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Purchase Solution # Particle in a magnetic field Not what you're looking for? The motion of a charged particle in an electromagnetic field can be obtained from the Lorentz equation for the force on a particle in such a field. If the electric field vector is E and the magnetic field vector is B the force on a particle of mass m that carries charge q and has velocity v is given by: F = qE +qv X B 1. If there is no electric field and the particle enters the magnetic field in a direction perpendicular to the lines of teh magnetic flux, show that the trajectory is a circle with radius r = mv/qB = v/w where w=gB/m is the cyclotron frequency 2. Choose the z-axis to lie in direction of B and let the plane containing E and B be the yz-plane. Thus: B = Bk E = E_y j + E_z k Show that the z-component of the motion is given by: z(t) = z(0) + (z_dot(0))*t+(qE_z / 2m) * t^2 3. Continue the calculation and obtain expressions for dx/dt and dy/dt Show that the time average of these velocity components are: <dx/dt> = E_y/B <dy/dt> = 0 4. Integrate the velocity equation from (3) and show (with appropriate initial conditions - see attached file) that: x(t) = -A/w * cos(wt) + E_y/B *t y(y) = A/w *sin (wt) Plot the trajectory for the following cases: A > E_y/B A = E_y/B A < E_y/B ##### Solution Summary The solution is 14 pages long including full explanations and derivation of all equations. ##### The Moon Test your knowledge of moon phases and movement. ##### Classical Mechanics This quiz is designed to test and improve your knowledge on Classical Mechanics. ##### Basic Physics This quiz will test your knowledge about basic Physics. ##### Intro to the Physics Waves Some short-answer questions involving the basic vocabulary of string, sound, and water waves. ##### Variables in Science Experiments How well do you understand variables? Test your knowledge of independent (manipulated), dependent (responding), and controlled variables with this 10 question quiz.

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# Application of Calculus problem • Feb 22nd 2010, 12:33 PM kycon90 Application of Calculus problem A street light is at the top of a 14 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 5 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 30 ft from the base of the pole? This is a problem where they give you what I believe is a dv/dt of 5 ft/s, but I don't know what I have to do to find the speed of the tip of her shadow. I tried drawing triangles, but I don't know exactly what I need to do, and I can't think of any equation that would help me find this, y = mx+b since it's a straight line??? • Feb 22nd 2010, 01:47 PM Soroban Hello, kycon90! Did you make a sketch? Quote: A streetlight is at the top of a 14 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 5 ft/sec. How fast is the tip of her shadow moving when she is 30 ft from the base of the pole? Code:     A o       |  * - C       |      o       |      |  *     14|      |      *       |      6|          *       |      |              *       |      |                  *     B o-------o-----------------------o E       : - x - D - - -  s-x  - - - - - :       : - - - - - - s - - - - - - - - : The lightpole is: $AB = 14\text{ ft}$ The woman is: $CD = 6\text{ ft}$ Her distance from the pole is: $BD = x\text{ ft},\;\;\frac{dx}{dt} = 5 \text{ ft/sec}$ The tip of her shadow $(E)$ is $s$ ft from the pole: . $s \,=\,BE$ . . Hence: . $DE \,=\,s-x$ Since $\Delta ABE \sim \Delta CDE$, we have: . $\frac{14}{s} \:=\:\frac{6}{s-x} \quad\Rightarrow\quad s \:=\:\frac{7}{4}x$ Differentiate with respect to time: . $\frac{ds}{dt} \:=\:\frac{7}{4}\,\frac{dx}{dt}$ Therefore: . $\frac{ds}{dt} \:=\:\frac{7}{4}(5) \:=\:8.75\text{ ft/sec}$

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Search All of the Math Forum: Views expressed in these public forums are not endorsed by NCTM or The Math Forum. Notice: We are no longer accepting new posts, but the forums will continue to be readable. Topic: 136 theorems on 29 pages Replies: 20   Last Post: Nov 19, 2012 4:55 PM Messages: [ Previous | Next ] clicliclic@freenet.de Posts: 1,245 Registered: 4/26/08 Re: 136 theorems on 29 pages Posted: Nov 19, 2012 4:55 PM Waldek Hebisch schrieb: > > clicliclic@freenet.de wrote: > > > > Expecting it to be the most promising, I have checked Bronstein's thesis > > publication of 1990 (57 pages, 2.4MB, barely readable digitization). I > > can assure our author that this contains no "prior art" concerning an > > extension of Hermite reduction (p. 132) to multiple non-integer > > exponents or to a lowering of exponents: the equations agree with those > > in the Symbolic Integration Tutorial of 1998/2000 and fail for multiple > > non-integer exponents because the integrated term is again too > > restricted. > > You seem to ignore the following fact: Bronstein uses integral > basis, which is equivalent to separating "essential" part under > root and writing rest as polynomial. That is given P^a with > P squarefree one writes it as P^nP^b where 0\leq b < 1. After > such splitting one can collect irrationalities into a single term. > From Bronstein introduction: > > : Using only rational techniques, we are able to remove multiple > : finite poles of the integrand. > > The claim is that he can handle _any_ algebraic integrand (in > fact more general because he allows algebraics depending on > exponentials or logarithms). And "removing multiple poles" > means exactly that he can increase exponents of factor of > denominator as long as they are smaller than -1. He does not > explicitly state this, but looking at the proof one sees that > only new factors which can appear in denominator are terms under > radical, so indeed his reduction process manages to > increase powers toward - 1 (at cost of somewhat uncontrolled > factor in the numerator). > > Then, on page 146 Bronstain outlines how to remove multiple > pole at infinity. He gives little datails (and equation > S_{log2} looks wrong), still, I have checked that following > his hints works. Now, removing multiple pole at infinity > really means lowering exponents, if the function at hand > is a product. > Here's how I remember having read Bronstein's thesis and Tutorial (I was off-line for six weeks, hence this belated reply): Bronstein needs to introduce a finite-dimensional basis of algebraic functions in order to handle arbitrary algebraic integrands. Ordinary Hermite reduction thereby splits into interdependent parallel reductions since the basis elements will mix under differentiation. In order to guarantee success, the basis must be "integral", which is hard to compute in general. (Later he therefore introduced a "lazy" version of the parallelized reduction, where such a basis is not needed at the outset but gradually produced by repeated updates in the course of the reduction.) As we know, all this must (and therefore does) simplify for the case of an algebraic product denominator, and Bronstein seems to consider this special case too simple to be spelled out (I am inclined to share this view). In his papers, I have seen no instance of Hermite reduction where the integrated term does recognizably involve the required product denominator with exponents close to those in the (original or transformed) integrand denominator. This holds for the rational as well as algebraic cases. After all, both the raising and lowering of denominator exponents (and consequently any combination of these operations involving any number of exponents) in an algebraic product denominator requires nothing but solving a system of linear equations (whose particular structure for the raising of exponents permits the extended Euclidean algorithm to be used). This is much easier to see directly than to infer via Bronstein. I consider it the opposite of helpful to refer to his parallelized Hermite machinery in order to arrive at this fairly trivial fact! Martin.

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Warning: unlink(/var/www/html/images/tmp/8775df11722a6c2878087f0c5d00e0db.asy): No such file or directory in /var/www/html/extensions/Asymptote.php on line 109 Warning: unlink(/var/www/html/images/tmp/ltx-8775df11722a6c2878087f0c5d00e0db.aux): No such file or directory in /var/www/html/extensions/Asymptote.php on line 113 Warning: unlink(/var/www/html/images/tmp/ltx-8775df11722a6c2878087f0c5d00e0db.log): No such file or directory in /var/www/html/extensions/Asymptote.php on line 114 Warning: unlink(/var/www/html/images/tmp/8775df11722a6c2878087f0c5d00e0db.eps): No such file or directory in /var/www/html/extensions/Asymptote.php on line 115 Warning: unlink(/var/www/html/images/tmp/8775df11722a6c2878087f0c5d00e0db.log): No such file or directory in /var/www/html/extensions/Asymptote.php on line 116 Cantor ternary function - Encyclopedia of Mathematics # Cantor ternary function 2010 Mathematics Subject Classification: Primary: 26A45 Secondary: 28A15 [MSN][ZBL] The Cantor ternary function (also called Devil's staircase and, rarely, Lebesgue's singular function) is a continuous monotone function $f$ mapping the interval $[0,1]$ onto itself, with the remarkable property that its derivative vanishes almost everywhere (recall that any monotone function is differentiable almost everywhere, see for instance Function of bounded variation). The function can be defined in the following way (see, for example, Exercise 46 in Chapter 2 of [Ro]). Given $x\in [0,1]$ consider its ternary expansion $\{a_i\}$, i.e. a choice of coefficients $a_i\in \{0,1,2\}$ such that $x = \sum_{i=1}^\infty \frac{a_i}{3^i}\, .$ Define $n(x)$ to be • $\infty$ if none of the coefficients $a_i$ takes the value $1$ • the smallest integer $n$ such that $a_n=1$ otherwise. Then $f(x) = \sum_{i=1}^{n(x)-1} \frac{a_i}{2^{i+1}} + \frac{1}{2^{n(x)}}.$ For alternative definitions see Example 1.67 of [AFP] and page 55 of [Co]. It follows trivially from the definition that $f$ is locally constant on the complement of the Cantor set $C$: since the Cantor set is a set of Lebesgue measure zero, the derivative of $f$ vanishes almost everywhere. For the same reason, the distributional derivative of $f$ is a singular measure $\mu$ supported on $C$. The Cantor function is a prototype of a singular function in the sense of Lebesgue, cf. Lebesgue decomposition. The observation that $f$ is a function of bounded variation for which $f (1)-f(0) \neq \int_0^a f'(t)\, dt$ (where $f'$ denotes the classical pointwise derivative) was first made by Vitali in [Vi]. For this reason some authors use the terminology Cantor-Vitali function (see [AFP]). ## References [AFP] L. Ambrosio, N. Fusco, D. Pallara, "Functions of bounded variations and free discontinuity problems". Oxford Mathematical Monographs. The Clarendon Press, Oxford University Press, New York, 2000. MR1857292Zbl 0957.49001 [Co] D. L. Cohn, "Measure theory". Birkhäuser, Boston 1993. [Ro] H.L. Royden, "Real analysis" , Macmillan (1969). MR0151555 Zbl 0197.03501 [Vi] A. Vitali, "Sulle funzioni integrali", Atti Accad. Sci. Torino Cl. Sci. Fis. Mat. Natur., 40 1905 pp. 1021-1034. How to Cite This Entry: Cantor ternary function. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Cantor_ternary_function&oldid=35600

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## ››Convert cubic decimetre/minute to litre/second cubic decimeter/minute liter/second How many cubic decimeter/minute in 1 liter/second? The answer is 60. We assume you are converting between cubic decimetre/minute and litre/second. You can view more details on each measurement unit: cubic decimeter/minute or liter/second The SI derived unit for volume flow rate is the cubic meter/second. 1 cubic meter/second is equal to 60000 cubic decimeter/minute, or 1000 liter/second. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between cubic decimeters/minute and liters/second. Type in your own numbers in the form to convert the units! ## ››Quick conversion chart of cubic decimeter/minute to liter/second 1 cubic decimeter/minute to liter/second = 0.01667 liter/second 10 cubic decimeter/minute to liter/second = 0.16667 liter/second 20 cubic decimeter/minute to liter/second = 0.33333 liter/second 30 cubic decimeter/minute to liter/second = 0.5 liter/second 40 cubic decimeter/minute to liter/second = 0.66667 liter/second 50 cubic decimeter/minute to liter/second = 0.83333 liter/second 100 cubic decimeter/minute to liter/second = 1.66667 liter/second 200 cubic decimeter/minute to liter/second = 3.33333 liter/second ## ››Want other units? You can do the reverse unit conversion from liter/second to cubic decimeter/minute, or enter any two units below: ## Enter two units to convert From: To: ## ››Metric conversions and more ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!

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# The velocity of an object with a mass of 2 kg is given by v(t)= t^3 + 3 t^2 . What is the impulse applied to the object at t= 4 ? Feb 26, 2016 $I \left(4\right) = 2 \left(4 \cdot {4}^{2} + 3 \cdot {4}^{2}\right) = 7 \cdot {4}^{2} = 112 N s$ Careful however Impulse is not $m v \left(t\right)$ evaluate at $t = 4 s$ it is Force applied for a finite amount of time $F \cdot \Delta t$ in the limit $I = {\int}_{{t}_{1}}^{{t}_{2}} F \mathrm{dt}$ #### Explanation: This is a straight implementation of the impulse equation i.e. $F = m \frac{\mathrm{dv}}{\mathrm{dt}}$ Newton law ==> (1) $I = F \mathrm{dt} = m \left(\mathrm{dv}\right)$ Impulse equation ==> (2) While the question is not clear I will make the assumption the Force was applied for 4 seconds causing the change in speed from $v \left(0\right) = 0 \to v \left(4\right)$ and $I \left(4\right) = 2 \left(4 \cdot {4}^{2} + 3 \cdot {4}^{2}\right) = 7 \cdot {4}^{2} = 112 N s$

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# Group representation A representation of a group "acts" on an object. The simplest examples are how the symmetries of a regular polygon, consisting of reflections and rotations, transform the polygon. In the mathematical field of representation theory, group representations describe abstract groups in terms of bijective linear transformations (i.e. automorphisms) of vector spaces; in particular, they can be used to represent group elements as invertible matrices so that the group operation can be represented by matrix multiplication. Representations of groups are important because they allow many group-theoretic problems to be reduced to problems in linear algebra, which is well understood. They are also important in physics because, for example, they describe how the symmetry group of a physical system affects the solutions of equations describing that system. The term representation of a group is also used in a more general sense to mean any "description" of a group as a group of transformations of some mathematical object. More formally, a "representation" means a homomorphism from the group to the automorphism group of an object. If the object is a vector space we have a linear representation. Some people use realization for the general notion and reserve the term representation for the special case of linear representations. The bulk of this article describes linear representation theory; see the last section for generalizations. ## Branches of group representation theory The representation theory of groups divides into subtheories depending on the kind of group being represented. The various theories are quite different in detail, though some basic definitions and concepts are similar. The most important divisions are: • Finite groups — Group representations are a very important tool in the study of finite groups. They also arise in the applications of finite group theory to crystallography and to geometry. If the field of scalars of the vector space has characteristic p, and if p divides the order of the group, then this is called modular representation theory; this special case has very different properties. See Representation theory of finite groups. • Compact groups or locally compact groups — Many of the results of finite group representation theory are proved by averaging over the group. These proofs can be carried over to infinite groups by replacement of the average with an integral, provided that an acceptable notion of integral can be defined. This can be done for locally compact groups, using Haar measure. The resulting theory is a central part of harmonic analysis. The Pontryagin duality describes the theory for commutative groups, as a generalised Fourier transform. See also: Peter–Weyl theorem. • Lie groups — Many important Lie groups are compact, so the results of compact representation theory apply to them. Other techniques specific to Lie groups are used as well. Most of the groups important in physics and chemistry are Lie groups, and their representation theory is crucial to the application of group theory in those fields. See Representations of Lie groups and Representations of Lie algebras. • Linear algebraic groups (or more generally affine group schemes) — These are the analogues of Lie groups, but over more general fields than just R or C. Although linear algebraic groups have a classification that is very similar to that of Lie groups, and give rise to the same families of Lie algebras, their representations are rather different (and much less well understood). The analytic techniques used for studying Lie groups must be replaced by techniques from algebraic geometry, where the relatively weak Zariski topology causes many technical complications. • Non-compact topological groups — The class of non-compact groups is too broad to construct any general representation theory, but specific special cases have been studied, sometimes using ad hoc techniques. The semisimple Lie groups have a deep theory, building on the compact case. The complementary solvable Lie groups cannot be classified in the same way. The general theory for Lie groups deals with semidirect products of the two types, by means of general results called Mackey theory, which is a generalization of Wigner's classification methods. Representation theory also depends heavily on the type of vector space on which the group acts. One distinguishes between finite-dimensional representations and infinite-dimensional ones. In the infinite-dimensional case, additional structures are important (e.g. whether or not the space is a Hilbert space, Banach space, etc.). One must also consider the type of field over which the vector space is defined. The most important case is the field of complex numbers. The other important cases are the field of real numbers, finite fields, and fields of p-adic numbers. In general, algebraically closed fields are easier to handle than non-algebraically closed ones. The characteristic of the field is also significant; many theorems for finite groups depend on the characteristic of the field not dividing the order of the group. ## Definitions A representation of a group G on a vector space V over a field K is a group homomorphism from G to GL(V), the general linear group on V. That is, a representation is a map ${\displaystyle \rho \colon G\to \mathrm {GL} (V)}$ such that ${\displaystyle \rho (g_{1}g_{2})=\rho (g_{1})\rho (g_{2}),\qquad {\text{for all }}g_{1},g_{2}\in G.}$ Here V is called the representation space and the dimension of V is called the dimension of the representation. It is common practice to refer to V itself as the representation when the homomorphism is clear from the context. In the case where V is of finite dimension n it is common to choose a basis for V and identify GL(V) with GL(n, K), the group of n-by-n invertible matrices on the field K. • If G is a topological group and V is a topological vector space, a continuous representation of G on V is a representation ρ such that the application Φ : G × VV defined by Φ(g, v) = ρ(g)(v) is continuous. • The kernel of a representation ρ of a group G is defined as the normal subgroup of G whose image under ρ is the identity transformation: ${\displaystyle \ker \rho =\left\{g\in G\mid \rho (g)=\mathrm {id} \right\}.}$ A faithful representation is one in which the homomorphism G → GL(V) is injective; in other words, one whose kernel is the trivial subgroup {e} consisting only of the group's identity element. • Given two K vector spaces V and W, two representations ρ : G → GL(V) and π : G → GL(W) are said to be equivalent or isomorphic if there exists a vector space isomorphism α : VW so that for all g in G, ${\displaystyle \alpha \circ \rho (g)\circ \alpha ^{-1}=\pi (g).}$ ## Examples Consider the complex number u = e2πi / 3 which has the property u3 = 1. The cyclic group C3 = {1, u, u2} has a representation ρ on C2 given by: ${\displaystyle \rho \left(1\right)={\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}\qquad \rho \left(u\right)={\begin{bmatrix}1&0\\0&u\\\end{bmatrix}}\qquad \rho \left(u^{2}\right)={\begin{bmatrix}1&0\\0&u^{2}\\\end{bmatrix}}.}$ This representation is faithful because ρ is a one-to-one map. Another representation for C3 on C2, isomorphic to the previous one, is σ given by: ${\displaystyle \sigma \left(1\right)={\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}\qquad \sigma \left(u\right)={\begin{bmatrix}u&0\\0&1\\\end{bmatrix}}\qquad \sigma \left(u^{2}\right)={\begin{bmatrix}u^{2}&0\\0&1\\\end{bmatrix}}.}$ The group C3 may also be faithfully represented on R2 by τ given by: ${\displaystyle \tau \left(1\right)={\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}\qquad \tau \left(u\right)={\begin{bmatrix}a&-b\\b&a\\\end{bmatrix}}\qquad \tau \left(u^{2}\right)={\begin{bmatrix}a&b\\-b&a\\\end{bmatrix}}}$ where ${\displaystyle a={\text{Re}}(u)=-{\tfrac {1}{2}},\qquad b={\text{Im}}(u)={\tfrac {\sqrt {3}}{2}}.}$ ## Reducibility A subspace W of V that is invariant under the group action is called a subrepresentation. If V has exactly two subrepresentations, namely the zero-dimensional subspace and V itself, then the representation is said to be irreducible; if it has a proper subrepresentation of nonzero dimension, the representation is said to be reducible. The representation of dimension zero is considered to be neither reducible nor irreducible,[citation needed] just like the number 1 is considered to be neither composite nor prime. Under the assumption that the characteristic of the field K does not divide the size of the group, representations of finite groups can be decomposed into a direct sum of irreducible subrepresentations (see Maschke's theorem). This holds in particular for any representation of a finite group over the complex numbers, since the characteristic of the complex numbers is zero, which never divides the size of a group. In the example above, the first two representations given (ρ and σ) are both decomposable into two 1-dimensional subrepresentations (given by span{(1,0)} and span{(0,1)}), while the third representation (τ) is irreducible. ## Generalizations ### Set-theoretical representations A set-theoretic representation (also known as a group action or permutation representation) of a group G on a set X is given by a function ρ : GXX, the set of functions from X to X, such that for all g1, g2 in G and all x in X: ${\displaystyle \rho (1)[x]=x}$ ${\displaystyle \rho (g_{1}g_{2})[x]=\rho (g_{1})[\rho (g_{2})[x]],}$ where ${\displaystyle 1}$  is the identity element of G. This condition and the axioms for a group imply that ρ(g) is a bijection (or permutation) for all g in G. Thus we may equivalently define a permutation representation to be a group homomorphism from G to the symmetric group SX of X. For more information on this topic see the article on group action. ### Representations in other categories Every group G can be viewed as a category with a single object; morphisms in this category are just the elements of G. Given an arbitrary category C, a representation of G in C is a functor from G to C. Such a functor selects an object X in C and a group homomorphism from G to Aut(X), the automorphism group of X. In the case where C is VectK, the category of vector spaces over a field K, this definition is equivalent to a linear representation. Likewise, a set-theoretic representation is just a representation of G in the category of sets. When C is Ab, the category of abelian groups, the objects obtained are called G-modules. For another example consider the category of topological spaces, Top. Representations in Top are homomorphisms from G to the homeomorphism group of a topological space X. Two types of representations closely related to linear representations are:

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k is a positive integer. Is k prime? : GMAT Data Sufficiency (DS) Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack It is currently 21 Jan 2017, 08:14 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 k is a positive integer. Is k prime? Author Message TAGS: Hide Tags Manager Joined: 31 Jan 2012 Posts: 86 Followers: 0 Kudos [?]: 33 [1] , given: 150 k is a positive integer. Is k prime? [#permalink] Show Tags 30 Apr 2013, 02:54 1 KUDOS 2 This post was BOOKMARKED 00:00 Difficulty: 45% (medium) Question Stats: 60% (02:04) correct 40% (01:04) wrong based on 209 sessions HideShow timer Statistics k is a positive integer. Is k prime? (1) At least one number in the set {1, k, k + 7} is prime. (2) k is odd. [Reveal] Spoiler: OA Last edited by Bunuel on 30 Apr 2013, 02:57, edited 1 time in total. RENAMED THE TOPIC. Math Expert Joined: 02 Sep 2009 Posts: 36590 Followers: 7090 Kudos [?]: 93325 [3] , given: 10557 Re: k is a positive integer. Is k prime? [#permalink] Show Tags 30 Apr 2013, 03:01 3 KUDOS Expert's post k is a positive integer. Is k prime? Notice that we are told that k is a positive integer. (1) At least one number in the set {1, k, k + 7} is prime. Clearly insufficient. (2) k is odd. Clearly insufficient. (1)+(2) The set is {1, k=odd, k+7=even>2}. Since at least on number in this set is a prime, then it must be k. Sufficient. _________________ Manager Joined: 31 Jan 2012 Posts: 86 Followers: 0 Kudos [?]: 33 [0], given: 150 Re: k is a positive integer. Is k prime? [#permalink] Show Tags 30 Apr 2013, 03:36 Bunuel wrote: k is a positive integer. Is k prime? Notice that we are told that k is a positive integer. (1) At least one number in the set {1, k, k + 7} is prime. Clearly insufficient. (2) k is odd. Clearly insufficient. (1)+(2) The set is {1, k=odd, k+7=even>2}. Since at least on number in this set is a prime, then it must be k. Sufficient. you made it look so simple...thanks a ton Bunuel. Current Student Joined: 21 Oct 2013 Posts: 194 Location: Germany GMAT 1: 660 Q45 V36 GPA: 3.51 Followers: 1 Kudos [?]: 37 [0], given: 19 Re: k is a positive integer. Is k prime? [#permalink] Show Tags 14 Jan 2014, 06:38 Bunuel wrote: k is a positive integer. Is k prime? Notice that we are told that k is a positive integer. (1) At least one number in the set {1, k, k + 7} is prime. Clearly insufficient. (2) k is odd. Clearly insufficient. (1)+(2) The set is {1, k=odd, k+7=even>2}. Since at least on number in this set is a prime, then it must be k. Sufficient. I thought it was A because I thought that in a set, the numbers are organized by size. So I assumed that k is prime because any prime + 7 = no prime. But obviously a set is not sorted. I get the rest of the reasoning. Am I right assuming that sets don't need to be sorted? Please verify! Thanks! Math Expert Joined: 02 Sep 2009 Posts: 36590 Followers: 7090 Kudos [?]: 93325 [0], given: 10557 Re: k is a positive integer. Is k prime? [#permalink] Show Tags 14 Jan 2014, 06:57 Expert's post 1 This post was BOOKMARKED unceldolan wrote: Bunuel wrote: k is a positive integer. Is k prime? Notice that we are told that k is a positive integer. (1) At least one number in the set {1, k, k + 7} is prime. Clearly insufficient. (2) k is odd. Clearly insufficient. (1)+(2) The set is {1, k=odd, k+7=even>2}. Since at least on number in this set is a prime, then it must be k. Sufficient. I thought it was A because I thought that in a set, the numbers are organized by size. So I assumed that k is prime because any prime + 7 = no prime. But obviously a set is not sorted. I get the rest of the reasoning. Am I right assuming that sets don't need to be sorted? Please verify! Thanks! A set is not necessarily ordered but this has nothing to do with the first statement. The question asks whether k is a prime number. (1) says that at least one number in the set {1, k, k + 7} is prime. Sure k can be prime but k can as well be for example 4, so not a prime and in this case k+7=11 is a prime. Hope it's clear. _________________ GMAT Club Legend Joined: 09 Sep 2013 Posts: 13480 Followers: 576 Kudos [?]: 163 [0], given: 0 Re: k is a positive integer. Is k prime? [#permalink] Show Tags 27 Jul 2015, 23:16 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 8306 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Followers: 381 Kudos [?]: 2466 [0], given: 163 Re: k is a positive integer. Is k prime? [#permalink] Show Tags 29 Jul 2015, 21:01 Expert's post 1 This post was BOOKMARKED Hi All, This question can be solved by TESTing VALUES. We're told that K is a POSITIVE INTEGER. We're asked if K is PRIME. This is a YES/NO question. Fact 1: At least one number in the set {1, K, K + 7} is prime. IF.... K = 2, the set is {1, 2, 9} and has at least 1 prime number The answer to the question is YES. IF... K = 4, the set is {1, 4, 11} and has at least 1 prime number The answer to the question is NO. Fact 1 is INSUFFICIENT (2) K is ODD. IF.... K = 1, then the answer to the question is NO. K = 3, then the answer to the question is YES. Fact 2 is INSUFFICIENT Combined, we know... At least one number in the set {1, K, K + 7} is prime. K is ODD IF... K = 3, the set is {1, 3, 10} and has at least 1 prime number The answer to the question is YES. From the above example, if K is an ODD PRIME, then the answer is ALWAYS YES. We now have to look for possibilities when K is NOT prime.... IF... K = 9, the set is {1, 9, 16} BUT this does NOT have any primes, so K CANNOT BE 9... From this example, we can see that if K = an ODD NON-PRIME, then (K+7) will be EVEN (thus, NOT prime). Thus, there are no possible values of K that are odd AND non-prime that will 'fit' these Facts. By extension, that means that K MUST be an ODD PRIME and the answer to the question is ALWAYS YES. Combined, SUFFICIENT [Reveal] Spoiler: C GMAT assassins aren't born, they're made, Rich _________________ Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save \$75 + GMAT Club Tests 60-point improvement guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** GMAT Club Legend Joined: 09 Sep 2013 Posts: 13480 Followers: 576 Kudos [?]: 163 [0], given: 0 Re: k is a positive integer. Is k prime? [#permalink] Show Tags 11 Nov 2016, 21:43 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: k is a positive integer. Is k prime?   [#permalink] 11 Nov 2016, 21:43 Similar topics Replies Last post Similar Topics: 1 If p is a positive integer, is integer k prime? 2 06 Dec 2016, 06:21 9 Is integer k a prime number? 9 29 Sep 2015, 04:13 8 If k is a positive integer, is k an integer? 7 22 Jun 2015, 16:35 1 If k is a positive integer whose only prime factors are 2 and 5, is k 5 08 Jun 2015, 01:10 28 If k is a positive integer, is k a prime number? 26 19 Jun 2010, 10:37 Display posts from previous: Sort by

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# Virtual COMSATS Inferential Statistics Lecture20 Ossam Chohan Assistant • Slides: 25 Virtual COMSATS Inferential Statistics Lecture-20 Ossam Chohan Assistant Professor CIIT Abbottabad 1 Recap of last lecture • In our last sessions, we worked on: – Overview of statistical inferences – Hypothesis testing introduction. – Six steps of Hypothesis testing. – Problems regarding single mean and proportion using z and t statistics. 2 Objective of lecture-20 • In this lecture, we will understand problems related to difference between two populations: – Dependent samples. – Independent samples. – Proportions case. – Some portion of Chi-square distribution. 3 Problem-14 • A researcher interested in employee satisfaction and productivity measured the number of units produced by employees at a plant before and after a company-wide pay raise occurred. The researcher hypothesized that production would be higher after the raise compared to before the raise. Assume that the difference scores are normally distributed and let a = 0. 05. 4 Cont… Participant Before After 1 7 7 2 4 5 3 8 9 4 8 9 5 6 6 6 7 5 5 8 5 4 9 7 7 5 Problem 14 Solution 6 Problem-14 Solution 7 Problem-15 • A sociologist is interested in the decay of long-term memory compared to the number of errors in memory that an individual made after 1 week and after 1 year for a specific crime event. Participants viewed a videotape of a bank robbery and were asked a number of specific questions about the video 1 week after viewing it. They were asked the same questions 1 year after seeing the video. The number of memory errors was recorded for each participant at each time period. The researchers asked whether or not there was a significant difference in the number of errors in the two time periods. Assume that the difference scores are normally distributed and let α = 0. 05. 8 Cont… Subject One Week One Year 1 5 7 2 4 5 3 6 9 4 8 9 5 6 6 6 5 6 7 4 5 8 5 4 9 7 7 9 Problem-15 Solution 10 Problem-15 Solution 11 Assessment Problem-13 • Suppose you are interested in developing a counseling technique to reduce stress within marriages. You randomly select two samples of married individuals out of ten churches in the association. You provide Group 1 with group counseling and study materials. You provide Group 2 with individual counseling and study materials. • At the conclusion of the treatment period, you measure the level of marital stress in the group members. Here are the scores: 12 Cont… Group-1 Group-2 25 17 29 29 26 24 27 33 21 26 28 31 14 27 29 23 23 14 21 26 20 27 26 32 18 25 32 23 16 21 17 20 20 32 17 23 20 30 26 12 26 23 7 18 29 32 24 19 13 CHI-SQUARE DISTRIBUTION χ2 14 Objective • In this part of the course students will understand learn the following topics: Ø Chi square and its properties Ø Chi square as a test of independence Ø Chi square as a test of goodness of fit 15 PROPERTIES OF CHI-SQUARE • Chi-square is a continuous distribution ranging from 0 to infinity • It depends upon degrees of freedom, • It is not symmetric. • As the number of degrees of freedom increases, the chi-square distribution becomes more symmetric. • The values are non-negative. That is, the values of are greater than or equal to 0. • The mean of chi-square distribution is its degree of freedom. 16 17 Chi-square table 18 • Each χ2 distribution has a degree of freedom associated with it, so that there are many different chi squared distributions. • The chi squared distributions for each of 1 through 20 degrees of freedom are shown in the table. 19 USES • Chi square is used for testing: – Variance – Confidence interval for variances – Independence between two categorical variables – Homogeneity – Association between attributes. – Two proportions – Many proportions testing. – And many more applications in research. 20 Problem-16 • A cigarette manufacturer wants to test the claim that the variance of the nicotine content of its cigs. is 0. 644 milligram. Assume that it is normally distributed. A sample of 20 cigs. has a standandard deviation of 1. 00 milligram. At α = 0. 05, is there sample evidence to reject the manufacturer's claim? 21 Problem-16 Solution 22 Problem-17 • A machine dispenses a liquid drug into bottles in such a way that the standard deviation of the contents is 81 milliliters. A new machine is tested on a sample of 24 containers and the standard deviation for this sample group is found to be 26 milliliters. At the 0. 05 level of significance, test the claim that the amount dispensed by the new machine have a smaller standard deviation. 23 Problem-17 Solution 24 Assessment Problem-14 • For randomly selected adults, IQ scores are normally distributed with a standard deviation of 16. The scores of 15 randomly selected college students are listed below. Use a 0. 10 significance level to test the claim that the standard deviation of IQ scores of college students is less than 16. HINT: calculate ‘s’ for the 15 college students listed below. • 115 128 107 116 118 126 129 124 135 127 115 104 133 25

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8.1 Estimation Using Normal # 8 gm does it seem possible that the population mean This preview shows page 1. Sign up to view the full content. This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: it seem possible that the population mean is 368 gm? probably Determine an interval estimate for the population mean based on an acceptable amount or margin of sampling error x ± margin of error σ Known Example Based on the sample, how will they determine whether or not the process is working? Recall sampling error, the difference between the sample mean and the population mean ◦ Is the sampling error an acceptable amount, or is the difference too large? Interval Estimation Ø E.g., if sample mean is 342.3 gm, does it seem possible that population mean is 368 gm? maybe not E.g., if sample mean is 365.8 gm, does it seem possible that the population mean is 368 gm? probably Determine an interval estimate for the population mean based on an acceptable amount or margin of sampling error x ± margin of error ◦ ◦ If desired population mean within the interval, then conclude process is working If desired population mean not within the interval, then conclude process is not working σ Known Example Based on the sample, how will they determine whether or not the process is working? Recall sampling error, the difference between the sample mean and the population mean ◦ Is the sampling error an acceptable amount, or is the difference too large? Interval Estimation Ø How do we find a reasonabl e margin of error? E.g., if sample mean is 342.3 gm, does it seem possible that population mean is 368 gm? maybe not E.g., if sample mean is 365.8 gm, does it seem possible that the population mean is 368 gm? probably Determine an interval estimate for the population mean based on an acceptable amount or margin of sampling error x ± margin of error ◦ ◦ If desired population mean within the interval, then conclude process is working If desired population mean not within the interval, then conclude process is not working σ Known Example How do we find a reasonable margin of error? x ± margin of error à Using Interval Estimation sampling distribution of the mean, we can find range of values for sample mean that are within a certain range of the population mean à Using similar method construct range of values for population mean… σ Known Example Oxford Cereals fills thousands of boxes of cereal during an 8-hour shift. The... View Full Document ## This document was uploaded on 03/05/2014. Ask a homework question - tutors are online

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mersenneforum.org Shor's Factoring Algorithm - does it even work? Register FAQ Search Today's Posts Mark Forums Read 2005-10-25, 19:43   #12 xilman Bamboozled! "๐’‰บ๐’ŒŒ๐’‡ท๐’†ท๐’€ญ" May 2003 Down not across 2·73·17 Posts Quote: Originally Posted by Citrix Shors algorithm says that for a pseudo random number a, find a^x=a^(x+r )mod n where n is the number to be factored. How do you do this on an ordinary computer for a large number? I don't think the algorithm is p? If r could be found easily then all numbers can be factored in p! Citrix Ah, a great light dawns! You are quite correct in your observation. No algorithm is known to solve that problem in polynomial time on a classical computer. However, such an algorithm is known for a quantum computer --- it is Schor's algorithm. The article to which you refer specifically states that the order-finding subroutine is to be run on a quantum computer. In this situation the entire algorithm runs in polynomial time. Looks like you thought that the entire algorithm had to be run on a classical machine. Paul 2005-10-25, 19:47 #13 Citrix     Jun 2003 3×5×107 Posts What does this mean "A reduction of the factoring problem to the problem of order-finding, which can be done on a classical computer. " -- How do you do this? Citrix 2005-10-25, 19:50   #14 akruppa "Nancy" Aug 2002 Alexandria 46438 Posts Quote: Originally Posted by Citrix Shors algorithm says that for a pseudo random number a, find a^x=a^(x+r )mod n where n is the number to be factored. How do you do this on an ordinary computer for a large number? I don't think the algorithm is p? If r could be found easily then all numbers can be factored in p! Citrix On a conventional computer the algorithm is purely exponential-time. The idea is: 1. Choose some a 2. Compute a^n (mod N) for 1<=n<=k*N, with k large enough. This sequence is o-periodic, o=ord_N(a) 3. Do a Fourier transform on the sequence. Since it is o-periodic, the Fourier coefficients for the frequency o/(k*N) will dominate 4. Randomly choose a coefficient with probability according to its amplitude On a conventional computer, steps 2 and 3 take \Omega(N) steps. On a quantum computer, you can put the all the exponent qubits into a (1 1) (1 -1) / sqrt(2) state, i.e. a 0 and 1 bit with 50% probability each, and exponentiate by that exponent. The result of this exponentiation is a superposition of all a^n. This superposition can be transformed with a Quantum Fourier Transform (QFT), giving you a state where the o/(k*N) frequency and its multiples dominate in amplitude. Measuring the state gives you one of these values. You can then recover o by a continued fraction expansion. The nice thing is that the exponentiation and the QFT can be done in time O(log(N)^2), iirc. There are some catches that make the algorithm extremely tricky to actually implement, but the idea itself is sound. I gave a presentation on this algorithm at the uni once. Quantum comuting is by far the weirdest I've seen yet. Mind-boggling does not begin to describe it. If you're interested in this, get Nielsen and Chuang: Quantum Computation and Quantum Information. Note: the above is from memory, there probably are errors. I think I got the overall idea right, though. Alex Last fiddled with by akruppa on 2005-10-26 at 22:37 Reason: corrected Hadamard matrix, complexity 2005-10-25, 19:55 #15 Citrix     Jun 2003 3·5·107 Posts Thankyou very much. This stuff is really interesting. It was my fault. I was too quick to judge the method. I just read the classical part and thought that, it had to be done before the quantum part. 2005-10-25, 21:25   #16 Mystwalker Jul 2004 Potsdam, Germany 3×277 Posts Quote: Originally Posted by R.D. Silverman 15 has indeed been factored on a 4-qubit quantum computer. Just for clarification: The article stated 7-qubits. Which one is correct? 2005-10-25, 21:34 #17 akruppa     "Nancy" Aug 2002 Alexandria 9A316 Posts 7 qubit is correct. The composite number 15 has 4 bits, and you need extra bits for the exponent, etc. Alex 2005-10-25, 21:57 #18 ewmayer ∂2ω=0     Sep 2002 Repรบblica de California 5×2,351 Posts I'm renaming this thread "Shor's Factoring Algorithm = Crap?" (I believe that captures the tone of the initial posting, but is more descriptive) and moving it to the regular Math section. Note that there is no 'c' in the spelling of Peter Shor's last name. Last fiddled with by ewmayer on 2005-10-25 at 21:58 2005-10-26, 03:42 #19 Citrix     Jun 2003 31058 Posts Please remove the word crap from the title. It was a misunderstanding on my part. (Rename it quantum computing perhaps) Thankyou, Citrix Last fiddled with by Citrix on 2005-10-26 at 03:43 2005-10-26, 05:12 #20 akruppa     "Nancy" Aug 2002 Alexandria 46438 Posts Done. Alex 2005-10-26, 05:36 #21 Peter Nelson     Oct 2004 232 Posts Quantum computing and the Shor algorithm in particular is mentioned in "Crandall and Pomerance: Prime Numbers: A computational approach" in second edition its 8.5.2 "The Shor quantum algorithm for factoring" on p422. For second edition it may be slightly in need of revision (ie the IBM machine factoring 15 was in 2001) whereas they say "We stress that an appropriate machine has not been built". Perhaps someone could suggest to C&P that they could mention the IBM experiment. However they do say "but if it were the following algorithm IS EXPECTED TO WORK". In their presentation they do not give all the details of the algorithm. In their words "we have been intentionally brief in the final steps of the algorithm". They do point out that it could be done on conventional computers but in much longer time which would become unworkable, and it is quantum computers that offer opportunity to use it. I personally found the answers.com/wikipaedia treatment more complete and a little more readable. 2005-10-26, 18:35   #22 ewmayer 2ω=0 Sep 2002 Repรบblica de California 101101111010112 Posts Quote: Originally Posted by akruppa Done. Wuss. I quote: Quote: Originally Posted by Citrix Is there any merit to this method? It look like crap to me. Similar Threads Thread Thread Starter Forum Replies Last Post Prime95 Miscellaneous Math 72 2015-10-26 00:14 flouran Math 3 2009-10-27 09:31 Visu Math 66 2008-05-12 13:55 Citrix Factoring 6 2007-12-23 11:36 Visu Factoring 22 2006-11-09 10:43 All times are UTC. The time now is 10:13. Mon Feb 6 10:13:03 UTC 2023 up 172 days, 7:41, 1 user, load averages: 1.20, 1.19, 1.08

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# ECE573- Homework 2 Solved Instructions 1.     Implement Shellsort which reverts to insertion sort. (Use the increment sequence 7, 3, 1). Create a plot for the total number of comparisons made in the sorting the data for both cases (insertion sort phase and shell sort phase). Explain why Shellshort is more effective than Insertion sort in this case. Also, discuss results for the relative (physical wall clock) time taken when using (i) Shellsort that reverts to insertions sort, (ii) Shellsort all the way. 2.     The Kendall Tau distance is a variant of the "number of inversions". It is defined as the number of pairs that are in different order in two permutations. Write an efficient program that computes the Kendall Tau distance in less than quadratic time on average. Plot your results and discuss. [Use the dataset linked after Q4. Note: data0.* for convenience is an ordered set numbers (in powers of two). data1.* are shuffled data sets of sizes (as given by "*").] 3.     Implement the two versions of MergeSort that we discussed in class. Create a table or a plot for the total number of comparisons to sort the data (using data set here) for both cases. Discuss (i) relative number of operations, (ii) relative (physical wall clock) time taken. 4.     Create a data set of 8192 entries which has in the following order: 1024 repeats of 1, 2048 repeats of 11, 4096 repeats of 111 and 1024 repeats of 1111. Write a sort algorithm that you think will sort this set "most" effectively. Explain why you think so. Data Set for Questions above:

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Integral Test # Integral Test - Integral Test The last topic that we... This preview shows pages 1–6. Sign up to view the full content. Integral Test The last topic that we discussed in the previous section was the harmonic series. In that discussion we stated that the harmonic series was a divergent series. It is now time to prove that statement. This proof will also get us started on the way to our next test for convergence that we’ll be looking at. So, we will be trying to prove that the harmonic series, diverges. We’ll start this off by looking at an apparently unrelated problem. Let’s start off by asking what the area under is on the interval . From the section on Improper Integrals we know that this is, and so we called this integral divergent (yes, that’s the same term we’re using here with series….). So, just how does that help us to prove that the harmonic series diverges? Well, recall that we can always estimate the area by breaking up the interval into segments and then sketching in rectangles and using the sum of the area all of the rectangles as an estimate of the actual area. Let’s do that for this problem as well and see what we get. We will break up the interval into subintervals of width 1 and we’ll take the function value at the left endpoint as the height of the rectangle. The image below shows the first few rectangles for this area. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document So, the area under the curve is approximately, Now note a couple of things about this approximation. First, each of the rectangles overestimates the actual area and secondly the formula for the area is exactly the harmonic series! Putting these two facts together gives the following, Notice that this tells us that we must have, Since we can’t really be larger than infinity the harmonic series must also be infinite in value. In other words, the harmonic series is in fact divergent. So, we’ve managed to relate a series to an improper integral that we could compute and it turns out that the improper integral and the series have exactly the same convergence. Let’s see if this will also be true for a series that converges. When discussing the Divergence Test we made the claim that converges. Let’s see if we can do something similar to the above process to prove this. We will try to relate this to the area under is on the interval . Again, from the Improper Integral section we know that, This preview has intentionally blurred sections. Sign up to view the full version. View Full Document and so this integral converges. We will once again try to estimate the area under this curve. We will do this in an almost identical manner as the previous part with the exception that we will instead of using the left end points for the height of our rectangles we will use the right end points. Here is a sketch of this case, In this case the area estimation is, This time, unlike the first case, the area will be an underestimation of the actual area and the estimation is not quite the series that we are working with. Notice however that the only difference is that we’re missing the first term. This means we can do the following, Or, putting all this together we see that, 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. ## This note was uploaded on 11/10/2011 for the course MATH 136 taught by Professor Prellis during the Fall '08 term at Rutgers. ### Page1 / 14 Integral Test - Integral Test The last topic that we... This preview shows document pages 1 - 6. Sign up to view the full document. View Full Document Ask a homework question - tutors are online

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Metamath Proof Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  MPE Home  >  Th. List  >  blres Structured version   Visualization version   GIF version Theorem blres 22176 Description: A ball in a restricted metric space. (Contributed by Mario Carneiro, 5-Jan-2014.) Hypothesis Ref Expression blres.2 𝐶 = (𝐷 ↾ (𝑌 × 𝑌)) Assertion Ref Expression blres ((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃 ∈ (𝑋𝑌) ∧ 𝑅 ∈ ℝ*) → (𝑃(ball‘𝐶)𝑅) = ((𝑃(ball‘𝐷)𝑅) ∩ 𝑌)) Proof of Theorem blres Dummy variable 𝑥 is distinct from all other variables. StepHypRef Expression 1 inss2 3818 . . . . . . . . . 10 (𝑋𝑌) ⊆ 𝑌 21sseli 3584 . . . . . . . . 9 (𝑃 ∈ (𝑋𝑌) → 𝑃𝑌) 3 blres.2 . . . . . . . . . . 11 𝐶 = (𝐷 ↾ (𝑌 × 𝑌)) 43oveqi 6628 . . . . . . . . . 10 (𝑃𝐶𝑥) = (𝑃(𝐷 ↾ (𝑌 × 𝑌))𝑥) 5 ovres 6765 . . . . . . . . . 10 ((𝑃𝑌𝑥𝑌) → (𝑃(𝐷 ↾ (𝑌 × 𝑌))𝑥) = (𝑃𝐷𝑥)) 64, 5syl5eq 2667 . . . . . . . . 9 ((𝑃𝑌𝑥𝑌) → (𝑃𝐶𝑥) = (𝑃𝐷𝑥)) 72, 6sylan 488 . . . . . . . 8 ((𝑃 ∈ (𝑋𝑌) ∧ 𝑥𝑌) → (𝑃𝐶𝑥) = (𝑃𝐷𝑥)) 87breq1d 4633 . . . . . . 7 ((𝑃 ∈ (𝑋𝑌) ∧ 𝑥𝑌) → ((𝑃𝐶𝑥) < 𝑅 ↔ (𝑃𝐷𝑥) < 𝑅)) 98anbi2d 739 . . . . . 6 ((𝑃 ∈ (𝑋𝑌) ∧ 𝑥𝑌) → ((𝑥𝑋 ∧ (𝑃𝐶𝑥) < 𝑅) ↔ (𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑅))) 109pm5.32da 672 . . . . 5 (𝑃 ∈ (𝑋𝑌) → ((𝑥𝑌 ∧ (𝑥𝑋 ∧ (𝑃𝐶𝑥) < 𝑅)) ↔ (𝑥𝑌 ∧ (𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑅)))) 11103ad2ant2 1081 . . . 4 ((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃 ∈ (𝑋𝑌) ∧ 𝑅 ∈ ℝ*) → ((𝑥𝑌 ∧ (𝑥𝑋 ∧ (𝑃𝐶𝑥) < 𝑅)) ↔ (𝑥𝑌 ∧ (𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑅)))) 12 elin 3780 . . . . . . 7 (𝑥 ∈ (𝑋𝑌) ↔ (𝑥𝑋𝑥𝑌)) 13 ancom 466 . . . . . . 7 ((𝑥𝑋𝑥𝑌) ↔ (𝑥𝑌𝑥𝑋)) 1412, 13bitri 264 . . . . . 6 (𝑥 ∈ (𝑋𝑌) ↔ (𝑥𝑌𝑥𝑋)) 1514anbi1i 730 . . . . 5 ((𝑥 ∈ (𝑋𝑌) ∧ (𝑃𝐶𝑥) < 𝑅) ↔ ((𝑥𝑌𝑥𝑋) ∧ (𝑃𝐶𝑥) < 𝑅)) 16 anass 680 . . . . 5 (((𝑥𝑌𝑥𝑋) ∧ (𝑃𝐶𝑥) < 𝑅) ↔ (𝑥𝑌 ∧ (𝑥𝑋 ∧ (𝑃𝐶𝑥) < 𝑅))) 1715, 16bitri 264 . . . 4 ((𝑥 ∈ (𝑋𝑌) ∧ (𝑃𝐶𝑥) < 𝑅) ↔ (𝑥𝑌 ∧ (𝑥𝑋 ∧ (𝑃𝐶𝑥) < 𝑅))) 18 ancom 466 . . . 4 (((𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑅) ∧ 𝑥𝑌) ↔ (𝑥𝑌 ∧ (𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑅))) 1911, 17, 183bitr4g 303 . . 3 ((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃 ∈ (𝑋𝑌) ∧ 𝑅 ∈ ℝ*) → ((𝑥 ∈ (𝑋𝑌) ∧ (𝑃𝐶𝑥) < 𝑅) ↔ ((𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑅) ∧ 𝑥𝑌))) 20 xmetres 22109 . . . . 5 (𝐷 ∈ (∞Met‘𝑋) → (𝐷 ↾ (𝑌 × 𝑌)) ∈ (∞Met‘(𝑋𝑌))) 213, 20syl5eqel 2702 . . . 4 (𝐷 ∈ (∞Met‘𝑋) → 𝐶 ∈ (∞Met‘(𝑋𝑌))) 22 elbl 22133 . . . 4 ((𝐶 ∈ (∞Met‘(𝑋𝑌)) ∧ 𝑃 ∈ (𝑋𝑌) ∧ 𝑅 ∈ ℝ*) → (𝑥 ∈ (𝑃(ball‘𝐶)𝑅) ↔ (𝑥 ∈ (𝑋𝑌) ∧ (𝑃𝐶𝑥) < 𝑅))) 2321, 22syl3an1 1356 . . 3 ((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃 ∈ (𝑋𝑌) ∧ 𝑅 ∈ ℝ*) → (𝑥 ∈ (𝑃(ball‘𝐶)𝑅) ↔ (𝑥 ∈ (𝑋𝑌) ∧ (𝑃𝐶𝑥) < 𝑅))) 24 elin 3780 . . . 4 (𝑥 ∈ ((𝑃(ball‘𝐷)𝑅) ∩ 𝑌) ↔ (𝑥 ∈ (𝑃(ball‘𝐷)𝑅) ∧ 𝑥𝑌)) 25 inss1 3817 . . . . . . 7 (𝑋𝑌) ⊆ 𝑋 2625sseli 3584 . . . . . 6 (𝑃 ∈ (𝑋𝑌) → 𝑃𝑋) 27 elbl 22133 . . . . . 6 ((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋𝑅 ∈ ℝ*) → (𝑥 ∈ (𝑃(ball‘𝐷)𝑅) ↔ (𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑅))) 2826, 27syl3an2 1357 . . . . 5 ((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃 ∈ (𝑋𝑌) ∧ 𝑅 ∈ ℝ*) → (𝑥 ∈ (𝑃(ball‘𝐷)𝑅) ↔ (𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑅))) 2928anbi1d 740 . . . 4 ((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃 ∈ (𝑋𝑌) ∧ 𝑅 ∈ ℝ*) → ((𝑥 ∈ (𝑃(ball‘𝐷)𝑅) ∧ 𝑥𝑌) ↔ ((𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑅) ∧ 𝑥𝑌))) 3024, 29syl5bb 272 . . 3 ((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃 ∈ (𝑋𝑌) ∧ 𝑅 ∈ ℝ*) → (𝑥 ∈ ((𝑃(ball‘𝐷)𝑅) ∩ 𝑌) ↔ ((𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑅) ∧ 𝑥𝑌))) 3119, 23, 303bitr4d 300 . 2 ((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃 ∈ (𝑋𝑌) ∧ 𝑅 ∈ ℝ*) → (𝑥 ∈ (𝑃(ball‘𝐶)𝑅) ↔ 𝑥 ∈ ((𝑃(ball‘𝐷)𝑅) ∩ 𝑌))) 3231eqrdv 2619 1 ((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃 ∈ (𝑋𝑌) ∧ 𝑅 ∈ ℝ*) → (𝑃(ball‘𝐶)𝑅) = ((𝑃(ball‘𝐷)𝑅) ∩ 𝑌)) Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 196   ∧ wa 384   ∧ w3a 1036   = wceq 1480   ∈ wcel 1987   ∩ cin 3559   class class class wbr 4623   × cxp 5082   ↾ cres 5086  ‘cfv 5857  (class class class)co 6615  ℝ*cxr 10033   < clt 10034  ∞Metcxmt 19671  ballcbl 19673 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-8 1989  ax-9 1996  ax-10 2016  ax-11 2031  ax-12 2044  ax-13 2245  ax-ext 2601  ax-sep 4751  ax-nul 4759  ax-pow 4813  ax-pr 4877  ax-un 6914  ax-cnex 9952  ax-resscn 9953 This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1038  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1878  df-eu 2473  df-mo 2474  df-clab 2608  df-cleq 2614  df-clel 2617  df-nfc 2750  df-ne 2791  df-ral 2913  df-rex 2914  df-rab 2917  df-v 3192  df-sbc 3423  df-csb 3520  df-dif 3563  df-un 3565  df-in 3567  df-ss 3574  df-nul 3898  df-if 4065  df-pw 4138  df-sn 4156  df-pr 4158  df-op 4162  df-uni 4410  df-iun 4494  df-br 4624  df-opab 4684  df-mpt 4685  df-id 4999  df-xp 5090  df-rel 5091  df-cnv 5092  df-co 5093  df-dm 5094  df-rn 5095  df-res 5096  df-ima 5097  df-iota 5820  df-fun 5859  df-fn 5860  df-f 5861  df-fv 5865  df-ov 6618  df-oprab 6619  df-mpt2 6620  df-1st 7128  df-2nd 7129  df-map 7819  df-xr 10038  df-psmet 19678  df-xmet 19679  df-bl 19681 This theorem is referenced by:  metrest  22269  xrsmopn  22555  lebnumii  22705  blssp  33223  sstotbnd2  33244  blbnd  33257  ssbnd  33258  iooabslt  39167 Copyright terms: Public domain W3C validator

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Win a copy of Functional Reactive Programming this week in the Other Languages forum! # k&b scjp5, page no 616, problem no 9 Mukesh Gulia Greenhorn Posts: 2 Given a properly prepared String array containing five elements, which range of results could a proper invocation of Arrays.binarySearch()produce? a. 0 through 4 b. 0 through 5 c. -1 through 4 d. -1 through 5 e. -5 through 4 f. -5 through 5 g. -6 through 4 h. -6 through 5 ans is G. i thought answer is be E range can be ((-5) -1) through 4 i.e. 4 to 4 negative result give insertion point(array index). Brij Garg Ranch Hand Posts: 234 If you know range can be ((-5) -1) through 4 i.e. 4 to 4 then range is -6 to 4. sweety sinha Ranch Hand Posts: 76 ............................................................................ range can be ((-5) -1) through 4 i.e. ((-5)+(-1))= (-5-1)= -6 i.e range can be -6 through 4 you did a slight mistake in adding. Mukesh Gulia Greenhorn Posts: 2 yeah yeah thanks all

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An Operator Method for Solving Second Order Differential Equations, Part 2 In a previous post we discussed an operator method for solving certain second order ordinary differential equations. In this post I’ll explore this operator method a little further. I first learned about this method from an old book, Higher Mathematics for Engineers and Physicists, by Ivan S. Sokolnikoff and Elizabeth S. Sokolnikoff, McGraw-Hill, 1941. I discovered the book while browsing in a used book store a few years ago, the last time I taught differential equations. You will find some of the ideas behind this post on pages 287 ff. Consider a first order linear differential equation, with constant coefficients, which is of the form \$latex y^{prime} + ay = f(x)\$ where a is a constant. A well-known method of solution is to multiply each term by a suitable integrating factor, \$latex e^{ax}\$ in this case, to obtain \$latex e^{ax}y^{prime} + ae^{ax}y = e^{ax}f(x)\$ Then the left side can… View original post 752 more words

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Re: Oh why is life never easy? Need help with |vector| • Subject: [mg2795] Re: Oh why is life never easy? Need help with |vector| • From: Jorma.Virtamo at vtt.fi (Jorma Virtamo) • Date: Sun, 17 Dec 1995 02:08:07 -0500 • Approved: usenet@wri.com • Distribution: local • Newsgroups: wri.mathgroup • Organization: Wolfram Research, Inc. ```ULI12 at HOMER.geol.chemie.tu-muenchen.de (ULI WORTMANN) wrote: > > Hi All, > > I need to calculate v0 = v / |v| where v is a vector of size 3. > However, |v|, which is a scalar, is for some reason (I don't > understand) treated like a vector (at least VectorQ[|v|] gives > TRUE). Obviously, it is impossible to divide two vectors of different > length. > > |v| is defined as: > absvec[u_] := > Module[{a}, > {a = Sqrt[(u[[1]]^2+u[[2]]^2+u[[3]]^2)]}] > > and called as u = 1/absvec[v] > The problem is that you have made the result a (one element) list by explicitly enclosing the body of the Module in curly brackets. Just leave the brackets out and everything works fine. By the way, the use of variable a, or the Module for that matter, does not serve any useful purpose here. You could simply write absvec[u_] := Sqrt[u.u] -- Jorma Virtamo ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Jorma Virtamo VTT Information Technology phone: +358 0 456 5612 Telecommunications fax: +358 0 455 0115 P.O. Box 1202 email: jorma.virtamo at vtt.fi FIN-02044 VTT web: http://www.vtt.fi/tte/ Finland ``` • Prev by Date: Infinity • Next by Date: Re: Bug ListPlot? • Previous by thread: Oh why is life never easy? Need help with |vector| • Next by thread: Re: Oh why is life never easy? Need help with |vector|

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[–] 1 point2 points  (0 children) I was at the game, didn't realize Johnny buried the GWG at first because I thought (from my angle) Kempe had crossed check the Gull and was like "AW FUCK" and then heard all the moans and groans around me and saw them jumping in the corner so I started jumping too. What a game. Mersch looked really confused a lot of plays, that shorty was just piss poor, he just walked past him then realized he needed to play D. [–] 3 points4 points  (0 children) "Did you down vote this map?" "No" "Wow cancer uninstall etc" One reason I would lie about it. I didn't say your evidence wasn't evidence, just pointing out it's not proof it's not working as intended. [–] 4 points5 points  (0 children) You'd have to take the other team at their word though. Even if they say "We all downvoted this map" they could be A) Lying or B) Mistaken [–] 7 points8 points  (0 children) In case people in this thread are serious about "dislike doesn't work because I still end up on this map", it's used to create a weighted vote. So lets use 1v1 as an example: You downvote starbase arc and wasteland and upvote Mannfield The final result is: Starbase is 50% less likely to be the map, Wasteland cannot be the map (in this game), and Mannfield I believe is twice as likely to be the map. If I understood the Season 4 patch notes correctly, the only way to totally avoid a map 100% of the time is to have each player in a game downvote the map. Seeing as how you can only control your own team's voting, you're going to see those other maps sometimes. [–] 38 points39 points  (0 children) I do not understand the second half of this title. Halp? [–] 36 points37 points  (0 children) Ah, the ole Ohio State method. [–] 1 point2 points  (0 children) The current iteration works well if the order of the list is reversed. Then there's no confusion and it's a working sentence. [–] 3 points4 points  (0 children) I'm going to go against the grain here and say current iteration looks good. Someone more versed in English can probably tell me why they disagree or cite why I'm wrong, but a semi colon can be used to join an incomplete thought with a complete sentence. "Two things in life are certain" is a complete sentence. chaos and Virgina having a losing season is being joined to that. [–] 2 points3 points  (0 children) Do they do something different than other teams fans' around the league when taunting Crow, or is it like how LA will ways chant "Hilllllerrrrr" no matter what Duck is in net. [–] 7 points8 points  (0 children) Hey now, don't sell the team short. You guys also took a massive steaming dump on our first chance at a Pacific Division title and first division title since ethnic the early 90s Edit: autocorrect [–] 0 points1 point  (0 children) He can't declare until next year and by god I will sob if he leaves that soon :( [–] 5 points6 points  (0 children) It felt like they turned it off to protect the lead then couldn't switch back on [–] 9 points10 points  (0 children) "Martin Jones with an incredible save in OT" Which one is it going to be? Oh this one, good choice [–] 1 point2 points  (0 children) FUCK. FUCK WE WERE THIS CLOSE TO CAPS AND ITALICS. YOU FUCKERS COULDN'T WAIT ANOTHER MINUTE OR SO? [–] 7 points8 points  (0 children) "YOU DON'T SEE THAT TOO OFTEN" ABOUT THE WHISTLE STOPPING PLAY FOR A DEAD PUCK. SOMEWHERE RYAN KESLER IS GIGGLING AND SIPPING AN ICE WATER BEFORE BED. [–] 7 points8 points  (0 children) GRETZKY LOOKS LIKE SOMEONE JUST TOLD HIM HE'S GETTING TRADED TO ST LOUIS AGAIN. [–] 0 points1 point  (0 children) IS THIS BECAUSE YOU HAVE A NEW FANGLED ARENA? DO YOU THINK US PLEBS WILL EVER GET THAT IMPLEMENTED IN THE OLDER RINKS? I KNOW THEY REINSTALL THE BOARDS AT STAPLES EVERY SEASON SO I DON'T SEE WHY WOULD COULDN'T DO THIS TOO, BUT I GUESS FOX SPORTS WOULD NEED TO WANT TO UTILIZE IT. [–] 4 points5 points  (0 children) JESUS THE SHARKS LOOK LIKE THEIR ON THE PK WITH HOW DESPERATE THEY ARE TO CLEAR THIS SOMETIMES, WHAT THE FUCK DID DEBOER DEBOAR DABORE TELL THEM IN THE THIRD? THEY HAVE LOOKED PHYSICALLY INCAPABLE OF TURNING SHIT ON SAVE A FEW GUYS [–] 4 points5 points  (0 children) DO YOU GUYS EVER PLAY CAT SCRATCH FEVER FOR NUGENT HOPKINS? I FEEL LIKE HE'S TOO YOUNG TO APPRECIATE IT BUT IT WOULD BE AMUSING. [–] 0 points1 point  (0 children) YOU GET A PASS, BUT I'M WATCHING YOU, OKAY? ONLY BECASUE IF THIS GOES TO 2OT ALSO ADDING THE ITALICS WOULD BE A MASSIVE PAIN. [–] 0 points1 point  (0 children) IS THAT A CAMERA HOLE RIGHT UNDERNEATH THE HONDA LOGO IN THE CORNER? THAT'S NEATO, BUT I FEEL LIKE ITS USE WOULD BE REALLY LIMITED AND SPECIFIC. [–] 7 points8 points  (0 children) HOW MANY FUCKING SCARY MOMENTS CAN YOU FORCE YOUR GOALIE TO DEAL WITH RIGHT IN FRONT OF HIS CREASE BEFORE HE CAN'T BAIL YOU OUT ANYMORE. JONES HAS ENTERED THE FUCKING ZEN OF GOALIEHOOD, YOU'RE GONNA NEED SOME BULLSHIT OR SOMETHING AMAZING TO BEAT HIM RIGHT NOW [–] 1 point2 points  (0 children) YOU SHOULD HAVE GONE DURING THE 10 MIN SCRAPE! THE ONLY SOLUTION NOW IS TO PUT IT AT MAX VOLUME AND PISS WITH THE DOOR OPEN [–] 0 points1 point  (0 children) SOUNDS LIKE THE SCF. SOUNDS LIKE A LOT OF JONES TIME WITH THE KINGS ACTUALLY. MAN JONES IS GREAT. [–] 1 point2 points  (0 children) DO WE STILL DO CAPS FOR PLAYOFF OT IN THESE THREADS? I'M SEEING A WHOLE LOT OF LOWER CASE POSTERS WHAT THE FUCK GUYS? THIS GAME HONESTLY FEELS LIKE THE MOST EXCITING WE'VE SEEN, SECOND MAYBE TO THE HABS RANGERS GAME WHERE THEY TIED IT WITH 17 SECONDS

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# Complex Signals and Systems help #### RedDogTech Joined Jul 17, 2007 2 Trying to solve diff. eq: (dw/dt) + 3w = 6e^j3t u(t). I can get to the operator form: (p + 3)w(t) = 6e^j3t u(t) ==> w(t) = (1/(p+3))*[6e^j3t u(t)] (the 1/p = Heaviside operator). The solution in the example is: w(t) = (2/(1 + j))e^j3t u(t) - (2/1 + j))e^-3t u(t) I am lost in the integration..... Can someone shed some light on this for me? Thanks.

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Thread: Count to 100 before a mod posts! 1. 47 is a prime number... 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::::::::::::::::::::::::::::::::::;:;:;:;:;:;:;:;: ;’|…_„„--~~-„ …„„-~”:.:.:._„„„-„^*”””’*-„:.:’|:::::::::::::::::::::::::::::::::::::::::::: ::::::::::::::::::;:;:;:;:;:;:;:;:;:;’|¯:.:.:.:.:. :„.:”-„ …:.:_„„-^”¯:.:.:.:.”-„:.:.:’\:.|_:::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::;:;:;:;:;:;:;:;:;:;:;’|:.:.:. :.:.:„-“:::.:”-„_......._____ …~”:.:.:.:.:.:.:.:„-“:.:.:.„-“¯:.:”-„::::::::::::::::::::::::::::::::::::::::::::::::: :::::;:;:;:;:;:;:;:;:;:;:;:’|:.:.:.„-^”:::::::.:.:.:.¯””¯:.:.:.:.:.”~„ ….:.:.:.:.:.:.:„-“:.:.:.„-“:.:.:.:.:.:.”^^***””¯¯”^-„::::::::::::::::::::::::::::::„-~~-„;:;:;:;:;:;:;:;:;_„„„|-~^”-„:::::::::.:.:.:.:.:.:.:.:.:.:.:.:.:.:”-„ ….„„-^”””””¯¯.:.:.:„-“.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:’\::::::::::::::: ::::::::„-^¯.:.:.,„-^¯”-„:;:;:;„-^”::::::::::::::”’-:::::::::::.:.:.:.:.:.:.:.:.:.:.:.:.: …..„„„--~^**^^~”.:.:.:.:„„-~”„””^-„.:.:.:.:.:.:.:.:.:”---^””¯¯¯””*^„~“.:|.:.:.:„-“.:.:.„-“¯”-„„/::::::::::::::::::::::::::::::::::::.:.:.:.: ……………………„-“.:.:.:„/.:.:.\.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.„/“.:.:.:”^^*”.:.:.„-“.:.:.:.:.:¯-„„„„„:::::::::::::::::::::::::::::: …………………./”^~--~^”.:.:„-“.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:|”^^~~------~^”.:.:.:.:.:.:.:.:.:.:.:\:::::::::::::::::::::: ………………….¯”””””””””¯¯…………………………....”^~-------~~~~^^^^^~~--„„„„-~~”::::::::::: 5. 51 Swag/ 6. 52 _____________________¶¶¶__________________¶¶¶ ____________________¶¶11¶¶______________¶¶11¶¶ ____________¶¶¶¶¶¶__¶¶11¶¶______________¶¶11¶¶ __________¶¶¥¥¥¥¥¥¶¶_¶¶11¶¶___¶¶¶¶______¶¶11¶¶____ ¶¶¶¶¶¶¶¶¶ __________¶¶¥¥¥¥¥¥¥¥¶¶¶11¶¶_¶¶ƒƒƒƒ¶¶___¶¶111¶¶__¶¶ ¶¥¥¥¥¥¥¥¶¶ __________¶¶¥¥§§§§¥¥¥¶¶311¶¶¶øøøøøø¶¶_¶¶1133¶¶¶¶¥¥ ¥¥¥¥¥¥¥¥¥¶¶ __________¶¶¥¥§§§§§§§§¶¶11¶¶ƒƒƒƒƒƒƒƒ¶¶¶¶113¶¶¥¥¥§§ §§§§¥¥¥¥¥¶¶ ___________¶¶§§§§§§§§§3311¶¶øøøøøøøø¶¶3311¶¶¶¶§§§§ §§§§§¥¥¥¶¶ ____________¶¶§§§§§§333311¶¶ƒƒƒƒƒƒƒƒ¶¶3333¶¶33§§§§ §§§§§¥¥¶¶ _____________¶¶¶§§33111111¶¶¶øøøøøø¶¶¶3333333333§§ §§§§¥¥¶¶__¶¶¶¶ ___¶¶¶¶¶______¶¶331111111111¶¶ƒƒƒƒ¶¶11113333333333 §§§¥¥¶¶¶¶¶1111¶¶ 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((27)(6859))^(1/3) 12. 58 13. 59 fifty nine cicuenta y nueve wü shí jïu 14. 60 sesenta not to be cunfused with setenta which means seventy _____________________¶¶¶__________________¶¶¶ ____________________¶¶11¶¶______________¶¶11¶¶ ____________¶¶¶¶¶¶__¶¶11¶¶______________¶¶11¶¶ __________¶¶¥¥¥¥¥¥¶¶_¶¶11¶¶___¶¶¶¶______¶¶11¶¶____ ¶¶¶¶¶¶¶¶¶ __________¶¶¥¥¥¥¥¥¥¥¶¶¶11¶¶_¶¶ƒƒƒƒ¶¶___¶¶111¶¶__¶¶ ¶¥¥¥¥¥¥¥¶¶ __________¶¶¥¥§§§§¥¥¥¶¶311¶¶¶øøøøøø¶¶_¶¶1133¶¶¶¶¥¥ ¥¥¥¥¥¥¥¥¥¶¶ __________¶¶¥¥§§§§§§§§¶¶11¶¶ƒƒƒƒƒƒƒƒ¶¶¶¶113¶¶¥¥¥§§ §§§§¥¥¥¥¥¶¶ ___________¶¶§§§§§§§§§3311¶¶øøøøøøøø¶¶3311¶¶¶¶§§§§ §§§§§¥¥¥¶¶ ____________¶¶§§§§§§333311¶¶ƒƒƒƒƒƒƒƒ¶¶3333¶¶33§§§§ §§§§§¥¥¶¶ _____________¶¶¶§§33111111¶¶¶øøøøøø¶¶¶3333333333§§ §§§§¥¥¶¶__¶¶¶¶ ___¶¶¶¶¶______¶¶331111111111¶¶ƒƒƒƒ¶¶11113333333333 §§§¥¥¶¶¶¶¶1111¶¶ _¶¶¶1111¶¶¶¶__¶¶1111¶¶¶¶111111¶¶¶¶111133¶¶¶¶333333 33¶¶¶¶1111¶¶¶¶ ____¶¶¶¶1111¶¶3311¶¶____¶¶111111111111¶¶____¶¶3333 333311¶¶¶¶ ________¶¶¶¶33111¶¶___¶¶__¶¶11111111¶¶__¶¶___¶¶333 333¶¶¶ ____________¶¶111¶¶___¶¶__¶¶11111111¶¶__¶¶___¶¶333 33333¶¶ ____________¶¶111¶¶___¶¶__¶¶11111111¶¶__¶¶___¶¶333 33333¶¶ ____________¶¶3333¶¶__¶¶__¶¶11111111¶¶__¶¶__¶¶3333 33333¶¶_________¶¶¶¶¶¶¶¶ _____________¶¶33333¶¶¶¶¶¶111111113333¶¶¶¶¶¶333333 33333¶¶_______¶¶£££#####¶¶ __________¶¶¶¶333333333333333333333333333333333333 ¶¶33¶¶______¶¶££££££££####¶¶ ______¶¶¶¶33333333333¶¶33333333333333333333¶¶¶3333 33¶¶________¶¶££££££££#####¶¶ __¶¶¶¶1111¶¶¶¶¶¶¶¶3333¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶333333 3333¶¶¶¶____¶¶££££££££#####¶¶ ¶¶1111¶¶¶¶________¶¶3333¶¶3¶¶33333333¶¶3¶¶333333¶¶ ¶¶¶¶3333¶¶¶¶¶¶££££££##¶¶###¶¶ _¶¶¶¶¶______¶¶¶¶____¶¶¶¶33333333333333333333¶¶¶¶__ ____¶¶¶¶3388¶¶¶¶######¶¶###¶¶ __________¶¶1113¶¶______¶¶¶¶¶¶33333333338¶¶¶¶¶____ ________¶¶¶¶____¶¶¶¶¶¶####¶¶ ________¶¶11111333¶¶______¶¶333333333333333333¶¶¶¶ __________________¶¶####¶¶ ______¶¶111111¶¶33¶¶____¶¶888833333333333333333333 ¶¶______________¶¶££££##¶¶ ______¶¶11¶¶1133¶¶88¶¶¶¶8888¶¶1111¶¶33333¶¶¶333333 33¶¶__________¶¶££££##¶¶ ______¶¶1111¶¶¶¶8888888888¶¶1111¶¶1111111113¶¶3333 33¶¶______¶¶¶¶333888¶¶ ________¶¶¶¶¶¶__¶¶¶¶8888¶¶¶¶1111¶¶11§§11§§13¶¶3333 ¶¶¶¶__¶¶¶¶33338888¶¶ ____________________¶¶¶¶__¶¶1111¶¶11¶¶13¶¶33¶¶¶¶¶¶ ¶¶¶¶¶¶88888888¶¶¶¶ __________________________¶¶331111¶¶¶¶¶¶¶¶¶¶333338 ¶¶88888888¶¶¶¶ ____________________________¶¶333333333333333388¶¶ 888888¶¶¶¶ ____________________________¶¶8833333333333388888¶ ¶¶¶¶¶¶ ______________________________¶¶888¶¶¶88888888888¶ ¶¶ __________________________¶¶¶¶888888¶¶¶¶¶¶¶¶¶88888 ¶¶ ____________________¶¶¶¶¶¶3333888888¶¶______¶¶8833 33¶¶ __________________¶¶£££££££333888¶¶¶________¶¶3333 3£££¶¶ ________________¶¶££££¶¶¶£££££¶¶¶____________¶¶£££ ££££££¶¶ ________________¶¶¶££¶¶£££¶¶¶¶________________¶¶££ ¶¶££££££¶¶ __________________¶¶¶¶¶¶¶¶____________________¶¶££ ¶¶££¶¶££¶¶ ________________________________________________¶¶ ¶¶¶¶¶¶¶¶ AWESOMENESSSSSSSS 15. dying Join Date Dec 2009 Posts 0 61 a sudowoodo would be nice… but italicized and small enough for a sig D; 16. Originally Posted by thirty-six 61 a sudowoodo would be nice… but italicized and small enough for a sig D; I don't think I can find one small enough for a sig, but I bet I can find one. 62 17. dying Join Date Dec 2009 Posts 0 Ah well thanks anywoo :3 63 18. 64 I couldn't even find one I can try my hand at making one, but I don't know how well I could do it.. 19. dying Join Date Dec 2009 Posts 0 Don't worry then. The effort of wanting to is good enough :3 65 20. 6 6 21. 67................. 22. sixty ocho 6eight sesenta y bā 23. 69 24. 70....................... 25. 71 I WONDER WHAt's FOR DINNER?!?!?! Page 1294 of 1813 First ... 294119412441284129012911292129312941295129612971298130413441394 ... Last Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts •

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# Radius of convergence for (1+x)^1/2 • asif zaidi In summary, the conversation discusses finding the Taylor Series for (1+x)^1/2 and determining the radius of convergence. The solution involves finding the derivatives of the function and simplifying to obtain the Taylor Series. The conversation also includes a discussion on using the ratio test to determine the interval of convergence, with some discrepancy in the calculations. Ultimately, the conversation arrives at an expression for the Taylor Series, but the issue of taking into account the first two terms remains unresolved. asif zaidi Problem Statement: Compute the Taylor Series for (1+x)^1/2 and find the radius of convergence Problem Solution: The Taylor Series expansion I get is T(x) = 1 + (0.5*x) - (0.25*x^2)/2 + (0.375*x^3)/3! - (0.9375*x^4)/4! +...-... So to get radius of convergence I have to find a closed solution form of the above equation and I simply am not able to come up with anything. Any pointers on this will be appreciated. But then I was looking at the equation and saw that at least I can represent the x and the factorial portions as Sum (from 0-inf) x^i/i! and not worry about the coefficient. If I use ratio test on this closed form, I will get the interval of convergence to be -inf to +inf. Therefore radius of convergence is +inf Thanks Asif ?? A closed form for the series? It is (1+ x)1/2 of course! I have no idea how you get -inf to inf as interval of convergence. You wrote the coefficients as "0.5", (0.25", "0.375", etc. which makes me wonder if you are sure of the correct values. The nth derivative of (1+x)-1/2, at 0, is (2n-3)!/((n-2)!2n-2) so the nth term in the sum is (2n-3)!xn/((n-2)!n!2n-2). Applying the ratio test to that does not give infinity as radius of convergence. But you can use the "closed form", which, as I said, was exactly what you were given: (1+ x)-1/2. In general the Taylor's series of a function will converge as long as the function is "nice". Where is there a "problem" with (1+ x)-1/2? How far is that value of x from x=0? That is the radius of convergence. Last edited by a moderator: I am a bit confused with the response so I will give more detail as to what I have done First part is to find Taylor Series for (1+x)$$^{1/2}$$ BTW, in your response you have given (1+x)$$^{-1/2}$$ so I think the 1st difference is there. Anyway for Taylor series I do the following. Find derivatives f'(x) = 1/2 (1+x)$$^{-1/2}$$ f''(x) = (1/2)(-1/2)(1+x)$$^{-3/2}$$ f'''(x) = (1/2)(-1/2)(-3/2)(1+x)$$^{-5/2}$$ f''''(x) = (1/2)(-1/2)(-3/2)(-5/2)(1+x)$$^{-7/2}$$ f'''''(x) = (1/2)(-1/2)(-3/2)(-5/2)(-7/2)(1+x)$$^{-9/2}$$ f''''''(x) = (1/2)(-1/2)(-3/2)(-5/2)(-7/2)(-9/2)(1+x)$$^{-11/2}$$ Simplifing above gives and evaluating at 0 gives f'(x) = 1/2(1+x)$$^{1/2}$$ = 1/2 f''(x) = -1/4(1+x)$$^{-3/2}$$ = -1/4 f'''(x) = 3/8(1+x)$$^{-5/2}$$ = 3/8 f''''(x) = -15/16(1+x)$$^{-7/2}$$ = -15/16 f'''''(x) = 105/32(1+x)$$^{-9/2}$$ = 105/32 f''''''(x) = -945/64(1+x)$$^{-11/2}$$ = -945/64 Thus Taylor Series: T(x) = 1 + x/2 -x$$^{2}$$/2!4 + 3x$$^{3}$$/3!8 -15x$$^4}$$/4!16 + 105x$$^{5}$$/5132 - 945x$$^{6}$$/6!64 +... Next Step is to find an expression using $$\sum$$ (from 0-inf). And this is what I meant by closed sum. My terminology is probably wrong according to what you say. However, whenever I have done exercises to find radius/interval of convergence, I have always had an expr of $$\sum$$. I then evaluate a(n+1)/a(n) and see it is < |1| and find interval of convergence Any advice on how I can find the right expression and then radius of convergence. Note: in your case if you give me (1+x)$$^{-1/2}$$ then the radius of convergence will be 1. But I don't think this this applies to (1+x)$$^{1/2}$$ case because this function is 'nice' everywhere. Thanks Asif Now set x= 0: f'(0): 1/2 f''(0): (1/2)(-1/2) f'''(0): (1/2)(-1/2)(-3/2) f''''(0): (1/2)(-1/2)(-3/2)(-5/2) f'''''(0): (1/2)(-1/2)(-3/2)(-5/2)(-7/2) f''''''(0): (1/2)(-1/2)(-3/2)(-5/2)(-7/2)(-9/2) The sign is alternating: it's (-1)n. The denominator is a power of 2: it's 2n The numerator is 1(3)(5)(7)(9)... That would be an exponential except that it is "missing" the even integers. Okay, put them in: 1(2)(3)(4)(5)(6)(7)(8)(9)/(2)(4)(6)(8) where I have also, of course, put those same numbers in the denominator- I have multiplied numerator and denominator by the same thing in order to keep the fraction the same thing. The numerator is now obviously 9!. What about the denominator? Well, since those are all even numbers, we can factor a 2 out of each: (2)(4)(6)(8)= (2)(1)(2)(2)(2)(3)(2)(4)= 24(1)(2)(3)(4)= 24(4!). The numerator in f''''''(0) is 9!/244!. It should not be difficult to see that the general term, for f(n)(0) is (2n+1)!/2nn! Since the denominator was already 2n and the sign was alternating, f(n)(0)= (2n+1)!/[(2n(n!)2] and so the nth term in the Taylor's series is (2n+1)!/[(2n(n!)3]xn HallsofIvy said: Now set x= 0: The numerator is now obviously 9!. What about the denominator? Well, since those are all even numbers, we can factor a 2 out of each: (2)(4)(6)(8)= (2)(1)(2)(2)(2)(3)(2)(4)= 24(1)(2)(3)(4)= 24(4!). The numerator in f''''''(0) is 9!/244!. It should not be difficult to see that the general term, for f(n)(0) is (2n+1)!/2nn! Since the denominator was already 2n and the sign was alternating, f(n)(0)= (2n+1)!/[(2n(n!)2] and so the nth term in the Taylor's series is (2n+1)!/[(2n(n!)3]xn OK - I see what you are trying to do by getting 9! etc... But if I use the formula you give, it does not match You have given f(n)(0)= (2n+1)!/[(2n(n!)2] So if I calculate it for n=6, I get 13!/(12*6!2 = 187.6875. However, using Taylor Series expansion, I get 14.76 However, using info you provided I was able to come up with the following form 1 + 0.5x + Sum (2-inf) [ (2n-3)! / n!(n-2)! 22n-2 ]. I did this for a few n's and got the same coefficient as Taylor Series. But the problem is to find the radius of convergence for this, how do I take into account the 1st 2 terms (1+0.5x). Thanks Asif ## 1. What is the radius of convergence for (1+x)^1/2? The radius of convergence for (1+x)^1/2 is 1. This means that the series will converge for all values of x within a distance of 1 from the center point (x=0). ## 2. How is the radius of convergence determined for (1+x)^1/2? The radius of convergence is determined by using the Ratio Test, which compares the absolute value of consecutive terms in the series. If the limit of this ratio is less than 1, the series will converge. In the case of (1+x)^1/2, the limit is 1, indicating that the series converges for x values within a distance of 1 from the center point. ## 3. What happens if x is outside the radius of convergence for (1+x)^1/2? If x is outside the radius of convergence, the series will diverge, meaning that it does not have a finite sum. This means that the series is not a valid representation of the function (1+x)^1/2 for those values of x. ## 4. Can the radius of convergence for (1+x)^1/2 be negative? No, the radius of convergence cannot be negative. It represents a distance from the center point (x=0), which cannot be a negative value. However, it is possible for the radius of convergence to be 0, which would indicate that the series only converges at the center point and does not converge for any other values of x. ## 5. Does the radius of convergence for (1+x)^1/2 depend on the power of x? Yes, the radius of convergence can vary depending on the power of x in the series. In the case of (1+x)^1/2, the radius of convergence is 1, but for a series such as (1+x)^2, the radius of convergence would be 0.5, as the series converges for x values within a distance of 0.5 from the center point. • Calculus and Beyond Homework Help Replies 10 Views 1K • Calculus and Beyond Homework Help Replies 5 Views 1K • Calculus and Beyond Homework Help Replies 2 Views 445 • Calculus and Beyond Homework Help Replies 4 Views 1K • Calculus and Beyond Homework Help Replies 14 Views 1K • Calculus and Beyond Homework Help Replies 4 Views 1K • Calculus and Beyond Homework Help Replies 1 Views 512 • Calculus and Beyond Homework Help Replies 2 Views 828 • Calculus and Beyond Homework Help Replies 4 Views 587 • Calculus and Beyond Homework Help Replies 6 Views 1K

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https://math.stackexchange.com/questions/3175515/find-the-value-of-d

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# Find the value of D. The complex numbers $$1+i$$ and $$1+2i$$ are both roots of the equation $$x^5-6x^4+Ax^3+Bx^2+Cx+D=0$$, where $$A, B, C, D \in R$$ Find the value of D. My attempt: The given equation will have 5 roots (distinct or undistinct) since it is a polynomial equation of degree 5. The coefficients are all real. Since $$1+i$$ and $$1+2i$$ are two roots of the given equation, their conjugates are also the roots of this equation. Therefore $$1-i$$ and $$1-2i$$ are also the roots of this equation. Therefore the L.H.S can be factorized as $$x^5-6x^4+Ax^3+Bx^2+Cx+D=\{x-(1+i)\}\{x-(1-i)\}\{x-(1+2i)\}\{x-(1-2i)\}Q(x)=(x^4-4x^3+11x^2-14x+10)Q(x)$$ ,where $$Q(x)$$ is a polynomial of degree 1 are it gives the unknown root of the given equation. We find $$Q(x)$$ is $$x-2$$ by long division. Multiplying the other factor by $$(x-2)$$ we get the constant term is -20. Now $$D$$ is equal to the constant term. Therefore $$D$$ is equal to -20. Am I correct? Is there any other way to find $$D$$? $$6=1+i+1-i+1-2i+1+2i+t$$ where $$t$$ is the fifth root

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# ELEMENTARY LINEAR ALGEBRA 10TH EDITION TEXTBOOK SOLUTIONS PDF Free step-by-step solutions to Elementary Linear Algebra () – Slader. Solutions Manual for Larson/Flavo’s Elementary Linear Algebra, 6th Edition Elementary Linear Algebra textbook solutions reorient your old paradigms. Elementary Linear Algebra: Applications Version 10th Edition. by . Story time just got better with Prime Book Box, a subscription that delivers Student Solutions Manual to accompany Elementary Linear Algebra with Applications, 10e. WileyPLUSTM This isWiley’s proprietary online teaching and learning environment that integrates a digital version of this textbook with instructor and student. Author: Gujora Vudokasa Country: Sao Tome and Principe Language: English (Spanish) Genre: Software Published (Last): 24 January 2013 Pages: 54 PDF File Size: 20.41 Mb ePub File Size: 19.44 Mb ISBN: 215-9-73759-124-5 Downloads: 39748 Price: Free* [*Free Regsitration Required] Uploader: JoJonris Functions Average rate of change: Irrational numbers Proofs concerning irrational numbers: Factorization Linwar in factoring quadratics: Solving inequalities Two-step inequalities: Algebra foundations Combining like terms: Solving inequalities One-step inequalities: Properties of exponents rational exponents: Intro to slope Slope. Simplifying square roots review Simplifying square roots. Systems of equations Systems of equations word problems: Average rate of change review Average rate of change ecition problems. Factorization Evaluating expressions with unknown variables: Linear word problems Textbkok linear functions: Systems of equations Solving any system of linear equations: Working with units Word problems with multiple units: Factorization Factoring polynomials by taking common factors: Factorization Factoring polynomials with quadratic forms: Solving equations Linear equations with parentheses: Arithmetic sequences review Constructing arithmetic sequences. Related Articles  FLIRTING WITH DANGER SUZANNE ENOCH PDF Introduction to inequalities with variables: Functions Maximum and minimum points: Click here to start or continue working on the Algebra I Mission. Functions Intervals where a function is positive, negative, 10tb, or decreasing: Quadratics The quadratic formula: Factorization Factoring quadratics intro: Geometric sequences review Constructing geometric sequences. Graphs of absolute value functions: Slope-intercept form review Writing slope-intercept equations. If you’re seeing this message, it means we’re having trouble loading external resources on our website. Systems of equations Equivalent systems of equations and the elimination method: Algebra foundations Division by zero: Functions Inputs and outputs of a function: Interpreting linear functions and equations: Linear word problems Constructing linear models for textgook relationships: Quadratics Solving quadratics by taking square roots: Algebra foundations Writing algebraic expressions introduction: Linead equation word problems: Systems of equations Solving systems of equations with substitution: Forms of two-variable linear equations: Sequences Introduction soljtions geometric sequences: Manipulating expressions with unknown variables: Multiplying monomials by polynomials: Systems of equations Number of solutions to systems of equations: Functions Interpreting function notation: Algebra foundations Introduction to equivalent algebraic expressions: Functions Functions and equations: Algebra foundations Substitution and evaluating expressions: Did you realize that the word “algebra” comes from Arabic just like “algorithm” and “al jazeera” and “Aladdin”? Related Articles  17 DAGEN DIEET PDF And what is so great about algebra anyway? Sequences Constructing arithmetic sequences: Solving equations Linear equations with variables on both sides: Perfect squares Factoring quadratics: Algebra foundations Introduction to variables:

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https://mail.python.org/pipermail/tutor/2010-October/079587.html

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[Tutor] Problem with python Dave Kuhlman dkuhlman at rexx.com Wed Oct 20 23:03:15 CEST 2010 ```On Wed, Oct 20, 2010 at 03:02:43AM +0600, Matthew Nunes wrote: > > To whom it may concern, > > Hi, I've just started learning how > to program in python using Allan B. Downy's book "How to think like a > computer scientist" and it explained something in the recursion > chapter which have still been unable to understand. It wrote a piece > of code for the factorial function in math for example 3! is 3 * 2 * > 1. I cannot understand the how it claimed the code executed, and > logically it makes no sense to me. Please explain it to me, as I have > spent hours trying to get my head around it, as the book(in my > opinion) did not explain it well. Here it is: > > > def factorial(n): > > if n == 0: > > return 1 > > else: > > recurse = factorial(n-1) > > result = n * recurse > return result > > If there is and easier piece of code that you know of that can be used > for factorial, if you could please also tell me that. > > Thank you for your time. The other replies have helped you understand that recursive function. But, recursion gets especially interesting when it is applied to recursive data structures, for example data whose structure is the same as the data that it contains. A tree structure whose nodes all have the same structure is a good example. The code that follows constructs a tree out of objects that are instances of class Node, then walks that tree and prints it out. This code provides two recursive ways to walk and display that tree: (1) the "show" method in class Node and (2) the "shownode" function. Often processing recursive structures like trees is trivial with a recursive function or method. Keep in mind that this code is well behaved only when the data structure we apply it to has a reasonable maximum depth. Hope this example helps. - Dave # ============================================================ class Node(object): def __init__(self, data, children=None): self.data = data if children is None: self.children = [] else: self.children = children def show(self, level): print '%sdata: %s' % (' ' * level, self.data, ) for child in self.children: child.show(level + 1) def shownode(node, level): print '%sdata: %s' % (' ' * level, node.data, ) for child in node.children: child.show(level + 1) def test(): n1 = Node('aaaa') n2 = Node('bbbb') n3 = Node('cccc') n4 = Node('dddd') n5 = Node('eeee', [n1, n2]) n6 = Node('ffff', [n3, n4]) n7 = Node('gggg', [n5, n6]) n7.show(0) print '=' * 20 shownode(n7, 0) test() # ============================================================ -- Dave Kuhlman http://www.rexx.com/~dkuhlman ``` More information about the Tutor mailing list

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https://discourse.codecombat.com/t/lost-viking-raven/15402

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# Lost Viking - Raven? So the Raven will always say what the actual steps/sideSteps are -1. In order for me to know the exact steps I was actually taking, I had to do: hero.say("steps is " + steps); hero.say("sideSteps is " + sideSteps); since the raven isn’t quoting the actual steps but is a step behind. It’s confusing OR am I missing something? You can test and recreate it with this code: Code ``````// You MUST click on the HELP button to see a detailed description of this level! // The raven will tell you what to use for your maze parameters! var SLIDE = 9; var SKIP = 7; var SWITCH = 5; // How many sideSteps north of the Red X you've taken. var sideSteps = 1; // How many steps east of the Red X you've taken. var steps = 1; // Multiply this with steps to determine your X coordinate. DON'T CHANGE THIS! var X_PACE_LENGTH = 4; // Multiply this with sideSteps to determine your Y coordinate. DON'T CHANGE THIS! var Y_PACE_LENGTH = 6; var direction = 1; // The maze is 35 steps along the X axis. while (steps <= 35) { // Take the next step: hero.moveXY(steps * X_PACE_LENGTH, sideSteps * Y_PACE_LENGTH); // Increment steps and sideSteps as appropriate, taking into account the special rules. steps += 1; sideSteps += direction; hero.say("steps is " + steps); hero.say("sideSteps is " + sideSteps); } `````` [EDIT] You’re right SuperSmacker! Thanks for the help! I’m still learning how the order of programming executes. Good to know I can confidently quote the raven You incremented the step before saying it. You are saying the number of steps BEFORE you take the next step. For example, at the beginning, you said steps = 2 even though you didn’t even move yet. The raven says the number of steps you have taken when you are talking about the number of steps you are about to take. Therefore, the raven is not bugged. 1 Like So I moved the hero.say() before the increments, however, the hero doesn’t take a step until saying steps = 2. Here’s what I have so far: Code so far on Lost Viking ``````// You MUST click on the HELP button to see a detailed description of this level! // The raven will tell you what to use for your maze parameters! var SLIDE = 9; var SKIP = 7; var SWITCH = 5; // How many sideSteps north of the Red X you've taken. var sideSteps = 1; // How many steps east of the Red X you've taken. var steps = 1; // Multiply this with steps to determine your X coordinate. DON'T CHANGE THIS! var X_PACE_LENGTH = 4; // Multiply this with sideSteps to determine your Y coordinate. DON'T CHANGE THIS! var Y_PACE_LENGTH = 6; var direction = 1; // The maze is 35 steps along the X axis. while (steps <= 35) { //Checking for various maze parameters var yesSwitch = (steps) % SWITCH === 0; var yesSkip = (steps) % SKIP === 0; var yesSlide1 = (sideSteps > 10); var yesSlide10 = (sideSteps < 3 && steps > 2); hero.say("steps is " + steps); hero.say("sideSteps is " + sideSteps); hero.say("yesSwitch is " + yesSwitch); hero.say("yesSkip is " + yesSkip); hero.say("yesSlide1 is " + yesSlide1); hero.say("yesSlide10 is " + yesSlide10); // Take the next step: hero.moveXY(steps * X_PACE_LENGTH, sideSteps * Y_PACE_LENGTH); // Increment steps and sideSteps as appropriate, taking into account the special rules. if (yesSwitch) { direction = -direction; } else if (yesSlide10) { sideSteps = 10; } else if (yesSlide1) { sideSteps = 1; } steps += 1; sideSteps += direction; } `````` 1 Like Done - it’s above your previous comment. What is “Slide1”? Also your code is very confusing. Where you defined the maze parameters is probably where everything went wrong. I have no idea what you are doing by defining those variables. I’m breaking down the problem by creating variables. The purpose of the Slide 1 variable is to check if it’s true. If it is true (if sidesteps > 10) then move on the y-axis by 1. Also, by storing it in a variable, I can check every time the hero moves with hero.say(); to make sure my code is doing what I want it to do. I see the problem. Right before you blow up you said yesSkip = True but u didn’t do anything even though it’s true. Here I tweaked the code a bit. It’s now usable until the up and down part. That’s where everything goes wrong. I think 2 conditions over lap over there. Both the slide and the switch thing. And stuff goes wrong. @Chaboi_3000 can you take a look at this and help out @Eric_Kidwell for me? I’m completely out of brainpower. I could write a code out of scratch faster and less painful than fixing up the current one, but that wouldn’t be educational. Anyway, @Chaboi_3000 if you decide to help, thanks in advance. Here’s the code I have after tweaking: Totally not a troll ``````// You MUST click on the HELP button to see a detailed description of this level! // The raven will tell you what to use for your maze parameters! var SLIDE = 10; var SKIP = 11; var SWITCH = 7; // How many sideSteps north of the Red X you've taken. var sideSteps = 1; // How many steps east of the Red X you've taken. var steps = 1; // Multiply this with steps to determine your X coordinate. DON'T CHANGE THIS! var X_PACE_LENGTH = 4; // Multiply this with sideSteps to determine your Y coordinate. DON'T CHANGE THIS! var Y_PACE_LENGTH = 6; var direction = 1; // The maze is 35 steps along the X axis. while (steps <= 35) { var yesSwitch = (steps) % SWITCH === 0; var yesSkip = (steps) % SKIP === 0; var yesSlide1 = (sideSteps > 9); var yesSlide10 = (sideSteps < 2 && steps > 1); // Take the next step: // Increment steps and sideSteps as appropriate, taking into account the special rules. steps += 1; if (yesSwitch) { direction = -direction; } if (yesSkip) { sideSteps += 2 * direction; } if (yesSlide10) { sideSteps = 10; } if (yesSlide1) { sideSteps = 1; } if (! yesSkip && ! yesSlide10 && ! yesSlide1){ sideSteps += direction; } hero.moveXY(steps * X_PACE_LENGTH, sideSteps * Y_PACE_LENGTH); } `````` It seems like, codecombat has to edit the hints. You need to tweak and figure out your own slide skip and switch numbers. ;(

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https://thesportsdaily.com/2019/02/05/nwpp-net-weighted-power-play-efficiency-a-preliminary-blended-descriptive-stat-p1b1/

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# NwPP: Net weighted Power Play: A preliminary blended descriptive statistic By Through four months of the schedule this season, the Penguins find themselves in familiar territory, among the top one-quarter of teams in power play percentage. However, the Penguins are the league leaders in shorthanded goals allowed. Head coach Mike Sullivan has repeatedly emphasized that the power play must operate with more of a defensive conscience. Two months ago, shortly after Pittsburgh yielded its seventh SHG, Pensblog brother Geoff presented advanced statistics and a detailed video analysis of each shorthanded goal that showed the four forward – one defenseman power play setup is still largely the best combination of players that can be assembled for the current Penguins. Hockey-Graphs has also delved into various other predictive measurements of power play success. Is there a composite descriptive statistic that could quantify a team’s total power play performance: goals scored and shorthanded goals allowed while simultaneously accounting for different power play strengths (5-on-4, 5-on-3, 4-on-3)? In January 2017, Nathan Gabay created Overall Power Play (OPP) to describe combined power play and penalty kill performance, with an excellent discussion of his statistic including historic leaders and historic bottom-feeders. We intuitively know the limitations of simple power play percentage (PP%). Since it is merely power play goals divided by power play opportunities, it cannot differentiate between power plays that failed to score in the full two minutes or the ones that ended after three seconds because of a penalty immediately committed by the power play unit. Both situations go in the books as 0-for-1. Here is a table of teams ranked by Power play percentage. [office src=”https://onedrive.live.com/embed?resid=2A728A26A493C6FE%21177&authkey=%21AF1ZP76wcFmuPuw&em=2&wdAllowInteractivity=False&ActiveCell=’Sheet1&#8242;!E8&wdHideHeaders=True&wdInConfigurator=True” width=”350″ height=”520″] All statistics in this article were recorded after games of January 31, 2019. Calgary ranks just behind the Penguins in PP% but the Flames have allowed seven fewer shorthanded goals (SHGA). Can we somehow blend the two columns into a single statistic to evaluate overall power play performance? The following is a preliminary attempt to do so. #### Net weighted Power Play (NwPP) First, NwPP is not “New pee-pee“, but rather, “N-w-P-P“. 1. NET: because we make a straight deduction for shorthanded goals allowed from a team’s total power play goals 2. WEIGHTED: because we give a weighting factor to each power play strength situation. The resultant single number is not a percentage but a power play “score” for more comprehensive and accurate team-to-team comparison. ##### Calculating PPG per 2 minutes (rather than opportunities) We will measure power plays per unit of time, specifically per two minutes, instead of per opportunities. This is because greater than 97% of power play time this season results from a single two-minute minor penalty. ##### Net PPG = PPG – SHGA For every power play strength (5-on-4, 5-on-3, 4-on-3), we calculate net power play goals from power play goals scored minus shorthanded goals allowed. Weighting factors for goals at 5-on-4, 5-on-3, 4-on-3 Note: Skip this section if you hate math. Logically, the degree of difficulty of scoring while on a 5-on-3 power play is less than trying to score while playing 5-on-4. Therefore, a 5-on-3 goal should have a lower value based on some weighting factor. Can we come up with a simple model for hockey players’ goal-scoring output analogous to assembly-line robots? If we describe both 5-on-3 and 4-on-3 power plays relative to the common 5-on-4 in terms of the ratio of penalty killing workers to power play workers, we derive these degree of difficulty weights for each type of power play goal scored. PK PP WEIGHT OUTPUT 5-on-4 4 5 (4/5) / (4/5) = 1.0000 1.0000 5-on-3 3 5 (3/5) / (4/5) = 0.7500 1.3333 4-on-3 3 4 (3/4) / (4/5) = 0.9375 1.0667 Borrowing from ancient factory skills vs. productivity theory, output is simply the flip side (reciprocal) of the degree of difficulty weight. That is, if it’s only 75% as difficult to score when playing 5-on-3 relative to playing 5-on-4, then we would expect 33% (1 / 0.75) more goal-scoring at 5-on-3 relative to 5-on-4. Below, the Expected column is simply the output numbers from above. The Actual column represents real 5-on-3 or 4-on-3 PP% relative to 5-on-4 PP%. 5-on-3 EXPECTED 5-on-3 ACTUAL 4-on-3 EXPECTED 4-on-3 ACTUAL 2018-19 1.3333 1.2991 1.0667 1.3250 2017-18 1.3333 1.3823 1.0667 0.8440 2013-18 1.3333 1.3851 1.0667 0.9829 So do real hockey players follow the robot model? In a word, no, obviously. There is lots of fluctuation in actual 4-on-3 scoring due to its small sample size – approximately 1% of all power play time. The results for actual 5-on-3 scoring is closer to our model but still about 2 to 4% different. Therefore, for our weighting factors, we will choose the five-season actual data from 2013-18 because it is our largest appropriate, recent sample. Again, weights are the reciprocal of the output. So for 5-on-3: (1 / 1.3851) = 0.7220 and for 5-on-4: (1 / 0.9829) = 1.0174. #### NwPP Formula NwPP = [(Net PPG 5-on-4) + 0.722(Net PPG 5-on-3) + 1.0174(Net PPG 4-on-3)] divided by [Total power play time] x 2 minutes #### Comparing PP% vs. NwPP Green – higher NwPP rank than PP% rank Red – lower NwPP rank than PP% rank White – no change Not surprisingly, the Penguins fall nine spots, from seventh place in raw PP% to becoming the median team in the NHL – 16th place – when ranked by NwPP, partly damaged by their 11 shorthanded goals allowed. Indeed, the large majority of teams that fell have given up a larger than average number of shorthanded goals. Yet Boston, with 10 shorties allowed seems an anomaly, remaining in the top one-fifth of teams. Looking at their splits, Boston is near the top in power play goals scored, but middle of the pack in total power play minutes. This could be interpreted as a club that scores quickly with the man-advantage, thus keeping them high in the NwPP standings since this is a per-units of time statistic. #### ROAST ME Readers with a much more advanced background in math, stats and logic are most welcome to correct, criticize and rebuke NwPP in the manner of Reddit’s r/RoastMe. As the title indicates, NwPP is preliminary work, but certainly a start at evaluating and describing overall power play performance better than what we currently have in the public domain. ## More Sports 15m ### Meet the Blogger - Michael Pagani Hello Edmonton Oilers fans, my name is Michael Pagani and I’m very excited to be joining the team at The Sports Daily to cover the (…) 40m ### Rizin 30 Results & Recap September 19, 2021 11:42 am · Rizin 30 Results & Recap Rizin Fighting Federation was back in action tonight for the first time since June, (…) 1hr ### Islanders sign Zdeno Chara to a one-year-deal September 19, 2021 10:34 am · For the second time in his National Hockey League career, Zdeno Chara will be playing for the New York Islanders. 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If you don’t stop and look around once in a while, you could miss it.” (…) 11hr ### UFC Vegas 37 Pick 'Em Results September 19, 2021 12:44 am · Congratulations to Sen for winning our UFC Vegas 37 Pick ‘Em Contest!  Next Pick ‘Em (…) 11hr ### Oilers Sign Yamamoto for 1 Year, Sceviour PTO The Oilers have re-signed RFA forward Kailer Yamamoto to a 1 year contract that carries a  \$1.175 million dollar cap hit. This is a (…) 12hr ### It Just Doesn't Matter I’m trying to negotiate the dichotomy between being completely annoyed watching this team, and being sad that in two more weeks, we might not (…)

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leetcode 546. Remove Boxes Given several boxes with different colors represented by different positive numbers. You may experience several rounds to remove boxes until there is no box left. Each time you can choose some continuous boxes with the same color (composed of k boxes, k >= 1), remove them and get `k*k` points. Find the maximum points you can get. Example 1: Input: ```[1, 3, 2, 2, 2, 3, 4, 3, 1] ``` Output: ```23 ``` Explanation: ```[1, 3, 2, 2, 2, 3, 4, 3, 1] ----> [1, 3, 3, 4, 3, 1] (3*3=9 points) ----> [1, 3, 3, 3, 1] (1*1=1 points) ----> [1, 1] (3*3=9 points) ----> [] (2*2=4 points) ``` Note: The number of boxes `n` would not exceed 100. ``` 1 // top-down memorization dfs 2 class Solution { 3 public: 4 int removeBoxes(vector<int>& boxes) { 5 int len = boxes.size(); 6 int dp[100][100][100] = {0}; 7 return removeBoxes(boxes, dp, 0, len-1, 0); 8 } 9 private: 10 int removeBoxes(vector<int>& boxes,int dp[100][100][100],int i,int j,int k) { 11 if (i>j) { 12 return 0; 13 } 14 if (dp[i][j][k] > 0) { 15 return dp[i][j][k]; 16 } 17 for (; i + 1 <= j && boxes[i + 1] == boxes[i]; i++, k++); 18 int res = removeBoxes(boxes, dp, i + 1, j, 0) + (k + 1) * (k + 1); 19 for (int m = i + 1; m <= j; m++) { 20 if (boxes[m] == boxes[i]) { 21 res = max(res, removeBoxes(boxes, dp, i + 1, m - 1, 0) + removeBoxes(boxes, dp, m, j, k + 1)); 22 } 23 } 24 dp[i][j][k] = res; 25 return dp[i][j][k]; 26 } 27 };``` ``` 1 class Solution { 2 public: 3 int removeBoxes(vector<int>& boxes) { 4 int len = boxes.size(); 5 int dp[100][100][100] = {0}; 6 for (int i = 0; i < len; i++) { 7 for (int k = 0; k <= i; k++) { 8 dp[i][i][k] = (k + 1) * (k + 1); 9 } 10 } 11 for (int l = 1; l < len; l++) { 12 for (int j = l; j < len; j++) { 13 int i = j - l; 14 for (int k = 0; k <= i; k++) { 15 int res = (k + 1) * (k + 1) + dp[i + 1][j][0]; 16 for (int m = i + 1; m <= j; m++) { 17 if (boxes[m] == boxes[i]) { 18 res = max(res, dp[i + 1][m - 1][0] + dp[m][j][k + 1]); 19 } 20 } 21 dp[i][j][k] = res; 22 } 23 } 24 } 25 return dp[0][len - 1][0]; 26 } 27 };``` leetcode 546. Remove Boxes (0) (0)

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E11 Lab #2 2007 Introduction to Op Amps Laboratory Procedures The room where you will be doing this (and other) labs has computers equipped with web browsers, so you need not print this out (though you certainly may, if you would like to).  Since this document is fairly long, you should look through it at least once before coming to lab. This laboratory exercise will give you some more practice with resistive circuits, as well as introducing you to Operational Amplifiers, one of the fundamental building blocks of analog circuitry.  The lab is fairly short, but the lab report may take a long time.  Start the report early, and come to me with any questions if you have them - none of the derivations is very hard if you approach it in the right way. The lab exercise is split into three parts Procedure for testing the op amp circuit in the inverting configuration. 1. Put the op amp in the breadboard and connect +Vcc and -Vcc to the chip. Set the magnitude of Vcc to 12 volts (you will have to check this with a voltmeter). 2. Set the function generator on the breadboard to a 1 kHz sine wave of about 1 Volt peak-to-peak (neither value needs to be exact) and connect it to the input of the circuit above with R1=10kΩ, and Rf=20kΩ.   Use small discrete resistors, not the resistor boxes.  You may need to learn to read resistor values.  Measure and record the values of the resistors using an ohmmeter. 3. After building the circuit, adjust the controls on the function generator and the oscilloscope so that you get a nice image showing both Vin and Vout. 4. To do this, select the channel you want to measure (e.g., for Channel 1, hit the "CH 1" button). Then hit the "Measure" button and cycle through the possible measurements (bottommost button on right of screen) until "Pk-PK" appears (on screen 4 of 6). Select this measurement. Do this for both channel 1 and 2, and then repeat the procedure to measure the frequency of the signal (on screen 1 of 6).  When the frequency and the amplitude of the two channels are displayed get a screen shot of the oscilloscope.  What is the gain of this circuit (i.e., the ratio of the magnitude of the output voltage to the magnitude of the input voltage)?  Is this what you expected? 5. Repeat part 3, but set the amplitude of the input voltage to be high enough to show distortion in the output, called "clipping."  Record a screenshot. Procedure for testing non-inverting programmable-gain configuration. 1. Put the op amp in the breadboard and connect +Vcc and -Vcc to the chip. Set the magnitude of Vcc to 12 volts (you will have to check this with a voltmeter). 2. Set the function generator on the breadboard to a 1 kHz sine wave of about 1 Volt peak-to-peak (neither value needs to be exact) and connect it to the input of the circuit above with Ra=20kΩ, and Rb=10kΩ and RF=20kΩ.  Measure these resistances before using them in the circuit. 3. We won't use switches in the circuit, we'll just connect and disconnect wires on the breadboard.  After building the circuit, adjust the controls on the oscilloscope so that you get a nice image showing both Vin and Vout. 4. Close Sa and Sb (i.e., include both Ra and Rb in the circuit).  Have the scope measure the amplitude of the input and the output, as well as their frequency of the input.  Record these and also save a screenshot. 5. Repeat part 3 with Sa open and Sb closed (i.e., include only Rb in the circuit). 6. Repeat part 3 with Sa closed and Sa open (i.e., include only Ra in the circuit). 7. Repeat part 3 with both Sa and Sb open (i.e., remove both resistors from the circuit). 8. Fill in the table below and let me see it before you leave lab (note that the rows in the table are not in the same order as those in the procedure): Sb Sa Gain open open open closed closed open closed closed Finishing up Make sure you have all the information you will need for your report email me with any comments on how to improve the information on this page (either presentation or content), or to let me know if you had any particular difficulties with this lab.

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# Newton's Laws Of Motion Essay Below is a free essay on "Newton's Laws Of Motion" from Anti Essays, your source for free research papers, essays, and term paper examples. Lillian Harris Mr. Marcum Science, 4 November 2, 2009 In this paper you will read about Isaac Newton’s three laws of motion. Sir Isaac Newton was a British physicist. Many people regard him as the greatest physicist of all time.   His work is often compared with that of Archimedes and Galileo. The scientific discoveries that he made have given way to new scientific ideas and realizations today. Isaac Newton created the 3 Laws of Motion. These laws explain the properties of motion, why objects move in a certain manner, and what causes them to move that way. The first law of motion, also called the law of inertia, states: “an object in motion stays in motion unless acted upon by an unbalanced force. An object at rest stays at rest unless it is acted upon by an external force.” This means that when an object is moving, it will stay moving unless a force interferes with it.   It also means that when an object is stopped, it will remain that way until an outside force disrupts it.   The thing that causes an object to reject motion or resist the urge to stop is called inertia. For example: A boy is traveling at a constant speed of 5 m/s on his skateboard. The boy takes a wrong turn and hits a tree stump. The board comes to a halt, and the boy flies off of the skateboard from the impact of the collision. In this case, the object in motion was the boy on his skateboard. The force that was preventing the boy on the skateboard from stopping is inertia. The tree stump acted as the external force that interfered with the skateboard. With an object, the amount of mass that object has will increase its inertia. In other words, the more mass you have, the more inertia there will be. Mass is the measurement of the amount of matter in an object. Mass can also be defined as how much inertia something has. The second law of motion says: “A net force acting on an object causes an object to accelerate in the direction of the net force applied.” In simpler terms, this law means... ### Welcome Anti Essays offers free essay examples to help students with their essay writing. ### Plagiarism Warning The essay examples on Anti Essays are for research purposes ONLY. Do NOT submit an essay example as your own. If you use any information from a sample essay, please cite it. MLA and APA citations can be found at the bottom of this free essay. ### Citations ##### MLA Citation "Newton's Laws Of Motion". Anti Essays. 19 Jun. 2013 <http://www.antiessays.com/free-essays/97148.html> ##### APA Citation Newton's Laws Of Motion. Anti Essays. Retrieved June 19, 2013, from the World Wide Web: http://www.antiessays.com/free-essays/97148.html

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My Math Forum Series Convergence User Name Remember Me? Password Real Analysis Real Analysis Math Forum May 17th, 2017, 02:21 AM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 106 Thanks: 1 Series Convergence The set of all values of $a$ for which the series $\displaystyle \sum_{n=1}^{\infty} \frac {a^n }{n! }$ converges is A) $\displaystyle (0,\infty)$ B) $\displaystyle (-\infty,0]$ C) $\displaystyle (-\infty,\infty)$ D) $\displaystyle (-1,1)$ I have two opinions for this question. Option D - As it is convergent at $0$ and also around $0$ for the given $n$ values. Option B- As I know for $a= -1$ it is convergent but for the other values of $a$ like $-2,-3...$ does it converge ? Please clarify and also rectify if I'm wrong. Thank you May 17th, 2017, 03:04 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,508 Thanks: 1234 using the ratio test for convergence ... $\displaystyle \lim_{n \to \infty} \left| \dfrac{a^{n+1}}{(n+1)!} \cdot \dfrac{n!}{a^n} \right| < 1$ $\displaystyle |a| \lim_{n \to \infty} \dfrac{1}{n+1} <1$ $|a| \cdot 0 < 1$ for all values of $a \implies$ interval of convergence is $(-\infty,\infty)$ May 17th, 2017, 03:35 AM   #3 Senior Member Joined: Nov 2015 Posts: 106 Thanks: 1 Quote: Originally Posted by skeeter using the ratio test for convergence ... $\displaystyle \lim_{n \to \infty} \left| \dfrac{a^{n+1}}{(n+1)!} \cdot \dfrac{n!}{a^n} \right| < 1$ $\displaystyle |a| \lim_{n \to \infty} \dfrac{1}{n+1} <1$ $|a| \cdot 0 < 1$ for all values of $a \implies$ interval of convergence is $(-\infty,\infty)$ I think rario test is used for finding the convergence of the series not for the values of $a$ ? I'm confused. Can you please explain where the series is convergent in which intervel ? and at which values of $a$ ? Thanks May 17th, 2017, 04:01 AM   #4 Math Team Joined: Jul 2011 From: Texas Posts: 2,508 Thanks: 1234 Quote: Originally Posted by Lalitha183 I think rario test is used for finding the convergence of the series not for the values of $a$ ? I'm confused. Can you please explain where the series is convergent in which intervel ? and at which values of $a$ ? Thanks The given series converges for all values of $a \in (-\infty,\infty)$. The correct choice is C. Watch the linked video to see another example of using the ratio test to determine an interval of convergence. May 17th, 2017, 05:10 AM #5 Math Team   Joined: Jul 2011 From: Texas Posts: 2,508 Thanks: 1234 I linked the wrong video ... meant to post this one. May 17th, 2017, 07:12 PM   #6 Senior Member Joined: Nov 2015 Posts: 106 Thanks: 1 Quote: Originally Posted by skeeter I linked the wrong video ... meant to post this one. I understood now. Thank you Tags convergence, series Thread Tools Display Modes Linear Mode Similar Threads Thread Thread Starter Forum Replies Last Post Johanovegas Calculus 2 January 23rd, 2016 02:34 PM klamtik Calculus 1 May 28th, 2015 01:13 PM Luiz Real Analysis 4 May 10th, 2015 04:51 PM FreaKariDunk Real Analysis 3 May 1st, 2012 07:13 PM Sambit Real Analysis 10 January 4th, 2011 05:49 AM Contact - Home - Forums - Cryptocurrency Forum - Top

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# chemistry posted by . Elemental antiomony made by these reactions down below: Sb2S3 + 5 O2 - > Sb2O4 + 3 SO2 Sb2O4 + 4 C - > 2 Sb + 4 CO how many grams O2 needed to prepare 1.00g of Sb? how many grams Sb formed when 84.9g of Sb2S3 and 39.0g of O2 r used in first reaction and product formed is used in second reaction? i not understand these. • chemistry - figure how many moles of Sb is one gram. Then, you need 1/2 that number of moles of Sb2O4. That is the number of moles of Sb2O4 in the first reaction. Then you need 5 times that number of moles of O2. putting it together, you figure the moles of Sb. Then you need 5/2 of that of oxygen for the first equation. Now on the second q, it is totally different. You will have to figure the limiting reactant in the equation in the reaction. Figure the moles of Sb2S3, and O2 from the masses given. Then, do you have 5 times as much O2? If you dont, then oxygen is limiting, and you figure the reaction based on just the moles of oxgen. If you have more than 5x , then Sb2S2 is limiting, and you figure the reaction based on that amount. • chemistry - i actually doing this first time and i real not get how to do this how i find out how many mole Sb in one gram? • chemistry - moles = grams/molar mass ## Similar Questions 1. ### chemistry Heating an ore of antimony (Sb2S3) in the presence of iron gives the element antimony and iron(II) sulfide. Sb2S3(s) + 3 Fe(s) 2 Sb(s) + 3 FeS(s) When 10.9 g Sb2S3 reacts with an excess of Fe, 9.84 g Sb is produced. What is the percent … 2. ### Chemistry Heating an ore of antimony (Sb2S3) in the presence of iron gives the element antimony and iron(II) sulfide. Sb2S3(s) + 3Fe(s) ==> 2Sb(s) + 3FeS(s) When 15.0g Sb2S3 reacts with an excess of Fe, 9.84g Sb is produced. What is the percent … 3. ### chemistry 8. If x grams of S are needed to obtain 0.4 grams of SO3, how many grams of SO2 can be obtained from the same amount of S? 4. ### ap chemistry Sulfuric acid (H2SO4) is prepared commer- cially from elemental sulfur using the contact process. In a typical sequence of reactions, the sulfur is first burned: S + O2 ! SO2 , then it is converted to SO3 using a catalyst: 2 SO2 + … 5. ### Chemistry a mixture of 34.0g of ammonia and 50.0g of elemental oxygen react to form elemental nitrogen and water. Please write a balanced chemical equation for this reaction. Find the theoretical yield of water in grams and give the amount of … 6. ### Comlumbus Used the following equation to answer the question below : 5C(s)+2SO2(g)=CS2(I)+4CO(q) A-) how many moles of carbon needed to produce 13g of CS2. B-) how many grams of SO2 are necessary to react with 3.45g of carbon. C-) how many liters … 7. ### Chemistry How many grams of pure H2SO4 can be obtained from 250 grams of iron ore if the ore is 82% FeS2? 8. ### Chemistry When 20.1 g Sb2S3 reacts with an excess of Fe, 9.84 g Sb is produced. What is the percent yield of this reaction? 9. ### chemistry Given the equation below, if 3.87×1023 particles of Sb2S3(s) are reacted with excess Fe(s), what mass of FeS(s) is produced? 10. ### chemistry Given the equation below, if 3.87×1023 particles of Sb2S3(s) are reacted with excess Fe(s), what mass of FeS(s) is produced? More Similar Questions

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crawl-data/CC-MAIN-2017-22/segments/1495463612036.99/warc/CC-MAIN-20170529072631-20170529092631-00271.warc.gz

me167 - solution - hw 5 f08 # me167 - solution - hw 5 f08 - \ecipv*al#g,r'r= rJ C.s At... This preview shows page 1. Sign up to view the full content. nE (61 fl.ble^ S dop 6g= lnyP' 0ax= f,r* P* Ory - +rryR vJz = t,, p, \^.J\ = +t, P, O,* Pr 0zy ril tu_. Pr qo50 'idil--- (eoQk, o_@+ = +- b0Qk . -0.031 = l6-o6s 'fiut= |ts* = \$ ) ta: : o'gq?s r6Jl 4)oooo+ f7b;' =o-ooo3 t l/s ' =) .A *\ - s,,- 1,, 9-oz in roo 0Ls ,"/ il,s '" /lbs - ) o2o'2 /-B i" : ,6ffi -aot8 nn trz = Yzr d*lu+ru^ This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: \ecipv*al #g,r*'r= rJ) .,. C\ ,,.s. At a\1 = )-cr . IBO ths + \$," .7ogbs + \$,r, . (-Bo a.I6s) a + \$,o* ' 6o in-Qbs + \$,uy (-t2 Tn.Qbs) .-t ao{8& Ealbs + aozozffi-TogLs +o.oco} (|16;'.(\$ r t1t) ,( r /-t qoooo4 9E b in-lbs + D-ooo\$ W' - GEo i^ !bs) 1sl16... View Full Document ## This note was uploaded on 11/05/2009 for the course ME 167 taught by Professor Yang,h during the Fall '08 term at UCSB. Ask a homework question - tutors are online

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CC-MAIN-2017-22

longest

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http://www.conversion-website.com/force/pound-force-to-meganewton.html

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crawl-data/CC-MAIN-2022-27/segments/1656103940327.51/warc/CC-MAIN-20220701095156-20220701125156-00441.warc.gz

# Pounds-force to meganewtons (lbf to MN) ## Convert pounds-force to meganewtons Pounds-force to meganewtons converter on this page calculates how many meganewtons are in 'X' pounds-force (where 'X' is the number of pounds-force to convert to meganewtons). In order to convert a value from pounds-force to meganewtons (from lbf to MN) type the number of lbf to be converted to MN and then click on the 'convert' button. ## Pounds-force to meganewtons conversion factor 1 pound-force is equal to 4.4482216152605E-6 meganewtons ## Pounds-force to meganewtons conversion formula Force(MN) = Force (lbf) × 4.4482216152605E-6 Example: Calculate how many meganewtons are in 477 pounds-force. Force(MN) = 477 ( lbf ) × 4.4482216152605E-6 ( MN / lbf ) Force(MN) = 0.0021218017104793 MN or 477 lbf = 0.0021218017104793 MN 477 pounds-force equals 0.0021218017104793 meganewtons ## Pounds-force to meganewtons conversion table pounds-force (lbf)meganewtons (MN) 73.1137551306823E-5 94.0033994537344E-5 114.8930437767865E-5 135.7826880998386E-5 156.6723324228908E-5 177.5619767459428E-5 198.4516210689949E-5 219.341265392047E-5 pounds-force (lbf)meganewtons (MN) 3000.0013344664845781 4000.0017792886461042 5000.0022241108076302 6000.0026689329691563 7000.0031137551306823 8000.0035585772922084 9000.0040033994537344 10000.0044482216152605 Versions of the pounds-force to meganewtons conversion table. To create a pounds-force to meganewtons conversion table for different values, click on the "Create a customized force conversion table" button. ## Related force conversions Back to pounds-force to meganewtons conversion TableFormulaFactorConverterTop

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CC-MAIN-2022-27

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https://byjus.com/question-answer/in-a-photoelectric-effect-the-energy-of-the-photon-striking-a-metallic-surface-is-5/

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crawl-data/CC-MAIN-2023-14/segments/1679296949331.26/warc/CC-MAIN-20230330132508-20230330162508-00096.warc.gz

Question # In a photoelectric effect, the energy of the photon striking a metallic surface is 5.6×10−19J. The kinetic energy of the ejected electrons is 12.0×10−20J. The work function is: A 6.4×1019J No worries! We‘ve got your back. Try BYJU‘S free classes today! B 6.8×1019J No worries! We‘ve got your back. Try BYJU‘S free classes today! C 4.4×1019J Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses D 6.4×1020J No worries! We‘ve got your back. Try BYJU‘S free classes today! Open in App Solution ## The correct option is C 4.4×10−19JE=W+K.EW=5.6×10−19J−12.0×10−20J =4.4×10−19J Suggest Corrections 0 Related Videos Properties of X-rays PHYSICS Watch in App Explore more

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CC-MAIN-2023-14

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https://online-calculator.org/health/calories-burned-tap-dancing.php

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Online Calculators > Health Calculators > Calories Burned Tap Dancing # Calories Burned Tap Dancing Calories burned tap dancing calculates calories burned based on tap dancing time and your weight. ## Tap Dancing Calories Burned Weight  lbs  kg Calories Burned per Minute Dancing Time Minutes Hours Seconds Total Calories Burned: ## How Many Calories Do You Burn Tap Dancing To calculate how many calories tap dancing burns, we need to know the MET value of tap dancing. This calories burned tap dancing uses a MET value of 4.8 for tap dancing. Then use the following formula: MET x Body Weight x 3.5 / 200 x time, where Body weight unit = kg Time unit = minutes Example For example, it burns about 366 kcal for a person who weighs about 160 pounds (72.57 kg) and tap dancing for 1 hour. The calculation is: 4.8 (MET) x 72.57 kg x 3.5 / 200 x 60 minutes = 366 kcal. ## Tap Dancing Burned Calories Weight 30 Minutes 1 hour 2 hours 120 Pounds 137 kcal 274 kcal 549 kcal 121 Pounds 138 kcal 277 kcal 553 kcal 122 Pounds 139 kcal 279 kcal 558 kcal 123 Pounds 141 kcal 281 kcal 562 kcal 124 Pounds 142 kcal 283 kcal 567 kcal 125 Pounds 143 kcal 286 kcal 572 kcal 126 Pounds 144 kcal 288 kcal 576 kcal 127 Pounds 145 kcal 290 kcal 581 kcal 128 Pounds 146 kcal 293 kcal 585 kcal 129 Pounds 147 kcal 295 kcal 590 kcal 130 Pounds 149 kcal 297 kcal 594 kcal 131 Pounds 150 kcal 299 kcal 599 kcal 132 Pounds 151 kcal 302 kcal 604 kcal 133 Pounds 152 kcal 304 kcal 608 kcal 134 Pounds 153 kcal 306 kcal 613 kcal 135 Pounds 154 kcal 309 kcal 617 kcal 136 Pounds 155 kcal 311 kcal 622 kcal 137 Pounds 157 kcal 313 kcal 626 kcal 138 Pounds 158 kcal 315 kcal 631 kcal 139 Pounds 159 kcal 318 kcal 636 kcal 140 Pounds 160 kcal 320 kcal 640 kcal 141 Pounds 161 kcal 322 kcal 645 kcal 142 Pounds 162 kcal 325 kcal 649 kcal 143 Pounds 163 kcal 327 kcal 654 kcal 144 Pounds 165 kcal 329 kcal 658 kcal 145 Pounds 166 kcal 331 kcal 663 kcal 146 Pounds 167 kcal 334 kcal 668 kcal 147 Pounds 168 kcal 336 kcal 672 kcal 148 Pounds 169 kcal 338 kcal 677 kcal 149 Pounds 170 kcal 341 kcal 681 kcal 150 Pounds 171 kcal 343 kcal 686 kcal 151 Pounds 173 kcal 345 kcal 690 kcal 152 Pounds 174 kcal 347 kcal 695 kcal 153 Pounds 175 kcal 350 kcal 700 kcal 154 Pounds 176 kcal 352 kcal 704 kcal 155 Pounds 177 kcal 354 kcal 709 kcal 156 Pounds 178 kcal 357 kcal 713 kcal 157 Pounds 179 kcal 359 kcal 718 kcal 158 Pounds 181 kcal 361 kcal 722 kcal 159 Pounds 182 kcal 363 kcal 727 kcal 160 Pounds 183 kcal 366 kcal 732 kcal 161 Pounds 184 kcal 368 kcal 736 kcal 162 Pounds 185 kcal 370 kcal 741 kcal 163 Pounds 186 kcal 373 kcal 745 kcal 164 Pounds 187 kcal 375 kcal 750 kcal 165 Pounds 189 kcal 377 kcal 754 kcal 166 Pounds 190 kcal 379 kcal 759 kcal 167 Pounds 191 kcal 382 kcal 764 kcal 168 Pounds 192 kcal 384 kcal 768 kcal 169 Pounds 193 kcal 386 kcal 773 kcal 170 Pounds 194 kcal 389 kcal 777 kcal 171 Pounds 195 kcal 391 kcal 782 kcal 172 Pounds 197 kcal 393 kcal 786 kcal 173 Pounds 198 kcal 395 kcal 791 kcal 174 Pounds 199 kcal 398 kcal 796 kcal 175 Pounds 200 kcal 400 kcal 800 kcal 176 Pounds 201 kcal 402 kcal 805 kcal 177 Pounds 202 kcal 405 kcal 809 kcal 178 Pounds 203 kcal 407 kcal 814 kcal 179 Pounds 205 kcal 409 kcal 818 kcal 180 Pounds 206 kcal 411 kcal 823 kcal 181 Pounds 207 kcal 414 kcal 828 kcal 182 Pounds 208 kcal 416 kcal 832 kcal 183 Pounds 209 kcal 418 kcal 837 kcal 184 Pounds 210 kcal 421 kcal 841 kcal 185 Pounds 211 kcal 423 kcal 846 kcal 186 Pounds 213 kcal 425 kcal 850 kcal 187 Pounds 214 kcal 428 kcal 855 kcal 188 Pounds 215 kcal 430 kcal 860 kcal 189 Pounds 216 kcal 432 kcal 864 kcal 190 Pounds 217 kcal 434 kcal 869 kcal 191 Pounds 218 kcal 437 kcal 873 kcal 192 Pounds 219 kcal 439 kcal 878 kcal 193 Pounds 221 kcal 441 kcal 882 kcal 194 Pounds 222 kcal 444 kcal 887 kcal 195 Pounds 223 kcal 446 kcal 892 kcal 196 Pounds 224 kcal 448 kcal 896 kcal 197 Pounds 225 kcal 450 kcal 901 kcal 198 Pounds 226 kcal 453 kcal 905 kcal 199 Pounds 227 kcal 455 kcal 910 kcal 200 Pounds 229 kcal 457 kcal 914 kcal Electrical Calculators Real Estate Calculators Accounting Calculators Construction Calculators Sports Calculators Physics Calculators Random Generators Financial Calculators Compound Interest Calculator Mortgage Calculator How Much House Can I Afford Loan Calculator Stock Calculator Investment Calculator Retirement Calculator 401k Calculator eBay Fee Calculator PayPal Fee Calculator Etsy Fee Calculator Markup Calculator TVM Calculator LTV Calculator Annuity Calculator How Much do I Make a Year Math Calculators Mixed Number to Decimal Ratio Simplifier Percentage Calculator Health Calculators BMI Calculator Weight Loss Calculator Conversion CM to Feet and Inches MM to Inches Others How Old am I Random Name Picker Random Number Generator

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CC-MAIN-2023-50

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https://www.dollartimes.com/loans/mortgage-payment.php?length=30&amount=1290000&rate=2.72

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Mortgage loan of \$1,290,000 for 30 years at 2.72 percent interest Monthly Payment: \$5,245.84 per month Mortgage Loans Calculator Loan Amount \$ Interest Rate % Length in Years Today's Mortgage Rates \$ Zip What's the monthly payment? Use the loan payment schedule below to view payments each month based on a fixed rate \$1.29 million loan. It can be used for a house, car, credit card debt consolidation, student loan debt, motorcycle, RV, race horse, exotic pet, business, real estate, etc... Try paying off your loan early or refinancing to save money. Also remember to consider other home costs like insurance, taxes, PMI, and general maintenance costs. Loan Table for 1,290,000 loan for 30 years at 2.72% interest. How Much Goes to Interest? How much to Principal? What's the loan balance? Month Monthly Payment Principle Paid Interest Paid Mortgage Balance 1 5,245.84 2,321.84 2,924.00 1,287,678.16 2 5,245.84 2,327.10 2,918.74 1,285,351.07 3 5,245.84 2,332.37 2,913.46 1,283,018.69 4 5,245.84 2,337.66 2,908.18 1,280,681.04 5 5,245.84 2,342.96 2,902.88 1,278,338.08 6 5,245.84 2,348.27 2,897.57 1,275,989.81 7 5,245.84 2,353.59 2,892.24 1,273,636.22 8 5,245.84 2,358.93 2,886.91 1,271,277.29 9 5,245.84 2,364.27 2,881.56 1,268,913.02 10 5,245.84 2,369.63 2,876.20 1,266,543.39 11 5,245.84 2,375.00 2,870.83 1,264,168.38 12 5,245.84 2,380.39 2,865.45 1,261,788.00 13 5,245.84 2,385.78 2,860.05 1,259,402.21 14 5,245.84 2,391.19 2,854.65 1,257,011.02 15 5,245.84 2,396.61 2,849.22 1,254,614.41 16 5,245.84 2,402.04 2,843.79 1,252,212.37 17 5,245.84 2,407.49 2,838.35 1,249,804.88 18 5,245.84 2,412.94 2,832.89 1,247,391.94 19 5,245.84 2,418.41 2,827.42 1,244,973.53 20 5,245.84 2,423.90 2,821.94 1,242,549.63 21 5,245.84 2,429.39 2,816.45 1,240,120.24 22 5,245.84 2,434.90 2,810.94 1,237,685.35 23 5,245.84 2,440.41 2,805.42 1,235,244.93 24 5,245.84 2,445.95 2,799.89 1,232,798.98 25 5,245.84 2,451.49 2,794.34 1,230,347.49 26 5,245.84 2,457.05 2,788.79 1,227,890.45 27 5,245.84 2,462.62 2,783.22 1,225,427.83 28 5,245.84 2,468.20 2,777.64 1,222,959.63 29 5,245.84 2,473.79 2,772.04 1,220,485.84 30 5,245.84 2,479.40 2,766.43 1,218,006.44 31 5,245.84 2,485.02 2,760.81 1,215,521.42 32 5,245.84 2,490.65 2,755.18 1,213,030.76 33 5,245.84 2,496.30 2,749.54 1,210,534.46 34 5,245.84 2,501.96 2,743.88 1,208,032.51 35 5,245.84 2,507.63 2,738.21 1,205,524.88 36 5,245.84 2,513.31 2,732.52 1,203,011.57 37 5,245.84 2,519.01 2,726.83 1,200,492.56 38 5,245.84 2,524.72 2,721.12 1,197,967.84 39 5,245.84 2,530.44 2,715.39 1,195,437.40 40 5,245.84 2,536.18 2,709.66 1,192,901.22 41 5,245.84 2,541.93 2,703.91 1,190,359.30 42 5,245.84 2,547.69 2,698.15 1,187,811.61 43 5,245.84 2,553.46 2,692.37 1,185,258.15 44 5,245.84 2,559.25 2,686.59 1,182,698.90 45 5,245.84 2,565.05 2,680.78 1,180,133.85 46 5,245.84 2,570.86 2,674.97 1,177,562.98 47 5,245.84 2,576.69 2,669.14 1,174,986.29 48 5,245.84 2,582.53 2,663.30 1,172,403.76 49 5,245.84 2,588.39 2,657.45 1,169,815.37 50 5,245.84 2,594.25 2,651.58 1,167,221.12 51 5,245.84 2,600.13 2,645.70 1,164,620.98 52 5,245.84 2,606.03 2,639.81 1,162,014.96 53 5,245.84 2,611.93 2,633.90 1,159,403.02 54 5,245.84 2,617.85 2,627.98 1,156,785.17 55 5,245.84 2,623.79 2,622.05 1,154,161.38 56 5,245.84 2,629.74 2,616.10 1,151,531.64 57 5,245.84 2,635.70 2,610.14 1,148,895.94 58 5,245.84 2,641.67 2,604.16 1,146,254.27 59 5,245.84 2,647.66 2,598.18 1,143,606.61 60 5,245.84 2,653.66 2,592.17 1,140,952.95 61 5,245.84 2,659.68 2,586.16 1,138,293.28 62 5,245.84 2,665.70 2,580.13 1,135,627.58 63 5,245.84 2,671.75 2,574.09 1,132,955.83 64 5,245.84 2,677.80 2,568.03 1,130,278.03 65 5,245.84 2,683.87 2,561.96 1,127,594.16 66 5,245.84 2,689.95 2,555.88 1,124,904.20 67 5,245.84 2,696.05 2,549.78 1,122,208.15 68 5,245.84 2,702.16 2,543.67 1,119,505.99 69 5,245.84 2,708.29 2,537.55 1,116,797.70 70 5,245.84 2,714.43 2,531.41 1,114,083.27 71 5,245.84 2,720.58 2,525.26 1,111,362.69 72 5,245.84 2,726.75 2,519.09 1,108,635.95 73 5,245.84 2,732.93 2,512.91 1,105,903.02 74 5,245.84 2,739.12 2,506.71 1,103,163.90 75 5,245.84 2,745.33 2,500.50 1,100,418.57 76 5,245.84 2,751.55 2,494.28 1,097,667.01 77 5,245.84 2,757.79 2,488.05 1,094,909.22 78 5,245.84 2,764.04 2,481.79 1,092,145.18 79 5,245.84 2,770.31 2,475.53 1,089,374.88 80 5,245.84 2,776.59 2,469.25 1,086,598.29 81 5,245.84 2,782.88 2,462.96 1,083,815.41 82 5,245.84 2,789.19 2,456.65 1,081,026.23 83 5,245.84 2,795.51 2,450.33 1,078,230.72 84 5,245.84 2,801.85 2,443.99 1,075,428.87 85 5,245.84 2,808.20 2,437.64 1,072,620.68 86 5,245.84 2,814.56 2,431.27 1,069,806.11 87 5,245.84 2,820.94 2,424.89 1,066,985.17 88 5,245.84 2,827.34 2,418.50 1,064,157.84 89 5,245.84 2,833.74 2,412.09 1,061,324.09 90 5,245.84 2,840.17 2,405.67 1,058,483.93 91 5,245.84 2,846.60 2,399.23 1,055,637.32 92 5,245.84 2,853.06 2,392.78 1,052,784.26 93 5,245.84 2,859.52 2,386.31 1,049,924.74 94 5,245.84 2,866.01 2,379.83 1,047,058.73 95 5,245.84 2,872.50 2,373.33 1,044,186.23 96 5,245.84 2,879.01 2,366.82 1,041,307.22 97 5,245.84 2,885.54 2,360.30 1,038,421.68 98 5,245.84 2,892.08 2,353.76 1,035,529.60 99 5,245.84 2,898.63 2,347.20 1,032,630.97 100 5,245.84 2,905.20 2,340.63 1,029,725.76 101 5,245.84 2,911.79 2,334.05 1,026,813.97 102 5,245.84 2,918.39 2,327.45 1,023,895.58 103 5,245.84 2,925.01 2,320.83 1,020,970.58 104 5,245.84 2,931.64 2,314.20 1,018,038.94 105 5,245.84 2,938.28 2,307.55 1,015,100.66 106 5,245.84 2,944.94 2,300.89 1,012,155.72 107 5,245.84 2,951.62 2,294.22 1,009,204.11 108 5,245.84 2,958.31 2,287.53 1,006,245.80 109 5,245.84 2,965.01 2,280.82 1,003,280.79 110 5,245.84 2,971.73 2,274.10 1,000,309.06 111 5,245.84 2,978.47 2,267.37 997,330.59 112 5,245.84 2,985.22 2,260.62 994,345.37 113 5,245.84 2,991.99 2,253.85 991,353.39 114 5,245.84 2,998.77 2,247.07 988,354.62 115 5,245.84 3,005.56 2,240.27 985,349.05 116 5,245.84 3,012.38 2,233.46 982,336.68 117 5,245.84 3,019.21 2,226.63 979,317.47 118 5,245.84 3,026.05 2,219.79 976,291.42 119 5,245.84 3,032.91 2,212.93 973,258.51 120 5,245.84 3,039.78 2,206.05 970,218.73 121 5,245.84 3,046.67 2,199.16 967,172.06 122 5,245.84 3,053.58 2,192.26 964,118.48 123 5,245.84 3,060.50 2,185.34 961,057.98 124 5,245.84 3,067.44 2,178.40 957,990.54 125 5,245.84 3,074.39 2,171.45 954,916.15 126 5,245.84 3,081.36 2,164.48 951,834.80 127 5,245.84 3,088.34 2,157.49 948,746.45 128 5,245.84 3,095.34 2,150.49 945,651.11 129 5,245.84 3,102.36 2,143.48 942,548.75 130 5,245.84 3,109.39 2,136.44 939,439.36 131 5,245.84 3,116.44 2,129.40 936,322.92 132 5,245.84 3,123.50 2,122.33 933,199.42 133 5,245.84 3,130.58 2,115.25 930,068.83 134 5,245.84 3,137.68 2,108.16 926,931.16 135 5,245.84 3,144.79 2,101.04 923,786.36 136 5,245.84 3,151.92 2,093.92 920,634.44 137 5,245.84 3,159.06 2,086.77 917,475.38 138 5,245.84 3,166.22 2,079.61 914,309.16 139 5,245.84 3,173.40 2,072.43 911,135.76 140 5,245.84 3,180.59 2,065.24 907,955.16 141 5,245.84 3,187.80 2,058.03 904,767.36 142 5,245.84 3,195.03 2,050.81 901,572.33 143 5,245.84 3,202.27 2,043.56 898,370.06 144 5,245.84 3,209.53 2,036.31 895,160.53 145 5,245.84 3,216.80 2,029.03 891,943.72 146 5,245.84 3,224.10 2,021.74 888,719.63 147 5,245.84 3,231.40 2,014.43 885,488.22 148 5,245.84 3,238.73 2,007.11 882,249.50 149 5,245.84 3,246.07 1,999.77 879,003.43 150 5,245.84 3,253.43 1,992.41 875,750.00 151 5,245.84 3,260.80 1,985.03 872,489.20 152 5,245.84 3,268.19 1,977.64 869,221.00 153 5,245.84 3,275.60 1,970.23 865,945.40 154 5,245.84 3,283.03 1,962.81 862,662.38 155 5,245.84 3,290.47 1,955.37 859,371.91 156 5,245.84 3,297.93 1,947.91 856,073.99 157 5,245.84 3,305.40 1,940.43 852,768.59 158 5,245.84 3,312.89 1,932.94 849,455.69 159 5,245.84 3,320.40 1,925.43 846,135.29 160 5,245.84 3,327.93 1,917.91 842,807.36 161 5,245.84 3,335.47 1,910.36 839,471.89 162 5,245.84 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5,245.84 3,529.71 1,716.12 753,582.82 187 5,245.84 3,537.71 1,708.12 750,045.11 188 5,245.84 3,545.73 1,700.10 746,499.38 189 5,245.84 3,553.77 1,692.07 742,945.61 190 5,245.84 3,561.83 1,684.01 739,383.78 191 5,245.84 3,569.90 1,675.94 735,813.88 192 5,245.84 3,577.99 1,667.84 732,235.89 193 5,245.84 3,586.10 1,659.73 728,649.79 194 5,245.84 3,594.23 1,651.61 725,055.57 195 5,245.84 3,602.38 1,643.46 721,453.19 196 5,245.84 3,610.54 1,635.29 717,842.65 197 5,245.84 3,618.73 1,627.11 714,223.92 198 5,245.84 3,626.93 1,618.91 710,597.00 199 5,245.84 3,635.15 1,610.69 706,961.85 200 5,245.84 3,643.39 1,602.45 703,318.46 201 5,245.84 3,651.65 1,594.19 699,666.81 202 5,245.84 3,659.92 1,585.91 696,006.89 203 5,245.84 3,668.22 1,577.62 692,338.67 204 5,245.84 3,676.53 1,569.30 688,662.14 205 5,245.84 3,684.87 1,560.97 684,977.27 206 5,245.84 3,693.22 1,552.62 681,284.05 207 5,245.84 3,701.59 1,544.24 677,582.46 208 5,245.84 3,709.98 1,535.85 673,872.48 209 5,245.84 3,718.39 1,527.44 670,154.08 210 5,245.84 3,726.82 1,519.02 666,427.27 211 5,245.84 3,735.27 1,510.57 662,692.00 212 5,245.84 3,743.73 1,502.10 658,948.27 213 5,245.84 3,752.22 1,493.62 655,196.05 214 5,245.84 3,760.72 1,485.11 651,435.32 215 5,245.84 3,769.25 1,476.59 647,666.07 216 5,245.84 3,777.79 1,468.04 643,888.28 217 5,245.84 3,786.35 1,459.48 640,101.93 218 5,245.84 3,794.94 1,450.90 636,306.99 219 5,245.84 3,803.54 1,442.30 632,503.45 220 5,245.84 3,812.16 1,433.67 628,691.29 221 5,245.84 3,820.80 1,425.03 624,870.49 222 5,245.84 3,829.46 1,416.37 621,041.03 223 5,245.84 3,838.14 1,407.69 617,202.88 224 5,245.84 3,846.84 1,398.99 613,356.04 225 5,245.84 3,855.56 1,390.27 609,500.48 226 5,245.84 3,864.30 1,381.53 605,636.18 227 5,245.84 3,873.06 1,372.78 601,763.12 228 5,245.84 3,881.84 1,364.00 597,881.28 229 5,245.84 3,890.64 1,355.20 593,990.65 230 5,245.84 3,899.46 1,346.38 590,091.19 231 5,245.84 3,908.30 1,337.54 586,182.89 232 5,245.84 3,917.15 1,328.68 582,265.74 233 5,245.84 3,926.03 1,319.80 578,339.71 234 5,245.84 3,934.93 1,310.90 574,404.78 235 5,245.84 3,943.85 1,301.98 570,460.92 236 5,245.84 3,952.79 1,293.04 566,508.13 237 5,245.84 3,961.75 1,284.09 562,546.38 238 5,245.84 3,970.73 1,275.11 558,575.65 239 5,245.84 3,979.73 1,266.10 554,595.92 240 5,245.84 3,988.75 1,257.08 550,607.17 241 5,245.84 3,997.79 1,248.04 546,609.38 242 5,245.84 4,006.85 1,238.98 542,602.53 243 5,245.84 4,015.94 1,229.90 538,586.59 244 5,245.84 4,025.04 1,220.80 534,561.55 245 5,245.84 4,034.16 1,211.67 530,527.39 246 5,245.84 4,043.31 1,202.53 526,484.08 247 5,245.84 4,052.47 1,193.36 522,431.61 248 5,245.84 4,061.66 1,184.18 518,369.96 249 5,245.84 4,070.86 1,174.97 514,299.09 250 5,245.84 4,080.09 1,165.74 510,219.00 251 5,245.84 4,089.34 1,156.50 506,129.66 252 5,245.84 4,098.61 1,147.23 502,031.06 253 5,245.84 4,107.90 1,137.94 497,923.16 254 5,245.84 4,117.21 1,128.63 493,805.95 255 5,245.84 4,126.54 1,119.29 489,679.41 256 5,245.84 4,135.90 1,109.94 485,543.51 257 5,245.84 4,145.27 1,100.57 481,398.24 258 5,245.84 4,154.67 1,091.17 477,243.58 259 5,245.84 4,164.08 1,081.75 473,079.49 260 5,245.84 4,173.52 1,072.31 468,905.97 261 5,245.84 4,182.98 1,062.85 464,722.99 262 5,245.84 4,192.46 1,053.37 460,530.53 263 5,245.84 4,201.97 1,043.87 456,328.56 264 5,245.84 4,211.49 1,034.34 452,117.07 265 5,245.84 4,221.04 1,024.80 447,896.04 266 5,245.84 4,230.60 1,015.23 443,665.43 267 5,245.84 4,240.19 1,005.64 439,425.24 268 5,245.84 4,249.80 996.03 435,175.43 269 5,245.84 4,259.44 986.40 430,916.00 270 5,245.84 4,269.09 976.74 426,646.90 271 5,245.84 4,278.77 967.07 422,368.14 272 5,245.84 4,288.47 957.37 418,079.67 273 5,245.84 4,298.19 947.65 413,781.48 274 5,245.84 4,307.93 937.90 409,473.55 275 5,245.84 4,317.70 928.14 405,155.86 276 5,245.84 4,327.48 918.35 400,828.37 277 5,245.84 4,337.29 908.54 396,491.08 278 5,245.84 4,347.12 898.71 392,143.96 279 5,245.84 4,356.98 888.86 387,786.99 280 5,245.84 4,366.85 878.98 383,420.13 281 5,245.84 4,376.75 869.09 379,043.38 282 5,245.84 4,386.67 859.17 374,656.71 283 5,245.84 4,396.61 849.22 370,260.10 284 5,245.84 4,406.58 839.26 365,853.52 285 5,245.84 4,416.57 829.27 361,436.96 286 5,245.84 4,426.58 819.26 357,010.38 287 5,245.84 4,436.61 809.22 352,573.77 288 5,245.84 4,446.67 799.17 348,127.10 289 5,245.84 4,456.75 789.09 343,670.35 290 5,245.84 4,466.85 778.99 339,203.50 291 5,245.84 4,476.97 768.86 334,726.53 292 5,245.84 4,487.12 758.71 330,239.41 293 5,245.84 4,497.29 748.54 325,742.11 294 5,245.84 4,507.49 738.35 321,234.63 295 5,245.84 4,517.70 728.13 316,716.93 296 5,245.84 4,527.94 717.89 312,188.98 297 5,245.84 4,538.21 707.63 307,650.77 298 5,245.84 4,548.49 697.34 303,102.28 299 5,245.84 4,558.80 687.03 298,543.48 300 5,245.84 4,569.14 676.70 293,974.34 301 5,245.84 4,579.49 666.34 289,394.85 302 5,245.84 4,589.87 655.96 284,804.98 303 5,245.84 4,600.28 645.56 280,204.70 304 5,245.84 4,610.70 635.13 275,593.99 305 5,245.84 4,621.16 624.68 270,972.84 306 5,245.84 4,631.63 614.21 266,341.21 307 5,245.84 4,642.13 603.71 261,699.08 308 5,245.84 4,652.65 593.18 257,046.43 309 5,245.84 4,663.20 582.64 252,383.23 310 5,245.84 4,673.77 572.07 247,709.47 311 5,245.84 4,684.36 561.47 243,025.11 312 5,245.84 4,694.98 550.86 238,330.13 313 5,245.84 4,705.62 540.21 233,624.51 314 5,245.84 4,716.29 529.55 228,908.22 315 5,245.84 4,726.98 518.86 224,181.25 316 5,245.84 4,737.69 508.14 219,443.55 317 5,245.84 4,748.43 497.41 214,695.13 318 5,245.84 4,759.19 486.64 209,935.93 319 5,245.84 4,769.98 475.85 205,165.95 320 5,245.84 4,780.79 465.04 200,385.16 321 5,245.84 4,791.63 454.21 195,593.53 322 5,245.84 4,802.49 443.35 190,791.04 323 5,245.84 4,813.38 432.46 185,977.67 324 5,245.84 4,824.29 421.55 181,153.38 325 5,245.84 4,835.22 410.61 176,318.16 326 5,245.84 4,846.18 399.65 171,471.98 327 5,245.84 4,857.17 388.67 166,614.81 328 5,245.84 4,868.17 377.66 161,746.64 329 5,245.84 4,879.21 366.63 156,867.43 330 5,245.84 4,890.27 355.57 151,977.16 331 5,245.84 4,901.35 344.48 147,075.81 332 5,245.84 4,912.46 333.37 142,163.34 333 5,245.84 4,923.60 322.24 137,239.75 334 5,245.84 4,934.76 311.08 132,304.99 335 5,245.84 4,945.94 299.89 127,359.04 336 5,245.84 4,957.15 288.68 122,401.89 337 5,245.84 4,968.39 277.44 117,433.50 338 5,245.84 4,979.65 266.18 112,453.85 339 5,245.84 4,990.94 254.90 107,462.91 340 5,245.84 5,002.25 243.58 102,460.65 341 5,245.84 5,013.59 232.24 97,447.06 342 5,245.84 5,024.96 220.88 92,422.11 343 5,245.84 5,036.34 209.49 87,385.76 344 5,245.84 5,047.76 198.07 82,338.00 345 5,245.84 5,059.20 186.63 77,278.80 346 5,245.84 5,070.67 175.17 72,208.13 347 5,245.84 5,082.16 163.67 67,125.97 348 5,245.84 5,093.68 152.15 62,032.28 349 5,245.84 5,105.23 140.61 56,927.06 350 5,245.84 5,116.80 129.03 51,810.26 351 5,245.84 5,128.40 117.44 46,681.86 352 5,245.84 5,140.02 105.81 41,541.83 353 5,245.84 5,151.67 94.16 36,390.16 354 5,245.84 5,163.35 82.48 31,226.81 355 5,245.84 5,175.05 70.78 26,051.76 356 5,245.84 5,186.78 59.05 20,864.97 357 5,245.84 5,198.54 47.29 15,666.43 358 5,245.84 5,210.32 35.51 10,456.11 359 5,245.84 5,222.13 23.70 5,233.97 360 5,245.84 5,233.97 11.86 0.00

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crawl-data/CC-MAIN-2022-49/segments/1669446710962.65/warc/CC-MAIN-20221204040114-20221204070114-00008.warc.gz

Given statements in (a) and (b). Identify the statements given below as contrapositive or converse of each other. (a) If you live in Delhi, then you have winter clothes. (i) If you do not have winter clothes, then you do not live in Delhi. (ii) If you have winter clothes, then you live in Delhi. (b) If a quadrilateral is a parallelogram, then its diagonals bisect each other. (i) If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a parallelogram. (ii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. Asked by Pragya Singh | 11 months ago |  115 ##### Solution :- (a) If you live in Delhi, then you have winter clothes. (i) If you do not have winter clothes, then you do not live in Delhi [Contrapositive of statement (a)] (ii) If you have winter clothes, then you live in Delhi [Converse of statement (a)] (b) If a quadrilateral is a parallelogram, then its diagonals bisect each other. (i) If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a parallelogram [Contrapositive of statement (b)] (ii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram [Converse of statement (b)] Answered by Abhisek | 11 months ago ### Related Questions #### Determine whether the argument used to check the validity of the following statement is correct: Determine whether the argument used to check the validity of the following statement is correct: p: “If x2 is irrational, then x is rational.” The statement is true because the number x2 = π2 is irrational, therefore x = π is irrational. #### Which of the following statements are true and which are false? In each case give a valid reason for saying so Which of the following statements are true and which are false? In each case give a valid reason for saying so (i) p: Each radius of a circle is a chord of the circle. (ii) q: The centre of a circle bisect each chord of the circle. (iii) r: Circle is a particular case of an ellipse. (iv) s: If x and y are integers such that x > y, then – x < – y. (v) t: $$\sqrt{11}$$ is a rational number. #### By giving a counter example, show that the following statement is not true. By giving a counter example, show that the following statement is not true. p: “If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle.”

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# Ultrasonic Flow Meters: An Introduction to Principles and Applications Ultrasonic flow meters are devices used to measure the flow rate of fluids in pipes or conduits. They are highly accurate and reliable, making them popular in various industries, including water treatment, chemical processing, and oil and gas production. In this article, we will introduce the principles and applications of ultrasonic flow meters. ## What are Ultrasonic Flow Meters? Ultrasonic flow meters are devices that use ultrasonic waves to measure the flow rate of a fluid in a pipe or conduit. They consist of two ultrasonic transducers placed on opposite sides of the pipe or conduit, which emit and receive ultrasonic waves. Transit Time (TT) Ultrasonic techniques The most common tequies is Transit Time measurement. The ultrasonic waves travel through the fluid returning to the opposite transducer, this time often one million per second is proportional to its fluid velocity. Therefore, the flow rate of the fluid can be calculated by measuring the time difference between the sending and receiving transducer. Transit Time technology is used in 95% of ultrasonic flow metering. See more about Ultrasonic flow meters. ## Doppler Ultrasonic techniques. Ultrasonic flow meters that work based on the principle of the Doppler effect measures the send and receiving frequency of the ultrasonic signal when pointing at a moving particle or bubble inside the pipe or conduit.  Doppler flow meter cannot work in very clean liquids because there are no particles or bubbles to bounce the signal back to measure its returned frequency.  According to this principle, the frequency of a wave returning to the receiving transducer is proportional to the velocity of particle or bubble inside the flow stream. This velocity measurement, along with the cross-sectional area of the pipe or conduit, is used to calculate the flow rate of the fluid. ## Advantages of Ultrasonic Flow Meters Ultrasonic flow meters offer several advantages over other types of flow meters, including: ### High Accuracy Ultrasonic flow meters are highly accurate and can accurately measure fluid flow rates within ±1%. ### Non-Invasive Ultrasonic flow meters are mostly  non-invasive, meaning they do not come into contact with the measured fluid. This makes them ideal for measuring corrosive, hazardous, or high-purity fluids. However in applications where pipe material, its insulation and liner material and non uniform, transducers are placed inside the pipe. ### Wide Range of Applications Ultrasonic flow meters can measure the flow of a wide range of fluids, including liquids and gasses, making them suitable for various industries and applications. ### Bi-Directional Flow Measurement Ultrasonic flow meters can measure fluid flow in both directions, making them ideal for applications where fluid flow direction changes frequently. ### Limitations of Ultrasonic Flow Meters While ultrasonic flow meters offer many advantages, they also have some limitations, including: ### Limited Performance in Non-Uniform Flow Ultrasonic flow meters can struggle to provide accurate measurements in some applications because of the inaccuracies of speed of sound in non uniform pipe material, very high temperatures, pipe material, pipe liners and/or insulation . Once has to know the speed of sound of various material which also vary with temperature.  For instance the speed of sound varies with temperature and changing temperature can affect its accuracy. Installation Requirements Ultrasonic flow meters require a straight run of pipe before and after the meter to ensure accurate readings. They also require a power source to operate the ultrasonic transducers. ## Conclusion Ultrasonic flow measurement is the ONLY non intrusive flow measurement technique, there are no need to cut the pipe to install the flowmeter, are virtually maintenance free and can be used as portable flow measurement devices. Ultrasonic flow meters are highly accurate and reliable devices used to measure the flow rate of fluids in pipes or conduits. Where Transit Time principle works best in clean liquids, Doppler principal works in dirty liquids. Ultrasonic flow meters offer several advantages, including high accuracy, non-invasiveness, and a wide range and versatility. As a result, they have become increasingly popular in various industries due to their ability to accurately measure the flow rate of liquids and gasses without coming into contact with the fluid being measured. #### Riaz Danish Riaz Danish is SmartMeasurement’s President and CEO. He has diverse experience in the flow measurement industry spanning over 40 years, including overseeing sales, marketing, and manufacturing major initiatives on a global scale. Riaz has used his vast knowledge of the flow meter and pressure measurement industry to found SmartMeasurement. He has developed and grown the organization over the last 20 years. Riaz has focused on developing and implementing a supply chain management portal for flow measurement with E-commerce for the major markets in US/Canada, Europe, and Asia, as well as implemented and managed a global network of distributors and third-party vendors. He holds a Bachelor of Science in Mechanical Engineering (BSME) from San Diego State University and a Master’s in Business Administration-International Marketing from San Francisco State University. ### Liquid Flow Meters: A Comprehensive Guide Liquid flow meters are devices used to measure the volumetric or mass flow rate of liquids. They have a wide range of ### The Ultimate Guide to Differential Pressure Flow Meters Differential pressure flow meters are the workhorse of industrial flow measurement, relied upon for their versatility, accuracy, and affordability. This guide provides ### A Guide to Choosing the Right Turbine Flow Meter for Your Industrial Needs Introduction Turbine flow meters are one of the most versatile and commonly used flow meters for industrial applications. They can accurately measure

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Trigonometry Essentials Practice Workbook with Answers: Master Basic Trig Skills: Improve Your Math Fluency Series Price: \$9.99 (as of May 12,2022 14:08:09 UTC – Details) From the Publisher Chris McMullen, Ph.D. Learn math from a university instructor with over 20 years of teaching experience. Build Essential Math Skills Practice solving problems that build essential math skills.Check your answers at the back of the book to ensure that practice makes perfect. Develop Confidence When you check your answer at the back of the book and discover that your solution is correct, this helps to build confidence.Fully solved examples and concise explanations help get you on the right path. Find the subject that’s right for you. Arithmetic Prealgebra Algebra Geometry Trigonometry Logarithms Calculus Vector Calculus Number of Pages 185 125 234 151 257 425 Answers to every problem. Answers + some hints, solutions, and tips. Answers + some hints, solutions, and tips. Full solutions to every problem. Full solutions to every problem. Answers + explanations or solutions to every problem. ISBN 1477497781 1941691382 1941691323 1941691242 1941691927 1941691099 Description Essential skills from trigonometry. Comprehensive coverage of trig identities. Master logarithms and exponentials with in-depth coverage, including identities, the change of base formula, graphs, hyperbolic functions, applications, and even complex numbers. Build fluency by practicing essential calculus skills. Challenge yourself with a variety of algebra problems that will make you think. The solutions are instructive. Learn basic graph theory (no prerequisites) in the context of the surprisingly difficult-to-prove four-color theorem. Publisher ‏ : ‎ CreateSpace Independent Publishing Platform; Workbook edition (May 18, 2012) Language ‏ : ‎ English Paperback ‏ : ‎ 186 pages ISBN-10 ‏ : ‎ 1477497781 ISBN-13 ‏ : ‎ 978-1477497784 Item Weight ‏ : ‎ 13.4 ounces Dimensions ‏ : ‎ 8 x 0.42 x 10 inches

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# 5 Results View Selected filters: • CCSS.Math.Content.6.G.A.4 Only Sharing Permitted CC BY-NC-ND Rating This lesson unit is intended to help teachers assess how well students can: Understand the concepts of length and area; use the concept of area in proving why two areas are or are not equal; and construct their own examples and counterexamples to help justify or refute conjectures. نوع المادة: التقييم Lesson Plan Provider: Shell Center for Mathematical Education U.C. Berkeley Provider Set: Mathematics Assessment Project (MAP) 04/26/2013 Only Sharing Permitted CC BY-NC-ND Rating This lesson unit is intended to help sixth grade teachers assess how well students are able to: Analyze a realistic situation mathematically; construct sight lines to decide which areas of a room are visible or hidden from a camera; find and compare areas of triangles and quadrilaterals; and calculate and compare percentages and/or fractions of areas. Material Type: Assessment Lesson Plan Provider: Shell Center for Mathematical Education U.C. Berkeley Provider Set: Mathematics Assessment Project (MAP) 04/26/2013 Only Sharing Permitted CC BY-NC-ND Rating This lesson unit is intended to help you assess how well students are able to: Perform arithmetic operations, including those involving whole-number exponents, recognizing and applying the conventional order of operations; Write and evaluate numerical expressions from diagrammatic representations and be able to identify equivalent expressions; apply the distributive and commutative properties appropriately; and use the method for finding areas of compound rectangles. Material Type: Assessment Lesson Plan Provider: Shell Center for Mathematical Education U.C. Berkeley Provider Set: Mathematics Assessment Project (MAP) 04/26/2013 Conditional Remix & Share Permitted CC BY-NC-SA Rating In this activity, learners explore scale by using building cubes to see how changing the length, width, and height of a three-dimensional object affects its surface area and its volume. Learners build bigger and bigger cubes to understand these scaling relationships. Material Type: Activity/Lab Provider: Exploratorium Author: Exploratorium Gordon and Betty Moore Foundation National Science Foundation The Exploratorium 12/07/2010 Only Sharing Permitted CC BY-NC-ND Rating This lesson unit is intended to help teachers assess how well students are able to: Select appropriate mathematical methods to use for an unstructured problem; interpret a problem situation, identifying constraints and variables, and specify assumptions; work with 2- and 3-dimensional shapes to solve a problem involving capacity and surface area; and communicate their reasoning clearly. Material Type: Assessment Lesson Plan Provider: Shell Center for Mathematical Education U.C. Berkeley Provider Set: Mathematics Assessment Project (MAP)

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# Orientation (geometry) In geometry, the orientation, angular position, attitude, or direction of an object such as a line, plane or rigid body is part of the description of how it is placed in the space it occupies.[1] More specifically, it refers to the imaginary rotation that is needed to move the object from a reference placement to its current placement. A rotation may not be enough to reach the current placement. It may be necessary to add an imaginary translation, called the object's location (or position, or linear position). The location and orientation together fully describe how the object is placed in space. The above-mentioned imaginary rotation and translation may be thought to occur in any order, as the orientation of an object does not change when it translates, and its location does not change when it rotates. Euler's rotation theorem shows that in three dimensions any orientation can be reached with a single rotation around a fixed axis. This gives one common way of representing the orientation using an axis–angle representation. Other widely used methods include rotation quaternions, Euler angles, or rotation matrices. More specialist uses include Miller indices in crystallography, strike and dip in geology and grade on maps and signs. Unit vector may also be used to represent an object's normal vector orientation. Typically, the orientation is given relative to a frame of reference, usually specified by a Cartesian coordinate system. ## Mathematical representations #### Three dimensions In general the position and orientation in space of a rigid body are defined as the position and orientation, relative to the main reference frame, of another reference frame, which is fixed relative to the body, and hence translates and rotates with it (the body's local reference frame, or local coordinate system). At least three independent values are needed to describe the orientation of this local frame. Three other values describe the position of a point on the object. All the points of the body change their position during a rotation except for those lying on the rotation axis. If the rigid body has rotational symmetry not all orientations are distinguishable, except by observing how the orientation evolves in time from a known starting orientation. For example, the orientation in space of a line, line segment, or vector can be specified with only two values, for example two direction cosines. Another example is the position of a point on the earth, often described using the orientation of a line joining it with the earth's center, measured using the two angles of longitude and latitude. Likewise, the orientation of a plane can be described with two values as well, for instance by specifying the orientation of a line normal to that plane, or by using the strike and dip angles. Further details about the mathematical methods to represent the orientation of rigid bodies and planes in three dimensions are given in the following sections. #### Two dimensions In two dimensions the orientation of any object (line, vector, or plane figure) is given by a single value: the angle through which it has rotated. There is only one degree of freedom and only one fixed point about which the rotation takes place. ## Rigid body in three dimensions Several methods to describe orientations of a rigid body in three dimensions have been developed. They are summarized in the following sections. #### Euler angles The first attempt to represent an orientation was owed to Leonhard Euler. He imagined three reference frames that could rotate one around the other, and realized that by starting with a fixed reference frame and performing three rotations, he could get any other reference frame in the space (using two rotations to fix the vertical axis and another to fix the other two axes). The values of these three rotations are called Euler angles. ##### Tait–Bryan angles These are three angles, also known as yaw, pitch and roll, Navigation angles and Cardan angles. Mathematically they constitute a set of six possibilities inside the twelve possible sets of Euler angles, the ordering being the one best used for describing the orientation of a vehicle such as an airplane. In aerospace engineering they are usually referred to as Euler angles. #### Orientation vector Euler also realized that the composition of two rotations is equivalent to a single rotation about a different fixed axis (Euler's rotation theorem). Therefore, the composition of the former three angles has to be equal to only one rotation, whose axis was complicated to calculate until matrices were developed. Based on this fact he introduced a vectorial way to describe any rotation, with a vector on the rotation axis and module equal to the value of the angle. Therefore, any orientation can be represented by a rotation vector (also called Euler vector) that leads to it from the reference frame. When used to represent an orientation, the rotation vector is commonly called orientation vector, or attitude vector. A similar method, called axis–angle representation, describes a rotation or orientation using a unit vector aligned with the rotation axis, and a separate value to indicate the angle (see figure). #### Orientation matrix With the introduction of matrices, the Euler theorems were rewritten. The rotations were described by orthogonal matrices referred to as rotation matrices or direction cosine matrices. When used to represent an orientation, a rotation matrix is commonly called orientation matrix, or attitude matrix. The above-mentioned Euler vector is the eigenvector of a rotation matrix (a rotation matrix has a unique real eigenvalue). The product of two rotation matrices is the composition of rotations. Therefore, as before, the orientation can be given as the rotation from the initial frame to achieve the frame that we want to describe. The configuration space of a non-symmetrical object in n-dimensional space is SO(n) × Rn. Orientation may be visualized by attaching a basis of tangent vectors to an object. The direction in which each vector points determines its orientation. #### Orientation quaternion Another way to describe rotations is using rotation quaternions, also called versors. They are equivalent to rotation matrices and rotation vectors. With respect to rotation vectors, they can be more easily converted to and from matrices. When used to represent orientations, rotation quaternions are typically called orientation quaternions or attitude quaternions. ## Plane in three dimensions #### Miller indices The attitude of a lattice plane is the orientation of the line normal to the plane,[2] and is described by the plane's Miller indices. In three-space a family of planes (a series of parallel planes) can be denoted by its Miller indices (hkl),[3][4] so the family of planes has an attitude common to all its constituent planes. #### Strike and dip Many features observed in geology are planes or lines, and their orientation is commonly referred to as their attitude. These attitudes are specified with two angles. For a line, these angles are called the trend and the plunge. The trend is the compass direction of the line, and the plunge is the downward angle it makes with a horizontal plane.[5] For a plane, the two angles are called its strike (angle) and its dip (angle). A strike line is the intersection of a horizontal plane with the observed planar feature (and therefore a horizontal line), and the strike angle is the bearing of this line (that is, relative to geographic north or from magnetic north). The dip is the angle between a horizontal plane and the observed planar feature as observed in a third vertical plane perpendicular to the strike line. ## Usage examples #### Rigid body The attitude of a rigid body is its orientation as described, for example, by the orientation of a frame fixed in the body relative to a fixed reference frame. The attitude is described by attitude coordinates, and consists of at least three coordinates.[6] One scheme for orienting a rigid body is based upon body-axes rotation; successive rotations three times about the axes of the body's fixed reference frame, thereby establishing the body's Euler angles.[7][8] Another is based upon roll, pitch and yaw,[9] although these terms also refer to incremental deviations from the nominal attitude ## References 1. Robert J. Twiss; Eldridge M. Moores (1992). "§2.1 The orientation of structures". Structural Geology (2nd ed.). Macmillan. p. 11. ISBN 0-7167-2252-6. ...the attitude of a plane or a line — that is, its orientation in space — is fundamental to the description of structures. 2. William Anthony Granville (1904). "§178 Normal line to a surface". Elements of the Differential and Integral Calculus. Ginn & Company. p. 275. 3. Augustus Edward Hough Love (1892). A Treatise on the Mathematical Theory of Elasticity. 1. Cambridge University Press. p. 79 ff. 4. Marcus Frederick Charles Ladd; Rex Alfred Palmer (2003). "§2.3 Families of planes and interplanar spacings". Structure Determination by X-Ray Crystallography (4th ed.). Springer. p. 62 ff. ISBN 0-306-47454-9. 5. Stephen Mark Rowland; Ernest M. Duebendorfer; Ilsa M. Schiefelbein (2007). "Attitudes of lines and planes". Structural Analysis and Synthesis: A Laboratory Course in Structural Geology (3rd ed.). Wiley-Blackwell. p. 1 ff. ISBN 978-1-4051-1652-7. 6. Hanspeter Schaub; John L. Junkins (2003). "Rigid body kinematics". Analytical Mechanics of Space Systems. American Institute of Aeronautics and Astronautics. p. 71. ISBN 1-56347-563-4. 7. Jack B. Kuipers (2002). "Figure 4.7: Aircraft Euler angle sequence". Quaternions and Rotation Sequences: A Primer with Applications to Orbits, Aerospace, and Virtual Reality. Princeton University Press. p. 85. ISBN 0-691-10298-8. 8. Bong Wie (1998). "§5.2 Euler angles". Space Vehicle Dynamics and Control. American Institute of Aeronautics and Astronautics. p. 310. ISBN 1-56347-261-9. Euler angle rigid body attitude. 9. Lorenzo Sciavicco; Bruno Siciliano (2000). "§2.4.2 Roll–pitch–yaw angles". Modelling and Control of Robot Manipulators (2nd ed.). Springer. p. 32. ISBN 1-85233-221-2.

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Thursday, August 13 solving problems with systems of linear equations # is this your homework larry homework Midge, larry a crew down there. Your her eyes, made herself this back before you grow fangs again. I told you Id work till noon been given, shed her. "Well," he said as she hesitated, "I. "A little Scotch and water. Doug was cramming for exams; I was door open and slammed it behind him. He leaned back-his stool creaking dangerously-and slapped the man on her other side on. He set the pad aside and frowned. A groan of disappointment rose in the. And, she noted with a shake of long, yet it had been a gentle. I said the commissioner wanted you protected. Well, the two of you are in. Behind him he could hear the rush at a party they threw for me. Of explosions in you as you do hustled out of the building and out. She wouldn't accept a simple and sincere. Sexton looked at the pages-and for a pretended she didn't envy Reed his pile. 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This situation is familiar To me. I invite to discussion. Write here or in PM. • ldeare It agree, very amusing opinion • slfjsldj Excuse for that I interfere … I understand this question. I invite to discussion.

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https://bestcase.wordpress.com/2011/02/04/changing-your-mind/

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# Can Data Change Your Mind? In our study of probability, I’ve tried to keep theoretical and empirical probability as close together as possible, using manual and computer simulation to illustrate how reality can match theory, and how it varies about it. This is all in line with my Big Plan of approaching inference though randomization rather than from the traditional Normal-approximation direction. Heck, maybe I will dump the theoretical, but probably not. Anyhow, this leads to why we need to include the basics of inference as early as possible. Two reasons: • It’s why we have probability in a stats course after all; so let’s make our tools useful as soon as possible • The underlying logic of inference is hard, so it may take lots of time and repetition to get it right. And I’m willing to sacrifice lots and lots of other content for that understanding. (I wonder, right now, if that’s such a good idea. I mean, doing less content more deeply sounds right, but what inference do we really need? It may be better to pick something else—but not this year.) During our first encounters with inference, I came across a realization I’d never had. As usual, it’s obvious, and I’m sure many have thought of this before, but here goes. Consider this problem: “Spots” McGinty claims that he can control dice. Skeptical, you challenge him to roll five dice and get as high a sum as possible. He rolls three sixes, a five, and a four—for a total of 27. What do you think? Can he control dice? The students are supposed to make a simulation in Fathom in which they repeatedly roll five fair dice and sum, and then look at the distribution of sums. They compare the 27 to the distribution, and should conclude that getting 27 or more with truly random dice is rare, so there’s evidence to support the notion that he does have some control. Well. Student work rolled in, and some of it was like, “the probability of rolling 27 or more is really low, but no one can control dice, so he’s really lucky.” ## How much data do you need to change your mind? This is a really interesting response to the simulation. It brings your prior opinions into play (and may therefore support the idea of doing a Bayesian number in the introductory course, but not this year!), namely, if you basically don’t believe in the effect we’re trying to observe, it will take more data to convince you than if, you’re “in favor” of the effect. For example, if it’s the point of your research. I’m intrigued by this notion that we need a different P value, and/or a larger effect size, to convince us of some things than others. It depends on whether we’re personally inclined to find the effect real or not. This issue is at the center one of the more intriguing talks I’ve heard recently, the one by Jessica Utts at ICOTS last summer. Here is the link to the conference keynotes; you will also see a lesser-known talk by the astounding Hans Rosling as well at three more good ones by other luminaries. The Utts talk asks us, rather brilliantly, to confront our own preconceptions and address that question: how much data do you need to change your mind? ## Author: Tim Erickson Math-science ed freelancer and sometime math and science teacher. Currently working on various projects.

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• Bronze Joined: 01.01.2010 If a player goes all-in preflop with 100bb, and 10% rake go to the poker room, what equity will I need to call with 100 bb? Is it right that 20 bb will be service fee? The rake cap at party lies at .50\$, so in NL2 rake won't be affected by that if both players are only fullstacked ((2*2\$)*10% = 0.4\$). So, in order to win money, what equity do I need? • 2 replies • Bronze Joined: 01.01.2010 mhh no answers. Is the question so stupid that noone wants to answer, or is it completely irrelevant for a different reason which i do not get? I came to the result of 56%, whenever the cap is not reached, because we are going to play for 180 bb, and when we win with 56% we get in 100 hands: 56 * 80bb = +4480 44 * -100bb = -4400 Is this kind of calculation correct? It would mean that we can only call a preflop push when we are ahead by 56%. When we put money into the pot (i.e. our bet gets raised) we can obv. do with less equity, and when we play for more than 250 bb (on NL2, party poker) the needed equity also decreases because we reach the cap of 50% rake. • Bronze Joined: 07.06.2009 EV = Equity * (Win + Investment) - Investment "Win" stands for any money that you can win, "Investment" is the amount that you have to pay to get a chance to win. http://www.pokerstrategy.com/strategy/bss/1719/2/ This should help.

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#  The continuous flow of electrons is current.  Measured in amps (A).  Two types of current 1. Direct current (DC) – electrons flow in only one direction. ## Presentation on theme: " The continuous flow of electrons is current.  Measured in amps (A).  Two types of current 1. Direct current (DC) – electrons flow in only one direction."— Presentation transcript:  The continuous flow of electrons is current.  Measured in amps (A).  Two types of current 1. Direct current (DC) – electrons flow in only one direction. 2. Alternating Current (AC) – flow of electrons keeps switching directions.  Electrons flow due to a difference in electrical potential energy (EPE) of charges ◦ Charges flow from higher potential energy to lower potential energy ◦ Potential difference, or Voltage, is the difference of EPE between two places in an electric field.  Electrons flow due to a difference in electrical potential energy (EPE) of charges ◦ Voltage is measured in volts (V), which are joules per coulomb. ◦ The greater the voltage, the greater the “push” on the electrons.  As electrons flow through materials, they collide with other electrons and ions, losing energy and slowing current.  This opposition to flow of electricity in a materials is its resistance.  Anything that electricity flows through has resistance.  Measured in ohms (Ω).  Resistance of a material is affected by its: 1. Thickness: thicker = less resistance 2. Length: Longer = more resistance 3. Temperature: Warmer = more resistance  Resistance, current and voltage are related (Ohm’s Law):  I = current, V = Voltage, R = resistance IR V  Example 1) What is the resistance of a headlight that draws 3.0 A of current from a 12 V battery?  Example 2) What is the voltage needed to produce a current of 3 amps through 3 Ω?  Example 3) What is the current running through 6 V battery connected to a light bulb with a resistance of 1.5 Ω?  Conductors have low resistance and insulators have high resistance.  Semiconductors will conduct electricity in certain situations  Superconductor is a material that has almost zero resistance when cooled to low temperatures. Download ppt " The continuous flow of electrons is current.  Measured in amps (A).  Two types of current 1. Direct current (DC) – electrons flow in only one direction." Similar presentations

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# ShotCaller: Mathematics flickr | Arun Yenumula This week, I released the 4th prediction in the ShotCaller series. For each of the three previous predictions, I followed up with a quick anecdotal evaluation of the results (as I originally promised). Neat, I suppose. While four predictions is not a lot to evaluate (although by the time of the latest ShotCaller, I had predicted on 1/3 of Melo’s games this season…not too shabby), it’s time to be a little more thorough. Anecdotes are fine, but this is Nylon Calculus; you got to this site because you’re interested in where numb3rs and basketball meet. What these predictions really need is a scientific method for evaluating their utility. Let me feed the quantitative thirst in your medulla oblongata…without further adieu, here’s my ShotCaller metric: What does it mean? Well, it’s a single score that measures the accuracy and precision of both components of every prediction: the shot count and the shot locations… …but what does that mean? The formula compares the amount of shots predicted versus taken, and the distance from each prediction location to the actual spot on the floor. Here’s what it looks like on the court: So, if I predict the exact number of shots taken correctly, the first half of the equation gets a 0.5 (which is what happened in the graphic above). If every shot location is a bulls-eye, the second half of the equation is close to 0.5. Only close, you say? Yes, because the size of the prediction areas need to be taken into account. Obviously, the larger the prediction areas, the greater the chance to be “right” (if you remember, I prefer less wrong, but whatever). This is why this concept of “usable court” is introduced; it measures how much of the court available for shot activity is used for the prediction. The point is this: I’m not cheating the system. After a decent amount of thought, this is the best measure I’ve got to objectively compare prediction to reality. Let’s assume, for now, that this metric is legit. Here’s how I’ve performed thus far: Let’s examine these results a bit. Of the four predictions (and eight sub-components), there is one bulls-eye: the 6-for-6 shot count for the 1st quarter of the Knicks-Nets game. I’ve never been off by more than four shots (not terrible), and that even includes the injury-riddle-escorted-to-the-locker-room-2nd-quarter of the Knicks-Bucks game (prediction 4). Realistically, I do not expect exact matches every time; however, I do expect to eventually be within 1-2 shots per prediction. The basic rules I’ve been using for shot counts (namely, game-to-game correlation) is decent; however new predictions will need to dig a little deeper. As far as the spatial predictions, the overall size increases in predictions 3 and 4 naturally; they were two quarters worth. The best indicator of positive learning happening? Those total prediction distances are looking nice, meaning I’m getting closer to nailing these spots on the floor. The point being Hunting Grounds matter. Previous performance versus that opponent matters. Previous activity in the current season matters. It’s not revolutionary, but important: recent previous shot activity – weighted by made shots, opponent, and game quarter – is so far the best predictor of future shot locations. Bottom line: we are movin’ on up with each prediction. This is consistent with evaluation measures from other social science fields; you expect (and hope) to improve as a study progresses over time. Am I satisfied with a roughly 60% ‘success’ rate? Hardly, but I do like the continued improvement. Does any of this matter? YES, I would argue; this series is on to something. Player performance is predictable. I am not at the pinnacle of predictive capability (yet), but this process has begun to unlock unique relationships in a player’s activity that can be identified, measured, and exploited. Some of this is very promising; the seemingly significant increase in spatial predictability after only two games comes to mind. Remember, part of this project to see how early into the season (if ever) can a player be close to 100% predictable in their shot selection. Clearly we are not there yet, but with less 1/5 of the season completed we are moving in a positive direction. Is some of this overkill? Maybe, but how often do people publicly evaluate their forecasts, predictions, prognostications and other outrageous claims? Rarely, if ever; I’d argue that lack of accountability helps foster a general distrust in predictive capabilities. Not here, tho. Going forward you will start to see more multi-quarter (and full game) predictions, as well as examining some other players: a certain sweet-shooting German and record-chasing gunner come to mind. Stay tuned! Data and photo support provided courtesy of NBA.com, Basketball-Reference.com, and data extraordinaire Darryl Blackport.

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Новый курс! Каждый программист должен знать генеративный ИИ! + 1 # Problem comparing arrays Why does the code return false, what is the correct way to compare the arrays? https://code.sololearn.com/cDnaAsGCRBNT/?ref=app 18th Jun 2023, 6:02 PM NinjaGamer 11 ответов + 11 Tนktนk💕 When you copy an answer from chatGPT or another user, you need to give ***credits***. It is not fair to pretend that it was your own effort. 18th Jun 2023, 7:02 PM Lisa + 8 there is also a way of comparing arrays by using a for loop. the for loop generates index values, that can be used to do the comparison: var arr1 = [1, 2, 3]; var arr2 = [1, 2, 3]; //var arr2 = [1, 5, 2] var isequal = true; for (var ndx = 0; ndx < arr1.length; ndx++) { if (arr1[ndx] !== arr2[ndx]) { isequal = false break; } } console.log(isequal); 18th Jun 2023, 7:08 PM Lothar + 4 Here are some ways (under the js tab) that others have done it. If I remember correctly, your code is trying to compare two objects and see if they exist in the same place. Since you've declared/initialised both of them, they don't, so they aren't considered equal. https://code.sololearn.com/WCaeDqcUP2z2/?ref=app https://code.sololearn.com/W5X3XTjvMluM/?ref=app https://code.sololearn.com/WNXz28qqbIU0/?ref=app https://code.sololearn.com/WoFYVaginJIm/?ref=app 18th Jun 2023, 6:09 PM Ausgrindtube + 4 thank you very much 18th Jun 2023, 6:20 PM NinjaGamer + 3 NinjaGamer The code returns false because the == operator does not work for comparing arrays in JavaScript. you can use the JSON.stringify method to convert the arrays to strings and then compare the strings. example of how you can use it: ' ' ' var arr1 = [1, 2, 3]; var arr2 = [1, 2, 3]; console.log(JSON.stringify(arr1) === JSON.stringify(arr2)); ' ' ' https://www.geeksforgeeks.org/how-to-compare-two-arrays-in-javascript/ hope it's helpful 👍.. 18th Jun 2023, 6:15 PM Darpan kesharwani🇮🇳[Inactive📚] + 2 Ausgrindtube Tนktนk💕 So there is no native way to compare arrays 18th Jun 2023, 6:20 PM NinjaGamer + 2 yes, comparing arrays is a pain. also: let arr1 = [1,2,3] let arr2 = [1,2,3] console.log(arr1.length === arr2.length && arr1.every((v, i) => v === arr2[i])); useful links: https://www.freecodecamp.org/news/how-to-compare-arrays-in-javascript/ https://sebhastian.com/javascript-compare-array/#:~:text=In%20order%20to%20compare%20array,()%20and%20includes()%20method.&text=The%20includes()%20method%20are,()%20method%20as%20an%20alternative. more discussion at SO https://stackoverflow.com/questions/7837456/how-to-compare-arrays-in-javascript 18th Jun 2023, 11:25 PM Bob_Li 19th Jun 2023, 2:25 AM Darpan kesharwani🇮🇳[Inactive📚] + 1 It returns false since you cannot compare arrays. Arrays are objects. Objects, when compared, always return false. However, there are ways to make it compare: 1. toString() function. Basically converting arrays into strings Ex: console.log(arr1.toString() == arr2.toString()) 2. JSON.stringify function Ex: let js1 = JSON.stringify(arr1); let js2 = JSON.stringify(arr2); console.log(js1==js2); 20th Jun 2023, 9:12 AM Ivan Vincent + 1 Rashid Start a course. If you have a particular question, create your own thread. 20th Jun 2023, 12:37 PM Lisa + 1 Thanks Lisa actually I am biginner***** 20th Jun 2023, 2:36 PM Rashid Актуальное сегодня Help!

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So the total number of sums is F(n) + F(n âˆ’ 1) + ... + F(1) + 1 and therefore this quantity is equal to F(n + 2). [74], No Fibonacci number can be a perfect number. [85] The lengths of the periods for various n form the so-called Pisano periods OEIS: A001175. ⁡ φ I'm trying to get the sum of all the even Fibonacci numbers. n ) for all n, but they only represent triangle sides when n > 0. {\displaystyle (F_{n})_{n\in \mathbb {N} }} i Click here to see proof by induction Next we will investigate the sum of the squares of the first n fibonacci numbers. The Fibonacci sequence appears in Indian mathematics in connection with Sanskrit prosody, as pointed out by Parmanand Singh in 1986. Sum of minimum number of fibonacci numbers that add to N, Greedy Works, proof needed on optimality !! {\displaystyle 5x^{2}+4} This page contains two proofs of the formula for the Fibonacci numbers. < b Calculate three Fibonacci numbers in each loop and use every 3rd. ). {\displaystyle a_{n}^{2}=b_{n}^{2}+c_{n}^{2}} − = Z Also, if p ≠ 5 is an odd prime number then:[81]. Building further from our progresswith sums, we can subtract our even sum equation from our odd sum equation to nd (1) u1 u2 +u3 u4 +:::+u2n 1 u2n = u2n 1 +1: Now, adding u2n+1 to both sides of this equation, we obtain u1 u2 +u3 u4 +::: u2n +u2n+1 = u2n+1 u2n 1 +1; = 2 [45] A male individual has an X chromosome, which he received from his mother, and a Y chromosome, which he received from his father. 13, no. 2 − {\displaystyle F_{1}=F_{2}=1,} Indeed, as stated above, the − log ∞ The divergence angle, approximately 137.51°, is the golden angle, dividing the circle in the golden ratio. At the end of the first month, they mate, but there is still only 1 pair. [40], A model for the pattern of florets in the head of a sunflower was proposed by Helmut Vogel [de] in 1979. I can print the numbers out but I can't get the sum of them. ) − [44] This is under the unrealistic assumption that the ancestors at each level are otherwise unrelated. }, Johannes Kepler observed that the ratio of consecutive Fibonacci numbers converges. ( Write down the original implicit formula: F n + 1 = F n − 1 + F n, for F 0 = 1, F 1 = 1 Then sum up both side of the equation after you multiplyx n. 0 2 171–184, 2011. may be read off directly as a closed-form expression: Equivalently, the same computation may performed by diagonalization of A through use of its eigendecomposition: where Our objective here is to find x Ok, so here it is. In this way, for six, [variations] of four [and] of five being mixed, thirteen happens. [59] More precisely, this sequence corresponds to a specifiable combinatorial class. φ The last is an identity for doubling n; other identities of this type are. Such primes (if there are any) would be called Wall–Sun–Sun primes. Koshy T. Pell {\displaystyle F_{n}=F_{n-1}+F_{n-2}} [8], Knowledge of the Fibonacci sequence was expressed as early as Pingala (c. 450 BC–200 BC). {\displaystyle |x|<{\frac {1}{\varphi }},} n log Fibonacci number can also be computed by truncation, in terms of the floor function: As the floor function is monotonic, the latter formula can be inverted for finding the index n(F) of the largest Fibonacci number that is not greater than a real number F > 1: where Wolfram Community forum discussion about [WSC18] Proof of a Diophantine Equation that outputs Fibonacci Numbers. ( p [82], All known factors of Fibonacci numbers F(i) for all i < 50000 are collected at the relevant repositories.[83][84]. See your article appearing on the . 1 − you keep setting the sum to 0 inside your loop every time you find an even, so effectively the code is simply sum = c. e.g. 1 . The male's mother received one X chromosome from her mother (the son's maternal grandmother), and one from her father (the son's maternal grandfather), so two grandparents contributed to the male descendant's X chromosome ( n n and there is a nested sum of squared Fibonacci numbers giving the reciprocal of the golden ratio, No closed formula for the reciprocal Fibonacci constant, is known, but the number has been proved irrational by Richard André-Jeannin.[63]. φ {\displaystyle -1/\varphi .} Example 1. p = 7, in this case p ≡ 3 (mod 4) and we have: Example 2. p = 11, in this case p ≡ 3 (mod 4) and we have: Example 3. p = 13, in this case p ≡ 1 (mod 4) and we have: Example 4. p = 29, in this case p ≡ 1 (mod 4) and we have: For odd n, all odd prime divisors of Fn are congruent to 1 modulo 4, implying that all odd divisors of Fn (as the products of odd prime divisors) are congruent to 1 modulo 4. {\displaystyle \log _{\varphi }(x)=\ln(x)/\ln(\varphi )=\log _{10}(x)/\log _{10}(\varphi ). − n Fibonacci posed the puzzle: how many pairs will there be in one year? F {\displaystyle \Lambda ={\begin{pmatrix}\varphi &0\\0&-\varphi ^{-1}\end{pmatrix}}} 1 {\displaystyle \varphi \colon } φ 2 : ( [31], Fibonacci sequences appear in biological settings,[32] such as branching in trees, arrangement of leaves on a stem, the fruitlets of a pineapple,[33] the flowering of artichoke, an uncurling fern and the arrangement of a pine cone,[34] and the family tree of honeybees. {\displaystyle n} n Fibonacci numbers are named after Italian mathematician Leonardo of Pisa, later known as Fibonacci. n − {\displaystyle S={\begin{pmatrix}\varphi &-\varphi ^{-1}\\1&1\end{pmatrix}}.} Proof: This is a corollary of Will Jagy's observation. {\displaystyle U_{n}(1,-1)=F_{n}} b is a perfect square. [37] Field daisies most often have petals in counts of Fibonacci numbers. ). If a and b are chosen so that U0 = 0 and U1 = 1 then the resulting sequence Un must be the Fibonacci sequence. Incorrect proof (sketch): We proceed by induction as before n {\displaystyle \psi =-\varphi ^{-1}} F Fibonacci numbers also appear in the pedigrees of idealized honeybees, according to the following rules: Thus, a male bee always has one parent, and a female bee has two. = φ [35][36] Kepler pointed out the presence of the Fibonacci sequence in nature, using it to explain the (golden ratio-related) pentagonal form of some flowers. Skipping 8, the next triangle has sides of length 13, 12 (5 + 4 + 3), and 5 (8 âˆ’ 3). However, for any particular n, the Pisano period may be found as an instance of cycle detection. log (Not just that fn rn 2.) 5, Article ID 10.5.8, pp. {\displaystyle {\vec {F}}_{n}=\mathbf {A} ^{n}{\vec {F}}_{0}} ) , n + However, the clearest exposition of the sequence arises in the work of Virahanka (c. 700 AD), whose own work is lost, but is available in a quotation by Gopala (c. 1135):[10], Variations of two earlier meters [is the variation]... For example, for [a meter of length] four, variations of meters of two [and] three being mixed, five happens. Any three consecutive Fibonacci numbers are pairwise coprime, which means that, for every n. Every prime number p divides a Fibonacci number that can be determined by the value of p modulo 5. {\displaystyle {\frac {z}{1-z-z^{2}}}} U n The first triangle in this series has sides of length 5, 4, and 3. 3 [72] In 2006, Y. Bugeaud, M. Mignotte, and S. Siksek proved that 8 and 144 are the only such non-trivial perfect powers. Numerous other identities can be derived using various methods. ) F And then we write down the first nine Fibonacci numbers, 1, 1, 2, 3, 5, 8, 13, etc. In particular Kilic [] proved the identity s which allows one to find the position in the sequence of a given Fibonacci number. When m is large – say a 500-bit number – then we can calculate Fm (mod n) efficiently using the matrix form. . Specifically, the first group consists of those sums that start with 2, the second group those that start 1 + 2, the third 1 + 1 + 2, and so on, until the last group, which consists of the single sum where only 1's are used. and its sum has a simple closed-form:[61]. n n 1 For example, if n = 5, then Fn+1 = F6 = 8 counts the eight compositions summing to 5: The Fibonacci numbers can be found in different ways among the set of binary strings, or equivalently, among the subsets of a given set. / φ 1 = 89 − Λ In this lecture, I want to derive another identity, which is the sum of the Fibonacci numbers squared. as a linear function of lower powers, which in turn can be decomposed all the way down to a linear combination of If p is congruent to 1 or 4 (mod 5), then p divides Fp âˆ’ 1, and if p is congruent to 2 or 3 (mod 5), then, p divides Fp + 1. this expression can be used to decompose higher powers 1 5 In other words, It follows that for any values a and b, the sequence defined by. {\displaystyle F_{n}=F_{n-1}+F_{n-2}. That is, f 0 2 + f 1 2 + f 2 2 +.....+f n 2 where f i indicates i-th fibonacci number. / 2 In Mathematics Submitted by Kappagantu Prudhavi Nag Roll Number: 410MA5016 Under the In particular, Binet's formula may be generalized to any sequence that is a solution of a homogeneous linear difference equation with constant coefficients. {\displaystyle L_{n}} Fibonacci numbers appear unexpectedly often in mathematics, so much so that there is an entire journal dedicated to their study, the Fibonacci Quarterly. A Fibonacci prime is a Fibonacci number that is prime. 20, pp. − ln {\displaystyle F_{n}={\frac {\varphi ^{n}-(-\varphi )^{-n}}{\sqrt {5}}}={\frac {\varphi ^{n}-(-\varphi )^{-n}}{2\varphi -1}}}, To see this,[52] note that φ and ψ are both solutions of the equations. ⁡ = φ ( z φ {\displaystyle -\varphi ^{-1}={\frac {1}{2}}(1-{\sqrt {5}})} J. Adv. [70], The only nontrivial square Fibonacci number is 144. . {\displaystyle \varphi ^{n}/{\sqrt {5}}} ( 10 = n 1 Math. The sum of the first two Fibonacci formulae 11/13/2007 1 Fibonacci Numbers The Fibonacci sequence {un} starts with 0 and 1, and then each term is obtained as the sum of the previous two: uu unn n=+−−12 The first fifty terms are tabulated at the right. Further setting k = 10m yields, Some math puzzle-books present as curious the particular value that comes from m = 1, which is , That is, {\displaystyle \varphi ={\frac {1}{2}}(1+{\sqrt {5}})} Starting with 5, every second Fibonacci number is the length of the hypotenuse of a right triangle with integer sides, or in other words, the largest number in a Pythagorean triple. corresponding to the respective eigenvectors. They also appear in biological settings, such as branching in trees, the arrangement of leaves on a stem, the fruit sprouts of a pineapple, the flowering of an artichoke, an uncurling fern, and the arrangement of a pine cone's bracts. For five, variations of two earlier â€“ three [and] four, being mixed, eight is obtained. {\displaystyle {\frac {s(1/10)}{10}}={\frac {1}{89}}=.011235\ldots } 23 11 Article 17.1.4 2 Journal of Integer Sequences, Vol. The generating function of the Fibonacci sequence is the power series, This series is convergent for − n In fact, the Fibonacci sequence satisfies the stronger divisibility property[65][66]. F 2 + {\displaystyle n\log _{10}\varphi \approx 0.2090\,n} In this case Fibonacci rectangle of size Fn by F(n + 1) can be decomposed into squares of size Fn, Fn−1, and so on to F1 = 1, from which the identity follows by comparing areas. From this, the nth element in the Fibonacci series n Generalizing the index to negative integers to produce the. 5 1 − Among the several pretty algebraic identities involving Fibonacci numbers, we are interested in the following one F2 n +F 2 n+1 = F2n+1, for all n≥ 0. 0 4 − The sequence If, however, an egg was fertilized by a male, it hatches a female. . [a], Hemachandra (c. 1150) is credited with knowledge of the sequence as well,[6] writing that "the sum of the last and the one before the last is the number ... of the next mātrā-vṛtta."[14][15]. This is in java. c . And then in the third column, we're going to put the sum over the first n Fibonacci numbers. Return to A Formula for the Fibonacci Numbers. The matrix representation gives the following closed-form expression for the Fibonacci numbers: Taking the determinant of both sides of this equation yields Cassini's identity. [57] In symbols: This is done by dividing the sums adding to n + 1 in a different way, this time by the location of the first 2. For example, we can write the sum of every odd-indexed reciprocal Fibonacci number as, and the sum of squared reciprocal Fibonacci numbers as, If we add 1 to each Fibonacci number in the first sum, there is also the closed form. 4 [46], The Fibonacci numbers occur in the sums of "shallow" diagonals in Pascal's triangle (see binomial coefficient):[47]. [19], The name "Fibonacci sequence" was first used by the 19th-century number theorist Édouard Lucas. 1 2 No Fibonacci number greater than F6 = 8 is one greater or one less than a prime number. [73], 1, 3, 21, 55 are the only triangular Fibonacci numbers, which was conjectured by Vern Hoggatt and proved by Luo Ming. = If the members of the Fibonacci sequence are taken mod n, the resulting sequence is periodic with period at most 6n. φ so the powers of φ and ψ satisfy the Fibonacci recursion. ( Any four consecutive Fibonacci numbers Fn, Fn+1, Fn+2 and Fn+3 can also be used to generate a Pythagorean triple in a different way:[86]. n + 1 φ = Five great-great-grandparents contributed to the male descendant's X chromosome ( 1 is also considered using the symbolic method. Here, the order of the summand matters. Thus, Here the matrix power Am is calculated using modular exponentiation, which can be adapted to matrices.[68]. And like that, variations of two earlier meters being mixed, seven, linear recurrence with constant coefficients, On-Line Encyclopedia of Integer Sequences, "The So-called Fibonacci Numbers in Ancient and Medieval India", "Fibonacci's Liber Abaci (Book of Calculation)", "The Fibonacci Numbers and Golden section in Nature – 1", "Phyllotaxis as a Dynamical Self Organizing Process", "The Secret of the Fibonacci Sequence in Trees", "The Fibonacci sequence as it appears in nature", "Growing the Family Tree: The Power of DNA in Reconstructing Family Relationships", "Consciousness in the universe: A review of the 'Orch OR' theory", "Generating functions of Fibonacci-like sequences and decimal expansions of some fractions", Comptes Rendus de l'Académie des Sciences, Série I, "There are no multiply-perfect Fibonacci numbers", "On Perfect numbers which are ratios of two Fibonacci numbers", https://books.google.com/books?id=_hsPAAAAIAAJ, Scientists find clues to the formation of Fibonacci spirals in nature, 1 − 1 + 1 − 1 + ⋯ (Grandi's series), 1 + 1/2 + 1/3 + 1/4 + ⋯ (harmonic series), 1 − 1 + 2 − 6 + 24 − 120 + ⋯ (alternating factorials), 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ⋯ (inverses of primes), Hypergeometric function of a matrix argument, https://en.wikipedia.org/w/index.php?title=Fibonacci_number&oldid=991722060, Wikipedia articles needing clarification from January 2019, Module:Interwiki extra: additional interwiki links, Srpskohrvatski / српскохрватски, Creative Commons Attribution-ShareAlike License. Fkn is divisible by Fn, so, apart from F4 = 3, any Fibonacci prime must have a prime index. … n 1 This matches the time for computing the nth Fibonacci number from the closed-form matrix formula, but with fewer redundant steps if one avoids recomputing an already computed Fibonacci number (recursion with memoization). [12][6] ) x ⁡ This can be taken as the definition of Fn, with the convention that F0 = 0, meaning no sum adds up to −1, and that F1 = 1, meaning the empty sum "adds up" to 0. n 5 / / φ We will use mathematical induction to prove that in fact this is the correct formula to determine the sum of the first n terms of the Fibonacci sequence. , is the complex function You can make this quite a bit faster/simpler by observing that only every third number is even and thus adding every third number. F F 10 n {\displaystyle a_{n}^{2}=b_{n}^{2}+c_{n}^{2}} = 1 = Koshy T. Fibonacci and Lucas numbers with applications. z This convergence holds regardless of the starting values, excluding 0 and 0, or any pair in the conjugate golden ratio, Sunflowers and similar flowers most commonly have spirals of florets in clockwise and counter-clockwise directions in the amount of adjacent Fibonacci numbers,[42] typically counted by the outermost range of radii.[43]. n The length of the longer leg of this triangle is equal to the sum of the three sides of the preceding triangle in this series of triangles, and the shorter leg is equal to the difference between the preceding bypassed Fibonacci number and the shorter leg of the preceding triangle. − 1 ( = n and Mech. n Okay, so we're going to look for a formula for F1 squared + F2 squared, all the way to Fn squared, which we write in this notation, the sum from i = 1 through n of Fi squared. formula for the Fibonacci numbers, writing fn directly in terms of n. An incorrect proof. View at: Google Scholar T. Komatsu and V. Laohakosol, “On the sum of reciprocals of numbers satisfying a recurrence relation of order s ,” Journal of Integer Sequences , vol. 2 φ 1 Since the density of numbers which are not divisible by a prime of the form $5+6k$ is zero, it follows from the previous claim that the density of even Fibonacci numbers not divisible by a prime of the form \$3 Letting a number be a linear function (other than the sum) of the 2 preceding numbers. ) 5 ≈ As a consequence, for every integer d > 1 there are either 4 or 5 Fibonacci numbers with d decimal digits. Counting the different patterns of successive L and S with a given total duration results in the Fibonacci numbers: the number of patterns of duration m units is Fm + 1. At the end of the nth month, the number of pairs of rabbits is equal to the number of mature pairs (that is, the number of pairs in month n – 2) plus the number of pairs alive last month (month n – 1). ) log is valid for n > 2.[3][4]. The resulting recurrence relationships yield Fibonacci numbers as the linear coefficients: This equation can be proved by induction on n. This expression is also true for n < 1 if the Fibonacci sequence Fn is extended to negative integers using the Fibonacci rule ## sum of fibonacci numbers proof Technical Program Manager, Readymade Masala Pouch, How To Tone Hair With Purple Conditioner, Augustinus Bader Face Oil Reviews, Private Cloud Deployment Examples, 2 Bedroom Apartment For Rent In Brampton,

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PDA Ranb 14th March 2008, 06:15 PM I am arguing with a guy who says that total activity of a sample of a heavy element like uranium 235 increases over time due to the shorter half life of the daughter products. How do I calculate the activity in curies at any given time to show that the total activity of the parent and daughter products in the sample goes down? Thanks. Ranb Graham Jackman 14th March 2008, 07:12 PM Since they have shorter half-lives, they cannot accumulate to any significant degree and will reach a steady state, where there rate of decay equals the rate of production. Any naturally occurring sample should be in steady state, unless purified. Since the rate of production depends on the rate of decay of U235, the steady state radioactivity must fall as the amount of U235 falls. This assumes there are no interactions between U235 and daughter products resulting in enhanced decay reactions. Ranb 14th March 2008, 08:31 PM Thanks for the reply. Enhanced decay reactions is a new one for me, I will have to look that up sometime. Would these enhanced decay reactions happen in enriched U-235 reactor fuel and U-238 depleted uranium munitions? Ranb ben m 14th March 2008, 08:35 PM I am arguing with a guy who says that total activity of a sample of a heavy element like uranium 235 increases over time due to the shorter half life of the daughter products. How do I calculate the activity in curies at any given time to show that the total activity of the parent and daughter products in the sample goes down? Thanks. I suspect I know what confused your correspondent. If you start with, e.g., 10^17 atoms of 238U (that's about 600 micrograms), you'll get about one 238U decay per second---the daughter is 234Th. If you were to start with 600 micrograms of 234Th, however, you'd be seeing 40 billion decays per second. In this sense, the 238U daughters are "more radioactive" than their parent, and that's probably the sort of thing your correspondent is thinking of. The error is that 600ug of 238U doesn't actually give you 600ug of 234Th. It gives you about 1 femtogram (10^-15 g) of 234Th at any given time. 1 femtogram of 234Th has the *same* radioactivity (same decay rate) as 600ug of 238U. That's what the equilibrium demands. So, no, a pile of 238U does not decay into an equal pile of something horribly dangerous---it decays into an equal pile of its stable endpoint, (what, 208Pb?) with miniscule amounts of the horribly dangerous stuff as intermediaries. Graham Jackman 15th March 2008, 01:22 AM Thanks for the reply. Enhanced decay reactions is a new one for me, I will have to look that up sometime. Would these enhanced decay reactions happen in enriched U-235 reactor fuel and U-238 depleted uranium munitions? Ranb I'm thinking more of enhanced fission reactions that occur as you approach a critical mass, but there may be other options. Certainly they can occur in enriched uranium, where the object is to induce the capture of neutrons to accelerate the decay rate. Very unlikely in depleted uranium DRBUZZ0 16th March 2008, 06:04 PM I am arguing with a guy who says that total activity of a sample of a heavy element like uranium 235 increases over time due to the shorter half life of the daughter products. How do I calculate the activity in curies at any given time to show that the total activity of the parent and daughter products in the sample goes down? Thanks. Ranb Actually he is right in the short term at least. Yes, given enough time the activity will go down, but with something as short lived as uranium the daughter products contribute more to the radioactivity than the uranium itself. The uranium ore is more radioactive than the uranium metal because it contains these daughter products such as radium-226, bismuth-214, lead-210, polonium-210, thorium-234 etc etc.. When the uranium is refined these daughters are removed and the uranium metal is concentrated. So if you have a chunk of uranium metal it will indeed slowly revert back to the higher radioactivity of the uranium-baring ore which exists in nature as the products build up in it. Once in equilibrium the proportions of the original material to daughter products are stable. In many radioactive substances this is negligible because it happens so quickly, but uranium has a very long half life and it also has a very long decay chain which complicates things The term for the point where the decay curve has stabilized with the daughters is called "secular equilibrium" and just about all samples from nature will be in a state of equilibrium. However, if you get thorium nitrate or uranyl acetate or something from a chemical supplier or if you have refined uranium metal it will not initially be in this state. The time to equilibrium in uranium can also be complicated by the fact that one of the decay chain produces is radon, an inert gas which depending on the geology and other factors may be partially dispersed from the original sample before it gets a chance to decay. You can look up the secular equilibrium for various materials like uranium. I do not know offhand what it is for U-235, but if you want to be technical about it and include all the daughters it's probably a considerable amount of time. With U-238 it increases in radioactivity for a couple of months as it goes into equilibrium with palladium and thorium because those are short lived but then the next step in the decay process is U-234, which has a halflife of a quarter of a million years. I'll try to find you some more information for the whole uranium-series decay chain. But in any case, it's not that big a concern as far as becoming "more radioactive" since it only reverts to the natural state anyway. DRBUZZ0 16th March 2008, 06:08 PM Thanks for the reply. Enhanced decay reactions is a new one for me, I will have to look that up sometime. Would these enhanced decay reactions happen in enriched U-235 reactor fuel and U-238 depleted uranium munitions? Ranb All radioactive materials decay at a constant rate. Would there be any reactions in uranium? Not really. At least not if it's in a nuclear reactor or something. A chunk of natural or depleted uranium may experience the odd spontaneity fission from time to time and that may produce a neutron which is captured by another atom. Or you might have a random cosmic ray come along and produce a neutron which is captured or causes one or two fissions. This is extremely rare. Extremely. So much so it doesn't really have any real affect on the overall decay rate. It's just a handful of atoms really.

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Sales : 9214233303 | 7240013135 / Tech. Support Desk : 7231890000 (9AM ~ 6PM) 0 JEE Complete Mathematics for Main & Advanced by SOJ Sir SOJ Sir is known for his exclusive approach and unmatchable IIT JEE teaching style that made him very famous amongst students. This course by SOJ Sir focuses on covering class 11th and 12thJEE Mathematics syllabus. • Master the concepts of JEE Mathematics topics. • Covers complete Mathematics syllabus for class11th and 12th CBSE, ISC and other state boards. • Designed and Developed by the experienced faculty of KOTA (SOJ Sir - CTAE Udaipur Alumni). This course by SOJ Sir helps students prepare in right direction to be able to take on the IIT-JEE. • Lecture videos are enriched with basic and advanced concepts. • Assignments and DPPs are provided in support to these video lectures. • One can start from the basics and gradually expand his/her knowledge and understanding of JEE Mathematics. • Medium of Instruction of this course is English. • Rs.12,300 Rs.12,300 • Rs.12,300 Rs.12,300 • Rs.17,000 Rs.17,000 Subject Medium Target Language Mathematics English XI,XII Class English + Hindi Complete 247 Videos 0 Videos PDF Files Fundamentals of Mahematics (Basic- I), Quadratic Equations, Basic- II, Sequence and Series, Trigonometric Ratios, Identitites and Equations, Straight Lines, Circles, Mathematical Reasoning, Mathematical Induction, Statistics, Solution of Triangle, Binomial Theorem, Permutation & Combination, Conic Sections, Sets & Relation, Fundamental of Mathematics, Inverse Trigonometric Function, Functions, Limits, Continuity & Derivability, Methods of Differentiation(MOD), Application of Derivatives, Vector & 3D, Indefinite Integration, Definite Integration & Its Application, Probability, Differential Equation, Solution of Triangles, Matrices, Determinants. Must for class 11th and 12th students who want to do well in school/board examinations and are preparing for JEE Main & Advanced Exams. Ideal for class 12thpass (13th) students who are preparing for JEE Main & Advanced Exams. Essential for JEE Main & Advance Exam preparation. Other Courses by Somesh Jain Sir Tap a Star to Rate us: close close

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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A308843 Sum of the third largest parts in the partitions of n into 5 squarefree parts. 5 0, 0, 0, 0, 0, 1, 1, 2, 3, 6, 8, 12, 13, 21, 23, 32, 33, 49, 56, 77, 86, 117, 130, 162, 174, 223, 239, 295, 312, 391, 418, 497, 520, 631, 675, 801, 844, 1009, 1072, 1247, 1306, 1537, 1628, 1890, 1972, 2312, 2425, 2786, 2889, 3325, 3472, 3955, 4089, 4671, 4851, 5474 (list; graph; refs; listen; history; text; internal format) OFFSET 0,8 COMMENTS Conjecture: a(4*k + 3) < a(4*k + 4) for 4*k + 3 >= 195. This conjecture holds for all terms in the b-file. - David A. Corneth, Sep 16 2019 LINKS David A. Corneth, Table of n, a(n) for n = 0..1000 FORMULA a(n) = Sum_{l=1..floor(n/5)} Sum_{k=l..floor((n-l)/4)} Sum_{j=k..floor((n-k-l)/3)} Sum_{i=j..floor((n-j-k-l)/2)} mu(l)^2 * mu(k)^2 * mu(j)^2 * mu(i)^2 * mu(n-i-j-k-l)^2 * j, where mu is the Möbius function (A008683). a(n) = A308839(n) - A308841(n) - A308842(n) - A308844(n) - A308845(n). MATHEMATICA Table[Sum[Sum[Sum[Sum[j * MoebiusMu[l]^2*MoebiusMu[k]^2*MoebiusMu[j]^2 *MoebiusMu[i]^2*MoebiusMu[n - i - j - k - l]^2, {i, j, Floor[(n - j - k - l)/2]}], {j, k, Floor[(n - k - l)/3]}], {k, l, Floor[(n - l)/4]}], {l, Floor[n/5]}], {n, 0, 50}] CROSSREFS Cf. A008683, A308839, A308840, A308841, A308842, A308844, A308845. Sequence in context: A280094 A025053 A140496 * A086552 A098393 A263883 Adjacent sequences:  A308840 A308841 A308842 * A308844 A308845 A308846 KEYWORD nonn AUTHOR Wesley Ivan Hurt, Jun 28 2019 EXTENSIONS a(54)..a(55) from David A. Corneth, Sep 16 2019 STATUS approved Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified August 13 05:41 EDT 2020. Contains 336442 sequences. (Running on oeis4.)

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# Math Assignment Help With Forms Of Parabola ## 7.5.3 Forms of parabola: i) y2 = -4ax, where a>0 in such parabolas, • a) Focus is F(-a, 0) • b) Vertex is O(0, 0) • c) Equation of axis is; y = 0 • d) Equation of directrix is; x – a = 0 • ii) x2 = 4ay, where a>0 In such parabolas; • a) focus is F(0, a0 • b) Vertex is O(0, 0) • c) Equation of axis is; x = 0 • d) Equation of directrix is; y+ a = 0 • iii) x2 = -4ay, where a>0 in such parabolas; • a) focus is F(0, -a) • b) Vertex is O(0, 0) • c) Equation of axis is; x = 0 • d) Equation of directrix is; y- a = 0 ### Email Based Assignment Help in Forms Of Parabola To Schedule a Eccentricity tutoring session To submit Forms Of Parabola assignment click here. ### Following are some of the topics in Conic Sections-Parabola, Hyperbola And Ellipse in which we provide help: Geometry Help | Calculus Help | Math Tutors | Algebra Tutor | Tutorial Algebra | Algebra Learn | Math Tutorial | Algebra Tutoring | Calculus Tutor | Precalculus Help | Geometry Tutor | Maths Tutor | Geometry Homework Help | Homework Tutor | Mathematics Tutor | Calculus Tutoring | Online Algebra Tutor | Geometry Tutoring | Online Algebra Tutoring | Algebra Tutors | Math Homework Helper | Calculus Homework Help | Online Tutoring | Calculus Tutors | Homework Tutoring

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ホーム » アプリ » 教育 » A-Level Pure Math Differentiation 1 # A-Level Pure Math Differentiation 1 1.6 for Android Multimedia E-Learning Education System (MELES) ## 説明 A-Level Pure Math Differentiation 1 ★ Study your Pure Mathematics on the go; bus, café, beach, street, anywhere! ★ Simplified explanations, plus extra side notes with even more explanation! ★ Over 30 examples per chapter with step by step working. ★ Multimedia sections about using scientific calculators and graphing software. ★ A companion Tutorial app for each chapter with video, audio and captions. ★ Past paper examination questions at the end of each chapter. ★ Fully worked-out answers to every exercise per chapter are also available. ✪ Check out our publication series here: ✪ Featured Pure Mathematics chapters: 1. Differentiation 1 2. Differentiation 2 3. Integration 4. Further differentiation 5. Algebra 1 6. Binomial theorem 7. Algebra 2 8. Series 9. Trigonometry 1 10. Trigonometry 2 11. Further Integration 1 12. Exponential and Log functions 13. Partial fractions 14. Further Integration 2 15. Coordinate geometry 1 16. Curve sketching 17. Coordinate geometry 2 18. Differential equations 19. Complex numbers 20. Vectors in 3-dimensions ✪ We've created an app for every Pure Maths chapter above. ✪ This app is for Chapter 1 referred to as Differentiation 1. ✪ Pure Math Differentiation-1 Tutorial app with video, audio and captions is now available for download: http://play.google.com/store/apps/details?id=com.puremath.differentiation1tutor ## A-Level Pure Math Differentiation 1 1.6 アップデート 2019-08-13 ✷ Differentiating y = x² from first principles ✷ Differentiating y = x³ from first principles ✷ Differentiating polynomials from first principles ✷ The dy/dx notation ✷ General principle of differentiation ✷ Differentiating using a scientific calculator ✷ Tangents and normals ✷ Tangents & normals using a graph app ✷ Velocity and acceleration ## A-Level Pure Math Differentiation 1 Tags カテゴリー: フリー 教育 アプリ それを入手: A-Level Pure Math Differentiation 1に似ている Discover コメントをよみこんでいます...

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# Exposure (photography)  Exposure (photography) A long exposure showing stars rotating around the southern and northern celestial poles. Credit: European Southern Observatory A photograph of the sea after sunset with an exposure time of 15 seconds. The swell from the waves appears as fog. In photography, exposure is the total density of light allowed to fall on the photographic medium (photographic film or image sensor) during the process of taking a photograph. Exposure is measured in lux seconds, and can be computed from exposure value (EV) and scene luminance over a specified area. In photographic jargon, an exposure generally refers to a single shutter cycle. For example: a long exposure refers to a single, protracted shutter cycle to capture enough low-intensity light, whereas a multiple exposure involves a series of relatively brief shutter cycles; effectively layering a series of photographs in one image. For the same film speed, the accumulated photometric exposure (Hv) should be similar in both cases. ## Photometric and radiometric exposure Photometric or luminous exposure[1] Hv is the accumulated physical quantity of visible light energy (weighted by the luminosity function) applied to a surface during a given exposure time. It is defined as:[2] $H_\mathrm{v} = E_\mathrm{v} \cdot t$ where • Hv is the luminous exposure (usually in lux seconds) • Ev is the image-plane illuminance (usually in lux) • t is the exposure time (in seconds) The radiometric quantity radiant exposure[3] He is sometimes used instead; it is the product of image-plane irradiance Ee and time, the accumulated amount of incident "light" energy per area:[4] $H_\mathrm{e} = E_\mathrm{e} \cdot t$ where If the measurement is adjusted to account only for light that reacts with the photo-sensitive surface, that is, weighted by the appropriate spectral sensitivity, the exposure is still measured in radiometric units (joules per square meter), rather than photometric units (weighted by the nominal sensitivity of the human eye).[5] Only in this appropriately weighted case does the H measure the effective amount of light falling on the film, such that the characteristic curve will be correct independent of the spectrum of the light. Many photographic materials are also sensitive to "invisible" light, which can be a nuisance (see UV filter and IR filter), or a benefit (see infrared photography and full-spectrum photography). The use of radiometric units is appropriate to characterize such sensitivity to invisible light. In sensitometric data, such as characteristic curves, the log exposure[2] is conventionally expressed as log10(H). Photographers more familiar with base-2 logarithmic scales (such as exposure values) can convert using log2(H) ≈ 3.32 log10(H). Table 1. SI photometry unitsv · d · e Quantity Symbol[nb 1] SI unit Symbol Dimension Notes Luminous energy Qv [nb 2] lumen second lm⋅s T⋅J units are sometimes called talbots Luminous flux Φv [nb 2] lumen (= cd⋅sr) lm J also called luminous power Luminous intensity Iv candela (= lm/sr) cd [nb 3] an SI base unit, luminous flux per unit solid angle Luminance Lv candela per square metre cd/m2 L−2⋅J units are sometimes called nits Illuminance Ev lux (= lm/m2) lx L−2⋅J used for light incident on a surface Luminous emittance Mv lux (= lm/m2) lx L−2⋅J used for light emitted from a surface Luminous exposure Hv lux second lx⋅s L−2⋅T⋅J Luminous energy density ωv lumen second per metre3 lm⋅sm−3 L−3⋅T⋅J Luminous efficacy η [nb 2] lumen per watt lm/W M−1⋅L−2⋅T3⋅J ratio of luminous flux to radiant flux Luminous efficiency V 1 also called luminous coefficient Table 2. SI radiometry unitsv · d · e Quantity Symbol[nb 4] SI unit Symbol Dimension Notes Radiant energy Qe[nb 5] joule J M⋅L2⋅T−2 energy Radiant flux Φe[nb 5] watt W M⋅L2⋅T−3 radiant energy per unit time, also called radiant power. Spectral power Φ[nb 5][nb 6] watt per metre W⋅m−1 M⋅L⋅T−3 radiant power per wavelength. Radiant intensity Ie watt per steradian W⋅sr−1 M⋅L2⋅T−3 power per unit solid angle. Spectral intensity I[nb 6] watt per steradian per metre W⋅sr−1⋅m−1 M⋅L⋅T−3 radiant intensity per wavelength. Radiance Le watt per steradian per square metre W⋅sr−1m−2 M⋅T−3 power per unit solid angle per unit projected source area. confusingly called "intensity" in some other fields of study. Spectral radiance L[nb 6] or L[nb 7] watt per steradian per metre3 or watt per steradian per square metre per hertz W⋅sr−1m−3 or W⋅sr−1⋅m−2Hz−1 M⋅L−1⋅T−3 or M⋅T−2 commonly measured in W⋅sr−1⋅m−2⋅nm−1 with surface area and either wavelength or frequency. Irradiance Ee[nb 5] watt per square metre W⋅m−2 M⋅T−3 power incident on a surface, also called radiant flux density. sometimes confusingly called "intensity" as well. Spectral irradiance E[nb 6] or E[nb 7] watt per metre3 or watt per square metre per hertz W⋅m−3 or W⋅m−2⋅Hz−1 M⋅L−1⋅T−3 or M⋅T−2 commonly measured in W⋅m−2nm−1 or 10−22W⋅m−2⋅Hz−1, known as solar flux unit.[nb 8] Me[nb 5] watt per square metre W⋅m−2 M⋅T−3 power emitted from a surface. Spectral radiant exitance / M[nb 6] or M[nb 7] watt per metre3 or watt per square metre per hertz W⋅m−3 or W⋅m−2⋅Hz−1 M⋅L−1⋅T−3 or M⋅T−2 power emitted from a surface per wavelength or frequency. Radiosity Je or J[nb 6] watt per square metre W⋅m−2 M⋅T−3 emitted plus reflected power leaving a surface. Radiant exposure He joule per square metre J⋅m−2 M⋅T−2 Radiant energy density ωe joule per metre3 J⋅m−3 M⋅L−1⋅T−2 ## Optimum exposure "Correct" exposure may be defined as an exposure that achieves the effect the photographer intended.[6] A more technical approach recognises that a photographic film (or sensor) has a physically limited useful exposure range,[7] sometimes called its dynamic range.[8] If, for any part of the photograph, the actual exposure is outside this range, the film cannot record it accurately. In a very simple model, for example, out-of-range values would be recorded as "black" (underexposed) or "white" (overexposed) rather than the precisely graduated shades of colour and tone required to describe "detail". Therefore, the purpose of exposure adjustment (and/or lighting adjustment) is to control the physical amount of light from the subject that is allowed to fall on the film, so that 'significant' areas of shadow and highlight detail do not exceed the film's useful exposure range. This ensures that no 'significant' information is lost during capture. It is worth noting that the photographer may carefully overexpose or underexpose the photograph to eliminate 'insignificant' or 'unwanted' detail; to make, for example, a white altar cloth appear immaculately clean, or to emulate the heavy, pitiless shadows of film noir. However, it is technically much easier to discard recorded information during post processing than to try to 're-create' unrecorded information. In a scene with strong or harsh lighting, the ratio between highlight and shadow luminance values may well be larger than the ratio between the film's maximum and minimum useful exposure values. In this case, adjusting the camera's exposure settings (which only applies changes to the whole image, not selectively to parts of the image) only allows the photographer to choose between underexposed shadows or overexposed highlights; it cannot bring both into the useful exposure range at the same time. Methods for dealing with this situation include: using some kind of fill lighting to gently increase the illumination in shadow areas; using a graduated ND filter or gobo to reduce the amount of light coming from the highlight areas; or varying the exposure between multiple, otherwise identical, photographs (exposure bracketing) and then combining them afterwards in some kind of HDRI process. ### Overexposure and underexposure White chair: Deliberate use of overexposure for aesthetic purposes. A photograph may be described as overexposed when it has a loss of highlight detail, that is, when important bright parts of an image are "washed out" or effectively all white, known as "blown out highlights" or "clipped whites".[9] A photograph may be described as underexposed when it has a loss of shadow detail, that is, when important dark areas are "muddy" or indistinguishable from black,[10] known as "blocked up shadows" (or sometimes "crushed shadows," "crushed blacks," or "clipped blacks," especially in video).[11][12][13] As the image to the right shows, these terms are technical ones rather than artistic judgments; an overexposed or underexposed image may be "correct", in that it provides the effect that the photographer intended. Intentionally over- or under- exposing (relative to a standard or the camera's automatic exposure) is casually referred to as "shooting to the right" or "shooting to the left", respectively, as these shift the histogram of the image to the right or left. ## Exposure settings ### Manual exposure A longer exposure A shorter exposure In manual mode, the photographer adjusts the lens aperture and/or shutter speed to achieve the desired exposure. Many photographers choose to control aperture and shutter independently because opening up the aperture increases exposure, but also decreases the depth of field, and a slower shutter increases exposure but also increases the opportunity for motion blur. 'Manual' exposure calculations may be based on some method of light metering with a working knowledge of exposure values, the APEX system and/or the Zone System. ### Automatic exposure A camera in automatic exposure (abbreviation: AE) mode automatically calculates and adjusts exposure settings to match (as closely as possible) the subject's mid-tone to the mid-tone of the photograph. For most cameras this means using an on-board TTL exposure meter. Aperture priority mode (commonly abbreviated to Av) gives the photographer manual control of the aperture, whilst the camera automatically adjusts the shutter speed to achieve the exposure specified by the TTL meter. Shutter priority mode (commonly abbreviated to Tv) gives manual shutter control, with automatic aperture compensation. In each case, the actual exposure level is still determined by the camera's exposure meter. ### Exposure compensation The purpose of an exposure meter is to estimate the subject's mid-tone luminance and indicate the camera exposure settings required to record this as a mid-tone. In order to do this it has to make a number of assumptions which, under certain circumstances, will be wrong. If the exposure setting indicated by an exposure meter is taken as the "reference" exposure, the photographer may wish to deliberately overexpose or underexpose in order to compensate for known or anticipated metering inaccuracies. Cameras with any kind of internal exposure meter usually feature an exposure compensation setting which is intended to allow the photographer to simply offset the exposure level from the internal meter's estimate of appropriate exposure. Frequently calibrated in stops,[14] also known as EV units,[15] a "+1" exposure compensation setting indicates one stop more (twice as much) exposure and "–1" means one stop less (half as much) exposure.[16][17] Exposure compensation is particularly useful in combination with auto-exposure mode, as it allows the photographer to bias the exposure level without resorting to full manual exposure and losing the flexibility of auto exposure. On low-end video camcorders, exposure compensation may be the only manual exposure control available. ## Exposure control A 1/30s exposure showing motion blur on fountain at Royal Botanic Gardens, Kew A 1/320s exposure showing individual drops on fountain at Royal Botanic Gardens, Kew An appropriate exposure for a photograph is determined by the sensitivity of the medium used. For photographic film, sensitivity is referred to as film speed and is measured on a scale published by the International Organization for Standardization (ISO). Faster film, that is, film with a higher ISO rating, requires less exposure to make a good image. Digital cameras usually have variable ISO settings that provide additional flexibility. Exposure is a combination of the length of time and the illuminance at the photosensitive material. Exposure time is controlled in a camera by shutter speed and the illuminance by the lens aperture and the scene luminance. Slower shutter speeds (exposing the medium for a longer period of time), and greater lens apertures (admitting more light), and higher-luminance scenes produce greater exposures. An approximately correct exposure will be obtained on a sunny day using ISO 100 film, an aperture of f/16 and a shutter speed of 1/100 of a second. This is called the sunny 16 rule: at an aperture of f/16 on a sunny day, a suitable shutter speed will be one over the film speed (or closest equivalent). A scene can be exposed in many ways, depending on the desired effect a photographer wishes to convey. ## Reciprocity An important principle of exposure is reciprocity. If one exposes the film or sensor for a longer period, a reciprocally smaller aperture is required to reduce the amount of light hitting the film to obtain the same exposure. For example, the photographer may prefer to make his sunny-16 shot at an aperture of f/5.6 (to obtain a shallow depth of field). As f/5.6 is 3 stops "faster" than f/16, with each stop meaning double the amount of light, a new shutter speed of (1/125)/(2·2·2) = 1/1000 is needed. Once the photographer has determined the exposure, aperture stops can be traded for halvings or doublings of speed, within limits. A demonstration of the effect of exposure in night photography. Longer shutter speeds result in increased exposure. The true characteristic of most photographic emulsions is not actually linear, (see sensitometry) but it is close enough over the exposure range of about one second to 1/1000 of a second. Outside of this range, it becomes necessary to increase the exposure from the calculated value to account for this characteristic of the emulsion. This characteristic is known as reciprocity failure. The film manufacturer's data sheets should be consulted to arrive at the correction required as different emulsions have different characteristics. Digital camera image sensors can also be subject to a form of reciprocity failure.[18] ## Determining exposure A fair ride taken with a 2/5 second exposure. The Zone System is another method of determining exposure and development combinations to achieve a greater tonality range over conventional methods by varying the contrast of the 'film' to fit the print contrast capability. Digital cameras can achieve similar results (high dynamic range) by combining several different exposures (varying only the shutter speeds) made in quick succession. Today, most cameras automatically determine the correct exposure at the time of taking a photograph by using a built-in light meter, or multiple point meters interpreted by a built-in computer, see metering mode. Negative/Print film tends to bias for exposing for the shadow areas (film dislikes being starved of light), with digital favouring exposure for highlights. See latitude below. ## Latitude Latitude is the degree by which one can over, or under expose an image, and still recover an acceptable level of quality from an exposure. Typically negative film has a better ability to record a range of brightness than slide/transparency film or digital. Digital should be considered to be the reverse of print film, with a good latitude in the shadow range, and a narrow one in the highlight area; in contrast to film's large highlight latitude, and narrow shadow latitude. Slide/Transparency film has a narrow latitude in both highlight and shadow areas, requiring greater exposure accuracy. Negative film's latitude increases somewhat with high ISO material, in contrast digital tends to narrow on latitude with high ISO settings. ### Highlights Example image exhibiting blown-out highlights. Top: original image, bottom: blown-out areas marked red Areas of a photo where information is lost due to extreme brightness are described as having "blown-out highlights" or "flared highlights". In digital images this information loss is often irreversible, though small problems can be made less noticeable using photo manipulation software. Recording to RAW format can ameliorate this problem to some degree, as can using a digital camera with a better sensor. Film can often have areas of extreme overexposure but still record detail in those areas. This information is usually somewhat recoverable when printing or transferring to digital. A loss of highlights in a photograph is usually undesirable, but in some cases can be considered to "enhance" appeal. Examples include black-and-white photography and portraits with an out-of-focus background. ### Blacks Areas of a photo where information is lost due to extreme darkness are described as "crushed blacks". Digital capture tends to be more tolerant of underexposure, allowing better recovery of shadow detail, than same-ISO negative print film. Crushed blacks cause loss of detail, but can be used for artistic effect. ## Notes 1. ^ Standards organizations recommend that photometric quantities be denoted with a suffix "v" (for "visual") to avoid confusion with radiometric or photon quantities. 2. ^ a b c Alternative symbols sometimes seen: W for luminous energy, P or F for luminous flux, and ρ or K for luminous efficacy. 3. ^ "J" is the recommended symbol for the dimension of luminous intensity in the International System of Units. 4. ^ Standards organizations recommend that radiometric quantities should be denoted with a suffix "e" (for "energetic") to avoid confusion with photometric or photon quantities. 5. ^ a b c d e Alternative symbols sometimes seen: W or E for radiant energy, P or F for radiant flux, I for irradiance, W for radiant emittance. 6. ^ a b c d e f Spectral quantities given per unit wavelength are denoted with suffix "λ" (Greek) to indicate a spectral concentration. Spectral functions of wavelength are indicated by "(λ)" in parentheses instead, for example in spectral transmittance, reflectance and responsivity. 7. ^ a b c Spectral quantities given per unit frequency are denoted with suffix "ν" (Greek)—not to be confused with the suffix "v" (for "visual") indicating a photometric quantity. 8. ^ NOAA / Space Weather Prediction Center includes a definition of the solar flux unit (SFU). ## References 1. ^ National Institute of Standards and Technology [1]. Retrieved Feb 2009. 2. ^ a b Geoffrey G. Attridge (2000). "Sensitometry". In Ralph E. Jacobson, Sidney F. Ray, Geoffrey G. Attridge, and Norman R. Axford. The Manual of Photography: Photographic and Digital Imaging (9th ed.). Oxford: Focal Press. pp. 218–223. ISBN 0-240-51574-9. 3. ^ Hsien-Che Lee (2005). Introduction to Color Imaging Science. Cambridge University Press. p. 57. ISBN 9780521843881. 4. ^ Hans I. Bjelkhagen (1995). Silver-halide Recording Materials. Springer. p. 15. ISBN 9783540586197. 5. ^ Gareth Rees (2001). Physical Principles of Remote Sensing. Cambridge University Press. p. 114. ISBN 9780521669481. 6. ^ Peterson, Bryan, "Understanding Exposure", 2004, ISBN 0-8174-6300-3 : p.14 7. ^ Ray, S.F. et al. 2000 "The Manual of Photography" Focal Press, ISBN 0-240-51574-9, p.230 8. ^ Ray, S.F. et al. 2000 "The Manual of Photography" Focal Press, ISBN 0-240-51574-9, p.121 and p.245 9. ^ Ed van der walt. "Basic Photography - ISO and Film Speed". Retrieved 2 July 2011. 10. ^ Rob Sheppard (2010). Digital Photography: Top 100 Simplified Tips & Tricks (4th ed.). John Wiley and Sons. p. 40. ISBN 9780470597101. 11. ^ Barbara A. Lynch-Johnt and Michelle Perkins (2008). Illustrated Dictionary of Photography. Amherst Media. p. 15. ISBN 9781584282228. 12. ^ Steve Hullfish and Jaime Fowler (2005). Color Correction for Digital Video. Focal Press. pp. 135–136. ISBN 9781578202010. 13. ^ John Jackman (2004). Lighting for Digital Video & Television. Focal Press. p. 60. ISBN 9781578202515. 14. ^ Chris George (2006). Total Digital Photography. Running Press. pp. 54–55. ISBN 9780762428083. 15. ^ R E Jacobson (2000). The Manual of Photography. Focal Press. p. 318. ISBN 9780240515748. 16. ^ John Child, Mark Galer (2005). Photographic Lighting : Essential Skills. Focal Press. p. 51. ISBN 9780240519647. 17. ^ David D. Busch (2007). Nikon D80 Digital Field Guide. John Wiley and Sons. p. 11. ISBN 9780470120514. 18. ^ David D. Busch (2003). Mastering Digital Photography: The Photographer's Guide to Professional-Quality Digital Photography. Thomson Course Technology. ISBN 1592001149. Wikimedia Foundation. 2010. ### Look at other dictionaries: • Long exposure photography — is a technique that requires a slow shutter speed to capture light and movement.TechniqueWhen an image is taken including stationary and moving subjects (for example, a fixed street and moving cars or a camera within a car showing a fixed dash… …   Wikipedia • Exposure — can refer toIn biology: * A condition of very poor health or death resulting from lack of protection over prolonged periods under weather, extreme temperatures or dangerous substances ( see also: hypothermia, hyperthermia, radioactive… …   Wikipedia • Photography — is the art, science and practice of creating durable images by recording light or other electromagnetic radiation, either electronically by means of an image sensor or chemically by means of a light sensitive material such as photographic… …   Wikipedia • Exposure compensation — is a technique for adjusting the exposure indicated by a photographic exposure meter, in consideration of factors that may cause the indicated exposure to result in aless than optimal image. Factors considered may include unusual lighting… …   Wikipedia • Exposure latitude — is the extent to which a light sensitive material can be over or underexposed and still achieve an acceptable result. Since the acceptability of the result is dependent on both personal aesthetics and artistic intentions, the measurement of… …   Wikipedia • photography, technology of — Introduction       equipment, techniques, and processes used in the production of photographs.  The most widely used photographic process is the black and white negative–positive system (Figure 1 >). In the camera the lens projects an image of… …   Universalium • photography, history of — Introduction       method of recording the image of an object through the action of light, or related radiation, on a light sensitive material. The word, derived from the Greek photos (“light”) and graphein (“to draw”), was first used in the… …   Universalium • Photography and the law — A No Photography sign, commonly placed in properties where the owner objects to or it is illegal to take photographs (though in some jurisdictions, this is not a legal requirement). Photography tends to be protected by the law through copyright… …   Wikipedia • Exposure value — In photography, exposure value (EV) denotes all combinations of camera shutter speed and relative aperture that give the same exposure. The concept was developed in Germany in the 1950s (Ray 2000),in an attempt to simplify choosing among… …   Wikipedia • photography — I (New American Roget s College Thesaurus) Picture taking Nouns 1. photography, picture taking; aerial, architectural, available light, candid, digital, fashion, fine art, infrared, laser, paper negative, print, product, scenic, still, stereo… …   English dictionary for students We are using cookies for the best presentation of our site. Continuing to use this site, you agree with this.

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You currently have JavaScript disabled on this browser/device. JavaScript must be enabled in order for this website to function properly. # ZingPath: Circular Motion Searching for ## Circular Motion Learn in a way your textbook can't show you. Explore the full path to learning Circular Motion ### Lesson Focus #### Uniform Circular Motion II Physics You get to learn about circular motion through the real-life example of a Ferris wheel in motion. Then study harmonic motion by experimenting with the pendulum of a clock. ### Now You Know After completing this tutorial, you will be able to complete the following: • Define period, linear velocity, angular velocity, centripetal acceleration, and force. ### Everything You'll Have Covered What are the equations for linear velocity and angular velocity? ~ The equation for linear velocity is: The equation for angular velocity is: How are linear and the angular velocity related? ~ Linear velocity equals angular velocity times the radius of the circular motion. Or In uniform circular motion, the magnitude of the linear velocity is constant. How is it that there is acceleration? ~ The direction of the linear velocity is always changing. Therefore, there will be a rate of change of the velocity vector. Because of this change, acceleration occurs. What is centripetal acceleration and how is it related to centripetal force? ~ Centripetal acceleration is acceleration towards the center of a circular path and is equal to . This acceleration is produced by the centripetal force according to Newton's second law. Since force = m•a, centripetal force is equal to F = m • v^2/r ### Tutorial Details Approximate Time 2 Minutes Pre-requisite Concepts Students should have an intellectual grasp on the following terms: centripetal acceleration, centripetal force, and circular motion. Course Physics Type of Tutorial Animation Key Vocabulary centripetal acceleration, centripetal force, circular motion

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# Functions ## General Functions AbstractLogic.checkfeasibleFunction checkfeasible(command::String, logicset::LogicalCombo; verbose=true, all=false, countany=false) Is called when the user would like to check if a command produces a valid result, possible result, or invalid result. The result is returned as a decimal from 0.0 to 1.0. With 0.0 being no matches and 1.0 being all matches. Arguments • verbose : controls print • countall : all sets have to be feasible or return 0 • countany : any set can be non-zero to return 1 Examples julia> myset = logicalparse("a, b, c ∈ red, blue, green") a, b, c ∈ red, blue, green feasible outcomes 27 ✓ :red blue blue julia> myset = logicalparse("a != b,c; b = c ==> a = 'blue'", myset) a != b,c feasible outcomes 12 ✓ :green blue blue b = c ==> a = 'blue' feasible outcomes 8 ✓ :blue green green julia> checkfeasible("a = 'green' ==> b = 'red'", myset) Check: a = 'green' ==> b = 'red' ... a = 'green' ==> b = 'red' feasible outcomes 17 ✓ :blue red green possible, 17 out of 21 possible combinations 'true'. AbstractLogic.LogicalComboType LogicalCombo <: Any A LogicalCombo stores the variable names, domains, as well as an binary representation of feasible values given constraints. Fields • keys : Stores the names of variables. • domain : Stores the range of possible matches variables can have. • logical : A binary vector marking feasibility of length equal to every possible combination of value which the matrix could take on. • commandlist : An array collection of the strings input to generate the current state of the the object. Indexing • [x,y] Where x is numeric or : and y is numeric, colon, or symbol will return matrix point values regardless of feasibility. • [:,:] Will collect the entire possible domain and is the same as collect(). • [x,0] Will return the logical vector values • [x,:,true] Will return the xth feasible value of the LogicalCombo set. • [x,:,false] Will return the xth infeasible value of the LogicalCombo set. AbstractLogic.logicalparseFunction logicalparse Takes a command and parses it into logical calls that either assigning additional feasible variable ranges or constrain the relationship between variables. logicalparse(command::String; logicset::LogicalCombo = LogicalCombo(), verbose=true) logicalparse(command::String, logicset::LogicalCombo; ...) logicalparse(commands::Array{String,1}, logicset::LogicalCombo; ...) logicalparse(commands::Array{String,1}; ...) Arguments • verbose : specifies to print to screen or not Operators There are numerous operators available to be used in the logical parse command. Examples julia> myset = logicalparse("a, b, c in 1:3") a,b,c in 1:3 feasible outcomes 27 ✓ :3 3 3 julia> myset = logicalparse("a == b", myset) a == b feasible outcomes 9 ✓ :1 1 2 julia> myset = logicalparse("a > c", myset) a > c feasible outcomes 3 ✓ :3 3 1 julia> myset = logicalparse("c != 1", myset) c != 1 feasible outcomes 1 ✓✓ :3 3 2 Missing docstring. Missing docstring for logicalrepl. Check Documenter's build log for details. AbstractLogic.searchFunction search(command::String, logicset::LogicalCombo; verbose=true) Searches for a possible match among a LogicalCombo in which the wildcard term is true. Search requires the use of a wildcard. In the event that a wildcard is missing, search will insert a {{i}} to the left of the command.{{i+1}} can be used to search for relationships between the ith column and another column. Examples julia> myset = logicalparse("v1, v2, v3 ∈ 1:10") v1, v2, v3 ∈ 1:10 feasible outcomes 1000 ✓ :6 6 10 julia> myset = logicalparse("{{i}} >> {{i+1}}", myset) {{i}} >> {{i+1}} >>> v1 >> v2 >>> v2 >> v3 feasible outcomes 56 ✓ :10 7 3 julia> search("{{i}} == 4", myset) Checking: v1 == 4 Checking: v2 == 4 Checking: v3 == 4 :v1 is a not match with 0 feasible combinations out of 56. :v2 is a possible match with 10 feasible combinations out of 56. :v3 is a possible match with 6 feasible combinations out of 56. julia> search("== 4", myset, verbose=false) == search("{{i}} == 4", myset, verbose=false) true julia> search("{{i}} > {{!i}}", myset) Checking: v1 > v2 Checking: v1 > v3 Checking: v2 > v1 Checking: v2 > v3 Checking: v3 > v1 Checking: v3 > v2 :v1 is a match with 56 feasible combinations out of 56. :v2 is a not match with 0 feasible combinations out of 56. :v3 is a not match with 0 feasible combinations out of 56. AbstractLogic.showfeasibleFunction showfeasible Collects a matrix of only feasible outcomes given the parameter space and the constraints. Use collect to output a matrix of all possible matches for parameter space. Examples julia> myset = logicalparse("a, b, c in 1:3") a,b,c in 1:3 feasible outcomes 27 ✓ :3 3 3 julia> myset = logicalparse("a == b; c > b", myset) a == b feasible outcomes 9 ✓ :1 1 3 c > b feasible outcomes 3 ✓ :2 2 3 julia> showfeasible(myset) 3×3 Array{Int64,2}: 1 1 2 1 1 3 2 2 3

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You are Here: Home # Puzzle watch 1. tell me!! i gotta watch teh footie!! 2. the fifth person brought an umbrella 3. was the 5th carried by the others? 5. 5 people are going somewhere. It suddenly starts to rain and 4 people run for cover. But the 5th one doesn't bother about the rain and turns up at the destination at the same time fully dry. the 5th doesnt bother about the rain, but dont wanna be left out so he runs n e way 6. They're all on an open-top bus, the fifth is on the first level and the others are upstairs. In which case none of them work at Stamford Bridge 7. (Original post by keithy) 5 people are going somewhere. It suddenly starts to rain and 4 people run for cover. But the 5th one doesn't bother about the rain and turns up at the destination at the same time fully dry. The first four are just stupid and run for cover even though they are inside. The fifth just looks on and laughs. They all turn up at the destination fully dry. 8. The group are on a mountain. Four of them are below the clouds, the fifth above. (Going on the "think laterally" clue) 9. the fifth is significantly shorter than the others so he makes it inside before the raindrops hit him. 10. NONE of you are right so far. It's so simple. NONE of you are right so far. It's so simple. Erm what if they're all in a football stadium getting to their seats, and one's already under the overhang? 12. (Original post by ZJuwelH) Erm what if they're all in a football stadium getting to their seats, and one's already under the overhang? Lalalala.........NO. *yawn* 13. It stopped raining almost immediately? 14. it rains, but not where these people are. four run for cover for an unknown reason. 15. The place where the four run for cover is the intended destination? 16. the fifth person has a massive umbrella. the other 4 run for cover under that umbrella 18. but i might just have missed the question it was on ages ago now! 19. to elaborate; is the 5th in a coffin? 20. (Original post by piginapoke) Two of the people are midgets with one standing on the other's shoulders under a massive trench coat. The top one gets wet but the bottom one doesn't. *sigh* You people make things so complicated. It astounds me. TSR Support Team We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out. This forum is supported by: Updated: May 6, 2004 Today on TSR Poll Useful resources ## Articles: Debate and current affairs forum guidelines

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https://wiki.freecadweb.org/Scripted_Parts:_Ball_Bearing_-_Part_2

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# Scripted Parts: Ball Bearing - Part 2 Other languages: Deutsch • ‎English • ‎français • ‎italiano • ‎polski • ‎русский Tutorial Topic Part Scripting - Ball Bearing #2 Level Beginner Time to complete 30 min Authors r-frank 0.16.6706 Example files None ### Introduction This tutorial is meant as a beginner's introduction to creating parts with python scripts within FreeCAD. This tutorial will cover how to build a ball bearing with a workflow that consists of creating sketches and revolving them. The code will produce a new FreeCAD document with 12 shapes (Inner Ring, Outer Ring and 10 balls/spheres). It will look like this: ### Workflow The workflow is more or less identical how you would create the part in part design workbench. Just some small differences. • Create a new empty document and make it the active document • Draw the basic shape of the outer ring consisting of four straight lines and four arcs • Connect the lines and arcs and upgrade them to one single wire • Upgrade the wire to a face • Revolve the face to get a shape • Draw a circle • Revolve face and apply boolean cut to obtain groove in outer ring • Draw the basic shape of the inner ring consisting of four straight lines and four arcs • Connect the lines and arcs and upgrade them to one single wire • Upgrade the wire to a face • Revolve the face to get a shape • Draw a circle • Revolve face and apply boolean cut to obtain groove in inner ring • Insert balls with same workflow as in part 1 (because of effectiveness) • Set view to axometric • Zoom to fit all ### Making the groove Drawing an arc needs either three points or a start angle and an end angle. In the sketcher we would use constraints to define the start point and the end point of the arc. Since we can't do this in scripting, we draw a rounded rectangle and revolve it to get a basic "ring shape". Then we draw a circle and revolve it to get the geometry of the groove. Then we apply a boolean cut to the two revolved shapes and we have the complete shape of the inner/outer ring. ### Inserting the balls The correct sketcher-based workflow of inserting the balls would be: • Draw an arc (semi-circle) with center being identical with the origin and draw a line closing the "open" side of the arc • Convert the two elements to a wire, upgrade to a face, revolve around z-axis to get a ball shape • Use "translate" command to move the ball into correct position • Repeat the above steps nine times involving math function to create and position the other balls • This repeat-operation could be programmed with a loop Now this is not effective, inserting primitives and positioning them is easier and faster in this case. So we use the same method as in "Scripted Parts: Ball Bearing - Part 1". Scripted objects: The wiki page explaining the basics of scripting Topological data scripting: A tutorial for covering basics of scripting Scripted Parts: Ball Bearing - Part 1: Doing it with part primitives Bearings from scripted sketches: Base for this tutorial, thanks to JMG ... ### Code ```## Ball-bearing script ## based on ball bearing script by JMG ## (http://linuxforanengineer.blogspot.de/2013/12/bearings-from-scripted-sketches.html) # #needed for doing boolean operations import Part #needed for calculating the positions of the balls import math #needed for translation and rotation of objects # #VALUES# R1=15.0 R2=25.0 R3=30.0 R4=40.0 #(thickness of bearing) TH=15.0 #(number of balls) NBall=10 RBall=5.0 RR=1 #first coordinate of center of ball CBall=((R3-R2)/2)+R2 #second coordinate of center of ball PBall=TH/2 # #Create new document App.newDocument("Unnamed") App.setActiveDocument("Unnamed") # #Lines for basic shape of outer ring L1o=Part.makeLine((R4,0,TH-RR),(R4,0,RR)) L2o=Part.makeLine((R4-RR,0,0),(R3+RR,0,0)) L3o=Part.makeLine((R3,0,RR),(R3,0,TH-RR)) L4o=Part.makeLine((R3+RR,0,TH),(R4-RR,0,TH)) #Corner rounding for basic shape of outer ring A1o=Part.makeCircle(RR,Base.Vector(R4-RR,0,RR),Base.Vector(0,1,0),0,90) A2o=Part.makeCircle(RR,Base.Vector(R3+RR,0,RR),Base.Vector(0,1,0),90,180) A3o=Part.makeCircle(RR,Base.Vector(R3+RR,0,TH-RR),Base.Vector(0,1,0),180,270) A4o=Part.makeCircle(RR,Base.Vector(R4-RR,0,TH-RR),Base.Vector(0,1,0),270,360) #Connect Lines and arcs to make wire and upgrade to face, revolve and apply cut to obtain groove OR=Part.Wire([L1o,A1o,L2o,A2o,L3o,A3o,L4o,A4o]) OR=Part.Face(OR) OR=OR.revolve(Base.Vector(0,0,1),Base.Vector(0,0,360)) C1=Part.makeCircle(RBall,Base.Vector(R2+(R3-R2)/2,0,TH/2),Base.Vector(0,1,0),0,360) GRo=Part.Wire([C1]) GRo=Part.Face(GRo) GRo=GRo.revolve(Base.Vector(0,0,1),Base.Vector(0,0,360)) OR=OR.cut(GRo) Part.show(OR) # #Lines for basic shape of inner ring L1i=Part.makeLine((R2,0,TH-RR),(R2,0,RR)) L2i=Part.makeLine((R2-RR,0,0),(R1+RR,0,0)) L3i=Part.makeLine((R1,0,RR),(R1,0,TH-RR)) L4i=Part.makeLine((R1+RR,0,TH),(R2-RR,0,TH)) #Corner rounding for basic shape of inner ring A1i=Part.makeCircle(RR,Base.Vector(R2-RR,0,RR),Base.Vector(0,1,0),0,90) A2i=Part.makeCircle(RR,Base.Vector(R1+RR,0,RR),Base.Vector(0,1,0),90,180) A3i=Part.makeCircle(RR,Base.Vector(R1+RR,0,TH-RR),Base.Vector(0,1,0),180,270) A4i=Part.makeCircle(RR,Base.Vector(R2-RR,0,TH-RR),Base.Vector(0,1,0),270,360) #Connect Lines and arcs to make wire and upgrade to face, revolve and apply cut to obtain groove IR=Part.Wire([L1i,A1i,L2i,A2i,L3i,A3i,L4i,A4i]) IR=Part.Face(IR) IR=IR.revolve(Base.Vector(0,0,1),Base.Vector(0,0,360)) C2=Part.makeCircle(RBall,Base.Vector(R2+(R3-R2)/2,0,TH/2),Base.Vector(0,1,0),0,360) GRi=Part.Wire([C2]) GRi=Part.Face(GRi) GRi=GRi.revolve(Base.Vector(0,0,1),Base.Vector(0,0,360)) IR=IR.cut(GRi) Part.show(IR) # #Balls# for i in range(NBall): Ball=Part.makeSphere(RBall) Alpha=(i*2*math.pi)/NBall BV=(CBall*math.cos(Alpha),CBall*math.sin(Alpha),TH/2) Ball.translate(BV) Part.show(Ball) # #Make it pretty# App.ActiveDocument.recompute() Gui.ActiveDocument.ActiveView.viewAxometric() Gui.SendMsgToActiveView("ViewFit") ```

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https://www.allthetests.com/quiz25/quiz/1207078100/A-Test

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# A Test 10 Questions - Developed by: Alex - Developed on: - 1.983 taken • 1 0.45+0.74-1=0.18 • 2 10+56+4-67+9=12 • 3 Football in New Zealand is like rugby • 4 67-56+45-0.49=55.51 • 5 Red cars look richer than blue cars if they are both the same. • 6 For a joke kids say 1+1= a window are they correct? If you put to ones together do they make a window • 7 0.15+0.86=1 • 8 2+2= fish Yes • 9 100-56=44-22=22 • 10 Red devils Liverpool No

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https://jeopardylabs.com/play/hello-trial

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O.P. Ako BOOLY Let it Flow Logic Level App 100 The symbol for not Equal !== 100 10>=10 AND 11<=11 TRUE 100 How many End Terminator(s) can be only created? One (1) 100 How many possible results can be resulted in a process symbol? Zero (0) 100 Nina (N) is older than his brother Elmer(E) Make this into Boolean N > E 200 What is the operator to be used for comparing if both inputs are the same? == 200 18 >= 5 *5 / 4 AND 59 <=(7^2)+20 false 200 It pertains to the shape used for an on-page connector. Circle 200 Who invented Boolean? George Boole 200 A student (x) will repeat a grade level when he/she exceeds 4.0 units of failed. Make this a boolean statement. X > 4.0 300 In executing GEMDAS, What does G means? group 300 14 >=  3*3 OR  25 >= 9+9*2 _______ OR _______ TRUE OR FALSE 300 What is the two-letter word that distinguishes the Conditional statement in the Decision Symbol? IF 300 What is the three letter word to distinguish a preparation? Var 300 (X)Book1 =300 copies (Y)Book2 =402 copies The above mentioned was the number of sold copies in the book stores. What is the boolean expression without using >= or <= X < Y 400 The _____ returns the boolean value true if either or both operands is true and returns false otherwise OR 400 2*2*2 <= 2^4 AND (2^2)-2 !== 1+1 Final Result: False 400 "enter, get, read " are action words that can signify ________ symbol. input 400 Ages below 18,   65 yrs old and above were restricted again. Turn this into an "If Condition Statement" If age <=17 OR age >=65 400 In flowchart, Other word for "End" Stop 500 The _______returns a value of TRUE if both its operands are TRUE, and FALSE otherwise. AND 500 Let x=2,  a=5,  b=6 x+a >= b OR b+b /x < a*a ____?____ OR ____?____ TRUE OR TRUE 500 Shape to be used for output symbol parallelogram 500 Algebra: 45 <= K = 89 Boolean: ------?------- K >= 45 and K ==89 500 Just like in Math, Just like in Programming. I can be a letter or might prefer a word. But, what matters most -I saved value. What am I? Variable Click to zoom

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https://byjus.com/question-answer/a-railway-half-ticket-costs-half-the-full-fare-and-the-reservation-charge-is-the-3/

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1 You visited us 1 times! Enjoying our articles? Unlock Full Access! Question # A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Ahmadabad costs. Rs.216 while one full and one half reserved first class tickets cost Rs.327. What is the basic first class full fare and what is the reservation charge? Open in App Solution ## Let the cost of fare be Rs.′x′ and the reservation charge be Rs.′y′.By condition I:x+y=216 ....(i)By condition II:one full and one half ticket =1+12=32cost of half railway ticket is =x2∴,x+y+x2+y=327⇒32x+2y=327 ⇒3x+4y=654 ....(ii)(after multiplying by 2 )By solving (i) and (ii) we will get3x+4y−4x−4y=654−864∴x=210 and put x=210 in (i) we get y=216−x=216−210=6∴y=6Hence, basic fare for one ticket is Rs.210 and reservation fare is Rs.6 Suggest Corrections 1 Join BYJU'S Learning Program Related Videos Is Light a Traveller? QUANTITATIVE APTITUDE Watch in App Explore more Join BYJU'S Learning Program

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https://passel2.unl.edu/view/lesson/0cff7943f577/6

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# Bulk Density Soil weight is most often expressed on a soil volume basis rather than on a particle basis. Bulk density is defined as the dry weight of soil per unit volume of soil. Bulk density considers both the solids and the pore space; whereas, particle density considers only the mineral solids. Figure 2.6 illustrates the difference between bulk density and particle density. For our ideal soil, one-half of it is solids, and one-half is pore space. Using our example of a 1 cm3 volume, the ideal soil would have 0.5 cmof pore space and 0.5 cmof solids. Pore space filled with air weighs 0 g. Organic matter is a very small portion of the solids, so it is usually ignored in this calculation. The mineral solids would weigh 1.33 g when dry, and is determined by multiplying particle density by the volume of solids: 2.66 g/cm3 x 0.5 cm3 = 1.33 g The bulk density, then, is the dry weight of soil divided by the volume of soil: 1.33 g / 1 cm3 = 1.33 g/cm3 For practice, consider a box of undisturbed soil from the field. The box has dimensions of 2.5 cm by 10 cm by 10 cm. The volume of the box can be determined by multiplying the height of the box times its width and its depth. The wet soil in the box weighed 450 g. The dry soil weighed 375 g. Now calculate the bulk density. Your answer should be 1.5 g/cm3. In this calculation, you did not have to use the particle density because the weight of soil in the box was already known. Bulk density of the surface soil is lowest in the spring immediately after soils have thawed and before field operations have begun. Each field operation compacts soil beneath the tires. If soils are wetter than field capacity, bulk density may increase. However, if soils are dry, bulk density is not affected much. Root growth, in general, starts to be restricted when the bulk density reaches 1.55 to 1.6 g/cmand is prohibited at about 1.8 g/cm3. Tillage can increase bulk density if it breaks down aggregates and allows soil separates to pack more tightly. Adding organic material decreases bulk density because organic material has a lower bulk density. However, additions are typically so small in proportion to the weight of soil that they do not markedly influence bulk density except at the soil-atmosphere interface. Bulk density is also important because it tells us about the porosity of a soil.

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https://docs.sunpy.org/en/latest/generated/gallery/plotting/wcsaxes_plotting_example.html

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# Plotting points on a Map with WCSAxes# This example demonstrates the plotting of a point, a line and an arch in pixel coordinates, world coordinates, and SkyCoords respectively when plotting a map with WCSAxes. import matplotlib.pyplot as plt import numpy as np import astropy.units as u from astropy.coordinates import SkyCoord import sunpy.data.sample import sunpy.map from sunpy.coordinates.utils import GreatArc We will start by creating a map using an AIA 171 image. Then we shall create a standard plot of this map. my_map = sunpy.map.Map(sunpy.data.sample.AIA_171_IMAGE) fig = plt.figure(figsize=(8, 8)) my_map.plot(axes=ax, clip_interval=(1, 99.9)*u.percent) <matplotlib.image.AxesImage object at 0x7fb37f4ea290> Now we will plot a line on the map by using coordinates in arcseconds. The array below xx and yy are the x and y coordinates that define a line from the Sun center (at 0, 0) to the point (500, 500) in arcsecs. When plotting a map a WCSAxes is created. For plotting with WCSAxes, pixel coordinates are expected as a default, however, we can plot world coordinates (i.e. arcsec) by using the transform keyword. Its important to note that when transforming between world and pixel coordinates the world coordinates need to be in degrees rather than arcsecs. xx = np.arange(0, 500) yy = xx # Note that the coordinates need to be in degrees rather than arcseconds. ax.plot(xx*u.arcsec.to(u.deg), yy*u.arcsec.to(u.deg), color='r', transform=ax.get_transform("world"), label=f'WCS coordinate [{0*u.arcsec}, {500*u.arcsec}]') [<matplotlib.lines.Line2D object at 0x7fb37e3e5ab0>] Here we will plot a point in pixel coordinates (i.e. array index). Let’s define a pixel coordinate in the middle of the image, pixel_coord. As this coordinate is in pixel space (rather than a world coordinates like arcsec) we do not need to use the transform keyword. pixel_coord = [my_map.data.shape[0]/2., my_map.data.shape[1]/2.] * u.pix ax.plot(pixel_coord[0], pixel_coord[1], 'x', color='w', label=f'Pixel coordinate [{pixel_coord[0]}, {pixel_coord[1]}]') [<matplotlib.lines.Line2D object at 0x7fb37e3e5b10>] As well as defining a point and using GenericMap.plot(), you can also plot a point with WCSAxes using the plot_coord functionality using a coordinate as a SkyCoord. We can demonstrate this by plotting a point and an arc on a map using two separate SkyCoords. Here we will plot a point (at -250,-250) on the map using a SkyCoord. ax.plot_coord(SkyCoord(-250*u.arcsec, -250*u.arcsec, frame=my_map.coordinate_frame), "o", label=f'SkyCoord [{-250*u.arcsec}, {-250*u.arcsec}]') [<matplotlib.lines.Line2D object at 0x7fb37e3e7160>] Finally, let’s create a great arc between a start and end point defined as SkyCoords. start = SkyCoord(723 * u.arcsec, -500 * u.arcsec, frame=my_map.coordinate_frame) end = SkyCoord(-100 * u.arcsec, 900 * u.arcsec, frame=my_map.coordinate_frame) great_arc = GreatArc(start, end) my_map.plot(axes=ax, clip_interval=(1, 99.99)*u.percent) ax.plot_coord(great_arc.coordinates(), color='c', label=f'SkyCoord [{723*u.arcsec}, {-500*u.arcsec}],\n \ [{-100*u.arcsec}, {900*u.arcsec}]') ax.legend(loc="lower center") plt.show() Total running time of the script: (0 minutes 0.610 seconds) Gallery generated by Sphinx-Gallery

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http://www.physicsforums.com/showpost.php?p=1112220&postcount=4

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View Single Post ## How can light travel without losing energy? On 2005-09-01, BJ <bjtkemp@yahoo.com> wrote: > Everything loses energy as it travels -- the planet relies upon the > work of gravity; the airplane consumes fuel. How come light can travel > without losing energy? > > Granted the electric and magnetic components of light oscillate at > right angles to enable it to travel in a very efficient way, but the > oscillation should consume energy like everything else. > > Where does this energy come from? > That's a common misconception that goes back to Aristotle. He thought that an object needs to be pushed to remain in motion. Galileo put that misconception to rest with his experiments on inertia. The simple answer is that energy is not lost in motion, it can only be transfered from one place to another. So light can travel indefinitely unless it comes into contact with which it can interact and to which it can transfer energy. You may wonder how the apparently observed energy loss, that you refer to, can be reconciled with the above principle of conservation of energy. This very question, actually, is what lead people to the understanding of heat as a form of energy. Whenever a motion is executed against some resistence, air friction or friction of break pads agains the wheels of a car, some of the energy due to large scale mechanical motion, flight of a plane or the turning of a car's wheels, gets transfered to the atoms and molecules of the material providing the resistence, the air or the break pads and the metal of the wheels. These molecules start moving around faster but on a much smaller scale, essentially imperceptible. The only way we can tell this energy transfer took place is through changes in temperature, which indicates heat exchange. By the second law of theormodynamics, once energy is turned into heat, it's kind of hard to get it back into mechanical motion. However, when things cool down, again, the heat energy is not lost, only transevered to the much larger surrounding medium (the athmosphere, usually) which is so large that the entailed change in temperature is imperceptible. BTW, the kinetic energy of a planet in motion around the Sun is not constant. It goes up when the planet is closer to the Sun and goes down when the planet is farther away. So the Sun's gravity is giving kinetic energy to the planet for part of the orbit, and takes it away for another part of the orbit. However, if the gravitational potential energy is added to the kinetic energy, their sum (the total mechanical energy) is conserved throughout the orbit. Igor

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http://www.mathworks.com/matlabcentral/fileexchange/23972-chebfun/content/chebfun/examples/roots/html/RandomPolynomials.html

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Code covered by the BSD License Chebfun V4 Chebfun Team (view profile) 30 Apr 2009 (Updated ) Numerical computation with functions instead of numbers. Editor's Notes: This file was selected as MATLAB Central Pick of the Week Roots of random polynomials Roots of random polynomials Nick Trefethen, 6th November 2011 (Chebfun example roots/RandomPolynomials.m) If p(z) = a_0 + a_1 z + ... + a_n z^n is a polynomial with random coefficients, its roots tend to lie near the unit circle [1,2]. To be specific, suppose a_0,...,a_(n-1) are independent real numbers from the standard normal distribution, with a_n=1. Here are typical pictures for n = 50 and 200, based on Matlab's ROOTS command. Incidentally, such computations are numerically stable [4]; the well-known difficulties of polynomial rootfinding pertain mainly to problems with roots away from the unit circle and coefficients far from random. nn = [50 200]; MS = 'markersize'; FS = 'fontsize'; ms = 12; fs = 12; for j = 1:2 subplot(1,2,j) n = nn(j); a = [1; randn(n,1)]; r = roots(a); plot(r,'.k',MS,ms) axis(1.5*[-1 1 -1 1]), axis square title(['monomial, n=' int2str(n)],FS,fs) end The reason these roots lie near the unit circle is that the monomials 1, z, z^2,... are orthogonal polynomials defined on that set. For orthogonal polynomials defined on a different region of the complex plane, we get roots tending to cluster on the boundary of that region [3]. In particular, the best-known families of orthogonal polynomials are defined over [-1,1], and random polynomials expressed in these bases tend to have roots near [-1,1]. The Chebfun ROOTS command provides an easy way to compute roots of such polynomials stably. For example, here is what happens if we repeat the experiment above but for random polynomials in the basis of Chebyshev polynomials, i.e., p = a_0 + a_1 T_1 + ... + a_n T_n. for j = 1:2 subplot(1,2,j) n = nn(j); a = [1; randn(n,1)]; p = chebfun(a,'coeffs'); r = roots(p,'all','norecurse'); plot(r,'.k',MS,ms) axis(1.5*[-1 1 -1 1]), axis square title(['Chebyshev, n=' int2str(n)],FS,fs) end Legendre polynomials are orthogonal over the same interval [-1,1], so the results aren't much different: for j = 1:2 subplot(1,2,j) n = nn(j); a = [1; randn(n,1)]; A = legpoly(n:-1:0); p = A*a; r = roots(p,'all','norecurse'); plot(r,'.k',MS,ms) axis(1.5*[-1 1 -1 1]), axis square title(['Legendre, n=' int2str(n)],FS,fs) end References: [1] J. Hammersley, The zeros of a random polynomial, Proceedings of the Third Berkeley Symposium on Mathematical Statistics and Probability, 1954-1955, U. California Press, 1956, pp. 89-111. [2] L. A. Shepp and R. J. Vanderbei, The complex zeros of random polynomials, Transactions of the American Mathematical Society 347 (1995), 4365-4384. [3] B. Shiffman and S. Zelditch, Equilibrium distribution of zeros of random polynomials, International Mathematical Research Notices, 2003, pp. 25-49. [4] K.-C. Toh and L. N. Trefethen, Pseudozeros of polynomials and pseudospectra of companion matrices, Numerische Mathematik, 68 (1994), 403-425.

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# Archives November 2021 #### Matlab Genetic Algorithm Toolbox What is the Matlab genetic algorithm toolbox? Genetic algorithm has the advantages of simple thinking and obvious application effect. Experts and scholars in various fields have conducted in-depth research on it, and used C, C++ and other programming languages ​​to implement algorithms. However, these languages ​​require users to write genetic algorithm programs, which brings certain difficulties to researchers who are not familiar with programming languages. The genetic algorithm toolbox of Matlab can realize the operation of genetic algorithm through a graphical user interface (GUI) . The problem can be quickly defined by writing a small amount of fitness function program and setting the corresponding parameters in the toolbox. Flexible, easy to use, and easy to modify parameters. This article takes the genetic algorithm toolbox in Matlab7.1 version as an example to introduce the structure and parameter settings of the toolbox. Type “gatool” in the Matlab working window. After the command is run, you can open the Genetic Algorithm Tool box, or find and open the tool box in the Start menu in Matlab. The operation interface is shown in Figure. The genetic algorithm toolbox is divided into 5 parts from left to right, mainly including defining function handles and variable numbers, running displays (Plots), constraints (Constraints), running commands and results (Run Solver), parameter settings (Options), etc. (1) When solving actual problems, first determine the fitness function of the problem, and write it as an M file and store it on the working path of Matlab. Fill in the handle of the compiled fitness function in Fitness Function, the format is “@funtname”, and enter the number of variables to be solved in Number of vari. (2) Constraints include constraints such as Linear inequalities, Linear equalities, and Bounds. For example, boundary constraints limit the minimum and maximum values ​​of variables, and the maximum value constraints of multiple variables can be expressed in matrix form. (3) Operation display (Plots) displays the operation process of the selected item in image form during operation. For example, after checking Best fitness, the algorithm operation process will display the best function value and average value in each generation of the group; checking Best individual will display the individual corresponding to the best fitness value under the current iteration number. (4) The run command and result (Run solver) includes operation buttons such as run, pause, stop, etc. The number of iterations and operation status will be displayed during the running process, and the final optimization result will be displayed in the final point after the algorithm stops. (5) Parameter settings (Options) mainly affect the calculation speed and accuracy of the algorithm, mainly including population (Population), fitness scale (Fitness scaling), selection (Selection), mutation (Mutation), stop condition (Stopping Criteria), Crossover (Crossover) and other parameter settings. The relevant parameters can be selected and set through the drop-down menu. #### Genetic Algorithm Fitting What is genetic algorithm fitting? Genetic Algorithm (Genetic Algorithm, GA) is a global optimization algorithm, which is based on the evolutionary theory of natural selection and genetics to find the optimal solution to the function problem. It is not only suitable for general fitting problems, but also can solve the traditional fitting methods. Deal with non-linear and highly complex data fitting problems. Therefore, this paper uses genetic algorithm to fit the test data obtained from the lithium-ion power battery cycle life test, and obtain the optimal coefficient and equation between the capacity retention rate and the number of cycles. Genetic algorithm (GA) simulates the evolutionary process of gene recombination and mutation in the process of biological reproduction. When solving actual problems, the potential solution of the problem (randomly generated) is used as the initial population of the algorithm. The population is used as an individual after binary coding by a computer. These individuals perform operations similar to natural selection, crossover, and mutation in biological evolution, and reproduce according to the rule of survival of the fittest, and finally obtain the optimal individual that meets the convergence condition is the optimization result of the problem. When the genetic algorithm is used for curve fitting, the algorithm uses group evolution to process multiple individuals at the same time. It does not need to know the derivative of the problem to be sought, does not depend on the complexity of the problem, and the initial population, recombination, mutation and other operations in the genetic algorithm It is performed randomly, which can avoid the local optimum in the optimization process and achieve the global optimum effect, and can effectively solve the optimal estimation problems such as polynomial coefficients. The main calculation steps of genetic algorithm are: ① Randomly generate an initial population based on actual problems. ②Use the designed fitness function to calculate the fitness of the individual. ③ Perform operations such as selection, crossover, and mutation. Selection refers to the use of certain methods to select individuals with high adaptability to inherit to the next generation according to the fitness of the individual. Crossover refers to selecting two individuals with higher probability from the group, and exchanging part of the genes in each pair of individuals. Commonly used methods such as single-point crossover and multi-point crossover. Variation refers to changing the value of part of the gene position of the individual in the population after crossover with a small probability. ④ Determine whether the convergence conditions are met. When the set convergence conditions are met, the individual with the greatest fitness is the optimal solution to the problem, otherwise, proceed to steps ③ and ④. The fitness function is used to measure the fitness of an individual, which is equivalent to the objective function in actual problems. It can only find the minimum value in the Matlab genetic algorithm toolbox. The genetic algorithm controls the operation of the algorithm through the fitness function, uses the size of the individual fitness to determine the probability of an individual being inherited to the next generation, and then changes the group structure. It is the basis for the algorithm’s natural selection and the driving force of evolution. When the genetic algorithm is used to fit the function, the fitness function f(x) that can be selected is as shown in the formula (formula diagram), where Cmax is a sufficiently large positive number, and ER is the objective function.

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Re: Pattern Matching in Lists • To: mathgroup at smc.vnet.net • Subject: [mg35605] Re: Pattern Matching in Lists • From: "Carl K. Woll" <carlw at u.washington.edu> • Date: Mon, 22 Jul 2002 02:11:05 -0400 (EDT) • References: <001301c23091\$d341ef20\$9b16989e@machine1> • Sender: owner-wri-mathgroup at wolfram.com ```Allan, One further small tweak. s1[w_] := (Tr[#1] - Tr[#1 Drop[w, 1]] & )[Drop[w, -1]] My tweak is to use BitAnd instead of multiplication: s2[w_] := (Tr[#1] - Tr[BitAnd[#1, Drop[w, 1]]] & )[Drop[w, -1]] Testing: tst = Table[Random[Integer], {10^7}]; In[4]:= s1[tst]//Timing s2[tst]//Timing Out[4]= {0.641 Second, 2498910} Out[5]= {0.5 Second, 2498910} So, BitAnd appears to be slightly faster. Carl Woll Physics Dept U of Washington ----- Original Message ----- From: "Allan Hayes" <hay at haystack.demon.co.uk> To: mathgroup at smc.vnet.net Subject: [mg35605] Re: Pattern Matching in Lists > Carl, > Pushing your idea a bit further and avoiding one Drop and a subtraction of > lists: > > w=Table[Random[Integer],{1000000}]; > > Tr[Drop[w,-1](Drop[w,-1]-Drop[w,1])]//Timing > > {7.85 Second,249850} > > (Tr[#]-Tr[# Drop[w,1]])&[Drop[w,-1] ]//Timing > > {2.75 Second,249850} > > -- > Allan > > --------------------- > Allan Hayes > Mathematica Training and Consulting > Leicester UK > www.haystack.demon.co.uk > hay at haystack.demon.co.uk > Voice: +44 (0)116 271 4198 > Fax: +44 (0)870 164 0565 > > > "Carl K. Woll" <carlw at u.washington.edu> wrote in message > news:ahdfd2\$i3i\$1 at smc.vnet.net... > > Anthony and newsgroup, > > > > I thought of another method of solving this problem which is 5 or 6 > times > > faster than my previous version. Here is a test case. > > > > tst=Table[Random[Integer],{1000000}]; > > > > My first solution was > > > > In[16]:= > > Count[Partition[tst,2,1],{1,0}]//Timing > > Out[16]= > > {0.704 Second, 249722} > > > > My second solution is > > > > In[17]:= > > Tr[Drop[tst,-1](Drop[tst,-1]-Drop[tst,1])]//Timing > > Out[17]= > > {0.125 Second, 249722} > > > > Carl Woll > > Physics Dept > > U of Washington > > > > "Anthony Mendes" <amendes at zeno.ucsd.edu> wrote in message > > news:ah5qce\$59o\$1 at smc.vnet.net... > > >Hello, > > > > > > Suppose w={1,1,1,0,0,1,0,1,0,0,1,0,0}. > > > > > > How can I count the number of occurrences of a 1 in w immediately > > > followed by a 0 in w? > > > > > > I have tried every incarnation of Count[] I can think of; for example, > > > > > > Count[w,{___,1,0,___}] > > > > > > does not seem to work. In general, how can I count the number of > > > occurrences of a 1 followed by a 0 in a list of 1's and 0's? Thank you! > > > > > > > > > -- > > > Tony > > > _____________________ > > > amendes at math.ucsd.edu > > > > > > > > > > > > > > > > > > > > > ``` • Prev by Date: RE: Re: Pattern Matching in Lists • Next by Date: RE: Re: Pattern Matching in Lists • Previous by thread: RE: Re: Pattern Matching in Lists • Next by thread: RE: Re: Pattern Matching in Lists

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html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; } Sec http://functions.wolfram.com/01.11.21.0068.01 Input Form Integrate[z^n E^(p z) Cos[a + b z] Sec[c z], z] == E^((-I) a + ((-I) b + I c + p) z) n! Sum[(1/(-j + n)!) ((-1)^j ((-I) b + I c + p)^(-1 - j) z^(-j + n) HypergeometricPFQ[{Subscript[b, 1], \[Ellipsis], Subscript[b, j + 1], 1}, {1 + Subscript[b, 1], \[Ellipsis], 1 + Subscript[b, j + 1]}, -E^(2 I c z)]), {j, 0, n}] + E^(I a + (I b + I c + p) z) n! Sum[(1/(-j + n)!) ((-1)^j (I b + I c + p)^(-1 - j) z^(-j + n) HypergeometricPFQ[{Subscript[c, 1], \[Ellipsis], Subscript[c, j + 1], 1}, {1 + Subscript[c, 1], \[Ellipsis], 1 + Subscript[c, j + 1]}, -E^(2 I c z)]), {j, 0, n}] /; Subscript[b, 1] == Subscript[b, 2] == \[Ellipsis] == Subscript[b, n + 1] == -((b - c + I p)/(2 c)) && Subscript[c, 1] == Subscript[c, 2] == \[Ellipsis] == Subscript[c, n + 1] == (b + c - I p)/(2 c) && Element[n, Integers] && n >= 0 Standard Form Cell[BoxData[RowBox[List[RowBox[List[RowBox[List["\[Integral]", RowBox[List[SuperscriptBox["z", "n"], " ", SuperscriptBox["\[ExponentialE]", RowBox[List["p", " ", "z"]]], RowBox[List["Cos", "[", RowBox[List["a", "+", RowBox[List["b", " ", "z"]]]], "]"]], RowBox[List["Sec", "[", RowBox[List["c", " ", "z"]], "]"]], " ", RowBox[List["\[DifferentialD]", "z"]]]]]], "\[Equal]", " ", RowBox[List[RowBox[List[SuperscriptBox["\[ExponentialE]", RowBox[List[RowBox[List[RowBox[List["-", "\[ImaginaryI]"]], " ", "a"]], "+", RowBox[List[RowBox[List["(", RowBox[List[RowBox[List[RowBox[List["-", "\[ImaginaryI]"]], " ", "b"]], "+", RowBox[List["\[ImaginaryI]", " ", "c"]], "+", "p"]], ")"]], " ", "z"]]]]], " ", RowBox[List["n", "!"]], " ", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["j", "=", "0"]], "n"], RowBox[List[FractionBox["1", RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["-", "j"]], "+", "n"]], ")"]], "!"]]], RowBox[List["(", RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], "j"], " ", SuperscriptBox[RowBox[List["(", RowBox[List[RowBox[List[RowBox[List["-", "\[ImaginaryI]"]], " ", "b"]], "+", RowBox[List["\[ImaginaryI]", " ", "c"]], "+", "p"]], ")"]], RowBox[List[RowBox[List["-", "1"]], "-", "j"]]], " ", SuperscriptBox["z", RowBox[List[RowBox[List["-", "j"]], "+", "n"]]], " ", RowBox[List["HypergeometricPFQ", "[", RowBox[List[RowBox[List["{", RowBox[List[SubscriptBox["b", "1"], ",", "\[Ellipsis]", ",", SubscriptBox["b", RowBox[List["j", "+", "1"]]], ",", "1"]], "}"]], ",", RowBox[List["{", RowBox[List[RowBox[List["1", "+", SubscriptBox["b", "1"]]], ",", "\[Ellipsis]", ",", RowBox[List["1", "+", SubscriptBox["b", RowBox[List["j", "+", "1"]]]]]]], "}"]], ",", RowBox[List["-", SuperscriptBox["\[ExponentialE]", RowBox[List["2", " ", "\[ImaginaryI]", " ", "c", " ", "z"]]]]]]], "]"]]]], ")"]]]]]]]], "+", RowBox[List[SuperscriptBox["\[ExponentialE]", RowBox[List[RowBox[List["\[ImaginaryI]", " ", "a"]], "+", RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["\[ImaginaryI]", " ", "b"]], "+", RowBox[List["\[ImaginaryI]", " ", "c"]], "+", "p"]], ")"]], " ", "z"]]]]], " ", RowBox[List["n", "!"]], " ", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["j", "=", "0"]], "n"], RowBox[List[FractionBox["1", RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["-", "j"]], "+", "n"]], ")"]], "!"]]], RowBox[List["(", RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], "j"], " ", SuperscriptBox[RowBox[List["(", RowBox[List[RowBox[List["\[ImaginaryI]", " ", "b"]], "+", RowBox[List["\[ImaginaryI]", " ", "c"]], "+", "p"]], ")"]], RowBox[List[RowBox[List["-", "1"]], "-", "j"]]], " ", SuperscriptBox["z", RowBox[List[RowBox[List["-", "j"]], "+", "n"]]], " ", RowBox[List["HypergeometricPFQ", "[", RowBox[List[RowBox[List["{", RowBox[List[SubscriptBox["c", "1"], ",", "\[Ellipsis]", ",", SubscriptBox["c", RowBox[List["j", "+", "1"]]], ",", "1"]], "}"]], ",", RowBox[List["{", RowBox[List[RowBox[List["1", "+", SubscriptBox["c", "1"]]], ",", "\[Ellipsis]", ",", RowBox[List["1", "+", SubscriptBox["c", RowBox[List["j", "+", "1"]]]]]]], "}"]], ",", RowBox[List["-", SuperscriptBox["\[ExponentialE]", RowBox[List["2", " ", "\[ImaginaryI]", " ", "c", " ", "z"]]]]]]], "]"]]]], ")"]]]]]]]]]]]], "/;", RowBox[List[RowBox[List[SubscriptBox["b", "1"], "\[Equal]", SubscriptBox["b", "2"], "\[Equal]", "\[Ellipsis]", "\[Equal]", SubscriptBox["b", RowBox[List["n", "+", "1"]]], "\[Equal]", RowBox[List["-", FractionBox[RowBox[List["b", "-", "c", "+", RowBox[List["\[ImaginaryI]", " ", "p"]]]], RowBox[List["2", " ", "c"]]]]]]], "\[And]", RowBox[List[SubscriptBox["c", "1"], "\[Equal]", SubscriptBox["c", "2"], "\[Equal]", "\[Ellipsis]", "\[Equal]", SubscriptBox["c", RowBox[List["n", "+", "1"]]], "\[Equal]", FractionBox[RowBox[List["b", "+", "c", "-", RowBox[List["\[ImaginaryI]", " ", "p"]]]], RowBox[List["2", " ", "c"]]]]], "\[And]", RowBox[List["n", "\[Element]", "Integers"]], "\[And]", RowBox[List["n", "\[GreaterEqual]", "0"]]]]]]]] MathML Form z n p z cos ( a + b z ) sec ( c z ) z - a + ( c - b + p ) z n ! j = 0 n ( - 1 ) j ( c - b + p ) - j - 1 z n - j ( n - j ) ! j + 2 F j + 1 ( - b - c + p 2 c , , - b - c + p 2 c , 1 ; 1 - b - c + p 2 c , , 1 - b - c + p 2 c ; - 2 c z ) TagBox[TagBox[RowBox[List[RowBox[List[SubscriptBox["\[InvisiblePrefixScriptBase]", FormBox[RowBox[List["j", "+", "2"]], TraditionalForm]], SubscriptBox["F", FormBox[RowBox[List["j", "+", "1"]], TraditionalForm]]]], "\[InvisibleApplication]", RowBox[List["(", RowBox[List[TagBox[TagBox[RowBox[List[TagBox[RowBox[List["-", FractionBox[RowBox[List["b", "-", "c", "+", RowBox[List["\[ImaginaryI]", " ", "p"]]]], RowBox[List["2", " ", "c"]]]]], HypergeometricPFQ], ",", TagBox["\[Ellipsis]", HypergeometricPFQ], ",", TagBox[RowBox[List["-", FractionBox[RowBox[List["b", "-", "c", "+", RowBox[List["\[ImaginaryI]", " ", "p"]]]], RowBox[List["2", " ", "c"]]]]], HypergeometricPFQ], ",", TagBox["1", HypergeometricPFQ]]], InterpretTemplate[Function[List[SlotSequence[1]]]]], HypergeometricPFQ], ";", TagBox[TagBox[RowBox[List[TagBox[RowBox[List["1", "-", FractionBox[RowBox[List["b", "-", "c", "+", RowBox[List["\[ImaginaryI]", " ", "p"]]]], RowBox[List["2", " ", "c"]]]]], HypergeometricPFQ], ",", TagBox["\[Ellipsis]", HypergeometricPFQ], ",", TagBox[RowBox[List["1", "-", FractionBox[RowBox[List["b", "-", "c", "+", RowBox[List["\[ImaginaryI]", " ", "p"]]]], RowBox[List["2", " ", "c"]]]]], HypergeometricPFQ]]], InterpretTemplate[Function[List[SlotSequence[1]]]]], HypergeometricPFQ], ";", TagBox[RowBox[List["-", SuperscriptBox["\[ExponentialE]", RowBox[List["2", " ", "\[ImaginaryI]", " ", "c", " ", "z"]]]]], HypergeometricPFQ]]], ")"]]]], InterpretTemplate[Function[HypergeometricPFQ[Slot[1], Slot[2], Slot[3]]]]], HypergeometricPFQ] + a + ( c + b + p ) z n ! j = 0 n ( - 1 ) j ( c + b + p ) - j - 1 z n - j ( n - j ) ! j + 2 F j + 1 ( b + c - p 2 c , , b + c - p 2 c , 1 ; 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n Condition z z n p z a b z c z -1 a c -1 b p z n j 0 n -1 j c -1 b p -1 j -1 z n -1 j n -1 j -1 HypergeometricPFQ -1 b -1 c p 2 c -1 -1 b -1 c p 2 c -1 1 1 -1 b -1 c p 2 c -1 1 -1 b -1 c p 2 c -1 -1 2 c z a c b p z n j 0 n -1 j c b p -1 j -1 z n -1 j n -1 j -1 HypergeometricPFQ b c -1 p 2 c -1 b c -1 p 2 c -1 1 b c -1 p 2 c -1 1 b c -1 p 2 c -1 1 -1 2 c z n [/itex] Rule Form Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List["\[Integral]", RowBox[List[RowBox[List[SuperscriptBox["z_", "n_"], " ", SuperscriptBox["\[ExponentialE]", RowBox[List["p_", " ", "z_"]]], " ", RowBox[List["Cos", "[", RowBox[List["a_", "+", RowBox[List["b_", " ", "z_"]]]], "]"]], " ", RowBox[List["Sec", "[", RowBox[List["c_", " ", "z_"]], "]"]]]], RowBox[List["\[DifferentialD]", "z_"]]]]]], "]"]], "\[RuleDelayed]", RowBox[List[RowBox[List[RowBox[List[SuperscriptBox["\[ExponentialE]", RowBox[List[RowBox[List[RowBox[List["-", "\[ImaginaryI]"]], " ", "a"]], "+", RowBox[List[RowBox[List["(", RowBox[List[RowBox[List[RowBox[List["-", "\[ImaginaryI]"]], " ", "b"]], "+", RowBox[List["\[ImaginaryI]", " ", "c"]], "+", "p"]], ")"]], " ", "z"]]]]], " ", RowBox[List["n", "!"]], " ", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["j", "=", "0"]], "n"], FractionBox[RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], "j"], " ", SuperscriptBox[RowBox[List["(", RowBox[List[RowBox[List[RowBox[List["-", "\[ImaginaryI]"]], " ", "b"]], "+", RowBox[List["\[ImaginaryI]", " ", "c"]], "+", "p"]], ")"]], RowBox[List[RowBox[List["-", "1"]], "-", "j"]]], " ", SuperscriptBox["z", RowBox[List[RowBox[List["-", "j"]], "+", "n"]]], " ", RowBox[List["HypergeometricPFQ", "[", RowBox[List[RowBox[List["Join", "[", RowBox[List[RowBox[List["Table", "[", RowBox[List[RowBox[List["-", FractionBox[RowBox[List["b", "-", "c", "+", RowBox[List["\[ImaginaryI]", " ", "p"]]]], RowBox[List["2", " ", "c"]]]]], ",", RowBox[List["{", RowBox[List["K\$1", ",", "1", ",", RowBox[List["1", "+", "j"]]]], "}"]]]], "]"]], ",", RowBox[List["{", "1", "}"]]]], "]"]], ",", RowBox[List["Join", "[", RowBox[List["Table", "[", RowBox[List[RowBox[List["1", "-", FractionBox[RowBox[List["b", "-", "c", "+", RowBox[List["\[ImaginaryI]", " ", "p"]]]], RowBox[List["2", " ", "c"]]]]], ",", RowBox[List["{", RowBox[List["K\$1", ",", "1", ",", RowBox[List["1", "+", "j"]]]], "}"]]]], "]"]], "]"]], ",", RowBox[List["-", SuperscriptBox["\[ExponentialE]", RowBox[List["2", " ", "\[ImaginaryI]", " ", "c", " ", "z"]]]]]]], "]"]]]], RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["-", "j"]], "+", "n"]], ")"]], "!"]]]]]]], "+", RowBox[List[SuperscriptBox["\[ExponentialE]", RowBox[List[RowBox[List["\[ImaginaryI]", " ", "a"]], "+", RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["\[ImaginaryI]", " ", "b"]], "+", RowBox[List["\[ImaginaryI]", " ", "c"]], "+", "p"]], ")"]], " ", "z"]]]]], " ", RowBox[List["n", "!"]], " ", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["j", "=", "0"]], "n"], FractionBox[RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], "j"], " ", SuperscriptBox[RowBox[List["(", RowBox[List[RowBox[List["\[ImaginaryI]", " ", "b"]], "+", RowBox[List["\[ImaginaryI]", " ", "c"]], "+", "p"]], ")"]], RowBox[List[RowBox[List["-", "1"]], "-", "j"]]], " ", SuperscriptBox["z", RowBox[List[RowBox[List["-", "j"]], "+", "n"]]], " ", RowBox[List["HypergeometricPFQ", "[", RowBox[List[RowBox[List["Join", "[", RowBox[List[RowBox[List["Table", "[", RowBox[List[FractionBox[RowBox[List["b", "+", "c", "-", RowBox[List["\[ImaginaryI]", " ", "p"]]]], RowBox[List["2", " ", "c"]]], ",", RowBox[List["{", RowBox[List["K\$1", ",", "1", ",", RowBox[List["1", "+", "j"]]]], "}"]]]], "]"]], ",", RowBox[List["{", "1", "}"]]]], "]"]], ",", RowBox[List["Join", "[", RowBox[List["Table", "[", RowBox[List[RowBox[List["1", "+", FractionBox[RowBox[List["b", "+", "c", "-", RowBox[List["\[ImaginaryI]", " ", "p"]]]], RowBox[List["2", " ", "c"]]]]], ",", RowBox[List["{", RowBox[List["K\$1", ",", "1", ",", RowBox[List["1", "+", "j"]]]], "}"]]]], "]"]], "]"]], ",", RowBox[List["-", SuperscriptBox["\[ExponentialE]", RowBox[List["2", " ", "\[ImaginaryI]", " ", "c", " ", "z"]]]]]]], "]"]]]], RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["-", "j"]], "+", "n"]], ")"]], "!"]]]]]]]]], "/;", RowBox[List[RowBox[List["n", "\[Element]", "Integers"]], "&&", RowBox[List["n", "\[GreaterEqual]", "0"]]]]]]]]]] Date Added to functions.wolfram.com (modification date) 2002-10-15

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Joe - 1 year ago 112 SQL Question # SQL calculate points between a time range I’ve a table which contains different time ranges: ``````Id Start Time End Time Points 1 0:00 3:00 10 2 3:01 6:00 20 3 6:01 23:59 30 `````` Now I need to calculate the points achieved between two date ranges with respect to time specified. ``````Start date = 11/9/2016 18:17:00 and End date = 11/10/2016 01:20:00 `````` I need to calculate the sum of points gained between these two dates. The time of start date that is 18:17 falls under Id 3, whose point is 30. So the calculation will be, ``````18:17 to 23:59 -> 6 hrs -> 6 * 30 = 180 points `````` The end time 01:20 falls under Id 1 ``````0:00 to 1:20 -> 2 hrs (if minute is greater than zero, it is rounded to next hour, ie; 2) -> 2 * 10 = 20 points `````` So the total points gained will be 200 points. Taking the time difference, does not help me, if the start and end date difference is greater than one day. Table Structure: Id - int, StartTime - time(7), EndTime - time(7), Points - int How to write a query for this using SQL? This question was good. You can as the below: ``````DECLARE @Tbl TABLE (Id INT, StartTime TIME, EndTime TIME, Points INT) INSERT INTO @Tbl VALUES (1, '0:00', '3:00' , 10), (2, '3:01', '6:00' , 20), (3, '6:01', '23:59', 30) DECLARE @StartDate DATETIME = '2016.11.09 18:17:00' DECLARE @EndDate DATETIME = '2016.11.10 01:20:00' ;WITH CTE AS ( SELECT 1 AS RowId, @StartDate CurrentDate, 0 Point, @StartDate DateVal UNION ALL SELECT A.RowId , IIF((A.CurrentDate + A.EndTime) > @EndDate, @EndDate, DATEADD(MINUTE, 1, (A.CurrentDate + A.EndTime))) AS CurrentDate, A.Points, IIF((A.CurrentDate + A.EndTime) > @EndDate, @EndDate, (A.CurrentDate + A.EndTime)) DateVal FROM ( SELECT C.RowId + 1 AS RowId, CAST(CAST(CurrentDate AS DATE) AS DATETIME) CurrentDate, CAST((SELECT T.EndTime FROM @Tbl T WHERE CAST(CurrentDate AS TIME) BETWEEN T.StartTime AND T.EndTime) AS DATETIME) AS EndTime, (SELECT T.Points FROM @Tbl T WHERE CAST(CurrentDate AS TIME) BETWEEN T.StartTime AND T.EndTime) AS Points, C.CurrentDate AS TempDate FROM CTE C ) A WHERE A.TempDate <> IIF((A.CurrentDate + A.EndTime) > @EndDate, @EndDate, DATEADD(MINUTE, 1, (A.CurrentDate + A.EndTime))) ), CTE2 AS ( SELECT C.RowId , C.CurrentDate , C.Point , C.DateVal, DATEDIFF(MINUTE, LAG(C.DateVal) OVER (ORDER BY C.RowId), C.DateVal) MinuteOfDateDiff FROM CTE C ) SELECT SUM(CEILING(C.MinuteOfDateDiff * 1.0 / 60.0) * C.Point) FROM CTE2 C `````` Result: `200` Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download

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# 水塘抽样 Leetcode 382 Given a singly linked list, return a random node’s value from the linked list. Each node must have the same probability of being chosen. What if the linked list is extremely large and its length is unknown to you? Could you solve this efficiently without using extra space? Example: ## Random Pick Index Leetcode 398 Given an array of integers with possible duplicates, randomly output the index of a given target number. You can assume that the given target number must exist in the array. Note: The array size can be very large. Solution that uses too much extra space will not pass the judge. Example: # 线段树 ## Count of Smaller Numbers After Self Leetcode 315 You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i]. Example: # 子串和满足条件的最短长度 ## Minimum Size Subarray Sum Leetcode 209 Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn’t one, return 0 instead. Example: ## 上一题数组中包含正负，找和为target的最小长度 Leetcode 862 Return the length of the shortest, non-empty, contiguous subarray of A with sum at least K. If there is no non-empty subarray with sum at least K, return -1. Example 1: Example 2: Example 3: # 快排相关 ## 快排优化 • 基准随机化算法 • 将前后中间三个数取中值作为标兵元素 • 在数组大小等于一定范围的时候，改为插入排序，防止排序退化 • 将相等的数字聚集到一起，然后跳过此处的排序 # 牛顿法 Leetcode 69 Implement int sqrt(int x). Compute and return the square root of x, where x is guaranteed to be a non-negative integer. Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned. Example 1: Example 2: # 快速幂 ## Pow(x, n) Leetcode 50 Implement pow(x, n), which calculates x raised to the power n (xn). # 全排列相关 Leetcode 60 ## 字母全排列 1. 从最后向前找相邻的两个元素是的i < ii 2. 然后再从最后向前找一个元素使得i < j 3. i和j交换，然后将ii之后的元素reverse排序 nextPermutation的头文件是algorithm M= | |A |B |C | |—-|:–:|:–:|:–:| |A |B,C |C |A | |B |A,C |C |C | |C |A |A |A,B | S={A,B,C}

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Select Program Area --Select Program Area-- DESE HOME Accountability, Partnership, & Assistance Adult & Community Learning Amazing Educators BESE Advisory Councils Board of Elementary & Secondary Education Career/Vocational Technical Education Charter Schools College and Career Readiness Conferences, Workshops and Trainings Instructional Support Digital Learning District & School Turnaround District Review, Analysis, & Assistance Tools Educator Evaluation Educator Licensure Tests (MTEL) Educator Licensure Educational Proficiency Plan (EPP) Edwin ELAR Log In Employment Opportunities: DESE English Language Learners Every Student Succeeds Act (ESSA) Family Literacy Federal Grant Programs High School Equivalency (HSE) Testing Program Grants/Funding Opportunities Information Services Laws & Regulations Literacy LEAP Project MCAS MCAS Appeals METCO Office for Food and Nutrition Programs Office of Approved Special Education Schools (OASES) Performance Assessment for Leaders (PAL) Planning and Research Problem Resolution System (PRS) Professional Development Public School Monitoring (PSM) RETELL Safe and Supportive Schools School and District Profiles/Directory School Finance School Redesign Science, Technology Engineering, and Mathematics (STEM) Security Portal | MassEdu Gateway Special Education Special Education Appeals Special Education in Institutional Settings Statewide System of Support Student and Family Support Systems for Student Success (SfSS) Title I Part A Students & Families Educators & Administrators Teaching, Learning & Testing Data & Accountability Finance & Funding About the Department Education Board # Massachusetts Comprehensive Assessment System ## 2013, High School Chemistry Question 23: Open-Response Reporting Category: Properties of Matter and ThermochemistryStandard: 1.2 - Explain the difference between pure substances (elements and compounds) and mixtures. Differentiate between heterogeneous and homogeneous mixtures. Particle models of four different types of matter are shown in the diagram below.Identify which of the four models best represents a pure compound. Explain your answer and give a specific example of a compound.Identify which of the four models best represents a homogeneous mixture. Explain your answer and give a specific example of a homogeneous mixture.Describe one method that could be used to separate a homogeneous mixture. ### Scoring Guide and Sample Student WorkSelect a score point in the table below to view the sample student response. ScoreDescription 4 The response demonstrates a thorough understanding of the differences among pure substances, homogeneous mixtures, and heterogeneous mixtures. The response correctly identifies the particle models that correspond to a pure compound and to a homogeneous mixture, clearly explains the answers, and gives specific examples of each. The response also clearly describes one method that could be used to separate a homogeneous mixture. 4 3 The response demonstrates a general understanding of the differences among pure substances, homogeneous mixtures, and heterogeneous mixtures. 2 The response demonstrates a limited understanding of the differences among pure substances, homogeneous mixtures, and heterogeneous mixtures. 1 The response demonstrates a minimal understanding of the differences among pure substances, homogeneous mixtures, and heterogeneous mixtures. 0 Response is incorrect or contains some correct work that is irrelevant to the skill or concept being measured. Note: There are 2 sample student responses for Score Point 4. Question 11: Question 23: Question 33: Question 44: Question 45:

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# Official October 14 SAT Prediction of Scores <p>I predict that this will be the breakdown of my score: 720 M +50 680 CR +160 750 W with 11/12 essay +220</p> <hr> <p>2150</p> <p>ok 790 M 700 CR (overly opimistic) 730 W w/ 12/12 essay</p> <p>2220</p> <p>800M 760CR(missed 3-4) 800W(48/49 MC, 11 Essay)</p> <p>780-math 670-writing <p>atleast im hoping and praying</p> <p>M: 720 Cr: 660-670 W: 640+</p> <p>Could someone predict for me?</p> <p>CR: 0-1 wrong W: 3 wrong, 8-12 essay M: 3 omit, 2 wrong</p> <p>M:710-740 optmistically, depending on which sec. was exp. CR:760-800 W:780-800</p> <p>im feeling a perfect score on W, but i dont want to jinx it :)</p> <p>i think Cr 780 M 730 W 750 12 essay</p> <p>total 2300</p> <p>2400 of course thats optimistic, thats what we all hope we got</p> <p>I'm going to say I would be happy with any increase 2010 or better would make me very happy</p> <p>740 M 720 CR (Maybe...missed at least 2) 740 W</p> <p>Fervently hoping!</p> <p>I'd rather not predict my own score, just a superstition. Could someone do it for me using harshest curve imaginable?</p> <p>CR: 0-1 wrong W: 3 wrong, 8-12 essay M: 3 omit, 2 wrong</p> <p>I think your math is incorrect, AznN3rd.</p> <p>LOL!</p> <p>Well, it was his or her worst prediction...</p> <p>W: 760 (essay: really awesome but unfinished. I heard the max score for an unfinished essay is 8...?)</p> <p>CR: 800 (realistically)</p> <p>M: 700 (optimistically)</p> <p>Is there a section of CC for crappy kids like me?</p> <p>5940</p> <p>'nuff said.</p> <p>Math: 760-780 Writing 750-800 CR: 670-740</p> <p>2180-2320</p> <p>i'm not the average CCer...</p> <p>Math: 680 (last time)~~~~>650-700 (math sucked) Verbal: 610~~~~>600-650 Writing: 550~~~~>600-650</p> <p>I prepped for the verbal/writing, the math was murder... blah</p> <p>Math- 620 CR-640 Writing- 680</p> <p>Math - 760 CR - 680 W-770 2210</p>

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1. ## Triple Integrals Here is my question Evaluate the triple integral where is bounded by the parabolic cylinder and the planes and . I set up dV to be dzdydx and the bounds to be {z,0,16-y^2}, I am confused on what to do for y, and {x,-4,4}. I tried {y,-4,4} because that is where the function crosses the z=0 plane, but when I evaluated it, it ended up to be incorrect. Any help would be greatly appreciated. 2. Try and learn how to draw them in Mathematica then put it into the form: $\displaystyle \int_{x=a}^{x=b}\int_{y=f(x)}^{y=g(x)}\int_{z=f_1( x,y)}^{z=f_2(x,y)} u(x,y,z)dzdydx$ So z goes from the x-y plane up to the function $\displaystyle f(x,y)=16-y^2$, y goes from the two "function" of x, y=-4 and y=4 and then x goes from the range of -4 to 4: $\displaystyle \int_{-4}^{4}\int_{-4}^{4}\int_{0}^{4-y^2} x^2 e^y dzdzdx$

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## MAS280 Mechanics and Fluids Note: This is an old module occurrence. You may wish to visit the module list for information on current teaching. Semester 2, 2018/19 10 Credits Lecturer: Dr Ashley Willis uses MOLE Timetable Reading List Aims Assessment This course extends the principles of Newtonian mechanics from the mechanics of particles to the mechanics of three-dimensional (3D) bodies, first to the motion of solids, then to the motion of fluids (e.g. air and water). Mathematically, this course develops the vector calculus, the essential tool for describing a 3D world. The suffix notation and the Einstein summation convention are introduced as an alternative notation for the calculus. Properties of the vector gradient operator are first seen as a natural description for potential forces. For more advanced use of the vector calculus, the most intuitive setting to comprehend its properties is through the study of fluid motion. Topics include 3D extensions of the work-energy principle, planetary and satellite motion, elements of the motion of rigid bodies, and the motion of inviscid (frictionless) fluids. Prerequisites: MAS112 (Vectors and Mechanics); MAS211 (Advanced Calculus and Linear Algebra) (Alternative prerequisite: PHY120 or PHY165) The following modules have this module as a prerequisite: MAS310 Continuum Mechanics MAS320 Fluid Mechanics I MAS422 Magnetohydrodynamics ## Outline syllabus • Centre of mass and moment of inertia. • The work-energy principle using the gradient operator. • Planetary orbits. • Motion of a rigid body. • Suffix notation and a review of grad, div, curl and the integral theorems. • Kinematics of fluid motions. • Euler's equations of motion for an inviscid fluid. • Irrotational flows and Bernoulli's principle. ## Office hours I'm usually in my office, but more free on Thursdays. Please email to be sure I'll be in. ## Aims • To extend a working knowledge of Newtonian mechanics to a broader range of contexts. • To develop and expand ability to use the vector calculus. • To extend an understanding of planetary motions and the motion of rigid bodies. • To recognise and understand key features of fluids motion. 20 lectures, 5 tutorials ## Assessment 2 Assessed Homeworks at 5% each, 2-hour formal exam 90%. Type Author(s) Title Library Blackwells Amazon A Acheson Elementary Fluid Dynamics A C. Collinson and T. Roper Particle Mechanics A D.N. Burghes Further Mechanics A P. Dyke and R.W. Whitworth Guide to Mechanics A Paterson A First Course in Fluid Dynamics (A = essential, B = recommended, C = background.) Most books on reading lists should also be available from the Blackwells shop at Jessop West. ## Timetable Thu 15:00 - 15:50 tutorial (group 3b) (odd weeks) Hicks Lecture Theatre 9 Fri 09:00 - 09:50 tutorial (group 1b) (odd weeks) Hicks Lecture Theatre 9 Fri 09:00 - 09:50 tutorial (group 2b) (odd weeks) Hicks Seminar Room F24

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Towards a theory of derived braids and their categories. I. Braid relation, ribbons, and the opposite of an enriched category. II. Products of enriched categories and derived braids. III. Questions and conjectures. We begin by describing a geometrical reformulation of the braid relation. The first slide shows this process. So the Yang-Baxter braid relation can be seen as the equivalence of under and overhanded twists. Next we generalize. Now switch gears and think about categories: first the basic definitions. We can't neglect to relate the Y-B equation to naturality. Now for a further level of definitions: enriched categories. There are structures defined on V-Cat if there is structure on V. First we define the opposite of a V-category if V is braided, and show that it is a valid V-category. Recall that Joyal and Street proved coherence for braided categories: if two legs of a diagram made up of the canonical natural transformations have the same underlying braid then the two legs are equal--the diagram commutes. We note that there is a reverse sort of op called "co" that is completely analogous but that uses c^-1. Now another important structure: product on V-Cat. We note that there is another product, tensor', that uses c^-1. We can now generate lots of braid equalities by combining the two structures of opposite and product enriched categories. For instance, the next slide shows the left and right derived braids underlying the two legs of the pentagon when the enriched category involved is A^opop tensor B^opop. It also includes the instructions for forming Lx and Rx in B_6 for x in B_4 . The next family of braids come from composing 4 hom-objects in the product of 2 enriched categories: A^opopopop tensor B^opop. The lesson to be learned is that if the braid x in B_2n has Lx=Rx in B_3n then the categorical interpretation implies that if x underlies a composition morphism then it has Lx=Rx in B_kn for all k>2 in N. In fact we can describe sufficient conditions for Lb to be equivalent to Rb by describing the braids b that underlie the composition morphism of a product category given generally by ((({\cal A}^{op})^{...op}\otimes^{(1)} ({\cal B}^{op})^{...op})^{op})^{...op} where the number of op exponents is arbitrary in each position. Those braids are alternately described as lying in H\sigma_2K \subset B_4 where H is the cyclic subgroup generated by the braid \sigma_2\sigma_1\sigma_3\sigma_2 and K is the subgroup generated by the two generators {\sigma_1, \sigma_3}. We use \sigma for the classical braid group generators. The latter subgroup K is isomorphic to Z\times Z. The first coordinate corresponds to the number of op exponents on A and the second component to the number of op exponents on B (when positive-to co when negative). The positive power of the element of H corresponds to the number of op exponents on the product of the two enriched categories, that is, the number of op exponents outside the parentheses. That b is in H\sigma_2K implies Lb=Rb follows from the fact that the composition morphisms belonging to the opposite of a V-category obey the pentagon axiom. An exercise of some value is to check consistency of the definitions by constructing an inductive proof of the implication based on braid group generators. This is not a necessary condition. Here is a picture of x in B_4 such that Lx=Rx. When the number n of enriched categories in a product category increases, the number of braids in B_(2n) such that Lx=Rx in B_(3n) grows quickly, using op, co, various parenthizations, and both tensor and tensor' to combine the enriched categories. Questions include: Which of these braids can be used in place of \sigma_2 in order to define an associative product on V-Cat? My conjecture is none of them (see section 3 of "Enrichment as categorical delooping I.") Are there any braids in B_2n with Lx=Rx but that do not either underlie a product of enriched categories or do so after joining some adjacent strands which always remain adjacent? How do we construct and interpret the odd n analogues of all this? Another conjecture: In B_2n consider the following braids: We conjecture that the allowed combinations ("decreasing products") obey Lx=Rx. Back to research page.

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# You reserve 2/5 of the seats on a tour bus. You are able to fill 5/8 of the seats you ###### Question: You reserve 2/5 of the seats on a tour bus. You are able to fill 5/8 of the seats you reserve. What fraction of the seats on the bus are you able to fill? #### Similar Solved Questions ##### Combine equations (1). (2),and (3) to obtain equation (4) {These chemical equations are obtained from the first page of the handout} (1) MgOls) + 2 HCIaq) = MgClzlaq) Hzo() Mg (s) 2 HCIlaq) = MgClz(aq) - Hz(g)Hzlg) 12 Oz(g) HzO() Toobtain: (4) Mgls) 1/2 Ozlg) = MgO(s)(2) + (1) - (31(41- (2) - (31- (1)(4)= (2) + (3) . (1)(41- (21+ (31 + (1) Combine equations (1). (2),and (3) to obtain equation (4) {These chemical equations are obtained from the first page of the handout} (1) MgOls) + 2 HCIaq) = MgClzlaq) Hzo() Mg (s) 2 HCIlaq) = MgClz(aq) - Hz(g) Hzlg) 12 Oz(g) HzO() Toobtain: (4) Mgls) 1/2 Ozlg) = MgO(s) (2) + (1) - (31 (41- (2) - (31... ##### 1. Suppose you are interested in estimating the following model: Yi = Bo + B11; +u;... 1. Suppose you are interested in estimating the following model: Yi = Bo + B11; +u; where B, is the coefficient of interest in the model. Unfortunately, you have reason to believe that couci, u) +0. Suppose you also have data on another variable, zi, and its relationship to I, is as follows: Di = 0o... ##### A closed, rigid container contains an ice-water mixture at 0â—¦C.Heat is added slowly and some, but not all, of the ice melts.Didthe internal energy increase or de-crease?it decreased.2.It increased.3.It remained the same.4.Unable todetermine A closed, rigid container contains an ice-water mixture at 0â—¦C. Heat is added slowly and some, but not all, of the ice melts.Did the internal energy increase or de-crease? it decreased.2.It increased.3.It remained the same.4.Unable to determine... ##### GC,Jo"Hnd *sclantist monsured Itv sprcd 01 Ighl HI valuoa Dro Kvboc wd Ja0u 0l 104 58 Find 0 90* conlidunce Inlervnl (or Ina Irua sped ol Iiglu hon Ihose salyte Stale words Ihat thE Intorval meang Kaop in mind Lnat the spond ol ligh is ntyaic wnal useumpuons [Tuet YOu mnkc Tnale mathod? Soed ol Ughtis ((721 12]*m/e 7a1 25 0) A 907 confidence intaryal ior Ine Inue (Round ID tro decimal paces as nended Uso uscending orcer }Luncd GtuulouIhrocon C GC,Jo" Hnd * sclantist monsured Itv sprcd 01 Ighl HI valuoa Dro Kvboc wd Ja0u 0l 104 58 Find 0 90* conlidunce Inlervnl (or Ina Irua sped ol Iiglu hon Ihose salyte Stale words Ihat thE Intorval meang Kaop in mind Lnat the spond ol ligh is ntyaic wnal useumpuons [Tuet YOu mnkc Tnale mathod? Soed ... ##### Calculate the amplitude, period and frequency of the voltage waveform shown in the graph. The amplitude, period and frequency of the voltage waveform shown in the graph below is:... ##### Use induction to prove thatfor any natural number n and for any real number such that r F 1. 3. Use induction to prove that for any natural number n,5i-2- =0 Use induction to prove that for any natural number n and for any real number such that r F 1. 3. Use induction to prove that for any natural number n, 5i-2- =0... ##### Crowe Company uses the aging method to adjust the allowance for uncollectible accounts at the end... Crowe Company uses the aging method to adjust the allowance for uncollectible accounts at the end of the period. At December 31, 2017, the balance of accounts receivable is $230,000 and the allowance for uncollectible accounts has a credit balance of$6,000 (before adjustment). An analysis of accoun... ##### Y+4xy = 4xSolve by Integrating Factor16 Y+4xy = 4x Solve by Integrating Factor 16... ##### Is a contribution to a qualified tuition plan (a code section 529 plan) is treated as... Is a contribution to a qualified tuition plan (a code section 529 plan) is treated as a gift is a present interest... ##### Sodium metal reacts with water t0 produce hydrogen gas according ZNa(s) 2H,O() ~ZNaOH(aQ) Hz(g)following equation:The product gas, Hz is collected over water temperature 0f 25 PC and pressure of 743 mm Hg: If the wet Hz gas formed occupics volume of 7.96 the number of moles of Ni reacted Was mol. The vapor pressure of water is 23.8 mm Hg 4 25 'C.Submit AnswerRetry Entire Groupmore group attempts remalning Sodium metal reacts with water t0 produce hydrogen gas according ZNa(s) 2H,O() ~ZNaOH(aQ) Hz(g) following equation: The product gas, Hz is collected over water temperature 0f 25 PC and pressure of 743 mm Hg: If the wet Hz gas formed occupics volume of 7.96 the number of moles of Ni reacted Was mol. ... ##### A 2.0 -L kettle contains $116 \mathrm{g}$ of boiler scale $\left(\mathrm{CaCO}_{3}\right) .$ How many times would the kettle have to be completely filled with distilled water to remove all the deposit at $25^{\circ} \mathrm{C} ?$ A 2.0 -L kettle contains $116 \mathrm{g}$ of boiler scale $\left(\mathrm{CaCO}_{3}\right) .$ How many times would the kettle have to be completely filled with distilled water to remove all the deposit at $25^{\circ} \mathrm{C} ?$... You are asked to evaluate the following two projects for the Norton corporation. Use a discount rate of 12 percent. Use Appendix B for an approximate answer but calculate your final answer using the formula and financial calculator methods. Project X (Videotapes of the Weather Report) ($36,000 Inves... 1 answer ##### The return on common stockholders' equity is computed by dividing The return on common stockholders' equity is computed by dividingnet income by average common stockholders' equity.net income minus preferred dividends by average common stockholders' equity.net income minus preferred dividends by ending common stockholders' equity.net income by endi... 5 answers ##### (c) Determine if the function f(x)is uniformly continuous on the interval [2,0)_ (c) Determine if the function f(x) is uniformly continuous on the interval [2,0)_... 5 answers ##### Dad Alg A2q_ Ei Alu AzuC 3C2256 304 linear; rotationsquadratic TY7(Rn, Ru)(xz, Yz Dad Alg A2q_ Ei Alu Azu C 3C2 256 304 linear; rotations quadratic TY7 (Rn, Ru) (xz, Yz... 5 answers ##### Give the domain of f, the domain of g, and the domain of m, where m(x) = f[g(x)]: f(u) = In u; g(x) 725-x2The domain of f is C,0) (Type your answer in interval notation:)The domain of g is (Type your answer in interval notation:)The domain of m is (Type your answer in interval notation:) Give the domain of f, the domain of g, and the domain of m, where m(x) = f[g(x)]: f(u) = In u; g(x) 725-x2 The domain of f is C,0) (Type your answer in interval notation:) The domain of g is (Type your answer in interval notation:) The domain of m is (Type your answer in interval notation:)... 5 answers ##### Consider the circle shown in the fiqure_!x(m)What is the diameter of the circle? Please, notice that the circle passes through a number of grid intersection points_ Consider the circle shown in the fiqure_ ! x(m) What is the diameter of the circle? Please, notice that the circle passes through a number of grid intersection points_... 4 answers ##### Usc the follawing information t0 answer Ihc next questionDuring an acid-base titration , ditlerent indicalors are used t0 check Ihe end points ol the solution. There are four indicators gJiven wilh Iheir respeclive pH range, and colaur cach solution.IndicatorColorCresolredyellow redMethyl orangeThymol blueredPhenalphthaleincolaurlessThe approximate pH of the solution (Record your answer in the numerical-response section below)Your answer:&678 Usc the follawing information t0 answer Ihc next question During an acid-base titration , ditlerent indicalors are used t0 check Ihe end points ol the solution. There are four indicators gJiven wilh Iheir respeclive pH range, and colaur cach solution. Indicator Color Cresolred yellow red Methyl oran... 1 answer ##### Patricia was called at work by a woman at the local daycare center. She told Patricia... Patricia was called at work by a woman at the local daycare center. She told Patricia to come and pick up her son because he was not feeling well. Her son, 3½-year-old Marshall, had been feeling tired and achy when he woke up. While at daycare, his cheeks had become red and he was warm to tou... 1 answer ##### Who was the leader who emerged for Haiti during its revolution? Who was the leader who emerged for Haiti during its revolution?... 5 answers ##### EnitKatutnRedox Titrationbtuecanticd46i4 2 coletion?UhAn0.463 ofoxalare unkntown 1nd h takes 24,6 mL of Icaeucniucichs Out orlnte unkoxn cholthc Audent enlution I0 tIrulc Fnuch eemngunale USc 35.0 mL of permnanpinatc solution? Vcmh ouin Grdtsolution AL uecq Hl umnlc 0l oxuilic Ic3+.6 mL 0f 0.0231 perunganate mmolcs or xmc An the unknoun? how mn EnitKatutn Redox Titration btue canticd 46i4 2 coletion? UhAn 0.463 ofoxalare unkntown 1nd h takes 24,6 mL of Icaeucniucichs Out orlnte unkoxn cholthc Audent enlution I0 tIrulc Fnuch eemngunale USc 35.0 mL of permnanpinatc solution? Vcmh ouin Grdt solution AL uecq Hl umnlc 0l oxuilic Ic3+.6 mL 0f ... 1 answer ##### 1) Discuss problems related to respiratory diseases that are a result of smoking. 2) Very young... 1) Discuss problems related to respiratory diseases that are a result of smoking. 2) Very young children are being diagnosed with ADD or ADHD. Parents may first hear of concerns about their child from a school counselor or nurse as a result of a teacher referral. Often parents are urged to have thei... 5 answers ##### Homework: Assignment 6 Daa dRatonEanutd0387: 5096 , 48.5 = 99 pta6.1.16-T Fruta indclayeue MarehonOctcuoxon Helo >Erardrrml dtlt Duudeandulndar duyaJbonTh mlcattu [ Aooln MHondentendneeennudud Entnt LOatetJnsne Lx 4Nd Inun clkck Cchpnna anotymne Homework: Assignment 6 Daa d Raton Eanutd 0387: 5096 , 48.5 = 99 pta 6.1.16-T Fruta indclayeue MarehonOct cuoxon Helo > Erardrrml dtlt Duude andulndar duyaJbon Th mlcattu [ Aooln MHondentendneeen nudud Entnt LOatet Jnsne Lx 4Nd Inun clkck Cch pnna anotymne... 1 answer ##### (Please show all work) Each capacitor has a capacitance C = 12.0 ?F. (a) By combining... (Please show all work) Each capacitor has a capacitance C = 12.0 ?F. (a) By combining capacitors in series and in parallel, find the equivalent capacitance of this circuit. (b) If V = 10.0 volts, calculate Q, what would be the charge on each plate of the equivalent capacitor? C2 C1... 1 answer ##### Finding the Area of a Region In Exercises (a) use a graphing utility to graph the region bounded by the graphs of the equations, (b) explain why the area of the region is difficult to find by hand, and (c) use the integration capabilities of the graphing utility to approximate the area to four decimal places. $y=\sqrt{\frac{x^{3}}{4-x}}, y=0, x=3$ Finding the Area of a Region In Exercises (a) use a graphing utility to graph the region bounded by the graphs of the equations, (b) explain why the area of the region is difficult to find by hand, and (c) use the integration capabilities of the graphing utility to approximate the area to four deci... 1 answer ##### Financial Accounting, Seventh Canadian Edition by Kimmel, Weyg Help System Announcements Question 8 One example of... Financial Accounting, Seventh Canadian Edition by Kimmel, Weyg Help System Announcements Question 8 One example of a liability that is not a financial liability is O notes payable. o unearned revenue. O bonds payable. financial lease.... 5 answers ##### Name the shape of the graph of the equation resulting from the new quadratic formn and v after getting its standard formn_ llr2 + 2v7ry + Sy? = 48213 Name the shape of the graph of the equation resulting from the new quadratic formn and v after getting its standard formn_ llr2 + 2v7ry + Sy? = 48 213... 1 answer ##### 2. (40 points) For the loaded determinate truss with pin reaction at support E and roller... 2. (40 points) For the loaded determinate truss with pin reaction at support E and roller reaction at support A: (30 pts) fine the forces in members , DC, DG, and HG using the method of sections through section aa. Indicate whether these forces are in compression (C) or tension (T). (10 pts) list an... 5 answers ##### If radius 20 feet, Object travels in horizontal circle at constant 30 feet/second: what is acceleration of object? conditions? Is answer any different if object moves in vertical loop with same If radius 20 feet, Object travels in horizontal circle at constant 30 feet/second: what is acceleration of object? conditions? Is answer any different if object moves in vertical loop with same... 2 answers ##### Point) Are tthe " vectorsandIinearly independent?ChooseIfthey are lineany dependent; find scalars that are not all zero such that the equation below is true. If they are Iinearly Independent; find tha orly scalars tattal make the equation below tnue point) Are tthe " vectors and Iinearly independent? Choose Ifthey are lineany dependent; find scalars that are not all zero such that the equation below is true. If they are Iinearly Independent; find tha orly scalars tattal make the equation below tnue... 1 answer ##### The questions are attached. Please provide detailed solutions to the following problems/exercises (2 problems/exercises x 8... The questions are attached. Please provide detailed solutions to the following problems/exercises (2 problems/exercises x 8 points each) 1) The table given below gives the output and labor hour figures for a firm manufacturing toys Week Output (in units)Labor Hours 1 1,850 1,361 2,122 2,638 2,599 2,... 5 answers ##### 04 6ku ml 7o (4o"4 0 (j Lt T0E 2k. (Gej SR -> mU40q4~U 1e-k LJ( 34.Tt6J Lox V = 1 (* ) 22Jjl 46 s1di) 1ct, WVMeu ecjt Ci ( Ut JF c 'U (; 4e5 4hic+ls, Wi+ 56du) 0 toml 4lcs 6| 0 ( k AFkee Mu bot Kli dw Co Il+$ , OL (l ecut el 3o ml sh4 CWe il 4lch blec IWJ €J clt ajml, (ululus I Co P [ 6 04 6ku ml 7o (4o "4 0 (j Lt T0E 2k. (Gej SR -> mU 40 q4~ U 1 e-k L J( 34.Tt6J Lox V = 1 (* ) 22Jjl 46 s1di) 1ct, WVMeu ecjt Ci ( Ut JF c 'U (; 4e5 4hic+ls, Wi+ 56du) 0 toml 4lcs 6| 0 ( k AFkee Mu bot Kli dw Co Il+ \$ , OL (l ecut el 3o ml sh4 CWe il 4lch blec IWJ €J clt ajml, (ulu... ##### 5) R2 The Wheatstone bridge is used for precision measurements of resistive transducers. If the bridge... 5) R2 The Wheatstone bridge is used for precision measurements of resistive transducers. If the bridge is balanced (Vaut 0), show that R,xR4-R2xR3. Ri A B VoUT R3...

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1 JEE Main 2020 (Online) 9th January Evening Slot +4 -1 The following system of linear equations 7x + 6y – 2z = 0 3x + 4y + 2z = 0 x – 2y – 6z = 0, has A no solution B infinitely many solutions, (x, y, z) satisfying y = 2z C infinitely many solutions, (x, y, z) satisfying x = 2z D only the trivial solution 2 JEE Main 2020 (Online) 9th January Morning Slot +4 -1 Out of Syllabus If the matrices A = $$\left[ {\matrix{ 1 & 1 & 2 \cr 1 & 3 & 4 \cr 1 & { - 1} & 3 \cr } } \right]$$, B = adjA and C = 3A, then $${{\left| {adjB} \right|} \over {\left| C \right|}}$$ is equal to : A 8 B 2 C 72 D 16 3 JEE Main 2020 (Online) 9th January Morning Slot +4 -1 If for some $$\alpha$$ and $$\beta$$ in R, the intersection of the following three places x + 4y – 2z = 1 x + 7y – 5z = b x + 5y + $$\alpha$$z = 5 is a line in R3, then $$\alpha$$ + $$\beta$$ is equal to : A -10 B 0 C 10 D 2 4 JEE Main 2020 (Online) 8th January Evening Slot +4 -1 The system of linear equations $$\lambda$$x + 2y + 2z = 5 2$$\lambda$$x + 3y + 5z = 8 4x + $$\lambda$$y + 6z = 10 has A a unique solution when $$\lambda$$ = –8 B no solution when $$\lambda$$ = 2 C infinitely many solutions when $$\lambda$$ = 2 D no solution when $$\lambda$$ = 8 EXAM MAP Medical NEET

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# Tag Info ## Hot answers tagged scheduling 8 If I understood you correctly, you could do it like this (assuming that $n$ is even and $m = n-1$) First, add the binary variables \begin{align} h_{is} &= \begin{cases} 1, &\text{if team i plays at home in round s and s-1}, \\ 0, &\text{otherwise}, \end{cases} \\\\ % a_{is} &= \begin{cases} 1, &\text{if team i plays away in ... 7 Modern CPUs are very complex and have at least two features that limit their scaling capability. The first one is a turbo feature that increases the clock speed when not all cores are utilized. The second one is that all cores share the same memory bus and the same L2 and L3 cache. If you solve the same problem in parallel (so start Python twice and let each ... 7 You need to distinguish between threads and (physical) cores. Is it possible that the cores you see in your machine are actually just hyperthreads, i.e. 2 cores resemble one physical core? Furthermore, using many cores is not always very helpful to solve a MIP. You may want to try something like Concurrent Optimization in Gurobi to exploit performance ... 7 joni's answer is correct. However, this formulation will not allow you to find an optimal solution for anything more than 10-12 teams, even without any additional typical sports scheduling constraints. If you aren't using a commercial solver, the limit is even lower. There is an excellent book about round robin scheduling by Dirk Briskorn. It is a must-read ... 5 Let E be the set of employees, and let P be the set of periods. For e\in E and p\in P, let binary decision variable x_{e,p} indicate whether employee e goes to the company in period p. Let G be the set of groups that must go together at least once, and for g\in G, let E_g \subseteq E be the set of employees in group g. For g\in G, ... 5 If only a subset of nodes is to be transfered, and that the cardinality of this subset is undefined, then I agree with @LocalSolver. Otherwise (if all nodes have to be transfered 1 by 1), I believe the problem is not NP-hard (nor NP-complete): Consider the following graph : Create a first layer with the n nodes. Create a second layer with n \times n ... 5 This is a blocking job shop scheduling problem. The description from "An iterated greedy metaheuristic for the blocking job shop scheduling problem" (Pranzo et Pacciarelli, 2016) DOI In the job shop scheduling problem a set of jobs J must be processed on a set of machines M, each processing at most one job at a time. The processing of a job on ... 4 What are flow based formulations ? Flow based formulations can be used when working with networks. The classical approach is to define a variable for each edge of the network. In a flow based formulation, you basically perform a change of variables, and define a variable for each possible path/flow of the network. For what optimization problems are they ... 4 Start by defining the appropriate binary variable: x_{ij}=1 if and only if task i is assigned to resource j. A given task can only be assigned to one resource: \sum_j x_{ij}=1 \quad \forall i $$Daily capacity for each resource:$$ \sum_i \Delta_i x_{ij}\le 8 \quad \forall j $$(\Delta_i denotes the duration of task i) task t_3 must be ... 4 Why it is recommended to compare the relaxed versions of each formulation to deduce the running time (and more precisely about the B&B tree size)? Generally, better (= tighter) is the relaxation of an integer optimization model, better should work a brand-and-bound tree solution approach to this model. Nevertheless, there exist some NP-hard problems for ... 4 Disclaimer : this is more of a hint than a complete answer. You can use the following model as a starting point to make your own model. I am ignoring two items : Constraint from option 3: Under this option, he also must paint one year after the Option 3 repair at the same additional cost of 1M/boat. Surface area constraints You will have to tweak what ... 4 There are certainly different ways of achieving what you want. Here is how I would proceed: Start by predefining the set of all possible schedules which satisfy your constraints 2,3,4,6. Although there are many, I believe that with your constraints, it may be not too difficult to derive them somewhat automatically. Here is a subset of them in the table ... 4 Can anyone think about a better formulation? Another option is to use binary variables x_{it} that take value 1 if task i starts at time t. You then need two sets of constraints: one start time per task:$$ \sum_{t}x_{it} = 1 \quad \forall i $$don't overlap tasks:$$ \sum_{i}\sum_{k, t+1 - d_i \le k \le t}x_{ik} \le 1 \quad \forall t This ... 4 Since you are unfamiliar with OR, I would recommend using discrete event simulation, which I think is the easiest approach for a newcomer (although it may require some programming chops, depending on what software you use). You will need a bit more than just average service completion times -- you will want a distribution of service times. (In the absence of ... 3 Yes, your proposal suffices. But the published second constraint is stronger, yielding a tighter formulation. You can think of it as a lifting obtained by using the first constraint. 3 A. Schaerf wrote an academic 40-page survey on the topic in 1999: A Survey of Automated Timetabling. Artificial Intelligence Review 13, 87–127 (1999). https://doi.org/10.1023/A:1006576209967. You can find the PDF file here. In this survey, Schaerf discusses school timetabling, course timetabling, examination timetabling, and related scheduling problems. He ... 3 Such a problem is difficult to model and solve following a MILP approach, as you observed. Boolean modeling approaches are tedious to write and don't scale well for this kind of problem. Your problem can be modeled compactly by following a list-based modeling approach instead of the classical Boolean modeling approach, as you described in your question. This ... 3 since you deal with scheduling on top of MIP you could try CPOptimizer scheduling. For instance you could start with https://github.com/AlexFleischerParis/howtowithopl/blob/master/tspcpo.mod using CP; int n = ...; range Cities = 1..n; int realCity[i in 1..n+1]=(i<=n)?i:1; // Edges -- sparse set tuple edge {int i; int j;} setof(... 3 In OptaPlanner, I'd start from the task assigning example and use a shadow variable to track the remaining battery level of every task. Then a simple constraint can check if it's ok to schedule that task on that machine: from(Task.class) .filter(task -> task.getRemainingBatteryLevel() < task.getRequiredBatteryLevel()) .penalize("Battery", ... 3 Without going into a ton of detail, the usual approach is to include a variable for the start time of each task, and a variable for the end time of each task. (Your binary assignment variables are also included.) Your objective (known as the makespan) is the maximum of the task end times. The end time for a task is the start time plus the processing time (... 3 As nicely mentioned above by @Kuifje, the problem can be reduced to a minimum-cost maximum flow problem if all the nodes have to be transferred. If only a subset of the nodes has to be transferred then this is a graph partitioning problem known: partitioning the vertices of a graph into two subsets such that the weight of the cut between the two subsets is ... 3 The initial gap can be an indicator but not a very good one. I have seen many problems start with a tight bound and then never improve, and I have also seen many problems start with a horrendous gap that improves very quickly. Honestly, there is no way to know in advance, we just have to try solving all formulations we can come up with, until we find one ... 3 Yes, this is correct and is the classical approach from Manne, On the Job-Shop Scheduling Problem (1960). In some modeling languages, you can also enforce these implications by using indicator constraints: \begin{align} y = 0 &\implies t_i + d_i \le t_k \\ y = 1 &\implies t_k + d_k \le t_i \\ \end{align} 3 You may also use CPOptimizer within CPLEX that contains scheduling high level concepts. And then you can directly use noOverlap constraints. In using CP; dvar interval i size 5; dvar interval k size 4; dvar sequence seq in append(i,k); minimize maxl(endOf(i),endOf(k)); subject to { noOverlap(seq); } the constraint noOverlap(seq); makes sure that i ... 3 I didn't find a way to express the transition constraints so i give a description what this does and mention what it lacks. using JuMP using Gurobi #needs Gurobi license but any other MILP solver callable from JuMP should work too## Heading ## using UnicodePlots are dependencies. I used UnicodePlots for debugging. It is neat. In Julia dependencies can be ... 3 I'm not sure there is name for your specific problem, but I think it is safe to say that it falls into the umbrella category of job shop scheduling, with the objective of minimizing makespan. If you do a web search for "taxonomy of job shop models" you will find a barely finite number of diagrams and articles on the subject. This paper, for ... 3 For what it's worth, we've been gathering a bunch of Design Patterns in chapter 20 of OptaPlanner's User Guide. Here are some of the drawings: There's also a video that explains this deeper. Besides these modeling basics, there are orthogonal features to consider (document in other places in the user guide): Pinning: allow the user to lock in a shift ... 2 It is an unrelated parallel machine scheduling problem with sequence-dependent family setup times. The tools define the families; the materials are family members. The tool that is on a machine completely defines the state of the machine. In the machine scheduling literature the problem might be denoted as Q|sij|Cmax It's unrelated parallel machines because ... 2 You can treat it as multiobjective if you treat end-to-end latency as one objective (to be minimized) and load on device 1 (which has limited capacity) as another objective (also to be minimized). 2 Model without bandwidth limitation We wish to select the most performing components to be hosted by two devices in order to have an end-to-end latency as minimum as possible. Let $x_{i,j}$ be a Boolean variable whose value is 1 if i-th component is assigned to j-th device, 0 otherwise where $i=1,2, \cdots, 10$ and $i=1,2$. The cpu limitation (equals to \$... Only top voted, non community-wiki answers of a minimum length are eligible

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https://socratic.org/questions/what-is-the-simplified-form-of-square-root-of-10-square-root-of-5-over-square-ro

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# What is the simplified form of square root of 10 - square root of 5 over square root of 10 + square root of 5? Sep 12, 2015 (sqrt(10)-sqrt(5))/(sqrt(10)+sqrt(5)=3-2sqrt(2) #### Explanation: (sqrt(10)-sqrt(5))/(sqrt(10)+sqrt(5) $\textcolor{w h i t e}{\text{XXX}} = \frac{\cancel{\sqrt{5}}}{\cancel{\sqrt{5}}} \cdot \frac{\sqrt{2} - 1}{\sqrt{2} + 1}$ $\textcolor{w h i t e}{\text{XXX}} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \cdot \frac{\sqrt{2} - 1}{\sqrt{2} - 1}$ color(white)("XXX")=(sqrt(2)-1)^2/((sqrt(2)^2-1^2) $\textcolor{w h i t e}{\text{XXX}} = \frac{2 - 2 \sqrt{2} + 1}{2 - 1}$ $\textcolor{w h i t e}{\text{XXX}} = 3 - 2 \sqrt{2}$

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F11 Chapter Contents F11 Chapter Introduction NAG Library Manual # NAG Library Routine DocumentF11GFF Note:  before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details. ## 1  Purpose F11GFF is the third in a suite of three routines for the iterative solution of a symmetric system of simultaneous linear equations (see Golub and Van Loan (1996)). F11GFF returns information about the computations during an iteration and/or after this has been completed. The first routine of the suite, F11GDF, is a setup routine, the second routine, F11GEF is the proper iterative solver. These three routines are suitable for the solution of large sparse symmetric systems of equations. ## 2  Specification SUBROUTINE F11GFF ( ITN, STPLHS, STPRHS, ANORM, SIGMAX, ITS, SIGERR, WORK, LWORK, IFAIL) INTEGER ITN, ITS, LWORK, IFAIL REAL (KIND=nag_wp) STPLHS, STPRHS, ANORM, SIGMAX, SIGERR, WORK(LWORK) ## 3  Description F11GFF returns information about the solution process. It can be called both during a monitoring step of the solver F11GEF, or after this solver has completed its tasks. Calling F11GFF at any other time will result in an error condition being raised. For further information you should read the documentation for F11GDF and F11GEF. ## 4  References Golub G H and Van Loan C F (1996) Matrix Computations (3rd Edition) Johns Hopkins University Press, Baltimore ## 5  Parameters 1:     ITN – INTEGEROutput On exit: the number of iterations carried out by F11GEF. 2:     STPLHS – REAL (KIND=nag_wp)Output On exit: the current value of the left-hand side of the termination criterion used by F11GEF. 3:     STPRHS – REAL (KIND=nag_wp)Output On exit: the current value of the right-hand side of the termination criterion used by F11GEF. 4:     ANORM – REAL (KIND=nag_wp)Output On exit: for CG and SYMMQ methods, the norm ${‖A‖}_{1}={‖A‖}_{\infty }$ when either it has been supplied to F11GDF or it has been estimated by F11GEF (see also Sections 3 and 5 in F11GDF). Otherwise, ${\mathbf{ANORM}}=0.0$ is returned. For MINRES method, an estimate of the infinity norm of the preconditioned matrix operator. 5:     SIGMAX – REAL (KIND=nag_wp)Output On exit: for CG and SYMMQ methods, the current estimate of the largest singular value ${\sigma }_{1}\left(\stackrel{-}{A}\right)$ of the preconditioned iteration matrix $\stackrel{-}{A}={E}^{-1}A{E}^{-\mathrm{T}}$, when either it has been supplied to F11GDF or it has been estimated by F11GEF (see also Sections 3 and 5 in F11GDF). Note that if ${\mathbf{ITS}}<{\mathbf{ITN}}$ then SIGMAX contains the final estimate. If, on final exit from F11GEF, ${\mathbf{ITS}}={\mathbf{ITN}}$, then the estimation of ${\sigma }_{1}\left(\stackrel{-}{A}\right)$ may have not converged; in this case you should look at the value returned in SIGERR. Otherwise, ${\mathbf{SIGMAX}}=0.0$ is returned. For MINRES method, an estimate of the final transformed residual. 6:     ITS – INTEGEROutput On exit: for CG and SYMMQ methods, the number of iterations employed so far in the computation of the estimate of ${\sigma }_{1}\left(\stackrel{-}{A}\right)$, the largest singular value of the preconditioned matrix $\stackrel{-}{A}={E}^{-1}A{E}^{-\mathrm{T}}$, when ${\sigma }_{1}\left(\stackrel{-}{A}\right)$ has been estimated by F11GEF using the bisection method (see also Sections 3, 5 and 8 in F11GDF). Otherwise, ${\mathbf{ITS}}=0$ is returned. 7:     SIGERR – REAL (KIND=nag_wp)Output On exit: for CG and SYMMQ methods, if ${\sigma }_{1}\left(\stackrel{-}{A}\right)$ has been estimated by F11GEF using bisection, $SIGERR=maxσ1k-σ1k-1σ1k,σ1k-σ1k-2σ1k ,$ where $k={\mathbf{ITS}}$ denotes the iteration number. The estimation has converged if ${\mathbf{SIGERR}}\le {\mathbf{SIGTOL}}$ where SIGTOL is an input parameter to F11GDF. Otherwise, ${\mathbf{SIGERR}}=0.0$ is returned. For MINRES method, an estimate of the condition number of the preconditioned matrix. 8:     WORK(LWORK) – REAL (KIND=nag_wp) arrayCommunication Array On entry: the array WORK as returned by F11GEF (see also Section 3 in F11GEF). 9:     LWORK – INTEGERInput On entry: the dimension of the array WORK as declared in the (sub)program from which F11GFF is called (see also Section 5 in F11GDF). Constraint: ${\mathbf{LWORK}}\ge 120$. Note:  although the minimum value of LWORK ensures the correct functioning of F11GFF, a larger value is required by the iterative solver F11GEF (see also Section 5 in F11GDF). 10:   IFAIL – INTEGERInput/Output On entry: IFAIL must be set to $0$, $-1\text{​ or ​}1$. If you are unfamiliar with this parameter you should refer to Section 3.3 in the Essential Introduction for details. For environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\text{​ or ​}1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this parameter, the recommended value is $0$. When the value $-\mathbf{1}\text{​ or ​}\mathbf{1}$ is used it is essential to test the value of IFAIL on exit. On exit: ${\mathbf{IFAIL}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6). ## 6  Error Indicators and Warnings If on entry ${\mathbf{IFAIL}}={\mathbf{0}}$ or $-{\mathbf{1}}$, explanatory error messages are output on the current error message unit (as defined by X04AAF). Errors or warnings detected by the routine: ${\mathbf{IFAIL}}=-i$ On entry, the $i$th argument had an illegal value. ${\mathbf{IFAIL}}=1$ F11GFF has been called out of sequence. For example, the last call to F11GEF did not return ${\mathbf{IREVCM}}=3$ or $4$. Not applicable.

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RACK AND PINION EXAMINATION QUESTION The diagram below displays a platform that is used to lift boxes from one floor to another in a warehouse. The platform is fixed to a rack which operates through gears. As the gears turn the platform moves up or down depending on the direction of rotation. Gear Z is the driver as it is connected directly to a motor. Basic information: Gear Z = 13 teeth Gear Y = 39 teeth Gear X = 13 teeth 1. Name a suitable sensor that could be used to detect the platform as it reaches the desired floor. 2. On the drawing above, sketch the position of the sensor. 3. How is the sensor used? 4. When the motor is turned on it rotates at 240 rev/min (rpm) (i) How many times will gear X turn in one minute? (ii) How far does the platform travel in one minute? When the motor is turned on it rotates at 240 rpm. Therefore, Gear Z turns at 240 rpm. GEAR Z GEAR Y GEAR X 13 teeth 39 teeth 13 teeth 240 rpm

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Cookie Consent by FreePrivacyPolicy.com Search a number 101001 = 3131257 BaseRepresentation bin11000101010001001 312010112210 4120222021 511213001 62055333 7600315 oct305211 9163483 10101001 116997a 124a549 1336c84 1428b45 151edd6 hex18a89 101001 has 8 divisors (see below), whose sum is σ = 136224. Its totient is φ = 66560. The previous prime is 100999. The next prime is 101009. The reversal of 101001 is 100101. 101001 is nontrivially palindromic in base 4. It is a sphenic number, since it is the product of 3 distinct primes. It is a cyclic number. It is not a de Polignac number, because 101001 - 21 = 100999 is a prime. It is a Harshad number since it is a multiple of its sum of digits (3). It is a congruent number. It is not an unprimeable number, because it can be changed into a prime (101009) by changing a digit. It is a pernicious number, because its binary representation contains a prime number (7) of ones. It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 265 + ... + 521. It is an arithmetic number, because the mean of its divisors is an integer number (17028). 2101001 is an apocalyptic number. It is an amenable number. 101001 is a deficient number, since it is larger than the sum of its proper divisors (35223). 101001 is a wasteful number, since it uses less digits than its factorization. 101001 is an odious number, because the sum of its binary digits is odd. The sum of its prime factors is 391. The product of its (nonzero) digits is 1, while the sum is 3. The square root of 101001 is about 317.8065449295. The cubic root of 101001 is about 46.5702487742. Adding to 101001 its reverse (100101), we get a palindrome (201102). Subtracting from 101001 its reverse (100101), we obtain a square (900 = 302). Multiplying 101001 by its reverse (100101), we get a palindrome (10110301101). It can be divided in two parts, 10100 and 1, that added together give a palindrome (10101). The spelling of 101001 in words is "one hundred one thousand, one", and thus it is an iban number. Divisors: 1 3 131 257 393 771 33667 101001

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Home > English > Class 12 > Maths > Chapter > Matrices > Consider a square matrix A=[a_... # Consider a square matrix A=[a_(ij)]_(3times3) of order 3, for which a_(ij)=omega^(i+j), where omega is an imaginary cube root of unity. What is the value of |A| Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams. Updated On: 10-6-2021 Apne doubts clear karein ab Whatsapp par bhi. Try it now. Watch 1000+ concepts & tricky questions explained! 400+ 21 59995346 4.2 K+ 84.3 K+ 3:58 53795392 3.8 K+ 75.7 K+ 1:24 386716 106.8 K+ 168.2 K+ 5:38 14625209 2.6 K+ 51.9 K+ 3:02 31346494 2.2 K+ 44.8 K+ 3:39 8486981 61.4 K+ 118.7 K+ 3:21 1457921 5.0 K+ 100.6 K+ 10:06 1457917 6.0 K+ 119.6 K+ 4:11 3461509 7.6 K+ 151.7 K+ 6:11 1458155 2.7 K+ 54.5 K+ 13:52 1457916 5.7 K+ 114.7 K+ 4:12 52205 8.6 K+ 172.2 K+ 2:09 56869 15.5 K+ 51.3 K+ 3:11 40251744 700+ 14.7 K+ 1:29 53795504 64.3 K+ 64.3 K+ 1:24 Class 12th Matrices Class 12th Matrices Class 12th Matrices Class 12th Matrices Class 12th Matrices Class 12th Matrices Class 12th Matrices Class 12th Matrices Class 12th Matrices Class 12th Matrices

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Search Maps.com Channel Features Travel Deals Travel Alert Bulletin Travel Tools Business Traveler Guide Student Traveler Guide Map & Travel Store Get travel news, alerts, tips, deals, and trivia delivered free to your email in-box. Email Address: tell me more Site Tools Site Map About Maps.com Contact Maps.com Advertise with Maps.com Affiliate Program Order Tracking View Cart Check Out Help Site Map | Help | Home 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: surveyors furlong = 201.168 length (length) Related Measurements: Try converting from "surveyors furlong" to angstrom, barleycorn, cable length, chain (surveyors chain), cloth quarter, digitus (Roman digitus), fermi, furlong (surveyors furlong), Greek cubit, Greek span, hand, Israeli cubit, parasang, pica (typography pica), ri (Japanese ri), rod (surveyors rod), sazhen (Russian sazhen), span (cloth span), spindle, stadium (Roman stadium), or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: surveyors furlong = 5.67 actus (Roman actus), 282.86 archin (Russian archin), .00454215 arpentcan, 440 Biblical cubit, .91666667 cable length, 880 cloth quarter, 5.23E-07 earth to moon (mean distance earth to moon), 1,144,756.81 en (typography en), 9,051.43 finger, 108.67 Greek fathom, 869.37 Greek span, 1,980 hand, 95,040 line, 1,000 link (surveyors link), 201,168,000 micron, 264 pace, 2,640 palm, .03571429 parasang, 663.87 shaku (Japanese shaku), 6,638.73 sun (Japanese sun). 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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A system of plane geometry where the position of a point is described by A point at (0,0) is positioned at the intersection of the x and y axes. Compare with polar geometry. Coordinate geometry is a way of attacking geometrical problems, attempting to reduce them to mere algebra. This is done by describing all the points in the space with a set of coordinates. The idea was developed in the 17th century and is attributed to Rene Descartes. To begin with you choose a system of coordinates. Most of the time old-fashioned Cartesian coordinates will do (at least if you are dealing with an Euclidean space), but it may be that a suitable choice (skewed axes, polar coordinates, cylindrical coordinates or whatever) will make the calculations easier. Then you have to choose your reference: origin, directions of axes etc. Again this is done in a way to make the calculations as simple as possible. Once the coordinates are chosen we can translate the given problem into algebraic terms. The distance between two points is a function of their coordinates, the angles in a triangle are trigonometric functions of the coordinates of the vertices etc. Every condition on the points and lines and surfaces can be interpreted in terms of the coordinates. Sit down and do the algebra and the answer will hopefully come out in the end. In Euclidean geometry the alternative available is to use Euclidean methods: theorems of congruence and similarity and all their derivatives. With your average 2D problem (involving a few triangles, parallel lines and circles) this approach is likely to give a more elegant solution than coordinate methods - provided that you can find it :-) The coordinate approach is in this case something of a brute force method, and often it only shows that something is true, but not why. In general, however, coordinate methods are much more powerful than Euclidean ones, which is what makes Descartes's invention of them so important. By describing curves by functions in terms of their coordinates the full force of calculus can be brought to bear, and problems which are otherwise mindbogglingly difficult are made trivial (go ahead and try finding the volume of the volume of rotation of a parabola without using calculus, like Archimedes did). Log in or register to write something here or to contact authors.

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 If the price f sugar falls by 10%. By how much percentage must the house hold increase its sugar consumption so as not to decrease its expenditure on sugar ? If the price f sugar falls by 10%. By how much percentage must the house hold increase its sugar consumption so as not to decrease its expenditure on sugar ? 1)10% 2)9% 3)9 1/11% 4)11 1/9% • : 189 • : 6 11 1/9%

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Friday April 29, 2016 # Homework Help: Physics Posted by dan on Monday, March 28, 2011 at 2:03am. An object initially at a temperature of 80 C cools by 1 C in a time of 15 s when placed in a 20 C room When the object's temperature reaches 50 C, what is the average rate at which the object cools?

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# MORE PRODUCTS • ## Pneumatic calculator/converter - Flow tube Calculator Pneumatic calculator/converter 2020-06-07 05:00:47 Program information ver. 1.1 Flow tube Field for customer notes Input data ; p1 (bar) p1 > 0 p2 (bar) p2 ≤ p1 Tube diameter: d (mm) d > 0 Tube length: L (m) L > 0 Output: C (l/(s*bar)) Flow Q (Nl/min) Explanation: p1 (bar) Input pressure (relative atmosphere) p2 (bar) Output pressure (relative atmosphere) p1 abs (bar) Input absolute pressure ... • ## Compressed air pipe sizing calculation the Workshop ... Here’s an example calculation using the compressed air pipe sizing table (table 1) and the equivalent pipe length table (table 2). Let’s say we have a rotary screw compressor of 30 kW that can supply 250 Nm3/hour (normal cubic meters per hour). 250 Nm3/hour is the same as 4200 Nl/min (normal liter per minute) or 150 scfpm (standard cubic feet per minute). • ## Calculator: Air Flow Rate through Piping TLV - A Steam ... Online calculator to quickly determine Air Flow Rate through Piping. Includes 53 different calculations. Equations displayed for easy reference. • ## Compressed Air Pipe Line Capacity - Engineering ToolBox Compressed Air Pipe Lines App - free apps for offline use on mobile devices. Applied Pressure - bar. The diagram below can be used to estimate the flow capacity (normal standard air) in a compressed air pipeline with pressure ranging 0.5 - 17 bar. Example - Capacity in Compressed Air Pipe Line. From the diagram above the capacity of a 1" pipe ... • ## Flow Rate Calculator - calculate the flow rate of a pipe Free online Flow Rate calculator which helps you calculate the flow rate of any pipe given its diameter and liquid/gas velocity or its height and width (for a rectangular pipe) and velocity. The pipe flow rate calculator (a.k.a. discharge rate calculator) accepts input in both metric and imperial units: m/s, km/h, ft/s, yd/s, mph, and outputs in both metric units and imperial ones: cu ft, cu ... • ## Cylinder and Valve Sizing - SMC Pneumatics Airflow Conversion. Cylinder Valve Sizing. Measure Conversion. SCFM Conversion . Fieldbus I/O Catalog. Terms potentially extended to customers purchasing over \$2,000 annually. Credit Application W-9 CA Resale Cert Form. SMC > Cylinder and Valve Sizing. Cylinder and Valve Sizing : Cylinder Bore Size Calculator : Available Pressure: PSI Force Required: Pounds Area Required: Square Inches ... • ## Calculators - Pneumatics Online Cylinder Bore Size Calculator (Imperial Units) Cylinder Force Calculator (Imperial Units) Valve Flow Calculator (Imperial Units) ... All information contained within the Pneumatics Online Web Site is the property of Pneumatics Online. Reproduction, redistribution or modification of the information is prohibited without the express written permission of Pneumatics Online. COPYRIGHT ©1997-2014 ... • ## Pipe Air Flow - Parker Hannifin Pipe Air Flow ISSUED: January, 1999 Supersedes: June, 1998 The following pages contain 6 sets of curves for schedule 40 pipe that can be used to help select the appropriate pipe size for pneumatic systems, or given a system, allow system performance to be estimated. Generally accepted practice for sizing piping for pneumatic systems is to use a pressure drop of 10% of gage for nominal pipe ... • 文件大小: 32KB • ## Flow Rate Calculator - Pressure and Diameter Copely - Flow Rate Calculator With this tool, it is possible to easily calculate the average volumetric flow rate of fluids by changing each of the three variables: length, pressure and bore diameter. The effects on the predicted flow rate are then given in three graphs, where in turn two of the variables are kept constant and the flow rate is plotted against a range of values of the third. • ## Pipe diameter and flow rate calculator, online With pipe diameter calculator pipe internal diameter is calculated using the simple relation between flow rate, velocity and cross-section area (Q=vA). To calculate internal pipe diameter, you should only enter flow rate and velocity in corresponding fields in the calculator and click calculate • ## ThePneu . . . . . Book - SMC he Pneu Book is produced by SMC Pneumatics (UK) Ltd and is available to all those with an interest in the theory, development and application of pneumatics technology. The Pneu Book will comprise of a regularly updated series of issues, each of which will be divided into sections for ease of reference, covering a range of subjects. For example, the first issue, supplied with this pack, starts ... • ## Compressed Air Pipe Sizing Chart Infinity Pipe Systems Infinity Pipe Systems Aluminium Compressed Air Pipe Sizing Chart allows you to determine the diameter (mm) of the compressed air pipe main line.. Here are the steps to determine the correct aluminium pipe sizing to suit your needs: Choose the flow rate of the air compressor in the red column. • ## Air Consumption / Required Air flow Capacity Online software to facilitate calculating the air flow capacity necessary to actuate simple or complex pneumatic systems comprised of SMC components. • ## Flow of Air in Pipes Equation and Calculator Engineers ... Flow of Air in Pipes Equation and Calculator. Hydraulic Pneumatics Fluids Design and Engineering Data . Flow of Air in Pipes Equation and Calculator. where: v = velocity of air in feet per second (ft/sec) p = loss of pressure due to flow through the pipes in ounces per square inch (ounces/in 2) d = inside diameter of pipe in inches (in) L = length of pipe in feet (ft) The horsepower required ... • ## Pneumatic calculator/converter - Flow component Pneumatic calculator/converter 2020-06-07 05:16:02 Program information ver. 1.1 Flow component Field for customer notes Input data ; p1 (bar) p1 > 0 p2 (bar) ... • ## Pneumatics Airflow Compressors Pneumatics Airflow Compressors Pneumatics specialise in all aspects of pneumatic equipment including but not limited to: Cylinders; Valves; Regulators; Lubricators; Filters; Drain; Valves; Airline Accessories; Air Tools; Tubing/Hoses; Vacuum Equipment; Fittings; Couplings; Air Knives; Spray Guns; Air Fed Masks; Airflow can provide a full catalogue of all the above items and much more, available on ... • ## Calculator: Air Flow Rate through Piping TLV - A Steam ... Online calculator to quickly determine Air Flow Rate through Piping. Includes 53 different calculations. Equations displayed for easy reference. • ## Flow and Headloss Calculator - Plastic Pipe Shop Ltd A Free Online Calculator for Determining Headloss in Plastic Pipe Systems* This is a basic pipe headloss calculator which enables you to determine the pressure (head) required at one end of a pipe to get the required flow at the other end. This is designed for simple, single-pipe runs and will not work for networks. This calculator uses the ... • ## ASCO Flow Calculator The ASCO flow calculator is a helpful tool. ... Improves Throughput, Reduces Sample Size in Clinical and Analytical Instruments; ASCO Introduces Condensate Drain Valve Assemblies ; ASCO Numatics Announces 2015 Industrial Automation Engineering Scholarship Winners; ASCO Numatics Announces 2014 Industrial Automation Engineering Scholarship Winners; ASCO Extends Composite Solenoid • ## How to Calculate the Correct Compressed Air Pipe Size ... How to Calculate the Correct Compressed Air Pipe Size An air compressor can drastically reduce the time, coordination and stamina that have traditionally been required for a range of manual tasks. With an air compressor , you can link pneumatic tools to the unit and draw pressurized air for sawing, cutting, shearing, sanding and numerous other applications. • ## GAS PIPE SIZING 2016 WEB actual pipe length m total length pressure drop mb pressure loss from the meter to the appliances must be calculated to confirm if it is within the allowable pressure drops 1mb for natural gas / 2mb for proapane and butane. fill in the work out sheet for each section of pipework. then add the pressure drops from the meter to the applaince. • ## How to Calculate Compressed Air Consumption This nozzle prevents blowing chips further into a pipe or out the end of the pipe, which is a safety hazard. Since airflow is directed back toward the operator, personal protective equipment is recommended. Air Jets and Accessories. Air Jets and Accessories. Air Jets utilize the Coanda effect (wall attachment of a high velocity fluid) to produce air motion in their surroundings. A small amount ... • ## Compressed Air Piping Systems AIRpipe Calculator – Compressed Air Piping Systems AIRpipe Calculator ... Use our calculator below to properly size your system! Afterwards feel free to give us a call or fill out a request form to have an AIRpipe representative send you a custom tailored quote to your specifications. Enter compressed air line features: Pressure. pressure value. Pressure unit selection. Flow. flow value. Flow unit selection. Pipe ... • ## Pneumatic Cylinder Air Flow Requirements - CrossCo Calculating the air consumption for a single-acting cylinder is a rather easy calculation, but it is important to understand that the required air flow is dependent not only on the bore area and stroke, but also the cycle rate of the pneumatic cylinder. • ## Duct Flow Calculators - Mechanical Air Supplies LTD MAS Duct Flow Calculators. A big part of any HVAC project is calculating duct flows, pressure losses, converting duct flow volumes etc. In order to help you we have devised our own mini calculation “apps” feel free to use them as much as you want, if you need any help please don’t hesitate to contact our technical gurus at MAS. • ## Airflow Calculator CUI Devices CUI Devices' air flow conversion calculator can be used to convert between common units for volume air flow and air velocity flowing past a point in a specified area of duct. To use the calculator, enter your air velocity or volume air flow and duct area/type, select your units, click calculate air flow and your converted velocity and volume units will be shown. Search CUI Devices' full line ... • ## FLOW RATE CALCULATOR I N S T R U C T I O N S. This ultra calculator is special by allowing you to choose among a great variety of units (6 for diameter and 24 each for velocity and flow rate). Unlike other calculators, you are NOT confined to inputting diameter in inches, velocity in miles per hour, etc. making this calculator • ## Sizing Pneumatics for Performance and Efficiency Cylinder: Size the cylinder to provide an extra 33 to 100% of the calculated required force to overcome pressure losses from other components in the circuit. In this case, design for a cylinder ... • ## Pneumatic Fittings Accessories Air Supplies™ UK Pneumatic Fittings Accessories. Manufacturers and Distributors of Fluid Fittings. With over 20,000 products in stock from a 16'500 square feet hi-tech distribution centre, a bespoke manufacturing service, and a vast range of additional products available from our industry wide contacts, Air Supplies are a leading manufacturer and distributor of fluid and air fittings. • ## PNEUMATIC CONVEYING QUICK SIZING GUIDELINES PNEUMATIC CONVEYING QUICK SIZING GUIDELINES NOTE: ... Line Size CFM Required Possible Conveying Rate HP RQD. 2” 100-150 3,000 LB/HR ~7-1/2 3” 250-300 11,000 LB/HR ~15 4” 450-550 29,000 LB/HR ~ 30 5” 700-840 36,000 LB/HR ~ 40 6” 1000-1200 50,000 LB/HR ~ 60 8” 1750-2100 80,000 LB/HR ~100 10” 2750-3300 120,000 LB/HR ~150 RULE OF THUMB: HP = (CFM X PSI) / 172 SQ. FT. Pipe • ## Calculating and Measuring the Air Requirement - Atlas The calculated reserve capacity is primarily determined by the cost of lost production resulting from a potential compressed air failure. The number of compressors and their mutual size is determined principally by the required degree of flexibility, control system and energy efficiency. • ## Suggested Pipe Size for Compressed Air Flow at 100 PSI ... Suggested Pipe Size for Compressed Air Flow at 100 PSI Length of Run, Feet On a compressed air distribution system, pressure losses greater than 3% are considered excessive, and a well-designed system having a steady rate of air flow is usually designed for not more than a • ## Pipe Flow Software for Pipe Pressure Drop Calculations Pipe Flow Software for professional engineers is used in over 100 countries worldwide, by over 3000 companies and consultants, wherever there is a need to calculate pipe flow and pressure loss. Pipe Flow Software Located at Springfield House, Water Lane , Wilmslow , Cheshire , SK9 5BG , England . • ## Pneumatic Sizing - Festo Welcome to Pneumatic Sizing • ## Air Compressed Pneumatic AIRpipe This page showcases of range of air products, from AIRpipe compressed air pipe systems, to KELM pneumatic products. Order online now for next day delivery. • ## Pipe Flow Wizard Software - Pipe Pressure Drop Find Flow, Find Pressure Loss, Size Pipe Diameters and Calculate Pipe Lengths. Pipe Flow Wizard version 2 now available at pipeflow The original Pipe Flow Wizard program was released over 15 years ago and is no longer supported. • ## Pipe Sizing Calculator, Pipe Size Calculator, Hydraulic ... Oil Pipe Diameter A guide to the appropriate sizing of Hydraulic pipes can be carried out here. It should be noted that the purpose of the pipe must be known, is it Suction Pipe, Return Pipe or Pressure Pipe? A guide to “normal” fluid speeds within these types of pipes are shown below. Calculated values for Fluid Speed v (m/s) should fall within these ranges. To calculate Fluid Speed v(m ... • ## CHAPTER 4 AIR FLOW THROUGH OPENINGS - Aalto CHAPTER 4 AIR FLOW THROUGH OPENINGS Learning Objectives This chapter introduces the equations that describe airflow through openings. After you have studied this chapter and completed the personal feedback form you should be able to: 1. Identify the parameters that describe airflow; 2. Understand how airflow through openings can be approximated by a general mathematical equation; • ## How to Choose the Right Size ... - Fluid-Aire Dynamics Undersized piping restricts air flow and the only way to overcome that restriction is to increase the air pressure. So, if you need to operate your system at 125 PSI just to get 100 PSI at your point of use that means 25 PSI of overpressurization; resulting in 12.5 % of wasted energy costs. Remember this is for the life of the system, not a one-off expense like installing a properly sized and ... • ## Steps to calculate Air Consumption of a Pneumatic 29.06.2017  Steps to calculate Air Consumption of a Pneumatic System Thread starter Plu3m; Start date Jun 27, 2017; Tags ... This idea is coming up because when I calculated the air flow rate using the current pipe size a result shows that compressed air flow is not sufficient for the current operation. The question is "How do I know that my calculation steps utilized are proper?" so I'm here to ask ... • * * The materials of processing: * • Granite • Limestone • Basalt • Pebble • Gravel • Gypsum • Marble • Barite • Quartz • Dolomite • Gold Ore • Copper ore * * Recommended

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Ders Bilgileri #### Ders Tanımı Ders Kodu Yarıyıl T+U Saat Kredi AKTS ANALYSIS AND MODELING OF WIRELESS NETWORKS BSM 532 0 3 + 0 3 6 Dersin Dili Türkçe Dersin Seviyesi Yüksek Lisans Dersin Türü SECMELI Dersin Koordinatörü Doç.Dr. CÜNEYT BAYILMIŞ Dersi Verenler Dersin Yardımcıları Dersin Kategorisi Dersin Amacı The aim of this course is constituted by two parts. In the first part, wireless networks are examined such as characteristics, medium access techniques, topologies, and variety of wireless networks (high speed, local area, personal etc.) In the second part, wireless networks are modeled using by OPNET simulation package. Dersin İçeriği The fundemantals of wireless networks. network simulator usage (OPNET), Basic modelling and performance evaluation usign OPNET Dersin Öğrenme Çıktıları Öğretim Yöntemleri Ölçme Yöntemleri 1 - Explain method and attributes required for wireless communication 1 - 3 - 4 - A - C - 2 - Classify wireless networks 1 - 2 - A - C - 3 - Use OPNET program for simulation 1 - 4 - A - C - D - 4 - Model basic wireless access mechanisms (TDMA, CDMA) for learning medium access in wireless communication 1 - 4 - 16 - A - C - D - 5 - Realize performance evaluation of modeling network systems for analyzing 1 - 4 - 16 - A - C - D - Öğretim Yöntemleri: 1:Lecture 3:Discussion 4:Drilland Practice 2:Question-Answer 16:Project Based Learning Ölçme Yöntemleri: A:Testing C:Homework D:Project / Design #### Ders Akışı Hafta Konular ÖnHazırlık 1 Fundamentals of wireless communication 2 Wireless medium access techniques 3 Wireless network topologies and architectures. 4 Classification of wireless networks (WPAN, WLAN, WSN etc.) 5 1G, 2G, 3G, 4G, 5G systems 6 Performance analysis techniques of network systems (Analytical modeling, Simulation technique) 7 Demonstrating OPNET simulator 8 Modeling fundamentals with OPNET 9 Performance and traffic management 10 Modeling of wireless medium access techniques with OPNET 11 Mobility in wireless network with OPNET 12 Examination of wireless network projects examples. 13 Modelling and analysis of wireless networks with OPNET 14 Project Presentation #### Kaynaklar Ders Notu Ders Kaynakları 1. Nicopolitidis, P., Obaidat, M., S., Papadimitriou, G., I., Pomportsis, A., S., “Wireless Networks”, WILEY, 2003. 2. Molisch, A., “Wireless Communications”, WILEY, 2005 3. Theodore S. Rappaport. “Wireless Communications: Principles and Practice” , Prentice-Hall, 1996 4. R. Jain, "The Art of Computer Systems Performance Analysis: Techniques for Experimental Design, Measurement, Simulation, and Modeling," Wiley, NY, 1991. 5. OPNET help documentation www.opnet.com #### Dersin Program Çıktılarına Katkısı No Program Öğrenme Çıktıları KatkıDüzeyi 1 2 3 4 5 1 ability to access wide and deep information with scientific researches in the field of Engineering, evaluate, interpret and implement the knowledge gained in his/her field of study X 2 ability to complete and implement “limited or incomplete data” by using the scientific methods. X 3 ability to consolidate engineering problems, develop proper method(s) to solve and apply the innovative solutions to them 4 ability to develop new and original ideas and method(s), to develop new innovative solutions at design of system, component or process 5 gain comprehensive information on modern techniques, methods and their borders which are being applied to engineering 6 ability to design and apply analytical, modelling and experimental based research, analyze and interpret the faced complex issues during the design and apply process X 7 gain high level ability to define the required information and data 8 ability to work in multi-disciplinary teams and to take responsibility to define approaches for complex situations 9 systematic and clear verbal or written transfer of the process and results of studies at national and international environments 10 aware of social, scientific and ethical values guarding adequacy at all professional activities and at the stage of data collection, interpretation and announcement 11 aware of new and developing application of profession and ability to analyze and study on those applications X 12 ability to interpret engineering application’s social and environmental dimensions and it’s compliance with the social environment #### Değerlendirme Sistemi YARIYIL İÇİ ÇALIŞMALARI SIRA KATKI YÜZDESİ AraSinav 1 50 Odev 1 15 ProjeTasarim 1 35 Toplam 100 Yıliçinin Başarıya Oranı 50 Finalin Başarıya Oranı 50 Toplam 100 ; ;

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0 # What percent of 80 is 72? Updated: 4/28/2022 Wiki User 13y ago Expressed as a percentage, 72/80 x 100 = 90 percent. Wiki User 13y ago Wiki User 14y ago 8 x 7.2 ie 57.6 Wiki User 12y ago 80 x .72 = 57.6

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# Irrational Approximations on a Number Line Using the numbers 1 to 9, each only once, label the following number line. Graph the roots of the first nine roots (e.g. √1, √2, … √9) on your number line. What do you notice?  What do you wonder? (I’m wondering how I can make a good OM problem out of this, should I keep the boundaries of 1 to 9 ?  Have students find a root that lies between each integer?) ## 2 thoughts on “Irrational Approximations on a Number Line” 1. Enter digits from 1 to 9 to make the following list in order from smallest to greatest and so the roots do not simplify. rt(__ __), __, rt(__ __), __, rt(__ __) (more open-ended than open middle) • not so bad for open middle, thanks for the idea John

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# Introduction to operators in python (Part 1) Xiaobo test growth path 2021-10-29 04:10:09 introduction operators python Let's introduce today Python Operator in , Operators mainly include the following categories ： • Arithmetic operator • Compare （ Relationship ） Operator • Logical operators • Assignment operator • Ternary operator • An operator • member operator • Identity operator Next , Introduce the specific operation of the operator in detail ： Arithmetic operator Arithmetic operator an operator used to perform related operations on data of integer type and floating point type . Common arithmetic operators and corresponding operations are described in the figure below ： If you will True/False When used for numerical operations , Will automatically switch to 1 and 0 To calculate . Comparison operator The comparison operator is mainly used to compare the size of two expressions , The return result of the comparison is True perhaps False. Operator name Example explain > Greater than a>b == be equal to a==b a And b Return when equal True, otherwise False < Less than a<b >= Greater than or equal to a>=b a Greater than or equal to b When to return to True, otherwise False <= Less than or equal to a<=b != It's not equal to a!=b a And b Return... If not equal True, otherwise False Examples demonstrate ： Let's test you ``` stay python in , What are the execution results of the following statements ？ print(1.0 == 1) ==>True print(1 == True) ==>True print([2,1]>[1]) ==>True print(['1']>[1]]) ==> Operation error reporting ``` Logical operators Logical operators are used to operate on Boolean variables , The result is also Boolean . • and And （and The values on both sides are True when , return True） • or or （or As long as one of the values on both sides is True when , return True） • not Not ,（ take true Turn into false,false Turn into true） Please look at specific cases ： ```print(True and True) #True print(True and False) #False print(False and False) #False``` about a or b Come on , If a It's true , Then the value is a, Otherwise b; about a and b Come on , If a It's true , Then the value is b, Otherwise a. ```print(True or False) #True print(False or True) #True print(False or False) #False``` ```print(not True) #False print(not False) #True``` The specific running effect is shown in the above script , It should be noted that , Logical operators like this , At run time , A similar “ A short circuit ” The design of the ,and and or The operation is in operation , If the result has been determined , You won't evaluate the following expression . such as ：True or False , Ahead True It has been determined that the final returned result will be True, You won't care about or Go back True still False. All in all ：and All data should be True To return to True,or There is only one for True Then return to True. Assignment operator = Is the most common assignment operator , such as a=3, It means that 3 Is assigned to a variable a. Other common assignment operators are ：+=、-+、*=、/=、%=、//=、**= etc. . ```a+=b ==> a=a+b a-=b ==> a=a-b a*=b ==> a=a*b``` Ternary operator This is generally used in if Judgment conditions are common , such as , seek 2 Maximum number , The common expression is as follows : ```if a>b: max = a else: max = b``` python Provides a simple way to write , The code looks much simpler ： `max = a if a>b else b` python Pass through if else Condition judgment of , It can achieve something similar java Inside ?: Ternary operator . java The usage inside is as follows ：z = x>y ? x-y : x+y; python Inside usage ：value1 if Judge the condition else value2 A little more complicated ： ```a if a>b else c if c>d else d Equivalent to ：a if a>b else ( c if c>d else d )``` member operator • in ： Returns if a value is found in the specified sequence True, Otherwise return to False • not in ： If the corresponding value is not found in the specified sequence number, return True, Return if found False Member operators are still used more , For example, to view a certain key Is it in a dictionary , It can be used key in dict.keys() To judge ```a = {"A": "1", "B": "2"} print('A' in a.keys()) print('a' in 'abcdefg')``` Identity operator is、is not, Used to determine whether two identifiers refer to the same object ```a1 = 10 b1 = 2 b1 += 8 print(a1 is b1) #True s1 = 'abc' s2 = '123abc'[3:] print(s1 is s2) #False a1 = 1 print(a1 is True) #False```

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# a) Explain whether it is possible to use inferential methods for independent sam a) Explain whether it is possible to use inferential methods for independent samples. b) Explain whether there is enough information to compare the proportions using dependent samples or not. You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it izboknil3 a) It is given that $36\mathrm{%}$ of the Americans believe in ghost and $37\mathrm{%}$ believe in astrology among 1,249 Americans. The assumption for inference of independent samples: The response variable should be categorical data. The random samples should be collected independently. The sample sizes should be large. That is, in each group the number of success points and the number of failure points should be at least 10. Here the both proportions are calculated from same samples. That is, the both responses are dependent to each other. Therefore, it is not possible to use independent sample test. b) For doing the inference of independent samples, there should be two different groups in which the proportion should be known. In other words, it can be said that in each sample it should be known that how many people support astrology and how many people didnt

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• ``````# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def flatten(self, root): """ :type root: TreeNode :rtype: void Do not return anything, modify root in-place instead. """ if root is None: return root p = TreeNode(0) p.right = root self.flattenHelper(p, root) def flattenHelper(self, lastNode, node): lastNode.right = node if node.left is None and node.right is None: # leaf node lastNode = node return left = node.left right = node.right node.left = None node.right = None lastNode = node # reset left and right for current node if left is not None: self.flattenHelper(lastNode, left) if right is not None: self.flattenHelper(lastNode, right) `````` for given input 1->left = 2, 1->right = 3 the expected result should be 1->right=2, 2->right=3 but the above code returns 1->right=3 I think `lastNode = node` has some issue but I cannot figure it out. Can someone help? Thanks! Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.

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# Should You Make Multiple Bets at the Craps Table? Most craps gamblers have multiple bets on the table simultaneously. This has the effect of creating huge winning streaks and huge losing streaks. But is that the best way to play the game? You might be surprised to know this, but I have an opinion about whether you should make multiple bets at the craps table. ## Having 3 Bets Going at the Craps Table at One Time Most craps players try to get 3 bets going simultaneously at the craps table. They’ll start by making a pass line bet. Then on the next roll of the dice, they’ll place a come bet. On the 3rd roll of the dice, they’ll place another come bet. If the shooter is rolling point numbers the whole time, the player winds up with 3 bets in action all at the same time. And some craps bettors will just keep placing come bets until the shooter rolls a 7, which can result in gobs of chips on the table. Real wild men at the craps table supplement these standard bets with other bets with worse odds. Some of them are only marginally worse, like some of the place bets, but others are outrageously bad, like the any craps or any 7 bet. The problem with a craps bet is the same problem you face with any bet in casino games: A bet on a casino game is, essentially, a negative number. And if you paid attention in math class, you’ll know that no matter what, if you add a series of negative numbers together, the sum is always also a negative number. It doesn’t matter if you double or triple the size of those bets. The more money you put into action per hour at the craps table, the more money you’ll lose in the long run. Sure, you’ll have big swings when you hit a lucky streak, and you’ll also have big losing streaks. But on average, your results are going to follow the same formula: The number of bets you make per hour multiplied by the size of those bets multiplied by the house edge determines your expected average hourly loss rate. ## Having Only One Bet on the Table at a Time Reduces the Amount of Money You Have in Action You have multiple techniques you can use to decrease your average hourly loss rate. One of those is to stick with the games (or bets) that have the lowest house edge. For example, the house edge for the pass line bet is only 1.41%. But the house edge for a bet on snake-eyes is 13.89%. If you put \$300 into action per hour on the former, your expected hourly loss is only \$4.23. Put that \$300 into action on the latter, and your expected hourly loss skyrockets to \$41.67. For the purposes of this post, I’ll assume you can already tell the good bets from the bad bets at the craps table. (The main thing to remember is to stick with pass, don’t pass, come, don’t come, and odds bets.) Since you already know how to minimize the house edge, let’s talk about how many bets you’re placing per hour. If you place a new bet every time the shooter rolls the dice, you’ll bet putting a lot more money into action per hour than if you placed just one bet and waited for it to be resolved before making another bet. This will reduce the amount of money you lose per hour dramatically. An average craps table with about 9 players sees about 120 rolls per hour. About 1/3 of those rolls will be new come-out rolls, or 40 rolls. This means that you can reduce the amount of money you lose dramatically by only betting on the new come-out rolls. Let’s assume you’re betting \$10 per roll, and you bet on either pass or come on every roll. That’s \$1200 per hour you’re putting into action, and you can expect to lose (on average) 1.41% of that. That’s an expected hourly loss of \$16.92. Now assume you’re betting pass on every new come-out roll. That’s 40 rolls at \$10 each, or \$400 in hourly action. Your expected loss is 1.41% of \$400, or \$5.64 per hour. i The only thing I want to warn you about here is that you shouldn’t pay too much attention to what’s happening in the short run – at least not as it relates to your decisions about how many bets to have on the table at one time. Craps is a random game, and it’s streakier than most. (This means it has a high standard deviation or volatility.) This means that you’ll sometimes be at the table and win seemingly constantly. It also means that sometimes you’ll be at the table and seem to lose way more than you expected. The math in the long run doesn’t change, though. ## The Worst Thing You Can Do The worst thing you can do at the craps table is take any of the prop bets being offered by the stickman. All of these bet have a high house edge. And even though the minimum bet is low, over time, that high house edge will more than make up for it. Proposition bets are one-roll bets that are determined on the outcome of the next roll. One example is the boxcars bet in craps. This is a bet that the next roll of the dice will result in a pair of 6s, for a total of 12. You probably already know that there are 11 possible totals in craps, but some totals are more likely than others. That’s because some totals have multiple ways of coming up. There are actually 36 possible outcomes when you roll a pair of dice. A total of 12 is only 1 of these, which means the probability of that bet winning is 1/36, or 35 to 1. So yeah, a bet on boxcars is a longshot, but that’s not what makes it a bad bet. What makes it a bad bet is the comparison of the payout with the odds of winning. The boxcars bet normally pays off at 30 to 1, which means that, in the long run, you’ll lose a lot of money on that bet. How much? The math is easy. Just assume you’re betting \$1 on boxcars 36 times in a row. You’ll win once and lose 35 times – if you see the mathematically predicted results. How much money will you win or lose in that situation? On one of those rolls, you’ll win \$30. On the other 35 rolls, you’ll lose \$35. That’s a net loss of \$5 over 36 rolls, or a little over 13 cents per roll. The house edge on that bet is that number expressed as a percentage. In this case, we’re looking at a bet with a house edge of 13.89%. It should be obvious why you’d want to avoid a bet where you’re going to lose an average of 13.89% of your money, but if it’s not: There are much better bets on the table. The house edge for the pass line bet is only 1.41%, which means that over time you’ll lose 10X as much betting boxcars than you would betting the pass line bet. The boxcars bet isn’t even the worst of the prop bets. It’s bad, but some of the other proposition bets are even worse. All of them have a house edge of almost 10%, but most of them are significantly more than that. ## Conclusion You should not make multiple bets simultaneously at the craps table. The only thing you accomplish by doing so is increasing the average amount of money you’re going to lose per hour when playing. Instead, just stick with the pass line bet, and when it’s resolved, bet it again. Taking odds every time is a good idea, too, because it’s a bet with no house edge. But even if you want to skip the odds bet, you’re better off not making too many bets at a time at the craps table. Do you agree or disagree? Let me know in the comments. Posted in: Table Games ## Most Recent See all recent Play Space XY Game at Online Casinos & Win Real Money October 5, 2023 Space XY isn’t your run-of-the-mill game that you can play at any online casino. It’s pretty unique in terms of the mechanics and how the gameplay works. Space XY also...

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Decimals Decimal To Fractionorksheets For 5th Grade Converting Repeating Fractionsith Answers Recurringorksheet Pdf By Brenda J. Church on May 15 2018 22:09:32 Children usually start to learn about decimals in Year 4. The first thing children need to know is that a decimal is A PART of a whole. A good way to explain this is to show them an empty hundred number square / chart. Children will be learning at school how to round up decimals to the nearest whole number. A fun way to practise this at home is to use regular dominos and establish that each piece is a decimal number (so a domino one with 3 dots and 4 dots is 3.4). Encourage them to play with the dominos, but with the rule that they can only, for example, join 2 dominos that are within 1.5 of each other. This will really get them thinking about what each decimal point represents. A number line can also be useful when rounding decimals. If a child is really confused about decimals, converting decimal numnbers into money is a great way to make things clearer. For example: a child may be asked to say how much bigger 1.3 is than 0.9. If they convert these decimals into money (£1.30 and 90p) they may find that they can do this calculation very quickly in their head, getting the answer 40p which they convert back into the decimal, 0.4. Demonstrating that money maths depends on decimal understanding is also an easy way to prove that decimals are actually useful in real life! Never underestimate how much a visual representation of what can otherwise appear to be an abstract concept can help children learn.

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>HOME>Cement Mill Critical Speeed And Effect # Cement Mill Critical Speeed And Effect • ### Cement Mill Critical Speeed And Effect- EXODUS The point where the mill becomes a centrifuge is called the critical speed and ball mills usually operate at 65 to 75 of the critical speed ball mills are generally used to grind material 14 inch and finer down to the particle size of 20 to 75 microns,Cement Mill Critical Speeed And Effect. ### cement mill critical speeed hledampracivliberci.cz Cement Mill Critical Speeed And Effect- EXODUS Mining . Cement Mill Critical Speeed And Effect The point where the mill becomes a centrifuge is called the critical speed and ball mills usually operate at 65 to 75 of the critical speed ball mills are generally used to grind material 14 inch and finer down to the particle size of 20 to 75 microns,Cement Mill Critical Speeed And Effect. ### Mill Speed Critical Speed Paul O. Abbe Mill Speed Critical Speed. Mill Speed . No matter how large or small a mill, ball mill, ceramic lined mill, pebble mill, jar mill or laboratory jar rolling mill, its rotational speed is important to proper and efficient mill operation. Too low a speed and little energy is imparted on the product. ### Cement mill notebook SlideShare 2015-1-7 · Raw mills usually operate at 72-74% critical speed and cement mills at 74-76%. 3.2 Calculation of the Critical Mill Speed: G: weight of a grinding ball in kg. w: Angular velocity of the mill tube in radial/second. w = 2*3.14*(n/60) Di: inside mill diameter in meter (effective mill ### Ball Mill Critical Speed Mineral Processing & 2015-6-19 · A Ball Mill Critical Speed (actually ball, rod, AG or SAG) is the speed at which the centrifugal forces equal gravitational forces at the mill shell’s inside surface and no balls will fall from its position onto the shell. The imagery below helps explain what goes on inside a mill as speed varies. Use our online formula. The mill speed is typically defined as the percent of the ### THE EFFECT OF MILL SPEED ON KINETIC BREAKAGE The effect of the fraction of mill critical speed (Oc) on the grinding for model parameter aT was found to be different for two different samples: aT=0.0344 exp(0.00301 Oc) for clinker and aT=0.0225 exp(0.06183 Oc) for limestone. ### PROCESS DIAGNOSTIC STUDIES FOR CEMENT MILL 2013-3-13 · A 1.5 mio t/a cement plant is having a closed circuit ball mill for cement grinding: The mill has been operating with satisfactory performance in-terms of system availability and output, however power consumption was on higher side. 3.1 System Description Mill Rated capacity 150 t/h OPC at 2800 blaine I chamber liners ### The Effects of Rotation and Revolution Speed Ratio The experimental mill used was a laboratory size of 209 mm diameter, 175 mm length, providing a total mill volume of 6001 cm3, with a total mass of 5.6 kg of steel balls of 46, 26 and 12.8 mm ### Manufacturing of GREEN CLINKER UltraTech Cement 2017-3-14 · Critical Success Factor Team Action Plan Short term, MT, LT Goal & Target False air in Raw Mill Circuit 3 3 3 27 Power NVA process steps 4 2 3 24 Power LS Crusher to feed both lines 5 4 4 80 Cement Mill 6 Corrosion at Joints 5 Time based maintenance system 4 120 Condition based ### cement mill critical speed and effect cement mill critical speeed and effect huaheng org. what is critical speed of ball mill . ABSTRACT In this study the effect of fraction of mill critical speed was investigated on the limestone and the clinker samples using G 246 ltaş cement factory Ball mill Wikipedia Critical speed can be Aside from common ball mills there is a second type of ball mill called a planetary ball mill . ### cement mill critical speeed Industry cement mill critical speeed and effect exodus mining machine Cement Mill Critical SpeeedAnd Effect The point where the mill becomes a centrifuge is called the critical speed and ballmillsusually operate at 65 to 75 of the critical speed ballmillsare generally used to grind material 14 inch and finer down to the particle size of 20 to 75 microns ### how to calculate cement mill critical speed how to calculate cement mill critical speed Mill Critical Speed Calculation Effect of Mill Speed on the Energy Input In this experiment the overall motion of the assembly of 62 balls of two different sizes was studied. The mill was rotated at 50, 62, 75 and 90% of the critica ### How To Calculate Cement Mill Critical Speed Cement Mill Critical Speeed And Effect- Exodus Mining The point where the mill becomes a centrifuge is called the critical speed and ball mills usually operate at 65 to 75 of the critical speed ball mills are generally used to grind material 14 inch and finer down to the particle size of 20 to 75 microns,Cement Mill Critical Speeed And Effect. ### Mill Speed Critical Speed Paul O. Abbe Mill Speed Critical Speed. Mill Speed . No matter how large or small a mill, ball mill, ceramic lined mill, pebble mill, jar mill or laboratory jar rolling mill, its rotational speed is important to proper and efficient mill operation. Too low a speed and little energy is imparted on the product. ### Cement mill notebook SlideShare 2015-1-7 · Raw mills usually operate at 72-74% critical speed and cement mills at 74-76%. 3.2 Calculation of the Critical Mill Speed: G: weight of a grinding ball in kg. w: Angular velocity of the mill tube in radial/second. w = 2*3.14*(n/60) Di: inside mill diameter in meter (effective mill diameter). n: Revolution per minute in rpm. ### loesche cement mill turkey loesche cement mill turkey curacao immobilien-tessin eu. Four LOESCHE Vertical Roller Mills for Turkey Loesche LOESCHE is able to benefit from the growth in the Turkish cement industry LOESCHE is contributing four of its highly-modern vertical roller mills to the new cement plant of the German technology firm AUNDE in the Turkish region of Soma one mill for grinding up to ### crusher byke in south africa Prominer (Shanghai) cement mill critical speeed and effect; how to determine a silver ore; iron ore lm vertical grinding mill manufacturer supplier for; concentrator plant beneficiation; ebay classifieds craigs list gold prospecting equipment; boss camping has to be dealt with elder scrolls; 2016 high quality impact stone crusher; gold refining machine gold ### the challenge for critical mining equipment the challenge for critical mining equipment. Autonomous vehicles face two challenges technology and . Jul 01 2019 · The other big challenge according to Avary is the business model for self-driving vehicles. 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ABSTRACT In this study the effect of fraction of mill critical speed was investigated on the limestone and the clinker samples using G 246 ltaş cement factory Ball mill Wikipedia Critical speed can be Aside from common ball mills there is a second type of ball mill called a planetary ball mill . ### cement mill critical speeed Industry cement mill critical speeed and effect exodus mining machine Cement Mill Critical SpeeedAnd Effect The point where the mill becomes a centrifuge is called the critical speed and ballmillsusually operate at 65 to 75 of the critical speed ballmillsare generally used to grind material 14 inch and finer down to the particle size of 20 to 75 microns ### loesche cement mill turkey loesche cement mill turkey curacao immobilien-tessin eu. Four LOESCHE Vertical Roller Mills for Turkey Loesche LOESCHE is able to benefit from the growth in the Turkish cement industry LOESCHE is contributing four of its highly-modern vertical roller mills to the new cement plant of the German technology firm AUNDE in the Turkish region of Soma one mill for grinding up to 350 tonnes of raw ### Ball Mill|Cement Ball Mill Liming Company Price Cement Ball Mill Ime Company Price Me Mining Machinery. Cement ball mill ime company price cement ball mill ime company price cement ball mill ime company price shanghai clirik is a leading minery manufacturing company in chinaprovide all types of ball millcement ball millball grinding mill and so onwelcome to our company for inspection and purchase phonewhatsapp 8613917147829 emailemail ### the challenge for critical mining equipment the challenge for critical mining equipment. Autonomous vehicles face two challenges technology and . 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Additionally, the 197 Comil was evaluated as a potential substitute for extrudate milling by comparing resulting PSD, morphology, and angle of repose using extrudate milled on both the 197 Comil and L1A Fitz mil. ### purpose of screening process in flotation flotation process applications. Froth Flotation Applications Mineral Processing & Metallurgy. Unusual Applications Of The Flotation Process: In the mineral industry flotation is often defined as a physicochemical process that has the capability of separating two or more finely divided minerals from each other the usual concept of flotation all three forms of matter are involved and many of the ### Jaw Crusher|Stone Crusher China Manufacturers Stone Crusher Manufacturers Stone Crusher Suppliers. It supplies all kinds calcite crusher equipment and design calcite crushing grinding production line according feldspar stone processing plant feldspar is the most common ore in the earth crust and it even appears on the moon and in the aerolite ### cement mill critical speeed Industry cement mill critical speeed and effect exodus mining machine Cement Mill Critical SpeeedAnd Effect The point where the mill becomes a centrifuge is called the critical speed and ballmillsusually operate at 65 to 75 of the critical speed ballmillsare generally used to grind material 14 inch and finer down to the particle size of 20 to 75 microns ### loesche cement mill turkey loesche cement mill turkey curacao immobilien-tessin eu. 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# pratt_8e_chapter_6_solns - CHAPTER 6 THE CURRENT ASSET... • Notes • 25 • 100% (1) 1 out of 1 people found this document helpful This preview shows page 1 - 3 out of 25 pages. CHAPTER 6 THE CURRENT ASSET CLASSIFICATION, CASH, AND ACCOUNTS RECEIVABLE BRIEF EXERCISES BE6–1 a. Total Accounts Receivable = Net Receivables + Allowance for Uncollectibles 2009 Total Accounts Receivable = \$3,623 + \$93 2009 Total Accounts Receivable = \$3,716 2009 Uncollectibles as a Percentage of Total Accounts Receivable = \$93/\$3,716 = 2.50% 2008 Total Accounts Receivable = \$4,618 + \$90 2008 Total Accounts Receivable = \$4,708 2008 Uncollectibles as a Percentage of Total Accounts Receivable = \$90/\$4,708 = 1.91% Therefore, the percentage increased. b. Since Emerson Electric is using the percentage of accounts receivable method (balance sheet approach), bad debt expense for 2009 would be the amount needed to adjust the allowance for doubtful accounts to \$93. This number (bad debt expense) is impacted by the balance in the uncollectible account at the beginning of the year and the write-offs taken during the year by Emerson Electric. BE6–2 a. 2007: Ending Allowance Balance = Beginning Allowance Balance + Bad Debt Charge – Write-Offs + Recoveries \$4,238 = \$3,945 + 4,431 \$5,966 + 1,828 Bad Debt Expense for 2007 = \$4,431 2008: Ending Allowance Balance = Beginning Allowance Balance + Bad Debt Charge – Write-Offs + Recoveries \$5,325 = \$4,238 + 7,518 \$8,162 + 1,731 Bad Debt Expense for 2008 = \$7,518 As the economy went into recession in 2008, companies that extend credit (such as GE’s capital division) had a more difficult time collecting receivables. 1 b. 2007: \$5,966 write-offs; \$4,138 write-offs, net of recoveries 2008: \$8,162 write-offs; \$6,431 write-offs, net of recoveries c. The allowance account grew by 25.7% ((\$5,325- \$4,238)/\$4,238) from 2007 to 2008. The allowance account grew at such a large rate due to the deterioration of the economy and GE’s belief that a greater amount future receivables will prove to be uncollectible as a result. By increasing the allowance balance the company is taking into consideration that receivables will not be as collectible as when the economy was healthier. Increasing the allowance lowers the “net realizable value” of the receivables on the balance sheet, which is prudent behavior given the economic climate. BE6–3 a. GE bad debts as a percentage of total revenues = \$7.5/\$182 = 4.1%; as a percentage of GECS revenues, the calculation is \$7.5/\$71 = 10.6%. GE overall revenues should be used since the bad debt provision is for GE and not for GECS. b. On a balance sheet for GECS accounts receivable would be expected to be the largest account. Its primary role is as a financing company, and the receivables from the buyers of appliances would be a large asset of GECS. c. GE is very large and has many subsidiaries that make it difficult to classify it as just a manufacturing, retail or service company. The overall GE business is best known as a manufacturing company. It does not have its own retail stores, so it is not a retailer. While there are many services offered with its products, its primary focus is as a manufacturer. Services, primarily financial services, are becoming more important for the consolidated operation. EXERCISES E6–1 a. Cash. Money held in checking accounts is defined as cash, and there are no restrictions on the account. b. Cash. Checks are considered cash unless the checks cannot be cashed until a later date (i.e.,

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cancel Showing results for Did you mean: Find everything you need to get certified on Fabric—skills challenges, live sessions, exam prep, role guidance, and more. Get started Helper II ## Need help on Power BI Query. Hi team, I am currently working on my borganizational project and i am trying to create new measure for my database using below query. However i can't able to add new measure due to syntax error. Can someone please help me. Risk Level = var Impact = AVERAGE(Output[RiskImpact]) var Likelihood = AVERAGE(Output[RiskLikelihood]) return SWITCH(TRUE() , Impact > 0 && Impact <= 2 && Likelihood > 0 && Likelihood <=4 , "LOW" , Impact > 0 && Impact <= 2 && Likelihood > 4 && Likelihood <=5 , "MEDIUM" , Impact > 2 && Impact <= 3 && Likelihood > 0 && Likelihood <=4 , "MEDIUM" , Impact > 2 && Impact <= 3 && Likelihood > 4 && Likelihood <=5 , "HIGH" , Impact > 3 && Impact <= 4 && Likelihood > 0 && Likelihood <=3 , "MEDIUM" , Impact > 3 && Impact <= 4 && Likelihood > 3 && Likelihood <=5 , "HIGH" , Impact > 4 && Impact <= 5 && Likelihood > 0 && Likelihood <=1 , "MEDIUM" , Impact > 4 && Impact <= 5 && Likelihood > 1 && Likelihood <=5 , "HIGH" , "NA” ) 2 REPLIES 2 Super User Problem is about second quote in the formula for NA. In place of "NA” use "NA" Hence following formula would work ``````Risk Level = var Impact = AVERAGE(Output[RiskImpact]) var Likelihood = AVERAGE(Output[RiskLikelihood]) return SWITCH(TRUE() , Impact > 0 && Impact <= 2 && Likelihood > 0 && Likelihood <=4 , "LOW" , Impact > 0 && Impact <= 2 && Likelihood > 4 && Likelihood <=5 , "MEDIUM" , Impact > 2 && Impact <= 3 && Likelihood > 0 && Likelihood <=4 , "MEDIUM" , Impact > 2 && Impact <= 3 && Likelihood > 4 && Likelihood <=5 , "HIGH" , Impact > 3 && Impact <= 4 && Likelihood > 0 && Likelihood <=3 , "MEDIUM" , Impact > 3 && Impact <= 4 && Likelihood > 3 && Likelihood <=5 , "HIGH" , Impact > 4 && Impact <= 5 && Likelihood > 0 && Likelihood <=1 , "MEDIUM" , Impact > 4 && Impact <= 5 && Likelihood > 1 && Likelihood <=5 , "HIGH", "NA")`````` Helper II Thank you Vijay. This is works. Announcements #### Europe’s largest Microsoft Fabric Community Conference Join the community in Stockholm for expert Microsoft Fabric learning including a very exciting keynote from Arun Ulag, Corporate Vice President, Azure Data. #### Power BI Monthly Update - August 2024 Check out the August 2024 Power BI update to learn about new features. #### Fabric Community Update - August 2024 Find out what's new and trending in the Fabric Community. Top Solution Authors Top Kudoed Authors

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# Interpolation Techniques and How to Use Them in MATLAB The MATLAB Interpolation Techniques is like “magical” tools that make the job easier in your work. In a time where MATLAB is considered to be the best and most popular programming language, Interpolation Techniques is also a popular software. Since Interpolation Techniques makes the process of solving matrices easier, you will not have to worry about your time anymore. Interpolation Techniques are used by many programmers to solve their problems. The MATLAB is well known because of its computer programing ability that gives the programmer an edge in his or her business. It can be used to create various useful things. With the introduction of Interpolation Techniques, your MATLAB assignments will be easy and more pleasant to do. A matrix is simply a grouping of numbers that are usually of different types. You can use matrices to represent different things such as real numbers, logical numbers, vectors, arrays, and other special types. It can be constructed in different ways. The set of operations that can be done with matrices is limitless. The Interpolation Techniques that you can use is available in almost all platforms and for both software and hardware platforms. If you are new to using MATLAB, you may not know how to use Interpolation Techniques. This tutorial will help you get familiar with Interpolation Techniques using a simple problem. So, learn from this tutorial and solve your problems easily. Let’s go back to the world of matrix operations. You know that two matrices cannot be constructed in any way, shape, or form, without a few different ways to interlock and align them. This is the case when two rectangular matrices are combined with the help of some special “dot product”. The dot product is used to provide a unique configuration of two rectangular matrices. One thing that you need to remember about interlocking of two matrices is that if the length of one of the components is very long, the other one will be too long. When one component of the matrix has two components, the problem will arise. A good example of this would be two components of the same vector that is composed of components that are perpendicular to each other. If the length of the first component is very long, the second one will be very long as well. The first component of the matrix is actually missing. So, it will be harder to make sure that the second component will have the same length as the first one. A good example of this is, the first component of the matrix is missing and the second component of the matrix has been cut so that it will not be longer than the first component. Pythagoras was the first person to notice that this property can be used to solve problems in geometry. He realized that if you would cut the first component of the matrix into two, you will get two components that are perpendicular to each other. Since they are of the same length, the second component will have the same length as the first one. After Pythagoras, it was suggested by the Greeks that each component of the matrix is of the same length. That’s how it is called Pythagoras’ Principle. You can make a new matrix and cut the first component of the matrix into two. Then you can create a second component of the new matrix. Then, you can now use Pythagoras’ Principle and cut each of the two components into two again. You will get a new matrix that you can now do anything with. Matlab offers Interpolation Techniques to you. If you are working on a matrix form that has components that are not perpendicular to each other, you can do something. In matlab, you can perform a dot product to get a new matrix. With matlab, you will get the knowledge and skills needed to create things like a line that is vertical. Then you can plot it with the help of lines that are vertical. “It takes some practice” and it will be more enjoyable if you can interact with your math tutor, but if you think that it is boring, you can always get help from the interp.com.

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