.1. Min-max scaler Scales all property values into the range `[0, 1]` where the minimum value(s) get the scaled value `0` and the maximum value(s) get the scaled value `1`, according to this formula: 1.2. Max scaler Scales all property values into the range `[-1, 1]` where the absolute maximum value(s) get the scaled value `1`, according to this formula: 1.3. Mean scaler Scales all property values into the range `[-1, 1]` where the average value(s) get the scaled value `0`. 1.4. Log scaler Transforms all property values using the natural logarithm. 1.5. Standard Score Scales all property values using the Standard Score (Wikipedia). 1.6. Center Transforms all properties by subtracting the mean. 1.7. L1 Norm Scales all property values into the range `[0.0, 1.0]`. 1.8. L2 Norm Scales all property values using the L2 Norm (Wikipedia). 2. Syntax This section covers the syntax used to execute the Scale Properties algorithm in each of its execution modes. We are describing the named graph variant of the syntax. To learn more about general syntax variants, see Syntax overview. Scale Properties syntax per mode Run Scale Properties in stream mode on a named graph. ``````CALL gds.alpha.scaleProperties.stream( graphName: String, configuration: Map ) YIELD nodeId: Integer, scaledProperty: List of Float`````` Table 1. Parameters Name Type Default Optional Description graphName String `n/a` no The name of a graph stored in the catalog. configuration Map `{}` yes Configuration for algorithm-specifics and/or graph filtering. Table 2. Configuration Name Type Default Optional Description nodeLabels List of String `['*']` yes Filter the named graph using the given node labels. relationshipTypes List of String `['*']` yes Filter the named graph using the given relationship types. concurrency Integer `4` yes The number of concurrent threads used for running the algorithm. jobId String `Generated internally` yes An ID that can be provided to more easily track the algorithm’s progress. logProgress Boolean `true` yes If disabled the progress percentage will not be logged. nodeProperties List of String `n/a` no The names of the node properties that are to be scaled. All property names must exist in the projected graph. scaler String `n/a` no The name of the scaler applied for the properties. Supported values are `MinMax`, `Max`, `Mean`, `Log`, `L1Norm`, `L2Norm` and `StdScore`. Table 3. Results Name Type Description nodeId Integer Node ID. scaledProperty List of Float Scaled values for each input node property. Run Scale Properties in mutate mode on a named graph. ``````CALL gds.alpha.scaleProperties.mutate( graphName: String, configuration: Map ) YIELD preProcessingMillis: Integer, computeMillis: Integer, mutateMillis: Integer, postProcessingMillis: Integer, nodePropertiesWritten: Integer, configuration: Map`````` Table 4. Parameters Name Type Default Optional Description graphName String `n/a` no The name of a graph stored in the catalog. configuration Map `{}` yes Configuration for algorithm-specifics and/or graph filtering. Table 5. Configuration Name Type Default Optional Description mutateProperty String `n/a` no The node property in the GDS graph to which the scaled properties is written. nodeLabels List of String `['*']` yes Filter the named graph using the given node labels. relationshipTypes List of String `['*']` yes Filter the named graph using the given relationship types. concurrency Integer `4` yes The number of concurrent threads used for running the algorithm. jobId String `Generated internally` yes An ID that can be provided to more easily track the algorithm’s progress. nodeProperties List of String `n/a` no The names of the node properties that are to be scaled. All property names must exist in the projected graph. scaler String `n/a` no The name of the scaler applied for the properties. Supported values are `MinMax`, `Max`, `Mean`, `Log`, `L1Norm`, `L2Norm` and `StdScore`. Table 6. Results Name Type Description preProcessingMillis Integer Milliseconds for preprocessing the data. computeMillis Integer Milliseconds for running the algorithm. mutateMillis Integer Milliseconds for adding properties to the projected graph. postProcessingMillis Integer Unused. nodePropertiesWritten Integer Number of node properties written. configuration Map Configuration used for running the algorithm. 3. Examples In this section we will show examples of running the Scale Properties algorithm on a concrete graph. The intention is to illustrate what the results look like and to provide a guide in how to make use of the algorithm in a real setting. We will do this on a small hotel graph of a handful nodes connected in a particular pattern. The example graph looks like this: The following Cypher statement will create the example graph in the Neo4j database: ``````CREATE (:Hotel {avgReview: 4.2, buildYear: 1978, storyCapacity: [32, 32, 0], name: 'East'}), (:Hotel {avgReview: 8.1, buildYear: 1958, storyCapacity: [18, 20, 0], name: 'Plaza'}), (:Hotel {avgReview: 19.0, buildYear: 1999, storyCapacity: [100, 100, 70], name: 'Central'}), (:Hotel {avgReview: -4.12, buildYear: 2005, storyCapacity: [250, 250, 250], name: 'West'}), (:Hotel {avgReview: 0.01, buildYear: 2020, storyCapacity: [1250, 1250, 900], name: 'Polar'}), (:Hotel {avgReview: 3.3, buildYear: 1981, storyCapacity: [240, 240, 0], name: 'Beach'}), (:Hotel {avgReview: 6.7, buildYear: 1984, storyCapacity: [80, 0, 0], name: 'Mountain'}), (:Hotel {avgReview: -1.2, buildYear: 2010, storyCapacity: [55, 20, 0], name: 'Forest'})`````` With the graph in Neo4j we can now project it into the graph catalog to prepare it for algorithm execution. We do this using a native projection targeting the `Hotel` nodes, including their properties. Note that no relationships are necessary to scale the node properties. Thus we use a star projection ('*') for relationships. In the examples below we will use named graphs and native projections as the norm. However, Cypher projections can also be used. The following statement will project a graph using a native projection and store it in the graph catalog under the name 'myGraph'. ``````CALL gds.graph.project( 'myGraph', 'Hotel', '*', { nodeProperties: ['avgReview', 'buildYear', 'storyCapacity'] } )`````` In the following examples we will demonstrate how to scale the node properties of this graph. 3.1. Stream In the `stream` execution mode, the algorithm returns the scaled properties for each node. This allows us to inspect the results directly or post-process them in Cypher without any side effects. Note that the output is always a single list property, containing all scaled node properties in the input order. For more details on the `stream` mode in general, see Stream. The following will run the algorithm in `stream` mode: ``````CALL gds.alpha.scaleProperties.stream('myGraph', { nodeProperties: ['buildYear', 'avgReview'], scaler: 'MinMax' }) YIELD nodeId, scaledProperty RETURN gds.util.asNode(nodeId).name AS name, scaledProperty ORDER BY name ASC`````` Table 7. Results name scaledProperty "Beach" [0.3709677419354839, 0.3209342560553633] "Central" [0.6612903225806451, 1.0] "East" [0.3225806451612903, 0.35986159169550175] "Forest" [0.8387096774193549, 0.12629757785467127] "Mountain" [0.41935483870967744, 0.4679930795847751] "Plaza" [0.0, 0.5285467128027681] "Polar" [1.0, 0.17863321799307957] "West" [0.7580645161290323, 0.0] In the results we can observe that the first element in the resulting `scaledProperty` we get the min-max-scaled values for `buildYear`, where the `Plaza` hotel has the minimum value and is scaled to zero, while the `Polar` hotel has the maximum value and is scaled to one. This can be verified with the example graph. The second value in the `scaledProperty` result are the scaled values of the `avgReview` property. 3.2. Mutate The `mutate` execution mode enables updating the named graph with a new node property containing the scaled properties for that node. The name of the new property is specified using the mandatory configuration parameter `mutateProperty`. The result is a single summary row containing metrics from the computation. The `mutate` mode is especially useful when multiple algorithms are used in conjunction. For more details on the `mutate` mode in general, see Mutate. In this example we will scale the two hotel properties of `buildYear` and `avgReview` using the Mean scaler. The output is a list property which we will call `hotelFeatures`, imagining that we will use this as input for a machine learning model later on. The following will run the algorithm in `mutate` mode: ``````CALL gds.alpha.scaleProperties.mutate('myGraph', { nodeProperties: ['buildYear', 'avgReview'], scaler: 'Mean', mutateProperty: 'hotelFeatures' }) YIELD nodePropertiesWritten`````` Table 8. Results nodePropertiesWritten 8 The result shows that there are now eight new node properties in the in-memory graph. These contain the scaled values from the input properties, where the scaled `buildYear` values are in the first list position and scaled `avgReview` values are in the second position. To find out how to inspect the new schema of the in-memory graph, see Listing graphs in the catalog. 3.3. List properties The `storyCapacity` property models the amount of rooms on each story of the hotel. The property is normalized so that hotels with fewer stories have a zero value. This is because the Scale Properties algorithm requires that all values for the same property have the same length. In this example we will show how to scale the values in these lists using the Scale Properties algorithm. We imagine using the output as feature vector to input in a machine learning algorithm. Additionally, we will include the `avgReview` property in our feature vector. The following will run the algorithm in `mutate` mode: ``````CALL gds.alpha.scaleProperties.stream('myGraph', { nodeProperties: ['avgReview', 'storyCapacity'], scaler: 'StdScore' }) YIELD nodeId, scaledProperty RETURN gds.util.asNode(nodeId).name AS name, scaledProperty AS features ORDER BY name ASC`````` Table 9. Results name features "Beach" [-0.17956547594003253, -0.03401933556831381, 0.00254261210704973, -0.5187592498702616] "Central" [2.172199255871029, -0.3968922482969945, -0.3534230828799124, -0.2806402499298136] "East" [-0.0447509371737933, -0.5731448059080679, -0.526320706159294, -0.5187592498702616] "Forest" [-0.8536381697712284, -0.513529970245499, -0.5568320514438908, -0.5187592498702616] "Mountain" [0.32973389273242665, -0.4487312358296632, -0.6076842935848854, -0.5187592498702616] "Plaza" [0.5394453974799097, -0.609432097180936, -0.5568320514438908, -0.5187592498702616] "Polar" [-0.672387512096618, 2.583849534831454, 2.5705808402272767, 2.542770749364069] "West" [-1.2910364511016934, -0.00809984180197948, 0.027968733177547028, 0.3316657499170525] The resulting feature vector contains the standard-score scaled value for the `avgReview` property in the first list position. We can see that some values are negative and that the maximum value sticks out for the `Central` hotel. The other three list positions are the scaled values for the `storyCapacity` list property. Note that each list item is scaled only with respect to the corresponding item in the other lists. Thus, the `Polar` hotel has the greatest scaled value in all list positions.

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CC-MAIN-2023-14

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Uniswap V3 initializing pools, adding liquidity and tools for calculations - Swiss knife post Warning, this post is only useful if you tried out adding liquidity both with UI and with code, read about V3 most of the info, and want to have more clearer idea of what is what. There are lots of books and posts about how to use Uniswap V3 and I will reference the best at the end of the article, yet some info and code on how to set proper values I didn’t find and had to dig out in various places. So let’s begin with some short snippets on how to get ticks and SqrtPriceX96 values depending on what you want to set and your liquidity plans. I could recode this in some solidity but actually, I use this with chatGPT and Python just to get quick results ### Price to tick formula import math def price_to_tick(p): return math.floor(math.log(p, 1.0001)) price_to_tick(1000000) ### Price to sqrtp q96 = 2**96def price_to_sqrtp(p): return int(math.sqrt(p) * q96) price_to_sqrtp(5000) ### From sqrtp to price def sqrtp_to_price(sqrtp): return (sqrtp / q96) ** 2 ### Sqrtp to tick q96 = 2**96 def sqrtp_to_tick(sqrtp): return math.floor(2 * math.log(sqrtp / q96, 1.0001)) So with some practical examples Price = y/x  or token1/token0 so if we want to add 1 eth and 1 000 000 of meme token then it is 1/1000000 or  0.00000001 and if we want to find a tick value for that price we need to put that into the formula above which will become −138,164  and if we convert that to sqrtp value it will be 79224253767016489810214999 when we put it into formula. This 1 / 1000 000 will also give us starting liquidity values, so we put 1 eth and 1 million meme tokens with that starting price, we want to put 2 eth, then calculate 2 / 1000000 and this value will be used in the createAndInitializePoolIfNecessary of NonfungiblePositionManager contract pool = nfpm.createAndInitializePoolIfNecessary( token0, token1, fee, sqrtPriceX96 ); also, we can just use this solidity function to calculate sqrtPriceX96 from amounts of tokens we want to add function calculateSqrtPriceX96(uint256 tokenAmount, uint256 wethAmount ) Code for that you can find here // SPDX-License-Identifier: MIT pragma solidity ^0.8.0; library SqrtPricex96 { uint256 constant Q96 = 2 ** 96; function calculateSqrtPriceX96( uint256 tokenAmount, uint256 wethAmount ) internal pure returns (uint160 sqrtPriceX96) { // Ensure non-zero values to prevent division by zero require( tokenAmount > 0 && wethAmount > 0, "Amounts must be greater than zero" ); // First, multiply wethAmount by Q96^2 to scale it up before division // Then, divide by tokenAmount and finally calculate the square root // The multiplication by Q96^2 is done first to avoid loss of precision before the division uint256 scaledWethAmount = wethAmount * Q96 ** 2; uint256 priceRatio = scaledWethAmount / tokenAmount; // Now compute the square root sqrtPriceX96 = uint160(sqrt(priceRatio)); } // Basic integer square root function function sqrt(uint256 y) internal pure returns (uint256 z) { if (y > 3) { z = y; uint256 x = y / 2 + 1; while (x < z) { z = x; x = (y / x + x) / 2; } } else if (y != 0) { z = 1; } else { z = 0; } return z; } // This function is to illustrate how you would reverse the calculation to find the price function calculatePriceFromSqrtPriceX96( uint256 sqrtPriceX96 ) internal pure returns (uint256 price) { uint256 squaredPrice = uint256(sqrtPriceX96) ** 2; uint256 scaledPrice = squaredPrice / Q96 ** 2; price = uint160(scaledPrice); } } After that when doing the minting of V3 NFT and adding first liquidity w nfpm.mint(INonfungiblePositionManager.MintParams({.... we will probably use this parameter to get the full range of ticks and depending on the fee pool type we choose to set these settings int24 MIN_TICK = -887272; int24 MAX_TICK = 887272; uint24 fee = 3000; int24 TICK_SPACING = 60; int24 minTick = (MIN_TICK / TICK_SPACING) * TICK_SPACING; int24 maxTick = (MAX_TICK / TICK_SPACING) * TICK_SPACING; tick spacing and fees are correlated with this table ### Some other useful info Uniswap pool will be created with multicall if you use their UI, to read that multicall use this The first line is the init function The third line is returning of extra eth

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# Solve problems with mixed operations II Lesson When we want to solve a problem that is given to us in words, it can help to write it out as a number problem.  Finding the numbers we need, and how we treat those numbers, helps us solve the problem more easily. Written problems give us clues as to the types of mathematical operations we may need.  For example: • if we need to add, the problem might use words like more or extra • if we need to subtract, the problem might use words like less or gave away • if we need to multiply the problem might say groups of • if we need to divide the problem might say share. There are many other types of words that occur in problems. As you watch this video, take note of the words and what operation they connect to.  Can you think of others? #### Worked examples ##### Question 1 Laura has saved $£60$£60 over the course of the last month. If she earns $£10$£10 this week from doing chores, and finds another $£20$£20 while out for a walk, how much does she have? ##### question 2 Write a division number sentence to match the word problem. You do not need to solve the question. 1. A lecture hall has $391$391 seats. Each row has $17$17 seats. How many rows of seats are there? ##### question 3 Miss Oakley has $37$37 chocolates to give out to $3$3 groups of students on a school excursion. She firstly takes away $4$4 chocolates to save for students who are absent, then she divides the remaining chocolates equally among the $3$3 groups. 1. If $N$N stands for the number of chocolates each group receives, find $N$N.

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http://forum.allaboutcircuits.com/threads/simple-led-timer-circuit.9847/

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# Simple LED Timer Circuit Discussion in 'The Projects Forum' started by muhammadk, Mar 2, 2008. Mar 2, 2008 5 1 Hello, This is my first post here and at initial glances, this website seems like an amazing resource. I need help with a very simple LED timer circuit. I am making a collection box for my GCSE Design & Technology course. I want to make it so that when money is inserted into the hole, it will press down a copper contact that will complete the circuit and light up six blue LEDs. Because the money falls into the box right away, the circuit will be broken in less than a second as the copper contact will spring back into place. However, I want the LEDs to remain turned on so that they illuminate the 'thank you' sign for at least 10 seconds in order for the donor to notice it. I do not know where to take it on from here. I want to use a 9 or 12 V battery as the power supply and 6 blue LEDs (that I am told require 5mA). I initially thought I could use a capacitor to store the energy and release it to the LEDs for 10 seconds. Do you'll think this is viable? If so would it be possible to point me in the right direction in terms of calculating the measurements because I haven't had any luck? (I've been told I may also need a resistor to protect the capacitor.) Alternatively, I was told that I could use a time-delay circuit but I have no idea where to begin with that. Would you advice me to take this option, if so can you guide me to any good resources? Furthermore, are there any other routes I can use to complete this circuit? Thank You So Much, 2. ### gee_emm Active Member Jan 16, 2008 34 0 Mar 2, 2008 5 1 Hmm... The circuit seems quite complex at a first look because the timer has 8 pins. I would prefer something a little bit simpler, but I'll give it a shot. Thanks! Just to clarify, if I used the formula R=T/1.1*C and I wanted 10 seconds using a 2200 microfarad capacitor, I would do: 10/(1.1*2200)=4.1*10(to the power of -3). Multiply by 1000 to get 4.1 K Ohms. Would this work with a 9 V battery and LEDs that require 30mA? Sorry, I am little but confused here. jumail likes this. 4. ### thingmaker3 Retired Moderator May 16, 2005 5,072 6 2200 uF is an awfully large capacitor for a timing circuit. It will draw a lot of current and your battery will not last as long. Better to use a 22uF capacitor and a 410 Kohm resistor. Do you know the forward voltage drop for your LEDs? Mar 2, 2008 5 1 Unfortunately I am not too sure. I will try and ask the electronic shop where I purchased them. Thanks. Mar 2, 2008 5 1 I couldn't find out the forward voltage drop because the store keepers did not know it, is it really essential because it vary by a lot? I also just wanted to check one more thing. For the monostable 555 timer circuit, what do the values of the other resistors and capacitors do because they vary from diagram to diagram (-I am guessing that they vary by the input voltage by the battery)? I started following this diagram: http://www.eleinmec.com/article.asp?4* . Do you think the values for the other capacitor (220 uF & 0.01 uF) and resistor (10 kohms) will fit my needs? Or do I need to change those as well? Thanks, Edit: *Note: Referring to figure 2 in the link. 7. ### thingmaker3 Retired Moderator May 16, 2005 5,072 6 We use the LED forward voltage drop and the current specification to select the correct current limiting resistor. We could have also compared the forward voltage drop with 1/6 of 9V to know if we could string the LEDs in series or not. Blue LEDs at 5mA will drop anywhere from 2.9V to 3.6V. If you have a good voltmeter, you can test yours by putting one in series with a 120 Ohm resistor and measuring across the LED. This will put us in the right "ball park." Those values will work just fine. 8. ### SgtWookie Expert Jul 17, 2007 22,182 1,728 I'm afraid that using a 120 Ohm resistor in series with a blue LED (with a Vf of 3v) across 9V will rapidly result in a burned-out LED, as it will be passing 50mA instead of 5mA. I suggest that for a 9v supply, a 510 Ohm resistor in series with the LED would be a much better value for current limiting purposes during test. It should safely pass sufficient current to light up most diodes, from a standard red LED (resulting in 14.3mA @ Vf=1.7v) to a white LED (9.8mA @ 4.0V). The use of a simple current limiting resistor won't ensure that the current remains constant when the LED is changed. A better method would be to use/construct a constant current supply. Have a look at the attached schematic. It's a constant current supply that is easy to build from a few components. Note that you can use other PNP transistors than a 2N2907; a 2N3906, 2N4403, etc will work as well. D1 and D2 can be replaced with a single standard red LED. R1 controls the current; increasing R1 will decrease the current through Dtest. As shown, the circuit produces a constant 11.94mA source, regardless of the Vf of Dtest (in the range of 1.2v to 5v). However, semiconductors do vary somewhat in their Vf; so construction of the circuit will require adjustments to R1 to obtain the desired current. It's better to start out with a higher value (say, 100 Ohms) and work down. Values below 50 Ohms will likely result in current flow high enough to rapidly burn out standard LEDs. • ###### LEDconstantcurrent.PNG File size: 5.7 KB Views: 84 Mar 2, 2008 5 1 Cool, thanks! I'll try the monostable 555 circuit timer first, if it doesn't work then I'll give a go at your method SgtWookie. 10. ### thingmaker3 Retired Moderator May 16, 2005 5,072 6 Has anyone seen a decimal point laying around? I seem to have misplaced one.

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CC-MAIN-2016-50

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https://speakerdeck.com/natmchugh/hack-the-hash

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Nathaniel McHugh June 05, 2015 8.7k # Hack The Hash Short 15 minute over view of hash collision attacks June 05, 2015 ## Transcript 1. ### Hack the Hash Hack the Hash / Nathaniel McHugh @natmchugh Hash Functions http://localhost:8000/index.html?print-pdf#/ 1 of 11 05/06/15 12:10 4. ### Collisions Collisions When H(m1) = H(m2) and m1≠m2 Forge Signatures, distribute �les di�erent behaviors, predict future not HMAC not pre-image Plenty in MD4, MD5, SHA0 None in full SHA1, SHA2 Hash Functions http://localhost:8000/index.html?print-pdf#/ 4 of 11 05/06/15 12:10 5. ### Brute Force Brute Force n ≈ √(-2 * ln(1-p) * √d If p=0.5 then n= 1.177 * √d √365 = 19 √(2^128) = 2^64 Hash Functions http://localhost:8000/index.html?print-pdf#/ 5 of 11 05/06/15 12:10 6. ### Wang Attack Wang Attack Start with random message 1. Create another message M’ with small di�s 2. Modify message so that certain bitwise conditions hold in intermediate state 3. Test for collision if not found go to 1 4. Hash Functions http://localhost:8000/index.html?print-pdf#/ 6 of 11 05/06/15 12:10 7. ### Δm1 = 2 , Δm2 = 2 − 2 , Δm12 = −2 Wang MD4 Wang MD4 M = M − M’ = (Δm0, Δm1, ......, Δm15) 31 31 28 16 Hash Functions http://localhost:8000/index.html?print-pdf#/ 7 of 11 05/06/15 12:10 9. ### Live Demo Live Demo Hash Functions http://localhost:8000/index.html?print-pdf#/ 9 of 11 05/06/15 12:10 12:10 11. ### Collision attack in Wild Collision attack in Wild Hash Functions http://localhost:8000/index.html?print-pdf#/ 11 of 11 05/06/15 12:10

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Re: creating exact numbers, strange behavior • To: mathgroup at smc.vnet.net • Subject: [mg53054] Re: creating exact numbers, strange behavior • From: "Maxim A. Dubinnyi" <maxim at nmr.ru> • Date: Wed, 22 Dec 2004 04:52:44 -0500 (EST) • References: <200412180859.DAA02167@smc.vnet.net> • Sender: owner-wri-mathgroup at wolfram.com ```Quotient[7*10^100, 10^100] 7 7.*10^100 is not exact integer number 7*10^100 is. Matthias Gottschalk wrote: >Hi, > >I try to convert machine number into an exact number. I do this the >following way: > >In[] = Quotient[7.*10^30,10^30] > >Out[] = 7 > >that is fine. > > >In[] = Quotient[7.*10^99, 10^99] > >Out[] = 7 > >still fine. > > >In[] = Quotient[7.*10^100,10^100] > >Out[] = 6 > >????????, I think that is bad. >Or am I doing something wrong? > > > > ``` • Prev by Date: Re: Questions about "collecting" complex exponentials and simplifying expressions • Next by Date: Why doesn't DSolve solve this? • Previous by thread: Re: creating exact numbers, strange behavior • Next by thread: Re : simply problem (not for me) about axes scale

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## SELECT within SELECT This tutorial looks at how we can use SELECT statements within SELECT statements to perform more complex queries. ## Exercises 1a. List each country name where the population is larger than 'Russia'. `bbc(name, region, area, population, gdp)` SELECT name FROM bbc WHERE population> (SELECT population FROM bbc WHERE name='Russia') Results 1b. List the `name` and `region` of countries in the regions containing 'India', 'Iran'. Results 1c. Show the countries in Europe with a per capita GDP greater than 'United Kingdom'. Per Capita GDP Results 1d. Which country has a population that is more than Canada but less than Algeria? Results To gain an absurdly detailed view of one insignificant feature of the language, read on. We can use the word `ALL` to allow >= or > or < or <=to act over a list. 2a. Which countries have a GDP greater than any country in Europe? [Give the name only.] Results We can refer to values in the outer SELECT within the inner SELECT. We can name the tables so that we can tell the difference between the inner and outer versions. 3a. Find the largest country in each region, show the region, the name and the population: SELECT region, name, population FROM bbc x WHERE population >= ALL (SELECT population FROM bbc y WHERE y.region=x.region AND population>0) Results The following questions are very difficult: 3b. Find each country that belongs to a region where all populations are less than 25000000. Show name, region and population. Results 3c. Some countries have populations more than three times that of any of their neighbours (in the same region). Give the countries and regions. Results

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# R/mix.R In mix: Estimation/Multiple Imputation for Mixed Categorical and Continuous Data #### Documented in dabipf.mixda.mixecm.mixem.mixgetparam.miximp.mixloglik.mixmi.inferenceprelim.mixrngseed ```#************************************************************************* # Initializes random number generator seed. Argument should be a # positive integer rngseed <- function(seed) { seed <- as.integer(seed) if(seed<=0)stop("'seed' must be a positive integer") .Fortran("rngs", seed, PACKAGE = "mix") invisible() } #************************************************************************* # Changes NA's to single precision missing value code .na.to.snglcode <- function(x,mvcode) { x[is.na(x)] <- mvcode x } #************************************************************************* # Changes NA's to integer missing value code .na.to.icode <- function(x,mvcode) { x[is.na(x)] <- as.integer(mvcode) x } #************************************************************************* # Changes missing value code to NA .code.to.na <- function(x,mvcode) { x[x==mvcode] <- NA x } #************************************************************************* # Perform preliminary manipulations on matrix of mixed data. # The categorical variables must be in columns 1:p of x, # and must be coded as positive integers. The other columns are # assumed to be continuous. prelim.mix <- function(x,p) { x <- as.matrix(x) # get dimensions of x, separate into w and z n <- nrow(x); p <- as.integer(p); q <- as.integer(ncol(x)-p) w <- x[,(1:p)]; storage.mode(w) <- "integer" z <- x[,(p+1):(p+q)]; storage.mode(z) <- "double" # store names if(!is.null(dimnames(x)[[2]])){ wnames <- dimnames(x)[[2]][1:p] znames <- dimnames(x)[[2]][(p+1):(p+q)] } else{ wnames <- NULL znames <- NULL } rnames <- dimnames(x)[[1]] w <- matrix(w,n,p); z <- matrix(z,n,q) # get dimensions of contingency table d <- integer(p) for(j in 1:p) d[j] <- as.integer(max(w[!is.na(w[,j]),j])) # missingness indicators for z rz <- 1*is.na(z) nmisz <- as.integer(apply(rz,2,sum)) mdpz <- as.integer((rz%*%(2^((1:q)-1)))+1) rz <- 1-rz; storage.mode(rz) <- "integer" # get missing data patterns for w rw <- 1*is.na(w) nmisw <- as.integer(apply(rw,2,sum)) nmis <- c(nmisw,nmisz) names(nmis) <- c(wnames,znames) mdpw <- as.integer((rw%*%(2^((1:p)-1)))+1) rw <- 1-rw; storage.mode(rw) <- "integer" # calculate the known part of the cell number for rows of w cumd <- as.integer(round(exp(cumsum(log(d))))) mobs <- 1+((.na.to.icode(w,1)-1)*rw)%*%(cumd/d) # do row sort ro <- order(mdpz,mdpw,mobs) w <- matrix(w[ro,],n,p); mdpw <- mdpw[ro]; mobs <- mobs[ro]; rw <- matrix(rw[ro,],n,p) z <- matrix(z[ro,],n,q); mdpz <- mdpz[ro]; rz <- matrix(rz[ro,],n,q) ro <- order(ro) # compress missing data patterns mdpzst <- as.integer(seq(along=mdpz)[!duplicated(mdpz)]) mdpz <- unique(mdpz); npattz <- length(mdpz) mdpzfin <- as.integer(c(mdpzst[2:npattz]-1,n)) if(npattz==1)mdpzfin <- n mdpzgrp <- numeric(); mdpwst <- numeric(); mdpwtmp <- numeric() for(i in 1:npattz){ tmp <- mdpw[mdpzst[i]:mdpzfin[i]] mdpwst <- c(mdpwst, seq(along=tmp)[!duplicated(tmp)]+mdpzst[i]-1) tmp <- unique(tmp) mdpzgrp <- c(mdpzgrp,length(tmp)) mdpwtmp <- c(mdpwtmp,tmp)} mdpw <- as.integer(mdpwtmp); npattw <- length(mdpw) storage.mode(mdpzgrp) <- "integer" storage.mode(mdpwst) <- "integer" mdpwfin <- as.integer(c(mdpwst[2:npattw]-1,n)) mdpwgrp <- numeric(); mobsst <- numeric(); mobstmp <- numeric() for(i in 1:npattw){ tmp <- mobs[mdpwst[i]:mdpwfin[i]] mobsst <- c(mobsst, seq(along=tmp)[!duplicated(tmp)]+mdpwst[i]-1) tmp <- unique(tmp) mdpwgrp <- c(mdpwgrp,length(tmp)) mobstmp <- c(mobstmp,tmp)} mobs <- as.integer(mobstmp); ngrp <- length(mobs) storage.mode(mdpwgrp) <- "integer" storage.mode(mobsst) <- "integer" # create r-matrix for display purposes r <- cbind(rw,rz)[mdpwst,] tmp <- as.character(c(mdpwst[2:npattw],n+1)-mdpwst) dimnames(r) <- list(tmp,c(wnames,znames)) ncells <- cumd[p]; jmp <- as.integer(cumd/d) rz <- matrix(rz[mdpzst,],npattz,ncol(rz)); rw <- rw[mdpwst,] # form matrix of packed storage indices npsi <- as.integer(q*(q+1)/2) psi <- .Fortran("mkpsi",as.integer(q-1),matrix(as.integer(0),q,q), PACKAGE="mix")[[2]] # center and scale the columns of z mvcode <- max(z[!is.na(z)])+1000 z <- .na.to.snglcode(z,mvcode) tmp <- .Fortran("ctrsc",z,n,q,numeric(q),numeric(q),mvcode, PACKAGE="mix") z <- tmp[[1]]; xbar <- tmp[[4]]; sdv <- tmp[[5]] z <- .code.to.na(z,mvcode) # return list nmobs <- as.integer(c(mobsst[-1],n+1)-mobsst) list(w=w,n=n,p=p,d=d,jmp=jmp,z=z,q=q, r=r,rz=rz,rw=rw,nmis=nmis,ro=ro, mdpzgrp=mdpzgrp,mdpwgrp=mdpwgrp, mobs=mobs,mobsst=mobsst,nmobs=nmobs,ncells=ncells,ngrp=ngrp, npattz=npattz,npattw=npattw,rnames=rnames,npsi=npsi,psi=psi, xbar=xbar,sdv=sdv,wnames=wnames,znames=znames,rnames=rnames) } #************************************************************************* em.mix <- function(s,start,prior=1,maxits=1000,showits=TRUE,eps=.0001) { if(length(prior)==1) prior <- rep(prior,s\$ncells) prior <- as.double(prior) w <- !is.na(prior) prior[!w] <- -999.0 s\$z <- .na.to.snglcode(s\$z,999) tp <- integer(s\$p); tq <- integer(s\$q) sigma <- numeric(s\$npsi); pii <- numeric(s\$ncells) mu <- matrix(0,s\$q,s\$ncells) tmp <- .Fortran("tobsm",s\$q,s\$psi,s\$npsi,sigma,s\$ncells,mu,pii, s\$npattz,s\$rz,s\$mdpzgrp,s\$npattw,s\$p,s\$rw,s\$mdpwgrp, s\$ngrp,s\$mobs,s\$mobsst,s\$nmobs,s\$n,s\$z,tp,tq, PACKAGE="mix") kn1 <- tmp[[4]];kn2 <- tmp[[6]];kn3 <- tmp[[7]] if(missing(start)){ tmp <- .Fortran("stvlm",s\$q,s\$psi,s\$npsi,sigma,s\$ncells,mu, PACKAGE="mix") sigma <- tmp[[4]];mu <- tmp[[6]] pii <- rep(1,s\$ncells) pii[!w] <- 0} if(!missing(start)){ sigma <- start\$sigma; mu <- start\$mu; pii <- start\$pi if(any(pii[w]==0)){ warning("Starting value on the boundary")} if(any(!w)){ if(any(pii[!w]!=0)){ stop("starting value has nonzero elements for structural zeros")}}} converged <- FALSE; it <- 0 if(showits) cat(paste("Steps of EM:","\n")) t1 <- sigma; t2 <- mu; t3 <- pii while((!converged)&(it<maxits)){ it <- it+1 if(showits) cat(paste(format(it),"...",sep="")) if(it>1){sigma <- t1; mu <- t2; pii <- t3} tmp <- .Fortran("estepm",s\$q,s\$psi,s\$npsi,s\$ncells,sigma,mu,pii,kn1, kn2,kn3,t1,t2,t3,s\$npattz,s\$rz,tq,tq,s\$mdpzgrp,s\$npattw, s\$p,s\$rw,tp,s\$mdpwgrp,s\$ngrp, s\$mobs,s\$mobsst,s\$nmobs,s\$n,s\$z,s\$d,s\$jmp,tp,kn3, PACKAGE="mix") t1 <- tmp[[11]];t2 <- tmp[[12]];t3 <- tmp[[13]] tmp <- .Fortran("mstepm",s\$q,s\$psi,s\$npsi,s\$ncells,t1,t2,t3,s\$n,prior, PACKAGE="mix") t1 <- tmp[[5]]; t2 <- tmp[[6]]; t3 <- tmp[[7]] if(any(t3<0)) stop("Estimate outside the parameter space. Check prior.") t3[w][t3[w]<(.0000001/sum(w))] <- 0 c3 <- all(abs(t3-pii)<=(eps*abs(pii))) c2 <- all(abs(t2-mu)<=(eps*abs(mu))) c1 <- all(abs(t1-sigma)<=(eps*abs(sigma))) converged <- c3&c2&c1} if(showits) cat("\n") list(sigma=t1,mu=t2,pi=t3) } #************************************************************************* loglik.mix <- function(s,theta) { s\$z <- .na.to.snglcode(s\$z,999) tp <- integer(s\$p); tq <- integer(s\$q) ll <- .Fortran("lobsm",s\$q,s\$psi,s\$npsi,s\$ncells,theta\$sigma, theta\$mu,theta\$pi,s\$npattz,s\$rz,tq,tq,s\$mdpzgrp,s\$npattw,s\$p, s\$rw,tp,s\$mdpwgrp,s\$ngrp,s\$mobs,s\$mobsst,s\$nmobs, s\$n,s\$z,s\$d,s\$jmp,integer(s\$p),ll=numeric(1), PACKAGE="mix")\$ll ll } #************************************************************************* # Retrieves list of parameters from theta. If corr=F, returns a list # containing an array of cell probabilities, a matrix of cell means, and # a covariance matrix. If corr=T, returns a list containing an array of # cell probabilities, a matrix of cell means, a vector of standard # deviations, and a correlation matrix. getparam.mix <- function(s,theta,corr=FALSE) { pii <- array(theta\$pi,s\$d) if(!is.null(s\$wnames)){ pinames <- as.list(1:s\$p) for(i in 1:s\$p){ pinames[[i]] <- paste(s\$wnames[i],"=",format(1:s\$d[i]),sep="")} dimnames(pii) <- pinames} mu <- theta\$mu*s\$sdv + s\$xbar dimnames(mu) <- list(s\$znames,NULL) sigma <- theta\$sigma[s\$psi] sigma <- matrix(sigma,s\$q,s\$q) tmp <- matrix(s\$sdv,s\$q,s\$q) sigma <- sigma*tmp*t(tmp) dimnames(sigma) <- list(s\$znames,s\$znames) mu[,pii==0] <- NA if(corr){ sdv <- sqrt(diag(sigma)) names(sdv) <- s\$znames tmp <- matrix(sdv,s\$q,s\$q) r <- sigma/(tmp*t(tmp)); dimnames(r) <- list(s\$znames,s\$znames) result <- list(pi=pii,mu=mu,sdv=sdv,r=r)} else result <- list(pi=pii,mu=mu,sigma=sigma) result } #************************************************************************* da.mix <- function(s,start,steps=1,prior=.5,showits=FALSE) { if(length(prior)==1) prior <- rep(prior,s\$ncells) prior <- as.double(prior) w <- !is.na(prior) prior[!w] <- -999.0 s\$z <- .na.to.snglcode(s\$z,999) s\$w <- .na.to.icode(s\$w,999) tp <- integer(s\$p); tq <- integer(s\$q) tmp <- .Fortran("tobsm",s\$q,s\$psi,s\$npsi,numeric(s\$npsi),s\$ncells, matrix(0,s\$q,s\$ncells),numeric(s\$ncells), s\$npattz,s\$rz,s\$mdpzgrp,s\$npattw,s\$p,s\$rw,s\$mdpwgrp, s\$ngrp,s\$mobs,s\$mobsst,s\$nmobs,s\$n,s\$z,tp,tq, PACKAGE="mix") kn1 <- tmp[[4]];kn2 <- tmp[[6]];kn3 <- tmp[[7]] sigma <- start\$sigma; pii <- start\$pi; mu <- start\$mu t1 <- sigma; t2 <- mu; t3 <- pii if(showits) cat(paste("Steps of Data Augmentation:","\n")) for(i in 1:steps){ if(showits) cat(paste(format(i),"...",sep="")) tmp <- .Fortran("istepm",s\$q,s\$psi,s\$npsi,s\$ncells,sigma,mu,pii,kn1, kn2,kn3,t1=t1,t2=t2,t3=t3,s\$npattz,s\$rz,tq,tq,s\$mdpzgrp, s\$npattw,s\$p,s\$rw,tp,s\$mdpwgrp,s\$ngrp, s\$mobs,s\$mobsst,s\$nmobs,s\$n,s\$z,s\$d,s\$jmp,tp,kn3,kn1,s\$w, numeric(s\$q), PACKAGE="mix") t1 <- tmp\$t1; t2 <- tmp\$t2; t3 <- tmp\$t3 tmp <- .Fortran("pstepm",s\$q,s\$psi,s\$npsi,s\$ncells,sigma=t1,mu=t2,pi=t3, s\$n,s\$p,as.double(prior),numeric(s\$npsi),matrix(0,s\$q,s\$q), numeric(s\$q),tq,err=numeric(1), PACKAGE="mix") {if(tmp\$err==1) stop("Improper posterior--empty cells") else{ sigma <- tmp\$sigma; mu <- tmp\$mu; pii <- tmp\$pi}}} if(showits) cat("\n") mu[,!w] <- NA list(sigma=sigma,mu=mu,pi=pii) } #************************************************************************* imp.mix <- function(s,theta,x) { sigma <- theta\$sigma mu <- theta\$mu pii <- theta\$pi z <- .na.to.snglcode(s\$z,999) w <- .na.to.icode(s\$w,999) tp <- integer(s\$p); tq <- integer(s\$q) tmp <- .Fortran("istepm",s\$q,s\$psi,s\$npsi,s\$ncells,sigma,mu,pii, sigma,mu,pii,sigma,mu,pii,s\$npattz,s\$rz,tq,tq,s\$mdpzgrp, s\$npattw,s\$p,s\$rw,tp,s\$mdpwgrp,s\$ngrp, s\$mobs,s\$mobsst,s\$nmobs,s\$n,z=z,s\$d,s\$jmp,tp,pii,sigma,w=w, numeric(s\$q), PACKAGE="mix") w <- tmp\$w[s\$ro,] z <- tmp\$z*matrix(s\$sdv,s\$n,s\$q,TRUE)+matrix(s\$xbar,s\$n,s\$q,TRUE) z <- z[s\$ro,] if(!missing(x)){ zorig <- x[,(s\$p+1):(s\$p+s\$q)] z[!is.na(zorig)] <- zorig[!is.na(zorig)]} ximp <- cbind(w,z) dimnames(ximp) <- list(s\$rnames,c(s\$wnames,s\$znames)) ximp } #************************************************************************* ecm.mix <- function(s,margins,design,start,prior=1,maxits=1000, showits=TRUE, eps=.0001) { if(length(prior)==1) prior <- rep(prior,s\$ncells) prior <- as.double(prior) w <- !is.na(prior) prior[!w] <- -999.0 storage.mode(design) <- "double"; storage.mode(margins) <- "integer" s\$z <- .na.to.snglcode(s\$z,999) tp <- integer(s\$p); tq <- integer(s\$q) sigma <- numeric(s\$npsi); pii <- numeric(s\$ncells) mu <- matrix(0,s\$q,s\$ncells) tmp <- .Fortran("tobsm",s\$q,s\$psi,s\$npsi,sigma,s\$ncells,mu,pii, s\$npattz,s\$rz,s\$mdpzgrp,s\$npattw,s\$p,s\$rw,s\$mdpwgrp, s\$ngrp,s\$mobs,s\$mobsst,s\$nmobs,s\$n,s\$z,tp,tq, PACKAGE="mix") kn1 <- tmp[[4]];kn2 <- tmp[[6]];kn3 <- tmp[[7]] if(missing(start)){ tmp <- .Fortran("stvlm",s\$q,s\$psi,s\$npsi,sigma,s\$ncells,mu, PACKAGE="mix") sigma <- tmp[[4]];mu <- tmp[[6]] pii <- rep(1,s\$ncells) pii[!w] <- 0} if(!missing(start)){ sigma <- start\$sigma; mu <- start\$mu; pii <- start\$pi if(any(pii[w]==0)){ warning("Starting value on the boundary")} if(any(!w)){ if(any(pii[!w]!=0)){ stop("Starting value has nonzero elements for structural zeros")}}} eps1 <- .0000001*s\$n/sum(w) r <- ncol(design); npsir <- as.integer(r*(r+1)/2); wk <- numeric(npsir) tr <- integer(r); wkr <- numeric(r); wkd <- numeric(s\$ncells) beta <- matrix(0,r,s\$q) psir <- .Fortran("mkpsi",as.integer(r-1),matrix(as.integer(0),r,r), PACKAGE="mix")[[2]] converged <- FALSE; it <- 0 if(showits) cat(paste("Steps of ECM:","\n")) t1 <- sigma; t2 <- mu; t3 <- pii while((!converged) && (it < maxits) ) { it <- it+1 if(showits) cat(paste(format(it),"...",sep="")) if(it>1){sigma <- t1; mu <- t2; pii <- t3} tmp <- .Fortran("estepm",s\$q,s\$psi,s\$npsi,s\$ncells,sigma,mu,pii,kn1, kn2,kn3,t1,t2,t3,s\$npattz,s\$rz,tq,tq,s\$mdpzgrp,s\$npattw, s\$p,s\$rw,tp,s\$mdpwgrp,s\$ngrp, s\$mobs,s\$mobsst,s\$nmobs,s\$n,s\$z,s\$d,s\$jmp,tp,kn3, PACKAGE="mix") t1 <- tmp[[11]];t2 <- tmp[[12]];t3 <- tmp[[13]] tmp <- .Fortran("mstepcm",s\$q,s\$psi,s\$npsi,s\$ncells,t1,t2,t3,sigma=t1, mu=t2,s\$n,r,design,wk,tr,psir,npsir,wkr,wkd,beta, PACKAGE="mix") t1 <- tmp\$sigma; t2 <- tmp\$mu t3[w] <- t3[w]+prior[w]-1 if(any(t3[w]<0)) stop("Estimate outside the parameter space. Check prior.") t3 <- .Fortran("ipf",s\$ncells,t3,pii,length(margins),margins, s\$p,s\$d,s\$jmp,tp,tp,tp,eps1, PACKAGE="mix")[[3]] t3[w] <- t3[w]/sum(t3[w]) c3 <- all(abs(t3-pii)<=(eps*abs(pii))) c2 <- all(abs(t2-mu)<=(eps*abs(mu))) c1 <- all(abs(t1-sigma)<=(eps*abs(sigma))) converged <- c3&c2&c1} if(showits) cat("\n") list(sigma=t1,mu=t2,pi=t3) } #************************************************************************* dabipf.mix <- function(s,margins,design,start,steps=1,prior=.5,showits=FALSE) { if(length(prior)==1) prior <- rep(prior,s\$ncells) prior <- as.double(prior) w <- !is.na(prior) prior[!w] <- -999.0 storage.mode(design) <- "double"; storage.mode(margins) <- "integer" s\$z <- .na.to.snglcode(s\$z,999) s\$w <- .na.to.icode(s\$w,999) tp <- integer(s\$p); tq <- integer(s\$q) r <- ncol(design); npsir <- as.integer(r*(r+1)/2); wk <- numeric(npsir) tr <- integer(r); wkr <- numeric(r); wkd <- numeric(s\$ncells) beta <- matrix(0,r,s\$q) psir <- .Fortran("mkpsi",as.integer(r-1),matrix(as.integer(0),r,r), PACKAGE="mix")[[2]] tmp <- .Fortran("tobsm",s\$q,s\$psi,s\$npsi,numeric(s\$npsi),s\$ncells, matrix(0,s\$q,s\$ncells),numeric(s\$ncells), s\$npattz,s\$rz,s\$mdpzgrp,s\$npattw,s\$p,s\$rw,s\$mdpwgrp, s\$ngrp,s\$mobs,s\$mobsst,s\$nmobs,s\$n,s\$z,tp,tq, PACKAGE="mix") kn1 <- tmp[[4]];kn2 <- tmp[[6]];kn3 <- tmp[[7]] sigma <- start\$sigma; pii <- start\$pi; mu <- start\$mu t1 <- sigma; t2 <- mu; t3 <- pii if(showits) cat(paste("Steps of Data Augmentation-Bayesian IPF:","\n")) for(i in 1:steps){ if(showits) cat(paste(format(i),"...",sep="")) tmp <- .Fortran("istepm",s\$q,s\$psi,s\$npsi,s\$ncells,sigma,mu,pii,kn1, kn2,kn3,t1=t1,t2=t2,t3=t3,s\$npattz,s\$rz,tq,tq,s\$mdpzgrp, s\$npattw,s\$p,s\$rw,tp,s\$mdpwgrp,s\$ngrp, s\$mobs,s\$mobsst,s\$nmobs,s\$n,s\$z,s\$d,s\$jmp,tp,kn3,kn1,s\$w, numeric(s\$q), PACKAGE="mix") t1 <- tmp\$t1; t2 <- tmp\$t2; t3 <- tmp\$t3 tmp <- .Fortran("pstepcm",s\$q,s\$psi,s\$npsi,s\$ncells,t1,t2,t3, sigma=t1,mu=t2,s\$n,r,design,wk,tr,psir,npsir, wkr,wkd,tq,numeric(s\$npsi),beta,matrix(0,s\$q,s\$q), PACKAGE="mix") sigma <- tmp\$sigma; mu <- tmp\$mu tmp <- .Fortran("bipf",s\$ncells,t3,pii=pii,prior, length(margins),margins,s\$p,s\$d,s\$jmp,tp,tp,tp, err=as.integer(0), PACKAGE="mix") if(tmp\$err==1) stop("Improper posterior - Check prior.") pii <- tmp\$pii; pii[w] <- pii[w]/sum(pii[w])} if(showits) cat("\n") list(sigma=sigma,mu=mu,pi=pii) } #************************************************************************* # multiple imputation inference mi.inference <- function(est,std.err,confidence=.95) { qstar <- est[[1]] for(i in 2:length(est)){qstar <- cbind(qstar,est[[i]])} qbar <- apply(qstar,1,mean) u <- std.err[[1]] for(i in 2:length(std.err)){u <- cbind(u,std.err[[i]])} dimnames(u)[[1]] <- dimnames(qstar)[[1]] u <- u^2 ubar <- apply(u,1,mean) bm <- apply(qstar,1,var) m <- dim(qstar)[2] tm <- ubar+((1+(1/m))*bm) rem <- (1+(1/m))*bm/ubar nu <- (m-1)*(1+(1/rem))**2 alpha <- 1-(1-confidence)/2 low <- qbar-qt(alpha,nu)*sqrt(tm) up <- qbar+qt(alpha,nu)*sqrt(tm) pval <- 2*(1-pt(abs(qbar/sqrt(tm)),nu)) fminf <- (rem+2/(nu+3))/(rem+1) result <- list(est=qbar,std.err=sqrt(tm),df=nu,signif=pval,lower=low, upper=up,r=rem,fminf=fminf) result } #*********************************************************************** ``` ## Try the mix package in your browser Any scripts or data that you put into this service are public. mix documentation built on June 20, 2017, 9:13 a.m.

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# Intermediate Algebra 5th edition Alan S. Tussy and R. David Gustafson Publisher: Cengage Learning ## Personal Study Plan Module Your students can use chapter and section assessments to gauge their mastery of the material and generate individualized study plans that include various online, interactive multimedia resources. ## Textbook Resources Additional instructional and learning resources are available with the textbook, and might include testbanks, slide presentations, online simulations, videos, and documents. Access is contingent on use of this textbook in the instructor's classroom. Academic Term Homework Homework and eBook Higher Education Single Term N/A \$80.00 High School N/A \$35.00 Online price per student per course or lab, bookstore price varies. Access cards can be packaged with most any textbook, please see your textbook rep or contact WebAssign • Chapter 1: A Review of Basic Algebra • 1.1: The Language of Algebra (41) • 1.2: The Real Numbers (48) • 1.3: Operations with Real Numbers (101) • 1.4: Simplifying Algebraic Expressions Using Properties of Real Numbers (110) • 1.5: Solving Linear Equations Using Properties of Equality (116) • 1.6: Solving Formulas; Geometry (42) • 1.7: Using Equations to Solve Problems (39) • 1.8: More about Problem Solving (37) • 1: Chapter Summary and Review (77) • 1: Chapter Test (11) • Chapter 2: Graphs, Equations of Lines, and Functions • 2.1: Graphs (57) • 2.2: Graphing Linear Equations in Two Variables (77) • 2.3: Rate of Change and the Slope of a Line (55) • 2.4: Writing Equations of Lines (69) • 2.5: An Introduction to Functions (76) • 2.6: Graphs of Functions (59) • 2: Chapter Summary and Review (43) • 2: Chapter Test (21) • 2: Cumulative Review (22) • Chapter 3: Systems of Equations • 3.1: Solving Systems of Equations by Graphing (79) • 3.2: Solving Systems of Equations Algebraically (78) • 3.3: Solving Systems of Equations in Three Variables (71) • 3.4: Solving Systems of Equations Using Matrices (66) • 3.5: Solving Systems of Equations Using Determinants (94) • 3.6: Problems Solving Using Systems of Two Equations (37) • 3.7: Problem Solving Using Systems of Three Equations (32) • 3: Chapter Summary and Review (27) • 3: Chapter Test (17) • 3: Cumulative Review (15) • Chapter 4: Inequalities • 4.1: Solving Linear Inequalities in One Variable (91) • 4.2: Solving Compound Inequalities (50) • 4.3: Solving Absolute Value Equations and Inequalities (94) • 4.4: Linear Inequalities in Two Variables (66) • 4.5: Systems of Linear Inequalities (67) • 4: Chapter Summary and Review (26) • 4: Chapter Test (17) • 4: Cumulative Review (23) • Chapter 5: Exponents, Polynomials, and Polynomial Functions • 5.1: Exponents (104) • 5.2: Scientific Notation (93) • 5.3: Polynomials and Polynomial Functions (79) • 5.4: Multiplying Polynomials (75) • 5.5: The Greatest Common Factor and Factoring by Grouping (72) • 5.6: Factoring Trinomials (112) • 5.7: The Difference of Two Squares; the Sum and Difference of Two Cubes (82) • 5.8: Summary of Factoring Techniques (46) • 5.9: Solving Equations by Factoring (68) • 5: Chapter Summary and Review (58) • 5: Chapter Test (23) • 5: Cumulative Review (20) • Chapter 6: Rational Expression and Equations • 6.1: Rational Functions and Simplifying Rational Expressions (67) • 6.2: Multiplying and Dividing Rational Expressions (45) • 6.3: Adding and Subtracting Rational Expressions (77) • 6.4: Simplifying Complex Fractions (45) • 6.5: Dividing Polynomials (90) • 6.6: Synthetic Division (87) • 6.7: Solving Rational Equations (43) • 6.8: Problem Solving Using Rational Equations (32) • 6.9: Proportion and Variation (51) • 6: Chapter Summary and Review (37) • 6: Chapter Test (19) • 6: Cumulative Review (34) • Chapter 7: Radical Expressions and Equations • 7.2: Rational Exponents (138) • 7.3: Simplifying and Combining Radical Expressions (117) • 7.4: Multiplying and Dividing Radical Expressions (126) • 7.5: Solving Radical Equations (60) • 7.6: Geometric Applications of Radicals (62) • 7.7: Complex Numbers (110) • 7: Chapter Summary and Review (85) • 7: Chapter Test (30) • 7: Cumulative Review (24) • Chapter 8: Quadratic Equations, Functions, and Inequalities • 8.1: The Square Root Property and Completing the Square (108) • 8.2: The Quadratic Formula (80) • 8.3: The Discriminant and Equations That Can Be Written in Quadratic Form (69) • 8.4: Quadratic Functions and Their Graphs (99) • 8.5: Quadratic and Other Nonlinear Inequalities (52) • 8: Chapter Summary and Review (8) • 8: Chapter Test (10) • 8: Cumulative Review (44) • Chapter 9: Exponential and Logarithmic Functions • 9.1: Algebra and Composition of Functions (93) • 9.2: Inverse Functions (60) • 9.3: Exponential Functions (55) • 9.4: Logarithmic Functions (110) • 9.5: Base-e Exponential and Logarithmic Functions (116) • 9.6: Properties of Logarithms (107) • 9.7: Exponential and Logarithmic Equations (73) • 9: Chapter Summary and Review (72) • 9: Chapter Test (29) • 9: Cumulative Review • Chapter 10: Conic Sections; More Graphing • 10.1: The Circle and the Parabola (80) • 10.2: The Ellipse (52) • 10.3: The Hyperbola (58) • 10.4: Solving Nonlinear Systems of Equations (45) • 10: Chapter Summary and Review (24) • 10: Chapter Test (18) • Chapter 11: Miscellaneous Topics • 11.1: The Binomial Theorem (68) • 11.2: Arithmetic Sequences and Series (53) • 11.3: Geometric Sequences and Series (45) • 11: Chapter Summary and Review (31) • 11: Chapter Test (22) • 11: Cumulative Review (65) The main focus of Intermediate Algebra, 5th edition, is to address the fundamental needs of today's developmental math students. Offering a uniquely modern, balanced program, Intermediate Algebra, 5th edition, integrates conceptual understanding with traditional skill and practice reinforced through visual and interactive practice in Enhanced WebAssign, available exclusively from Cengage Learning. By helping students understand the language of algebra and the "why" behind problem solving through instructional approaches and worked examples, they are better equipped to succeed at the "how." Practice is essential in making these connections and it is emphasized in Intermediate Algebra, 5th edition, with additional practice problems both in the text and Enhanced WebAssign. Give your students confidence by showing them how Algebra is not just about the x…it's also about the why. ### New for Fall 2019! • New MindTap Reader eBook now supported by HTML5 (non-flashed based) includes embedded media assets for a more integrated study experience • Coming soon! An all new, (non-flashed based) interactive graphing tool! • New WebAssign Student User Experience that empowers learning at all levels with an upgraded, modern student interface #### Take a Fresh Look at WebAssign Coming this Fall, WebAssign is updating to better address the needs and expectations of today's students. Learn about the changes coming to WebAssign-which have been developed to ensure support across changing course models and teaching curricula. ## Features • Read It links under each question quickly jump to the corresponding section of a complete, interactive eBook that lets students highlight and take notes as they read. Videos are also integrated into the eBook Read It links for a just-in-time learning approach. • A Personal Study Plan lets your students gauge their mastery of the material through chapter and section assessments and generates individualized study plans that include various online, interactive multimedia resources. • Are You Ready (AYR) problems review crucial prerequisite skills students need to be successful with application questions. • Master It Tutorials (CMI) show students how to solve a similar problem in multiple steps by providing direction along with derivation, so the student understands the concepts. #### Take a Fresh Look at WebAssign Coming this Fall, WebAssign is updating to better address the needs and expectations of today's students. Learn about the changes coming to WebAssign-which have been developed to ensure support across changing course models and teaching curricula. ## Questions Available within WebAssign Most questions from this textbook are available in WebAssign. The online questions are identical to the textbook questions except for minor wording changes necessary for Web use. Whenever possible, variables, numbers, or words have been randomized so that each student receives a unique version of the question. This list is updated nightly. ##### Question Group Key AYR - Are You Ready Questions CMI - Master It E - Example Problem E.XP - Extra Example Problem TP - Test Preparation XP - Extra Problem ##### Question Availability Color Key BLACK questions are available now GRAY questions are under development Group Quantity Questions Chapter 1: A Review of Basic Algebra 1.R 77 001b 003 004 005b 006a.TP 006b.TP 007a.TP 007b 008b 009b 010b 012.TP 015a.CMI 016a 017.CMI 018.TP 019 020 022 023 024 025.CMI 026 028 029 030 032.CMI 033 034 035 036 037 038 040 042 044 048 049 050 052 053.TP 054.TP 056 057 058 060a 060b 061.TP 062 063 065.CMI 066 067 068 071.CMI 072 074 076 077 078 080 082 084 086 087 088 090 091.TP 092.TP 093 095 096b 097 100.TP 102 104 111 1.T 11 002b 004b 004d 007 009 013 014b 016 020 022.TP 030 1.1 41 AYR.001 AYR.002 AYR.003 AYR.004 001.TP 002.TP 003.TP 004 005 006 007d.TP 008 009.CMI 014 016 018 022 026 029 033 034 036 038 039.TP 040.CMI 041.TP 043 044.CMI 045.TP 046.CMI 047 051.CMI 052 053 054.CMI 056 061 063 066ab.CMI 070 072 1.2 48 AYR.001 AYR.002 AYR.003 AYR.004 001 002 003 004.TP 005.TP 006.TP 007 008 009 016 017 018.TP 020 030 031 036 037.TP 038.TP 039.TP 040 047.TP 052.TP 053 054.TP 059.CMI 062 064 065.TP 066 067.TP 069.CMI 071.TP 074.CMI 075.TP 076.CMI 077.CMI 079.TP 080.CMI 082 084 086 088 092 096 1.3 101 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 001 002 003 004 005 006 007 008 012 013 015 016 017 018 019 019.TP 020 020.TP 021 022 023 024 025 026 027 027.TP 028 028.TP 029 030 031 031.TP 032 032.TP 033 034 035 036 037 038 039 039.TP 040 041 042 043 044 045 046 047 048 049 049.TP 050 051 052 053 054 055 056 057 058 059 060 061 061.TP 062 063 064 065 065.TP 066 067 068 069 070 071.TP 072.CMI 076 082 087.TP 089.TP 092 093.TP 096.TP 097 099.TP 104 107 109 110 112 114 116 121 1.4 110 AYR.001 AYR.002 AYR.003 AYR.004 001 002 003 004 005 006 007 008 012 015 016 018 021 023.TP 024 025 027.TP 028 029.TP 030 035 036 037 038 039 040 041 042 043 044 045 045.TP 046 047 048 049 049.TP 050 051 052 053 054 055 056 057 058 059 060 061 062 063 063.TP 064 064.TP 065 066 067 068 069 070 070.TP 071 071.TP 072 073 074 075 076 077 078 079 079.TP 080 081 081.TP 082 083 083.TP 084 085 086 087 088 089 090 091 092 093 093.TP 094 095 096 097 098 099 100 101 102 103 104 105 106 112 116 120 125 1.5 116 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 001 002 003 004 005 006 010 014 016.CMI 019 020 021 022 023 024 025 026 027 028 029 030 031 031.TP 032 033 034 035 036 037 038 039 040 041 042 043 043.TP 044 045 046 047 047.TP 048 049 050 050.TP 051 052 053 054 055 056 057 058 059 060 060.TP 061 062 062.TP 063 064 065 066 067 068 068.TP 069 070 071 072 072.TP 073 074 075 076 077 078 079 080 081 081.TP 082 083 083.TP 084 087 087.TP 088 089 090 091 092 093 094 095 096 097 098 099 100 101 101.TP 102 103 104 105 106 108 114 119 1.6 42 AYR.001 AYR.002 AYR.003 AYR.004 001 002 003 004 011.CMI 012.CMI 013 014 016 018 022 024 026 029 034 037 040 042.TP 044 046.TP 050 055 061 062.TP 063 064 065 066 068.TP 071.TP 083 087 090 094 096 098 099 109 1.7 39 AYR.001.TP AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 001 002 003.TP 004 005.TP 006.TP 007 008.TP 010 012 013.CMI 014.TP 018 019.TP 020 021.TP 022 023.TP 024.TP 025.CMI 026 027.TP 031 034 036 039.TP 040 042 043 044 046 048 051 1.8 37 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 001 002 003 004 012 013 020 022 024 028.CMI 029.TP 030 032 034.CMI 036.CMI 037.TP 038 042.CMI 043.TP 044.CMI 046 047.TP 048 049.TP 051 052.CMI 059.TP 061.TP 064 070 073 Chapter 2: Graphs, Equations of Lines, and Functions 2.C 22 001b 001c.TP 001d.TP 001f 009b 011 012 013 014 015 021 026 027 028 029 030 031 034 038.TP 039 042 043 2.R 43 001a 001b 001c 004 007.TP 008 009 010 011 012 013 014 017 019 020 021 022 023.CMI 024 027 028 030 031 033a 034 037 038a 038b 041 042.TP 046 047 049 056 057 058 059 061 063 064 067 072 073 2.T 21 001a 001b 001c 001d 001e 004 005 006 007 009 010.TP 012.CMI 013 014 015 019a.TP 019b 019c 020 025 032 2.1 57 AYR.001 AYR.002 AYR.003 AYR.004 E01 E02 E04 001 002 003.TP 004.TP 005 006 007.TP 008 010 011.TP 012 014 019 019.TP 020 020.TP 021 021.TP 022 023 024 025 026 028.CMI 031.TP 033.TP 036 037.CMI 039ab.CMI 041 042 043 044 044.TP 045 045.TP 046 047 048 049.TP 050.CMI 052 054 057 058 061a.TP 062 064 066 078 2.2 77 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E01 E03 E06 001 002 003 004 005 006.TP 008 010 011a.TP 012 016 018 020 021.TP 022 023.TP 024 025 026 027 027.TP 028 029 030 031 031.TP 032 033 033.TP 034 034.TP 035 035.TP 036 037 037.TP 038 039 040 041 042 042.TP 043 044 045 045.TP 046 047 048 053 054 055 056 057 058 059 060 061 062 063 064 080 082 085.CMI 088 090 100 103 2.3 55 AYR.001 AYR.002 AYR.003 AYR.004 E.XP.02 E03 E05 E06 001 002 003 004 005 006 007 009 011a.TP 012 013 014 017.TP 018.CMI 020 025 025.TP 026 027 028 029 029.TP 030 031 032 033 033.TP 034 035 036 037 038 039 040 041.TP 042.CMI 044.CMI 045.TP 050 052 055 056 059.TP 061 064 065 074 2.4 69 AYR.001 AYR.002 AYR.003 AYR.004.TP AYR.005 AYR.006 E02 E05 E06 001 002 004 007 015 016 016.TP 017 018 020 021 023 024 025 026 028 030 031 031.TP 032 033 034 036 037.CMI 039 040 041 042 042.TP 043 044 045 045.TP 046 047.TP 048.CMI 049 051.TP 054 055.TP 056 059.TP 060 062 063.TP 065 066 068.TP 076 078 079.TP 080.TP 088 094 095 098 104 108 112 114 2.5 76 AYR.001 AYR.002 AYR.003 AYR.004 E.XP.06 E04abc E04d 001 002 003 004 005 006 016 020 021.TP 023 025.TP 026.TP 028 030 032 033.TP 034 035.TP 036 037.TP 038.TP 046 047 048 049 050 052 055 056 056.TP 057 059 059.TP 060 061 061.TP 062 063 064 066 068 070 072 075 076 078 081 083 084 085 086 087 088 089 090 092 094 096 098 100 101 104 106 107 108 109 110 118 123 2.6 59 AYR.001 AYR.002 AYR.003 AYR.004 E04 E05 E06 E08 001 002 003 004 010 015b 016 017 018 020c.TP 022 026 028 029 030 031 032 033 036 037 038 039 041.TP 044 049 049.TP 050 050.TP 051 052 053 054 055 056 057 057.TP 058 059 060 061 062 063 064 065 066 067 068 071 073 082 092 Chapter 3: Systems of Equations 3.C 15 002a.TP 002b 005 006 007 008 011 018 019 020 021 026 032 037a 037b 3.R 27 001 011 013 014 015 023 026 030 031 033.TP 034 035a 035c 036b 036c 037 039 041 042 043 044 045 047 048 050 051 052.TP 3.T 17 001a 001c 001d.TP 001e.TP 002 003 007 008 012 013 014 019 023 024 027 028 029 3.1 79 AYR.001 AYR.002 AYR.003 AYR.004 E03 E04 E06 001 002.TP 003 004 005.TP 006.TP 007a.TP 009.TP 010.TP 013 014 015 015.TP 016 016.TP 017 018 019 020 021 021.TP 022 023 023.TP 024 025 026 027 028 029 030 031 032 033 034 035 036 038 040 041 042 043 044 045 046 047 048 049 049.TP 050 051 052 053 054 055 056 057 057.TP 058 058.TP 059 060 061 062 063 064 066 069 070 072 074 085 3.2 78 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E01 E05 001 002 003 004 005 006 011 013 014 014.TP 015 016 016.TP 017 017.TP 018 019 020 021 022 023 024 025 025.TP 026 026.TP 027 028 029 030 031 032 033 034 035 036 037 037.TP 038 039 039.TP 040 041 041.TP 042 042.TP 043 044 045 046 047 048 049 049.TP 050 051 052 053 054 055 055.TP 056 060 061 062 068 076 078 079 080 3.3 71 AYR.001 AYR.002 AYR.003 AYR.004 E02 E06 001 002 003 004.TP 005 006 008 011.TP 012 014 015 015.TP 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034 035 036 037 038 039 040 040.TP 041 042 043 043.TP 044 045 046 047 048 048.TP 049 050 051 052 053 053.TP 054 055 056 057 057.TP 058 060 061 064 070 071 3.4 66 AYR.001 AYR.002 AYR.003 AYR.004 E.XP.03 E.XP.04 E.XP.05a E01 001.TP 002.TP 003 004.TP 005 006 009 010b.TP 011 013.TP 014 015 016.TP 017 020 022 023 025 025.TP 026 026.TP 027 027.TP 028 029 029.TP 030 031 031.TP 032 033 033.TP 034 035 035.TP 036 037 038 039 040 041 042 043 044 045 046 047 048 049 049.TP 050 051 052 054 057 061 066 068 3.5 94 AYR.001 AYR.002 AYR.003 AYR.004 E.XP.04 E.XP.06 E01 E03 001 002 003.TP 004 008 009 010.TP 011 012.TP 014 015 016 017 017.TP 018 019 020 021 022 023 024 025 026 027 028 029 029.TP 030 030.TP 031 032 033 034 035 036 037 038 038.TP 039 039.TP 040 040.TP 041 042 043 044 045 046 047 047.TP 048 049 050 050.TP 051 051.TP 052 053 054 055 055.TP 056 056.TP 057 058 059 060 061 062 063 064 065 066 067 067.TP 068 068.TP 069 070 071 072 075 077 084 090 091 3.6 37 AYR.001 AYR.002 AYR.003 AYR.004 E04 E08 001 002 005 009 010 011 012 013 016 020 021 023 024 026 028 029 031 033.CMI 034.TP 036 038 039.TP 040 042 045 048 049 051 058 060 073 3.7 32 AYR.001 AYR.002 AYR.003 AYR.004 E03 001 002 003 004 007.CMI 010 011 012 013 014 015 016 017.TP 018 019 020 021 023 024 025 026 027.TP 028.CMI 029 030 031.TP 042 Chapter 4: Inequalities 4.C 23 001.TP 002 005a.TP 005b 008 010 012 016.CMI 017 019 020 021 023 025 029 031 037 038 040 043 045 046 047 4.R 26 013 024.CMI 028 031 033 034 037 038 040 041 043 044 046 048 049 058 059.TP 060 062 063 064 066 067 070 073.CMI 074 4.T 17 001a 001c 001e 012 013 014a 021 023 025 027 029 031 033 035 038.CMI 039 042 4.1 91 AYR.001 AYR.002 AYR.003 AYR.004 E01a E02 E06 001 002 003 004 005 006 011 018 021 022 023 024 025 025.TP 026 027 028 029 030 031 032 033 034 035 036 037 038 039 040 041.CMI 044 045 045.TP 046 047 048 049 050 051 051.TP 052 053 053.TP 054 054.TP 055 055.TP 056 056.TP 057 057.TP 058 059 060 061 062 063 063.TP 064 065 066 067 068 069 069.TP 070 070.TP 071 072 073 074 075 076 080 082 090 094 099 101.TP 102 103.CMI 104 106 115 4.2 50 AYR.001 AYR.002 AYR.003 AYR.004 E03 E04 E05 E07 001 002 003 004 007a 010 019 024 025.TP 027.TP 029.CMI 032 033.TP 035 036.TP 037.TP 039.TP 040.CMI 041 045.TP 046 047.TP 048 049.TP 052 053.TP 055.TP 057.TP 059 064.TP 070 078 081 082 083 084 085 086 088 090 094 097 4.3 94 AYR.001 AYR.002 AYR.003 AYR.004 E.XP.10 E03 E05 E06 001.TP 002 003 004 005.TP 006 007 012 019 020 021 022 023 023.TP 024 025 026 027 028 029 030 031 031.TP 032 033 034 035 036 036.TP 037 038 039 040 041 042 043 044 045 046 047 048 049 050 051 052 052.TP 053 054 055 056 057 058 059 060 060.TP 061 062 063 063.TP 064 065 065.TP 066 067 068 069 069.TP 070 072 075.TP 082 083 090 091 092 096 100 103 104 105 106 107 108 110 113 116 4.4 66 AYR.001 AYR.002 AYR.003 AYR.004 E.XP.01 E.XP.02 E.XP.03 E.XP.05 001 002 003 004 005.TP 007 009 011 012 013 014 015 015.TP 016 016.TP 017 018 019 019.TP 020 021 022 023 023.TP 024 025 025.TP 026 027 027.TP 028 029 029.TP 030 031 032 033 034 035 036 037 038 039 040 041 042 044 052 053 054 055 056 058 062 067a.TP 067b.TP 068a 068b 4.5 67 AYR.001 AYR.002 AYR.003 AYR.004 E01 E02 E03 E05 001 002 003 004 005 006 009 009.TP 010 011 011.TP 012 012.TP 013 014 014.TP 015 015.TP 016 017.TP 018.CMI 019.TP 020 021 022 023 023.TP 024 024.TP 025 026 027 028 029 030 030.TP 031 032 033 034 034.TP 035 036 037 038 039 040 044 045 046 048 049.CMI 050.CMI 051 052 054 057 058 062 Chapter 5: Exponents, Polynomials, and Polynomial Functions 5.C 20 003 007.CMI 010 012 014 016 020 023 024 025 027 029 032 034 038.CMI 039 040 041 042 045 5.R 58 002 003.CMI 004 005 008 009 010 011 013 014 016 018.TP 019 020 023 024 025 026 041 042 043 045 046 056 059.TP 060 068 069 072 073 074 077 078 080 083 085 087 088 089 092 094 096 098 100 101 102 105 106.CMI 109 110 112 116.TP 123 124 125 130 131 134 5.T 23 001a 001c 001d 001e 002 008 013 014 017 018 019 025.TP 027 028.TP 031 033.TP 034 035 036 037.CMI 041 042 045 5.1 104 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E02 E03 E06 E09 001 002 003 004 005 006 008 010 014 015a 015b 016a 016b 019 020 021 022 022.TP 023 024 025 026 027 028 029 029.TP 030 030.TP 031 032 033 034 035 036 036.TP 037 038 039 040 040.TP 041 042 043 044 045 046 047 048 049 050 051 051.TP 052 053 053.TP 054 055 056 057 058 061 063 064 065 065.TP 066 066.TP 067 068 069 070 074 076.TP 078 079.TP 082 085.TP 086 088.TP 093.TP 095.TP 096 097 101.TP 108 114 119 120 121 122 124 128 130 134 5.2 93 AYR.001 AYR.002 AYR.003 AYR.004 E02 E03 E04 E06 001 002 004 006 008 009 009.TP 010 011 011.TP 012 013 014 015 016 017 018 019 020 021 022 023 024 025 025.TP 026 027 027.TP 028 029 029.TP 030 031 031.TP 032 033 034 034.TP 035 036 037 038 039 039.TP 040 041 041.TP 042 043 044 045 046 047 047.TP 048 049 049.TP 050 051 052 053 054 055 056 057 058 059 059.TP 060 061 063.TP 065.TP 066.TP 067 070 074 076 078 080 082 084 086 090 094 098 5.3 79 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E02 E05ab 001.TP 002 003 004.TP 005 006 007 008 009 010a.TP 010b.TP 012b 014d.TP 018 020 021 024 026 029 029.TP 030 031 032 033 034 035 036 037 037.TP 038 038.TP 039 040 041 042 043 044 045 046 046.TP 047 048 048.TP 049 050 051 052 053 054 055 056 058 060 063.TP 064.TP 065.TP 072 082 085 086 087 088 090.CMI 092 096 100 104 107 110.TP 111.TP 112 5.4 75 AYR.001.CMI AYR.002.CMI AYR.003.CMI AYR.004 AYR.005 AYR.006 E11 001 002 003 004 005b 009 011 012 013 014 015 016 017 018 019 020 021 022 023 023.TP 024 025 026 027 028 029 030 031 031.TP 032 033 033.TP 034 036 037 040.CMI 041.TP 042.TP 043 044 045 046 047 048 049 050 051 052 053 054 056 059 065 070.CMI 085.TP 086 093.TP 094 100 103 110 115 116 118 120 121 128 131 5.5 72 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E.XP.07 E01 E05 001 002 003 004 005 006 008 011 012 013 014 015 016 017 018 019 020 023 024 025 026 027 028 029 029.TP 030 031 032 034.TP 035 036 037 038 042.TP 043.TP 044 047.TP 048.TP 052.CMI 058 059.TP 063.TP 065 068 069 072 073.TP 078 082 085.TP 087.TP 093.TP 094 101.TP 104 106 107 108 109 110 115 118 128 5.6 112 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E01b E06 E10 001 002 003 004 009 011 014 016 017 018 019 020 021 022 023 024 025 026 027 028 028.TP 029 029.TP 030 031 032 033 034 035 036 037 038 038.TP 039 040 041 042 043 044 045 046 047 048 048.TP 049 049.TP 050 051 052 053 054 055 056 057 058 059 059.TP 060 060.TP 061.TP 062 068 069 070.TP 074 076 081 082 083 084 084.TP 085 085.TP 086 087 088 088.TP 089 090 091 092 093 094 095 096 097 098 099 100 100.TP 101 102 102.TP 103 104 105 106 107 108 113 114 118 126 5.7 82 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E02 E04 001 002 003 005 008 009 009.TP 010 011 011.TP 012 013 014 015 016 017 018 019 020 021 021.TP 022 023 024 025 026 027 028 029.TP 031 032.TP 033.TP 036 037 037.TP 038 039 040 040.TP 041 042 043 044 045.TP 048.CMI 049.TP 053 054 055 055.TP 056 057 058 059 060 061 066 069.TP 070.TP 071 073 079 084.TP 086 096 099.TP 102 104 108 111 112 113 122 124 5.8 46 AYR.001 AYR.002 AYR.003 AYR.004 E01 E02 E03 E05 001 002 003 004 006 008 012 013.TP 014.TP 015.TP 016 018 019 020 021.TP 022.TP 023.TP 024 025.TP 028.TP 032 034.TP 035.TP 037.TP 038 040 043.TP 044 046 050 052 063 065 068 070 074 077 078 5.9 68 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E07a E09 001 002 003 004 005c.TP 006a.TP 009 014 015 016 017 017.TP 018 019 020 021 022 023 024 024.TP 025 026 027 028 028.TP 029 030 031.TP 035 039.TP 040 041.TP 043.TP 044 045.TP 047.TP 049.TP 050 051.TP 053 058 059.TP 069 072.TP 077 082 088 091.TP 092 094 096 100 105 108 109.TP 110 114 118 120 124 Chapter 6: Rational Expression and Equations 6.C 34 003 008 010 011.TP 017 018 019 021 022 025 027.TP 028 029 030 032.TP 033 038 042 043 044 045 047 048.TP 049 051 052 053 057.TP 060 064 065 066 069 073.TP 6.R 37 005 009.TP 010 011.TP 020 025 026 027 028 030 031 033 036 037 040 042.TP 047.TP 048 051 054 056 057 058 059 060 065.TP 066.CMI 067 068.TP 075 079 086 087.CMI 089.TP 091 092 096 6.T 19 001a.TP 001b.TP 001c.TP 001d 001e.TP 002 003 005 007d.TP 010 012.TP 020 022 026.CMI 028.TP 029 030 032 037.TP 6.1 67 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E02 E03 E04b E07 001.TP 002 003 004 005 006 007 008 009c.TP 010 012 013 017 020 021.TP 022 025.TP 026 027 028 029 030 031 032 033 033.TP 034 037 040.TP 045.TP 047.CMI 050.TP 052 054 056 058 060 063.TP 064.CMI 066 071 073 074.TP 078.TP 080 084.TP 085.TP 095 096 097 098 100 104 106 110 111 114 6.2 45 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E01 E02b E05 E07 001 002.TP 003 004 008 010 012 016 018 020 023.CMI 027.TP 029.TP 034.CMI 038 042 044 048 052.CMI 060 066 070.CMI 071 072 075.TP 076 080 082 088 090 092 095 100 101d 104 6.3 77 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E.XP.03 E06 E07 001 002.TP 003 004 012 014 017 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034 037.TP 038 039 039.TP 040 041 042 044 046 048 049.CMI 051 052 053 054 055 055.TP 056 060 063.TP 064 065 066.TP 067.TP 068 072.CMI 073.TP 074 075 076.TP 078 080 082.TP 085.TP 093.TP 102 105.TP 106.TP 108 110 112 116 120 124 126 6.4 45 AYR.001 AYR.002 AYR.003 AYR.004 E03 E04 E05 E06 001.TP 002 004 006 009 013.TP 014 018 022 023.TP 024 026 030 032 033.TP 035 036.TP 037.TP 041.TP 043.TP 050 051.TP 052 054 055.TP 059.TP 060 063.TP 067.TP 068 071 072 073 074 076 082 089 6.5 90 AYR.001 AYR.002 AYR.003 AYR.004 E02 E04 E05 E07 001.TP 002 003 004 006 012 014 015.TP 018 021.TP 022 023 024.TP 025.TP 026.TP 027 027.TP 028 029 030 031 032 033 034 035 036 037 038 038.TP 039 040 041 041.TP 042 043 044 045 046 047 048 049 049.TP 050 052.CMI 053.TP 055 056 057 057.TP 058 058.TP 059 060 061 062 063 063.TP 064 067 068 069 070 071 072 073 075 076 077 078 079 080 081 082 084 088 089 090 091 092 096 100 102 6.6 87 AYR.001 AYR.002 AYR.003 AYR.004 E01 E03 E04 E05 001 002 003 004 005 006.TP 007 008 012 013 014 015 015.TP 016 017 018 019 020 021 022 023 023.TP 024 025 026 027 028 029 030 031 032 033 034 035 036 038 043.TP 044.CMI 045.CMI 052.CMI 054.CMI 057.TP 059.TP 060 061.TP 062 064 065.TP 066.CMI 067.TP 068.CMI 069 069.TP 070 071 071.TP 072 073 074 075 076 077 078 078.TP 079 080 081 081.TP 082 083 084 084.TP 085 086 087 088 092 096 100 6.7 43 AYR.001 AYR.002 AYR.003 AYR.004 E01 E02 001.TP 002 006 007 012 014.CMI 016.CMI 019 021.TP 024 026 030 032.CMI 035 038.CMI 044 048.TP 049 051 052 054.TP 056.TP 061.CMI 069 071.TP 077 081.TP 084 085.TP 087 088.TP 089 094 098 102 105 108 6.8 32 AYR.001.TP AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E03 001 002 004 008 010 011.CMI 014.CMI 015 017 020 022 023 026 027 028 031.CMI 032 033 035 036 039 041 042 048 055 6.9 51 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E01 E04 E06 E08 001.TP 002.TP 003 004 005 006.TP 007 008 009 010 020 023.TP 024.CMI 028.CMI 029.TP 030.CMI 031.TP 033.TP 036.CMI 037.TP 039.CMI 041.CMI 048 051 058 064 072 074.CMI 076 078.TP 081 086.TP 087 089.TP 091.TP 092 093.TP 096 102 104 109 Chapter 7: Radical Expressions and Equations 7.C 24 003 007 008 009 012 015 016 017 019.TP 020 021 022.TP 024.TP 026 027 032 033 035 045 046.CMI 051 053 054 057 7.R 85 002a 003 004 005 006 007 008 009 010 011 012 014 015 018 021 022 025 027 028 030 032 033 034 036 038 039 041 042 043 044 045 046 048 050 052 053 056 058 059 060 061 062 068 069 070 071 080 081 083 084 086 088 089 090 092 093 100 102 104 108 109 114.TP 116 117 118 119 120 121.TP 122.TP 123.CMI 124.TP 125 126 127 128 130 132a 132b 133 134 136 137 140 141 142 7.T 30 001b.TP 001d 001e.TP 001f.TP 002 004 007 008 010 011 012 013 014 015.TP 016 018 020 022.TP 026 027 028 039 041.CMI 043 044 045 046 047 048 050 7.1 106 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E06 E09 001 002 003 004 005 006.TP 007 008 009.TP 010 013 021b 023 023.TP 024 025 025.TP 026 027 027.TP 028 029 030 031 031.TP 032 033 034 036 039 039.TP 040 041 042 043 044 045 046 047 048 049 050 052.TP 054 056 058 062 064 065a 066 068 070 071 072 073 073.TP 074 075 076 077 078 079 079.TP 080 081 082 083.TP 084 086 087 088 089 090 091 091.TP 092 094 095 095.TP 096 097 098 099 100 101 102 109.TP 120 126 127 128 129 130 131 132 133 138 140 7.2 138 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E07 E08 E09 001 002 003 004 005 006 009 010 014 016 017 017.TP 018 019 020 021 022 023 023.TP 024 025 025.TP 026 027 028 029 030 031 032 033 034 035 036 037 038 039 040 041 042 042.TP 043 044 045 046 047 048 049 049.TP 050 051 052 053 054 055 056 057.TP 058 060 062 064.TP 065 065.TP 066 067 068 069 070 071 071.TP 072 073 074 075 076 077 078 079 079.TP 080 081 082 083 084 085 086 086.TP 087 088 090 092 094 096 100 102 103 106 109 110 111 112 113 113.TP 114 115 116 117 118 119 120 121 122 124 125 126 127 128 129 130 131 132 134 138 139 140 141 142 143 146 148 7.3 117 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E01 E02 E06a E06b 001.TP 002 003 004 006 007 008 011 012 013 013.TP 014 017 017.TP 018 019 020 020.TP 021 021.TP 022 023 023.TP 024 025.TP 026 027 027.TP 028 029 029.TP 030 031 032 033 034 035 036 037 038 039 040 041 042 043 044 045 046 047 048 049 050 051 052 053 054 055 056 057 059 059.TP 060 061.TP 063 064 065 065.TP 066 067 068 069 069.TP 070 072 075 076 078 079 080 083 084 085 086 087 088 089 090 094 097 098 099 100 102 104 106 112 113 114 115 116 117 118 120 121 124 128 130 7.4 126 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E02 E04abd E09 E10 001 002 003.TP 004 005 006 012 014 015 016 017 017.TP 018 019 019.TP 020 021 022 023 023.TP 024 025 026 027 028 029 029.TP 030 031 032 032.TP 033 034 035 035.TP 036 036.TP 037 038 039 040 041 042 043 044 045 046 047 047.TP 048 049 050 052 053 054 055 055.TP 056 057 058 059 060 061 061.TP 062 063 063.TP 064 066 067 068 069 070 071 072 073 074 075 076 077 078 079 080 081 082 083 084 085 086 087 088 089 090 091.TP 094 095.TP 096.CMI 098 100 102 107 109 110 119 120 121 122 126 129 130 131 132 136 138 140 142 7.5 60 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E.XP.05 E01 E04 E08 001.TP 002 003 004.TP 005 006 008 018.TP 019 021 024 026.CMI 027 032 034 035.TP 038 040 044 050 052 054 058 059 061.TP 063.TP 064 066.TP 068 074.TP 077.TP 078 080 081 088 092 095.TP 100 102 105 106 107 108 109 110 112 113 124 126 128 7.6 62 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E01 E02 E07a 001.TP 002 003 004 005.TP 006 011.TP 014 015.TP 019.TP 020 021 022 023.TP 024 026 027 028.TP 030 031.TP 032 034 036 037 040.TP 041 043.TP 044 045.TP 046.CMI 047.TP 048.CMI 050 052.TP 056.TP 057 058 060 061 062 063 064 065 066 067 068 069 070 072 073 076 078 080 7.7 110 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E01 E04 E09 E10 001.TP 002.TP 003 004.TP 005 014 016 017 017.TP 018 019 020 021 021.TP 022 023 024 025 026 027 027.TP 028 030 032 033 037 037.TP 038 039 040 041 041.TP 042 043 043.TP 044 044.TP 045 046 046.TP 047 047.TP 048 049 050 051 052 053 054 055 056 057 058 059 059.TP 060 061 062 063 064 066 068 069 070 071 072 073 074 075 076 077 077.TP 078 079 080 081 082 083 084 086 088 089 089.TP 090 091 092 093 094 095 096 098 110 112 114 118 121 122 124 128 130 Chapter 8: Quadratic Equations, Functions, and Inequalities 8.C 44 001 003.TP 004 010 011 012 017 018 019 020 023 026 027 028 029 030 031.TP 034.TP 037 039 040 041.TP 042 043.TP 044 047.TP 051 052 056 058 060 061.TP 063.CMI 064 065 066 067 068 069 071 073 075 076.TP 078.CMI 8.R 8 003 004 005 006 030 039 054 060 8.T 10 010 011 015 018 024 026 035.CMI 036 038 040 8.1 108 AYR.001 AYR.002.TP AYR.003 AYR.004 AYR.005 AYR.006 E.XP.01 E03 001.TP 002 003 004 008 014 017 019 019.TP 020 020.TP 021 022 022.TP 023 023.TP 024 026 027 028 030 031 032 033 034 034.TP 035 036 037 041.CMI 042 044 048 050 052 055 056 057 058 059 060 061 061.TP 062 063 064 064.TP 065 066 071 072 073 073.TP 074 074.TP 075 075.TP 076 077 078 079 079.TP 080 081 082 083 084 086 087 088 089 090 091 092 093 094 096 102 103 104 105 108 112 120 501.XP 502.XP.TP 503.XP.TP 504.XP.TP 505.XP 506.XP.TP 507.XP 508.XP.TP 509.XP 510.XP 511.XP 512.XP 512.XP.TP 513.XP 514.XP 515.XP.CMI 8.2 80 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E01 E02 E03 E04abc 001 002 003b.TP 004 007.TP 009 012 013 014 015 016 017 018 018.TP 019 020 020.TP 021 022 023 023.TP 024 024.TP 025 026 027 028 029 030 031 031.TP 032 033 034 034.TP 035 036 039 042 047.TP 049.TP 050.TP 052 056.TP 058 059.TP 060 065.TP 066 070 072 074 077 081 083.TP 084 086 087 088 089 090 091 092 095 096 098 100 102 103 111 8.3 69 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E01 E02 E06 E07 001.TP 002 004 006 011 012 013 014 015 016 017 018 019 020 021 022 023 023.TP 024 025 026 028.CMI 029.TP 031.TP 032 033.TP 035 037.CMI 038 040 041.TP 042 044 045 048 052 053 057.TP 058 061.TP 064 067 074 076 077.TP 082 084 087 088 090 091 094 096 101 501.XP 502.XP 503.XP 504.XP 505.XP 8.4 99 AYR.001 AYR.002 AYR.003 AYR.004 E06 E07 E08 E10 001.TP 002.TP 003.TP 004 005c.TP 008 010 012 014 016 018.CMI 020.CMI 021 022.CMI 023 024 025 025.TP 026 027 028 029 029.TP 030 031 032 033 033.TP 034 035 036 037 038 039 039.TP 040 040.TP 041 042 043 044 044.TP 045 046 047 048 049 050 051 052 053 054 055 056 057 058 059 059.TP 060 060.TP 061 062 063 063.TP 064 065 066 067 068 069 069.TP 070 071 072 073 074 074.TP 084.TP 086 087 088 089 090 092 093 094 096 104 106 501.XP 502.XP 8.5 52 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E02 E03 E06 E07 001.TP 002.TP 003 004 008 010 014 015.TP 016.CMI 018.CMI 020 022.CMI 024.CMI 025.TP 026.CMI 027.TP 028.CMI 030.CMI 031.TP 032 034 035.TP 036.TP 038.CMI 039 039.TP 040 041 041.TP 042 049.TP 050 051 052.TP 055.TP 056 063 064 066 068 070 074 Chapter 9: Exponential and Logarithmic Functions 9.R 72 001 002 003 004 005 006 007 008 010 012.TP 013 014 017 018 019 020 022 023 024 033.TP 034 036 037 038 039 042 044 045 046 047 048 050 052 053.TP 054 056 057.CMI 058 065 069.TP 070 071 072 073 074 076 078 079 080 082 084 086 087 088 089 090 099 101 102 103.CMI 104 106 107.TP 108.TP 109.CMI 110.CMI 113 115.TP 116 118 122 124 9.T 29 001b 005 006 009a 009b 009d.TP 010.CMI 011.CMI 013.TP 016 017 022 023 025 027 028 029 030 031 032 033 036 038.CMI 039 043.TP 044.TP 046 050.TP 052 9.1 93 AYR.001 AYR.002 AYR.003 AYR.004 E01ab E03 E06 001.TP 002 003 004.TP 005 006 008 011 012.CMI 013 013.TP 014 014.TP 015 016 017 017.TP 018 019 020 021 022 023 024 025 026 027 028 030 032 034 035 037 037.TP 038 039 040 041 042 043 044 045 046 050 052 054 056 057 058 059 059.TP 060 060.TP 061 062 063 064 064.TP 065 065.TP 066 067 067.TP 068 068.TP 069 070 071 072 073 074 075 076 078 080 081a 082 084 085 087 088 089 090 092 096 098 9.2 60 AYR.001 AYR.002 AYR.003 AYR.004 E.XP.02 E05 001.TP 002 003 004 007 010.TP 013 018 019.TP 020 022.CMI 023.TP 025.TP 026 027.TP 029.TP 030 032 035 036 036.TP 037 038 039 040 041 042 043 044 045 046 047 048 049 050 051 051.TP 052 053 054 055.TP 056 060.CMI 061.TP 062.CMI 066 067.TP 068 069 070 074 082 083 086 9.3 55 AYR.001 AYR.002 AYR.003 AYR.004 E05 001 002 003 004 005a.TP 017 019 020 021 022 023 024 025 025.TP 026 027 028 028.TP 029 029.TP 030 030.TP 031 032 033 034 036 038 039 040 041.TP 046 047 048 051 052 053.TP 054 056 057 058 060 064 070 072 075 076.CMI 501.XP 502.XP 503.XP 9.4 110 AYR.001 AYR.002 AYR.003 AYR.004 001 002 003 004 005e 011 012.TP 013.TP 016 017.TP 021 022 023 023.TP 024 025 026 027 028 029 030 032 036 039 040 041 042 043 043.TP 044 045 046 048 050 051 052 053 053.TP 054 055 056 057 058 059 060 061 062 063 063.TP 064 064.TP 065 065.TP 066 067 068 069 070 071 072 073 074 075 075.TP 076 077 078 079 080 081 082 082.TP 083 083.TP 084 085 086 090 091 092 093 094 095 096 097 098 099.TP 100 101.TP 103.TP 106 107.TP 109 111.CMI 112 113 120 121 122 124 128 130 132 133 501.XP 502.XP 9.5 116 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006.TP E01 E02 E03 E04 E05 E06 001 002 003 004 005a.TP 009 010.TP 011.TP 014 015a.TP 016a.TP 017.TP 019 021 022 023 025 026 026.TP 027 027.TP 028 028.TP 029 029.TP 030 030.TP 031 032 033 033.TP 034 035 036 037 038 039 040 041 042 042.TP 043 044 044.TP 045 046 046.TP 047 048 048.TP 049 050 051 052 053 053.TP 054 055 056 057 058 059 059.TP 060 061 061.TP 062 063 064 065 066 067 068 072 073.CMI 074 075 076 077 078 080 081 083 084 086.TP 088 090 091 092 096.CMI 097 098 099 100 106 110 112 113.TP 114.TP 116 121 124 501.XP.TP 502.XP 9.6 107 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E.XP.10 E07 E09 001 002 003 004 005c 006b 016 019 019.TP 020 021 021.TP 022 023 024 025 026 027 027.TP 028 029 030 031 031.TP 032 033 034 035 036 037 038 039 040 041 042 043 044 045 046 046.TP 047 048 049 050 051 052 053 054 055 056 057 058 059 060 061 062 063 063.TP 064 065 065.TP 066 067 067.TP 068 068.TP 069 070 072 074 075 075.TP 076 076.TP 077 077.TP 078 079 080 081 083 084 088 089.TP 090 092 094 096 097 098 099.CMI 104 107 108 112 120 124 501.XP 9.7 73 AYR.001 AYR.002 AYR.003 AYR.004 AYR.005 AYR.006 E04 E08 E11 E12 001.TP 002 005 009 016.TP 020 022 023 025 026 027.TP 029.TP 031.TP 032 033.TP 035.CMI 038 039.CMI 042 044 046 048 050.CMI 052.TP 053.TP 054 056 058 059 061 062 068 072 077.TP 078.TP 084.TP 086 090 094 101.CMI 102 103.TP 104 105.TP 106 109.TP 112 114 116 118 122 124 126 128 130 501.XP 502.XP 503.XP 504.XP 505.XP 506.XP 507.XP 508.XP Chapter 10: Conic Sections; More Graphing 10.R 24 002.TP 006.TP 008 009.TP 010 012 013.TP 014.TP 015.TP 016 017 019 020 021 022.TP 023 024.TP 025 030c 031 035.CMI 036 037 038 10.T 18 001a.TP 001b.TP 001d 004 007 008 009.TP 010.TP 011 012 015.TP 016 017.TP 020 021.TP 023.CMI 025 027.CMI 10.1 80 AYR.001 AYR.002 AYR.003 AYR.004 E02 E03 E04 001.TP 002.TP 003.TP 004 008 010 013.TP 014 015 016 017 017.TP 018 018.TP 019 020 020.TP 021 022 023 024 025 025.TP 026 027 028 029 029.TP 030 031 032 033 034 035 035.TP 036 037 037.TP 038 039 040 041 041.TP 042 043 044 044.TP 045 046 047 048 049 049.TP 050 052.TP 064 073.TP 076 079 080.TP 084 086 087 088 089 090 091 092 093 094 096 100 105 10.2 52 AYR.001 AYR.002 AYR.003 AYR.004 001 002.TP 003.TP 004.TP 005.TP 006 010c.TP 011c.TP 012 014 016 017 017.TP 018 018.TP 019 020 020.TP 021 021.TP 022 023 024 025 026 026.TP 027 027.TP 028 029 030 030.TP 031 032 039 040 043 052 054 057 057b 058 060 061 062 066 068 073 10.3 58 AYR.001 AYR.002 AYR.003 AYR.004 001.TP 002 003.TP 004 005 006 010c.TP 011b.TP 012 014 016 017 018 018.TP 019 020 021 022 023 024 024.TP 025 026 026.TP 027 027.TP 028 029 030 031 032 033 034 035 036 037 038 039 043 046 047.TP 048.TP 049.TP 050.TP 051 059 061 062 063 066 070 072 080 082 10.4 45 AYR.001 AYR.002 AYR.003 AYR.004 E01 E02 E03 E04 001 002 003 004 005.TP 006 007cd 008.TP 009.TP 012 015.CMI 018.CMI 023.CMI 024 027.CMI 029.CMI 032.TP 034 035.CMI 036.CMI 042 045.TP 047.TP 057 060.TP 065.TP 066 067 068 069 070 071 072 073 074 076 082 Chapter 11: Miscellaneous Topics 11.C 65 001a 001b.TP 001c.TP 002 003 007a 007b 010 013 014c.TP 017 019 020 021 023 025 026 030 035 036 038 044 047.TP 048.TP 051 052.TP 053.TP 054.TP 055.TP 057 058 061 062 063 064 065 067 069 074.TP 077 078 079 080 081 083 084 088 089.CMI 090 091 093 094 095 096 100 102 105 106.TP 107 111 112 115 117 119.TP 122.TP 11.R 31 002 004 005.TP 006.TP 008.TP 009.TP 010 012 013.TP 014 015 016 017 022 023 024 025.TP 026 027 028 029 030 032 033.TP 034 035 036.TP 038 040.TP 041.TP 042 11.T 22 001a.TP 001b 001c.TP 001d.TP 001e.TP 002 003 004 005 006 007 008.TP 009 011 012 013 014 015 018.TP 019.TP 020 022 11.1 68 AYR.001 AYR.002 AYR.003 AYR.004 E02 E03 E07 001.TP 002 003.TP 004.TP 005.TP 006 010 014 015 018 023.TP 024 025 028 029 030 030.TP 031 032 033 033.TP 034 035 036 037 038 038.TP 039 039.TP 040 041 042 043 044 045 045.TP 046 047 047.TP 048 049 050 051 052 055.TP 057.TP 058 061.TP 062 067.TP 068 072 075.TP 076 078.TP 080 084 092 094 098 104 11.2 53 AYR.001 AYR.002 AYR.003 AYR.004 E05 E06 001.TP 002 003.TP 004 005.TP 006.TP 008 014 020 025 029.TP 031.TP 032 038 041.TP 043 049 053.TP 054 055.TP 057.TP 058 061 062 063 064 065 066 067 068 069 070 071 072 076 078 087.TP 088 092 093.TP 094 097 098 100 106 107.TP 109 11.3 45 AYR.001 AYR.002 AYR.003 AYR.004 E02 E07 E09 001.TP 002 003.TP 004.TP 006 008 012 015.TP 017.TP 018 022 025 027.TP 028 030 032 035.TP 036 040 043.TP 046.TP 048 051 053.TP 058 060.TP 063.TP 069.TP 075.TP 076 079.TP 080 082 084 086 090 094 098 Total 6114

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https://mail.python.org/pipermail/python-list/2010-November/591082.html

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# factorial of negative one (-1) Bj Raz whitequill.bj at gmail.com Mon Nov 1 09:19:26 CET 2010 ```On Fri, Oct 29, 2010 at 1:02 AM, Chris Rebert <clp2 at rebertia.com> wrote: > On Thu, Oct 28, 2010 at 9:41 PM, Bj Raz <whitequill.bj at gmail.com> wrote: > > I am working with differential equations of the higher roots of negative > > one. (dividing enormous numbers into other enormous numbers to come out > with > > very reasonable numbers). > > I am mixing this in to a script for Maya (the final output is graph-able > as > > a spiral.) > > I have heard that Sage, would be a good program to do this in, but I'd > like > > to try and get this to work in native python if I can. > > The script I am trying to port to Python is; > http://pastebin.com/sc1jW1n4. > > Unless your code is really long, just include it in the message in the > future. > So, for the archive: > indvar = 200; > q = 0; > for m = 1:150 > lnanswer = (3 * m) * log(indvar) - log(factorial(3 * m)) ; > q(m+1) = q(m)+ ((-1)^m) * exp(lnanswer); > end > q > > Also, it helps to point out *what language non-Python code is in*. I'm > guessing MATLAB in this case. > > Naive translation attempt (Disclaimer: I don't know much MATLAB): > > from math import log, factorial, exp > indvar = 200 > q = [0] > for m in range(1, 151): > lnanswer = (3 * m) * log(indvar) - log(factorial(3 * m)) > q.append(q[-1] + (1 if m % 2 == 0 else -1) * exp(lnanswer)) > print(q) > > Cheers, > Chris > -- > http://blog.rebertia.com >

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https://www.physicsforums.com/threads/edwin-hubble-and-the-evidence-for-universe-expansion.1010881/page-2

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# Edwin Hubble and the Evidence for Universe Expansion • I Gold Member If you want the same set of objects with modern distance/velocity values, you can plot it yourself. Use the tables in that paper to find each object in any public database. Hi @Bandersnatch: I will do the plot myself you recommend if I need to. I would prefer to not have to do this. What would help me is help with reading the 2 X labels and the 2 Y labels: The X labels seem to be 106 parsecs and 2 x 106 parsecs. Is that correct? The Y labels are more difficult to read. Regards, Buzz labels x: 1 and 2 Mpc y: 500 and 1000 km/s Buzz Bloom Gold Member I will do the plot myself you recommend if I need to. I would prefer to not have to do this. What would help me is help with reading the 2 X labels and the 2 Y labels: The X labels seem to be 106 parsecs and 2 x 106 parsecs. Is that correct? The Y labels are more difficult to read. I plotted the v vs. r data shown in Table 1 of Hubble’s 1929 paper and compared it with his Fig. 1. I noticed that some of the data points are not placed correctly in his figure. This does not change things; the Hubble constant determined from his figure, about 500 km/s/Mpc, is too high anyway. Buzz Bloom (...) Fig. 1. I noticed that some of the data points are not placed correctly in his figure. Those data need to be corrected for solar motion, as discussed in the paper. You can see when comparing the two graphs that the differences are on the order of +/- 200 km/s. In any case, I think Buzz wanted to plot the original set of galaxies but with modern data for their distances and velocities. Buzz Bloom Gold Member Those data need to be corrected for solar motion, as discussed in the paper. You can see when comparing the two graphs that the differences are on the order of +/- 200 km/s. If not in Table 1 and Table 2, where in the paper are the plotted, corrected v vs. r data? If not in Table 1 and Table 2, where in the paper are the plotted, corrected v vs. r data? It's not given explicitly in the paper. There is the raw data in table 1, values for solar motion are extracted (under the first equation), and the corrected data plotted. Gold Member It's not given explicitly in the paper. There is the raw data in table 1, values for solar motion are extracted (under the first equation), and the corrected data plotted. I am confused. Can you calculate for one case to demonstrate the use of the following equation? Do we know the angles needed? Gold Member It's not given explicitly in the paper. There is the raw data in table 1, values for solar motion are extracted (under the first equation), and the corrected data plotted. Referring to Table 1 in his paper, Hubble writes, The data in the table indicate a linear correlation between distances and velocities, whether the latter are used directly or corrected for solar motion, according to the older solutions.​ As you said, the raw v vs. r data are listed in Table 1. However, the corrected data are used in Fig. 1. The methods of correction are discussed in the paper, but the results are not listed in it. This explains the discrepancy shown in Post #38. I went and graphed Hubble's table 1 data corrected for his assessment of solar motion (individual objects only): Looks almost the same as the graph in the paper. He did mention also adding residuals to the final graph, so maybe that's where the remaining differences are. But I don't know how it's done. Took me way too long to figure out how to remove the solar motion correctly (hopefully), since apparently I completely forgot how to work in spherical coordinates. For comparison, here's the same graph using modern data (from Cosmicflows-3 & 1) for the same set of objects: The order of magnitude underestimation of distances in the original work is immediately apparent. Here's a link to the spreadsheet, in case anyone wants to play with it (you'll have to make a copy to edit). Gold Member I went and graphed Hubble's table 1 data corrected for his assessment of solar motion . . . For comparison, here's the same graph using modern data (from Cosmicflows-3 & 1) for the same set of objects . . . That’s very impressive! I put the original data from Hubble’s Table 1 and your corrected data in one figure: The Hubble constant estimated from your graph using modern data is about 69.4 km/s/Mpc. Just a note, I replaced the values for right ascension, declination, and solar velocity that I thought Hubble was using (the rounded average between the two solutions) with values he derived from the 24 objects only, and the graph now looks identical to what's in the paper. As far as I can tell, at least. Only the trend line better matches the solution for groups of objects, for some reason. If anyone knows what is meant by adding residuals to the solution, which doesn't seem to have any visible effect anyway - do tell. I thought he meant uncertainties, but there's no error bars on the graph, nor can I see any way in which they'd 'average 150 and 100 km/s'. Gold Member For comparison, here's the same graph using modern data (from Cosmicflows-3 & 1) for the same set of objects: View attachment 296757 I have a question about the distances used in the graph. Are they the distances at emissions of the received photons or the present distances of the galaxies? They can be quite different at higher z values (e.g., z = 0.1). I have a question about the distances used in the graph. Are they the distances at emissions of the received photons or the present distances of the galaxies? They can be quite different at higher z values (e.g., z = 0.1). With these objects you don't even get to z=0.01, so which distance is reported makes no appreciable difference. Especially seeing how actual error bars on the measurements have been omitted. This database reports either comoving or luminosity distance - I'd have to check. But, again, this close, and at this level of imprecision, it doesn't matter much which one it is. They're all approximately equal. Gold Member With these objects you don't even get to z=0.01, so which distance is reported makes no appreciable difference. If we are only talking about Hubble’s 1929 paper, I agree with you. However, in the reference you gave in Post #12, the distances extend to about 400 Mpc (z values extend to about 0.1). As can be seen from the output of Jorrie’s calculator, the difference can be significant:

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# What are the different types of angles on parallel lines? Angles associated with Parallel Lines ## What are the different types of angles on parallel lines? Angles associated with Parallel Lines • z and x (or u and v) are alternate angles. • x and y are corresponding angles. • u and x (or z and v) are allied (or co-interior) angles. • y and z are vertically opposite angles. ## Are adjacent angles equal to 180? a and b are adjacent angles. Adjacent angles add up to 180 degrees. These add up to 180 degrees (e and c are also interior). Any two angles that add up to 180 degrees are known as supplementary angles. ## Can parallel lines be on top of each other? Lines that coincide lie one on top of the other. They are the SAME line with the equations expressed in different forms. If two coincident lines form a system, every point on the line is a solution to the system. Lines in a plane that are parallel, do not intersect. ## Does parallel lines meet? Parallel lines do not meet at a point. This section of Wikipedia worth a lot here: In geometry, parallel lines are lines in a plane which do not meet; that is, two lines in a plane that do not intersect or touch each other at any point are said to be parallel. ## Do parallel lines never meet? Parallel lines are lines in a plane that are always the same distance apart. Parallel lines never intersect. Perpendicular lines are lines that intersect at a right (90 degrees) angle. ## Can 2 adjacent angles be supplementary? Also These angles are adjacent, according to the definition of adjacent angles, and these pairs of angles sum to 180 degree such that ∠AOB+∠BOC=, forming a supplementary pair of angles. Therefore, It is possible that two adjacent angles form supplementary angles. ## Are same side interior angles supplementary? Same side interior angles are two angles that are on the same side of the transversal and on the interior of (between) the two lines. Same Side Interior Angles Theorem: If two parallel lines are cut by a transversal, then the same side interior angles are supplementary. ## Which angles are adjacent angles? Adjacent angles are two angles that have a common vertex and a common side but do not overlap. In the figure, ∠1 and ∠2 are adjacent angles. They share the same vertex and the same common side. ## Do 2 parallel lines intersect? In the Euclidean plane, parallel lines don’t intersect. Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction. If they intersect, then you don’t call them parallel. ## Which angle is an adjacent interior angle? Angle 3 is the interior adjacent angle. Any two interior angles that share a common side are called the adjacent interior angles. They have the same vertex and have a common arm. ## Why do parallel lines have the same slope? Since slope is a measure of the angle of a line from the horizontal, and since parallel lines must have the same angle, then parallel lines have the same slope — and lines with the same slope are parallel. So perpendicular lines have slopes which have opposite signs. ## What are adjacent angles in parallel lines? Adjacent angles are two angles sharing a common vertex and a common side. They appear in many places but are prominent in parallel lines cut by transversals. ## How many solutions do parallel lines have? When the lines are parallel, there are no solutions, and sometimes the two equations will graph as the same line, in which case we have an infinite number of solutions. ## Why do parallel lines have to be on the same plane? Two lines are parallel if and only if they lie in the same plane and do not intersect. Parallel lines never cross. Parallel lines are always the same distance apart, which is referred to as being “equidistant”. For our purposes, parallel lines will always be straight lines that go on indefinitely. ## What type of slope do parallel lines have? The slope of the parallel line is undefined and the slope of the perpendicular line is 0. ## Why are same side interior angles supplementary? The same-side interior angles theorem states that same-side interior angles are supplementary when the lines intersected by the transversal line are parallel. 2) Since the lines A and B are parallel, the same-side interior angles theorem states that same-side interior angles will be supplementary. ## How do you prove two lines are parallel without angles? If two lines have a transversal which forms alternative interior angles that are congruent, then the two lines are parallel. If two lines have a transversal which forms corresponding angles that are congruent, then the two lines are parallel. ## How do you find adjacent angles? The two angles are said to be adjacent angles when they share the common vertex and side. The endpoints of the ray form the side of an angle is called the vertex of a angle. Adjacent angles can be a complementary angle or supplementary angle when they share the common vertex and side.

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Intuitionistic Logic Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  ILE Home  >  Th. List  >  relun Unicode version Theorem relun 4668 Description: The union of two relations is a relation. Compare Exercise 5 of [TakeutiZaring] p. 25. (Contributed by NM, 12-Aug-1994.) Assertion Ref Expression relun Proof of Theorem relun StepHypRef Expression 1 unss 3257 . 2 2 df-rel 4558 . . 3 3 df-rel 4558 . . 3 42, 3anbi12i 456 . 2 5 df-rel 4558 . 2 61, 4, 53bitr4ri 212 1 Colors of variables: wff set class Syntax hints:   wa 103   wb 104  cvv 2691   cun 3076   wss 3078   cxp 4549   wrel 4556 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2123 This theorem depends on definitions:  df-bi 116  df-tru 1335  df-nf 1438  df-sb 1738  df-clab 2128  df-cleq 2134  df-clel 2137  df-nfc 2272  df-v 2693  df-un 3082  df-in 3084  df-ss 3091  df-rel 4558 This theorem is referenced by:  funun  5179 Copyright terms: Public domain W3C validator

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Community Profile Hà Thị Thắm VNU University of Engineering and Technology Last seen: 1 month ago Active since 2022 K65MCLC2 Programming Languages: C++ Spoken Languages: English Professional Interests: Content Feed View by Solved Find vampire numbers A <http://en.wikipedia.org/wiki/Vampire_number vampire number> is a number v that is the product of two numbers x and y such th... 1 month ago Solved Divisible by 10 Pursuant to the <http://www.mathworks.com/matlabcentral/cody/problems/42404-divisible-by-2 first problem> in this series, this o... 1 month ago Solved Draw 'I' Given n as input, draw a n-by-n matrix 'I' using 0 and 1. example: n=3 ans= [0 1 0 0 1 0 0 1 0] n=... 1 month ago Solved Generalised Hamming Number Inspired by Project Euler n°204 and Problem 1308 by James A generalised Hamming number of type n, has no prime factor larger ... 1 month ago Solved Divisible by n, Truncated-number Divisors Some divisors only require a few numbers at the end of the number in question to determine divisibility, no matter how long. Exa... 1 month ago Solved Divisible by n, Composite Divisors Pursuant to <http://www.mathworks.com/matlabcentral/cody/problems/42453-divisible-by-n-prime-vs-composite-divisors Divisible by ... 1 month ago Solved Divisible by n, prime divisors from 20 to 200 Pursuant to the previous problem (linked below), this problem requires a function that checks for divisibility of large numbers ... 1 month ago Solved Divisible by n, prime divisors - 11, 13, 17, & 19 Divisibility checks against prime numbers can all be accomplished with the same routine, applied recursively, consisting of add ... 1 month ago Solved Fibonacci-Sum of Squares Given the Fibonacci sequence defined by the following recursive relation, * F(n) = F(n-1) + F(n-2) * where F(1) = 1 and F(1)... 1 month ago Solved Fibonacci Decomposition Every positive integer has a unique decomposition into nonconsecutive Fibonacci numbers f1+f2+ ... Given a positive integer n, r... 1 month ago Solved Proper Factors Generate the proper factors of input integer x and return them in ascending order. For more information on proper factors, refer... 1 month ago Solved Numbers with prime factors 2, 3 and 5. Make a function which takes one positive integer n and returns a matrix with the numbers of the form (2^i)*(3^j)*(5^k) which are... 1 month ago Solved Multiples of a Number in a Given Range Given an integer factor _f_ and a range defined by _xlow_ and _xhigh_ inclusive, return a vector of the multiples of _f_ that fa... 1 month ago Solved Smith numbers Return true if the input is a Smith number in base ten. Otherwise, return false. Read about Smith numbers at <http://en.wikipedi... 1 month ago Solved (Linear) Recurrence Equations - Generalised Fibonacci-like sequences This problem is inspired by problems <http://uk.mathworks.com/matlabcentral/cody/problems/2187-generalized-fibonacci 2187>, <htt... 1 month ago Solved Factorize THIS, buddy List the prime factors for the input number, in decreasing order. List each factor only once, even if the factorization includes... 2 months ago Solved Prime factor digits Consider the following number system. Calculate the prime factorization for each number n, then represent the prime factors in a... 2 months ago Solved How many Fibonacci numbers? Find the number of unique Fibonacci numbers (don't count repeats) in a vector of positive integers. Example: x = [1 2 3 4... 2 months ago Solved Get all prime factors List the prime factors for the input number, in decreasing order. List each factor. If the prime factor occurs twice, list it as... 2 months ago Solved Divisible by n, prime divisors (including powers) For this problem, you will be provided an array of numbers (not necessarily in order). Return the array of numbers with only pri... 2 months ago Solved Divisible by n, prime vs. composite divisors In general, there are two types of divisibility checks; the first involves composite divisors and the second prime divisors, inc... 2 months ago Solved Divisible by 8 Pursuant to the <http://www.mathworks.com/matlabcentral/cody/problems/42404-divisible-by-2 first problem> in this series, this o... 2 months ago Solved Divisible by 4 Pursuant to the <http://www.mathworks.com/matlabcentral/cody/problems/42404-divisible-by-2 first problem> in this series, this o... 2 months ago Solved Divisible by 16 Write a function to determine if a number is divisible by 16. This can be done by a few different methods. Here are two: # If... 2 months ago Solved Divisible by 15 Write a function to determine if a number is divisible by 15. If a number is <http://www.mathworks.com/matlabcentral/cody/proble... 2 months ago Solved Divisible by 14 Write a function to determine if a number is divisible by 14. If a number is <http://www.mathworks.com/matlabcentral/cody/proble... 2 months ago Solved Divisible by 13 Write a function to determine if a number is divisible by 13. Similar to the number seven, this can be done by a few different m... 2 months ago Solved Divisible by 7 Pursuant to the <http://www.mathworks.com/matlabcentral/cody/problems/42404-divisible-by-2 first problem> in this series, this o... 2 months ago Solved Animated GIF Creator This Challenge is to execute the Function Template which has a fully functional Animated GIF creator of a shape related to a Zer... 2 months ago Solved Find the area of the circle geometry 2 months ago

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Question # On January 1, a company issued 10%, 10-year bonds payable with a par value of \$720,000.... On January 1, a company issued 10%, 10-year bonds payable with a par value of \$720,000. The bonds pay interest on July 1 and January 1. The bonds were issued for \$817,860 cash, which provided the holders an annual yield of 8%. Prepare the journal entry to record the first semiannual interest payment, assuming it uses the straight-line method of amortization. Par value of bonds = \$720,000 Stated interest rate= 10% Issue price of bonds = \$817,860 Premium on bonds payable = Issue price of bonds - Par value of bonds = 817,860-720,000 = \$97,860 Semi annual interest payment = Par value of bonds x Stated interest rate x 6/12 = 720,000 x 10% x 6/12 = \$36,000 Semi annual amortization of bond premium = Premium on bonds payable / Semi annual interest payment periods = 97,860/20 = \$4,893 Date General Journal Debit Credit July 1 Interest expense \$31,107 Premium on bonds payable \$4,893 Cash \$36,000 ( To record first semi annual interest payment) Kindly comment if you need further assistance. Thanks‼! #### Earn Coins Coins can be redeemed for fabulous gifts.

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698 views ### Why is $\sqrt{4} = 2$ and Not $\pm 2$? [duplicate] I've always been told that if $\ x^2 = 4,$ $=>x = \pm2$ But recently, Prof. mentioned that if $x = \sqrt{4}$, Then $x = +2(only)$ I am very skeptical about this because they both mean the ... 22k views ### Is sqrt(x) a function? Does it matter if a domain is given? [duplicate] Possible Duplicates: Reason why the even root of a number always positive Square roots — positive and negative I saw the following during a practice exam: $f(x) = \sqrt x$ for $x ≥ 2$ ... 7k views 647 views ### Why does Wolfram Alpha say that $\sqrt{1}=-1$? [duplicate] Why does Wolfram Alpha say that $\sqrt{1}=-1$? Is this a mistake or what? Can anyone help? Thanks in advance. 383 views ### Why does the square root of a square involve the plus-minus sign? If $\sqrt{x^2}$ can be simplified as follows: $\sqrt{x^2} = (x^2)^\frac{1}{2} = x^{\frac{2}{1}\times\frac{1}{2}} =x^\frac{2}{2} = x^1 = x$ Then why would $\sqrt{x^2} = \pm x$? 68 views ### Square roots: $-\sqrt{a} = \sqrt{b}$ [duplicate] [CLOSED] Thanks GoodDeeds and Henry [1] I understood the fundamental problem. As √9 = ± 3 If , √A = √B Thus, ** ±a = ±b** And so a = b; -a = b; and a = -b;; Thus, √9 = +3 OR -3 Let, √A = ±a ... ### Find $f(g(x))=-\sqrt{x}$ where $f(x) =\sqrt {x}$ If we have $f(x)=\sqrt{x}$ Find $g(x)$ where $f(g(x))=-f(x) =-\sqrt{x}$ I know that the square root can have plus and minus value in the same time but I what that $f(g(x))$ value is negative when \$...

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## Do My Affine Group Lab As discussed above, I made use of to compose an easy as well as uncomplicated math lab with only Affine Group Nevertheless, the less complex you make your lab, the much easier it becomes to obtain stuck at completion of it, after that at the start. This can be very aggravating, and all this can occur to you since you are using Affine Group and/or Modular Equations inaccurately. With Modular Equations, you are currently using the wrong equation when you obtain stuck at the beginning, if not, then you are likely in a stumbling block, and there is no possible escape. This will just get worse as the problem becomes extra intricate, but then there is the question of how to proceed with the trouble. There is no chance to appropriately deal with fixing this sort of math problem without having the ability to right away see what is taking place. It is clear that Affine Group as well as Modular Equations are challenging to learn, and also it does take practice to establish your own sense of intuition. But when you wish to fix a mathematics issue, you need to make use of a device, and also the devices for finding out are used when you are stuck, as well as they are not used when you make the wrong action. This is where lab Assist Service is available in. As an example, what is wrong with the inquiry is incorrect concepts, such as obtaining a partial value when you do not have sufficient functioning components to finish the entire work. There is a great reason that this was wrong, and it refers reasoning, not instinct. Reasoning allows you to adhere to a step by step treatment that makes good sense, and also when you make a wrong move, you are usually compelled to either attempt to move forward and also fix the error, or attempt to go backward and also do an in reverse action. One more instance is when the student does not comprehend an action of a process. These are both logical failings, and also there is no way around them. Also when you are embeded an area that does not allow you to make any type of type of relocation, such as a triangle, it is still important to comprehend why you are stuck, so that you can make a far better relocation and go from the step you are stuck at to the following area. With this in mind, the most effective way to fix a stuck circumstance is to just take the advance, as opposed to attempting to go backward. Both procedures are different in their approach, yet they have some standard similarities. However, when they are tried together, you can promptly tell which one is better at fixing the issue, and also you can likewise tell which one is extra effective. Let's discuss the initial instance, which connects to the Affine Group math lab. This is not too challenging, so allow's very first talk about exactly how to begin. Take the complying with procedure of attaching a component to a panel to be utilized as a body. This would require three dimensions, and also would be something you would need to connect as part of the panel. Now, you would have an additional dimension, yet that does not suggest that you can simply maintain that measurement and also go from there. When you made your first step, you can easily forget the measurement, and then you would have to go back as well as retrace your steps. Nevertheless, instead of remembering the added dimension, you can use what is called a "mental shortcut" to help you remember that extra dimension. As you make your initial step, envision on your own taking the measurement and affixing it to the component you wish to connect to, and after that see how that makes you really feel when you duplicate the process. Visualisation is a really powerful technique, and also is something that you need to not skip over. Imagine what it would certainly feel like to really attach the part as well as have the ability to go from there, without the dimension. Currently, allow's look at the 2nd example. Allow's take the exact same procedure as in the past, but now the trainee has to keep in mind that they are going to move back one step. If you tell them that they have to move back one step, but after that you get rid of the concept of having to return one action, after that they will not know just how to proceed with the issue, they won't know where to seek that step, and also the procedure will certainly be a mess. Rather, make use of a mental shortcut like the mental representation to psychologically reveal them that they are going to move back one step. as well as place them in a setting where they can progress from there. without having to consider the missing an action. ## Pay Me To Do Your Affine Group Lab " Affine Group - Need Aid With a Mathematics lab?" Unfortunately, lots of trainees have actually had a problem realizing the concepts of direct Affine Group. Fortunately, there is a new format for linear Affine Group that can be used to instruct direct Affine Group to students that deal with this principle. Pupils can utilize the lab Aid Service to help them learn new techniques in linear Affine Group without dealing with a hill of issues and also without having to take an examination on their principles. The lab Assist Service was produced in order to aid struggling trainees as they move from university and also senior high school to the university and also task market. Many pupils are unable to handle the stress of the discovering procedure and also can have very little success in understanding the ideas of linear Affine Group. The lab Assist Service was established by the Educational Testing Service, that uses a variety of various online tests that students can take as well as practice. The Test Help Solution has helped numerous trainees boost their ratings and can assist you enhance your scores too. As pupils relocate from university and also secondary school to the university and job market, the TTS will certainly aid make your pupils' shift less complicated. There are a few different manner ins which you can make the most of the lab Assist Service. The primary manner in which trainees use the lab Assist Service is via the Solution Managers, which can assist trainees discover strategies in straight Affine Group, which they can utilize to help them prosper in their courses. There are a number of problems that students experience when they initially make use of the lab Help Service. Pupils are frequently overloaded and do not recognize just how much time they will certainly require to commit to the Service. The Answer Supervisors can assist the pupils evaluate their concept knowing and help them to assess every one of the material that they have actually already discovered in order to be gotten ready for their next training course job. The lab Aid Service works the same way that a teacher performs in regards to aiding trainees realize the principles of direct Affine Group. By giving your trainees with the devices that they require to discover the important concepts of direct Affine Group, you can make your trainees much more effective throughout their research studies. As a matter of fact, the lab Aid Service is so reliable that many pupils have switched over from traditional math course to the lab Aid Service. The Task Supervisor is designed to aid students manage their research. The Job Supervisor can be set up to set up how much time the trainee has available to complete their designated research. You can also establish a custom-made amount of time, which is a wonderful attribute for students who have an active schedule or an extremely active high school. This attribute can assist students prevent feeling overwhelmed with math projects. An additional useful attribute of the lab Assist Solution is the Pupil Assistant. The Student Aide helps pupils manage their job and gives them a place to upload their homework. The Student Assistant is practical for pupils that do not want to get bewildered with addressing numerous questions. As students obtain more comfortable with their tasks, they are encouraged to get in touch with the Task Supervisor as well as the Student Aide to get an on the internet support system. The on-line support group can help pupils keep their focus as they answer their assignments. All of the projects for the lab Aid Solution are included in the plan. Pupils can login and complete their appointed work while having the student aid available behind-the-scenes to help them. The lab Help Service can be a fantastic aid for your students as they start to navigate the tough university admissions as well as job hunting waters. Pupils must be prepared to get made use of to their jobs as quickly as feasible in order to reach their primary objective of entering into the college. They have to work hard sufficient to see results that will enable them to walk on at the next degree of their research studies. Obtaining utilized to the process of finishing their projects is really important. Trainees have the ability to discover various methods to help them find out exactly how to utilize the lab Assist Solution. Learning how to use the lab Assist Service is vital to students' success in college and work application. ## Pay Someone To Take My Affine Group Lab Affine Group is made use of in a lot of colleges. Some educators, however, do not utilize it very efficiently or utilize it improperly. This can have a negative influence on the pupil's learning. So, when designating jobs, use an excellent Affine Group aid service to help you with each lab. These solutions supply a variety of helpful solutions, consisting of: Tasks might require a great deal of evaluating and looking on the computer. This is when using an aid solution can be a fantastic advantage. It permits you to obtain more work done, enhance your comprehension, as well as avoid a lot of anxiety. These kinds of research services are a wonderful means to begin collaborating with the very best sort of assistance for your requirements. Affine Group is one of the most tough based on understand for pupils. Collaborating with a service, you can see to it that your demands are met, you are shown correctly, and also you comprehend the product effectively. There are numerous ways that you can teach yourself to function well with the course and achieve success. Make use of an appropriate Affine Group aid service to assist you and also obtain the work done. Affine Group is just one of the hardest courses to discover however it can be easily mastered with the appropriate aid. Having a homework solution also aids to boost the trainee's qualities. It enables you to include extra credit in addition to raise your Grade Point Average. Getting extra credit rating is commonly a huge benefit in lots of universities. Pupils that do not maximize their Affine Group course will end up moving ahead of the rest of the course. The bright side is that you can do it with a quick and also very easy solution. So, if you wish to continue in your class, use a good aid solution. One thing to remember is that if you really intend to raise your quality level, your program work needs to get done. As long as feasible, you need to recognize as well as collaborate with all your troubles. You can do this with a great aid service. One advantage of having a homework solution is that you can help yourself. If you don't feel confident in your capability to do so, then an excellent tutor will certainly be able to assist you. They will certainly have the ability to address the problems you encounter and also help you understand them so as to get a much better grade. When you finish from secondary school and also enter university, you will require to work hard in order to remain ahead of the other trainees. That means that you will require to work hard on your research. Making use of an Affine Group service can assist you get it done. Keeping your grades up can be challenging because you usually need to examine a lot and take a lot of tests. You do not have time to work on your qualities alone. Having a great tutor can be a fantastic help since they can aid you and your homework out. An assistance solution can make it less complicated for you to handle your Affine Group class. Furthermore, you can discover more regarding yourself and also help you prosper. Find the most effective tutoring solution and you will have the ability to take your research abilities to the following degree.

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Community Profile # Phodisho Makanatleng ##### Last seen: 9 dagar ago 47 total contributions since 2021 View details... Contributions in View by Solved Make a random, non-repeating vector. This is a basic MATLAB operation. It is for instructional purposes. --- If you want to get a random permutation of integer... Solved Which values occur exactly three times? Return a list of all values (sorted smallest to largest) that appear exactly three times in the input vector x. So if x = [1 2... Solved Return the largest number that is adjacent to a zero This example comes from Steve Eddins' blog: <http://blogs.mathworks.com/steve/2009/05/27/learning-lessons-from-a-one-liner/ Lear... Solved Back and Forth Rows Given a number n, create an n-by-n matrix in which the integers from 1 to n^2 wind back and forth along the rows as shown in the... Solved Maximum value in a matrix Find the maximum value in the given matrix. For example, if A = [1 2 3; 4 7 8; 0 9 1]; then the answer is 9. Solved Remove the vowels Remove all the vowels in the given phrase. Example: Input s1 = 'Jack and Jill went up the hill' Output s2 is 'Jck nd Jll wn... Solved Remove any row in which a NaN appears Given the matrix A, return B in which all the rows that have one or more <http://www.mathworks.com/help/techdoc/ref/nan.html NaN... Solved Create times-tables At one time or another, we all had to memorize boring times tables. 5 times 5 is 25. 5 times 6 is 30. 12 times 12 is way more th... Solved Arrange Vector in descending order If x=[0,3,4,2,1] then y=[4,3,2,1,0] Solved Who Has the Most Change? You have a matrix for which each row is a person and the columns represent the number of quarters, nickels, dimes, and pennies t... Solved Check if number exists in vector Return 1 if number _a_ exists in vector _b_ otherwise return 0. a = 3; b = [1,2,4]; Returns 0. a = 3; b = [1,... Solved Flip the vector from right to left Flip the vector from right to left. Examples x=[1:5], then y=[5 4 3 2 1] x=[1 4 6], then y=[6 4 1]; Request not ... Solved Reverse the vector Reverse the vector elements. Example: Input x = [1,2,3,4,5,6,7,8,9] Output y = [9,8,7,6,5,4,3,2,1] Solved Find the numeric mean of the prime numbers in a matrix. There will always be at least one prime in the matrix. Example: Input in = [ 8 3 5 9 ] Output out is 4... Solved Convert from Fahrenheit to Celsius Given an input vector |F| containing temperature values in Fahrenheit, return an output vector |C| that contains the values in C... Solved Sum all integers from 1 to 2^n Given the number x, y must be the summation of all integers from 1 to 2^x. For instance if x=2 then y must be 1+2+3+4=10. Solved Triangle sequence A sequence of triangles is constructed in the following way: 1) the first triangle is Pythagoras' 3-4-5 triangle 2) the s... Solved Project Euler: Problem 2, Sum of even Fibonacci Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 te... Solved Fibonacci-Sum of Squares Given the Fibonacci sequence defined by the following recursive relation, * F(n) = F(n-1) + F(n-2) * where F(1) = 1 and F(1)... Solved Number of Even Elements in Fibonacci Sequence Find how many even Fibonacci numbers are available in the first d numbers. Consider the following first 14 numbers 1 1 2... Solved Fibonacci sequence Calculate the nth Fibonacci number. Given n, return f where f = fib(n) and f(1) = 1, f(2) = 1, f(3) = 2, ... Examples: Inpu... Solved Return the 3n+1 sequence for n A Collatz sequence is the sequence where, for a given number n, the next number in the sequence is either n/2 if the number is e... Solved Remove the air bubbles Given a matrix a, return a matrix b in which all the zeros have "bubbled" to the top. That is, any zeros in a given column shoul... Solved ~~~~~~~ WAVE ~~~~~~~~~ |The WAVE generator| Once upon a time there was a river. 'Sum' was passing by the river. He saw the water of the river that w... Solved Draw 'C'. Given x as input, generate a x-by-x matrix 'C' using 0 and 1. example: x=4 ans= [0 1 1 1 1 0 0 0 ... Solved Draw 'J' Given n as input, generate a n-by-n matrix 'J' using 0 and 1 . Example: n=5 ans= [0 0 0 0 1 0 0 0 0 1 0 0 ... Solved Draw 'H' Draw a x-by-x matrix 'H' using 1 and 0. (x is odd and bigger than 2) Example: x=5 ans= [1 0 0 0 1 1 0 0 0 1 ... Solved Draw a 'X'! Given n as input Draw a 'X' in a n-by-n matrix. example: n=3 y=[1 0 1 0 1 0 1 0 1] n=4 y=[1 0 0... Solved Draw a 'N'! Given n as input, generate a n-by-n matrix 'N' using 0 and 1 . Example: n=5 ans= [1 0 0 0 1 1 1 0 0 1 1 0 ...

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Document # On Packing Low-Diameter Spanning Trees ## File LIPIcs.ICALP.2020.33.pdf • Filesize: 0.61 MB • 18 pages ## Cite As Julia Chuzhoy, Merav Parter, and Zihan Tan. On Packing Low-Diameter Spanning Trees. In 47th International Colloquium on Automata, Languages, and Programming (ICALP 2020). Leibniz International Proceedings in Informatics (LIPIcs), Volume 168, pp. 33:1-33:18, Schloss Dagstuhl – Leibniz-Zentrum für Informatik (2020) https://doi.org/10.4230/LIPIcs.ICALP.2020.33 ## Abstract Edge connectivity of a graph is one of the most fundamental graph-theoretic concepts. The celebrated tree packing theorem of Tutte and Nash-Williams from 1961 states that every k-edge connected graph G contains a collection 𝒯 of ⌊k/2⌋ edge-disjoint spanning trees, that we refer to as a tree packing; the diameter of the tree packing 𝒯 is the largest diameter of any tree in 𝒯. A desirable property of a tree packing for leveraging the high connectivity of a graph in distributed communication networks, is that its diameter is low. Yet, despite extensive research in this area, it is still unclear how to compute a tree packing of a low-diameter graph G, whose diameter is sublinear in |V(G)|, or, alternatively, how to show that such a packing does not exist. In this paper, we provide first non-trivial upper and lower bounds on the diameter of tree packing. We start by showing that, for every k-edge connected n-vertex graph G of diameter D, there is a tree packing 𝒯 containing Ω(k) trees, of diameter O((101k log n)^D), with edge-congestion at most 2. Karger’s edge sampling technique demonstrates that, if G is a k-edge connected graph, and G[p] is a subgraph of G obtained by sampling each edge of G independently with probability p = Θ(log n/k), then with high probability G[p] is connected. We extend this result to show that the diameter of G[p] is bounded by O(k^(D(D+1)/2)) with high probability. This immediately gives a tree packing of Ω(k/log n) edge-disjoint trees of diameter at most O(k^(D(D+1)/2)). We also show that these two results are nearly tight for graphs with a small diameter: we show that there are k-edge connected graphs of diameter 2D, such that any packing of k/α trees with edge-congestion η contains at least one tree of diameter Ω((k/(2α η D))^D), for any k,α and η. Additionally, we show that if, for every pair u,v of vertices of a given graph G, there is a collection of k edge-disjoint paths connecting u to v, of length at most D each, then we can efficiently compute a tree packing of size k, diameter O(D log n), and edge-congestion O(log n). Finally, we provide several applications of low-diameter tree packing in the distributed settings of network optimization and secure computation. ## Subject Classification ##### ACM Subject Classification • Theory of computation → Design and analysis of algorithms ##### Keywords • Spanning tree • packing • edge-connectivity ## Metrics • Access Statistics • Total Accesses (updated on a weekly basis) 0 ## References 1. Keren Censor-Hillel, Mohsen Ghaffari, George Giakkoupis, Bernhard Haeupler, and Fabian Kuhn. Tight bounds on vertex connectivity under vertex sampling. In Proceedings of the twenty-sixth annual ACM-SIAM symposium on Discrete algorithms, pages 2006-2018. Society for Industrial and Applied Mathematics, 2015. 2. Keren Censor-Hillel, Mohsen Ghaffari, and Fabian Kuhn. Distributed connectivity decomposition. In ACM Symposium on Principles of Distributed Computing, PODC '14, Paris, France, July 15-18, 2014, pages 156-165, 2014. 3. Keren Censor-Hillel, Mohsen Ghaffari, and Fabian Kuhn. A new perspective on vertex connectivity. In Proceedings of the twenty-fifth annual ACM-SIAM symposium on Discrete algorithms, pages 546-561. Society for Industrial and Applied Mathematics, 2014. 4. Mohit Daga, Monika Henzinger, Danupon Nanongkai, and Thatchaphol Saranurak. Distributed edge connectivity in sublinear time. In Proceedings of the 51st Annual ACM SIGACT Symposium on Theory of Computing, pages 343-354. ACM, 2019. 5. Michal Dory. Distributed approximation of minimum k-edge-connected spanning subgraphs. In Proceedings of the 2018 ACM Symposium on Principles of Distributed Computing, pages 149-158. ACM, 2018. 6. Mohsen Ghaffari. Distributed broadcast revisited: Towards universal optimality. In International Colloquium on Automata, Languages, and Programming, pages 638-649. Springer, 2015. 7. Mohsen Ghaffari. Improved Distributed Algorithms for Fundamental Graph Problems. PhD thesis, MIT, USA, 2017. URL: https://groups.csail.mit.edu/tds/papers/Ghaffari/PhDThesis-Ghaffari.pdf. 8. Mohsen Ghaffari and Bernhard Haeupler. Distributed algorithms for planar networks ii: Low-congestion shortcuts, mst, and min-cut. In Proceedings of the twenty-seventh annual ACM-SIAM symposium on Discrete algorithms, pages 202-219. SIAM, 2016. 9. Mohsen Ghaffari and Bernhard Haeupler. Distributed algorithms for planar networks II: low-congestion shortcuts, mst, and min-cut. In Proceedings of the Twenty-Seventh Annual ACM-SIAM Symposium on Discrete Algorithms, SODA 2016, Arlington, VA, USA, January 10-12, 2016, pages 202-219, 2016. 10. Mohsen Ghaffari and Fabian Kuhn. Distributed minimum cut approximation. In International Symposium on Distributed Computing, pages 1-15. Springer, 2013. 11. Mohsen Ghaffari and Krzysztof Nowicki. Faster algorithms for edge connectivity via random out contractions, 2019. 12. Alon Itai and Michael Rodeh. The multi-tree approach to reliability in distributed networks. Information and Computation, 79(1):43-59, 1988. 13. Tomáš Kaiser. A short proof of the tree-packing theorem. Discrete Mathematics, 312(10):1689-1691, 2012. 14. David R Karger. Random sampling in cut, flow, and network design problems. Mathematics of Operations Research, 24(2):383-413, 1999. 15. Naoki Kitamura, Hirotaka Kitagawa, Yota Otachi, and Taisuke Izumi. Low-congestion shortcut and graph parameters. In 33rd International Symposium on Distributed Computing (DISC 2019). Schloss Dagstuhl - Leibniz-Zentrum für Informatik, 2019. 16. Fabian Kuhn. A distributed perspective on graph connectivity and cuts. In 26th ACM Symposium on Parallelism in Algorithms and Architectures, SPAA '14, Prague, Czech Republic - June 23 - 25, 2014, page 1, 2014. 17. Zvi Lotker, Boaz Patt-Shamir, and David Peleg. Distributed MST for constant diameter graphs. Distributed Computing, 18(6):453-460, 2006. 18. Danupon Nanongkai and Hsin-Hao Su. Almost-tight distributed minimum cut algorithms. In International Symposium on Distributed Computing, pages 439-453. Springer, 2014. 19. CSJA Nash-Williams. Edge-disjoint spanning trees of finite graphs. Journal of the London Mathematical Society, 1(1):445-450, 1961. 20. Merav Parter and Eylon Yogev. Low congestion cycle covers and their applications. In Proceedings of the Thirtieth Annual ACM-SIAM Symposium on Discrete Algorithms, pages 1673-1692. SIAM, 2019. 21. David Peleg. Distributed Computing: A Locality-sensitive Approach. SIAM, 2000. 22. David Pritchard and Ramakrishna Thurimella. Fast computation of small cuts via cycle space sampling. ACM Transactions on Algorithms (TALG), 7(4):46, 2011. 23. Atish Das Sarma, Stephan Holzer, Liah Kor, Amos Korman, Danupon Nanongkai, Gopal Pandurangan, David Peleg, and Roger Wattenhofer. Distributed verification and hardness of distributed approximation. SIAM Journal on Computing, 41(5):1235-1265, 2012. 24. William Thomas Tutte. On the problem of decomposing a graph into n connected factors. Journal of the London Mathematical Society, 1(1):221-230, 1961. X Feedback for Dagstuhl Publishing

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src/HOLCF/Tools/domain/domain_axioms.ML author paulson Fri, 05 Oct 2007 09:59:03 +0200 changeset 24854 0ebcd575d3c6 parent 24712 64ed05609568 child 26012 f6917792f8a4 permissions -rw-r--r-- filtering out some package theorems ``` (* Title: HOLCF/Tools/domain/domain_axioms.ML ID: \$Id\$ Author: David von Oheimb Syntax generator for domain command. *) structure Domain_Axioms = struct local open Domain_Library; infixr 0 ===>;infixr 0 ==>;infix 0 == ; infix 1 ===; infix 1 ~= ; infix 1 <<; infix 1 ~<<; infix 9 ` ; infix 9 `% ; infix 9 `%%; infixr 9 oo; fun calc_axioms comp_dname (eqs : eq list) n (((dname,_),cons) : eq)= let (* ----- axioms and definitions concerning the isomorphism ------------------ *) val dc_abs = %%:(dname^"_abs"); val dc_rep = %%:(dname^"_rep"); val x_name'= "x"; val x_name = idx_name eqs x_name' (n+1); val dnam = Sign.base_name dname; val abs_iso_ax = ("abs_iso", mk_trp(dc_rep`(dc_abs`%x_name') === %:x_name')); val rep_iso_ax = ("rep_iso", mk_trp(dc_abs`(dc_rep`%x_name') === %:x_name')); val when_def = ("when_def",%%:(dname^"_when") == foldr (uncurry /\ ) (/\x_name'((when_body cons (fn (x,y) => Bound(1+length cons+x-y)))`(dc_rep`Bound 0))) (when_funs cons)); val copy_def = let fun idxs z x arg = if is_rec arg then (cproj (Bound z) eqs (rec_of arg))`Bound(z-x) else Bound(z-x); fun one_con (con,args) = foldr /\# (list_ccomb (%%:con, mapn (idxs (length args)) 1 args)) args; in ("copy_def", %%:(dname^"_copy") == /\"f" (list_ccomb (%%:(dname^"_when"), map one_con cons))) end; (* -- definitions concerning the constructors, discriminators and selectors - *) fun con_def m n (_,args) = let fun idxs z x arg = (if is_lazy arg then fn t => %%:upN`t else I) (Bound(z-x)); fun parms vs = mk_stuple (mapn (idxs(length vs)) 1 vs); fun inj y 1 _ = y | inj y _ 0 = %%:sinlN`y | inj y i j = %%:sinrN`(inj y (i-1) (j-1)); in foldr /\# (dc_abs`(inj (parms args) m n)) args end; val con_defs = mapn (fn n => fn (con,args) => (extern_name con ^"_def", %%:con == con_def (length cons) n (con,args))) 0 cons; val dis_defs = let fun ddef (con,_) = (dis_name con ^"_def",%%:(dis_name con) == list_ccomb(%%:(dname^"_when"),map (fn (con',args) => (foldr /\# (if con'=con then %%:TT_N else %%:FF_N) args)) cons)) in map ddef cons end; val mat_defs = let fun mdef (con,_) = (mat_name con ^"_def",%%:(mat_name con) == list_ccomb(%%:(dname^"_when"),map (fn (con',args) => (foldr /\# (if con'=con then %%:returnN`(mk_ctuple (map (bound_arg args) args)) else %%:failN) args)) cons)) in map mdef cons end; val pat_defs = let fun pdef (con,args) = let val ps = mapn (fn n => fn _ => %:("pat" ^ string_of_int n)) 1 args; val xs = map (bound_arg args) args; val r = Bound (length args); val rhs = case args of [] => %%:returnN ` HOLogic.unit | _ => foldr1 cpair_pat ps ` mk_ctuple xs; fun one_con (con',args') = foldr /\# (if con'=con then rhs else %%:failN) args'; in (pat_name con ^"_def", list_comb (%%:(pat_name con), ps) == list_ccomb(%%:(dname^"_when"), map one_con cons)) end in map pdef cons end; val sel_defs = let fun sdef con n arg = Option.map (fn sel => (sel^"_def",%%:sel == list_ccomb(%%:(dname^"_when"),map (fn (con',args) => if con'<>con then UU else foldr /\# (Bound (length args - n)) args) cons))) (sel_of arg); in List.mapPartial I (List.concat(map (fn (con,args) => mapn (sdef con) 1 args) cons)) end; (* ----- axiom and definitions concerning induction ------------------------- *) val reach_ax = ("reach", mk_trp(cproj (%%:fixN`%%(comp_dname^"_copy")) eqs n `%x_name === %:x_name)); val take_def = ("take_def",%%:(dname^"_take") == mk_lam("n",cproj (%%:iterateN \$ Bound 0 ` %%:(comp_dname^"_copy") ` UU) eqs n)); val finite_def = ("finite_def",%%:(dname^"_finite") == mk_lam(x_name, mk_ex("n",(%%:(dname^"_take") \$ Bound 0)`Bound 1 === Bound 1))); in (dnam, [abs_iso_ax, rep_iso_ax, reach_ax], [when_def, copy_def] @ con_defs @ dis_defs @ mat_defs @ pat_defs @ sel_defs @ [take_def, finite_def]) end; (* let *) fun infer_props thy = map (apsnd (FixrecPackage.legacy_infer_prop thy)); in (* local *) fun add_axioms (comp_dnam, eqs : eq list) thy' = let val comp_dname = Sign.full_name thy' comp_dnam; val dnames = map (fst o fst) eqs; val x_name = idx_name dnames "x"; fun copy_app dname = %%:(dname^"_copy")`Bound 0; val copy_def = ("copy_def" , %%:(comp_dname^"_copy") == /\"f"(foldr1 cpair (map copy_app dnames))); val bisim_def = ("bisim_def",%%:(comp_dname^"_bisim")==mk_lam("R", let fun one_con (con,args) = let val nonrec_args = filter_out is_rec args; val rec_args = List.filter is_rec args; val recs_cnt = length rec_args; val allargs = nonrec_args @ rec_args @ map (upd_vname (fn s=> s^"'")) rec_args; val allvns = map vname allargs; fun vname_arg s arg = if is_rec arg then vname arg^s else vname arg; val vns1 = map (vname_arg "" ) args; val vns2 = map (vname_arg "'") args; val allargs_cnt = length nonrec_args + 2*recs_cnt; val rec_idxs = (recs_cnt-1) downto 0; val nonlazy_idxs = map snd (filter_out (fn (arg,_) => is_lazy arg) (allargs~~((allargs_cnt-1) downto 0))); fun rel_app i ra = proj (Bound(allargs_cnt+2)) eqs (rec_of ra) \$ Bound (2*recs_cnt-i) \$ Bound (recs_cnt-i); val capps = foldr mk_conj (mk_conj( Bound(allargs_cnt+1)===list_ccomb(%%:con,map (bound_arg allvns) vns1), Bound(allargs_cnt+0)===list_ccomb(%%:con,map (bound_arg allvns) vns2))) (mapn rel_app 1 rec_args); in foldr mk_ex (Library.foldr mk_conj (map (defined o Bound) nonlazy_idxs,capps)) allvns end; fun one_comp n (_,cons) =mk_all(x_name(n+1),mk_all(x_name(n+1)^"'",mk_imp( proj (Bound 2) eqs n \$ Bound 1 \$ Bound 0, foldr1 mk_disj (mk_conj(Bound 1 === UU,Bound 0 === UU) ::map one_con cons)))); in foldr1 mk_conj (mapn one_comp 0 eqs)end ));

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# [seqfan] Re: Sequence needing more terms, A6156, squarefree ternary words Neil Sloane njasloane at gmail.com Sat Jul 28 02:37:17 CEST 2012 ```Concerning A006156, thanks to everyone who responded. Ron, could you enter the triangle T(n,k) when you reach a stopping point, and a second triangle giving the coefficients of the polynomials that generate the rows (even if they are only empirical)? Also the columns giving the number of squarefree words over alphabets of sizes 4, 5, 6, ... (assuming they aren't already in the OEIS) Thanks! Neil On Fri, Jul 27, 2012 at 2:50 PM, Ron Hardin <rhhardin at att.net> wrote: > It's possible I can add a term, but the thing hits an exponential-looking > cost > wall somewhere around there. > > If you make a T(n,k) table, the rows are polynomials and maybe somebody > can play > guess the coefficients > > current state: > > T(n,k)=Number of square-free words of length n in a (k+1)-ary alphabet > > Table starts > > .2...3.....4......5.......6........7.........8..........9.........10.........11 > > .2...6....12.....20......30.......42........56.........72.........90........110 > > .2..12....36.....80.....150......252.......392........576........810.......1100 > > .0..18....96....300.....720.....1470......2688.......4536.......7200......10890 > > .0..30...264...1140....3480.....8610.....18480......35784......64080.....107910 > > .0..42...696...4260...16680....50190....126672.....281736.....569520....1068210 > > .0..60..1848..15960...80040...292740....868560....2218608....5062320...10575180 > > .0..78..4848..59580..383520..1706250...5953248...17467128...44991360..104683590 > > .0.108.12768.222600.1838160..9946020..40806528..137522448..399866400.1036270620 > > .0.144.33480.830880.8807400.57970080.279692784.1082712960.3553806960........... > > Rows 1-10 > Empirical: a(k) = 1*k + 1 > Empirical: a(k) = 1*k^2 + 1*k > Empirical: a(k) = 1*k^3 + 1*k^2 > Empirical: a(k) = 1*k^4 + 1*k^3 - 1*k^2 - 1*k > Empirical: a(k) = 1*k^5 + 1*k^4 - 2*k^3 - 1*k^2 + 1*k > Empirical: a(k) = 1*k^6 + 1*k^5 - 3*k^4 - 2*k^3 + 2*k^2 + 1*k > Empirical: a(k) = 1*k^7 + 1*k^6 - 4*k^5 - 3*k^4 + 5*k^3 + 2*k^2 - 2*k > Empirical: a(k) = 1*k^8 + 1*k^7 - 5*k^6 - 4*k^5 + 8*k^4 + 4*k^3 - 4*k^2 - > 1*k > Empirical: a(k) = 1*k^9 + 1*k^8 - 6*k^7 - 5*k^6 + 12*k^5 + 8*k^4 - 9*k^3 - > 4*k^2 > + 2*k > Empirical: a(k) = 1*k^10 + 1*k^9 - 7*k^8 - 6*k^7 + 17*k^6 + 12*k^5 - > 17*k^4 - > 7*k^3 + 6*k^2 > > > > rhhardin at mindspring.com > rhhardin at att.net (either) > > > > ----- Original Message ---- > > From: Neil Sloane <njasloane at gmail.com> > > To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu> > > Cc: R. H. Hardin <rhhardin at mindspring.com> > > Sent: Fri, July 27, 2012 12:44:10 PM > > Subject: [seqfan] Sequence needing more terms, A6156, squarefree ternary > words > > > > A006156 is the subject of several research papers, so it would > > be nice to have a b-file. Ron, is this something that can be > > handled by your techniques? > > Neil > > > > -- > > Dear Friends, I have now retired from AT&T. New coordinates: > > > > Neil J. A. Sloane, President, OEIS Foundation > > 11 South Adelaide Avenue, Highland Park, NJ 08904, USA > > Email: njasloane at gmail.com > > > > _______________________________________________ > > > > Seqfan Mailing list - http://list.seqfan.eu/ > > > > _______________________________________________ > > Seqfan Mailing list - http://list.seqfan.eu/ > -- Dear Friends, I have now retired from AT&T. New coordinates: Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA

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{[ promptMessage ]} Bookmark it {[ promptMessage ]} # Quiz3sol - Math 107 Section 03 Quiz Sample{idiom 1 Solve... This preview shows pages 1–2. Sign up to view the full content. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: Math 107 Section 03 Quiz; Sample {idiom 1 Solve for :r. :L e -5 [:112a:3e“3I—4I35-3r=n 93% (2 1":(l- 190:0 (2 i 70 :1!me = _.L [b]e"-+-4—21c_"=fl 3‘ 0’ 1 Qifeh 4~2!€'*)=u =7» e”+r+e"— not: [c] lemma) -+-mg5(I—2) =1 :‘2 (6+1 H615) =39 exiq Lug; lxi'i'BX-lgjgl z") x2+514H32 efifi % . =‘> {JfiXflSw = ,-so§;;; " #1 nos 2 Fmd the domain ll? (DWP 39d [a] y = In£2 + a: — 1:2) [1)] y = inlet — 5| let-Ibo lxe H :2 Pram) ) Hid- [c]y=£og(:r+1)+iog-—m -- swim: _ X') "I " leaflet? J) (Sex {/11 :v _l<¥qfi Edi-m _ t 2. 1+‘K it :0 :‘> I‘X—lfo :3 (ramming :7 -l.=-;u<€2 r --.-. 3 Given P (—3333) is a. terminal point associated witii i. unit circle, and t is in quadrant II. Find ainfl), cosfi], imam, coifl}, tacit), seep). -11, 5?hl£]=fl20 since '{rn fencirej‘l ll. giniflgfli- are = i '1 emit}: :[Z csc{1)=-,L- = l: _:___ = :1 = same a 4- it) I . SEE“ 5mm 5! mute) 1E6}; q; 4 State tile am}: 11': de and period of lie or iowmg fur 1 11; nd graph 2 complete periods. Hm) = 3cos[-—2;r} : 5mg. [1 in Amflfiudf 2 3}, Par-pal ,1 5 State the amplitude, period, phase shift of the following functions. [a] y = —4sin[%(:r + fimtli‘l'wle : : %— : q E (AM-2. Slu'fl _« .. .EI [13] y = flsiHG-r - T7) = stn [-g- {x— £11] wrmte .. 2 Fermi = as, s plum-t1 s ‘ a} 5 State the period, phase shift and vertical asyniptotes of the fellow- ings: [a] y = ta.n.(2:r + tan Elf): - frfifl] Fulfil ‘ 12E Ell'ifi- ‘ “ii Len-Hal a m be; ': “E14111 _ s. -lmvgalh] y = rot 3.1: — 5 97 Fla 1 :1. g 2 Ito-t 5{x-—-2-1] =-§~1.EII. “5.2- ‘fl : I - .' . " 11' P9. ' Meet?» M‘ 3““? ill .MEZ 7 Find the length of an are that snbtends a central angle of in a circle of radius Sm. a: r- E) : 8%:27: 8 The area of a circle is 72m? Find the area of a sector of this circle that subtends a central angle of 3H". \$0°=%- were 12=r‘*% ma . 9 Express the length m in terms of m and tratlos of 5'. JE : mug) :‘2 3:. lh‘fimffl') AABC 1-3-9. 80, among-aim So if statics?» = saw) u-- .___'__..._._...._. _.._ __ 2 at: idea-n19] = m tam) - Sin-(.9) ... View Full Document {[ snackBarMessage ]} ### Page1 / 2 Quiz3sol - Math 107 Section 03 Quiz Sample{idiom 1 Solve... This preview shows document pages 1 - 2. Sign up to view the full document. View Full Document Ask a homework question - tutors are online

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# How to Calculate Force From Velocity ••• radsport image by JanUFotO from Fotolia.com Print #### Things You'll Need • Calculator • Compendium of physics formulas The first equation taught in dynamics is F= ma which is “force equals mass times acceleration.” This equation describes the force that must be exerted on an object of known weight or mass as it is accelerated or decelerated. If a racing bicycle with a rider traveling at 20 miles per hour must stop within a certain distance, you can calculate how much force will be applied to the caliper brake on the rim of the rear wheel. You can also prove that doubling velocity quadruples (squares) the force required to stop. ••• healthy life image by .shock from Fotolia.com Define the velocity to force application. In this example, the bicycle with its rider weighs 210 pounds. The rider notes a white stop line that is 30 feet in front of him when he applies the brake. Since you already know the velocity, you now have enough information to calculate the required braking force. ••• Car Brake image by Joelyn Pullano from Fotolia.com Solve for time T, which will enable you to calculate acceleration, or in this case, deceleration. The average velocity over the 30 feet is 20 mph divided by two, or 10 mph, which is 14.66-feet-per-second. If the 30 feet are covered at an average speed of 14.66-feet-per-second, then it takes 2.045 seconds to stop. Solve for acceleration using the 2.045 seconds to cover 30 feet. Since the distance calculation is D = v(0) x T +1/2 (a) T^2, the first term can be ignored since all distance covered is accounted for by the deceleration to zero. Therefore, 30 feet equals ½ a xT^2, which is 30 = ½ a x 2.045 ^2 or 30 = 1/2 a x 4.18. Re-arranging, a = 30 x 2/4.18 = 14.35 feet per second/sec. Solve for force using the F=ma basic equation. Force F = 210 x 14.35 feet per second/sec / 32.2 feet per second/sec (acceleration of gravity) or 93.58 pounds of force consistently applied by the brake to the rim for 2.045 seconds to stop the bike. This is probably right at the practical limit of this bicycle’s capability to stop. ••• bike race image by jeancliclac from Fotolia.com Prove that doubling the velocity quadruples the required force. A 40 mile per hour velocity would result in a stopping time of 1.023 seconds, one-half of 2.045 seconds in the first instance. The D = ½ x a x T^2 term would work out to an acceleration of a = 30 x 2/1.046, or 57.36 feet per second/sec. F = ma would therefore work out to F = 374.08 pounds, very unreasonable for a caliper brake on a skinny racing tire. This foolish rider would never stop from 40 mph in the 30-foot distance, and they would streak right past the stop sign. #### Tips • Always remember that stopping force quadruples as velocity doubles. #### Warnings • Accelerating quickly to a given velocity uses more force and far more fuel than smooth acceleration. Dont Go! We Have More Great Sciencing Articles!

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# ISEE Lower Level Math : Distributive Property ## Example Questions ### Example Question #71 : Isee Lower Level (Grades 5 6) Mathematics Achievement Simplify Explanation: To use the distributive property, you multiply the outside value by each one in the parentheses. and . ### Example Question #72 : Isee Lower Level (Grades 5 6) Mathematics Achievement Given that , simplify: Explanation: To solve this problem, first substitute  for : Then apply the distributive property: Where, Thus, the final solution is: ### Example Question #73 : Isee Lower Level (Grades 5 6) Mathematics Achievement Multiply: Explanation: To multiply these terms apply the Distributive property: , Where Thus, the solution is: is the final answer because this expression can't be simplified. ### Example Question #74 : Isee Lower Level (Grades 5 6) Mathematics Achievement Find an expression with an equivalent value to: Explanation: In order to find an equivalent expression to  consider the Distributive property: It is necessary to select an answer choice where the product of  and where the product of Therefore, the solution is  because ### Example Question #75 : Isee Lower Level (Grades 5 6) Mathematics Achievement Multiply: Explanation: To multiply these terms apply the Distributive property by applying the FOIL method. The FOIL method requires that we multiply each term and then combine like terms: Notice that in this problem Therefore, the solution is: ### Example Question #76 : Isee Lower Level (Grades 5 6) Mathematics Achievement Find an expression with an equivalent value to: Explanation: In order to find an equivalent expression to  consider the Distributive property: It is necessary to select an answer choice where the product of  and where the product of Therefore, the solution is  because: ### Example Question #77 : Isee Lower Level (Grades 5 6) Mathematics Achievement Multiply: Explanation: To multiply these terms apply the Distributive property by applying the FOIL method. The FOIL method requires that we multiply each term and then combine like terms: Note: FOIL stands for First, Outside, Inside, Last, where First Outside Inside Last However, in this particular problem we will not end up with any like terms that we can sum together. ### Example Question #78 : Isee Lower Level (Grades 5 6) Mathematics Achievement Given that evaluate: Explanation: To solve this problem, first substitute  for : or once you've substituted  for , apply the distributive property: Where, ### Example Question #79 : Isee Lower Level (Grades 5 6) Mathematics Achievement Multiply: Explanation: To evaluate these terms apply the Distributive property: Then, combine any like terms to simplify the solution. The solution is: ### Example Question #80 : Isee Lower Level (Grades 5 6) Mathematics Achievement Multiply: Explanation: To multiply these terms use the Distributive property by applying the FOIL method. The FOIL method requires that we multiply each term and then combine like terms: Note: FOIL stands for First, Outside, Inside, Last, where First Outside Inside Last In this problem, Thus, The final solution is:

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Had been usage in Kaplan : GMAT Sentence Correction (SC) Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack It is currently 20 Jan 2017, 06:15 ### 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 # Had been usage in Kaplan Author Message TAGS: ### Hide Tags Intern Joined: 14 Jul 2009 Posts: 44 Followers: 1 Kudos [?]: 12 [0], given: 0 ### Show Tags 22 Sep 2009, 22:37 00:00 Difficulty: (N/A) Question Stats: 0% (00:00) correct 0% (00:00) wrong based on 0 sessions ### HideShow timer Statistics Question: Having forfeited her severance package in order to keep the rights to her intellectual property, it was believed by the employee that she had won a moral victory. Incorrect: Having forfeited her severance package in order to keep the rights to her intellectual property, the employee believed that the moral victory was won. Option above was described as grammatically incorrect because was is the wrong tense. Instead, it should be had been. Can somebody please explain why had been is correct and was is not? If you have any questions New! Kaplan GMAT Prep Discount Codes Jamboree Discount Codes EMPOWERgmat Discount Codes Kaplan GMAT Instructor Joined: 25 Aug 2009 Posts: 644 Location: Cambridge, MA Followers: 83 Kudos [?]: 276 [0], given: 2 ### Show Tags 23 Sep 2009, 14:20 Off the top of my head, the incorrect answer is passive voice: victory was won [[by her]]. While technically correct grammar, the passive voice is rarely stylistically correct. _________________ Eli Meyer Kaplan Teacher http://www.kaptest.com/GMAT Prepare with Kaplan and save \$150 on a course! Kaplan Reviews Similar topics Replies Last post Similar Topics: The telecom minister had been 0 22 Jun 2014, 20:24 3 When do we use Had and Had been? 3 08 Nov 2012, 02:38 1 I am confused about the usage of these two idioms. Say I had 4 03 May 2011, 21:22 The family s mood, which had been enthusiastic at the 5 20 Dec 2008, 23:46 1 Sleeping pills had been showing up with regularity as a 9 01 Jul 2007, 08:25 Display posts from previous: Sort by

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## 2625 Days Before May 26, 2023 Want to figure out the date that is exactly two thousand six hundred twenty five days before May 26, 2023 without counting? Your starting date is May 26, 2023 so that means that 2625 days earlier would be March 18, 2016. You can check this by using the date difference calculator to measure the number of days before Mar 18, 2016 to May 26, 2023. March 2016 • Sunday • Monday • Tuesday • Wednesday • Thursday • Friday • Saturday 1. 1 2. 2 3. 3 4. 4 5. 5 1. 6 2. 7 3. 8 4. 9 5. 10 6. 11 7. 12 1. 13 2. 14 3. 15 4. 16 5. 17 6. 18 7. 19 1. 20 2. 21 3. 22 4. 23 5. 24 6. 25 7. 26 1. 27 2. 28 3. 29 4. 30 5. 31 March 18, 2016 is a Friday. It is the 78th day of the year, and in the 11st week of the year (assuming each week starts on a Sunday), or the 1st quarter of the year. There are 31 days in this month. 2016 is a leap year, so there are 366 days in this year. The short form for this date used in the United States is 03/18/2016, and almost everywhere else in the world it's 18/03/2016. ### What if you only counted weekdays? In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 2625 weekdays before May 26, 2023, you can count up each day skipping Saturdays and Sundays. Start your calculation with May 26, 2023, which falls on a Friday. Counting forward, the next day would be a Monday. To get exactly two thousand six hundred twenty five weekdays before May 26, 2023, you actually need to count 3675 total days (including weekend days). That means that 2625 weekdays before May 26, 2023 would be May 3, 2013. If you're counting business days, don't forget to adjust this date for any holidays. May 2013 • Sunday • Monday • Tuesday • Wednesday • Thursday • Friday • Saturday 1. 1 2. 2 3. 3 4. 4 1. 5 2. 6 3. 7 4. 8 5. 9 6. 10 7. 11 1. 12 2. 13 3. 14 4. 15 5. 16 6. 17 7. 18 1. 19 2. 20 3. 21 4. 22 5. 23 6. 24 7. 25 1. 26 2. 27 3. 28 4. 29 5. 30 6. 31 May 3, 2013 is a Friday. It is the 123rd day of the year, and in the 123rd week of the year (assuming each week starts on a Sunday), or the 2nd quarter of the year. There are 31 days in this month. 2013 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 05/03/2013, and almost everywhere else in the world it's 03/05/2013. ### Enter the number of days and the exact date Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date.

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# Upper bounding the variance of a sum I have random variables $$X_1, \ldots, X_N$$ and two functions: a bounded function $$|f|\leq M$$ and another function $$0 \le g \le 1$$. I would like to upper bound the variance $$\mathbb{V}\left[\sum_{i=1}^N f(X_i)g(X_i) \right]$$ by a bound that does not depend on $$f$$ (happy for it to depend on $$g$$). Is it possible? # Attempt I tried by using the usual variance formula, but I am unsure if I can do this type of bounds \begin{align} \mathbb{V}\left[\sum_{i=1}^N f(X_i)g(X_i) \right] &= \mathbb{E}\left[\left(\sum_{i=1}^N f(X_i)g(X_i)\right)^2\right] - \mathbb{E}\left[\sum_{i=1}^N f(X_i)g(X_i)\right]^2 \\ &\leq \mathbb{E}\left[\left(\sum_{i=1}^N M g(X_i)\right)^2\right]- \mathbb{E}\left[\sum_{i=1}^N f(X_i)g(X_i)\right]^2 \\ &= M^2 \mathbb{E}\left[\left(\sum_{i=1}^N g(X_i)\right)^2\right] - \mathbb{E}\left[\sum_{i=1}^N f(X_i)g(X_i)\right]^2 \\ &\leq M^2 \mathbb{E}\left[\left(\sum_{i=1}^N g(X_i)\right)^2\right] \end{align} but I was wondering if I can do anything better. One idea I had was to work directly on the formula for the variance of a sum $$\mathbb{V}\left[\sum_{i=1}^N f(X_i)g(X_i)\right] = \sum_{i=1}^N \mathbb{V}[f(X_i)g(X_i)] + 2 \sum_{i=1}^N \sum_{j=i+1}^N \text{Cov}\left(f(X_i)g(X_i), f(X_j)g(X_j)\right)$$ • That inequality isn't valid as written (what if $\sum g(X_i) = 0$?); you'd need $\sum M|g(X_i)|$ instead. Mar 21 at 15:58 • @GregMartin Mmmm I think I forgot to write a condition. Here I know that $g$ is non-negative as well Mar 21 at 16:00 • $(X_n)_{n=1,..,N}$ are i.i.d? – NN2 Mar 21 at 16:34 • @NN2 Yes, they are! $X_1, \ldots X_n \overset{\text{iid}}{\sim} \mu$ Mar 21 at 16:56 From the assumption that $$(X_i)_{i=1,..,N}$$ are i.i.d, we deduce that $$(f(X_i)g(X_i))_{i=1,..,N}$$ are also i.i.d. As a consequence \begin{align} L:=\mathbb{V}\left(\sum_{i=1}^N f(X_i)g(X_i) \right) &= n\cdot \mathbb{V}\left( f(X_1)g(X_1) \right) \\ &= n\cdot \mathbb{E}\left(f^2(X_1)g^2(X_1) \right)-n\cdot \mathbb{E}^2\left(f(X_1)g(X_1) \right)\\ \end{align} Without any further information about $$f,g$$ and $$X$$, a best upper bound is $$L\stackrel{(1)}{\le}n\cdot \mathbb{E}\left(f^2(X_1)g^2(X_1) \right)\stackrel{(2)}{\le}\color{red}{nM^2\cdot \mathbb{E}\left(g^2(X_1) \right)}$$ The equality of $$(1)$$ occurs if for example $$f$$ is an odd function, $$g$$ is an even function and $$X$$ follows a symmetric distribution. The equality of $$(2)$$ occurs if and only if $$|f(x)| = M$$ for all $$x$$.

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# Build a predictive model for number of hockey goals in a season for an individual player I'm looking to build a predictive model for hockey players individual statistics. My goal is to predict how many points a player can be expected to have at the end of the season. To do so, I thought I could base myself on the progression of the player's statistics over the year. I currently have 10 years worth of data and, for example, a player could have played in 5 of those 10 years. In this case, I'm able to see if this particular player is on an increasing or decreasing curve regarding his statistics. So, to resume, for some thousands of players the data goes like this: ### Player #1 Season 2005-2006, Points : X, Game played : y, ... Season 2006-2007, Points : X, Game played: Y, ... Season 2007-2008, Points : X, Game played: Y, ... ... ### Player #2 ... My main problem is that I don't know what kind of data this is so I can't find more information on how to build my model. In my title I've written "multiple small multivariate time series" for a lack of better name. I've obviously looked into multivariate time series to fit them in an RNN architecture, but it doesn't seem to fit because of the multiple time series that aren't really related. I've found this question which kind of relates but there were no answer and the related comments didn't help me. All in all, I'm trying to find some guidance as to what kind of model I could use to work with my data. I'd also like to know if there's a name for this kind of data, so I can inform myself better. "Predict how many points a player can be expected to have at the end of the season" could be framed as count-based regression problem (not time series). A player will have a collection of features (e.g., age, position, team, minutes plays). A machine learning model can learn how to weight to the features to make a prediction. In this case, the prediction is the number of goals by the end of the season (e.g., 0, 1, 2, 3, …). The data and features might have time component but the model does not have to be time-series. • I agree with this, I don't know about hockey but I would assume that the performance of players in a match are independent on other matches? Or maybe if a player is in form then they will continue being in form? If that is the case maybe the time component needs to be taken into account, or at least some way of capturing this, through feature augmentation? Mar 15 at 13:31 This can be done fairly well with a Bayesian model. Let's first make an unrealistic and simplifying solution: • Games are IID (False, poor performance may result in being moved to the farm team) • There is no variation in the expected goals per game by position (again False, defensemen are expected to score fewer goals than forwards). We will revisit the second assumption later on. Consider a scenario in which you only have data from the last season. From your post, it looks like you have the following information: • Player identifier $$i$$ • Games played $$n_i$$ • Goals scored $$y_i$$ You can estimate the number of goals per game for player $$i$$ as $$\lambda_i$$. Naturally, those players with fewer games played will have a noisier estimate than those who played every game. We can regularize those estimates by specifying a hierarchical model. Let's simulate some data to exposit the approach. library(tidyverse) n_player <- 100 # Simulate 100 player's goal rate # The average goal per game is 0.4 lam <- rgamma(n_player, 1, scale=0.4) # Simulate how many games they played n_games <- round(runif(n_player, 1, 80)) # The goals per game can be considered as a poisson rate # if games are iid, then simply muliply the rate by the number of games n_goals_per_season <- rpois(n_player, n_games*lam) d <- tibble(lam, n_goals_per_season, n_games, player = 1:n_player) ## Fitting a Model Now, we specify a Bayesian model for these data. We will assume that the $$\lambda_i$$ are draws from some larger population. Let's use the brms library to specify this model, as we will need it later. The model we fit is $$y_i \mid \lambda_i \sim \mbox{Poisson}(\lambda_i; n_i)$$ $$\log(\lambda_i) = \beta_{0,i} + \log(n_i)$$ $$\beta_{0, i} \sim \mbox{Normal}(., .)$$ Here $$\beta_{0, i}$$ are the log expected goals per game, and they are assumed to be draws from a normal distribution with some mean and variance. We get to specify that in our model. For now, I'll just use brms' defaults since I don't have good priors on what these should be. The model in brms is library(brms) fit <- brm(n_goals_per_season ~ 1 + (1|player) + offset(log(n_games)), data=d, family = poisson(), backend = 'cmdstanr') Let's plot the fitted vs estimated rates from the raw data per player. I will plot them on the log scale so we can see some interesting patterns and facet by the number of games played. d %>% ggplot(aes(log(n_goals_per_season/n_games), log(.epred/n_games))) + stat_pointinterval() + geom_abline() + labs( x=expression(log(y) - log(n)), y = expression(log(hat(lambda))) ) + facet_wrap(~cut(n_games, c(0, 10, 20, 40, 80))) Let's take a look at the top left facet -- those players playing between 1 and 10 games. Note how these estimates are regularized. Because we expect the lowest and highest average scorers to score more/less than we saw because we saw them in so few games. There is also some regularization in the other facets for the lowest scoring players. Ok, let's talk about how to make predictions. ## Making Predictions So we have an estimated goals per game $$\hat{\lambda}_i$$. How do we make predictions. Well, first we need some idea of how many games each player will play at the end of the season. This in itself may require a model since injuries and promotions/demotions from farm teams can modulate this. But let's say we are going to focus on the first 5 games of play. All we need to do then is draw samples from a poisson distribution with rate $$5\hat{\lambda}_i$$. Because our model is Bayesian, we get a distribution of expected goals. Luckily tidybayes has some functions to make this easy. I'll demonstrate how to do this with one player. tibble(player=1, n_games=5) %>% ggplot(aes(.prediction)) + geom_bar(aes(y = after_stat(count)/sum(after_stat(count)))) + labs(y = 'Probability', x = 'Goals Scored After 5 Games') This player is expected to probably score 0 or 1 total goals in the first 5 games of the next season. But your question was doing this over time. How do we extend this model? ## Model Extensions You have player data over seasons, not just for one season. We can extend this data so that the goals per game can evolve over time $$t$$. $$y_i \mid \lambda_i \sim \mbox{Poisson}(\lambda_i; n_i)$$ $$\log(\lambda_i) = \beta_{0,i} + \beta_{1,i} t + \log(n_{i, t})$$ This is essentially the same model, but now there is a linear effect of time (so players can get better and score more...or less). It is also easy to account for player position by adding appropriate covariates to the model. For an additive effect of position $$p$$, maybe we could specify $$y_i \mid \lambda_i \sim \mbox{Poisson}(\lambda_i; n_i)$$ $$\log(\lambda_i) = \beta_{0,i} + \beta_{1,i} t +\beta_{2, i}p + \log(n_{i, t})$$ with appropriate priors on all parameters. Or maybe the rate at which players improve depends on their position (interaction). Or maybe there are team level effects, or shift level effects, etc etc. there are many extensions to this model, but starting with a fairly simple model (predicting the next season from the last) is a good way to start. Then, we can extend that model slowly to the one we actually want.

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