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(a) A child stands at the centre of a turntable with his two arms outstretched. Question: (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction. (b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy? Solution: (a) 100 rev/min Initial angular velocity, $\omega_{1}=40 \mathrm{rev} / \mathrm{min}$ Final angular velocity $=\omega_{2}$ The moment of inertia of the boy with stretched hands $=I_{1}$ The moment of inertia of the boy with folded hands $=I_{2}$ The two moments of inertia are related as: $I_{2}=\frac{2}{5} I_{1}$ Since no external force acts on the boy, the angular momentum $L$ is a constant Hence, for the two situations, we can write: $I_{2} \omega_{2}=I_{1} \omega_{1}$ $\omega_{2}=\frac{I_{1}}{I_{2}} \omega_{1}$ $=\frac{I_{1}}{\frac{2}{5} I_{1}} \times 40=\frac{5}{2} \times 40$ $=100 \mathrm{rev} / \mathrm{min}$ (b)Final K.E. $=2.5$ Initial K.E. Final kinetic rotation, $E_{\mathrm{F}}=\frac{1}{2} I_{2} \omega_{2}^{2}$ Initial kinetic rotation, $E_{1}=\frac{1}{2} I_{1} \omega_{1}^{2}$ $\frac{E_{F}}{E_{1}}=\frac{\frac{1}{2} I_{2} \omega_{2}^{2}}{\frac{1}{2} I_{1} \omega_{1}^{2}}$ $=\frac{2}{5} \frac{I_{1}}{I_{1}} \frac{(100)^{2}}{(40)^{2}}$ $=\frac{2}{5} \times \frac{100 \times 100}{40 \times 40}$ $=\frac{5}{2}=2.5$ $\therefore E_{\mathrm{F}}=2.5 E_{\mathrm{l}}$ The increase in the rotational kinetic energy is attributed to the internal energy of the boy.
# Linear Equations and Inverse Matrices Save this PDF as: Size: px Start display at page: ## Transcription 1 Chapter 4 Linear Equations and Inverse Matrices 4. Two Pictures of Linear Equations The central problem of linear algebra is to solve a system of equations. Those equations are linear, which means that the unknowns are only multiplied by numbers we never see x or x times y. Our first linear system is deceptively small, only by. But you will see how far it leads Two equations x y Two unknowns x C y () We begin a row at a time. The first equation x y produces a straight line in the xy plane. The point x, y is on the line because it solves that equation. The point x, y is also on the line because. For x we find y. The slope of this line in Figure 4. is, because y increases by when x changes by. But slopes are important in calculus and this is linear algebra! y.; / x y x C y x Figure 4. Row picture The point.; / where the two lines meet is the solution. 2 98 Chapter 4. Linear Equations and Inverse Matrices The second line in this row picture comes from the second equation x Cy. You can t miss the intersection point where the two lines meet. The point x ; y lies on both lines. It solves both equations at once. This is the solution to our two equations. ROWS The row picture shows two lines meeting at a single point (the solution). Turn now to the column picture. I want to recognize the same linear system as a vector equation. Instead of numbers we need to see vectors. If you separate the original system into its columns instead of its rows, you get a vector equation Combination equals b x C y b () This has two column vectors on the left side. The problem is to find the combination of those vectors that equals the vector on the right. We are multiplying the first column by x and the second column by y, and adding vectors. With the right choices x and y (the same numbers as before), this produces.column / C.column / b. COLUMNS The column picture combines the column vectors on the left side of the equations to produce the vector b on the right side. (column ) 4 multiply by 4 b combination column column C Figure 4. Column picture A combination (column ) + (column ) gives the vector b. Figure 4. is the column picture of two equations in two unknowns. The left side shows the two separate columns, and column is multiplied by. This multiplication by a scalar (a number) is one of the two basic operations in linear algebra Scalar multiplication 3 4.. Two Pictures of Linear Equations 99 If the components of a vector v are v and v, then cv has components cv and cv. The other basic operation is vector addition. We add the first components and the second components separately. and C give the vector sum.; / as desired Vector addition C The right side of Figure 4. shows this addition. The sum along the diagonal is the vector b.; / on the right side of the linear equations. To repeat The left side of the vector equation is a linear combination of the columns. The problem is to find the right coefficients x and y. We are combining scalar multiplication and vector addition into one step. That combination step is crucially important, because it contains both of the basic operations on vectors multiply and add. Linear combination of the columns C Of course the solution x ; y is the same as in the row picture. I don t know which picture you prefer! Two intersecting lines are more familiar at first. You may like the row picture better, but only for a day. My own preference is to combine column vectors. It is a lot easier to see a combination of four vectors in four-dimensional space, than to visualize how four planes might possibly meet at a point. (Even one three-dimensional plane in four-dimensional space is hard enough...) The coefficient matrix on the left side of equation () is the by matrix A Coefficient matrix A This is very typical of linear algebra, to look at a matrix by rows and also by columns. Its rows give the row picture and its columns give the column picture. Same numbers, different pictures, same equations. We write those equations as a matrix problem Av b Matrix multiplies vector x y The row picture deals with the two rows of A. The column picture combines the columns. The numbers x and y go into the solution vector v. Here is matrix-vector multiplication, matrix A times vector v. Please look at this multiplication Av! ot products with rows Combination of columns Av b is () 4 Chapter 4. Linear Equations and Inverse Matrices Linear Combinations of Vectors Before I go to three dimensions, let me show you the most important operation on vectors. We can see a vector like v.; / as a pair of numbers, or as a point in the plane, or as an arrow that starts from.; /. The arrow ends at the point.; / in Figure 4.. v.; / column vector point.; / arrow to.; / Figure 4. The vector v is given by two numbers or a point or an arrow from.; /. A first step is to multiply that vector by any number c. If c then the vector is doubled to v. If c then it changes direction to v. Always the scalar c multiplies each separate component (here and ) of the vector v. The arrow doubles the length to show v and it reverses direction to show v v v v v v v column vectors arrows to.; / and. ; / Figure 4.4 Multiply the vector v.; / by scalars c and to get cv.c; c/. If we have another vector w. ; /, we can add it to v. Vector addition v C w can use numbers (the normal way) or it can use the arrows (to visualize v C w). The arrows in Figure 4. go head to tail At the end of v, place the start of w. v C w C w v C w v Figure 4. The sum of v.; / and w. ; / is v C w.; /. This is also w C v. Allow me to say, adding v C w and multiplying cv will soon be second nature. In themselves they are not impressive. What really counts is when you do both at once. 5 4.. Two Pictures of Linear Equations Multiply cv and also dw, then add to get the linear combination cv C dw. Linear combination v C w C This is the basic operation of linear algebra! If you have two -dimensional vectors like v.; ; ; ; / and w.; ; ; ; /, you can multiply v by and w by. You can combine to get v C w.; ; ; ; 4/. Every combination cv C dw is a vector in the big -dimensional space R. I admit that there is no picture to show these vectors in R. Somehow I imagine arrows going to v and w. If you think of all the vectors cv, they form a line in R. The line goes in both directions from.; ; ; ; / because c can be positive or negative or zero. Similarly there is a line of all vectors d w. The hard but all-important part is to imagine all the combinations cv C dw. Add all vectors on one line to all vectors on the other line, and what do you get? It is a -dimensional plane inside the big -dimensional space. I don t lose sleep trying to visualize that plane. (There is no problem in working with the five numbers.) For linear combinations in high dimensions, algebra wins. ot Product of v and w The other important operation on vectors is a kind of multiplication. This is not ordinary multiplication and we don t write vw. The output from v and w will be one number and it is called the dot product v w. EFINITION The dot product of v.v ; v / and w.w ; w / is the number v w v w v w C v w (4) The dot product of v.; / and w. ; / is v w./. / C././. Example The column vectors.; / and. ; / have a zero dot product ot product is zero Perpendicular vectors C In mathematics, zero is always a special number. For dot products, it means that these two vectors are perpendicular. The angle between them is 9 ı. The clearest example of two perpendicular vectors is i.; / along the x axis and j.; / up the y axis. Again the dot product is i j C. Those vectors i and j form a right angle. They are the columns of the by identity matrix I. The dot product of v.; / and w.; / is. Soon v w will reveal the angle between v and w (not 9 ı /. Please check that w v is also. 6 Chapter 4. Linear Equations and Inverse Matrices Multiplying a Matrix A and a Vector v Linear equations have the form Av b. The right side b is a column vector. On the left side, the coefficient matrix A multiplies the unknown column vector v (we don t use a dot for Av). The all-important fact is that Av is computed by dot products in the row picture, and Av is a combination of the columns in the column picture. I put those words combination of the columns in boldface, because this is an essential idea that is sometimes missed. One definition is usually enough in linear algebra, but Av has two definitions the rows and the columns produce the same output vector Av. The rules stay the same if A has n columns a ; ; a n. Then v has n components. The vector Av is still a combination of the columns, Av v a C v a C C v n a n. The numbers in v multiply the columns in A. Let me start with n. By rows Av.row / v.row / v By columns Av v.column /Cv.column / Example In equation () I wrote dot products with rows and combination of columns. Now you know what those mean. They are the two ways to look at Av ot products with rows Combination of columns a v C b v c v C d v v a c C v b d () You might naturally ask, which way to find Av? My own answer is this I compute by rows and I visualize (and understand) by columns. Combinations of columns are truly fundamental. But to calculate the answer Av, I have to find one component at a time. Those components of Av are the dot products with the rows of A. 4 v v v C v 4v C v v 4 C v Singular Matrices and Parallel Lines The row picture and column picture can fail and they will fail together. For a by matrix, the row picture fails when the lines from row and row are parallel. The lines don t meet and Av b has no solution A 4 v v 4v v Parallel lines no solution The row picture shows the problem and so does the algebra times equation produces 4v v. But equation requires 4v v. Notice that this line goes through the center point.; / because the right side is zero. 7 4.. Two Pictures of Linear Equations How does the column picture fail? Columns and point in the same direction. When the rows are dependent, the columns are also dependent. All combinations of the columns.; 4/ and.; / lie in the same direction. Since the right side b.; / is not on that line, b is not a combination of those two column vectors of A. Figure 4. (a) shows that there is no solution to the equation. 4 line of columns line of columns b not on line b b is on line b Figure 4. Column pictures (a) No solution (b) Infinity of solutions Example Same matrix A, now b.; /, infinitely many solutions to Av b A 4 v v 4v v In the row picture, the two lines are the same. All points on that line solve both equations. Two times equation gives equation. Those close lines are one line. In the column picture above, the right side b.; / falls right onto the line of the columns. Later we will say b is in the column space of A. There are infinitely many ways to produce.; / as a combination of the columns. They come from infinitely many ways to produce b.; / (choose any c). Add one way to produce b.; /.; 4/. c 4 C c 4 C () The vector v n.c; c/ is a null solution and v p.; / is a particular solution. Av n equals zero and Av p equals b. Then A.v p C v n / b. Together, v p and v n give the complete solution, all the ways to produce b.; / from the columns of A Complete solution to Av b v complete v p C v n c C c () 8 4 Chapter 4. Linear Equations and Inverse Matrices Equations and Pictures in Three imensions In three dimensions, a linear equation like x C y C z produces a plane. The plane would go through.; ; / if the right side were. In this case the moves us to a parallel plane that misses the center point.; ; /. A second linear equation will produce another plane. Normally the two planes meet in a line. Then a third plane (from a third equation) normally cuts through that line at a point. That point will lie on all three planes, so it solves all three equations. This is the row picture, three planes in three dimensional space. They meet at the solution. One big problem is that this row picture is hard to draw. Three planes are too many to see clearly how they meet (maybe Picasso could do it). The column picture of Av b is easier. It starts with three column vectors in threedimensional space. We want to combine those columns of A to produce the vector v.column / C v.column / C v.column / b. Normally there is one way to do it. That gives the solution.v ; v ; v / which is also the meeting point in the row picture. I want to give an example of success (one solution) and an example of failure (no solution). Both examples are simple, but they really go deeply into linear algebra. Example 4 Invertible matrix A, one solution v for any right side b. Av b is 4 4 v v v 4 (8) This matrix is lower triangular. It has zeros above the main diagonal. Lower triangular systems are quickly solved by forward substitution, top to bottom. The top equation gives v, then move down. First v. Then v C v gives v 4. Then v C v gives v 9. Figure 4. shows the three columns a ; a ; a. When you combine them with ; 4; 9 you produce b.; ; /. In reverse, v.; 4; 9/ must be the solution to Av b. c a 4 a 4 a 4 c c Figure 4. Independent columns a ; a ; a not in a plane. ependent columns c ; c ; c are three vectors all in the same plane. 9 4.. Two Pictures of Linear Equations Example Singular matrix no solution to C v b or infinitely many solutions (depending on b). w w b w C w b w C w b 4 4 w w 4 or w 4 or 4 (9) This matrix C is a circulant. The diagonals are constants, all s or all s or all s. The diagonals circle around so each diagonal has three equal entries. Circulant matrices will be perfect for the Fast Fourier Transform (FFT) in Chapter 8. To see if C w b has a solution, add those three equations to get b C b C b. Left side.w w / C. w C w / C. w C w / () C w b cannot have a solution unless b Cb Cb. The components of b.; ; / do not add to zero, so C w.; ; / has no solution. Figure 4. shows the problem. The three columns of C lie in a plane. All combinations Cw of those columns will lie in that same plane. If the right side vector b is not in the plane, then C w b cannot be solved. The vector b.; ; / is off the plane, because the equation of the plane requires b C b C b. Of course C w.; ; / always has the zero solution w.; ; /. But when the columns of C are in a plane (as here), there are additional nonzero solutions to C w. Those three equations are w w and w w and w w. The null solutions are w n.c; c; c/. When all three components are equal, we have C w n. The vector b.; ; / is also in the plane of the columns, because it does have b C b C b. In this good case there must be a particular solution to C w p b. There are many particular solutions w p, since any solution can be a particular solution. I will choose the particular w p.; ; / that ends in w C w p The complete solution is w complete w p C any w n Summary These two matrices A and C, with third columns a and c, allow me to mention two key words of linear algebra independence and dependence. This book will develop those ideas much further. I am happy if you see them early in the two examples a ; a ; a are independent c ; c ; c are dependent A is invertible C is singular Av b has one solution v C w has many solutions w n Eventually we will have n column vectors in n-dimensional space. The matrix will be n by n. The key question is whether Av has only the zero solution. Then the columns don t lie in any hyperplane. When columns are independent, the matrix is invertible. 10 Chapter 4. Linear Equations and Inverse Matrices Problem Set 4. Problems 8 are about the row and column pictures of Av b. With A I (the identity matrix) draw the planes in the row picture. Three sides of a box meet at the solution v.x; y; z/.; ; 4/ x C y C z x C y C z or 4 4 x y 4 x C y C z 4 z 4 raw the four vectors in the column picture. Two times column plus three times column plus four times column equals the right side b. If the equations in Problem are multiplied by ; ; 4 they become V B x C y C z 4 x C y C z 9 or V 4 4 x y B x C y C 4z 4 z Why is the row picture the same? Is the solution V the same as v? What is changed in the column picture the columns or the right combination to give B? If equation is added to equation, which of these are changed the planes in the row picture, the vectors in the column picture, the coefficient matrix, the solution? The new equations in Problem would be x, x C y, z 4. 4 Find a point with z on the intersection line of the planes x C y C z and x y C z 4. Find the point with z. Find a third point halfway between. The first of these equations plus the second equals the third x C y C z x C y C z x C y C z The first two planes meet along a line. The third plane contains that line, because if x; y; z satisfy the first two equations then they also. The equations have infinitely many solutions (the whole line L). Find three solutions on L. Move the third plane in Problem to a parallel plane x C y C z 9. Now the three equations have no solution why not? The first two planes meet along the line L, but the third plane doesn t that line. In Problem the columns are.; ; / and.; ; / and.; ; /. This is a singular case because the third column is. Find two combinations of the columns that give b.; ; /. This is only possible for b.4; ; c/ if c. 11 4.. Two Pictures of Linear Equations 8 Normally 4 planes in 4-dimensional space meet at a. Normally 4 vectors in 4-dimensional space can combine to produce b. What combination of.; ; ; /;.; ; ; /;.; ; ; /;.; ; ; / produces b.; ; ; /? Problems 9 4 are about multiplying matrices and vectors. 9 Compute each Ax by dot products of the rows with the column vector 4 (a) 4 4 to Compute each Ax in Problem 9 as a combination of the columns 9(a) becomes Ax 4 C 4 4 C How many separate multiplications for Ax, when the matrix is by? Find the two components of Ax by rows or by columns 4 4 and and 4. Multiply A times x to find three components of Ax 4 4 x y and 4 4 and z 4 (a) A matrix with m rows and n columns multiplies a vector with components to produce a vector with components. (b) The planes from the m equations Ax b are in -dimensional space. The combination of the columns of A is in -dimensional space. 4 Write xcy CzCt 8 as a matrix A (how many rows?) multiplying the column vector x.x; y; z; t/ to produce b. The solutions x fill a plane or hyperplane in 4-dimensional space. The plane is -dimensional with no 4 volume. Problems ask for matrices that act in special ways on vectors. (a) What is the by identity matrix? I times x y equals xy. (b) What is the by exchange matrix? P times x y equals yx. 12 8 Chapter 4. Linear Equations and Inverse Matrices (a) What by matrix R rotates every vector by 9 ı? R times x y is yx. (b) What by matrix R rotates every vector by 8 ı? Find the matrix P that multiplies.x; y; z/ to give.y; z; x/. Find the matrix Q that multiplies.y; z; x/ to bring back.x; y; z/. 8 What by matrix E subtracts the first component from the second component? What by matrix does the same? E and E What by matrix E multiplies.x; y; z/ to give.x; y; z C x/? What matrix E multiplies.x; y; z/ to give.x; y; z x/? If you multiply.; 4; / by E and then multiply by E, the two results are. / and. /. What by matrix P projects the vector.x; y/ onto the x axis to produce.x; /? What matrix P projects onto the y axis to produce.; y/? If you multiply.; / by P and then multiply by P, you get. / and. /. What by matrix R rotates every vector through 4 ı? The vector.; / goes to. p =; p =/. The vector.; / goes to. p =; p =/. Those determine the matrix. raw these particular vectors in the xy plane and find R. Write the dot product of.; 4; / and.x; y; z/ as a matrix multiplication Av. The matrix A has one row. The solutions to Av lie on a perpendicular to the vector. The columns of A are only in -dimensional space. In MATLAB notation, write the commands that define this matrix A and the column vectors v and b. What command would test whether or not Av b? A v b 4 4 If you multiply the 4 by 4 all-ones matrix A = ones(4) and the column v = ones(4,), what is Av? (Computer not needed.) If you multiply B = eye(4) + ones(4) times w = zeros(4,) + ones(4,), what is Bw? Questions review the row and column pictures in,, and 4 dimensions. raw the row and column pictures for the equations x y, x C y. For two linear equations in three unknowns x; y; z, the row picture will show ( or ) (lines or planes) in ( or )-dimensional space. The column picture is in ( or )- dimensional space. The solutions normally lie on a. For four linear equations in two unknowns x and y, the row picture shows four. The column picture is in -dimensional space. The equations have no solution unless the vector on the right side is a combination of. 13 4.. Two Pictures of Linear Equations 9 Challenge Problems 8 Invent a by magic matrix M with entries ; ; ; 9. All rows and columns and diagonals add to. The first row could be 8; ; 4. What is M times.; ; /? What is M 4 times.; ; ; / if a 4 by 4 magic matrix has entries ; ;? 9 Suppose u and v are the first two columns of a by matrix A. Which third columns w would make this matrix singular? escribe a typical column picture of Av b in that singular case, and a typical row picture (for a random b). Multiplying by A is a linear transformation. Those important words mean If w is a combination of u and v, then Aw is the same combination of Au and Av. It is this linearity Aw cau C dav that gives us the name linear algebra. If u and v then Au and Av are the columns of A. Combine w cu C dv. If w how is Aw connected to Au and Av? A 9 by 9 Sudoku matrix S has the numbers ; ; 9 in every row and column, and in every by block. For the all-ones vector v.; ; /, what is Sv? A better question is Which row exchanges will produce another Sudoku matrix? Also, which exchanges of block rows give another Sudoku matrix? Section 4. will look at all possible permutations (reorderings) of the rows. I see orders for the first rows, all giving Sudoku matrices. Also permutations of the next rows, and of the last rows. And block permutations of the block rows? Suppose the second row of A is some number c times the first row a b A ca cb Then if a, the second column of A is what number d times the first column? 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Inverses and Transposes................................. 0.3 Column Spaces and NullSpaces............................. ### Vector algebra Christian Miller CS Fall 2011 Vector algebra Christian Miller CS 354 - Fall 2011 Vector algebra A system commonly used to describe space Vectors, linear operators, tensors, etc. Used to build classical physics and the vast majority ### ( % . This matrix consists of \$ 4 5 " 5' the coefficients of the variables as they appear in the original system. The augmented 3 " 2 2 # 2 " 3 4& Matrices define matrix We will use matrices to help us solve systems of equations. A matrix is a rectangular array of numbers enclosed in parentheses or brackets. 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The chapter ### Mathematics of Cryptography Modular Arithmetic, Congruence, and Matrices. A Biswas, IT, BESU SHIBPUR Mathematics of Cryptography Modular Arithmetic, Congruence, and Matrices A Biswas, IT, BESU SHIBPUR McGraw-Hill The McGraw-Hill Companies, Inc., 2000 Set of Integers The set of integers, denoted by Z, ### Algebra 2 Chapter 1 Vocabulary. identity - A statement that equates two equivalent expressions. Chapter 1 Vocabulary identity - A statement that equates two equivalent expressions. verbal model- A word equation that represents a real-life problem. algebraic expression - An expression with variables. ### Basic Terminology for Systems of Equations in a Nutshell. E. L. Lady. 3x 1 7x 2 +4x 3 =0 5x 1 +8x 2 12x 3 =0. Basic Terminology for Systems of Equations in a Nutshell E L Lady A system of linear equations is something like the following: x 7x +4x =0 5x +8x x = Note that the number of equations is not required ### LINEAR ALGEBRA W W L CHEN LINEAR ALGEBRA W W L CHEN c W W L Chen, 1997, 2008 This chapter is available free to all individuals, on understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied, ### Equations Involving Lines and Planes Standard equations for lines in space Equations Involving Lines and Planes In this section we will collect various important formulas regarding equations of lines and planes in three dimensional space Reminder regarding notation: any quantity ### Situation 23: Simultaneous Equations Prepared at the University of Georgia EMAT 6500 class Date last revised: July 22 nd, 2013 Nicolina Scarpelli Situation 23: Simultaneous Equations Prepared at the University of Georgia EMAT 6500 class Date last revised: July 22 nd, 2013 Nicolina Scarpelli Prompt: A mentor teacher and student teacher are discussing ### MATH 240 Fall, Chapter 1: Linear Equations and Matrices MATH 240 Fall, 2007 Chapter Summaries for Kolman / Hill, Elementary Linear Algebra, 9th Ed. written by Prof. J. Beachy Sections 1.1 1.5, 2.1 2.3, 4.2 4.9, 3.1 3.5, 5.3 5.5, 6.1 6.3, 6.5, 7.1 7.3 DEFINITIONS ### 4. Matrix inverses. left and right inverse. linear independence. nonsingular matrices. matrices with linearly independent columns L. Vandenberghe EE133A (Spring 2016) 4. Matrix inverses left and right inverse linear independence nonsingular matrices matrices with linearly independent columns matrices with linearly independent rows ### 521493S Computer Graphics. Exercise 2 & course schedule change 521493S Computer Graphics Exercise 2 & course schedule change Course Schedule Change Lecture from Wednesday 31th of March is moved to Tuesday 30th of March at 16-18 in TS128 Question 2.1 Given two nonparallel, ### Chapter 4: Systems of Equations and Ineq. Lecture notes Math 1010 Section 4.1: Systems of Equations Systems of equations A system of equations consists of two or more equations involving two or more variables { ax + by = c dx + ey = f A solution of such a system is an ### FURTHER VECTORS (MEI) Mathematics Revision Guides Further Vectors (MEI) (column notation) Page of MK HOME TUITION Mathematics Revision Guides Level: AS / A Level - MEI OCR MEI: C FURTHER VECTORS (MEI) Version : Date: -9-7 Mathematics ### Lecture 21: The Inverse of a Matrix Lecture 21: The Inverse of a Matrix Winfried Just, Ohio University October 16, 2015 Review: Our chemical reaction system Recall our chemical reaction system A + 2B 2C A + B D A + 2C 2D B + D 2C If we write ### Section 1.4. Lines, Planes, and Hyperplanes. The Calculus of Functions of Several Variables The Calculus of Functions of Several Variables Section 1.4 Lines, Planes, Hyperplanes In this section we will add to our basic geometric understing of R n by studying lines planes. If we do this carefully, ### Inverses. Stephen Boyd. EE103 Stanford University. October 27, 2015 Inverses Stephen Boyd EE103 Stanford University October 27, 2015 Outline Left and right inverses Inverse Solving linear equations Examples Pseudo-inverse Left and right inverses 2 Left inverses a number ### APPLICATIONS OF MATRICES. Adj A is nothing but the transpose of the co-factor matrix [A ij ] of A. APPLICATIONS OF MATRICES ADJOINT: Let A = [a ij ] be a square matrix of order n. Let Aij be the co-factor of a ij. Then the n th order matrix [A ij ] T is called the adjoint of A. It is denoted by adj
#### Graphing Polynomials For this section, it might be a good idea to recall the basic facts of the polynomial, p(x) = anxn + an-1xn-1 +...+ a1x + a0 in the Polynomials and Roots section. Here p(x) has at most n roots. What does that mean? Well, when you are graphing the polynomial, the number of different roots are the number of times the polynomial crosses the x-axis, where a root is the place where the polynomial meets the x-axis. Let's start with a polynomial of degree 1. This is a straight line that will cross the x-axis 1 time. The sign of the coefficient of x tells us if the line will be sloping this way / or \. Let's take a look at some examples. When we have a polynomial of degree 2, it's graph forms a parabola. The coefficient of the x2 term now determines if the parobola will point up (if the coefficient is positive) or down (a negative coefficient). Let's have a look at these types of graphs. We will do one more specific example and then generalize. So, the next one is a third degree polynomial. Again, the leading coefficient (the coefficient of the highest power term - in this case the x3) determines which way the graph will point. If it is positive, then the graph will go from lower left to upper right, while a negative coefficient will give a graph going from upper left to lower right. Take a look at the graphs for these. You might be noticing a pattern now. If the degree is odd, the polynomial will have at least one root and up to as many as the degree of the polynomial. The leading coefficient will determine whether the graph is tending from lower left to upper right or from upper left to lower right on the grand scale. When we have even degree polynomials, say of degree n, the graph has anywhere from 0 to n roots. Here the leading coefficient determines whether the graph points up (if it is positive) or down (if it is negative) in the big picture. So by plotting the roots and using these two "rules", we can get an idea of what the polynomial looks like. These just give a sketch of the graph of the polynomials. To draw the polynomials accurately takes a lot more mathematics then what we are covering in this section. Now we will work through a few examples.
# Write the cubes of 5 natural numbers which are of the form 3n + 1 (e.g. 4, 7, 10, ...) and verify the following: Question: Write the cubes of 5 natural numbers which are of the form 3n + 1 (e.g. 4, 7, 10, ...) and verify the following: 'The cube of a natural number of the form 3n + 1 is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1'. Solution: Five natural numbers of the form $(3 n+1)$ could be written by choosing $n=1,2,3 \ldots$ etc. Let five such numbers be $4,7,10,13$, and 16 . The cubes of these five numbers are: $4^{3}=64,7^{3}=343,10^{3}=1000,13^{3}=2197$ and $16^{3}=4096$ The cubes of the numbers $4,7,10,13$, and 16 could expressed as: $64=3 \times 21+1$, which is of the form $(3 n+1)$ for $n=21$ $343=3 \times 114+1$, which is of the form $(3 n+1)$ for $n=114$ $1000=3 \times 333+1$, which is of the form $(3 n+1)$ for $n=333$ $2197=3 \times 732+1$, which is of the form $(3 n+1)$ for $n=732$ $4096=3 \times 1365+1$, which is of the form $(3 n+1)$ for $n=1365$ The cubes of the numbers $4,7,10,13$, and 16 could be expressed as the natural numbers of the form $(3 n+1)$ for some natural number $n$; therefore, the statement is verified.
# How to solve for x and y In this blog post, we will be discussing How to solve for x and y. Our website can solving math problem. ## How can we solve for x and y These can be very helpful when you're stuck on a problem and don't know How to solve for x and y. Solving for exponents can be a tricky business, but there are a few basic rules that can help to make the process a bit easier. First, it is important to remember that any number raised to the power of zero is equal to one. This means that when solving for an exponent, you can simply ignore anyterms that have a zero exponent. For example, if you are solving for x in the equation x^5 = 25, you can rewrite the equation as x^5 = 5^3. Next, remember that any number raised to the power of one is equal to itself. So, in the same equation, you could also rewrite it as x^5 = 5^5. Finally, when solving for an exponent, it is often helpful to use logs. For instance, if you are trying to find x in the equation 2^x = 8, you can take the log of both sides to get Log2(8) = x. By using these simple rules, solving for exponents can be a breeze. Any mathematician worth their salt knows how to solve logarithmic functions. For the rest of us, it may not be so obvious. Let's take a step-by-step approach to solving these equations. Logarithmic functions are ones where the variable (usually x) is the exponent of some other number, called the base. The most common bases you'll see are 10 and e (which is approximately 2.71828). To solve a logarithmic function, you want to set the equation equal to y and solve for x. For example, consider the equation log _10 (x)=2. This can be rewritten as 10^2=x, which should look familiar - we're just raising 10 to the second power and setting it equal to x. So in this case, x=100. Easy enough, right? What if we have a more complex equation, like log_e (x)=3? We can use properties of logs to simplify this equation. First, we can rewrite it as ln(x)=3. This is just another way of writing a logarithmic equation with base e - ln(x) is read as "the natural log of x." Now we can use a property of logs that says ln(ab)=ln(a)+ln(b). So in our equation, we have ln(x^3)=ln(x)+ln(x)+ln(x). If we take the natural logs of both sides of our equation, we get 3ln(x)=ln(x^3). And finally, we can use another property of logs that says ln(a^b)=bln(a), so 3ln(x)=3ln(x), and therefore x=1. So there you have it! Two equations solved using some basic properties of logs. With a little practice, you'll be solving these equations like a pro. A calculus solver can be a helpful tool for these students. By enterinng the equation they are trying to solve, the solver will provide step-by-step instructions on how to solve it. This can be a valuable resource for students who are struggling to understand the material or simply need extra practice. With the help of a calculus solver, students can improve their grades and get a better understanding of the subject. In a right triangle, the longest side is called the hypotenuse, and the other two sides are called legs. To solve for x in a right triangle, you will need to use the Pythagorean theorem. This theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. In other words, if you know the lengths of all three sides of a right triangle, you can solve for any one of them using this equation. To solve for x specifically, you will need to square both sides of the equation and then take the square root of each side. This will give you the length of side x. You can then use this information to calculate the other two sides if needed. ## We cover all types of math issues Great experience overall, but for some reason I am unable to subscribe to the app plus. I click on the app plus where it should prompt me to purchase a subscription but it just loads forever and doesn't open an option for me to subscribe. I really love the app and would love to use plus as well, if only I could access it. Gracelyn Campbell I will say this is the best math solution app with best system. Whenever I face a math problem or I get confuse by thinking that either the math is correct or not, immediately I use the app. Using this app, you can get more knowledge on math. Kyla Coleman Get more math problem solver How to solve tangent Math apps that show work Second order linear differential equation solver 4 variable system of equations solver
# “The Key to Multiplication: Understanding the Answer Name” As we learn multiplication in our early years of schooling, we often focus solely on memorizing the times tables and knowing how to multiply numbers quickly. While this is a crucial aspect of understanding multiplication, it is only part of the equation. The other essential component is understanding the answer name. What is an answer name, you may ask? An answer name is the name given to the result of a multiplication problem. For example, if we multiply 3 by 4, the answer name is 12. Understanding the answer name goes beyond simply knowing the numerical value; it is a crucial concept that allows us to understand the underlying meaning of multiplication. To begin to grasp the importance of the answer name, it is helpful to first examine a basic multiplication problem. Let’s consider the problem 3 x 4. When we solve this problem, we find that the answer is 12. However, the answer name “12” represents much more than just a numerical value. In fact, the answer name can help us understand the relationship between the two numbers being multiplied. In this case, the answer name “12” represents the total number of objects we would have if we had three groups of four objects. This concept of grouping is at the heart of multiplication. When we multiply, we are essentially taking a certain number of groups and adding them together to find the total number of objects. Understanding the answer name allows us to see this connection and can help us conceptualize more complex problems. Let’s take a slightly more complex problem: 6 x 5. The answer name for this problem is 30, but what does that mean? To understand, we need to think about this problem in terms of groups. If we have six groups of five objects, we would have a total of 30 objects. This concept of grouping can be extended to even larger numbers, allowing us to multiply larger quantities with ease. Understanding the answer name is also important when we begin to explore division. Division is essentially the inverse of multiplication, and understanding the answer name can help us better understand this relationship. For example, if we know that 6 x 5 = 30, we can also say that 30 ÷ 6 = 5 or 30 ÷ 5 = 6. This is because we know that the answer name “30” represents the total number of objects we have when we group six sets of five objects. Understanding the answer name can also help us make connections between different multiplication problems. For example, if we know that 6 x 5 = 30, then we can also say that 5 x 6 = 30. This is because we are still combining the same number of groups, just in a different order. In conclusion, while memorizing the times tables is certainly an important aspect of understanding multiplication, it is only the beginning. Understanding the answer name is a crucial concept that allows us to see the underlying meaning of multiplication and make connections between different problems. By focusing on the answer name, we can develop a deeper understanding of multiplication that can make it easier to solve even the most complex problems.
# Understanding Similar Triangles: Exploring Geometric Relationships Welcome to Warren Institute! Today, we'll delve into the fascinating world of similar triangles. Similar triangles are a fundamental concept in geometry, where two triangles have the same shape but not necessarily the same size. Understanding their properties and relationships is crucial in various mathematical applications. In this article, we will explore the definition of similar triangles, their properties, and how to identify them. Join us as we unravel the mysteries of these geometric wonders! ## Definition of Similar Triangles Similar triangles are two or more triangles that have the same shape but not necessarily the same size. This means that their corresponding angles are equal, and their corresponding sides are in proportion. In other words, if you were to enlarge or shrink one of the triangles, the resulting shape would be identical to the other triangle. ## Properties of Similar Triangles One of the key properties of similar triangles is that their corresponding angles are congruent. Additionally, the ratios of the lengths of corresponding sides in similar triangles are equal. This property is known as the "angle-angle" criterion for similarity. Another important property is that if two angles of one triangle are equal to two angles of another triangle, then the triangles are similar. ## Methods to Determine Similarity There are several methods to determine whether two triangles are similar. These include the angle-angle (AA) criterion, the side-side-side (SSS) criterion, and the side-angle-side (SAS) criterion. The AA criterion states that if two angles of one triangle are equal to two angles of another triangle, then the triangles are similar. The SSS criterion states that if the corresponding sides of two triangles are in proportion, then the triangles are similar. The SAS criterion states that if two sides of one triangle are in the same proportion to two sides of another triangle, and their included angles are equal, then the triangles are similar. ## Applications of Similar Triangles Similar triangles have numerous applications in mathematics and real-world scenarios. They are used in trigonometry to solve problems involving angles and distances. Engineers and architects use the concept of similar triangles to scale down large structures for models or to estimate distances that are difficult to measure directly. Similar triangles are also fundamental to understanding the concept of proportions and solving problems involving indirect measurement. ## Practical Examples and Exercises To further understand the concept of similar triangles, it is helpful to work through practical examples and exercises. These could include finding missing side lengths or angles in similar triangles, using the properties of similar triangles to solve real-world problems, and applying the concept of similarity to geometric constructions. Practice with similar triangles helps reinforce the understanding of their properties and applications. I hope this helps! Let me know if you need further assistance. ## frequently asked questions ### What are similar triangles and how are they used in Mathematics education? Similar triangles are triangles with the same shape but different sizes. In Mathematics education, they are used to teach proportionality, ratios, and the concept of similarity in geometry. ### How can students identify and prove similar triangles in Mathematics education? Students can identify and prove similar triangles in Mathematics education by using the AAA (Angle-Angle-Angle) or SSS (Side-Side-Side) similarity criteria, as well as by applying the properties of similar triangles such as proportional side lengths and congruent angles. ### What are the properties of similar triangles that are important for Mathematics education? The important properties of similar triangles in Mathematics education are corresponding angles are congruent and corresponding sides are proportional. ### How do teachers effectively teach about similar triangles in Mathematics education? Teachers can effectively teach about similar triangles in Mathematics education by using visual aids, real-world examples, and hands-on activities to help students understand the concept of similarity and apply it to solve problems. ### What are some real-life applications of similar triangles that can be used to engage students in Mathematics education? Architectural design, map reading, and photography are some real-life applications of similar triangles that can engage students in Mathematics education. In conclusion, understanding similar triangles is crucial in mathematics education as it lays the foundation for various geometric concepts and problem-solving skills. Recognizing the proportionality of corresponding sides and angles in similar triangles enables students to apply this knowledge to real-world scenarios and more advanced mathematical topics. By grasping the properties and applications of similar triangles, students can enhance their spatial reasoning and geometric thinking, setting the stage for a deeper understanding of geometry and trigonometry. Overall, the study of similar triangles serves as a fundamental building block in the development of mathematical proficiency and critical thinking skills. See also Exploring the Key Traits of a Triangle: Unveiling its Characteristics If you want to know other articles similar to Understanding Similar Triangles: Exploring Geometric Relationships you can visit the category Geometry. Michaell Miller Michael Miller is a passionate blog writer and advanced mathematics teacher with a deep understanding of mathematical physics. With years of teaching experience, Michael combines his love of mathematics with an exceptional ability to communicate complex concepts in an accessible way. His blog posts offer a unique and enriching perspective on mathematical and physical topics, making learning fascinating and understandable for all. Go up
Enable contrast version # Tutor profile: Noah D. Inactive Noah D. Tutor/Teacher for 4 years & graduate student of marine ecology and modelling Tutor Satisfaction Guarantee ## Questions ### Subject:Trigonometry TutorMe Question: Suppose you have a right triangle with a hypotenuse that has length 8 and angle between the hypotenuse and adjacent leg of the triangle of $$\theta=43.2$$ Calculate the length of the adjacent side and the opposite side of the triangle. Inactive Noah D. For any right triangle we have the following: $$sin(\theta)=\frac{opp}{hyp}$$ $$cos(\theta)=\frac{adj}{hyp}$$ $$tan(\theta)=\frac{opp}{adj}$$ Given these simple relations we can solve for any number of properties of a right triangle. Step one: Calculate the length of the opposite side: $$sin(43.2)=\frac{x}{8}$$ multiply each side by 8 $$8sin(43.2)=x$$ plug into calculator $$x=-5.6$$ (we can ignore the negative here since there are no negative lengths!) Step two: Adjacent side $$cos(43.2)=\frac{x}{8}$$ $$8cos(43.2)=x$$ $$x=-5.67$$ (we can again ignore the negative sign So we know the sides of our triangle are 8, 5.6 and 5.67 respectively. ### Subject:Calculus TutorMe Question: Find the maximum and minimum of the following function and subsequent interval: $$f(x)=x^{4}-x^{3}$$ on the interval $$[0,7]$$ Inactive Noah D. Step 1: Take the derivative of $$f(x)$$ $$f'(x)=4x^{3}-3x^{2}=x^{2}(4x-3))$$ Note that $$f'(x)=0$$ when $$x=0$$ and $$x=(\frac{3}{4})$$ These are noted as "critical points". Step 2: Evaluate the function at the zeros and the endpoints of the interval itself (i.e. 0, 7, and $$(\frac{3}{4})$$ $$f(0)=0$$ $$f(7)=7^{4}-7^{3}=2058$$ $$f(\frac{3}{4})=(\frac{3}{4})^{4}-(\frac{3}{4})^{3}=-\frac{27}{256}$$ Thus $$f(x)$$ has a maximum of 2058 at $$x=7$$ and a minimum of $$-\frac{27}{256}$$ at $$x=\frac{3}{4}$$ ### Subject:Algebra TutorMe Question: Erin and her colleagues are ecologists on an excursion working on modelling the spatial distribution of the Red-Tailed Deer and Cougars in Oregon, USA. Before the excursion Erin decided that she would take data from an area of $$1km^{2}$$. Erin knows from previous studies that for every 100 deer observed in a $$1km^2$$ of this region there is 1 cougar in the habitat. After a 2 days of recording data in $$1km^{2}$$ Erin had observed 224 deer. She also noted that there were $$4km^{2}$$ of habitat in total that she was interested in. Erin wanted to make a quick estimation as to how many cougars she could expect to see in $$4km^{2}$$ but first needed to set up an equation to do so. Using a set of algebraic equations to solve this, how many cougars would Erin have estimated there to be in the $$4km^{2}$$ habitat? (Note: the ecological statistics in this problem are likely not accurate but rather used as a basis for understanding algebraic systems) Inactive Noah D. Story problems like this one can often be confusing to dissect. What information is important and what information isn't? Sometimes it can helpful to write what we know and what we need to know, for example: What we know: In $$1km^{2}$$ there are 224 deer. We are given a ratio of $$\frac{100 deer}{1 cougar}$$ in $$1km^{2}$$ There is $$4km^{2}$$ of total habitat. What we want to know: How many cougars are there in the $$4km^{2}$$ habitat? We can approach this problem in a variety of ways but the problem itself specifically calls for us to set up a set of algebraic equations to solve the problem. The first step is to figure out what are the variables: We first need to figure out how many cougars there are in $$1km^{2}$$: $$\frac{224deer}{Xcougars}=\frac{100deer}{1cougar}$$ We need to isolate X in order to solve for it. First we multiple each side by $$X$$: $$(X)(\frac{224}{X})=(\frac{100}{1})(X)$$ $$\Rightarrow$$ $$224=100X$$ Now we divide each side by 100 to isolate X: $$\frac{224}{100}=\frac{100X}{100}$$ $$\Rightarrow$$ $$\frac{224}{100}=X$$ $$\Rightarrow$$ $$=2.24$$ Which because we can't have .24 of a cougar, we round down to the nearest integer: 2. Now we know that in $$1km^{2}$$ of habitat there were 2 cougars. The problem wants us to figure out how many cougars there would be in $$4km^{2}$$ of habitat using algebra. Thus we'd use the same method as above: $$\frac{2cougars}{1km^{2}}$$=$$\frac{Ycougars}{4km^{2}}$$ Once again we are trying to isolate the variable, this time Y: $$4(\frac{2}{1km^2})=4(\frac{Y}{4km^{2}})$$ $$\Rightarrow$$ $$8=Y$$ Thus we can say that Erin's estimate of how many cougars she would have recorded in the $$4km^{2}$$ is 8. ## Contact tutor Send a message explaining your needs and Noah will reply soon. Contact Noah
## Interest rate compounded half-yearly 6 Nov 2015 10000 at 12% rate of interest for 1 year, compounded half-yearly. Solution: Amount with CI = 10000 [1+ (12/2 * 100)]2 = Rs. 11236. Therefore, CI 10 Oct 2019 Given,Principal = Rs 10000Here rate is compounded half-yearly,So, rate of interest = R = 10/2 %= 5%Time = 2 yearsn = number of half yearsn An interest rate compounded more than once a year is called the nominal interest rate of 8% p.a. compounded half-yearly is actually an effective rate of 8, 16% 13 Dec 2019 Calculate the amount and Compound interest if the interest is compounded half yearly. (a). principal = ₹ 2560 rate = 12 1/2 time = 1 year A bank offers 5% compound interest calculated on half-yearly basis. A customer R = rate. n = no.of years. But in the problem we are dealing with half year. Interest is compounded half-yearly, therefore, Amount = P ( 1 + (R/2) /100 )2n - - - - - - - - - [Interest compounded Half-yearly] Given : Principal = Rs. 20000, Rate Compound Interest when Compounded Half Yearly Example 2: Find the compound interest on Rs 8000 for 3/2 years at 10% per annum, interest is payable half-yearly. Solution: Rate of interest = 10% per annum = 5% per half –year. Word problems on compound interest when interest is compounded half-yearly: 1. Find the amount and the compound interest on $8,000 at 10 % per annum for 1$$\frac{1}{2}$$ years if the interest is compounded half-yearly. To find compound interest when interest is compounded half yearly, we use the following formula. n = number of years. Examples : 1) Compute the compound interest on$12,000 for 2 years ate 20% p.a. when compounded half-yearly. Solution : Here, P = $12,000, R = 20% and n = 2 years. Compound Interest (CI) is the addition of Interest to the Initial principal value and also the accumulated interest of previous periods of a loan or any deposit. Use this online compound interest calculator to calculate C.I compounded for annually, half-yearly, quarterly. ## Compound interest is the addition of interest to the principal sum of a loan or deposit, or in other The frequency could be yearly, half-yearly, quarterly, monthly, weekly, daily, The yearly compounded rate is higher than the disclosed rate. Compound Interest when Compounded Half Yearly Example 2: Find the compound interest on Rs 8000 for 3/2 years at 10% per annum, interest is payable half-yearly. Solution: Rate of interest = 10% per annum = 5% per half –year. Word problems on compound interest when interest is compounded half-yearly: 1. Find the amount and the compound interest on$ 8,000 at 10 % per annum for 1$$\frac{1}{2}$$ years if the interest is compounded half-yearly. To find compound interest when interest is compounded half yearly, we use the following formula. n = number of years. Examples : 1) Compute the compound interest on $12,000 for 2 years ate 20% p.a. when compounded half-yearly. Solution : Here, P =$12,000, R = 20% and n = 2 years. Compound Interest (CI) is the addition of Interest to the Initial principal value and also the accumulated interest of previous periods of a loan or any deposit. Use this online compound interest calculator to calculate C.I compounded for annually, half-yearly, quarterly. ### 2. to calculate how much compound interest payable based on the half-yearly rate of interest over a period of time with either monthly, quarterly, half-yearly or Calculation of Compound Interest When the Rate is Compounded half Yearly. Let us calculate the compound interest on a principal, P kept for 1 year at Find out how much compound interest you could earn on your savings, and daily compounding; monthly compounding; quarterly compounding; half yearly and Multiply the principal amount by one plus the annual interest rate to the power Question 1: Calculate the amount and compound interest on Rate has been halved because interest is compounded half yearly, time has been doubled for the 13 Dec 2019 Calculate the amount and Compound interest if the interest is compounded half yearly. (a). principal = ₹ 2560 rate = 12 1/2 time = 1 year A bank offers 5% compound interest calculated on half-yearly basis. A customer R = rate. n = no.of years. But in the problem we are dealing with half year. Interest is compounded half-yearly, therefore, Amount = P ( 1 + (R/2) /100 )2n - - - - - - - - - [Interest compounded Half-yearly] Given : Principal = Rs. 20000, Rate ### Compound Interest (CI) is the addition of Interest to the Initial principal value and also the accumulated interest of previous periods of a loan or any deposit. Use this online compound interest calculator to calculate C.I compounded for annually, half-yearly, quarterly. Compound Interest (CI) is the addition of Interest to the Initial principal value and also the accumulated interest of previous periods of a loan or any deposit. Use this online compound interest calculator to calculate C.I compounded for annually, half-yearly, quarterly. Compound Interest Calculation from simple Interest where Interest is compounded half yearly. If the rate of interest is R% per annum and the interest is compounded half-yearly, then the rate of interest will be R/2% per half year. Compound interest (CI) calculator - formulas & solved example problems to calculate the total interest payable on a given principal sum at a certain rate of interest over a period of time with either one of monthly, quarterly, half-yearly or yearly compounding frequency, in different world currencies such as USD, GBP, AUD, JPY, INR, NZD, CHF, RMB etc. When compounding of interest takes place, the effective annual rate becomes higher than the overall interest rate. The more times the interest is compounded within the year, the higher the effective annual rate will be. More information on effective annual interest rate can be found in this article from Investopedia. To convert a yearly interest rate for annually compounding loans, you can simply divide the annual interest rate into 12 equal parts. So, for example, if you had a loan with a 12 percent interest rate attached to it, you can simply divide 12 percent by 12, or the decimal formatted 0.12 by 12, in order to determine that 1 percent interest is essentially being added on a monthly basis. ## Compound Interest (CI) is the addition of Interest to the Initial principal value and also the accumulated interest of previous periods of a loan or any deposit. Use this online compound interest calculator to calculate C.I compounded for annually, half-yearly, quarterly. PREVIOUS A sum of Rs. 25000 was given as loan on compound interest for 3 years compounded annually at 5% per annum during the first year, 6% per annum during the second year and 8% per annum during the third year. Fixed Deposits are a great way to invest for those who rate safety higher than returns. This Fixed Deposit (FD) Calculator helps you find out how much interest you can earn on an FD and the value of your invesment (Principal) on Maturity when compounding of interest is done on a monthly, quarterly, half-yearly or yearly basis. Compound Interest is calculated on the initial payment and also on the interest of previous periods. Example: Suppose you give \$100 to a bank which pays you 10% compound interest at the end of every year. After one year you will have \$100 + 10% = \$110, and after two years you will have \$110 + 10% = \$121. Our online tools will provide quick answers to your calculation and conversion needs. On this page, you can calculate compound interest with daily, weekly, monthly, quarterly, half-yearly, and yearly compounding. You can also use this calculator to solve for compounded rate of return, time period and principal. Usually, the compounding is done quarterly, half-yearly and annually which means a number of compounding per year of 4, 2 and 1 respectively. Step 3: Finally, the formula for effective interest rate can be derived by using the stated rate of interest (step 1) and a number of compounding periods per year (step 2) as shown below. Let us see calculation difference for simple interest formula and compound interest formula. Suppose a person wants to start a yearly recurring deposit of$500 for a period of 10 years for the interest rate of 5%. Then he calculates the same and gets the below values. Interest rates on Deposits upto ` 2 Crore Rate of Interest (p.a.) Period Monthly Income Plan Quarterly Option Half-Yearly Option Annual Income Plan Cumulative Option* * For cumulative option, Interest is compounded annually. Compound interest occurs when interest is added to the original deposit – or principal – which results in interest earning interest. Financial institutions often offer compound interest on deposits, compounding on a regular basis – usually monthly or annually. The compounding of interest grows your investment without any further deposits
# Make Maths EZEE 1,013 9 1 I have a big fan of MATHS and in my 15 years of life till now I have learnt some tricks from various book of vedic maths and few I found out myself .I want to share some 8 tricks that I remember................Hope you will like them ............. ## Step 1: Vedic Maths WHAT IS Vedic maths? "Welcome to the wonderful world of "Vedic" mathematics, a science that its founder claims was lost due to the advent of modern mathematics. Vedic mathematics is said by its founder to be a gift given to this world by the ancient sages of India, though there is no historical evidence whatsoever for this claim. It is a system for limited arithmetic and polynomial calculation which is simpler and more enjoyable than the equivalent algorithms of modern mathematics." this description is taken from Wikipedia. ## Step 2: Squaring a Numbers Ending in 5 Multiply the first digit from left with digit + 1 and write 25 beside it......... Thus 25 = 2 X 3 / 25 = 625. In the same way, 35 = 3 X (3+1) /25 = 3 X 4/ 25 = 1225; 65 = 6 X 7 / 25 = 4225; 105 = 10 X 11/25 = 11025; 135 = 13 X 14/25 = 18225; ## Step 3: Multiplying by 11 Multiplying by 11 Working from right to left for 627 * 11 first write the digit 7 (right most) then (7+2) after that (6+2) then write 6 (left most digit) 6,(6+2),(7+2),7 = 6,8,9,7 thus , 627 * 11 = 6897 ## Step 4: Multiplying by 5 This is a simple one , and most people can figure it out by themselves...........But still I will show it..... eg. 5 * 46 we can easily find it out by dividing 46 by 2 and then add a zero that is 230 for decimal 5 * 4.62 we will have to dividing it with 2 and then sifting the decimal one step right 2/ 4.62 = 2.31 and answer is 23.1 ## Step 5: Multiplying by 15 Multiplying by 15 can be broken down into a multiplication by 10 plus a multiplication by 5. Multiplication with 10 can easily be done by putting a 0 at the end and multiplication with 5 is shown in the last step........ 345 * 15 = 3450 + 1725 = 5175 ## Step 6: Vulgar Fractions Whose Denominators Are Numbers Ending in NINE : First we recognize the last digit of the denominator of the type 1 / a9. Here the last digit is 9. For a fraction of the form in whose denominator 9 is the last digit, we take the case of 1 / 19 as follows: For 1 / 19, 'previous' of 19 is 1. And one more than of it is 1 + 1 = 2. Therefore 2 is the multiplier for the conversion. We write the last digit in the numerator as 1 and follow the steps leftwards. Step. 1 : 1 12 Step. 2 : 21(multiply 1 by 2, put to left) Step. 3 : 421(multiply 2 by 2, put to left) Step. 4 : 8421(multiply 4 by 2, put to left) Step. 5 : 168421 (multiply 8 by 2 =16, 1 carried over, 6 put to left) Step. 6 : 1368421 ( 6 X 2 =12,+1 [carry over] = 13, 1 carried over, 3 put to left ) Step. 7 : 7368421 ( 3 X 2, = 6 +1 [Carryover] = 7, put to left) Step. 8 : 147368421 (as in the same process) Step. 9 : 947368421 ( Do – continue to step 18) Step. 10 : 18947368421 Step. 11 : 178947368421 Step. 12 : 1578947368421 Step. 13 : 11578947368421 Step. 14 : 31578947368421 Step. 15 : 631578947368421 Step. 16 : 12631578947368421 Step. 17 : 52631578947368421 Step. 18 : 1052631578947368421 Now from step 18 onwards the same numbers and order towards left continue. Thus 1 / 19 = 0.052631578947368421 ## Step 7: All From 9 and the Last From 10 The formula can be very effectively applied in multiplication of numbers, which are nearer to bases like 10, 100, 1000i.e., to the powers of 10 . The procedure of multiplication using the Nikhilam involves minimum number of steps, space, time saving and only mental calculation. The numbers taken can be either less or more than the base considered. The difference between the number and the base is termed as deviation. Deviation may be positive or negative. ## Step 8: Multiplying Single Digit No. Now the base is 10. Since it is near to both the numbers, 7 we write the numbers one below the other Take the deviations of both the numbers from the base and represent _ the minus sign before the deviations 7 -3 8 -2 remainders 3 and 2 implies that the numbers to be multiplied are both less than 10 The product or answer will have two parts, one on the left side and the other on the right. A vertical or a slant linei.e., a slash may be drawn for the demarcation of the two parts i.e., The R.H.S. of the answer is the product of the deviations of the numbers. It shall contain the number of digits equal to number of zeroes in the base. _ i.e., 7 3 8 2 (3x2) = 6 Since base is 10, 6 can be taken as it is. L.H.S of the answer is the sum of one number with the deviation of the other. It can be arrived at in any one of the four ways. i) Cross-subtract deviation 2 on the second row from the original number7 in the first row i.e., 7-2 = 5. ii) Cross–subtract deviation 3 on the first row from the original number8 in the 20 second row (converse way of(i)) i.e., 8 - 3 = 5 iii) Subtract the base 10 from the sum of the given numbers. i.e., (7 + 8) – 10 = 5 iv) Subtract the sum of the two deviations from the base. i.e., 10 – ( 3 + 2) = 5 Hence 5 is left hand side of the answer. _Thus 7 3 8 2 ‾‾‾‾‾‾‾‾‾‾‾‾ 5 / remember the first valve we found 6 and the second value is 5 so the result is 56 ## Step 9: Multiplying (5436457665873632635399999999999999999999) That seem tough, but it is not... notice that both have 19 digit ..... this process is valid only when no. of. digit are equal and the second digit has only nine repeated. lets start with a simple one 5245 9999 on the right sides write the number by following the "All from 9 and the last from 10" rule or subtract the (first number) from (second number + 1).......... 10000 - 5245 = 4755 "All from 9 and the last from 10" is a easier method for finding it.......... now subtract 1 from the first number that is 5245 - 1 = 5244 placing them one after the other 5244 4755 5245 * 9999 = 52444755 try out 5436457665873632635399999999999999999999 on your own LOGIC/REASON 5245 9999 = 5245 * (10000 - 1) = 52450000 - 5245 = 52430000 + 10000 - 5245 = 52430000 + (10000 - 5245) = 52430000 + 4755 = 52444755 PROVED......... ## Step 10: Participated in the Big or Small Challenge ## Recommendations • ### Lamps Class 9,399 Enrolled ## Discussions Great work. This method seems so easy to solve mathematical calculations, when practiced prefectly. If kids are not going coachings, they should concentrate on this method. It would improve their calculation speed.I think, Students attending coachings like eyelevelashburnsouth, are so nicely trained that they don't require extra tools for math learning. Still, if practiced with these kind of methods, it would be icing on the cake.
# Urgent Finding Coordinates • Oct 7th 2008, 03:26 PM kaley015 Urgent Finding Coordinates Three points have coordinates A(2,5), B(10,9) and C(6,2). Line L1 passes through A and B. Line L2 passes through C and is perpendicular to L1. Find the coordinates of the point of intersection of L1 and L2? With working. • Oct 7th 2008, 03:32 PM icemanfan The slope of AB = L1 is $\displaystyle \frac{9-5}{10-2} = \frac{4}{8} = \frac{1}{2}$. So the slope of the line through C (L2) will have opposite reciprocal slope, which is -2. Now use point-slope form ($\displaystyle y - y_0 = m(x - x_0)$) to find the equations for both lines and then solve the system of equations given by the lines. • Oct 7th 2008, 03:42 PM masters Quote: Originally Posted by kaley015 Three points have coordinates A(2,5), B(10,9) and C(6,2). Line L1 passes through A and B. Line L2 passes through C and is perpendicular to L1. Find the coordinates of the point of intersection of L1 and L2? With working. First, find the slope of the L1. Then define the equation. $\displaystyle m=\frac{y_2-y_1}{x_2-x_1}=\frac{4}{8}=\frac{1}{2}$ Use y=mx+b with (2, 5) and m = 1/2 $\displaystyle y=mx+b$ $\displaystyle 5=\frac{1}{2}(2)+b$ $\displaystyle 5=1+b$ $\displaystyle 4=b$ Equation of L1 = $\displaystyle \boxed{y=\frac{1}{2}x+4}$ Now use the negative reciprocal of the slope of L1 for your perpendicular. That would be -2. Use point C(6,2) and define the equation of your perpendicular as above. Use y=mx+b, m = -2, passing through C(6, 2) $\displaystyle 2=-2(6)+b$ $\displaystyle 2=-12+b$ $\displaystyle 14=b$ Equation of L2 = $\displaystyle \boxed{y=-2x+14}$ Now, you can solve the system of those two linear equations and find where they intersect. Have fun.
# Conditional probability regarding multiple choice Q: A multiple choice exam has 4 choices for each question. A student has studied enough so that the probability they will know the answer to a question is 0.5, the probability that they will be able to eliminate one choice is 0.25, otherwise all 4 choices seem equally plausible. If they know the answer they will get the question right. If not they have to guess from the 3 or 4 choices. As the teacher you want the test to measure what the student knows. If the student answers a question correctly what’s the probability they knew the answer? A: Stuck in the process of trying to answer this. The way I thought about it was using Baye's rule and treating it as a conditional probability. In that, $$P(answer correct \| knew answer) = \frac{P(know answer \| answer correct) \cdot P(answer correct)}{P(knew answer)}$$ The probability that the student knows the answer is 0.5 so that gives us the denominator. Figuring out probability for the answer being correct and knowing the answer given a correct answer is a bit more confusing for me. Would appreciate guidance. Let A represent the event that the question is answered correctly Let X represent the event that the student knows the correct answer Let Y represent the event that the student is able to eliminate one choice Let Z represent the event that the student is not able to eliminate any choice I interpret the statement .... the probability they will know the answer to a question is 0.5, the probability that they will be able to eliminate one choice is 0.25, otherwise all 4 choices seem equally plausible. ... to mean ... $$P(X)=0.5; P(Y)=.25;p(Z)=.25$$ So ... $$\begin{eqnarray} P(X\|A) &=& \frac{P(A \cap X)}{P(A)} \\&=& \frac{P(A \cap X)}{P(A\cap X)+P(A\cap Y)+P(A\cap Z)} \\&=& \frac{P(A \|X)P(X)}{P(A \|X)P(X)+P(A \|Y)P(Y)+P(A \|Z)P(Z)} \\&=& \frac{1 \cdot \frac 12}{1 \cdot \frac 12+\frac 14 \cdot \frac 13+\frac 14 \cdot \frac 14} \\&=& \frac {\frac 12} {\frac{24}{48}+\frac4{48}+\frac3{48}} =\frac{24}{31} \end{eqnarray}$$ • Ah, yes, I interpreted the problem statement differently in my Answer, but I bet you have the correct interpretation! +1 – Bram28 Jul 12 '17 at 19:36 With $A$: students answers correctly $B$: student knew answer what you want is $P(B\|A)$, rather than $P(A\|B)$ Now, we are given that: $$P(B)= \frac{1}{2}$$ and hence $$P(B^C)=1-\frac{1}{2}=\frac{1}{2}$$ Also: $P(A\|B)=1$ (if they knew the answer they give the correct answer) and $$P(A\|B^C)=\frac{1}{4}\cdot \frac{1}{3}+\frac{3}{4}\cdot \frac{1}{4}=\frac{1}{12}+\frac{3}{16}=\frac{4}{48}+\frac{9}{48}=\frac{13}{48}$$ (When they don't know the answer there is a $\frac{1}{4}$ probability they can eliminate one of the answers and thus have a $\frac{1}{3}$ probability of guessing correctly between the remaining 3, and there is a $\frac{3}{4}$ probability they can't eliminate any one answer in which case they have a $\frac{1}{4}$ probability of guessing correctly) Next, we have that: $$P(A \cap B)=P(A\|B)\cdot P(B)=1\cdot \frac{1}{2}=\frac{1}{2}$$ and $$P(A \cap B^C)=P(A\|B^C)\cdot P(B^C)=\frac{13}{48}\cdot \frac{1}{2}=\frac{13}{96}$$ and thus: $$P(A)=P(A \cap B)+P(A \cap B^C)=\frac{1}{2}+\frac{13}{96}=\frac{61}{96}$$ So finally: $$P(B\|A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{1}{2}}{\frac{61}{96}}=\frac{48}{61}$$
# Section 2.2 Instantaneous Rates of Change ## Presentation on theme: "Section 2.2 Instantaneous Rates of Change"— Presentation transcript: Section 2.2 Instantaneous Rates of Change MAT 213 Brief Calculus Section 2.2 Instantaneous Rates of Change “At the instant the horse crossed the finish line, it was traveling at 42 miles per hour.” There is a paradox in trying to study motion at a particular instant in time. By focusing on a single instant, we stop motion!!! How can we know the speed of a horse at the instant it crosses the finish line? Instantaneous Rate of Change The instantaneous rate of change at a point on a curve is the slope of the curve at that point. Slope is the measure of tilt of a LINE! In the last section we discussed average rate of change as the slope of the secant line between two points Now instantaneous rate of change gives us the slope of the tangent line at a single point Local Linearity If we look close enough near any point on a smooth curve, the curve will look like a straight line! Example Grapefruit This is called the tangent line at that point Local Linearity If we look close enough near any point on a smooth curve, the curve will look like a straight line! This is called the tangent line at that point The slope of a graph at a point is the slope of the tangent line at that point Example On this graph, indicate where the slope of the tangent line is: positive greatest negative zero Find two points on the curve where the slopes are about the same. Instantaneous Rate of Change Average Rate of Change Graphically, the average rate of change over the interval a ≤ x ≤ b is the slope of the secant line connecting (a,f(a)) with (b,f(b)) Instantaneous Rate of Change Instantaneous rate of change at a is the slope of the line TANGENT to the curve at a single point, (a,f(a)) Secant line becoming tangent line The Tangent Line The tangent line at a point Q on a smooth, continuous graph is the limiting position of the secant lines between point Q and point P as P approaches Q along the graph (if the limiting position exists) Except at an inflection point. Examples Concave up Concave down The Tangent Line General Rule Lines tangent to a smooth, nonlinear curve do not “cut through” the graph of the curve at the point of tangency and lie completely on one side of the graph near the point of tangency Except at an inflection point. Examples Concave up Concave down Where instantaneous rate of change DOES NOT exist… Point of discontinuity Sharp point Vertical tangent (no “run”) Let’s look at some examples Consider the graph of f(x) = \|x\| Is it continuous at x = 0? Is it differentiable at x = 0? Let’s zoom in at 0 No matter how close we zoom in, the graph never looks linear at x = 0 Therefore there is no tangent line there so it is not differentiable at x = 0 Example- Piecewise Defined Function Example Note: This is a graph of It has a vertical tangent at x = 0 In groups let’s try the following from the book 7, 25
Short and Sweet Calculus ## 3.3 Graphing the Derivative If we have the graph of $$y=f(x)$$, we can draw a sketch of the graph of $$y=f'(x)$$ by estimating the slope of the tangent to the graph of $$f$$ at each $$x$$ value. We then plot the points $$(x,f'(x))$$ in the $$xy$$-plane and connect them by a smooth curve whenever possible. This curve represents the graph of $$y=f'(x)$$. • When you seek to graph the derivative, it is often easiest to first identify the places where the slope of the tangent line is 0. Those are the places where you can hope the sign on the derivative may change. Example 3.4. The graph of $$y=f(x)$$ is shown below. Give a rough sketch of the graph of $$y=f'(x).$$ Solution The tangent to the curve is horizontal at $$x=0$$ and $$x=2.$$ $\begin{cases} x<0 & \text{tangent line makes an acute angle with positive \ensuremath{x} axis}\Rightarrow f'(x)>0\\ 0<x<2 & \text{tangent line makes an obtuse angle with positive \ensuremath{x} axis}\Rightarrow f'(x)<0\\ 2<x & \text{tangent line makes an acute angle with positive \ensuremath{x} axis}\Rightarrow f'(x)>0 \end{cases}$ Specifically, the slope of the tangent line at $$x=-1$$ is $m_{\text{tan}}=\frac{3}{1}\Rightarrow f'(-1)=3.$ The slope of the tangent line at $$x=1$$ is $m_{\text{tan}}=\frac{-1}{1}\Rightarrow f'(1)=-1.$ At $$x=3$$ $m_{\text{tan}}=\frac{-3}{-1}\Rightarrow f'(3)=3.$ If we connects these points: $(-1,3),(0,0),(1,-1),(2,0),(3,3),$ we can sketch the graph of $$f’$$ (Figure 3.3).
Calculating Volumes - Cylindrical Shell Method # Calculating Volumes - Cylindrical Shells Method We have just looked at the method of using disks/washers to calculate a solid of revolution. We are now going to look at a new technique involving cylindrical shells. Suppose that we had a function $y = f(x)$, and we wanted to find the volume of the solid of revolution formed from rotating the area trapped between $f(x)$, the $x$-axis, and the lines $x = a$ and $x = b$ where $0 ≤ a < b$. The idea behind cylindrical shells is to "stack" multiple cylindrical shells within each other to form the solid. Recall that the volume of a cylinder can be obtained by the formula $v = \pi r^2 h$. Now let's calculate an equation for the volume of a single cylindrical shell with inner radius $r_1$ and outer radius $r_2$ as the diagram below illustrates We can calculate this volume by taking the volume of the larger cylinder, $V_{\mathrm{larger cylinder}} = \pi r_2^2 h$ and subtracting the volume of the smaller cylinder, $V_{\mathrm{smaller cylinder}} = \pi r_1^2 h$. We will also do some algebraic manipulation to be able to recognize some important components of our equation as follows: (1) \begin{align} V_{\mathrm{shell}} = \pi r_2^2 h - \pi r_1^2h \\ V_{\mathrm{shell}} = \pi(r_2^2 - r_1^2)h \\ V_{\mathrm{shell}} = \pi(r_2 + r_1)(r_2 - r_1)h \\ V_{\mathrm{shell}} = 2\pi\frac{r_2 + r_1}{2}(r_2 - r_1)h \end{align} We will note that $\frac{r_2 + r_1}{2}$ represents the average radius of our cylindrical shell which we will now denote simply as $r$, while $r_2 - r_1$ represents the change in the radius, or rather, $r_2 - r_1 = \Delta r$. Thus we obtain the formula $V_{\mathrm{shell}} = 2\pi r h \Delta r$ for any cylindrical shell. From this formula, we can easily identify the circumference of the circle defining the shell ($2\pi r$), the height of the shell ($h$), and the width of the shell ($\Delta r$). Now if we convert this formula in terms of our problem with calculating the solid of revolution with cylindrical shells, we let $\bar{x}$ will represent the average radius of a shell, $f(\bar{x})$ represents the height of our shell, and $\Delta x$ will represent the change in thickness between our inner and outer radii. Thus, we get the volume formula of the $k^{\mathrm{th}}$ single shell from our solid as $V = 2\pi \bar{x}_k f(\bar{x}_k) \Delta x$. If we divide our interval $[a, b]$ into $n$ subintervals $[x_{k-1}, x_{k}$, then we will let $\bar{x}_k$ be the midpoint of this $k^{\mathrm{th}}$ subinterval. Now, the volume of the entire solid can be approximated by taking these $n$ cylindrical shells and summing their volumes, that is, $V \approx \sum_{i=1}^{n} 2\pi \bar{x}_k f(\bar{x}_k) \Delta x$. As $n \to \infty$ (that is, the number of subintervals of $[a, b]$ so to infinity and thus, the number of cylindrical shells composing our solid of revolution goes to infinity), our approximation gets better and better, and thus: (2) \begin{align} V = \lim_{n \to \infty} \sum_{i=1}^{n} 2\pi \bar{x}_k f(\bar{x}_k) \Delta x \\ V = \int_{a}^{b} 2\pi x f(x) \: dx \end{align} The diagram below illustrates the technique we are implementing: Remark: We will note that a more concise formula for using this method by cylindrical shells would be that $V = \int_{a}^{b} (2 \pi r) h \: dx$ where $r$ is the radius of our shell, and $h$ is the height of our shell. In our general formula, we had $r = x$ and $h = f(x)$. We will now look at some examples of applying this method. ## Example 1 Determine the solid of revolution determined by the region bounded by the function $f(x) = 2x^2 - x^3$ around the $y$-axis. First let's find the $x$-intercepts of the function $f$ which defines the rotated region of area: (3) $$f(x) = 2x^2 - x^3 0 = x^2(2 - x)$$ So there exists x-intercepts at $x = 0$ and $x = 2$. We can now apply the formula for calculating the volume of revolution by cylindrical shells: (4) \begin{align} \mathbf{Volume} = \int_0^2 2\pi \: x \: (2x^2 - x^3) \: dx \\ \mathbf{Volume} = 2 \pi \int_0^2 2x^3 - x^4 \: dx \\ \mathbf{Volume} = 2 \pi [ \frac{x^4}{2} - \frac{x^5}{5} ] \bigg \|_{0}^{2} \\ \mathbf{Volume} = 2 \pi [ \frac{2^4}{2} - \frac{2^5}{5} ] = 2 \pi (8 - \frac{32}{5}) = \frac{16 \pi}{5} \end{align} ## Example 2 Determine the volume of the solid of revolution formed by rotating the region of the function $f(x) = \sqrt{x}$ bounded by the $x$-axis on the interval $[0, 2]$ about the $y$-axis. Setting up the integral and evaluating we obtain: (5) \begin{align} \mathbf{Volume} = \int_0^2 2\pi x \sqrt{x} \: dx \\ \mathbf{Volume} = 2\pi \int_0^2 x^{3/2} \: dx \\ \mathbf{Volume} = 2\pi [ \frac{2}{5}x^{5/2} ] \bigg \|_{0}^{2} \\ \mathbf{Volume} = \frac{4 \pi \sqrt{32}}{5} \end{align}
# Integer Exponents & the Quotient Rule An error occurred trying to load this video. Try refreshing the page, or contact customer support. Coming up next: How to Add and Subtract Rational Expressions ### You're on a roll. Keep up the good work! Replay Your next lesson will play in 10 seconds • 0:03 Integer Exponents • 0:53 Positive Integer Exponents • 1:33 Negative Integer Exponents • 2:23 Zero Exponent • 3:08 Quotient Rule • 5:19 Lesson Summary Save Save Want to watch this again later? Timeline Autoplay Autoplay Speed Speed #### Recommended Lessons and Courses for You Lesson Transcript Instructor: Laura Pennington Laura received her Master's degree in Pure Mathematics from Michigan State University. She has 15 years of experience teaching collegiate mathematics at various institutions. In this lesson, we will learn what integer exponents are. We will look at positive integer exponents, negative integer exponents, the zero exponent, and the quotient rule for exponents. At the end of the lesson, you can test your knowledge with a quiz. ## Integer Exponents Integer exponents are exactly what the name implies. They are integers that are exponents. Thus, let's first recall what integers and exponents are. Integers are numbers with no fractional part. They consist of the counting numbers and their negatives, along with the number zero. Exponents are numbers that we raise another number to. In an exponential expression, the number being raised to the exponent is called the base, and the number we are raising the base to is called the exponent. Knowing the definition of an integer and of an exponent allows us to understand the definition of an integer exponent. It is a number that is an exponent and an integer. This image shows some examples of integer exponents: ## Positive Integer Exponents Positive integer exponents are positive integers that are exponents. When we have a positive integer exponent, the exponent is telling us how many times we want to multiply the base by itself. For instance, consider the expression 2 4. This expression has 2 as its base and 4 as its exponent. Therefore, it's telling us to multiply 2 by itself 4 times. 2 x 2 x 2 x 2 = 16, so 2 4 = 16 Here are some more examples of positive integer exponents: ## Negative Integer Exponents Negative integer exponents differ from the positive integer exponents in that they consist of negative integers. When we raise a base to a negative integer, the negative flips the numerator and denominator of the base, and then the integer tells us how many times to multiply that number by itself. To demonstrate this, suppose we want to calculate 3 -2. The negative would flip the numerator and denominator of 3 to give 1/3, and then we would multiply 1/3 by itself 2 times. 3 -2 = (1/3) 2 = (1/3)(1/3) = 1/9 We can observe some more examples here: ## Zero Exponent We've considered positive integers and negative integers as exponents, so what's left? That's right, zero! Zero is neither positive nor negative, so we have a separate rule for when we raise a number to the power of zero. This rule is that anytime we raise a number (no matter what number) to the exponent zero, we get 1. Here are a few examples: • 20 0 = 1 • 53 0 = 1 • 1,387,720 0 = 1 Get the picture? No matter what our base is, if we raise to the exponent zero, we get 1. To unlock this lesson you must be a Study.com Member. ### Register to view this lesson Are you a student or a teacher? #### See for yourself why 30 million people use Study.com ##### Become a Study.com member and start learning now. Back What teachers are saying about Study.com ### Earning College Credit Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.
# Square & Square Root of 160 Created by: Team Maths - Examples.com, Last Updated: April 28, 2024 ## Square 160 The square of 160 represents the result when the number 160 is multiplied by itself. In mathematical notation, this is expressed as 160². To determine the square of 160, you simply perform the multiplication of 160 by 160. Mathematically, it is shown as: 160² (160 × 160) = 25,600 This calculation involves straightforward arithmetic, requiring the multiplication of 160 by itself. It is a common operation found in various mathematical fields, particularly in geometry where squaring numbers can be related to determining areas, and in algebra where exponentiation plays a key role in solving equations and exploring numeric properties. ## Square Root of 160 √160 ≈12.6491106407 The square root of 160 refers to the number that, when multiplied by itself, yields the product 160. In mathematical terms, finding the square root is the inverse operation of squaring. The notation for the square root of 160 is √160. To calculate the square root of 160, you identify the number that produces 160 when squared. The calculation reveals that approximately 12.65 × 12.65 ≈ 160, indicating that 12.65 is the square root of 160. Calculating the square root of 160 involves approximating the value since 160 is not a perfect square. However, through mathematical techniques such as estimation or using calculators, we can determine an approximate value for the square root. In this case, the square root of 160 is approximately 12.6491106407, showcasing a fundamental mathematical principle applied in various contexts, including geometry, algebra, and engineering. Square Root of 160: 12.6491106407 Exponential Form: 160^½ or 160^0.5 ## Is the Square Root of 160 Rational or Irrational? The square root of 160 is an irrational number To comprehend whether the square root of 160 is rational or irrational, let’s first define these terms. Rational numbers are those expressible as a fraction of two integers, where the denominator is not zero, written as a/b. Examples include 1/2, -3, and 5. Irrational numbers, however, cannot be expressed as simple fractions of two integers. Their decimal representations are non-repeating and non-terminating. Examples include √2, π (pi), and √3. Square Root of 160 as Irrational: When we calculate the square root of 160, we find that it does not simplify to a fraction of two integers. Its decimal expansion continues infinitely without repeating a pattern, indicating its irrational nature. In summary, the square root of 160 is irrational because it cannot be expressed as a simple fraction of two integers, and its decimal expansion is non-repeating and non-terminating. ## Method to Find Value of Root 160 Finding the square root of 160 involves various methods, each offering a unique approach to obtain the value. These methods are essential in mathematics and provide different strategies for calculating square roots efficiently and accurately. 1. Guess and Check Method: This method involves guessing potential numbers that might be the square root of 160 and checking by squaring them. It’s straightforward and effective for perfect squares. 2. Prime Factorization Method: Breaking down 160 into its prime factors (2^5 * 5) helps determine the square root by taking half the power of each prime factor. 3. Long Division Method: Similar to long division, this traditional approach is useful for finding square roots, especially for non-perfect squares or when higher precision is required. 4. Using a Calculator: For quick and precise results, utilizing a calculator is the simplest method. You input 160 and use the square root function to obtain the value. Understanding and employing these methods not only facilitate the calculation of the square root of 160 but also enhance overall mathematical skills and problem-solving abilities. ## Square Root of 160 by Prime Factorization Method The prime factorization method to find the square root of 160 involves breaking down the number 160 into its basic prime factors. Prime factorization means expressing a number as the product of its prime numbers, which are numbers greater than 1 that have no divisors other than 1 and themselves. Here’s how to use the prime factorization method for 160: 1.Divide the Number: Start by dividing 160 by the smallest prime number that can evenly divide it, which is 2. Keep dividing the quotient by 2 until you cannot evenly divide by 2 anymore.Now, 5 is a prime number, so we stop here. • 160÷2=80 • 80÷2=4080÷2=40 • 40÷2=2040÷2=20 • 20÷2=1020÷2=10 • 10÷2=510÷2=5 2.Write Down the Prime Factors: The prime factors of 160, based on the divisions above, are 2⁵ and 5. 3.Calculate the Square Root: To find the square root, take the square root of each prime factor and multiply them. For each prime factor, divide the exponent by 2 and multiply the results. Therefore, the square root of 160 by the prime factorization method is 4 × √5. This method is not only useful for finding square roots but also helps in understanding the fundamental structure of numbers, enhancing one’s ability to perform various mathematical calculations. ## Square Root of 160 by Long Division Method The square root of 160 by the long division method consists of the following steps: Step 1: Starting from the right, we pair up the digits of 160 by putting a bar above 60 and 1 separately. We also pair the 0s in decimals in pairs of 2 from left to right. Step 2: Find a number that, when multiplied to itself, gives a product less than or equal to 160. The number 12 fits here as 12 squared gives 144. Dividing 160 by 12 with the quotient as 12, we get the remainder as 16. Step 3: Drag a pair of 0’s down and fill it next to to make the dividend 1600. Step 4: Double the divisor 12, and enter 24 below with a blank digit on its right. Guess the largest possible digit (X) to fill in the blank and the quotient for which the product of 24X and X results in a value less than or equal to 1600. Since 6 fits the value of X, we fill 6 in the quotient after the decimal point. Divide and write the remainder. Step 5: Repeat this process to get the desired number of decimal places. ## Is the square root of 160 a Natural Number? The square root of 160 is approximately 12.65, which is not a natural number. Natural numbers are whole numbers greater than zero, without any fractions or decimals. Since 12.65 contains a decimal component, it is not considered a natural number. ## What square root is between 160? The square root of 160 is between 12 and 13. This is because the perfect square closest to 160 is 144 (12^2 = 144), and the perfect square greater than 160 is 169 (13^2 = 169). Therefore, the square root of 160 falls between 12 and 13. ## Is the Square Root of 160 a Perfect Square? No, the square root of 160 is not a perfect square. A perfect square is a number that is the square of an integer. Since 160 is not the square of any integer, its square root is not a perfect square. Text prompt
Courses Courses for Kids Free study material Offline Centres More Store # Simplify the following integral:$\int{\dfrac{{{\left( {{x}^{4}}-x \right)}^{\dfrac{1}{4}}}}{{{x}^{5}}}dx}$(a) $\dfrac{2}{5}{{\left( 1-\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{-5}{4}}}+c$(b) $\dfrac{4}{15}{{\left( 1-\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{-5}{4}}}+c$(c) $\dfrac{4}{15}{{\left( 1-\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{5}{4}}}+c$(d) $\dfrac{2}{5}{{\left( 1-\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{5}{4}}}+c$ Last updated date: 16th Jul 2024 Total views: 402.3k Views today: 5.02k Verified 402.3k+ views Hint: We solve this problem by converting the integral part as power of some variable by assuming the variable as some function of $'x'$ that is $t=f\left( x \right)$ Then we differentiate that assumption to get the value of $'dx'$ in terms of $'dt'$ so as to substitute in the given integral to get in the standard form that is $\int{{{t}^{n}}dt}$ We use the standard formula of integration that is $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$ By using the above formula we get the required value of integral. Let us assume that the given integral as $\Rightarrow I=\int{\dfrac{{{\left( {{x}^{4}}-x \right)}^{\dfrac{1}{4}}}}{{{x}^{5}}}dx}$ Now by taking the common term out from the numerator and cancelling with denominator we get \begin{align} & \Rightarrow I=\int{\dfrac{{{\left( {{x}^{4}} \right)}^{\dfrac{1}{4}}}{{\left( 1-\dfrac{x}{{{x}^{4}}} \right)}^{\dfrac{1}{4}}}}{{{x}^{5}}}dx} \\ & \Rightarrow I=\int{\dfrac{{{\left( 1-\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{1}{4}}}}{{{x}^{4}}}dx}..........equation(i) \\ \end{align} Now, let us assume that the value inside the bracket as some other variable that is $\Rightarrow t=1-\dfrac{1}{{{x}^{3}}}$ Now, by differentiating with respect to $'x'$ on both sides we get $\Rightarrow \dfrac{dt}{dx}=\dfrac{d}{dx}\left( 1 \right)-\dfrac{d}{dx}\left( \dfrac{1}{{{x}^{3}}} \right)$ We know that the standard formulas of differentiation as \begin{align} & \dfrac{d}{dx}\left( \text{constant} \right)=0 \\ & \dfrac{d}{dx}\left( {{x}^{n}} \right)=n.{{x}^{n-1}} \\ \end{align} By using these formulas to above equation we get \begin{align} & \Rightarrow \dfrac{dt}{dx}=0-\left( -3.{{x}^{-3-1}} \right) \\ & \Rightarrow dt=\dfrac{3}{{{x}^{4}}}dx \\ & \Rightarrow \dfrac{1}{{{x}^{4}}}dx=\dfrac{dt}{3} \\ \end{align} Now, by substituting the required values in equation (i) we get \begin{align} & \Rightarrow I=\int{{{\left( t \right)}^{\dfrac{1}{4}}}\left( \dfrac{dt}{3} \right)} \\ & \Rightarrow I=\dfrac{1}{3}\int{{{t}^{\dfrac{1}{4}}}dt} \\ \end{align} We know that the standard formula of integration that is $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$ By using this formula in above equation we get \begin{align} & \Rightarrow I=\dfrac{1}{3}\left( \dfrac{{{t}^{\dfrac{1}{4}+1}}}{\dfrac{1}{4}+1} \right)+c \\ & \Rightarrow I=\dfrac{1}{3}\left( \dfrac{4}{5}{{t}^{\dfrac{5}{4}}} \right) \\ & \Rightarrow I=\dfrac{4}{15}{{t}^{\dfrac{5}{4}}} \\ \end{align} Now, by substituting the value of $'t'$ in terms of $'x'$ in above equation we get $\Rightarrow I=\dfrac{4}{15}{{\left( 1-\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{5}{4}}}+c$ Therefore the value of given integral is $\therefore \int{\dfrac{{{\left( {{x}^{4}}-x \right)}^{\dfrac{1}{4}}}}{{{x}^{5}}}dx}=\dfrac{4}{15}{{\left( 1-\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{5}{4}}}+c$ So, the correct answer is “Option C”. Note: Students may make mistakes in converting the integral of $'x'$ to integral of some other variable. We have the value of integration as $\Rightarrow I=\int{\dfrac{{{\left( 1-\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{1}{4}}}}{{{x}^{4}}}dx}$ Then we assume that $\Rightarrow t=1-\dfrac{1}{{{x}^{3}}}$ After this assumption we need to differentiate the above equation to get the value as $\Rightarrow \dfrac{1}{{{x}^{4}}}dx=\dfrac{dt}{3}$ But, some students miss this differentiation and take the value as$dx=dt$ which will be wrong. We need to differentiate the above equation to get $'dx'$ in terms of $'dt'$
# 10th Maths Chapter 1 Exercise 1.6 Guide 10th Standard Maths Chapter 1 Exercise 1.6 Relations and Functions Guide Book Back Answers Solutions. TN 10th SSLC Samacheer Kalvi Guide. 10th All Subject Guide – Click Here. Class 1 to 12 All Subject Book Back Answers – Click Here ## 10th Maths – Chapter 1 Relations and Functions – Exercise 1.6 Guide ### 1.If n(A × B) = 6 and A = {1, 3} then n(B) is (1) 1 (2) 2 (3) 3 (4) 6 Solu.: (3) 3 Hint: If n(A × B) = 6 A = {1, 1}, n(A) = 2 n(B) = 3 ### 2.A = {a, b,p}, B = {2, 3}, C = {p, q, r, s) then n[(A ∪ C) × B] is …………. (1) 8 (2) 20 (3) 12 (4) 16 Solu.: (3) 12 Hint: A ∪ C = [a, b, p] ∪ [p, q, r, s] = [a, b, p, q, r, s] n (A ∪ C) = 6 n(B) = 2 ∴ n [(A ∪ C)] × B] = 6 × 2 = 12 ### 3.If A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8} then state which of the following statement is true. (1) (A × C) ⊂ (B × D) (2) (B × D) ⊂ (A × C) (3) (A × B) ⊂ (A × D) (4) (D × A) ⊂ (B × A) Solu.: (1) (A × C) ⊂ (B × D)] Hint: A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6}, D ={5, 6, 7, 8} A × C ={(1,5), (1,6), (2, 5), (2, 6)} B × D = {(1, 5),(1, 6),(1, 7),(1, 8),(2, 5),(2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8)} ∴ (A × C) ⊂ B × D it is true ### 4. If there are 1024 relations from a set A = {1, 2, 3, 4, 5} to a set B, then the number of elements in B is …………………. (1) 3 (2) 2 (3) 4 (4) 8 Solu.: (2) 2 Hint: n(A) = 5 n(A × B) = 10 (consider 1024 as 10) n(A) × n(B) = 10 5 × n(B) = 10 n(B) = 105 = 2 n(B) = 2 ### 5. The range of the relation R = {(x, x2)\|x is a prime number less than 13} is (1) {2, 3, 5, 7} (2) {2, 3, 5, 7, 11} (3) {4, 9, 25, 49, 121} (4) {1, 4, 9, 25, 49, 121} Solu.: (3) {4, 9, 25, 49, 121}] Hint: R = {(x, x2)/x is a prime number < 13} The squares of 2, 3, 5, 7, 11 are {4, 9, 25, 49, 121} ### 6. If the ordered pairs (a + 2,4) and (5, 2a + 6) are equal then (a, b) is ……… (1) (2, -2) (2) (5, 1) (3) (2, 3) (4) (3, -2) Solu.: (4) (3, -2) Hint: The value of a = 3 and b = -2 ### 7. Let n(A) = m and n(B) = n then the total number of non-empty relations that can be defined from A to B is (1) mn (2) nm (3) 2mn – 1 (4) 2mn Solu.: (4) 2mn Hint: n(A) = m, n(B) = n n(A × B) = 2mn ### 8. If {(a, 8),(6, b)} represents an identity function, then the value of a and 6 are respectively (1) (8,6) (2) (8,8) (3) (6,8) (4) (6,6) Solu.: (1) (8,6) Hint: f = {{a, 8) (6, 6)}. In an identity function each one is the image of it self. ∴ a = 8, b = 6 ### 9. Let A = {1, 2, 3, 4} and B = {4, 8, 9, 10}. A function f : A → B given by f = {(1, 4),(2, 8),(3, 9),(4, 10)} is a (1) Many-one function (2) Identity function (3) One-to-one function (4) Into function Solu.: (3) One-to one function Hint: A = {1, 2, 3, 4), B = {4, 8, 9,10} ### 10. If f(x) = 2×2 and g (x) = 13x, Then fog is Solu.: (3) 29×2 Hint: f(x) = 2×2 g(x) = 13x fog = f(g(x)) = f(13x)=2(13x)2 = 2 × 19×2=29×2 ### 11. If f: A → B is a bijective function and if n(B) = 7, then n(A) is equal to ………….. (1) 7 (2) 49 (3) 1 (4) 14 Solu.: (1) 7 Hint: n(B) = 7 Since it is a bijective function, the function is one – one and also it is onto. n(A) = n(B) ∴ n(A) = 7 ### 12. Let f and g be two functions given by f = {(0, 1), (2, 0), (3, -4), (4, 2), (5, 7)} g = {(0, 2), (1, 0), (2, 4), (-4, 2), (7, 0)} then the range of fog is (1) {0, 2, 3, 4, 5} (2) {-4, 1, 0, 2, 7} (3) {1, 2, 3, 4, 5} (4) {0, 1, 2} Solu.: (4) {0, 1, 2} Hint: gof = g(f(x)) fog = f(g(x)) = {(0, 2),(1, 0),(2, 4),(-4, 2),(7, 0)} Range of fog = {0, 1, 2} ### 13. Let f (x) = 1+x2−−−−−√ then ……………….. (1) f(xy) = f(x) f(y) (2) f(xy) > f(x).f(y) (3) f(xy) < f(x). f(y) (4) None of these Solu.: (3) f(xy) < f(x) . f(y) ### 14. If g = {(1, 1),(2, 3),(3, 5),(4, 7)} is a function given by g(x) = αx + β then the values of α and β are (1) (-1, 2) (2) (2, -1) (3) (-1, -2) (4) (1, 2) Solu.: (2) (2,-1) Hint: g(x) = αx + β α = 2 β = -1 g(x) = 2x – 1 g(1) = 2(1) – 1 = 1 g(2) = 2(2) – 1 = 3 g(3) = 2(3) – 1 = 5 g(4) = 2(4) – 1 = 7 (1) linear (2) cubic (3) reciprocal
# Can I get some help finding the x - intercept please? Thanks! Jan 15, 2018 ${x}_{1} = - 1 + \frac{1}{\sqrt{6}}$ and ${x}_{2} = - 1 - \frac{1}{\sqrt{6}}$ #### Explanation: $6 {x}^{2} + 12 x + 5 = 0$ This can be solved using the quadratic formula: $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ $x = \frac{- 12 \pm \sqrt{{12}^{2} - 4 \left(6\right) \left(5\right)}}{2 \left(6\right)}$ $x = \frac{- 12 \pm \sqrt{144 - 120}}{12}$ $x = - 1 \pm \frac{\sqrt{24}}{12} = - 1 \pm \frac{\sqrt{6 \times 4}}{12}$ $x = - 1 + \frac{2 \sqrt{6}}{12} = - 1 \pm \frac{\sqrt{6}}{6}$ $x = - 1 \pm \frac{1}{\sqrt{6}}$ So the $x$ intercepts are: ${x}_{1} = - 1 + \frac{1}{\sqrt{6}}$ and ${x}_{2} = - 1 - \frac{1}{\sqrt{6}}$ Jan 15, 2018 $\frac{- 6 + \sqrt{6}}{6}$ and $\frac{- 6 - \sqrt{6}}{6}$, or $- 0.59$ and $- 1.41$ #### Explanation: Use the quadratic formula to find the x-values of the equation. x=(-(12)+-sqrt((12^2)-4(6)(5)))/(2(6) Simplify the radicand to get $\sqrt{144 - 120}$, which simplifies to $\sqrt{24}$, or $2 \sqrt{6}$. Simplify the rest of the equation to get $\frac{- 12 \pm 2 \sqrt{6}}{12}$ which further simplifies to $\frac{- 6 \pm \sqrt{6}}{6}$. Make sure you understand why you are doing this! You are basically making the function equal to zero because whenever the function crosses the x axis, the x value is 0. Therefore, you are trying to find for what values the x equals zero. Jan 15, 2018 This is effectively asking us to solve $6 {x}^{2} + 12 x + 5 = 0$ The first way of trying to solve a quadratic equation is to factorise. We need two numbers that sum to $12$ and multiply to $5 \times 6 = 30$. Try thinking of some. In fact, there are no such numbers. So next, we turn to our two other ways of solving quadratic equations. Method 1 - Completing the Square This method is good if you don't have a calculator, and ideal if the coefficient of ${x}^{2}$ is 1 or a factor of the $x$ coefficient. $6 {x}^{2} + 12 x + 5 = 0$ ${x}^{2} + 2 x + \frac{5}{6} = 0$ ${\left(x + 1\right)}^{2} - 1 + \frac{5}{6} = 0$ ${\left(x + 1\right)}^{2} - \frac{1}{6} = 0$ ${\left(x + 1\right)}^{2} = \frac{1}{6}$ $x + 1 = \pm \sqrt{\frac{1}{6}}$ $x + 1 = \pm \frac{1}{\sqrt{6}}$ $x + 1 = \pm \frac{\sqrt{6}}{6}$ $x = - 1 \pm \frac{\sqrt{6}}{6}$ Tidy up $x = \frac{- 6 \pm \sqrt{6}}{6}$ So $x = \frac{- 6 + \sqrt{6}}{6}$ and $x = \frac{- 6 - \sqrt{6}}{6}$ Method 2 - The Quadratic Formula A much better method if you have a calculator. Given the choice, the formula is a much better option. This states that, for a quadratic $a {x}^{2} + b x + c = 0$, the solutions are given by: $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ So for this example, $a = 6 , b = 12$ and $c = 5$ $x = \frac{- 12 \pm \sqrt{{12}^{2} - 4 \times 6 \times 5}}{2 \times 6}$ $x = \frac{- 12 \pm \sqrt{144 - 120}}{12}$ $x = \frac{- 12 \pm \sqrt{24}}{12}$ $x = \frac{- \cancel{12} \textcolor{red}{6} \pm \cancel{2} \sqrt{6}}{\cancel{12} \textcolor{red}{6}}$ $x = \frac{- 6 \pm \sqrt{6}}{6}$ So which method is better to use? Factorising is the best method, although it is not always possible to factorise. Personally, the quadratic formula is the better method if you cannot factorise, since there are fewer steps involved compared to completing the square. Can I factorise? Look at the quadratic formula again. $x = \frac{- b \pm \sqrt{\textcolor{red}{{b}^{2} - 4 a c}}}{2 a}$ The area under the square root sign can tell us about the nature of the solutions. This expression, ${b}^{2} - 4 a c$, is known as the discriminant. The symbol $\Delta$ (capital delta) may be used to refer to this. -if $\Delta > 0$, then there are two distinct real roots (2 solutions) -if $\Delta = 0$ then there is one repeated real root (1 solution) -if $\Delta < 0$ then there are no real roots (does not cross the x axis). This is because we are square rooting a negative, which can be a no-go. If a quadratic factorises, then ${b}^{2} - 4 a c$ will be a positive square number since when we square root it, there will be no surd. In our example: ${b}^{2} - 4 a c = {12}^{2} - 4 \times 6 \times 5 = 24$ When we square root this, we get $\sqrt{24} = 2 \sqrt{6}$. Since we have a surd, this won't factorise. This is why there are no numbers to make such an expression in the beginning.
United States of AmericaPA 5.07 Interpreting proportional relationships Lesson We've already learned about proportional relationships, where two quantities vary in such a way that one is a constant multiple of the other. In other words, they always vary by the same constant. Remember! Proportional relationships are a special kind of linear relationship that can be written generally in the form $y=kx$y=kx and always pass through the origin $\left(0,0\right)$(0,0). To interpret information from a graph, we need to look at pairs of coordinates. Coordinates tell us how one variable relates to the other. Each pair has an $x$x value and a $y$y value in the form $\left(x,y\right)$(x,y). • The $x$x-value tells us the value of the variable on the horizontal axis. • The $y$y-value tells us the value of the variable on the vertical axis. It doesn't matter what labels we give our axes, this order is always the same. Let's look at some examples and see this in action. Worked example Question 1 The number of eggs farmer Joe's chickens produce each day are shown in the graph. What does the point $\left(6,3\right)$(6,3) represent on the graph? Think: The first coordinate corresponds to the values on the $x$x-axis (which in this problem would represent the number of days) and the second coordinate corresponds to values on the $y$y-axis ( which in this problem would represent the number of eggs). Do: Using the given information in context, we can interpret this point to mean that in $6$6 days the chickens will produce $3$3 eggs. Reflect: How many days does it take for the chickens to lay $1$1 egg? Practice questions question 2 The number of liters of gas used by a fighter jet over a certain number of seconds is shown in the graph. What does the point on the graph represent? 1. $6$6 liters of gas are used by the fighter jet every $12$12 seconds. A $12$12 liters of gas are used by the fighter jet every $6$6 seconds. B $6$6 liters of gas are used by the fighter jet every $12$12 seconds. A $12$12 liters of gas are used by the fighter jet every $6$6 seconds. B question 3 The number of cupcakes eaten at a party is shown on the graph. 1. What does the point on the graph represent? $2$2 cupcakes are eaten by $4$4 guests. A $4$4 cupcakes are eaten by $2$2 guests. B $2$2 cupcakes are eaten by $4$4 guests. A $4$4 cupcakes are eaten by $2$2 guests. B 2. Danielle is having a party, and expects to have $10$10 guests. According to the rate shown on the graph, how many cupcakes should she buy? Outcomes CC.2.1.7.D.1 Analyze proportional relationships and use them to model and solve real-world and mathematical problems. M07.A-R.1.1.5 Explain what a point (x, y) on the graph of a proportional relationship means in terms of the situation, with special attention to the points (0, 0) and (1, r), where r is the unit rate
# Ordered pair 🏆Practice ordered pair As we have already learned, Coordinate System has two axes and, therefore, any value defined by this coordinate system must include two values. These two values are called "ordered pairs". ## What is an ordered pair? An ordered pair is a pair of numbers that "belong" to a function. An ordered pair is generally represented in parentheses $(X, Y)$ where the value on the left within the parentheses $X$ represents the solution, while the value on the right within the parentheses $Y$ represents the function value, that is, the result. • When the function is represented by an equation, then $X$ is the value placed in the equation, while $Y$ is the result obtained. • When the function is represented through a graph, each point on the graph is essentially an ordered pair. That is, if we move vertically from the point to the $X$ axis (the axis runs from left to right), we obtain the value on the left within the parentheses. Conversely, if we move from the point vertically to the $Y$ axis (the axis that goes from bottom to top), we obtain the correct value within the parentheses. An ordered pair represents virtually all the points on the function graph ## Test yourself on ordered pair! Which point is marked on the map? ## Let's consider two examples of an ordered pair ### Example 1 Given the equation: $Y=2X-3$ • $X$ is the independent variable, meaning, the value that we input into the equation. • $Y$ is the dependent variable, meaning, the result obtained after deciphering the value of $X$. • If we establish that$X = 2$, we obtain that $Y = 1$, that is, a point or the ordered pair $(2,1)$ is on the linear function $Y = 2X-3$. • If we establish that $X = 3$, we obtain that $Y = 3$, that is, a point or the ordered pair $(3,3)$ is on the linear function$Y = 2X-3$. • If we establish that $X = 10$, we obtain that $Y = 17$, that is, a point or the ordered pair$(10,17)$ is on the linear function $Y = 2X-3$. ### Example 2 Given the equation $Y=X+3$ represented by the following graph: These ordered pairs are represented not only in a calculable manner, but also in the graph. Any point on the graph of the function $Y=X+3$ can be represented by an ordered type of values that can be easily found by drawing two verticals to each of the coordinates. These verticals are represented in the graph by dashed lines. If any point is on one of the axes, it means that the value of the other axis is zero, as in the case of $(0,3)$ in the present example. If you are interested in more information about "graphs" you can find detailed information in the following articles: Data collection and organization - statistical research Graph Discrete graph Continuous graph On the Tutorela website, you will find a variety of articles with interesting explanations about mathematics ## Examples and exercises with solutions of ordered pairs ### Exercise #1 Which point is marked on the map? ### Video Solution $(5,3)$ ### Exercise #2 Which point is marked on the graph? ### Video Solution $(6,1)$ ### Exercise #3 Choose the appropriate drawing where the dots appear: $(1,3),\lparen2,5),(3,3)$ ### Exercise #4 Choose the appropriate drawing where the dots appear: $(1,0),\lparen-2,-2),(-5,4)$ ### Exercise #5 Choose the figure that shows the following points: $(2,0),(2,4),(6,-2)$
# Harrison Chapman ## Math 317: Homework 1 Due: Friday, September 6, 2019 You are welcome to use any result of Theorem 3.1 and 3.2 in the book if you find them helpful. 1. Prove that $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$. We prove this using induction. As a base case, we notice that $$1^3 = 1 = (1)^2$$. Suppose now that $$1^3 + 2^3 + \cdots + k^3 = (1 + 2 + \cdots k)^2$$. Then: So we are done by induction. 2. Guess a formula for the sum $1 + 3 + \cdots + (2n - 1)$ (try evaluating for $n = 1,2,3,4$ and look for patterns). Then prove that your formula is correct. We claim that $$1 + 3 + \cdots + (2n - 1) = n^2$$. We prove this using induction. As a base case, we notice that $$1 = 1^2$$. Suppose now that $$1 + 3 + \cdots + (2k - 1) = k^2$$. Then: So we are done by induction. 3. For each subset of $\mathbb R$ below, determine both the maximum and the minimum, if they exist. If either doesn’t exist, say so. You do not need to give a rigorous proof of your answer. 1. Maximum 5, minimum 1 2. Maximum 1/2, minimum -1 3. Maximum 1, minimum -1 4. Maximum DNE, minimum DNE 5. Maximum DNE, minimum 1/2 4. Don’t worry about writing out any formal proofs in this problem. Decide whether each of the following statements is true. If the statement is true, you don’t need to do anything more. If the statement is false, give a concrete example (that is, a counterexample) that shows the statement failing. 1. An irrational number is an element of $\mathbb R \setminus \mathbb Q$. If $r \ne 0$ is rational and $\alpha$ is irrational, then $r + \alpha$ is irrational. 2. $\vert \vert a \vert - \vert b \vert \vert \le \vert a - b \vert$ for all $a, b \in \mathbb R$. 3. A nonempty finite set always has a maximum. 1. True. If $$r + \alpha = p/q$$ for some integers $$p, q$$ then $$\alpha = p/q - r$$, a rational; this is a contradiction. 2. True. Without loss of generality, assume that $$\vert a \vert \ge \vert b \vert$$ so that $$\vert \vert a \vert - \vert b \vert \vert = \vert a \vert - \vert b \vert$$. We have that $$\vert a \vert = \vert (a-b) + b \vert \le \vert a-b \vert + \vert b \vert$$ by the Triangle inequality, which is equivalent to saying that $$\vert a \vert - \vert b \vert \le \vert a - b \vert$$. 3. True. This is a fact about maximums; since the set is finite, a maximum can always be found in finite time (just look through all elements). 5. Let $A$ be a nonempty set of real numbers which has a minimum. Let $-A$ be the set $\{ -x \;:\; x \in A \}$. Prove that Let $$m = \min A$$. So for each $$a \in A, m \le a$$. Then for any $$y=-a \in -A$$, we have that $$m \le a$$ and so $$-m \ge -a = y$$. So $$-m = \max (-A)$$ or equivalently that $$\min A = -\max(-A)$$.
Notes on Sphere and Hemisphere \| Grade 10 > Compulsory Mathematics > Mensuration \| KULLABS.COM • Note • Things to remember • Videos • Exercise • Quiz We are so much familiar with spherical objects. The sphere is also a solid object whose each point in the outer surface is equidistance from the fixed point inside it. Such a fixed point is called the centre of the sphere. The constant distance is called the radius of the sphere. A solid object such as a globe, volleyball, toy ball, table tennis ball, marble etc is the example of a sphere. The figure as shown alongside is a sphere. The fixed point 'O' inside it is the centre which is equidistance from each point P on the surface of the sphere. So, OP = r is the radius of the sphere. If we cut a sphere through its diameter, there are two half spheres called the hemisphere and the cross section is called the great circle. The radius of the sphere is same as the radius of the great circle. The centre of the sphere and its great circle. The centre of the sphere and its great circle is same. Surface area of sphere The surface area of a sphere is the area of its outer part, which is a smooth curved surface. The surface area of a sphere is given by SA = 4πr2 where r is the radius of the sphere. The total surface area of hemisphere = 2πr2+ πr2 = 3πr2 square unit. Note Curved surface area of hemisphere = 2$$\pi$$r2square units. The surface area of a sphere (SA) = $$\pi$$d2, if the diameter is given. Volume of sphere The volume of a sphere means the space that it occupies. We can measure the volume of sphere experimentally. Fill up the measuring cylinder with the water level in the cylinder. The difference of two levels is the volume of the sphere. Alternatively Choose a sphere of given diameter (d) =10 cm (say). Immerse of the whole sphere into the water in the measuring cylinder, the water level is raised by 523.33 ml. Therefore its volume is 523.33 cm3. From this experiment, The diameter (d) = 10cm The volume of sphere (V) = 523.33 cm3 Let us make the ratio \begin{align} 6V:d^3 &= \frac {6V} {d^3} \\ &= \frac {6 \times 523.33} {10^3} \\ &= \frac {3140} {1000} \\ &= 3.14 \end{align} $$(\because \pi = \frac {22} {7} = 3.14)$$ \begin{align} \therefore \frac {6V} {d^3} &= \pi \\ or, V &= \frac {\pi (2r)^3} {6} \\ &= \frac {4 \pi r^3} {3} \\ \end{align} $$\therefore \text {Volume of a sphere} (V) = \frac {4 \pi r^3} {3} = \frac {\pi d^3} {6} \text {cubic units.}$$ 1. Curved surface area of hemisphere = 2$$\pi$$r2square units. 2. The surface area of a sphere (SA) =d2, if the diameter is given. 3. Volume of a sphere(v) = $$\frac{4\ (\pi) r^3}{3}$$ = $$\frac{\ (\pi)d^3}{6}$$ cubic units. . Click on the questions below to reveal the answers Here, r = 15 cm \begin{align} \text{Total Surface Area of hemisphere} &= 3{\pi}r^2\\ &= 3 \times \frac {22}7 \times 15 \times 15\\ &= \frac {14850}{7}\\ &= 2121.43 cm^2_{Ans}\\ \end{align} Surface Area of the sphere (A) = 616 cm2 radius of the sphere (r) = ? By formula, A = 4$$\pi$$r2 or, r2 = $$\frac A{4\pi}$$ or, r2 = $$\frac {616 \times 7}{4 \times 22}$$ or, r2 = 49 ∴ r = 7cmAns Suppose, Then, Surface Area of Sphere = 4$$\pi$$r2 or, $$\pi$$ = 4$$\pi$$r2 or, r2 = $$\frac {\pi}{4\pi}$$ or, r2 = $$\frac 14$$ or, r = $$\sqrt {\frac 14}$$ ∴ r = $$\frac 12$$cmAns Here, r = $$\frac {42 cm}{2}$$ = 21 cm \begin{align} \text{Volume of the hemisphere (V)} &=\frac 23{\pi}r^3\\ &= \frac 23 \times \frac {22}7 \times (21)^3\\ &= 44 \times (21)^2\\ &= 19404 cm^3_{Ans}\\ \end{align} Here, Volume of marble (V) = $$\frac {\pi}6cm^3$$ or, $$\frac 43$$ $$\pi$$r3 = $$\frac {\pi}6cm^3$$ or, r3 = $$\frac {\pi}6$$ $$\times$$ $$\frac 3{4\pi}cm^3$$ or, r3 = $$\frac 1{2^3}cm^3$$ ∴ r = $$\frac 12cm$$ Hence, the diameter of the marble = 2 $$\times$$ r = 2 $$\times$$ $$\frac 12$$ cm = 1 cmAns Let r be the radius of the sphere. Volume = $$\frac 43$$$$\pi$$r3 or, 38808 = $$\frac 43$$ $$\times$$ $$\frac {22}7$$ $$\times$$ r3 or, r3 = 38808 $$\times$$ $$\frac {21}{88}$$ or, r3 = 441 $$\times$$ 21 or, r3 = (21)3 ∴ r = 21Ans Here, Volume of sphere = $$\frac {3773}{21}cm^3$$ or, $$\frac 43$$$$\pi$$r3 = $$\frac {3773}{21}cm^3$$ or, $$\frac 43$$ $$\times$$ $$\frac {22}7$$ $$\times$$ r3 = $$\frac {3773}{21}cm^3$$ or, $$\frac {88}{21}r^3$$ = $$\frac {3773}{21}cm^3$$ or, r3 = $$\frac {3773}{21}$$ $$\times$$ $$\frac {21}{88}cm^3$$ or, r3 = $$\frac {343}{8}cm^3$$ or, r3 = ($$\frac 73$$cm)3 ∴ r = $$\frac 72$$cm Hence, Circumference of sphere = 2$$\pi$$r = 2 $$\times$$ $$\frac {22}7$$ $$\times$$ $$\frac 72$$ cm = 22 cmAns Here, Volume of spherical solid = $$\frac 43$$$${\pi}cm^3$$ or, $$\frac 43$$$${\pi}r^3$$ = $$\frac 43$$$${\pi}cm^3$$ or, r3 = $$\frac {\frac {4\pi}{3}}{\frac {4\pi}{3}}cm^3$$ or, r3 = 1 cm3 ∴ r = 1 cm Hence, the radius of spherical solid is 1 cm.Ans Here, Volume of sphere (V) = 36$${\pi}cm^3$$ Diameter of a sphere (d) = ? We know that, V = $$\frac 16$$$${\pi}d^3$$ or, d = $$\sqrt [3]\frac{6V}{d}$$ or, d = $$\sqrt [3]\frac {6 \times 36\pi}{\pi}$$ or, d = $$\sqrt [3]{6 \times 6 \times 6}$$ ∴ d = 6cm \begin{align} \text{The Surface Area (A)} &= {\pi}d^2\\ &= \frac {22}7 \times (6cm)^2\\ &= \frac {22}7 \times 36 cm^2\\ &= 113.14cm^2_{Ans}\\ \end{align} Here, Volume (V) = $$\frac {1372\pi}{3}cm^3$$ Total Surface Area (S) = ? By Formula, V = $$\frac 43$$$${\pi}r^3$$ or, $$\frac {1372\pi}{3}$$ =$$\frac 43$$$${\pi}r^3$$ or, r3 =$$\frac {1372\pi}{3}$$ $$\times$$ $$\frac 3{4\pi}$$ or, r3 = 343 or, r3 = 73 ∴ r = 7 We know that, \begin{align} S &= 4{\pi}r^2\\ &= 4 \times \frac {22}7 \times 7^2 cm^2\\ &= 616 cm^2_{Ans}\\ \end{align} Here, Surface Area of sphere = 616 cm2 or, 4$${\pi}r^2$$ = 616 cm2 or, $${\pi}r^2$$ = $$\frac {616}4 cm^2$$ ∴$${\pi}r^2$$ = 154 cm2 We know that, \begin{align} \text{Total Surface Area of a hemisphere} &= 3{\pi}r^2\\ &= 3 \times 154 cm^2\\ &= 462 cm^2_{Ans}\\ \end{align} Here, Total Surface Area of hemisphere (S) = 243 $${\pi}cm^2$$ Volume (V) = ? We know that, S = 3$${\pi}r^2$$ or, 243$$\pi$$ = 3$${\pi}r^2$$ or, r2 = $$\frac {243\pi}{3\pi}$$ or, r2 = 81 ∴ r = 9 cm Now, \begin{align} V &= \frac 23 {\pi}r^3\\ &= \frac 23 \times \frac {22}7 \times 9^3\\ &= 1527.43 cm^3_{Ans}\\ \end{align} Here, 4$${\pi}r^2$$ = $$\frac 1{4\pi}$$ or, r2 = $$\frac 1{(4\pi)^2}$$ ∴r = $$\frac 1{4\pi}$$ Now, \begin{align} \text{Volume (V)} &= \frac 43 {\pi}r^3\\ &= \frac {4\pi}3 (\frac 1{4\pi})^3\\ &= \frac 13 \times \frac 1{16\pi^2}\\ &= \frac 1{44\pi^2}cm^3_{Ans}\\ \end{align} Here, r1 = $$\frac {6cm}2$$ = 3cm r2 = $$\frac {8cm}2$$ = 4cm r3 = $$\frac {10cm}{2}$$ = 5cm Suppose, radius of new sphere = R Now, Volume of new sphere = sum of the volume of first, second and third spheres or, $$\frac 43$$$${\pi}R^3$$ = $$\frac 43$$$${\pi}r_1^3$$ +$$\frac 43$$$${\pi}r_2^3$$ +$$\frac 43$$$${\pi}r_3^3$$ or, R3 = r13 + r23 + r33 or, R3 = (33 + 43 + 53) or, R3 = 27 + 64 + 125 or, R3 = 216 or, R = $$\sqrt [3]{216}$$ ∴ R = 6 cm Hence, the diameter of new sphere (d) = 2R = 2 $$\times$$ 6 cm = 12 cmAns \begin{align} \text{Volume of small ball} &= \frac 43 {\pi}r^3\\ &= \frac 43 {\pi}1^3\\ &= \frac 43 {\pi}cm^3\\ \end{align} \begin{align} \text{Radius of big ball} &= \frac {8cm}2\\ &= 4 cm\\ \end{align} \begin{align} \text{Volume of big ball} &= \frac 43 {\pi}r^3\\ &= \frac 43 {\pi}4^3\\ &= \frac 43 {\pi}64cm^3\\ \end{align} Now, \begin{align} \text{Number of small balls that can be made from the big ball} &= \frac {\frac 43 {\pi} \times 64}{\frac 43 {\pi}}\\ &= 64_{Ans}\\ \end{align} Let x be the radius of the first sphere. Then, Radius of second sphere is $$\frac r4$$ Volume of first sphere (V1) = $$\frac 43$$ $${\pi}r^3$$ Volume of second sphere (V2) = $$\frac 43$$ $$\pi$$ ($$\frac r4$$)3 = $$\frac {4\pi}{3}$$$$\times$$ $$\frac {r^3}{64}$$ Now, $$\frac {V_2}{V_1}$$ = $$\frac {\frac {4\pi}{3} \times \frac {r^3}{64}}{\frac 43 {\pi}r^3}$$ = $$\frac 1{64}$$ Hence, V1 :V2 = 1 : 64Ans 0% 65 cm 45 cm 75 cm 84 cm 12 cm 10 cm 7 cm 15 cm 10.5 cm 15.71 cm 12.5 cm 9.5 cm 22 cm 29 cm 19 cm 12 cm 5 cm 2 cm 3 cm 10 cm 16 cm 12 cm 9 cm 14 cm 3 cm 4 cm 1 cm 2 cm 10 cm 5 cm 7 cm 8 cm 7 cm 4 cm 6 cm 5 cm 850 m 864 m 800 m 855 m 30cm 35 cm 25 cm 32 cm • The diameter of a hemisphere is 84 cm, find its total surface area and volume. 15532 cm2,155243 cm3 16632 cm2,155232 cm3 13904 cm2, 155238 cm3 16635 cm2, 155132 cm3 4159 cm2 4158 cm2 4170 cm2 4055 cm2 1845cm2 1848 cm2 1849 cm2 1846cm2 ASK ANY QUESTION ON Sphere and Hemisphere Forum Time Replies Report Pukar poudel If S snd V be the Surface area and Volume respectively.Prove that :36S^3=V^2 Hemisphere Find the radius of the hemisphere having total surface area 27πsq. cm.\ sushila total surface area of a sphere
Anda di halaman 1dari 6 # Centre for Education ## in Mathematics and Computing Euclid eWorkshop # 6 Circle Geometry c 2014 UNIVERSITY OF WATERLOO Euclid eWorkshop #6 C IRCLE G EOMETRY C IRCLE G EOMETRY Geometry and more specifically the geometry of the circle represents an area of mathematics from which relatively difficult and interesting problems for the Euclid contest are frequently taken. Although this is a very broad content area, we present only a brief outline of some of the more elementary results of the geometry of the circle. We will assume that the student has some knowledge of the elementary properties of triangles, including congruence and similarity, as well as the properties of parallelograms, rhombi and trapezoids that follow from these properties of triangles. The Star Trek Theorem The central angle subtended by any arc is twice any of the inscribed angles on that arc. In other words, in the accompanying figure AOB = 2ACB. Demonstration This theorem is the lynchpin of all the results that follow. It is useful to give a brief demonstration of this property to see that if follows directly from the properties of isosceles triangles. O L A In the diagram we draw the line connecting C and O and let a point on CO extended be L. Since OA, OB and OC are radii we have that the triangles OAC and OBC are isosceles. Thus OAC = OCA. Furthermore 1 OAC + OCA = AOL, the exterior angle of the triangle. Therefore OCA = (AOL). Similarly 2 1 OCB = (BOL). 2 1 1 Adding these 2 results ACB = ACO + OCB = (AOL + LOB) = AOB. 2 2 Extensions See if you can extend this theorem to obtain each of the following: 1. Show that if the chord AB is a diameter then ACB = 90 . (The angle subtended by a diameter is a right angle). 2. Show that the result is still true if AOB is greater than 180 . 3. Show that the result is true in the case where the point C is chosen so that the segments AC and OB intersect, that is, when C is moved along the circle towards the point B. (This proof will use differences where the earlier proof uses sums). 4. Show that if C1 and C2 are 2 different choices for the position of the point C along the arc AB then AC1 B = AC2 B. This result is described as angles subtended by the same arc(or chord) are equal. (The two points must both lie on the same side of the chord). 5. If C1 and C2 are two points on the circle, one on the minor arc AB and the other on the major arc AB, prove that AC1 B + AC2 B = 180 (the opposite angles of a cyclic quadrilateral are supplementary). C ENTRE FOR E DUCATIONS IN M ATHEMATICS C OMPUTING Euclid eWorkshop #6 C IRCLE G EOMETRY ## The Crossed Chord Theorem If two chords AB and CD of a circle intersect at the point P as shown in the diagram, prove that (P A)(P B) = (P C)(P D). A P D Proof Draw the line segments joining the points AD and BC. Then we observe using the result (4) above that BCD = BAD and ADC = ABC. Thus ADP is similar to CBP by the angle-angle similarity. Thus PA PD = , and by cross-multiplying we arrive at (P A)(P B) = (P C)(P D). PC PB Problem Two perpendicular chords AB and CD intersect at P . If P A = 4, P B = 10 and CD = 13, calculate the length of N A O P M D Solution We let x represent the length PC. Then, using the Crossed Chord Theorem, x(13 x) = 4(10) and x = 5 or 10 + 4 8+5 13 8. Now we label the midpoints AB and CD as M and N . Thus M B = = 7 and N C = = . 2 2 2 13 13 3 Thus N P = 8 or 5 = . However the line segment joining the centre of a circle to the mid 2 2 2 point of a chord is perpendicular to the chord and the chords were given perpendicular, so OM P N is a rectan3 gle. Thus OM = N P = and we can use the theorem of Pythagoras in triangle OM B to calculate the radius 2 s 2 3 205 = r = OB = 72 + . 2 2 ## C ENTRE FOR E DUCATIONS IN M ATHEMATICS C OMPUTING Euclid eWorkshop #6 C IRCLE G EOMETRY Extension In the diagram, P AB and P CD are two secants of the same circle and they intersect at the point P outside the circle. Prove that (P A)(P B) = (P C)(P D). D T C Solution Consider 4P AD and 4P CB. Since both share AP C we need only have one more pair of angles equal to establish similarity. But ADC and ABC both are subtended by the arc AC and so are equal. Thus P AD is similar to P CB and using the same logic as above we arrive at (P A)(P B) = (P C)(P D). Now imagine a series of secants passing through P , each intersecting the circle at two points but with the chord that is within the circle getting shorter and shorter as the secant approaches the point where it intersects the circle at a single point. In this case the secant becomes a tangent to the circle at the limiting point of intersection which we label T . We notice that as this process takes place, P A approaches P T and P B approaches P T . Thus we have (P A)(P B) = (P C)(P D) = (P T )2 . Try to prove this last statement directly using similarity and result (IV) below! Other Important Properties of Tangents If P is a point outside of a circle and we draw the two tangents to the circle P T and P S then the following results follow: I. A tangent at a point on a circle is perpendicular to the radius drawn to the point. (OT is perpendicular to P T ) II. P S = P T : Tangents to a circle from an external point are equal. III. OP bisects the angle between the tangents. (T P S). O T IV. Tangent Chord Theorem: Given that T A is any chord of a circle and P T is tangent to the circle at T . If C is a point on the circle chosen to be on the side of the chord opposite to the tangent then T CA = P T A. C O Proof Since we know OT is perpendicular to T P we draw in the radii OT and OA. Since the tangent is perpendicular 1 1 to the radius, P T A = 90 AT O = (180 AT O OAT ) = (AOT ) = ACT . 2 2 ## C ENTRE FOR E DUCATIONS IN M ATHEMATICS C OMPUTING Euclid eWorkshop #6 C IRCLE G EOMETRY P ROBLEM S ET 1. In a circle with centre O, two chords AC and BD intersect at P . Show that AP B = 1 (AOB + COD). 2 2. If the points A, B, C and D are any 4 points on a circle and P , Q, R and S are the midpoints of the arcs AB, BC, CD and DA respectively, show that P R is perpendicular to QS. 3. Calculate the value of x. 3x 2x 4. The three vertices of triangle ABC lie on a circle. Chords AX, BY, CZ are drawn within the interior angles A, B, C of the triangle. Show that the chords AX, BY and CZ are the altitudes of triangle XY Z if and only if they are the angle bisectors of triangle ABC. 5. As shown in the diagram, a circle with centre A and radius 9 is tangent to a smaller circle with centre D and radius 4. Common tangents EF and BC are drawn to the circles making points of contact at E, B and C. Determine the length of EF . A E D F 6. In the diagram, two circles are tangent at A and have a common tangent touching them at B and C respectively. (a) Show that BAC = 90 . (Hint: with touching circles it is usual to draw the common tangent at the point of contact!) (b) If BA is extended to meet the second circle at D show that CD is a diameter. C B A D ## C ENTRE FOR E DUCATIONS IN M ATHEMATICS C OMPUTING Euclid eWorkshop #6 C IRCLE G EOMETRY ## 7. If ABCD is a quadrilateral with an inscribed circle as shown, prove that AB + CD = AD + BC. A D B C 8. In this diagram, the two circles are tangent at A. The line BDC is tangent to the smaller circle. Show that AD bisects BAC. C D 9. Starting at point A1 on a circle, a particle moves to A2 on the circle along chord A1 A2 which makes a clockwise angle of 35 to the tangent to the circle at A1 . From A2 the particle moves to A3 along chord A2 A3 which makes a clockwise angle of 37 to the tangent at A2 . The particle continues in this way. From Ak it moves to Ak+1 along chord Ak Ak+1 which makes a clockwise angle of (33 + 2k) to the tangent to the circle at Ak . After several trips around the circle, the particle returns to A1 for the first time along chord An A1 . Find the value of n. 350 A1 A2 370 A3 410 390 A4
# 2.4 The cross product (Page 2/16) Page 2 / 16 Notice what this means for the direction of $\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}.$ If we apply the right-hand rule to $\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u},$ we start with our fingers pointed in the direction of $\text{v},$ then curl our fingers toward the vector $\text{u}.$ In this case, the thumb points in the opposite direction of $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}.$ (Try it!) ## Anticommutativity of the cross product Let $\text{u}=⟨0,2,1⟩$ and $\text{v}=⟨3,-1,0⟩.$ Calculate $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}$ and $\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}$ and graph them. We have $\begin{array}{ccc}\hfill \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}& =\hfill & ⟨\left(0+1\right),\text{−}\left(0-3\right),\left(0-6\right)⟩=⟨1,3,-6⟩\hfill \\ \hfill \text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}& =\hfill & ⟨\left(-1-0\right),\text{−}\left(3-0\right),\left(6-0\right)⟩=⟨-1,-3,6⟩.\hfill \end{array}$ We see that, in this case, $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=\text{−}\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}\right)$ ( [link] ). We prove this in general later in this section. Suppose vectors $\text{u}$ and $\text{v}$ lie in the xy -plane (the z -component of each vector is zero). Now suppose the x - and y -components of $\text{u}$ and the y -component of $\text{v}$ are all positive, whereas the x -component of $\text{v}$ is negative. Assuming the coordinate axes are oriented in the usual positions, in which direction does $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}$ point? Up (the positive z -direction) The cross products of the standard unit vectors $\text{i},\text{j},$ and $\text{k}$ can be useful for simplifying some calculations, so let’s consider these cross products. A straightforward application of the definition shows that $\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}=\text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}=\text{k}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}=0.$ (The cross product of two vectors is a vector, so each of these products results in the zero vector, not the scalar $0.\right)$ It’s up to you to verify the calculations on your own. Furthermore, because the cross product of two vectors is orthogonal to each of these vectors, we know that the cross product of $\text{i}$ and $\text{j}$ is parallel to $\text{k}.$ Similarly, the vector product of $\text{i}$ and $\text{k}$ is parallel to $\text{j},$ and the vector product of $\text{j}$ and $\text{k}$ is parallel to $\text{i}.$ We can use the right-hand rule to determine the direction of each product. Then we have $\begin{array}{cccccccc}\hfill \text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}& =\hfill & \text{k}\hfill & & & \hfill \text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}& =\hfill & \text{−}\text{k}\hfill \\ \hfill \text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}& =\hfill & \text{i}\hfill & & & \hfill \text{k}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}& =\hfill & \text{−}\text{i}\hfill \\ \hfill \text{k}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}& =\hfill & \text{j}\hfill & & & \hfill \text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}& =\hfill & \text{−}\text{j}.\hfill \end{array}$ These formulas come in handy later. ## Cross product of standard unit vectors Find $\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}\right).$ We know that $\text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}=\text{i}.$ Therefore, $\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}\right)=\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}=0.$ Find $\left(\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{k}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}\right).$ $\text{−}\text{i}$ As we have seen, the dot product is often called the scalar product because it results in a scalar. The cross product results in a vector, so it is sometimes called the vector product . These operations are both versions of vector multiplication, but they have very different properties and applications. Let’s explore some properties of the cross product. We prove only a few of them. Proofs of the other properties are left as exercises. ## Properties of the cross product Let $\text{u},\text{v},$ and $\text{w}$ be vectors in space, and let $c$ be a scalar. $\begin{array}{cccccccc}\text{i.}\hfill & & & \hfill \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}& =\hfill & \text{−}\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}\right)\hfill & & \text{Anticommutative property}\hfill \\ \text{ii.}\hfill & & & \hfill \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{v}+\text{w}\right)& =\hfill & \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}+\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\hfill & & \text{Distributive property}\hfill \\ \text{iii.}\hfill & & & \hfill c\left(\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}\right)& =\hfill & \left(c\text{u}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(c\text{v}\right)\hfill & & \text{Multiplication by a constant}\hfill \\ \text{iv.}\hfill & & & \hfill \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0& =\hfill & 0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}=0\hfill & & \text{Cross product of the zero vector}\hfill \\ \text{v.}\hfill & & & \hfill \text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}& =\hfill & 0\hfill & & \text{Cross product of a vector with itself}\hfill \\ \text{vi.}\hfill & & & \hfill \text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)& =\hfill & \left(\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}\right)·\text{w}\hfill & & \text{Scalar triple product}\hfill \end{array}$ ## Proof For property $\text{i}.,$ we want to show $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=\text{−}\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}\right).$ We have $\begin{array}{cc}\hfill \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}& =⟨{u}_{1},{u}_{2},{u}_{3}⟩\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}⟨{v}_{1},{v}_{2},{v}_{3}⟩\hfill \\ & =⟨{u}_{2}{v}_{3}-{u}_{3}{v}_{2},\text{−}{u}_{1}{v}_{3}+{u}_{3}{v}_{1},{u}_{1}{v}_{2}-{u}_{2}{v}_{1}⟩\hfill \\ & =\text{−}⟨{u}_{3}{v}_{2}-{u}_{2}{v}_{3},\text{−}{u}_{3}{v}_{1}+{u}_{1}{v}_{3},{u}_{2}{v}_{1}-{u}_{1}{v}_{2}⟩\hfill \\ & =\text{−}⟨{v}_{1},{v}_{2},{v}_{3}⟩\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}⟨{u}_{1},{u}_{2},{u}_{3}⟩\hfill \\ & =\text{−}\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}\right).\hfill \end{array}$ Unlike most operations we’ve seen, the cross product is not commutative. This makes sense if we think about the right-hand rule. where we get a research paper on Nano chemistry....? nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review Ali what are the products of Nano chemistry? There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others.. learn Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level learn da no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts Bhagvanji hey Giriraj Preparation and Applications of Nanomaterial for Drug Delivery revolt da Application of nanotechnology in medicine what is variations in raman spectra for nanomaterials ya I also want to know the raman spectra Bhagvanji I only see partial conversation and what's the question here! what about nanotechnology for water purification please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment. Damian yes that's correct Professor I think Professor Nasa has use it in the 60's, copper as water purification in the moon travel. Alexandre nanocopper obvius Alexandre what is the stm is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.? Rafiq industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong Damian How we are making nano material? what is a peer What is meant by 'nano scale'? What is STMs full form? LITNING scanning tunneling microscope Sahil how nano science is used for hydrophobicity Santosh Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq Rafiq what is differents between GO and RGO? Mahi what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq Rafiq if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION Anam analytical skills graphene is prepared to kill any type viruses . Anam Any one who tell me about Preparation and application of Nanomaterial for drug Delivery Hafiz what is Nano technology ? write examples of Nano molecule? Bob The nanotechnology is as new science, to scale nanometric brayan nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale Damian Is there any normative that regulates the use of silver nanoparticles? what king of growth are you checking .? Renato What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ? why we need to study biomolecules, molecular biology in nanotechnology? ? Kyle yes I'm doing my masters in nanotechnology, we are being studying all these domains as well.. why? what school? Kyle biomolecules are e building blocks of every organics and inorganic materials. Joe can you provide the details of the parametric equations for the lines that defince doubly-ruled surfeces (huperbolids of one sheet and hyperbolic paraboloid). Can you explain each of the variables in the equations?
# Exponents-lesson We start with the simplest example, that is when the exponent is 1. Base can be any natural number. By the way, do you know what natural numbers are? Last time I've been in one city in Montenegro. On a bus station there I discovered they have platform 0. Is that natural to have? What do you think? Let's start with our lesson. I wish you good luck in understanding how exponents work and I expect you write me your feedback about what do you need to better understand exponents. We denote by "^", "to the power", because of Word document we use. Let's start. 2^1 = 2, 3^1 = 3, 4^1 = 4,.. and so on. We see that on the right hand side of these equations we have 1 base number (in our case 2, 3 and 4). Let's remember, on the left hand side of these equations we have exponent 1, and on the right hand side we have 1 base number. After 1 there goes 2. Let's see what happens when we put exponent 2 to our base numbers 2, 3 and 4. 2^2 = 2 x 2 = 4, 3^2 = 3 x 3 = 9, 4^2 = 4 x 4 = 16. We see that things changed. Can you tell what? Hmmm, I see that on the left hand side of these equations we have exponent 2, and on the right hand side we have 2 base numbers and between them multiplication sign, x. Do you agree with me? What do we remember here? Base number to the exponent 2 is the same as (base number x base number). I bet you can do it yourself for exponent 3. Try. From all of this we can conclude general rule for exponent e and base number b. That is b^e = b x b x..x b (on the right hand side we have b multiplied by itself e times). Let us see what happens if we put for our base number 0 instead of a natural number. Remember, our general rule applies here too. That is, we have 0^1 = 1, 0^2 = 0 x 0 = 0. What is next? I am sure you know. 0^3 = 0, right? I think it is time for to conclude that for any exponent, e, we have 0^e = 0. What is next? Of course, negative base. We start with -1. General rule still applies! :) (-1)^1 = -1, (-1)^2 = -1 x (-1) = 1, (-1)^3 = -1 x (-1) x (-1) = -1, (-1)^4 = 1. What can we say here? First difference from the case when the base is a natural number is a BRACKET! Why do we put negative base into brackets? The reason is that (-1)^2 is different than -1^2. If we apply our general rule to -1^2, what do we get? -1^2 = - (1 x 1) = -1. Let us see few more examples to better understand the difference. -1^3 = - (1 x 1 x 1) = -1, (-1)^3 = -1 x (-1) x (-1) = -1, but -1^4 = -(1 x 1 x 1 x 1) = -1, (-1)^4 = (-1) x (-1) x (-1) x (-1) = 1. When an exponent is 0, we get 1^0 = 1, 2^0 = 1, 3^0 = 1, (-1)^0 = 1, (-2)^0 = 1, so we will remember for now that every number to the power of 0 equals 1. At this stage we just understand power of 0 as a rule because we don't have enough of knowledge for that now. Sometimes in mathematics we have situations like this, and they are very good for developing our mental skills. At one point, and it is a matter of our mental growth, this and similar things become very obvious. You don't know how that happened and that is a beauty of mathematics. We don't need to be discouraged when we don't understand something. It is so common in math. Since we are talking about rules that we will understand at one point, here is one more to remember, 0^0 is not defined!
Learn What Is 0.5 as a fraction & More Numbers can be represented in mathematics in a variety of ways, such as decimals, fractions, and percentages. A typical example of such a number that is also frequently used in mathematics is 0.5. This article will check out what is 0.5 as a fraction and other important factors about fraction and decimal. What are Decimals? Decimals are commonly used to represent non-whole numbers in written form. Between two whole numbers, there are ten decimal places, and a decimal number, such as 12.5, falls between the whole numbers 12 and 13. Decimals and fractions are equivalent, but they are written in different forms. Using the example of 12.5, it is equivalent to the mixed number 12 1/2, no matter how complex the decimal is. Another example is 0.75, which is equivalent to 3/4 or 75%. A decimal point, also called a decimal, is a dot that separates whole numbers from decimals. When considering decimal places, all numbers to the left of the decimal point are whole numbers, such as units, tens, hundreds, and thousands. The decimal places on the right represent tenths, hundredths, and thousandths. Understanding place value is crucial in solving mathematical problems, such as converting 0.5 as a fraction. Types of Decimals Decimal numbers can be separated into two groups based on how many digits after the decimal point: Like decimals: Two decimal values are said to be “like” decimals if the number of digits after the decimal point is exactly the same. Because they both include two digits after the decimal point, the decimals 6.34 and 2.67, for example, are Like decimals. Unlike decimals: When two decimal values have different sums of digits after the decimal point, they are said to be “contrary to” one another. For instance, because they both have different numbers of digits after the decimal point, 5.3 and 6.873 are not decimals. What Is a Fraction? Fractions are described as being a portion of a whole. A fraction’s numerical value is a portion or division of any quantity, with any quantity serving as the fraction’s source. So a fraction is a portion or element of a whole. In the example here, a pizza is cut into 8 equal pieces, each of which represents a portion of the entire pie. As a result, the quantity of the pizza is represented by the pieces. 1/8 is used to represent the removal of one part in fractional form. Keep reading to know the conversion of 0.5 as a fraction and the differences between fraction and decimal. When deciding how to show decimals, examine how you use them most frequently in your daily life. One of the most typical applications for decimals is the use of money. A very little portion of the entire value is represented by the decimal in money. When working with dollars, for instance, the decimal point shows a small fraction of a dollar. When a whole is small or fragmented, the type of part it is called depends on how many equivalent parts it has been divided into. For instance, portions made by dividing a strip into three equivalent pieces are referred to as thirds, and portions made by dividing a strip into five equivalent sections are referred to as fifths. Differences between fraction and decimal As we know about decimal and fraction letâ€™s check out the differences to better understand 0.5 as a fraction conversion. Decimals and fractions are two different methods to express numbers.Decimals and fractions are two different methods to express numbers. Fractions typically express ratios of whole numbers and may not always divide into an easy-to-express decimal. For example, 1/3 is represented by the repeating decimal 0.33333. By reversing a fraction, it can be transformed into its reciprocal, which is the number that can be multiplied by the fraction to produce 1. On the other hand, decimals can be used to express long, complex, and potentially infinite quantities such as the number pi. Decimals can also be useful for representing whole numbers when a whole-number ratio required to build a fraction is not available. How To Reduce A Fraction? Reducing a fraction only needs one step if you can find the highest common factor that both integers have. If you break this down into factors smaller than the largest common factor, it can take several steps. The biggest common component reduction, however, is unquestionably the best option because it only calls for one reduction. When two numbers are really large, it may take some time to find their largest common component. How to find Common Factor? The most important thing we need to convert 0.5 as a fraction is a common factor. Divide the two numbers into their prime components to discover what the two numbers have in common. Multiply the shares of the prime factors to get all common factors. What characteristics, for example, do 135 and 225 have in common that are more than 1? First found are the prime factors of 135 and 225. 3x3x3x5 for a total of 135 and 3x3x5x5 for 225. The numbers have a shared attribute of 3 x 3 x 5. By multiplying 3, 3, and 5 in different ways, it is possible to find the other common elements besides 3 and 5: 33=9, 35=15, and 3x3x5=45. What is 0.5 as a fraction? If you want to change a decimal number into a fraction such as 0.5 as a fraction, follow the instructions below. Step one: Write the provided number as the numerator and add the necessary number of zeros after placing 1 in the denominator immediately below the decimal point. In this case, we put 10 in the denominator and get rid of the decimal point because there is a digit after the decimal. It is going to be a 5 out of 10. Step two: After that, this fraction can be made simpler. 5/10 = 1/2 Visit- https://www.gettoplists.com/mobile-app-development-agency-atlanta/
{[ promptMessage ]} Bookmark it {[ promptMessage ]} 6-4a - Chapter 6.4 Permutations and Combinations Example... This preview shows pages 1–3. Sign up to view the full content. Chapter 6.4: Permutations and Combinations Example: Suppose six friends wish to line up for a photo. How many ways can this be done? Using the rule of product, there are six tasks to complete: who is the first person from the left, who is the second, the third, fourth, fifth, and sixth. There are six choices for the first person. Then, there will be five choices for the second person, four for the third person, and so on. 6 × 5 × 4 × 3 × 2 × 1 = 720 —– —– —– —– —– —– 1st 2nd 3rd 4th 5th 6th There are 720 different arrangements. Example: Suppose only five friends wish to line up for a photo. How many different arrangements are there? We solve the problem in the same way as before: 5 × 4 × 3 × 2 × 1 = 120 —– —– —– —– —– 1st 2nd 3rd 4th 5th There are 120 different arrangements. Notice that when we wished to arrange 6 people that it could be done 6 × 5 × 4 × 3 × 2 × 1 ways, and that when we wished to arrange 5 people, it could be done 5 × 4 × 3 × 2 × 1 ways. In general, arranging n objects, can be done n × ( n - 1) × · · · × 2 × 1 ways. These products have a special name, and a symbol. 1 This preview has intentionally blurred sections. Sign up to view the full version. View Full Document Suppose there are n distinct objects we wish to arrange. The number of different arrangements is n !, which we read as “ n factorial ”. The symbol n ! means the following: n ! = n × ( n - 1) × · · · × 2 × 1. n ! is defined for positive integer values of n . We also define 0! = 1. This tool helps us tackle certain counting problems quickly. You should note that your calculator contains a factorial button, to help speed up the solution of problems. Example: How many different ways are there to arrange the letters in the word “OBJECTS”? Since there are seven distinct letters in the word, then there are 7! = 5040 different arrangements. Example: Ten people get in line for a movie. How many different ways are there for them to get in line? This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]}
# Proving a function is a one-to-one correspondence 1. Oct 27, 2007 ### Jacobpm64 1. The problem statement, all variables and given/known data Determine whether each of the following functions is one-to-one, onto, neither or both. $$f : (2, \infty) \rightarrow (1, \infty)$$, given by $$f(x) = \frac{x}{x-2}$$ 3. The attempt at a solution So, I think this is one-to-one and onto. So i need to prove it. Claim: If $$f : (2, \infty) \rightarrow (1, \infty)$$, given by $$f(x) = \frac{x}{x-2}$$, then f is a one-to-one correspondence. Proof: Assume $$f : (2, \infty) \rightarrow (1, \infty)$$, given by $$f(x) = \frac{x}{x-2}$$. First we must show that f is one-to-one. Let $$a,b \in (2, \infty)$$ such that $$f(a) = f(b)$$. Notice that $$f(a) = f(b)$$ $$\frac{a}{a-2} = \frac{b}{b-2}$$ $$a(b-2) = b(a-2)$$ $$ab - 2a = ab - 2b$$ $$-2a = -2b$$ $$a = b$$ Hence, f is one-to-one. Here, do I need to use the fact that the codomain is (1, infinity) or the domain is (2, infinity)? Now we need to show that f is onto. Let $$b \in (1, \infty)$$ and take a = .... Here I would have solved b = a / (a-2) for a, but I do not know how to solve this. Is there any other way to do this proof besides solving for a then substituting back in to show that it gives me f(a) = b? 2. Oct 27, 2007 ### ZioX I think what you're getting at is since a,b in (2,infty) then a/(a-2) make sense (the denominator is nonzero). For ontoness (not a word) solving for b=a/(a-2) is the way to go. A lot of times you don't get a very nice function where you can't solve and so you'll have to show some kind of existence. But this is a nice function, not only can you show that there must exist an a mapping to b, but you can actually find the value of a. NB: You must demonstrate that a is in the domain of the function. If you keep that same function and fudge around with the domain you can lose ontoness. eg if you define f the same, but change the domain to (3,infty) you get a very different function: the image of f under (3,infty) is (1,3) which is missing quite a bit of (1,infty)! 3. Oct 27, 2007 ### Jacobpm64 Thanks a lot.
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. ### Course: Get ready for Algebra 2>Unit 5 Lesson 4: Introduction to the trigonometric ratios # Trigonometric ratios in right triangles Learn how to find the sine, cosine, and tangent of angles in right triangles. The ratios of the sides of a right triangle are called trigonometric ratios. Three common trigonometric ratios are the sine (sin), cosine (cos), and tangent (tan). These are defined for acute angle $A$ below: In these definitions, the terms opposite, adjacent, and hypotenuse refer to the lengths of the sides. ## SOH-CAH-TOA: an easy way to remember trig ratios The word sohcahtoa helps us remember the definitions of sine, cosine, and tangent. Here's how it works: Acronym PartVerbal DescriptionMathematical Definition $SOH$$\text{S}$ine is ${\text{O}}$pposite over ${\text{H}}$ypotenuse$\mathrm{sin}\left(A\right)=\frac{{\text{Opposite}}}{{\text{Hypotenuse}}}$ $CAH$$\text{C}$osine is ${\text{A}}$djacent over ${\text{H}}$ypotenuse$\mathrm{cos}\left(A\right)=\frac{{\text{Adjacent}}}{{\text{Hypotenuse}}}$ $TOA$$\text{T}$angent is ${\text{O}}$pposite over ${\text{A}}$djacent$\mathrm{tan}\left(A\right)=\frac{{\text{Opposite}}}{{\text{Adjacent}}}$ For example, if we want to recall the definition of the sine, we reference $SOH$, since sine starts with the letter S. The ${\text{O}}$ and the ${\text{H}}$ help us to remember that sine is ${\text{opposite}}$ over ${\text{hypotenuse}}$! ## Example Suppose we wanted to find $\mathrm{sin}\left(A\right)$ in $\mathrm{△}ABC$ given below: Sine is defined as the ratio of the ${\text{opposite}}$ to the ${\text{hypotenuse}}$ $\left(SOH\right)$. Therefore: Here's another example in which Sal walks through a similar problem: Trigonometric ratios in right trianglesSee video transcript ## Practice Triangle 1: $\mathrm{△}DEF$ $\mathrm{cos}\left(F\right)=$ $\mathrm{sin}\left(F\right)=$ $\mathrm{tan}\left(F\right)=$ Triangle 2: $\mathrm{△}GHI$ $\mathrm{cos}\left(G\right)=$ $\mathrm{sin}\left(G\right)=$ $\mathrm{tan}\left(G\right)=$ Challenge problem In the triangle below, which of the following is equal to $\frac{a}{c}$? ## Want to join the conversation? • hey I have a question what if we have a triangle with no known sides but 2 angles(including one right angle) is given then how will we find the 3rd angle and 3 sides? is it possible? • If you know two angles of a triangle, it is easy to find the third one. Since the three interior angles of a triangle add up to 180 degrees you can always calculate the third angle like this: Let's suppose that you know a triangle has angles 90 and 50 and you want to know the third angle. Let's call the unknown angle x. x + 90 + 50 = 180 x + 140 = 180 x = 180 - 140 x = 40 As for the side lengths of the triangle, you need more information to figure those out. A triangle of side lengths 10, 14, and 9 has the same angles as a triangle with side lengths of 20, 28, and 18. • How is theta defined in accurate mathematical language? • theta is not defined in math language, it is a symbol used as a variable to generally represent an angle. • What is the etymology of sin, cos and tan? • From Wikipedia - Trigonometric Functions - Etymology The word sine derives from Latin sinus, meaning "bend; bay", and more specifically "the hanging fold of the upper part of a toga", "the bosom of a garment", which was chosen as the translation of what was interpreted as the Arabic word jaib, meaning "pocket" or "fold" in the twelfth-century translations of works by Al-Battani and al-Khwārizmī into Medieval Latin. The choice was based on a misreading of the Arabic written form j-y-b (جيب), which itself originated as a transliteration from Sanskrit jīvā, which along with its synonym jyā (the standard Sanskrit term for the sine) translates to "bowstring", being in turn adopted from Ancient Greek χορδή "string". The word tangent comes from Latin tangens meaning "touching", since the line touches the circle of unit radius, whereas secant stems from Latin secans—"cutting"—since the line cuts the circle. The prefix "co-" (in "cosine", "cotangent", "cosecant") is found in Edmund Gunter's Canon triangulorum (1620), which defines the cosinus as an abbreviation for the sinus complementi (sine of the complementary angle) and proceeds to define the cotangens similarly. • IS there ANY way to easily remember the SIN, COS and TAN formulas?? Any tips and tricks? • SOH CAH TOA. The sine of theta, θ, or sine(θ = opposite side divided by hypotenuse, cosine(θ = adjacent side divided by hypotenuse, and tangent(θ = opposite divided by adjacent side. Or SOH CAH TOA • Based on the first paragraph, "The ratios of the sides of a right triangle are called trigonometric ratios.", if in trigonometry the ratios of the sides of a triangle are called 'trigonometric ratios' then what if the triangle is not a right triangle. Will the ratios of the sides of that triangle have a different label. And based on my question, how will the mnemonic 'soh cah toa' help find the sides of the 'non- right triangle' triangle? Are there more methods to find the sides of a triangle relative to trigonometric functions or formula? • How to find the sin, cos and tan of the 90 degree angle? Will we follow the same procedure as we did with the other two angles? • If we consider the right angle, the side opposite is also the hypotenuse. So sin(90)=h/h=1. By pythagorean theorem, we get that sin^2(90)+cos^2(90)=1. So, substituting, 1+cos^2(90)=1 cos^2(90)=0 cos(90)=0 And we see that tan(90)=sin(90)/cos(90)=1/0. So tan(90) is undefined. • I've heard that there are other trigonometric functions out there, with names like versine. Who decided that sine, cosine, and tangent would be the ones we learn in school? What happened to the others? • I would guess that it's because these functions are technically more complex than the ones we learn in school. For example, versine(x) = 1 - cos(x). Applications of these functions seem to be applicable to navigation, especially across a spherical plane. However, with the progression of technology (I assume) these older functions have grown less practical and have fallen away in favor of manipulations of the more familiar 6 trig functions we study today. • What is the symbol theta • Theta is a Greek letter that is commonly used in Math to symbolize a variable that represents an angle :) • Can you explain the multiple choice question, The Khan explanation didn't really help me. • First, let's think about the rules of SOH-CAH-TOA: SOH -> Sine = Opposite / Hypotenuse CAH -> Cosine = Adjacent / Hypotenuse TOA -> Tangent = Opposite / Adjacent This is relative to the value of the angle inputted into each of these functions. The question is asking us which of the follow values is equal to a/c. Note that you can select multiple answers, which provides a hint as to how many answers you should get. Let's evaluate them, one by one: cos(20) -> CAH -> cos(20) = Adjacent / Hypotenuse -> b / c sin(20) -> SOH -> sin(20) = Opposite / Hypotenuse -> a / c tan(20) -> TOA -> tan(20) = Opposite / Adjacent -> a / b cos(70) -> CAH -> cos(70) = Adjacent / Hypotenuse -> a / c sin(70) -> SOH -> sin(70) = Opposite / Hypotenuse -> b / c tan(70) -> TOA -> tan(70) = Opposite / Adjacent -> b / a Out of these 6 answer choices, only sin(20) and cos(70) produce the desired result of a / c. Hence, they are the two answers. I hope this clarified the question for you. If not, feel free to comment and ask away!
How to Solve Polynomial Functions Key Terms o Polynomial function o Subscript o Coefficient o Degree (of a polynomial) o Nonlinear function o Parabola o Factor Objectives o Recognize a polynomial function o Know how many solutions a polynomial equation has o Learn how to factor quadratic expressions o Know how to use the quadratic formula Polynomial Functions Another type of function (which actually includes linear functions, as we will see) is the polynomial. A polynomial function is a function that is a sum of terms that each have the general form axn, where a and n are constants and x is a variable. Thus, a polynomial function p(x) has the following general form: Note that we use subscripts with the a factors (also referred to as coefficients) to make the representation of the function more uniform and lucid. When the function has a finite number of terms, the term with the largest value of n determines the degree of the polynomial: we say that the function is a polynomial of degree n (or an nth degree polynomial). Thus, a polynomial of degree n can be written as follows: Notice, then, that a linear function is a first-degree polynomial: f(x) = mx + b Polynomials of a degree higher than one are nonlinear functions; that is, they do not plot graphically as a straight line. Instead, polynomials can have any particular shape depending on the number of terms and the coefficients of those terms. Finding the zeros of a polynomial function (recall that a zero of a function f(x) is the solution to the equation f(x) = 0) can be significantly more complex than finding the zeros of a linear function. For simplicity, we will focus primarily on second-degree polynomials, which are also called quadratic functions. A quadratic function is a polynomial of degree two. Because it is common, we'll use the following notation when discussing quadratics: f(x) = ax2 + bx + c Let's take a look at the shape of a quadratic function on a graph. We'll just graph f(x) = x2. The graph shows that the function is obviously nonlinear; the shape of a quadratic is actually a parabola. The quadratic function can be oriented either up (when a > 0, as in the above graph) or down (when a < 0), and it can be translated to any position in the plane (through variation of b and c). Note also that, depending on its location, the parabola can cross the x-axis in two places, in one place (as in the above graph), or nowhere at all. Generally, however, a quadratic equation has two solutions, which may or may not correspond to the same number. Furthermore, the solutions to a quadratic equation may be complex numbers. (In fact, it is generally the case that an equation consisting of a polynomial of degree n has n solutions.) There are two essential approaches to solving a quadratic equation: factoring and the quadratic formula. A solution to any equation f(x) = 0 is the value or values of x for which f(x) is zero (that is, for which it crosses the x-axis). Let's look at the example quadratic function above: f(x) = x2 = (x)(x) What we have done here is factor the original expression. We can now see that this quadratic function has two zeros, both of which are at x = 0. But what if the function is more complicated? Let's first consider the following general quadratic expression. We can expand the expression by carefully applying the rule of distributivity. f(x) = (px + q)(rx + s) = px(rx + s) + q(rx + s) = prx2 + pxs + qrx + qs f(x) = prx2 + (ps + qr)x + qs Note that this is fundamentally the same form as f(x) = ax2 + bx + c, but different names are used for the coefficients. The zeros of the function are the x values for which either factor is equal to zero-thus, we can see that there are generally two solutions to a quadratic. We can find these solutions by setting each factor equal to zero and solving for x. Of course, going from the factored form to the standard form is much more difficult than the reverse process, but in many cases the factored form can be found without too much difficulty. The process of factoring a quadratic function is usually a process of trial and error, but with practice, you can learn to spot how to factor some quadratics. Practice Problem: Factor the expression x2 - 4. Solution: This is our first problem involving factoring; start with what you know. We can write a quadratic expression as follows: (px + q)(rx + s) = prx2 + (ps + qr)x + qs More simply, we can assume p = r = 1 and write (x + q)(x + s) = x2 + (s + q)x + qs Now, we see that s must be equal to –q and that the product of s and q must be –4. s + q = 0 s = –q And, sq = –4 –s2 = –4 s2 = 4 Now, s can be either 2 or –2, but it can only be one or the other. Let's just pick one: s = 2. Then, q = –2. Let's write the new expression and check our result: (x + 2)(x – 2) = x(x – 2) + 2(x – 2) = x2 – 2x + 2x – 4 = x2 – 4 This result checks out. Practice Problem: Factor the expression x2 + 3x + 2. Solution: We can follow the same approach here as in the previous problem: our solution should have the following form: (x + a)(x + b) = x2 + (a + b)x + ab = x2 + 3x + 2 So, we know that ab = 2 and a + b = 3. Here, trial and error is the best route. By inspection, a = 1 and b = 2 satisfies these equations. ab = (1)(2) = 2 a + b = 1 + 2 = 3 b = 2 a = 1 The factored form is thus (x + 1)(x + 2). (Note that the zeros of the function are then x = –1 and x = –2.) Practice Problem: Factor the expression x2 + 1. Solution: Here we run into a slight problem. Let's take a look. (x + q)(x + s) = x2 + (s + q)x + qs = x2 + 1 We see that q = –s, but also qs = 1. qs = –s2 = 1 s2 = –1 s = ± Thus, we see that complex numbers can be involved in factors and zeros of quadratic functions. We can still handle this problem recalling what we already know about complex numbers. s = i q = –i So, the factored expression is (x + i)(x – i). Let's expand this just to check. (x + i)(xi) = x2 + ix – ix – i2 = x2 – (–1) = x2 + 1 The zeros of this expression are then i and –i. (As it turns out, complex solutions exist when the graph of the quadratic function does not cross the x-axis anywhere--try graphing the expression in this problem to see.) A more general and direct way to find the zeros of a quadratic function (and, also, to find the factors) is through the quadratic formula. We will not derive the quadratic formula here, but suffice it to say you can derive it using algebra. Given a quadratic function ax2 + bx + c, the zeros of the function are at x = Practice Problem: Find the solutions to the equation x2 – 4 = 0. Solution: We can use the factoring approach, as we did in a previous practice problem, or we can use the quadratic formula with a = 1, b = 0, and c = –4. Let's try this latter approach to compare. x = The solutions to the equation are then x = 2 and x = –2. In some cases, the use of the quadratic equation is faster, even though factoring of the quadratic expression is still an option.
## Graphing Lines ### Transcript So now we'll talk about graphing lines. You may well remember from high school that a big topic of the x-y plane has to do with graphing lines and finding equations of lines. So for example, we might have a line like this. We might have to find the equation of a line, or we might be given an equation and have to produce this line, something along those lines. In this lesson, we'll start with the very basics. This in many ways will just be kind of a conceptual introduction to the idea of graphing lines in the plane. First we have to focus on a few big ideas, before we can actually get to the mechanics of actually how to graph a particular line. So big idea number one, every possible line in the x-y plane has its own unique equation. So there's this, you could say a one to one pairing between a unique line and a unique equation, so that's a big idea. Every line has it's own equation. Big idea number two, for any given line, all the points on the line have x and y coordinates that satisfy the equation of the line. So that's a really big idea. That's a deep idea and people don't appreciate how deep that idea is. On any line there's an infinite number of points. All infinity of those points, every single one of them, we can pick out any point at all on that line, find its x coordinate, y coordinate, plug it in, and it would satisfy the equation of the line. That is absolutely huge. And finally big idea number three, any linear equation that relates x to the first power to y to the first power, as long as there is no multiplication or division of variables, or something odd like square roots or something. As long as there's just ordinary x and ordinary y and a bunch of numbers, that must be the equation of some line in the x-y plane. So for example we look at this. Y is to the first power. X is to the first power. That has to be the equation of some line in the x-y plane, and that's exactly why these are called linear equations. You may remember back in algebra when we were referring to these as linear equations, we were referring to them because every single one of them corresponds to a unique line in the x-y plane. So let's look at this particular equation, suppose the problem gave us that equation. We could find values that satisfy that equation, and these would be points on the line. So for example, we could just plug in. If we plug in x = 0, then we'd see that we'd get 3y = 12, so y would equal 4. So that means that 0,4 has to be one point on the plane. Similarly, we could plug in y equals 0. If y equals 0, then we get negative 4 equals 12. We divide. We get x equals negative 3. So that must be another point on the plane, x equals negative 3, y equals 0. So we have two points. Technically, two points are enough to determine a line. So for example, if we plot those two points and just draw a straight line between them we get a graph of a line. We were able to graph the line simply by plugging in points, but this is not the most efficient way to plot a line. In the coming lessons of this module, we'll learn much more efficient ways to plot the line that represents a particular equation. So we'll get to that, but first we have to make sure that we understand all the basics here. For now, once again, the really big idea is that any point on the line has to satisfy the equation of the line. This idea can be tested in a variety of contexts. So here's a relatively easy practice problem. Pause the video, and then we'll talk about this. Okay. We're given the equation of the line, and the equation of the line has this variable in it, the variable K. So we don't know what K is, but we're told that the line must pass through the point (2,1). Well we know that, that point has to satisfy the equation of the line. So if we plug in x equals 2 and y equals 1, we're gonna have to get an equation that works. So we'll do that, we'll plug in x equals 2, y equals 1, what we get is 2K plus 3K is 5K, 5K = 17,and so K must equal 17/5. So that's the value of K. In summary, every line in the x-y plane has its own unique equation. Every point on the line satisfies the equation of the line, and we can figure out the graph of a line by plotting individual points. This is one option, and we will learn other options in the later videos of this module.
# Unit 11 Volume And Surface Area Answer Key – Everything You Need To Know! ## Introduction Are you struggling with the volume and surface area calculations of Unit 11? Are you looking for the answer key to help you prepare for your exams? If yes, then you have come to the right place! In this article, we will provide you with the complete answer key for Unit 11, along with some tips and tricks to help you ace your exams. ## What is Unit 11? Unit 11 is a part of the mathematics curriculum that deals with the calculations of volume and surface area. In this unit, you will learn how to calculate the volume and surface area of different shapes, such as cubes, spheres, cylinders, and cones. These calculations are essential in many fields, such as engineering, architecture, and physics. ## What is Volume? Volume is the amount of space occupied by an object or a substance. In Unit 11, you will learn how to calculate the volume of different shapes using their dimensions, such as length, width, and height. For example, to calculate the volume of a cube, you need to multiply its length, width, and height. ## What is Surface Area? Surface area is the total area that the surface of an object occupies. In Unit 11, you will learn how to calculate the surface area of different shapes using their dimensions, such as length, width, and height. For example, to calculate the surface area of a cylinder, you need to add the areas of its top and bottom circles and the area of its curved side. ## Answer Key for Unit 11 Now, let's move on to the answer key for Unit 11. Please note that these answers are for reference purposes only, and you should try to solve the problems on your own before checking the answers. 1. What is the volume of a cube with a side length of 5 cm? Answer: The volume of the cube is 125 cubic cm. 2. What is the surface area of a sphere with a radius of 7 cm? Answer: The surface area of the sphere is 616 square cm. 3. What is the volume of a cylinder with a radius of 4 cm and a height of 10 cm? Answer: The volume of the cylinder is 502.65 cubic cm. 4. What is the surface area of a cone with a radius of 5 cm and a slant height of 13 cm? Answer: The surface area of the cone is 282.74 square cm. ## Tips and Tricks Here are some tips and tricks to help you ace your Unit 11 exams: 1. Practice, practice, practice! The more you practice, the better you will get at calculating volume and surface area. 2. Understand the formulas. Make sure you understand the formulas for calculating volume and surface area of different shapes. 3. Draw diagrams. Drawing diagrams can help you visualize the shapes and their dimensions, which can make it easier to calculate their volume and surface area. 4. Check your answers. Always double-check your answers to make sure you have calculated the volume and surface area correctly. ## Conclusion In conclusion, Unit 11 volume and surface area calculations can be challenging, but with the help of this answer key and some tips and tricks, you can ace your exams. Remember to practice, understand the formulas, draw diagrams, and double-check your answers. Good luck!
# 1. Turtles and Strings and L-Systems¶ This section describes a much more interesting example of string iteration and the accumulator pattern. Even though it seems like we are doing something that is much more complex, the basic processing is the same as was shown in the previous sections. In 1968 Astrid Lindenmayer, a biologist, invented a formal system that provides a mathematical description of plant growth known as an L-system. L-systems were designed to model the growth of biological systems. You can think of L-systems as containing the instructions for how a single cell can grow into a complex organism. L-systems can be used to specify the rules for all kinds of interesting patterns. In our case, we are going to use them to specify the rules for drawing pictures. The rules of an L-system are really a set of instructions for transforming one string into a new string. After a number of these string transformations are complete, the string contains a set of instructions. Our plan is to let these instructions direct a turtle as it draws a picture. To begin, we will look at an example set of rules: A Axiom A -> B Rule 1 Change A to B B -> AB Rule 2 Change B to AB Each rule set contains an axiom which represents the starting point in the transformations that will follow. The rules are of the form: left hand side -> right hand side where the left hand side is a single symbol and the right had side is a sequence of symbols. You can think of both sides as being simple strings. The way the rules are used is to replace occurrences of the left hand side with the corresponding right hand side. Now let’s look at these simple rules in action, starting with the string A: A B Apply Rule 1 (A is replaced by B) AB Apply Rule 2 (B is replaced by AB) BAB Apply Rule 1 to A then Rule 2 to B ABBAB Apply Rule 2 to B, Rule 1 to A, and Rule 2 to B Notice that each line represents a new transformation for entire string. Each character that matches a left-hand side of a rule in the original has been replaced by the corresponding right-hand side of that same rule. After doing the replacement for each character in the original, we have one transformation. So how would we encode these rules in a Python program? There are a couple of very important things to note here: 1. Rules are very much like if statements. 2. We are going to start with a string and iterate over each of its characters. 3. As we apply the rules to one string we leave that string alone and create a brand new string using the accumulator pattern. When we are all done with the original we replace it with the new string. Let’s look at a simple Python program that implements the example set of rules described above. Try running the example above with different values for the numIters parameter. You should see that for values 1, 2, 3, and 4, the strings generated follow the example above exactly. One of the nice things about the program above is that if you want to implement a different set of rules, you don’t need to re-write the entire program. All you need to do is re-write the applyRules function. Suppose you had the following rules: A Axiom A -> BAB Rule 1 Change A to BAB What kind of a string would these rules create? Modify the program above to implement the rule. Now let’s look at a real L-system that implements a famous drawing. This L-system has just two rules: F Axiom F -> F-F++F-F Rule 1 This L-system uses symbols that will have special meaning when we use them later with the turtle to draw a picture. F Go forward by some number of units B Go backward by some number of units - Turn left by some degrees + Turn right by some degrees Here is the applyRules function for this L-system. def applyRules(ch): newstr = "" if ch == 'F': newstr = 'F-F++F-F' # Rule 1 else: newstr = ch # no rules apply so keep the character return newstr Pretty simple so far. As you can imagine this string will get pretty long with a few applications of the rules. You might try to expand the string a couple of times on your own just to see. The last step is to take the final string and turn it into a picture. Let’s assume that we are always going to go forward or backward by 5 units. In addition we will also assume that when the turtle turns left or right we’ll turn by 60 degrees. Now look at the string F-F++F-F. You might try to use the explanation above to show the resulting picture that this simple string represents. At this point its not a very exciting drawing, but once we expand it a few times it will get a lot more interesting. To create a Python function to draw a string we will write a function called drawLsystem The function will take four parameters: • A turtle to do the drawing • An expanded string that contains the results of expanding the rules above. • An angle to turn • A distance to move forward or backward def drawLsystem(aTurtle,instructions,angle,distance): for cmd in instructions: if cmd == 'F': aTurtle.forward(distance) elif cmd == 'B': aTurtle.backward(distance) elif cmd == '+': aTurtle.right(angle) elif cmd == '-': aTurtle.left(angle) Here is the complete program in activecode. The main function first creates the L-system string and then it creates a turtle and passes it and the string to the drawing function. Feel free to try some different angles and segment lengths to see how the drawing changes.
Subscribe to Posts ## Rock and Box Question You have three boxes. One is red, one is yellow, one is green. You also have six unique rocks. How many ways can you put the six rocks in the three boxes if you must put two rocks in each box? Explain how you arrive at your answer. ### 4 Responses to "Rock and Box Question" 1. The obvious answer is that you're choosing four things from a set of six (because the you don't have any choice about the last two). That would lead you to think that there should be (65)(43)=360 possibilities. That's wrong, though, because we don't care what order the rocks were put into the boxes. A box that contains rock A and rock B is the same as a box that contains rock B and rock A. The easiest way to get to the right answer is to take the simple answer above and adjust it to remove duplicates. There are 65=30 possible ways to choose two rocks in order from a set of six, and 21=2 ways to arrange two things, so there are (65)/2=15 ways to choose two rocks from a set of six irrespective of order. Continuing, we have: ((65)/2)((43)/2)((21)/2) = 156*1 = 90 You can put the rocks in the boxes 90 different ways. 2. I arrived to the same answer: 90. Here's how I got it: There are 6! = 720 ways to permute six rocks. You then take the first two and put them in the first box, the next two in the second box and the last two in the third box. Now you get the same arrangement if you swap the two rocks in any of the boxes. There are 2^3 = 8 ways to swap or not the rocks in the three boxes, so the final answer is 720 / 8 = 90. The second way I tried was to figure out that there are 15 distinct ways to make three pairs out of 6 rocks (if you arbitrarily put rock #1 in the first pair, there are 5 ways to make the first pair and then 3 ways to arrange the remaining 4 rocks into 2 pairs) and then 3! = 6 ways to permute these pairs into 3 boxes. 6 x 15 = 90. I asked the question because I don't think that a 7th grader would know how to solve this problem. I thought I might have overlooked some glaringly obvious solution. 4. 7th graders are 11 to 13 years old, is that right? Then yes, that seems way too tough. Two boxes and four rocks would be more appropriate to make them understand the concept of indistinct pairs, assuming the teacher helps with the reasoning. I remember solving a lot of logic problems at school when I was that age, but we were doing them during class, with help from the teacher and only half of the class headcount (about 10 kids). Nothing as difficult as this and we had chosen the scientific option. The puzzles I preferred were the ones where there are N people living in N houses, each with different characteristics, and you had to tell who lived where using incomplete propositions such as "the doctor lives next to the house of the man who has a cat".
1. ## Algebra word problem At the neighborhood fruit market, cherries are priced at 2 lbs. for $3.50 and strawberries are priced at 3lbs. for$4.71. Cheryl bought 3lbs of cherries and 4 lbs. of strawberries. If the sales tax is 2%, how much change did Cheryl get from a $20 bill? round to the nearest cent. 2. Originally Posted by chellyluv At the neighborhood fruit market, cherries are priced at 2 lbs. for$3.50 and strawberries are priced at 3lbs. for $4.71. Cheryl bought 3lbs of cherries and 4 lbs. of strawberries. If the sales tax is 2%, how much change did Cheryl get from a$20 bill? round to the nearest cent. First, find the price of cherries and strawberries per pound. 2 lbs of cherries = $3.50 = 350 cents 1 lb of cherry =$\displaystyle \frac{350}{2} = 175$cents =$1.75 3 lbs of strawberries = $4.71 = 471 cents 1 lb of strawberries =$\displaystyle \frac{471}{3} = 157 $cents =$1.57 So, she buys 3 lbs of cherries and 4 lbs of strawberries to pay: $\displaystyle (3 \times 1.75) + (4 \times 1.57) = 5.25+ 6.28 = 11.53$ Add a 2% tax to this amount: $\displaystyle 11.53 + [ \frac{2}{100} \times 11.53 ] = 11.7606$ Note: 2% tax on $11.53 =$\displaystyle \frac{2}{100} \times 11.53 = 0.23$. This is the tax that she pays(23 cents) Rounding off to the nearest cents gives you:$11.76 So she gives a $20 bill means she gets back,$20.00 - $11.76 =$8.24 3. Originally Posted by chellyluv At the neighborhood fruit market, cherries are priced at 2 lbs. for $3.50 and strawberries are priced at 3lbs. for$4.71. Cheryl bought 3lbs of cherries and 4 lbs. of strawberries. If the sales tax is 2%, how much change did Cheryl get from a $20 bill? round to the nearest cent. Edit: Moving a bit slow these days. Hi chellyluv, This is more of an arithmetic problem, than it is an algebra problem.$\displaystyle C=\frac{3.50}{2}=1.75$per pound$\displaystyle S=\frac{4.71}{3}=1.57$per pound She spent$\displaystyle 3C + 4S = 3(1.75)+4(1/57)=11.53$2% tax on 11.53 =$\displaystyle .02 \times 11.53 = .23$Total spent including tax =$\displaystyle 11.53+.23=11.76\$ You figure the change from a 20 dollar bill.
# Grade 7 Maths Ncert solutions Simple Equations NCERT books are recommended text book for CBSE school curriculum. Most of the govt aided and private schools in India follow NCERT text books as the syllabus books for school subjects. They are written by eminent academicians and scholars from leading universities and academic institutions in India, and help in building a strong base, as well help understand each concept by relevant examples and exercises. They are termed best in term of concept building, for their usage of real life practical examples, and good end of the chapter exercises. In this article, we have shared solved NCERT exercises for Class 7 maths Simple equation chapter. Simple equations is an important chapter, as it set a base for the algebra for the future classes, as well as build the arithmetic skills among the students, when they analyze various equations and solve them by performing required mathematical operations. These NCERT solutions will help students in solving their doubts and clarifying confusion regarding confusions regarding Simple equations chapter. Students can refer to NCERT solutions for simple equations chapter for their conceptual understanding, as well as matching their own answers with the right solutions of the exercises. ## Class 7 math ncert solution simple equations chapter Exercise 4.1 solutions Q1 – Complete the last column of the table View solution Q2 – Check whether the value given in the brackets is a solution to the given equation or not: (a) n + 5 = 19 (n = 1) (b) 7n + 5 = 19 (n = – 2) (c) 7n + 5 = 19 (n = 2)(d) 4p – 3 = 13 (p = 1) (e) 4p – 3 = 13 (p = – 4) (f) 4p – 3 = 13 (p = 0) View solution Q3 – Solve the following equations by trial and error method: (i) 5p + 2 = 17 (ii) 3m – 14 = 4 View solution Q4 – Write equations for the following statements: (i) The sum of numbers x and 4 is 9. (ii) 2 subtracted from y is 8. (iii) Ten times a is 70. (iv) The number b divided by 5 gives 6. (v) Three-fourth of t is 15. (vi) Seven times m plus 7 gets you 77. (vii) One-fourth of a number x minus 4 gives 4. (viii) If you take away 6 from 6 times y, you get 60. (ix) If you add 3 to one-third of z, you get 30. View solution Q5 – Write the following equations in statement forms View solution Q6 – Set up an equation in the following cases: (i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit’s marbles.) (ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be l.) (iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees). View solution Exercise 4.2 solutions Q1 – Give first the step you will use to separate the variable and then solve the equation: View solution Q2 – Give first the step you will use to separate the variable and then solve the equation: View solution Q3 – Give the steps you will use to separate the variable and then solve the equation: View solution Q4 – Solve the following equations: View solution Exercise 4.3 solutions Q1 – Solve the following equations: View solution Q2 – Solve the following equations: View solution Q3 – Solve the following equations: View solution Q4 – (a) Construct 3 equations starting with x = 2 (b) Construct 3 equations starting with x = – 2 View solution Exercise 4.4 solutions Q1 – Set up equations and solve them to find the unknown numbers in the following cases: (a) Add 4 to eight times a number; you get 60. (b) One-fifth of a number minus 4 gives 3. (c) If I take three-fourths of a number and add 3 to it, I get 21. (d) When I subtracted 11 from twice a number, the result was 15. (e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8. (f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8. (g) Anwar thinks of a number. If he takes away 7 from 5/2of the number, the result is 23. View solution Q2 – Solve the following: (a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score? (b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°). (c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score? View solution Q3 – Solve the following: (i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have? (ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age? (iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77? View solution Q4 – Solve the following riddle: I am a number, Tell my identity! Take me seven times over And add a fifty! To reach a triple century You still need forty! View solution ## End Note In the article, we have shared solutions for ncert exercises for class 7 maths for simple equations chapter. Students can click to respective chapters and refer to the right solution. We advise students to refer to them as clearing their concepts, and matching their solutions. Student should not use them to copy solutions for their homework and instead should try to solve homework on own, as it will impact their learning process. In our observation, many students struggle in working with equations, especially with higher order equations. These ncert solutions on simple equations chapter will help them clear their conceptual doubts too. In case you have any doubts regarding solution, or need further clarification, you can write to us. You can also reach out to us, in case you have doubt in other text books’ problems or school exam papers. We will be glad to assist you in helping with your doubt.
# Activity: Find an Approximate Value For Pi You can read about π (Pi) first You will need: A piece of card A compass and pencil A protractor A ruler A pair of scissors Glue and paper ## Step 1 Draw a circle on your card. The exact size doesn't matter, but let's use a radius of 5 cm (centimeters). Use your protractor to divide the circle up into twelve equal sectors. What is the angle for each sector? That's easy – just divide 360° (one complete turn) by 12: 360° / 12 = 30° So each of the angles must be 30° ## Step 2 Divide just one of the sectors into two equal parts – that's 15° for each sector. You now have thirteen sectors – number them 1 to 13: ## Step 3 Cut out the thirteen sectors using the scissors: ## Step4 Rearrange the 13 sectors like this (you can glue them onto a piece of paper): Now that shape resembles a rectangle: ## Step 5 What are the (approximate) height and width of the rectangle? Its height is the circle's radius: just look at sectors 1 and 13 above. When they are in the circle they are "radius" high. Its width (actually one "bumpy" edge), is half of the curved parts around the circle ... in other words it is about half the circumference of the original circle. We know that: Circumference = 2 × π × radius And so the width is: Half the Circumference = π × radius And so we have (approximately): With a radius of 5 cm, the rectangle should be: • 5 cm high • about 5π cm wide ## Step 6 Measure the actual length of your "rectangle" as accurately as you can using your ruler. Divide by the radius (5 cm) to get an approximation for π "Rectangle" Width Divide by 5 cm ≈ π Remember π is about 3.14159... how good was your answer? Note: You could probably get a better answer if you: • used a bigger circle • divided your circle into 25 sectors (23 with an angle of 15° and 2 with an angle of 7.5°). ## Optional Step You could work out the percentage error in your answer. You can find out how to do this on the page Percentage Difference vs Percentage Error.
Question # Find the least common multiple by listing multiples: 20 and 30. Factors and multiples Find the least common multiple by listing multiples: 20 and 30. 2021-08-01 Given pair of numbers: 20,30. Concept used: Least Common Multipleis also called smallest common multiple. Least common multiple of two or more numbers is the smallest multiple that is completely divisible by the corresponding numbers leaving remainder zero. Least Common Multiple can be find by two methods - one by multiples method while other by prime factors method. In multiples method, we list out multiples of given numbers and look out for the least multiple that is common for all the given numbers. In case, we don't find any of common multiple then we simply multiple the given numbers, the product obtained is the least common multiple of the given numbers. In prime factors method, we break the given numbers in terms of their prime factors and look out for common factors. Then, we multiply the each of the common factor, number of times it occurs in factorization of any of the number and the factors that are not common. The product thus, obtained is the least common multiple of the given numbers. Calculations: According to question, the given pair of numbers are 20, 30. Using multiples method, we have, $$\displaystyle\Rightarrow{20}:{20},{40},{60},{80},{100},{120},{140}\ldots..$$ and so on $$\displaystyle\Rightarrow{30}:{30},{60},{90},{120},{150},{180},{210}\ldots..$$ and so on From the above list of multiples we find that 60 is the least common multiple among all the multiples of 20,30. Hence, the least common multiple of pair of numbers 20,30 is 60
# Video: Using Number Lines to Model Multiplication within 100 Look at how Victoria used a number line to multiply. I start at 0 and make 4 jumps of 5. I land on 20. So, 4 × 5 = 20. What calculation has Liam shown on this number line? What is 3 × 6? 04:38 ### Video Transcript Look at how Victoria used a number line to multiply. I start at zero and make four jumps of five. I land on 20. So four times five equals 20. What calculation has Liam shown on this number line? And what is three times six? This problem is all about using number lines to model multiplication facts. We’ve given an example to start with. And the first part of the problem actually tells us to look at how Victoria used a number line to multiply. So let’s start of by doing that. The clue to how she uses her number line is in the speech bubble. Victoria says that she starts at zero. And then she makes four jumps of five. One, two, three, four. For each jump she can count in five. Five, 10, 15, 20. She starts on zero, makes four jumps of five, and ends up on the number 20. So she knows that four lots of five are 20 or four times five equals 20. And this is the multiplication fact that she shows using her number line. Four is the number of jumps, and five is the number that is being jumped each time, four jumps of five. Now we can use Victoria’s method to solve some multiplication of our own. Firstly we’re shown a number line. It already has some jumps marked on it. The question asks us which calculation has Liam shown on the number line? Well, firstly, we know that it’s going to be a multiplication. Something multiplied by something equals something. Let’s imagine Liam says the same sort of thing as Victoria. What do you say? I start at zero and make how many jumps? One, two, three. I start at zero and make three jumps of- of how many? We can see all the plus eights underneath the arrows. This tells us that Liam has added eight each time. He starts at zero and makes three jumps of eight. Where does he land? Eight, 16, 24. He lands on 24. Three jumps of eight takes him to 24. So we can complete Liam’s multiplication as three times eight equals 24. In the last question we need to solve, we’ve been given a multiplication. And now we need to use the number line for ourselves. What is three times six? Well, to model three times six on the number line in the same ways we’ve done already, we need to make three jumps of six. And we can see that one jump of six has been made already. So we only need to make another two jumps of six. What’s six plus six? Well, if we add another six to six, we get 12. We’ve now made two jumps of six, one more to go. Where do we end up if we make a jump of six from 12? Well, 12 add six equals 18. We’ve made three jumps of six. We started at zero six, 12, 18. If we start at zero and make three jumps of six, we land on 18. So three times six equals 18. The calculation that Liam shows on his number line is three times eight equals 24. And we’ve also used a number line to show that three times six equals 18.
# Features of linear relationships Lesson You will have covered the concepts here before, either in previous years or even this year. But I'll summaries the key features of linear equations below so that you have a handy reference point all in one place. ## Slope or Slope The slope of a line (also called the slope) that passes through two known points, say $\left(x_1,y_1\right)$(x1,y1) and $\left(x_2,y_2\right)$(x2,y2) on the cartesian plane can be found easily. Slope is a measure of steepness. It is the ratio of a line's rising (or falling) to its running. If, over a distance of $8$8 metres, a driveway rises $2$2 metres, then its slope is said to be the ratio $\frac{2}{8}=\frac{1}{4}=0.25$28=14=0.25. It is also defined as the tangent of the angle of rise as shown in this simple diagram. Consider the following example of a line passing through two points $\left(-3,7\right)$(3,7) and $\left(5,9\right)$(5,9) as shown here: Looking at the two $y$y values of the two points, the rise is clearly $2$2. We could either use a formula for rise which might look like $Rise=y_2-y_1=9-7=2$Rise=y2y1=97=2 or simply notice that there is a gap of $2$2 between the two values. Looking at the two $x$x values we could again either use the formula $Run=x_2-x_1=5-\left(-3\right)=8$Run=x2x1=5(3)=8 or simply notice that the gap between $-3$3 and $5$5 is $8$8 We also realise that the line is rising and this means that the slope is positive. The slope, often denoted by the letter $m$m is simply the ratio given by: $m=\frac{y_2-y_1}{x_2-x_1}=\frac{2}{8}=0.25$m=y2y1x2x1=28=0.25 From the fact that $\tan\theta=0.25$tanθ=0.25 we can use a scientific calculator to show that $\theta=\tan^{-1}\left(0.25\right)=14^\circ2'$θ=tan1(0.25)=14°2, which gives some sense to the steepness of the rise. Note that if the line is falling, the line's slope will be negative. In such cases the acute angle the line makes with the $x$x axis will be shown on the calculator as a negative angle. Adding $180^\circ$180° to this will reveal the obtuse angle of inclination the line makes with the axis. For example if the slope was given by $m=-0.25$m=0.25, then $\theta=180^\circ+\tan^{-1}\left(-0.25\right)$θ=180°+tan1(0.25), which simplifies to $\theta=165^\circ58'$θ=165°58 Suppose we consider the line given by $5x-2y=20$5x2y=20. By putting $x=0$x=0 we see that $y=-10$y=10 (note the negative sign here). Also, by putting $y=0$y=0, we find that $x=4$x=4. This means that the $x$x and $y$y intercepts are $4$4 and $-10$10 respectively. The situation is shown here: Note that the rise and run can be determined from the $x$x and $y$y intercepts. The positive slope of the line shown is given as $m=\frac{10}{4}=2.5$m=104=2.5 #### Worked Examples ##### Question 1 What is the slope of the line shown in the graph given that point A(3,3) and point B(6,5) both line on the line. ##### Question 2 What is the slope of the line going through A and B? ## Finding the Equation The line with equation $y=mx+b$y=mx+b has a slope $m$m and a y intercept $b$b . It is important to observe that this form of the line shows $y$y explicitly as a function of $x$x with $m$m and $b$b as constants, different values of $x$x will determine different values of $y$y . For example, the line, say $L_1$L1, given by $y=3x+3$y=3x+3 has a slope of $3$3 and a $y$y intercept of $3$3. The $y$y intercept can be determined by noting that at $x=0$x=0$y=3$y=3 The line $L_2$L2 given in general form as $2x+y-8=0$2x+y8=0 can be rearranged to $y=-2x+8$y=2x+8 and the slope $-2$2 and y intercept $8$8 can be easily determined. The line $L_3$L3 given by $5x+4y-29=0$5x+4y29=0 can be rearranged to $4y=29-5x$4y=295x and then to $y=\frac{29}{4}-\frac{5}{4}x$y=29454x with slope $m=-\frac{5}{4}$m=54 and $y$y intercept $b=7.25$b=7.25 We will now go through some of the skills in finding lines, intersections and midpoints by considering a number of questions relating to the lines $L_1,L_2$L1,L2 and $L_3$L3. As we answer the questions, check the sketch below to confirm your understanding of each answer. ##### Question 3 Is the point $A\left(1,6\right)$A(1,6) on $L_1$L1? By substituting $\left(1,6\right)$(1,6) into $y=3x+3$y=3x+3 we see that $6=3\times1+3$6=3×1+3. This is true and so the given point is on $L_1$L1. ##### Question 4 Find $P$P, the $x$x intercept of $L_2$L2. Since $L_2$L2 is given by $2x+y-8=0$2x+y8=0, the $x$x intercept is found by putting $y=0$y=0. Then $2x-8=0$2x8=0 and solving for $x$x, we see that $x=4$x=4. The point of intercept is thus $P\left(4,0\right)$P(4,0). ##### Question 5 Find the equation of the line $L_4$L4, which passes through $P$P and $M$M. With $P\left(4,0\right)$P(4,0) and $M\left(5,1\right)$M(5,1), we have two methods to find $L_4$L4. Both methods require finding the slope of the line given by $m=\frac{y_2-y_1}{x_2-x_1}=\frac{1-0}{5-4}=1$m=y2y1x2x1=1054=1 Then method 1 makes use of the point slope form of the line. Specifically we know that the equation we are looking for must have the form $y=1x+b$y=1x+b. Since $M\left(5,1\right)$M(5,1) is on this line, it must satisfy it. Thus we can write $1=1\times5+b$1=1×5+b and so with a little thought, $b$b must be $-4$4. the equation of $L_4$L4 must be $y=x-4$y=x4. The second method makes use of the point slope formula $y-y_1=m\left(x-x_1\right)$yy1=m(xx1). We know that the slope $m=1$m=1 and choosing one of the known points on the line, say $M\left(5,1\right)$M(5,1), we can determine the equation of $L_4$L4 as $y-5=1\left(x-4\right)$y5=1(x4), and this simplifies once again to $y=x-4$y=x4 ##### Question 6 A line passes through the point $A\left(-2,-9\right)$A(2,9) and has a slope of $-2$2. Using the point-slope formula, express the equation of the line in slope intercept form. ##### Question 7 A line passes through the point $\left(3,-5\right)$(3,5) and $\left(-7,2\right)$(7,2) a) Find the slope of the line b) Find the equation of the line by substituting the slope and one point into $y-y_1=m\left(x-x_1\right)$yy1=m(xx1) ##### Question 8 a) Find the equation, in general form, of the line that passes through $A\left(-12,-2\right)$A(12,2) and $B\left(-10,-7\right)$B(10,7) b) Find the $x$x-coordinate of the point of intersection of the line that goes through $A$A and $B$B, and the line $y=x-2$y=x2 c) Hence find the $y$y-coordinate of the point of intersection ## Intercepts of Horizontal and Vertical Lines ### Horizontal Horizontal lines are lines that follow the horizon. They look like this... Imagine now horizontal lines on the Cartesian plane. Horizontal lines are parallel to the $x$x axis, and as you move along a horizontal line, the $x$x value will change but the $y$y value will remain the same. A horizontal line will: • only have a $y$y intercept • have an equation of the form $y=b$y=b (every point on the line has a $y$y value of $b$b) • have a $y$y intercept = $b$b, no $x$x intercept ### Vertical Vertical Lines are lines that go up and down (they are perpendicular to horizontal lines). Imagine now vertical lines on the Cartesian plane. Vertical lines are parallel to the $y$y axis, and as you move along a vertical line, the $y$y value will change but the $x$x value will remain the same. A vertical line will: • only have an $x$x intercept • have an equation of the form $x=b$x=b (every point on the line has an $x$x value of $b$b) • have an $x$x intercept = $b$b, no $y$y intercept ## Parallel and perpendicular lines ### Parallel lines These occur when we have 2 lines that NEVER cross each other and have no points in common. For this to happen the two lines need to have exactly the same slope. If they have different slopes they will cross. Parallel lines occur often in the real world. Consider the line $y=x$y=x, with slope=1. What would happen if we shifted every point on the line $2$2 units upwards? We would get a new line that is parallel to $y=x$y=x, but with every point having a $y$y value that is two greater: $y=x+2$y=x+2 So parallel lines are just shifts of one another. Parallel lines on the Cartesian Plane have the same slope (slope). ### Perpendicular Lines The Leaning Tower of Pisa Perpendicular is the word used to describe when one object meets another at exactly 90°. So perpendicular lines are simply lines that cross each other at exactly 90°. To see how important the idea of perpendicular really is just think about your floor, walls and roof. If a builder does not take care to make the walls perpendicular to the floor and ceiling you'll end up with an unstable house. The leaning tower of Pisa is a famous example of perpendicular angles gone wrong! Prior to restoration work performed between 1990 and 2001, the tower leaned at an angle of 5.5°, but the tower now leans at about 3.99°. That means the acute angle made by the tower and the ground is 86.01°. Perpendicular lines on the Cartesian plane will have one point of intersection, and at that point of intersection the angle between them will be 90°. ## Intersections and concurrent lines Because lines extend forever in both directions, unless they are parallel they will intersect somewhere. Now when 3 or more lines all pass through the same point we give those lines a special name: they are called concurrent lines. The point of intersection is called the "point of concurrency", labelled point P below. . ### Intersections of two lines Where two lines intersect, they share a common point. The $x$x and $y$y values of this point satisfy the equations of both lines. #### Example If one line has equation $y=2x+3$y=2x+3 and another has equation $y=x+6$y=x+6 then the point of intersection is where both the $y$y's are the same value. If they are the same value, then we can say that: (at the point of intersection) $2x+3$2x+3 is equal to $x+6$x+6 $2x+3=x+6$2x+3=x+6 We can then solve for the $x$x value at the point of intersection. $2x-x=6-3$2xx=63 $x=3$x=3 Now that we have $x$x, we can find the $y$y value at the point of intersection. Which equation should we substitute back into? Well since the point is common to both lines, you can choose either equation. $y=x+6$y=x+6 $y=3+6$y=3+6 $y=9$y=9 So these lines cross at the point $\left(3,9\right)$(3,9). ##### Question 9 Consider the following linear equations: $y=2x+2$y=2x+2 and $y=-2x+2$y=2x+2. a) What are the intercepts of the line $y=2x+2$y=2x+2? b) What are the intercepts of the line $y=-2x+2$y=2x+2? c) Plot the lines of the two equations on the same graph. d) State the values of $x$x and $y$y that satisfy both equations. #### Worked Examples ##### Question 10 Examine the graph attached and assess: 1. the slope of the line. 2. the $y$y-intercept of the line. 3. the $x$x-intercept of the line ##### Question 11 Consider the graph of the linear function shown. 1. What is the slope of the line? 2. What is the $y$y-intercept? 3. What is the $x$x-intercept? 4. What is the equation of the line? 5. What is the zero of the function? ## Zeros The zeros of a function are the values of that function that make it equal to zero. For example for the line $y=2x+1$y=2x+1, the zero is the value of $x$x, that makes the whole function ($2x+1$2x+1) equal to zero. So we set $2x+1=0$2x+1=0 and solve for $x$x. $2x+1=0$2x+1=0 $2x=-1$2x=1 $x=-\frac{1}{2}$x=12 Does that process look familiar? It should. It's exactly the same process we use when we are finding the $x$x intercepts. This means that the phrase zero of a function, (and also sometimes root of a function) is actually asking for the $x$x-intercepts. ### Outcomes #### 10P.LR2.05 Graph lines by hand, using a variety of techniques
# Arithmetic Sequence – Formula, Definition With Examples Welcome to Brighterly, your reliable partner in the fascinating journey of learning mathematics. Here, we empower our young learners with knowledge, encouraging them to explore, understand, and love the fascinating world of numbers. Today, we will unfold the secrets of one of the most critical concepts in math – the Arithmetic Sequence. Let’s embark on this intriguing journey and unearth the formula, definition, and real-life examples that make Arithmetic Sequences come alive! ## What Is an Arithmetic Sequence? An arithmetic sequence, also known as an arithmetic progression, is a sequence of numbers in which the difference between any two successive members is a constant. This attribute is what sets this particular type of sequence apart in the expansive world of mathematics. For instance, the sequence 2, 4, 6, 8, and so on, exemplifies an arithmetic sequence where the common difference is 2. ## Importance of Arithmetic Sequences in Mathematics Arithmetic sequences permeate through multiple facets of our lives, driving their significance in mathematics. In the realm of mathematical modeling, arithmetic sequences are employed to model and predict various real-life scenarios such as predicting population growth, business forecasting, calculating loan payments, and more. Moreover, understanding arithmetic sequences is pivotal in learning advanced mathematical concepts like series, calculus, and even computer algorithms. Indeed, the foundational knowledge of arithmetic sequences often serves as the stepping stone to a broader mathematical comprehension. ## Definition of a Sequence in Mathematics A sequence, in the context of mathematics, is defined as a set of numbers arranged in a specific order. This order is typically determined by a certain rule or pattern. A sequence can be finite, containing a limited number of terms, or infinite, where the numbers continue indefinitely. Sequences find their application across various mathematical disciplines, from the simple number series we learn in primary school to the complex functions in calculus. ## Definition of an Arithmetic Sequence An arithmetic sequence, as a specific type of sequence, is a list of numbers where the difference between successive numbers remains constant. This constant difference, also known as the ‘common difference,’ is what characterizes an arithmetic sequence. To further illustrate this, consider the sequence 10, 15, 20, 25, where the common difference is 5. ## Properties of Sequences in Mathematics Sequences in mathematics, be it arithmetic, geometric, or harmonic, are all characterized by unique properties. Most importantly, each term in a sequence can be identified by a unique position or index. Additionally, a sequence can be finite (having a limited number of terms) or infinite (continuing indefinitely). While some sequences follow a specific rule or pattern (like arithmetic or geometric sequences), others might be more random. Further, sequences can be either increasing or decreasing. ## Properties of Arithmetic Sequences Arithmetic sequences possess several intriguing properties. Primarily, the difference between consecutive terms, known as the ‘common difference,’ is constant throughout. Secondly, each term of an arithmetic sequence can be calculated using a specific formula. Additionally, the sum of the terms in an arithmetic sequence, known as the arithmetic series, can also be calculated using a unique formula. These properties have profound implications in the mathematical modeling of various real-world phenomena. ## Difference Between a General Sequence and an Arithmetic Sequence A general sequence is merely an ordered list of numbers, whereas an arithmetic sequence is a special type of sequence where the difference between consecutive terms remains constant. This ‘common difference’ is the defining characteristic that sets an arithmetic sequence apart from a general sequence. For instance, while the sequence 1, 2, 3, 4 is both a general sequence and an arithmetic sequence, the sequence 1, 2, 4, 8 qualifies as a general sequence but not an arithmetic sequence. ## The Formula of an Arithmetic Sequence The formula for an arithmetic sequence is given by a_n = a_1 + (n – 1) * d, where ‘a_n’ represents the nth term of the sequence, ‘a_1’ represents the first term, ‘d’ represents the common difference, and ‘n’ signifies the position of the term in the sequence. This formula is instrumental in deriving any term of an arithmetic sequence without explicitly iterating through all the preceding terms. ## Understanding the Formula of an Arithmetic Sequence Unpacking the arithmetic sequence formula, the nth term ‘a_n’ is found by adding the product of the common difference ‘d’ and the preceding term (n – 1) to the first term ‘a_1’. This essentially means that to find any term in an arithmetic sequence, we just need to know the first term, the common difference, and the position of the term we are seeking. It essentially allows us to “skip” through the sequence to any term we desire. ## Writing Terms of an Arithmetic Sequence Using the Formula Utilizing the arithmetic sequence formula, we can derive any term in the sequence without listing all the terms. For example, in an arithmetic sequence with a first term of 3 and a common difference of 2, the fourth term can be calculated as: a_4 = a_1 + (4 – 1) * 2 = 3 + 6 = 9. Similarly, we can find the tenth or the hundredth term using this formula. ## Practice Problems on Arithmetic Sequences For those looking to test their understanding and hone their skills, here are a few practice problems: 1. If the first term of an arithmetic sequence is 5, and the common difference is 3, what is the 7th term? 2. Given an arithmetic sequence where the first term is 2 and the 5th term is 14, what is the common difference? 3. Find the sum of the first 10 terms of an arithmetic sequence where the first term is 1 and the common difference is 2. ## Conclusion Arithmetic sequences, with their simplicity and profound applications, form the bedrock of our understanding of numerous mathematical and real-life phenomena. From predicting population growth to calculating loan installments, the influence of these sequences is far-reaching. The inherent beauty of an arithmetic sequence lies in its predictability and its representation of order in a seemingly random world. At Brighterly, we aim to make this understanding accessible to all young learners, fostering a deep-seated appreciation for the intricate yet captivating world of mathematics. We hope this exploration of arithmetic sequences has been insightful and has stirred curiosity in your young minds. Continue exploring, questioning, and learning. After all, the world of mathematics is an endless frontier, and there’s always more to discover! ## Frequently Asked Questions on Arithmetic Sequences ### What is the common difference in an arithmetic sequence? The common difference in an arithmetic sequence is the constant value that separates two consecutive terms in the sequence. For example, in the sequence 2, 4, 6, 8, the common difference is 2. ### Why are arithmetic sequences important in mathematics? Arithmetic sequences are vital in mathematics because they offer a simple way of understanding patterns and making predictions. They also form a fundamental base for learning advanced mathematical concepts such as series and calculus. ### How is the nth term of an arithmetic sequence calculated? The nth term of an arithmetic sequence can be calculated using the formula a_n = a_1 + (n – 1) * d, where a_n is the nth term, a_1 is the first term, d is the common difference, and n is the position of the term in the sequence. ### What is the difference between an arithmetic sequence and a geometric sequence? The primary difference between an arithmetic sequence and a geometric sequence lies in their respective patterns. In an arithmetic sequence, the difference between consecutive terms is constant, while in a geometric sequence, the ratio of successive terms is constant. Information Sources Math Catch Up Program • Learn Math Simple - Fast - Effective • Overcome math obstacles and reach new heights with Brighterly. Simple - Fast - Effective Overcome math obstacles and reach new heights with Brighterly.
# What is the answer to the square root of 50 + square root of 8 - square root of 18? Jul 30, 2015 =color(green)(4sqrt2 #### Explanation: Prime factorising each of the numbers in order to simplify the terms: $\sqrt{50} = \sqrt{5 \cdot 5 \cdot 2} = 5 \sqrt{2}$ $\sqrt{8} = \sqrt{2 \cdot 2 \cdot 2} = 2 \sqrt{2}$ $\sqrt{18} = \sqrt{2 \cdot 3 \cdot 3} = 3 \sqrt{2}$ Now, $\sqrt{50} + \sqrt{8} - \sqrt{18} = 5 \sqrt{2} + 2 \sqrt{2} - 3 \sqrt{2}$ =color(green)(4sqrt2
Algebraic Expression Image Credit: Pixabay.com Algebra Concept of algebra originated as a problem solving technique in which unknown numbers (quantities to be found) are denoted by letters and together with known numbers are used to form expressions and equations that can be solved for the unknowns. Even complex mathematical statements can be easily put into equation using algebraic expressions. ❕ It is said that algebra as a branch of Mathematics began about 1550 BC, i.e.more than 3500 years ago, when people in Egypt started using symbols to denote unknown numbers. Diophantus ( lived around 250 AD) of Alexandria, sometimes called “the father of algebra”, was an Alexandrian Greek mathematician and the author of a series of books called Arithmetica, many of which are now lost. These texts deal with solving algebraic equations. The word ‘algebra’ is derived from the title of the book,‘Aljebar w’al almugabalah’, written about 825 AD by an Arab mathematician, Mohammed Ibn Al Khowarizmi of Baghdad. Variables The unknown quantities in mathematical expressions denoted by letters are called variables. (x, y etc.) e.g., 2x, 4+y, 5+3x etc. Here, 2, 4 5 and 3 are fixed values, such numbers are called constants but x, y can take any value, such numbers are called variable. ➢ Variable are used to write formulas and rules in general way applicable to any number. ➢ Variables are used to form expressions and equations . Example: The number of angles in a triangle = 3 The number of angles in two triangles = 3×2 = 6 The number of angles in three triangle = 3×3 = 9 So, we can say that the number of angles in any n triangles = 3×n Another example: Ram is three years older than Shyam, When Shyam is two years old, Ram’s age = 2+3 = 5 years When Shyam is five years old, Ram’s age = 5+3 = 8 years So, we can say that when Shyam is x years old, Ram’s age = (x + 3) years Here, x is a unknown number and hence a variable. Use of variable in common rules and formulas Some Rules from geometry Formula for Perimeter of a square A square has four equal sides Perimeter of square = sum of all sides of the square = 4×length of a side of the square Let, the length of each side be l unit, then Perimeter of square = 4×l p = l It is the formula for the perimeter of a square which is expressed as a relation between the perimeter and the length of the square. It is concise and easy to remember. Here, p and l both are variables, l can take any value but value of p will depend on l. Formula for Perimeter of a rectangle A rectangle has two pairs of equal sides Perimeter of rectangle = sum of all sides of the rectangle Let, the length and breadth be l and b unit respectively, then Perimeter of rectangle = 2×(l+b) p = 2×(l+b) It is the formula for the perimeter of a rectangle which is expressed as a relation between the perimeter and the length of the square. It is concise and easy to remember. Here, p, l and b are variables, l & b can take any value independent of each other but value of p will depend on both l & b. Some Rules from Arithmetic Properties of Numbers Properties of numbers can be expressed concisely as a general rule by using variables – Commutative Properties for addition and multiplication Let a and b are two variables representing any two numbers, then a+b = b+a a×b = b×a Let a, b and c are three variables representing any three numbers, then a×(b+c) = a×b + a×c a×(b−c) = a×b − a×c Algebraic expressions Arithmetic expression – Mathematical expression that contain numbers (constants) and operators (+, −, ×, ÷) are called arithmetic expressions. e.g., 2+5, 3×4-6 etc. Algebraic expression – Mathematical expression that contain both constants and variables together with operators (+, −, ×, ÷) are called algebraic expressions. e.g., 2 × x + 5, 3 × 4 – 6y etc. ● In algebraic expressions we can use raised dot ⋅ or parenthesis ( ) in place of multiplication operator × While multiplying a number with a variable or a variable with another variable we can omit the multiplication operator. x, 2(x), 2⋅ x and 2x are same expression. A number expression like (3× 4) − 6 can be immediately evaluated as (3 × 4) − 6 = 12 − 6 = 6 But an expression containing variable like (2x + 5), cannot be evaluated. Only if x is given some value, an expression like (2x + 5) can be evaluated. e.g., when x = 3, 2x + 5 = 2⋅3 + 5 = 6 + 5 = 11 Forming an Algebraic Expressions Verbal phrases can be transformed into algebraic expressions by combining number and variables with operators. (a) 7 added to p p + 7 (b) 7 subtracted from p p − 7 (c) p multiplied by 7 p × 7 (d) p divided by 7 p ÷ 7 (e) 7 subtracted from – m −m − 7 (f) – p multiplied by 5 −p × 5 (g) – p divided by 5 −p ÷ 5 (h) p multiplied by – 5 p(−5) (i) 11 added to 2m 2m + 11 (j) 11 subtracted from 2m 2m − 11 (k) 5 times y to which 3 is added 5y + 3 (l) 5 times y from which 3 is subtracted 5y − 3 (m) y is multiplied by – 8 y(−8) (n) y is multiplied by – 8 and then 5 is added to the result −8y +5 (o) y is multiplied by 5 and the result is subtracted from 16 16 − 5y (p) y is multiplied by – 5 and the result is added to 16. 16 − 5y ↓Class 7 An expression is sum of its terms. e.g., 2n has one term, 2n 5 + 3xy has two terms, 5 and 3xy 7x – 2y + 4 has three terms, 7x, −2y and 4. Factors of a term A term is product of its factors. e.g, factors of the term 2n are 2 and n factors of the term 3xy are 3, x and y factors of the term −2y are −2 and y factors of the term 9x2 are 9, and x 1 is not taken as separate factor. Coefficient A coefficient is the numerical factor of a term. e.g., coefficient of the term is . coefficient of the term is . coefficient of the term is When the coefficient of a term is , it is usually omitted. e.g., is written as ; is written as and so on. The coefficient is indicated only by the minus sign. Thus is written as ; is written as and so on. Sometimes a factor or product of factors in a term is called the coefficient of the remaining part of the term. e.g., coefficient of the in the term is coefficient of the in the term is . Like & Unlike terms Terms having same algebraic factors are called like terms. Thus, like terms have the same variables raised to the same power. e.g., are like terms. Terms having different algebraic factors are called unlike terms. Thus, unlike terms have different variables or same variables raised to different powers. e.g., are unlike terms. Monomials and Polynomials Polynomial An algebraic expression containing one or more terms is called a polynomial in which the exponent (power) of the variables must be a whole number. e.g., . is not a polynomial. is not a polynomial. Monomial A polynomial containing only one term is called monomial. e.g., Binomial A polynomial containing two unlike term is called binomial. e.g. Trinomial A polynomial containing three unlike terms is called trinomial. e.g., Value of an Expression The value of an expression depend on the value of variables it contain. We can find the value of an expression by putting the given value of variable in it. Let’s find the value of , for Putting the value of in , we get Addition & Subtraction of Algebraic Expressions We can only add or subtract like terms, unlike terms cannot be added or subtracted with each other. While adding or subtracting the like terms, the variables (the common algebraic factors) are taken out and numerical factors or coefficient is then added or subtracted inside a parenthesis.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # Square Roots and Irrational Numbers ## Simplify square roots by factoring Estimated12 minsto complete % Progress Practice Square Roots and Irrational Numbers MEMORY METER This indicates how strong in your memory this concept is Progress Estimated12 minsto complete % Square Roots and Irrational Numbers Suppose an elementary school has a square playground with an area of 3000 square feet. Could you find the width of the playground? Would the width be a rational or irrational number ### Square Roots and Irrational Numbers The square root of a number \begin{align}n\end{align} is any number \begin{align}s\end{align} such that \begin{align}s^2 = n\end{align}. Every positive number has two square roots, the positive and the negative. The symbol used to represent the square root is \begin{align}\sqrt{x}\end{align}. It is assumed that this is the positive square root of \begin{align}x\end{align}. To show both the positive and negative values, you can use the symbol \begin{align}\pm\end{align}, read “plus or minus.” For example: \begin{align}\sqrt{81}=9\end{align} means the positive square root of 81. \begin{align}-\sqrt{81}= -9\end{align} means the negative square root of 81. \begin{align}\pm \sqrt{81} = \pm 9\end{align} means the positive or negative square root of 81. #### Let's solve the following problem by using square roots: Human chess is a variation of chess, often played at Renaissance fairs, in which people take on the roles of the various pieces on a chessboard. The chessboard is played on a square plot of land that measures 324 square meters with the chess squares marked on the grass. How long is each side of the chessboard? The human chessboard measures 324 square meters. The area of a square is \begin{align}s^2 = Area\end{align}. The value of Area can be replaced with 324. \begin{align}s^2=324\end{align} The value of \begin{align}s\end{align} represents the square root of 324. \begin{align}s= \sqrt{324}=18\end{align} The chessboard is 18 meters long by 18 meters wide. #### Approximating Square Roots When the square root of a number is a whole number, this number is called a perfect square. 9 is a perfect square because \begin{align}\sqrt{9}=3\end{align}. Not all square roots are whole numbers. Many square roots are irrational numbers, meaning there is no rational number equivalent. For example, 2 is the square root of 4 because \begin{align}2 \times 2 = 4\end{align}. The number 7 is the square root of 49 because \begin{align}7 \times 7 = 49\end{align}. What is the square root of 5? There is no whole number multiplied by itself that equals five, so \begin{align}\sqrt{5}\end{align} is not a whole number. To find the value of \begin{align}\sqrt{5}\end{align}, we can use estimation. To estimate the square root of a number, look for the perfect integers less than and greater than the value, and then estimate the decimal. #### Let's estimate \begin{align}\sqrt{5}\end{align}: The perfect square below 5 is 4 and the perfect square above 5 is 9. Therefore, \begin{align}4<5<9\end{align}. Therefore, \begin{align}\sqrt{5}\end{align} is between \begin{align}\sqrt{4}\end{align} and \begin{align}\sqrt{9}\end{align}, or \begin{align}2< \sqrt{5}<3\end{align}. Because 5 is closer to 4 than 9, the decimal is a low value: \begin{align}\sqrt{5} \approx 2.2\end{align}. #### Identifying Irrational Numbers Real numbers have two categories: rational and irrational. If a square root is not a perfect square, then it is considered an irrational number. These numbers cannot be written as a fraction because the decimal does not end (non-terminating) and does not repeat a pattern (non-repeating). Although irrational square roots cannot be written as fractions, we can still write them exactly, without typing the value into a calculator. For example, suppose you do not have a calculator and you need to find \begin{align}\sqrt{18}\end{align}. You know there is no whole number squared that equals 18, so \begin{align}\sqrt{18}\end{align} is an irrational number. The value is between \begin{align}\sqrt{16}=4\end{align} and \begin{align}\sqrt{25}=5\end{align}. However, we need to find the exact value of \begin{align}\sqrt{18}\end{align}. Begin by writing the prime factorization of \begin{align}\sqrt{18}\end{align}. \begin{align}\sqrt{18} = \sqrt{9 \times 2}= \sqrt{9} \times \sqrt{2}\end{align}. \begin{align}\sqrt{9} = 3\end{align} but \begin{align}\sqrt{2}\end{align} does not have a whole number value. Therefore, the exact value of \begin{align}\sqrt{18}\end{align} is \begin{align}3 \sqrt{2}\end{align}. You can check your answer on a calculator by finding the decimal approximation for each square root. #### Let's find the exact value of \begin{align}\sqrt{75}\end{align}: \begin{align}\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25} \times \sqrt{3} = 5 \sqrt{3}\end{align} ### Examples #### Example 1 Earlier, you were told that an elementary school has a square playground with an area of 3000 square feet. What is the width of the playground? Is the width a rational or irrational number? Since the playground is a square, we can use the formula for the area of a square: \begin{align}a=s^2.\end{align} \begin{align}& a=s^2\\ &3000=s^2\\ &\sqrt{3000}=\sqrt{s^2}\\ &\sqrt{3000}=s \end{align} Now, simplify: \begin{align}&\sqrt{100\times 30}=s\\ &\sqrt{100}\times\sqrt{ 30}=s\\ &10\sqrt{30}=s\end{align} The width of the playground is \begin{align}10\sqrt{30}\end{align} feet. Since 30 is not a perfect square, the width is irrational. #### Example 2 The area of a square is 50 square feet. What are the lengths of its sides? Using the area of a square formula, \begin{align}a=s^2\end{align}: \begin{align}& a=s^2\\ &50=s^2\\ &\sqrt{50}=\sqrt{s^2}\\ &\sqrt{50}=s\end{align} Now we will simplify: \begin{align}\sqrt{50}=\sqrt{25\cdot 2}=5\sqrt{2}.\end{align} The length of each side of the square is \begin{align}5\sqrt{2}\end{align} feet. ### Review Find the following square roots exactly without using a calculator. Give your answer in the simplest form. 1. \begin{align}\sqrt{25}\end{align} 2. \begin{align}\sqrt{24}\end{align} 3. \begin{align}\sqrt{20}\end{align} 4. \begin{align}\sqrt{200}\end{align} 5. \begin{align}\sqrt{2000}\end{align} 6. \begin{align}\sqrt{\frac{1}{4}}\end{align} 7. \begin{align}\sqrt{\frac{9}{4}}\end{align} 8. \begin{align}\sqrt{0.16}\end{align} 9. \begin{align}\sqrt{0.1}\end{align} 10. \begin{align}\sqrt{0.01}\end{align} Use a calculator to find the following square roots. Round to two decimal places. 1. \begin{align}\sqrt{13}\end{align} 2. \begin{align}\sqrt{99}\end{align} 3. \begin{align}\sqrt{123}\end{align} 4. \begin{align}\sqrt{2}\end{align} 5. \begin{align}\sqrt{2000}\end{align} 6. \begin{align}\sqrt{0.25}\end{align} 7. \begin{align}\sqrt{1.35}\end{align} 8. \begin{align}\sqrt{0.37}\end{align} 9. \begin{align}\sqrt{0.7}\end{align} 10. \begin{align}\sqrt{0.01}\end{align} To see the Review answers, open this PDF file and look for section 2.12. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English Spanish Square Root The square root of a term is a value that must be multiplied by itself to equal the specified term. The square root of 9 is 3, since 3 * 3 = 9. approximate solution An approximate solution to a problem is a solution that has been rounded to a limited number of digits. Irrational Number An irrational number is a number that can not be expressed exactly as the quotient of two integers. Perfect Square A perfect square is a number whose square root is an integer. principal square root The principal square root is the positive square root of a number, to distinguish it from the negative value. 3 is the principal square root of 9; -3 is also a square root of 9, but it is not principal square root. rational number A rational number is a number that can be expressed as the quotient of two integers, with the denominator not equal to zero.
12:00 am Trig Indentity # What are Trigonometric Derivatives and What are They? Before understanding what Trigonometric Derivatives are, it is essential for a student to know what is meant by the derivative of a function. As a part of one of the fundamental concepts of mathematics, derivative occupies an important place. In calculus, students should know about the process of integration as well as the differentiation of a function. The procedure used for finding the derivative of a function is known as differentiation. Integration is considered as the inverse procedure of differentiation and is known as integration. ## Trigonometric Derivatives Coming to the question of what are trigonometric derivatives and what are they, the derivatives of trigonometric functions involve six numbers. They are cosecant (cscx), secant (secx), cotangent(cotx), tangent(tanx), cosine (cosx), and sine (sinx). These functions can be subjected to the process of differentiation and their corresponding derivatives can be found. These are known as continuous functions. It is possible to find the derivative of trigonometric function through the process of differentiation. With respect to the rate at which a variable can be changed is also referred to as the process of differentiation. It is possible to show from the first principles that derivatives of tangent, cosine, and sine functions are given as d(tanx)/dx=sec2x, d(cosx)/dx=sinx, and d(sinx)/dx=cosx. The derivative of a trigonometric function can be found by using algebra. At any given value of x and from the general expression of the slope of a curve, it is possible for a student to differentiate a function. This will enable to find the derivative of the particular function in question. The student should know that there are derivatives of circular trigonometric functions. It can be evaluated through the usage of cos(x) and sin(x). Here, a rule of quotient is applied in order to differentiate the function. The expression that results from this process, leads to the corresponding derivatives of trigonometry. ### Derivatives of Inverse Trigonometric Functions FAQs related to derivative trigonometry Q. what are trigonometric derivative ? A: Trigonometric derivatives are the derivatives of the trigonometric functions. In calculus, the derivative of a function is a measure of how the function changes as its input changes. The derivative of a trigonometric function is calculated using the rules of differentiation. Q. What are function of trigonometric derivatives ? • The derivative of sine is cosine. • The derivative of cosine is negative sine. • The derivative of tangent is secant squared. • The derivative of cotangent is negative cosecant squared. • The derivative of secant is secant multiplied by tangent. • The derivative of cosecant is negative cosecant multiplied by cotangent. Q: Why are Trigonometric Derivatives Important ? A: Trigonometric derivatives are important in various fields of science and engineering. They are used to model the behavior of waves, oscillations, and vibrations, which are common phenomena in physics and engineering Visited 54 times, 6 visit(s) today Close
## Exercises - Patterns and Conjectures 1. Letting $F_i$ denote the $i^{th}$ Fibonacci number (where $F_0 = F_1 = 1$), we might conjecture that $$\sum_{i=0}^n F_i = F_{n+2} - 1$$ We can establish this result via induction. First note that when $n=0$, the left side above is simply $\sum_{i=0}^n F_i = F_0 = 1$, while the right side evaluates to and $F_2 - 1 = 2 - 1 = 1$. As these agree in value, our basis step is complete. Then, to proceed with the inductive step -- we need to show that if $\sum_{i=0}^k F_i = F_{k+2} - 1$, then $\sum_{i=0}^{k+1} F_i = F_{k+3} - 1$. To this end, consider the following: $$\begin{array}{rcl} \sum_{i=0}^{k+1} F_i &=& \left[ \sum_{i=0}^k F_i \right] + F_{k+1}\\\\ &=& (F_{k+2} - 1) + F_{k+1}\\\\ &=& (F_{k+2} + F_{k+1}) - 1\\\\ &=& F_{k+3} - 1 \end{array}$$ Therefore by the principle of mathematical induction, the following holds for integer values of $n \ge 0$: $$\sum_{i=0}^n F_i = F_{n+2} - 1$$ 2. We conjecture that $$F_{n-1} \cdot F_{n+1} = F_n^2 + (-1)^{n+1}$$ We can prove this via strong induction... First, we deal with the basis step. When $n=1$, we have $F_0 \cdot F_2 = 1 \cdot 2 = 2$, and $F_1^2 + (-1)^2 = 1 + 1 = 2$. These values being equal, the basis step is established. Now, we turn our attention to the inductive step. We need to show that if $F_{n-1} \cdot F_{n+1} = F_n^2 + (-1)^{n+1}$ when $n \le k$, then $F_k \cdot F_{k+2} = F_{k+1}^2 + (-1)^{k+2}$. To this end, consider the following... $$\begin{array}{rcl} F_k \cdot F_{k+2} &=& (F_{k-2} + F_{k-1})(F_k + F_{k+1})\\\\ &=& F_{k-1} F_{k+1} + F_{k-1} F_k + F_{k-2} F_k + F_{k-2} F_{k+1}\\\\ &=& \left[ F_k^2 + (-1)^{k+1} \right] + F_{k-1} F_k + \left[ F_{k-1}^2 + (-1)^k \right] + F_{k-2} F_{k+1}\\\\ &=& F_k^2 + F_{k-1} F_k + F_{k-1}^2 + F_{k-2} F_{k+1}\\\\ &=& (F_k^2 + 2 F_{k-1} F_k + F_{k-1}^2) -F_{k-1} F_k + F_{k-2} F_{k+1}\\\\ &=& (F_k + F_{k-1})^2 -F_{k-1} F_k + F_{k-2} F_{k+1}\\\\ &=& F_{k+1}^2 -F_{k-1} F_k + F_{k-2} F_{k+1}\\\\ &=& F_{k+1}^2 - F_{k-1} (F_{k-2} + F_{k-1}) + F_{k-2} F_{k+1}\\\\ &=& F_{k+1}^2 + F_{k-2} (F_{k+1} - F_{k-1}) - F_{k-1}^2\\\\ &=& F_{k+1}^2 + F_{k-2} F_k - F_{k-1}^2\\\\ &=& F_{k+1}^2 + F_{k-1}^2 + (-1)^k - F_{k-1}^2\\\\ &=& F_{k+1}^2 + (-1)^k\\\\ &=& F_{k+1}^2 + (-1)^{k+2} \end{array}$$ Hence, by the principle of mathematical induction, the following holds for all integers $n \ge 1$: $$F_{n-1} \cdot F_{n+1} = F_n^2 + (-1)^{n+1}$$ QED. 3. Pascal's triangle can be constructed in the following way. First, we write a bunch of ones along the left and right edges, and then below each pair of adjacent numbers we write their sum. It is well known that the $n^{th}$ row (where the top row corresponds to $n=0$) yields the coefficients to the terms resulting from expanding $(x+y)^n$. The Binomial Theorem also gives a way to produce these coefficients, stating that $$(x+y)^n = x^n + {}_nC_1 x^{n-1} y + {}_nC_2 x^{n-2} y^2 + {}_nC_3 x^{n-3} y^3 + \cdots + {}_nC_{n-1} x y^{n-1} + y^n$$ where ${}_nC_k$ counts the number of ways to choose $k$ objects from a group of $n$ objects, which is given by $${}_nC_k = \frac{n!}{k! (n-k)!}$$ 1. To see why this works, consider the terms of the expansion of $$(x+y)^n = \underbrace{(x+y)(x+y)(x+y) \cdots (x+y)}_{n \textrm{ factors}}$$ Each term is formed by choosing either an $x$ or a $y$ from the first factor, and then choosing either an $x$ or a $y$ from the second factor, and then choosing an $x$ or a $y$ from the third factor, etc... up to finally choosing an $x$ or a $y$ from the $n^{th}$ factor, and then multiplying all of these together. As such, each of these terms will consist of some number of $x$'s multiplied by some number of $y$'s, where the total number of $x$'s and $y$'s is $n$. For example, choosing $y$ from the first two factors, and $x$ from the rest will produce the term $x^{n-2}y^2$. Alternatively, choosing $x$ from the first $7$ factors and $y$ from the rest results in the term $x^7 y^{n-7}$. Let's consider a specific example. Consider the terms we see from expanding the following expression (assuming we don't collect any "like terms" along the way): $$\begin{array}{rcl} (x+y)^4 &=& (x+y)(x+y)(x^2 + xy + xy + y^2)\\\\ &=&(x+y)(x^3 + x^2y + x^2y + xy^2 + x^2y + xy^2 + y^3)\\\\ &=&x^4 + x^3y + x^3y + x^2y^2 +x^3y+x^2y^2 + xy^3\\ & &+ x^3y + x^2 y^2 + x^2 y^2 + x y^3 + x y^3 + y^4 \end{array}$$ Do you see how every term above takes the form $x^a y^b$ with $a+b=4$? Now, when we finally "collect like terms", the resulting coefficient on $x^ay^b$ will be the number of times it appears in the expansion. As such, to figure out the coefficient on $x^ay^b$, we just need to figure out how many ways we can form a term that looks like $x^ay^b$. Consider the terms $xy^3$ above. Note these terms were formed by letting three of the four $(x+y)$ factors contribute a $y$ to the product, with the remaining factor contributing an $x$. As such, the number of these terms will be given by the number of ways we can take $4$ factors and choose $3$ of them to contribute a $y$. In the parlance of combinations, this is given by ${}_4C_3$. Likewise, the terms $x^2y^2$ were formed by letting $2$ of the $4$ factors contribute a $y$ to the product, with the remaining factors contributing a $x$. Consequently, the number of such terms will be equal to the number of ways we can can take $4$ factors and choose $2$ of them to contribute a $y$. Again, in terms of combinations, this is given by ${}_4C_2$. In general, we can form terms of the form $x^{n-k}y^k$ by taking $n$ of our factors and choosing $k$ of them to contribute a $y$, which is given in the language of combinations by ${}_nC_k$. Given that the non-collected terms of the expansion of $(x+y)^n$ can have as few as zero $y$'s or at most $n$ of them (with every integer possibility between), our possible terms are $$x^n, \quad x^{n-1} y, \quad x^{n-2} y^2, \quad \ldots, \quad x y^{n-1}, \quad y^n$$ Finally, noting that in the expansion of $(x+y)^n$, each $x^{n-k}y^k$ occurs ${}_nC_k$ times, we have: $$(x+y)^n = x^n + {}_nC_1 x^{n-1} y + {}_nC_2 x^{n-2} y^2 + {}_nC_3 x^{n-3} y^3 + \cdots + {}_nC_{n-1} x y^{n-1} + y^n$$ 2. Pascal's triangle can be constructed in the following way. First, we write a bunch of ones along the left and right edges, and then below each pair of adjacent numbers we write their sum. As another way to construct Pascal's triangle -- it is well known that the $n^{th}$ row (where the top row corresponds to $n=0$) yields the coefficients to the terms resulting from expanding $(x+y)^n$. The Binomial Theorem gives a way to produce these coefficients, stating that $$(x+y)^n = x^n + {}_nC_1 x^{n-1} y + {}_nC_2 x^{n-2} y^2 + {}_nC_3 x^{n-3} y^3 + \cdots + {}_nC_{n-1} x y^{n-1} + y^n$$ where ${}_nC_k$ counts the number of ways to choose $k$ objects from a group of $n$ objects, which is given by $${}_nC_k = \frac{n!}{k! (n-k)!}$$ How do we know the two techniques described for constructing Pascal's triangle yield the same numbers? Consider the numbers seen in the triangle constructed by looking at the coefficients seen in the expansion of $(x+y)^n$. These are given by the binomial theorem, and all take the form ${}_nC_k$ for some appropriate non-negative integers $n$ and $k$. Suppose we look at one such number, ${}_nC_k$. This number lies on the $n^{th}$ row. If it lies on an end of this row, then either $k=0$ or $k=n$. Note ${}_nC_0 = {}_nC_n = 1$. This explains the ones seen on the left and right sides of Pascal's triangle. Now suppose ${}_nC_k$ is not on the end of a row. Then ${}_nC_{k+1}$ would be the number to its right. Likewise, ${}_{n+1}C_{k+1}$ would be the number on the next row, directly below ${}_nC_k$ and ${}_nC_{k+1}$. As such, we seek to argue that ${}_nC_k + {}_nC_{k+1} = {}_{n+1}C_{k+1}$, as this will establish that the "adding-the-two-above" technique for constructing Pascal's triangle yields the same numbers as the triangle constructed from finding coefficients of $(x+y)^n$ expanded. We proceed directly... $$\begin{array}{rcl} {}_nC_k + {}_nC_{k+1} &=& \frac{n!}{k!(n-k)!} + \frac{n!}{(k+1)!(n-k-1)!}\\\\\\ &=& \frac{n!}{k!(n-k)!} \cdot \frac{(k+1)}{(k+1)} + \frac{n!}{(k+1)!(n-k-1)!} \cdot \frac{(n-k)}{(n-k)}\\\\\\ &=& \frac{k \cdot n! + n! + n \cdot n! - k \cdot n!}{(k+1)! (n-k)!}\\\\\\ &=& \frac{(n+1) n!}{(k+1)!(n-k)!} \quad = \quad {}_{n+1}C_{k+1} \end{array}$$ 4. A pyramid of balls consisting of five "layers" is shown below. Conjecture a formula for the number of balls needed to make a pyramid of $n$ layers. Can you prove your formula holds? Suppose we let $f(n)$ denote the number of balls in a pyramid of $n$ layers. Looking at a few small cases, we find: $$\begin{array}{rcl} f(1) &=& 1\\ f(2) &=& 4\\ f(3) &=& 10\\ f(4) &=& 20\\ f(5) &=& 35\\ f(6) &=& 56 \end{array}$$ We might guess $f(n)$ is cubic, and thus of the form $f(n) = an^3 + bn^2 + cn + d$. Under this assumption, we can use the above values to set up a system of equations to solve for $a$, $b$, $c$, and $d$: $$\begin{array}{rcl} 1 &=& a + b + c + d\\ 4 &=& 8a + 4b + 2c + d\\ 10 &=& 27a + 9b + 3c + d\\ 20 &=& 64a + 16b + 4c + d \end{array}$$ Solving, we find $a=1/6, \quad b=1/2, \quad c=1/3, \quad \textrm{ and } d=0$. This suggests that $$\begin{array}{rcl} f(n) &=& \frac{n^3}{6} + \frac{n^2}{2} + \frac{n}{3}\\\\ &=& \frac{n^3 + 3n^2 + 2n}{6} \end{array}$$ Of course, this is still just a guess at this point. We might check $f(5)$ and $f(6)$ to see if we are on the right track... $$\begin{array}{rcl} f(5) = 35 &\textrm{ and }& \frac{5^3 + 3 \cdot 5^2 + 2 \cdot 5}{6} = \frac{125 + 75 + 10}{6} = \frac{210}{6} = 35\\\\ f(6) = 56 &\textrm{ and }& \frac{6^3 + 3 \cdot 6^2 + 2 \cdot 6}{6} = 6^2 + 3 \cdot 6 + 2 = 36 + 18 + 2 = 56 \end{array}$$ Since these check out, we are much more confident that this is the formula we seek, but it remains to prove it... We proceed by induction... The table above establishes the basis step, so all that remains is the inductive step. We need to show that if $f(k) = (k^3 + 3k^2 + 2k) / 6$ balls, then $$f(k+1) = \frac{(k+1)^3 + 3(k+1)^2 +2(k+1)}{6}$$ To show this, first note that a pyramid of $k+1$ layers (with $f(k+1)$ balls) can be split into two pieces -- one consisting of the top $k$ layers, and the other consisting of the bottom layer alone. The first piece, by the inductive hypothesis, contains $f(k) = (k^3 + 3k^2 + 2k) / 6$ balls. The second piece (i.e., the bottom layer) is a triangular arrangement with $1 + 2 + 3 + \cdots + (k+1)$ balls. Further, we know that $$1 + 2 + 3 + \cdots + (k+1) = \frac{(k+1)(k+2)}{2}$$ Thus, $$\begin{array}{rcl} f(k+1) &=& f(k) + \left[ 1+2+3+\cdots+(k+1) \right]\\\\ &=& \frac{k^3 + 3k^2 + 2k}{6} + \frac{(k+1)(k+2)}{2}\\\\ &=& \frac{k^3+3k^2+2k+3k^2+9k+6}{6}\\\\ &=& \frac{(k^3+3k^2+3k+1)+(3k^2+6k+3)+(2k+2)}{6}\\\\ &=& \frac{(k^3+3k^2+3k+1)+3(k^2+2k+1)+2(k+1)}{6}\\\\ &=& \frac{(k+1)^3 + 3(k+1)^2 + 2(k+1)}{6} \end{array}$$ This completes the inductive step, so by the principle of mathematical induction, a pyramid of $n$ layers has $f(n)$ balls where $$f(n) = \frac{n^3 + 3n^2 + 2n}{6}$$ QED. 5. The first diagonal of Pascal's triangle is made entirely of ones. The second diagonal lists the natural numbers in order (1, 2, 3, 4, ... ). Is there any significance to the third diagonal (which begins with 1, 3, 6, ...) ? Can you find a pattern to the sequence? What about the sum of the numbers on each row? Is there a pattern? Make some conjectures, and prove them if you can. 6. Suppose we denote the number of ways to put $n$ cents in a machine if we use identical nickles, dimes, and quarters by $f(n)$. The trick here is to "think recursively" and tell yourself that you already know that when $k \lt n$, you can put $k$ cents in the machine in $f(k)$ ways. As you start dropping change in the machine, you might start by putting in either a nickel, a dime, or a quarter. Let us suppose you put in a nickel. Now you have to ask yourself how many ways can you put in the remaining $n-5$ cents. But $n-5 \lt n$, so this can be done in $f(n-5)$ ways. If instead, you had put in a dime first, then you have to ask yourself how many ways can you put in the remaining $n-10$ cents. Here again, $n-10 \lt n$, so this can be done in $f(n-10)$ ways. Lastly, if you first dropped a quarter in the machine, you similarly have $f(n-25)$ ways to put in the remaining change. Taking these cases together, we arrive at our recursive relationship: $$f(n) = f(n-5) + f(n-10) + f(n-25)$$ All that remains is to identify what happens in the first cases. Note that $f(0) = 1$ and $f(5) = 1$ by inspection. Further, $f(n) = 0$ if $n \lt 0$, as we would need "negative coins" otherwise. This completes our recursive definition for $f(n)$. 7. $n$ $F_n$ $F_{n+1} / F_n$ 0 1 1 1 1 2 2 2 3 3 1.5 4 5 1.666667 5 8 1.6 6 13 1.625 7 21 1.615385 8 34 1.619048 9 55 1.617647 10 89 1.618182 11 144 1.617978 12 233 1.618056 13 377 1.618026 14 610 1.618037 15 987 1.618033 16 1597 1.618034 17 2584 1.618034 18 4181 1.618034 19 6765 1.618034 20 10946 1.618034 Calculating the first 20 values of $F_{n+1}/F_{n}$ is straight-forward. The results are shown in the table on the left. You can clearly see that the ratio $F_{n+1} / F_n$ appears to be very close to a constant for large values of $n$. More specifically, it appears that we are multiplying $F_n$ by roughly $1.618034$ each time to find $F_{n+1}$. This suggests that we might be able to approximate $F_n$ with something of the following form: $$F_n = C \cdot 1.618034^n$$ where $C$ is some constant. Assuming the above is a good approximation and solving for $C$, we find $C \approx F_n / 1.618034^n$ when $n$ is large enough. For $n=20$, this yields $C \approx 0.723607$. So it appears that we can approximate $F_n$ with the following formula: $$F_n = 0.723607\cdot 1.618034^n$$ Indeed, this formula does a great job approximating $F_n$, predicting its value exactly (provided one rounds to the nearest integer) all the way up to $n=29$. Further, had we better approximated the apparent constant $1.618034...$, the resulting formula would have worked all that much longer. This begs the question, however -- what is the exact value of the constant $1.618034...$ that we seek, and is there anything significant about it? Hmmm.... 8. In the arts, the golden rectangle, often considered to be the rectangle with the most aesthetically pleasing proportions, has the property that if you remove a square whose edge length matches that of the smallest side of the rectangle (shown in blue below), you are left with a smaller rectangle (shown in red) with the same proportions as the original one. That is to say $(a+b)/a = a/b$. 1. We know that $$\frac{a+b}{a} = \frac{a}{b}$$ If we wish to solve for $a/b$, it would be helpful if this was the only unknown in the equation. That is to say, if $x=a/b$, can we rewrite our equation so that it is in terms of only $x$? We can turn the $a$ in the denominator into an $a/b$ by dividing it by $b$. So that we don't alter the value of the expression on the left, let us do this to the numerator as well, yielding $$\frac{a/b + 1}{a/b} = \frac{a}{b}$$ Now, we can replace each $a/b$ with $x$ as suggested above, and then simply solve for $x$ $$\frac{x+1}{x} = x$$ Clearing the fractions by multiplying by $x$, we have $$x+1 = x^2$$ This of course is quadratic, so its solution is routine. $$x^2-x-1=0$$ which yields $$x = \frac{1 \pm \sqrt{5}}{2}$$ Knowing that $x = a/b$ is a ratio of side lengths, it must be true that $x>0$, so we can eliminate the negative solution found. Thus, the remaining solution must be the value of the golden ratio, $\varphi$, that we seek: $$\varphi = \frac{1 +\sqrt{5}}{2}$$ 2. Suppose $\varphi$ denotes the golden ratio and $\phi$ denotes its negative reciprocal. Find the first few powers of each. What can you say about their differences, $\varphi^n - \phi^n$? Make a conjecture and prove it with induction. Recall, the golden ratio is given by $$\varphi = \frac{1+\sqrt{5}}{2}$$ First, we need to find the value of $\phi$, the negative reciprocal of the golden ratio, $\phi$, $$\begin{array}{rcl} \phi &=& -\frac{2}{1+\sqrt{5}}\\\\ &=& -\frac{2}{1+\sqrt{5}} \cdot \frac{1-\sqrt{5}}{1-\sqrt{5}}\\\\ &=& \frac{-2(1-\sqrt{5})}{-4}\\\\ &=& \frac{1-\sqrt{5}}{2} \end{array}$$ Now if we expand and simplify $\varphi^n - \phi^n$ for the first few positive integers $n$, we find: $$\begin{array}{cccc} \left( \frac{1 + \sqrt{5}}{2} \right)^1 - \left( \frac{1 - \sqrt{5}}{2} \right)^1 & = & \sqrt{5} \\\\ \left( \frac{1 + \sqrt{5}}{2} \right)^2 - \left( \frac{1 - \sqrt{5}}{2} \right)^2 & = & \sqrt{5} \\\\ \left( \frac{1 + \sqrt{5}}{2} \right)^3 - \left( \frac{1 - \sqrt{5}}{2} \right)^3 & = & 2\sqrt{5} \\\\ \left( \frac{1 + \sqrt{5}}{2} \right)^4 - \left( \frac{1 - \sqrt{5}}{2} \right)^4 & = & 3\sqrt{5} \\\\ \left( \frac{1 + \sqrt{5}}{2} \right)^5 - \left( \frac{1 - \sqrt{5}}{2} \right)^5 & = & 5\sqrt{5} \\\\ \left( \frac{1 + \sqrt{5}}{2} \right)^6 - \left( \frac{1 - \sqrt{5}}{2} \right)^6 & = & 8\sqrt{5} \\\\ \left( \frac{1 + \sqrt{5}}{2} \right)^7 - \left( \frac{1 - \sqrt{5}}{2} \right)^7 & = & 13\sqrt{5} \end{array}$$ We conjecture that $\varphi^n -\phi^n = F_n \sqrt{5}$ where $F_n$ is the $n^{th}$ term of the Fibonacci sequence. Recall the terms of the Fibonacci sequence are defined by the recursive relationship $F_n = F_{n-1} + F_{n-2}$ along with $F_0=F_1=1$.) We argue by strong induction... The basis step is already well established by the calculations in the list above. Proceeding with the inductive step, we need to show that if $\varphi^n -\phi^n = F_n \sqrt{5}$ for every $n \le k$, then $\varphi^{k+1} -\phi^{k+1} = F_{k+1} \sqrt{5}$. Working from the right side to the left, we have $$\begin{array}{rcl} F_{k+1} \sqrt{5} &=& \sqrt{5} (F_k + F_{k-1})\\\\ &=& \sqrt{5} F_k + \sqrt{5} F_{k-1}\\\\ &=& \sqrt{5} \left( \frac{\varphi^k - \phi^k}{\sqrt{5}} \right) + \sqrt{5} \left( \frac{\varphi^{k-1} - \phi^{k-1}}{\sqrt{5}} \right) \quad \quad \textrm{...by the inductive hypothesis}\\\\ &=& \varphi^k - \phi^k + \varphi^{k-1} - \phi^{k-1}\\\\ &=& (\varphi^k + \varphi^{k-1}) - (\phi^k + \phi^{k-1})\\\\ &=& \varphi^{k-1} (\varphi + 1) - \phi^{k-1} (\phi + 1)\\\\ &=& \varphi^{k-1} \varphi^2 - \phi^{k-1} \phi^2 \quad \quad \textrm{...see note below}\\\\ &=& \varphi^{k+1} - \varphi^{k+1} \end{array}$$ Note: the last two steps may not seem obvious at first blush, but remember we know where we are headed. We know we need to link up the left side and right side below: $$\varphi^{k-1} (\varphi + 1) - \phi^{k-1} (\phi + 1) = \cdots = \varphi^{k+1} - \varphi^{k+1}$$ So it sure would be nice if $\varphi + 1 = \varphi^2$ and $\phi + 1 = \phi^2$. Since we know the exact values of $\varphi$ and $\phi$, we can check to see if this is the case by simple evaluation -- and it is: $$\left( \frac{1 \pm \sqrt{5}}{2} \right) + 1 = \frac{3 \pm \sqrt{5}}{2}$$ and $$\left( \frac{1 \pm \sqrt{5}}{2} \right)^2 = \frac{1 + 2\sqrt{5} + 5}{4} = \frac{6 \pm 2\sqrt{5}}{4} = \frac{3 \pm \sqrt{5}}{2}$$ (Alternatively, if you recall that $\varphi$ and $\phi$ are the roots of the quadratic equation $x^2 -x - 1 = 0$, then isolating the $x^2$ on one side reveals the desired relationship immediately.) This completes the induction step, so by the second principle of mathematical induction (i.e., strong induction), the following holds for all positive integers $n$: $$\varphi^n -\phi^n = F_n \sqrt{5}$$ QED. 9. Let $C(n)$ be defined in the following way: $$C(n)=\left\{ \begin{array}{cl} \frac{n}{2} & \textrm{ if } n \textrm{ even}\\ 3n+1 & \textrm{ otherwise } \end{array} \right.$$ Now suppose a function $f(n)$ satisfies the following two properties: $f(1)=1$ and for $n \neq 1$, $f(n)=f(C(n))$. Find $f(n)$ for $n=1,2,3,\ldots$, until you feel confident in making a conjecture. (Interestingly, if you make the same conjecture most people do -- nobody on the planet knows whether or not it is true.) 10. 🔎 Into how many regions is the plane divided by $n$ lines where no two of the lines are parallel, and no three lines meet at a single point? 11. 🔎 Lockers in a row are numbered 1, 2, 3, ..., 1000. At first, all the lockers are closed. A person walks by, and opens every other locker, starting with locker #2. Another person walks by and changes the 'state' of every third locker starting with locker #3. Changing the 'state' means closing open lockers and opening closed lockers. Then another person changes the state of every fourth locker starting with locker #4. This process continues until no more lockers can be altered. Which lockers will be closed? 12. 🔎 Let $x = 1^1 + 2^2 + 3^3 + 4^4 + \cdots + 100^{100}$. Find the unit's digit of $x$. 13. 🔎 Let $N$ denote the natural numbers $\{1, 2, 3, 4, \ldots\}$. Consider a function $f : N \rightarrow N$ which satisfies $f(1) = 1$, $f(2n) = f(n)$, and $f(2n+1) = f(2n) + 1$ for all $n \in N$. Find a simple algorithm for computing $f(n)$. Your algorithm should be a single sentence long, at most. 14. 🔎 Consider $f_b f_c + f_{b+1}f_{c+1}$ for a variety of natural numbers $b$ and $c$, where $f_n$ is the $n^{th}$ Fibonacci number. (Recall the $n^{th}$ Fibonacci number is defined by $f_1 = 1$, $f_2 = 1$ and $f_n = f_{n-1} + f_{n-2}$ for all $n \gt 2$.) You should notice that each such sum seems to also always be a Fibonacci number. Conjecture which Fibonacci number the sum should be in terms of $b$ and $c$ and then prove your conjecture. 15. 🔎 Make a conjecture about the value $f_{b+1} f_{b-1} - f_{b}^2$ after considering a variety of natural numbers $b$. As always, $f_n$ represents the $n^{th}$ Fibonacci number defined by $f_1 = 1$, $f_2 = 1$ and $f_n = f_{n-1} + f_{n-2}$ for all $n \gt 2$. Then prove your conjecture. 16. 🔎 Suppose that $n$ quarters are stacked so that they are all "heads up". The top coin is removed, flipped over, and then replaced on the top of the stack. Then, the top two coins are removed, flipped over (as a group), and then replaced on the top of the stack. Then, this is done to the top three coins, and then the top four coins, etc..., until finally, the entire stack of $n$ coins is flipped over. Then, this process is repeated, starting again with flipping just the top coin. How many flips does it take to return the stack to "all heads up"?
#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 49 $A=\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right]$ Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix. Given: $A\left[\begin{array}{cc}1 & -2 \\ 1 & 4\end{array}\right]=6 I_{2}$ is identity matrix of order 2 $I_{2}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ Now, let $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ $\\\\ \Rightarrow\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{cc}1 & -2 \\ 1 & 4\end{array}\right]=6\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\\\\\\ \Rightarrow\left[\begin{array}{ll}a+b & -2 a+4 b \\ c+d & -2 c+4 d\end{array}\right]=\left[\begin{array}{ll}6 & 0 \\ 0 & 6\end{array}\right]$ Since, corresponding entries of equal matrices are equal, so $\\\\ \Rightarrow a+b=6 \quad \ldots(i) \\\\ \Rightarrow-2 a+4 b=0 \quad \ldots(i i)\\\\ \Rightarrow c+d=0 \quad \ldots(iii)\\\\ \Rightarrow-2 c+4 d=6 \quad \ldots(iv) \\$ Multiply equation i by 4 and subtract equation ii from i $\\\\ 4 a+4 b=24\\\\ -2 a+4 b=0\\\\ 6 a=24\\\\ a=\frac{24}{6}\\\\ \Rightarrow a=4$ Put a=4 in equation (i) $\\\\ \Rightarrow a+b=6\\\\ \Rightarrow 4+b=6\\\\ \Rightarrow b=6-4=2\\\\ \Rightarrow b=2$ Multiply equation iii by 2 and add equation iii and iv $\\\\ 2 c+2 d=0\\\\ -2 c+4 d=6\\\\ 6 d=6\\\\ \Rightarrow d=1$ Put d=1 in equation iii $\\\\ \Rightarrow c+d=0\\ \Rightarrow c=-1$ Hence $A=\left[\begin{array}{cc} 4 & 2 \\ -1 & 1 \end{array}\right]$ ## Crack CUET with india's "Best Teachers" • HD Video Lectures • Unlimited Mock Tests • Faculty Support
# Determining the value of required parameter for the equations to have a common root Determine the value of $a$ such that $x^2-11x+a=0$ and $x^2-14x+2a=0$ may have a common root. My attempt: Let the common root be $\alpha$ On substituting $\alpha$ in both equations and then subtracting, $a = -3\alpha$ How do I continue from here? What are the other conditions for them to have common roots? • The resultant (en.wikipedia.org/wiki/Resultant) would give a direct solution of the problem. In this case, the resultant of the two polynomials is $a^2 - 24a$ so they have a common root if and only if $a = 0$ or $a = 24$. – Daniel Schepler Sep 4 '17 at 16:04 Eliminate $a$. Then, $$-2a=2(x^2-11x)=x^2-14x$$ $$x^2-8x=0$$ Therefore, the common root should be either $0$ or $8$. If the common root is $0$, then $a=0$. If the common root is $8$, then $a=24$. Another way to get $a$ directly, without first determining the common root. • Eliminating $x^2$ between the equations by direct subtraction gives $3x-a=0 \iff x = \frac{a}{3}\,$. • Eliminating $x$ between the equations by multiplying the first one by $14$, the second one by $11$ then subtracting the two gives $3x^2-8a=0 \iff x^2 = \frac{8a}{3}$. Equating $(x)^2=x^2$ between the two previous expressions gives $\frac{a^2}{9}=\frac{8a}{3} \iff a \in \mathbb\{0, 24\}$. If you do the calculation correctly $a=3\alpha$. So,$x^2-11x+3\alpha=0$ is the first equation and it has root $\alpha$. So $\alpha^2-11\alpha+3\alpha=0$, that is $\alpha(\alpha-8)=0$. So, $\alpha=0,8$. If common root $\alpha=8$ then $a=3\alpha=3*8=24$. If $\alpha=0$ then $a=0$. • But OP wants the value of $a$. – John Wayland Bales Sep 4 '17 at 4:56 • I have edited the answer. – Arindam Sep 4 '17 at 5:08
Q.5. Factorise: (i) x3 - 2x2 - x + 2 (ii) x3 - 3x2 - 9x - 5 (iii) x3 + 13x2 + 32x + 20 (iv) 2y3 + y2 - 2y - 1 Solution: Using trial and error with factor theorem. (i) Let p(x) = x3 - 2x2 - x + 2 Factors of 2 are ±1 and ± 2 By trial and error method, we find that, p(-1) = (-1)3 - 2(-1)2 - (-1) + 2 = -1 -2 + 1 + 2 = 0 Therefore, (x+1) is a factor of p(x) Now, we can write, using long division, x3 - 2x2 - x + 2 =(x+1) (x2 - 3x + 2) = (x+1) (x2 - x - 2x + 2) [by midterm splitting] = (x+1) {x(x-1) -2(x-1)} = (x+1) (x-1) (x+2) (ii) Let p(x) = x3 - 3x2 - 9x - 5 Factors of 5 are ±1 and ±5 By trial and error method, we find that, p(5) = (5)3 - 3(5)2 - 9(5) - 5 = 125 - 75 - 45 - 5 = 0 Therefore, (x-5) is the factor of p(x) Now, we can write, using long division, x3 - 3x2 - 9x - 5 = (x - 5) (x2 + 2x + 1) = (x-5) (x2 + x + x + 1) [by midterm splitting] = (x-5) {x(x + 1) +1(x + 1)} = (x - 5) (x+1) (x+1) (iii) Let p(x) = x3 + 13x2 + 32x + 20 Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20 By trial and error method, we find that, p(-1) = (-1)3 + 13(-1)2 + 32(-1) + 20 = -1 + 13 - 32 + 20 = 0 Therefore, (x+1) is a factor of p(x) Now, we can write, using long division, x3 + 13x2 + 32x + 20 = (x+1) (x2 + 12x + 20) = (x+1) (x2 + 2x + 10x + 20) [by midterm splitting] = (x-5) {x(x+2) +10(x+2)} = (x-5) (x+2) (x+10) (iv) Let p(y) = 2y3 + y2 - 2y - 1 Factors of ab = 2×(-1) = -2 are ±1 and ±2 By trial and error method, we find that, p(1) = 2(1)3 + (1)2 - 2(1) - 1 = 2 +1 - 2 - 1 = 0 Therefore, (y-1) is a factor of p(y) Now, we can write, using long division, 2y3 + y2 - 2y - 1 = (y-1) (2y2 + 3y + 1) = (y-1) (2y2 + 2y + y + 1) [by midterm splitting] = (y-1) {2y(y+1) +1(y+1)} = (y-1) (2y+1) (y+1)
Such equations will have many possible combinations of x and y that work. Nonlinear equations appear curved when graphed and do not have a constant slope. Let's take a look at this graphically below. Linear equations are equations of the first order. Now a solution for the system, the system that has three equations, two of which are nonlinear, in order to … Several methods exist for determining whether an equation is linear or nonlinear, including graphing, solving an equation and making a … Linear equations are those equations that are of the first order. So that's just this line right over here. An equation that is not a straight line when it is graphed. y = 2x + 5 with a = 2 and b = 5, y = -3x + 2 with a = -3 and b = 2, and y = 4x + - 1 with a = 4 and b = -1 are other examples of linear equations. Therefore, they have the opposite properties of a linear function. Not in a straight line. A nonlinear equation will not match this equation. As we stated earlier, nonlinear functions are functions that are not linear functions. 2. Used of an equation. An equation for a straight line is called a linear equation. By Yang Kuang, Elleyne Kase . The substitution method we used for linear systems is the same method we will use for nonlinear systems. The general representation of the straight-line equation is y=mx+b, where m is the slope of the line and b is the y-intercept.. In a nonlinear system, at least one equation has a graph that isn’t a straight line — that is, at least one of the equations has to be nonlinear.Your pre-calculus instructor will tell you that you can always write a linear equation in the form Ax + By = C (where A, B, and C are real numbers); a nonlinear system is represented by any other form. Mathematics a. b. The linear equation is a sum of terms like "Ax" where x is a variable, and A is a number or a constant. Any equation that cannot be written in this form in nonlinear. Linear equations are often written with more than one variable, typically x and y. And the last one, the last one, x squared plus y squared is equal to five, that's equal to that circle. When those points (known as coordinate pairs) are plotted on an x-y axis, they will form a straight line. Of or relating to a system of equations whose effects are not proportional to their causes. Containing a variable with an exponent other than one. Recall that a linear equation can take the form [latex]Ax+By+C=0[/latex]. On the other hand if x or y was a constant (like e or pi), it could be treated as a number and the whole expression would become linear. The graph of a linear function is a line. Example: y = 2x + 1 is the equation can be represented on the graph as Here it represents a straight line so it is a linear equation. Occurring as a result of an operation that is not linear. If an equation gives a straight line then that equation is a linear equation. 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# Texas Go Math Grade 8 Module 4 Quiz Answer Key Refer to ourÂ Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 4 Quiz Answer Key. ## Texas Go Math Grade 8 Module 4 Quiz Answer Key Texas Go Math Grade 8 Module 4 Ready to Go On? Answer Key 4.1 Representing Linear Nonproportional Relationships Question 1. Complete the table using the equation y = 3x + 2. In the given equation y = 3x + 2 include the given values of x. x1 = -1 â†’ y = 3 â€¢ (-1) + 2 = -3 + 2 â†’ y = -1 x2 = 0 â†’ y = 3 â€¢ 0 + 2 = 0 + 2 â†’ y1 = 2 x3 = 1 â†’ y = 3 â€¢ 1 + 2 = 3 + 2 â†’ y1 = 5 x4 = 2 â†’ y = 3 â€¢ 2 + 2 = 6 + 2 â†’ y1 = 8 x5 = 3 â†’ y = 3 â€¢ 3 + 2 = 9 + 2 â†’ y1 = 11 Obtained values of y put in the tabla 4.2 Determining Slope and Y-intercept Grade 8 Module 4 End of Module Assessment Answer Key Question 2. Find the slope and y-intercept of the line in the graph. The following equation for finding the slope and y-intercept is y = mx + b The slope presents the ‘m’ in the equation, and the y-intercept represents b The slope is calculated by: m = $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$ From the given graph, find two points that will help you calculate the slope. Iâ€™ll take (0, 1) for (x1, y1) and (1, 4) for (x2, y2). Thus: m = $$\frac{4-1}{1-0}$$ = $$\frac{43}{1}$$ = 3 To find the y-intercept, we need to include values of x and y from one of these two points and include the obtained slope. So: y = mx + b 1 = 3 . 0 + b 1 = 0 + b b = 1 m = 3, b = 1 4.3 Graphing Linear Nonproportional Relationships Question 3. Graph the equation y = 2x – 3 using slope and y-intercept. y = 2x – 3 Slope = 2 y-intercept = -3 Plot the point that contains the y-intercept: (0, -3) The slope is m = $$\frac{2}{1}$$ Use the slope to find a second point. From (0, -3) count $2$ unit up and $1$ unit right. The new point is (1, -1) Draw a line through the points 4.4 Proportional and Nonproportional Situations Question 4. Does the table represent a proportional or a nonproportional linear relationship? To see if some linear relationship is proportional or nonproportional, we have to use ratio $$\frac{y}{x}$$, which needs to be constant for proportionality. From the given table, we have values for x and y, so just put them into the calculation. $$\frac{y}{x}$$ = $$\frac{4}{1}$$ = 4 = $$\frac{8}{2}$$ = 4 = $$\frac{12}{3}$$ = 4 = $$\frac{16}{4}$$ = 4 = $$\frac{20}{5}$$ = 4 Thus, the obtained values are constant so this table represents a proportional relationship. 4.5 Solving Systems of Linear Equations by Graphing Question 5. A school band ordered hats for $3 and large T-shirts for$5. They bought 150 items in all for \$590. Graph a system of equations to find how many hats and T-shirts the band ordered. ____________ The graph represents the nonproportional relationship because the obtained b, i.e. y-intercept, is bigger than 0 (b / = 0) and the line does not pass the origin. When the b = 0 and the line passes the origin, the linear relationship is proportional. Essential Question Grade 8 Math Module 4 Answer Key Question 6. How can you identify a linear nonproportional relationship from a table, a graph, and an equation? The graph represents the nonproportional relationship because the b, i.e. y-intercept from the given equation, is bigger than 0 (b â‰ 0) and the line does not pass the origin. When the b = 0 and the line passes the origin, the linear relationship is proportional. Texas Go Math Grade 8 Module 4 Mixed Review Texas Test Prep Answer Key Selected Response Question 1. The table below represents which equation? (A) y = -x – 10 (B) y = -6x (C) y = 4x – 6 (D) y = -4x + 2 (C) y = 4x – 6 Explanation: The table is represented by Option C From the table, you can see that the y-intercept (when x = 0) is b = -6. Comparable to y = mx + b y = 4x – 6 Question 2. The graph of which equation is shown below? (A) y = -2x + 3 (B) y = -2x + 1.5 (C) y = 2x + 3 (D) y = 2x + 1.5 (A) y = -2x + 3 Explanation: Option C is rejected Since the graph is slanting downwards, the slope is negative. The graph represents y = -2x + 3 Texas Go Math 8th Grade Module 4 Answer Key Pdf Question 3. The table below represents a linear relationship. What is the y-intercept? (A) -4 (B) -2 (C) 2 (D) 3 The y-intercept is -2. Option B is the correct answer. Question 4. Which equation represents a nonproportional relationship? (A) y = 3x + 0 (B) y = -3x (C) y = 3x + 5 (D) y = $$\frac{1}{3}$$x (C) y = 3x + 5 Explanation: Option C represents a non-proportional relationship y = 3x + 5 For a non-proportional relationship, the equation is y = mx + b and b â‰ 0. Go Math 8th Grade Pdf Module 4 Quiz Answer Key Question 5. Which statement describes the solution of a system of linear equations for two lines with the same slope and the same y-intercept? (A) one nonzero solution (B) infinitely many solutions (C) no solution (D) solution of 0
# Ordering Integers In ordering integers we will learn how to order the integers on a number line. An integer on a number line is always greater than every integer on its left. Thus, 3 is greater than 2, 2 > 1, 1 > 0, 0 > -1, -1 > -2 and so on. Similarly, an integer on a number line is always lesser than every integer on its right. Thus, -3 is less than -2, -2 < -1, -1 < 0, 0 < 1, 1 < 2 and so on. (i) Every positive integer is greater than every negative integer. (ii) Zero is less than every positive integer and is greater than every negative integer. (iii) The greater the number, the lesser is its opposite. i.e., 8 is greater than 5, but -8 is less than -5; similarly, -9 > -15 or, 9 < 15 and so on (iv) The lesser the number, the greater is its opposite. i.e., 6 is less than 7, but -6 is greater than -7; similarly, -8 < -5 or 8 > 5 and so on. Note: The symbol (-) is used to denote a negative integer as well as for subtraction. (i) The temperature at an Everest is -10°C. Here the symbol (-) indicates the negative integer (-10) and no subtraction is involved. (ii) On the other hand, 23 – 7 indicates the subtraction of 7 from 23. Solved examples on ordering integers: 1. Arrange the integers from greater to lesser: (i) 9, -2, 3, 0, -5, -7, 7, -1 (ii) -11, 17, -2, 2, -6, -15, 0, 1 (iii) 12, -21, -18, 14, -5, -1, 1, 10 Solution: (i) 9, 7, 3, 0, -1, -2, -5, -7 (ii) 17, 2, 1, 0, -2, -6, -11, -15 (iii) 14, 12, 10, 1, -1, -5, -18, -21 2. Arrange the integers from lesser to greater: (i) 0, 4, -4, 9, -10, -7, 12, -13 (ii) -14, 7, -25, -17, 20, 5, -9, -3 (iii) -6, 4, -18, 21, 29, -8, -16, 19 Solution: (i) -13, -10, -7, -4, 0, 4, 9, 12 (ii) -25, -17, -14, -9, -3, 5, 7, 20 (iii) -18, -16, -8, -6, 4, 19, 21, 29 `
## Introduction: Convert Decimal to Binary and Vice Versa This is a technique that helped me out a lot when learning about AND OR NOR gates, creating subnet masks and helped me pass time seeing how high I could get with the binary sequence (not very far). I was able to teach my son this method when he was 7 or 8. I plan to eventually make this a Computer Science Collection of Instructables to help teach kids. If anyone would like to help out, provide suggestions and/or write Instructables for this collection let me know. I keep seeing my Assembly Language Quick Reference Guide on my desk.... There's nothing quick about Assembly at least not for me. ## Step 1: Binary Sequence The key to this method is laying out the numbers binary sequence which is powers of 2 since binary is base 2. This that much more difficult than counting by 2's. Note that we use base 10 considering we have 10 digits and computers at their lowest level use base 2. People argue that some computers use base 16 but remember that a switch is either on or off, 1 or 0. The numbers will need to be from right to left the lowest number to the highest number. This is illustrated by the second picture in the collection. ## Step 2: Binary to Base 10 Here we will use the sequence table (2nd Image) to find the base 10 number. Use the visual (1st image) to get a better idea of how this works. So here you have a given binary number. 1. You will fill in the table from right to left with the 1's & 0's from the binary number. 2. You will find the decimal numbers that have a 1 and not a zero. Remember that 1 is on and 0 is off. 3. Add the decimal numbers you found in the previous step together to get the decimal value of the binary number that you were given. ## Step 3: Decimal to Binary Here we will use the sequence table (2nd Image) to find the base 2 number. Use the visual (1st image) to get a better idea of how this works. So here you have a given decimal (base 10) number. 1. Find the highest number that will go into your given decimal number. 2. Place a 1 in the table for each number that you were able to subtract. 3. Subtract that number from your decimal number. 4. Find the next highest number that will go into the number from number 3. 5. Repeat steps 1-5 until you have a zero remainder. 6. Your binary number will not start with zero but everywhere else that you have a blank fill in a zero (see example)
# Question Video: Simplifying and Solving Equations Involving πth Roots Mathematics Find the value (or values) of π₯ given that ((π₯ + 9)/5)Β² = β(144 Γ 3Β²). 04:25 ### Video Transcript Find the value or values of π₯ given that π₯ plus nine over five squared equals the square root of 144 times three squared. Letβs begin this question by seeing if we can evaluate the right-hand side of this equation. We could, of course, work out that three squared is nine, multiply nine by 144, and find the square root. But there is a more efficient way to work this out, particularly if this is a question where weβre not allowed to use a calculator. Letβs use the exponent rule, which tells us that the πth root of ππ is equal to the πth root of π times the πth root of π where the πth of π and the πth root of π are real numbers. And so, the right-hand side of our equation will be equal to the square root of 144 times the square root of three squared. We know that the square root of 144 is 12 and the square root of three squared is three. And multiplying 12 by three gives us 36. The complete equation is therefore π₯ plus nine over five squared equals 36. Next, we can perform the inverse operation of squaring, which is taking the square root. We will have π₯ plus nine over five equals plus or minus the square root of 36. Notice how we have included the plus or minus sign. This is because when weβre taking an even root, and we can include the small two in the square root, of a positive value, such as this 36, then we know that there will be two solutions. In this case, we will have the positive and the negative of the square root of 36. So, the right-hand side of this equation will simplify to plus or minus six. This means that when we continue solving the problem, we have two possibilities: either π₯ plus nine over five equals positive six or π₯ plus nine over five equals negative six. In each of these equations, the next step in solving them would be to multiply both sides of each equation by five. We would have π₯ plus nine equals six times five, which is 30. Or in the other equation, we would have π₯ plus nine is equal to negative six times five. Thatβs negative 30. Subtracting nine from both sides of both equations, we have π₯ is equal to 21 and π₯ is equal to negative 39. And so, we can give the answer that there are two values of π₯ which satisfy the given equation: π₯ is equal to 21 and π₯ is equal to negative 39. As a check of our answer, we can substitute these values into the given equation. That means that when π₯ is equal to 21, we need to check if 21 plus nine over five squared is equal to the square root of 144 times three squared. On the left-hand side, we would have 30 over five squared is equal to 36. We know that 30 over five is equal to six, so we have six squared is equal to 36. And since both sides of this equation are equal to 36, then we know that π₯ equals 21 must be a solution. In the same way, we can check if π₯ is equal to negative 39 is a correct solution by checking if negative 39 plus nine over five squared is equal to the square root of 144 times three squared. This time on the left-hand side, we have negative 30 over five squared. We know that if we square a negative value, we get a positive value. Once again, both sides of this equation are equal to 36. And so, we have verified that π₯ equals negative 39 is the second valid solution.
Question # The two concentric circles of radius 13 cm and 8 cm. AB is a diameter of the bigger circle and BD is a tangent to the smaller circle touching it at D and intersecting the larger Circle at P on producing. Find the length of AP. Hint – In order to solve this problem you need to draw the diagram and analyse it to get the length of AP. Similarity of triangles will play the final role in getting the answer.. Use the property that the diagonal of a circle creates a right angle at its circumference. Above figure is the diagram for this problem. We have joined the common centre O and the point D. It is given that the two concentric circles of radius 13 cm and 8 cm. AB is a diameter of the bigger circle and BD is a tangent to the smaller circle touching it at D and intersecting the larger Circle at P on producing. So, we can say from the diagram that AB = 13(2) = 26 (diameter of the larger) OD = 8 cm (Radius of smaller circle) OB = 13cm (Radius of larger circle) We know that the end points of the diameter make a right angle on the circle. So, we can say that angle APB is 90 degrees. And we also know that angle ODB is also 90 degrees, since OD is the radius and DB is the tangent. Therefore on considering the triangles APB and ODB we can say, $\angle {\text{ABP}} = \angle {\text{OBD}}$(Common angles) $\angle {\text{APB}} = \angle {\text{ODB}}$ (Right angles) Therefore, $\Delta {\text{APB}} \sim \Delta {\text{ODB}}$ (By angle-angle similarity criteria) So, we can say that there respective sides are proportional, So, $\dfrac{{{\text{AP}}}}{{{\text{OD}}}}{\text{ = }}\dfrac{{{\text{AB}}}}{{{\text{OB}}}}$ On putting the values we get, $\dfrac{{{\text{AP}}}}{{\text{8}}}{\text{ = }}\dfrac{{{\text{26}}}}{{{\text{13}}}} \\ {\text{AP = 16cm}} \\$ Hence, the value of AP is 16 cm. Note – In this problem we have used the concept that if we join the end point of the diameter on the circle then it will construct a right angle triangle right angle on the circle and diameter as hypotenuse. Then we have used the concept that if two triangles are similar then the sides of those triangles are proportional. Proceeding like this will provide you the right answer.
# Hard math equations with answers Here, we will show you how to work with Hard math equations with answers. We can solving math problem. ## The Best Hard math equations with answers Keep reading to understand more about Hard math equations with answers and how to use it. The sine function is used to find the angle between two lines. It takes the form of sin(x) where x is in radians, and is used to calculate the angle between two distinct lines, or theta. To solve for the angle, we use the cosine function (see below). The sine function can be used to find the values for other trigonometric functions as well as other angles. For example, if you know the value of one of these functions, you can use the sine function to determine the value of other trigonometric functions. This technique is known as triangulation. The following equation shows how this works: sin(A) = Acos(B) + Bsin(A) In this equation, sin(A) represents the value of one trigonometric function (e.g., tan, arc tangent), while A and B represent a pair of distinct lines (e.g., x-axis and y-axis). To solve for another trigonometric function in terms of sin(A), you simply plug in that value for sin(A). For example, if you know that tan(60°) = 1.5, you can use this equation to determine that 1.5 = cos(60°) + sin(60°). You can also use equations like this one to determine When the y-axis of the graph is horizontal and labeled "time," it's an asymptotic curve. Locally, these functions are just straight lines, but globally they cross over each other — which means they both increase and decrease with time. You can see this in the picture below: When you're searching for horizontal asymptotes, first look at the local behavior of your function near the origin. If you start dragging your mouse around the origin, you should begin to see where your function crosses zero or approaches infinity. The point at which your function crosses zero or approaches infinity is known as an asymptote (as in "asymptotic approach"). If your function goes from increasing to decreasing to increasing again before reaching infinity, then you have a horizontal asympton. If it crosses zero before going up or down more than once, then you have a vertical asymptote.
# How do you find all the critical points to graph -4x^2 + 9y^2 - 36 = 0 including vertices, foci and asymptotes? Feb 16, 2018 The center is located at the origin, vertices at $\left(0 , 2\right)$ and $\left(0 , - 2\right)$, and foci at $\left(0 , 3.6\right)$ and $\left(0 , - 3.6\right)$; the slopes of the asymptotes are ${y}^{2} = \pm \frac{2}{3} {x}^{2}$ $G r a p h :$ graph{y^2/4-x^2/9=1 [-10, 10, -5, 5]} #### Explanation: We can arrange this equation into $9 {y}^{2} - 4 {x}^{2} = 36$ in order to correct format. I also added $36$ to both sides as this will become necessary down the road. Next, I will divide both sides by $36$: $\frac{9 {y}^{2}}{36} - \frac{4 {x}^{2}}{36} = 1$. Simplify the remaining equation; ${y}^{2} / 4 - {x}^{2} / 9 = 1$ The next step is to find $a$, $b$, and $c$ (excluding the center; it is at the origin). In order to do so, we have to define where they are: ${y}^{2} / {a}^{2} - {x}^{2} / {b}^{2} = 1$ $a$ is always the root of the first denominator in hyperbolic equations, so it would be $2$. $b$ would be $3$ ($\sqrt{9} = 3$) $c$ can be found, only in hyperbolic equations, through Pythagora's Theorem: ${a}^{2} + {b}^{2} = {c}^{2}$ c=~3.6 To find the foci, add and subtract the value of $c$ to the variable being divided by $a$ (which is $y$). This value must be directly above or to the side of the center. The value of the foci are $\left(0 , 3.6\right)$ and $\left(0 , - 3.6\right)$ To find the vertices, add and subtract your $a$ value to the number being divided by it ($y$, again). Vertices: $\left(0 , 2\right)$ and $\left(0 , - 2\right)$. To find the slope depends on the format: if ${y}^{2} / {a}^{2} - {x}^{2} / {b}^{2}$, the slopes of the asymptotes are $\pm \frac{a}{b}$. (If your equation looks like this: ${x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = 1$, the slope will be $\pm \frac{b}{a}$) The slopes of the asymptotes are $\pm \frac{2}{3}$
Discuss the continuity and differentiability of f (x) = \|log \|x\|\|. Question: Discuss the continuity and differentiability of f (x) = \|log \|x\|\|. Solution: We have, f (x) = \|log \|x\|\| $\|x\|=\left\{\begin{array}{cl}-x & -\infty<x<-1 \\ -x & -1<x<0 \\ x & 0<x<1 \\ x & 1<x<\infty\end{array}\right.$ $\log \|x\|=\left\{\begin{array}{cc}\log (-x) & -\infty<x<-1 \\ \log (-x) & -1<x<0 \\ \log (x) & 0<x<1 \\ \log (x) & 1<x<\infty\end{array}\right.$ $\|\log \| x\|\|=\left\{\begin{array}{lc}\log (-x) & -\infty<x<-1 \\ -\log (-x) & -1<x<0 \\ -\log (x) & 0<x<1 \\ \log (x) & 1<x<\infty\end{array}\right.$ $(\mathrm{LHD}$ at $x=-1)=\lim _{x \rightarrow-1^{-}} \frac{f(x)-f(-1)}{x+1}$ $=\lim _{x \rightarrow-1^{-}} \frac{\log (-x)-0}{x+1}$ $=\lim _{h \rightarrow 0} \frac{\log (1+h)}{-1-h+1}$ $=-\lim _{h \rightarrow 0} \frac{\log (1+h)}{h}=-1$ $(\mathrm{RHD}$ at $x=-1)=\lim _{x \rightarrow-1^{+}} \frac{f(x)-f(-1)}{x+1}$ $=\lim _{x \rightarrow-1^{+}} \frac{-\log (-x)-0}{x+1}$ $=\lim _{h \rightarrow 0} \frac{-\log (1-h)}{-1+h+1}$ $=\lim _{h \rightarrow 0} \frac{-\log (1-h)}{h}=1$ Here, LHD ≠ RHD So, function is not differentiable at x = − 1 At 0 function is not defined. $(\mathrm{LHD}$ at $x=1)=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}$ $=\lim _{x \rightarrow 1^{-}} \frac{-\log (x)-0}{x-1}$ $=\lim _{h \rightarrow 0} \frac{-\log (1-h)}{1-h-1}$ $=-\lim _{h \rightarrow 0} \frac{\log (1-h)}{h}=-1$ $(\mathrm{RHD}$ at $x=1)=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}$ $=\lim _{x \rightarrow 1^{+}} \frac{\log (x)-0}{x-1}$ $=\lim _{h \rightarrow 0} \frac{\log (1+h)}{1+h-1}$ $=\lim _{h \rightarrow 0} \frac{\log (1+h)}{h}=1$ Here, LHD ≠ RHD So, function is not differentiable at x = 1 Hence, function is not differentiable at x = 0 and ± 1 At 0 function is not defined. So, at 0 function is not continuous. $(\mathrm{LHL}$ at $x=-1)=\lim _{x \rightarrow-1^{-}} f(x)$ $=\lim _{x \rightarrow-1^{-}} \log (-x)$ $=\log (1)=0$ $(\mathrm{RHL}$ at $x=-1)=\lim _{x \rightarrow-1^{+}} f(x)$ $=\lim _{x \rightarrow-1^{+}}-\log (-x)$ $=-\log 1=0$ $f(-1)=0$ Therefore, $f(x)=\|\log \| x\|\|$ is continuous at $x=-1$’ $(\mathrm{LHL}$ at $x=1)=\lim _{x \rightarrow 1^{-}} f(x)$ $=\lim _{x \rightarrow 1^{-}}-\log (x)$ $=-\log (1)=0$ $(\mathrm{RHL}$ at $x=1)=\lim _{x \rightarrow 1^{+}} f(x)$ $=\lim _{x \rightarrow 1^{+}} \log (x)$ $=\log 1=0$ $f(1)=0$ Therefore, at $x=1, f(x)=\|\log \| x\|\|$ is continuous. Hence, function $f(x)=\\| \log \|x\| \mid$ is not continuous at $x=0$
# What Math Lessons Can We Learn from the Game of Nim? Have you played the game of Nim before? Do you know what lessons we can pull from the game? Watch me play the game with two of my daughters Jules and Lucie. You can see right when the game gets to 4 left that each girl knows they lost! You can see it on their faces and even Lucie explains it to us by giving us all the options. Jules even wonders out loud “How did you do that” She knows there’s some trick here. So, let’s see how I’m the World Champion at this game. This game is a variation on the game of Nim “Nim is a mathematical game of strategy in which two players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects provided they all come from the same heap. The goal of the game is to avoid being the player who must remove the last object.” I also play this game with my students. The interesting thing about this game is that there is a winning strategy. And in our variation of the game if the player to go FIRST knows the winning strategy then they are guaranteed to win. So even though it looks like I’m the World Champion it just comes down to math. Let’s explain, Jules and Lucie know they have lost the game when they are left with 4 to choose. So as a player if you can leave 4 objects for your opponent to choose from you have won the game!! So now the game becomes who can leave 4 for their opponent to choose from. How can you always get to a position where you leave 4 for your opponent? Think about it for a moment. What number should I leave my opponent to choose from so that no matter what they do I can then leave them with 4 to choose from? Right..l Should leave them with 8 to choose from! And then where’s my next winning position? 12 then 16 then 20. Multiples of 4! So right from the start since there was 21 objects in the pile I can get to 20 on my first move by going first! And win the game every time. In my class I usually put some cash down on the table to enhance the experience. “Anyone who beats me at the game will get the cash!” Every time I play this game I’m reminded of my math education as a student. You see, in the game of Nim if you know the winning strategy you win every time. You know the path to follow. You see how it works. If you don’t know the strategy you are playing the game almost as if you are blind. You’re not sure how your choices will affect the final moves near the end. You are hoping the moves will pay off down the line. As a student most of my math educational experience was like the experience of the player in the game of Nim that doesn’t know the strategy. I followed the teacher (who does know the “strategy”) blindly. I wasn’t sure of how my “moves” would pay off in the end. I just followed the rules hoping for good outcome. I was such a good rule follower that sometimes it awarded me some success. In the fourth grade I remember earning one of those big puffy, stick off the page stickers for being a master multiplier. Yay go me!! But when it came to being pushed to show my understanding the wheels fell off. Here is a 4th grade test on multiplying (when I look it over now it looks like I must have fixed this up after getting it back). Math for me was like a series of tricks that I could memorize and then try to perform. Thinking back to playing the game with my daughters you can hear Jules, the first girl in the video ask right at the end “How did you do that?” She was thinking this is all a trick! The game was like a magic trick. How many of our students see their math education as a series of tricks? Lots of them I bet. We don’t want kids thinking math is just a series of tricks to memorize. If they do think the math they are learning is a trick then it’s our duty to uncover the trick. Show them how it works. Like in the game of Nim students should know why the first player has the winning strategy. This is what I want from my math lessons. Let’s continue to fuel sense making in our students instead of showing them just tricks. So, in your next class play the game of Nim with them. Blow their socks off and win 3 times in a row…..but don’t leave it as a trick. Uncover the math and strategy behind it together! A great read is Nix The Tricks by Tina Cardone https://nixthetricks.com/ You can read more about how to fuel sense making for students from Kyle Pearce as he describes a task around Donuts. https://tapintoteenminds.com/3act-math/donut-delight/ and also here from me showing a task about the defrost function on my microwave http://mrorr-isageek.com/fuel-sense-making-black-box-defrost/
Navigation Panel: Previous \| Up \| Forward \| Graphical Version \| PostScript version \| U of T Math Network Home # Applications of the Geometric Mean Asked by Senthil Manick on May 22, 1997: When would one use the geometric mean as opposed to arithmetic mean? What is the use of the geometric mean in general? The arithmetic mean is relevant any time several quantities add together to produce a total. The arithmetic mean answers the question, "if all the quantities had the same value, what would that value have to be in order to achieve the same total?" In the same way, the geometric mean is relevant any time several quantities multiply together to produce a product. The geometric mean answers the question, "if all the quantities had the same value, what would that value have to be in order to achieve the same product?" For example, suppose you have an investment which earns 10% the first year, 50% the second year, and 30% the third year. What is its average rate of return? It is not the arithmetic mean, because what these numbers mean is that on the first year your investment was multiplied (not added to) by 1.10, on the second year it was multiplied by 1.60, and the third year it was multiplied by 1.20. The relevant quantity is the geometric mean of these three numbers. The question about finding the average rate of return can be rephrased as: "by what constant factor would your investment need to be multiplied by each year in order to achieve the same effect as multiplying by 1.10 one year, 1.60 the next, and 1.20 the third?" The answer is the geometric mean (1.10 x 1.60 x 1.20)^(1/3). If you calculate this geometric mean you get approximately 1.283, so the average rate of return is about 28% (not 30% which is what the arithmetic mean of 10%, 60%, and 20% would give you). Any time you have a number of factors contributing to a product, and you want to find the "average" factor, the answer is the geometric mean. The example of interest rates is probably the application most used in everyday life. Here are some basic mathematical facts about the arithmetic and geometric mean: Suppose that we have two quantities, A and B. Taking their arithmetic mean we get the number (A+B)/2 which can be interpreted in a number of ways. One interpretation (probably the most common) is that this quantity is the midpoint of the two numbers viewed as points on a line. Now suppose that we have a rectangle with sides of lengths A and B. The arithmetic mean can also be interpreted as the length of the sides of a square whose perimeter is the same as our rectangle. Similarly, the geometric mean sqrt(AB) is the length of the sides of a square which has the same area as our rectangle. It is known that the geometric mean is always less than or equal to the arithmetic mean (equality holding only when A=B). The proof of this is quite short and follows from the fact that (sqrt(A)-sqrt(B))^2) is always a non-negative number. This inequality can be surprisingly powerful though and comes up from time to time in the proofs of theorems in calculus. Asked by G. Ellis, student, Southeast Bulloch High on January 16, 1997: Could you give the formula for the geometric mean for a series of numbers if I am trying to get the compound annual growth rate for a series of number that include negative numbers? In general, you can only take the geometric mean of positive numbers. The geometric mean of numbers a_1, a_2, ..., a_n is the nth root of the product a_1a_2...a_n. In your example, you are taking the mean of positive numbers. For example, if you're looking at an investment that increases by 10% one year and decreases by 20% the next, the simple rates of change are 10% and -20%, but that's not what you're taking the geometric mean of. At the end of the first year you have 1.1 times what you started with (the original plus another tenth of it). At the end of the second year you have 0.8 times what you started the second year with (the original minus one fifth of it). So, the numbers you are taking the geometric mean of are 1.1 and 0.8. This mean is approximately 0.938. This means that, on average, your investment is being multiplied by 0.938 (= 93.8%) each year, a 6.2% loss. So, the compound anual growth rate is (approximately) -6.2%. Asked by Paul van Esbroeck on October 5, 1997: (Question abridged from original posting) I am presently engaged in a dispute with a bank concerning a Stock Market Tracker GIC. I have found significantly different definitions for the terms "average percentage growth" and "average percentage growth rate", and in many instances I cannot distinguish between the use of the words "growth" vs. "growth rate" in the literature. It was in investigating growth that I came to your question page about the geometric mean. While I, like most people I asked, originally understood these GICs to be guaranteeing a rate of return equal to the rate of growth of the index, the bank has a different interpretation. Let's say a stock market index starts at 1000, and at the end of 1 month is 1010, at the end of 2 months is 1020, and so on, ending up at 1120 after one full year. The bank seems to be considering the growth by month end for each month (which is 10 for the first month, 20 for the second, and so on, with this growth being 120 by the end of the twelfth month), then averaging these growths, obtaining an "average growth" of 60 and an average percentage growth of 60/1000 or 6%. For other definitions there seems to be quite a bit of agreement that: • The real growth was 120/1000 = 12%. • The average percentage growth rate is 12% per year. • The annual compound growth rate is 12%. It seems to me that it is incorrect to average growth as done by the bank. Since an average should be of equal periods, in the banks formula we average growth for the first month with growth for the first 11 months. Since the growth was 10 index units each month, should the average growth not be 10 units per month, but how is this different from the average growth rate? What if not growth, do you call the plot of (Index value minus Index value at some starting time) ? What if any, is the correct distinction between growth and growth rate ? Do you find the term "average percentage growth" as used by the bank problematic? First let's try to get the words sorted out. Then we can address the situation you describe. The word "growth" is often used quite loosely to mean any of the above concepts. The best and most correct definition of growth would be a quantity that is associated to a particular period in time, and describes the index value at the end of the period minus the index value at the beginning of the period. In your example, the growth for the period consisting of the first month was 10, and the growth for the period consisting of the first 12 months was 120. "Growth rate" describes growth per unit time. It may vary with time. The "average growth rate" over a period is the growth divided by the length of the period. If the growth rate is constant over a period, then the average growth rate over the period will be the same as that constant value. In your example the growth rate was a constant 10 units per month. The average growth rate over every period was also 10 units per month. For example, the average growth rate during the first month was 10 units per 1 month = 10 units per month. The average growth rate during the entire year was 120 units per 12 months = 10 units per month. These concepts are not relevant to typical investements, stock market indices, etc., because the growth rate tends to be proportional to the index value. After all, if a 10 dollar investment grew by 1 dollar over a year (an average growth rate of 1 dollar per year), you'd expect a 10,000 dollar investment to grow by 1000 dollars per year: a very different growth rate! So instead, the relevant concepts are percentage growth and growth rates. The percentage growth over a period is the ratio of the growth to the starting value. In your example, the percentage growth during the first month was 1%. The percentage growth during the second month was about 0.99% (10/1010). The percentage growth during the entire year was 12% (120/1000). The percentage growth rate is the ratio of the growth rate to the current value. Thus, if an index value is 1000 and it is growing at 10 units per month, it is experiencing a percentage growth rate of 1% per month. If its value is 1010 and it is growing at 10 units per month, its percentage growth rate is about 0.99% per month. And so on. In your exaple, by the end of the year it is still growing at 10 units per month and its value is 1120, so the percentage growth rate by the end of the year has dropped to about 0.89%. The only tricky thing about percentage growth rate is in changing units: a percentage growth rate of 1% per month is not the same thing as 12% per year. If an index were growing at a constant percentage growth rate of 1% per month, after one month it would be 1.01 times its starting value, after two months it would be 1.01^2 times its starting value, and after 12 months it would be 1.01^(12) = 1.1268... times its starting value. Therefore, a percentage growth rate of 1% per month is the same as a percentage growth rate of about 12.7% per year. If the percentage growth is G over a period of T time units, then the average percentage growth rate over that period is (1+G)^(1/T) - 1. To figure out the average percentage growth rate of your example over the entire year: if your units are years, then T=1. G = 12%, so the average percentage growth rate is (1+G)^1 - 1 = G = 12% per year. If your units are months, the average percentage growth rate is (1+G)^(1/12) - 1 which is about 0.949% per month. These are each saying the same thing; an average percentage growth rate of 12% per year is the same as an average percentage growth rate of about 0.949% per month. Often, in investment circles, when people refer to "growth" or "growth rate" they are meaning percentage growth and percentage growth rate. Now, on to your particular situation. As you point out, it is not particularly meaningful to average the growths over time periods of wildly differing lengths. So, to average the growth during the first month with the growth during the first 12 months is not a reasonable thing to do. However, what is reasonable to do is to average the growths over overlapping time periods of similar lengths, to smooth out fluctuations in the index. For example, let's suppose the end of the year happened to be a really bad day on the stock market, and instead of rising to 1120 the index fell back to 1000, jumping back up to 1120 the next day. Would you really want to say that there was no growth at all during the year, just because the day you picked to evaluate the index happened to be a bad one? I think not. So, what is commonly done is to average the index value over a certain time period. For example, while what you describe is unreasonable, it could be perfectly reasonable to say that the GIC's rate of return from 1996 to 1997 will be the percentage growth of the index's average 1996 value to its average 1997 value. That way, fluctuations due to the particular day the index is measured will be smoothed out. Now, if the index were a constant 1000 during all of 1996, and rose to 1120 in the manner you describe during 1997, you would unfortunately only be getting a 6% return from average 1996 value (1000) to average 1997 value (1060). But that's because some of the 1996 behaviour is being factored in as well. Another thing that could be reasonable is to average the percentage growths over different 12-month periods ending in the same year. For example, you could take the January 1996-January 1997 percentage growth, the February 1996-February 1997 percentage growth, and so on, and average them. This is a reasonable sort of average to take, since the percentage growths are being averaged over equal but overlapping periods. These sorts of averages have the advantage of smoothing out wild fluctuations in the index. For example, if the index started 1996 at 1000 and rose at 10 units per month steadily, ending 1997 at almost 1240 but plummeting down to 1150 on the last day of 1997 due to a temporary correction in the markets, you would still get the benefit of the growths. Even though the December 31, 1996 to December 31, 1997 growth was from 1120 to 1150, just 2.7% on the entire year, the November 30, 1996 to November 30, 1997 growth was from 1110 to 1230, about 11%. So, by including these in the average, rather than just looking at the average percentage growth over a single year-long period, you get a much less volatile measure of performance. My suspicion is that the GIC you are concerned about probably employs some sort of legitimate averaging such as the above, and that the bank has miscommunicated the calculation method to you. Another possibility is that the bank may be measuring index value from start of investment to the average index value during the year preceding the end of investment. That means that during the first year of the investment you have the unreasonable calculation you described, but over the long term those effects are negligble. For instance, the 10-year return might be the average of the index return over the twelve periods all beginning on the date of invesment and ending on the twelve month-end dates during that 10th year. True, this is still averaging growth over periods of differing length, but the lengths range from 9 years to 10 years and this is much less of a difference than your example where they range from 1 month to 12 months! I will not venture to speculate further on what methods the bank may or may not be using for the GIC you have in mind. However, I hope this clears up for you some of the mathematical ideas involved. 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Hong Kong Stage 4 - Stage 5 # Multiplication of Polynomials Lesson In this chapter, we are going to look at how to multiply polynomials. The good news is that we've already started to do this without even knowing it when we learnt how to expand binomials using the distributive law! Remember, Not all questions are the same as binomial expansion though. We may be asked to evaluate the product of two polynomials. For example, if $P$P$\left(x\right)=12$(x)=12 and $Q$Q$\left(x\right)=9$(x)=9, the product of these polynomials is $12\times9$12×9, which equals $108$108. ie. $P$P$\left(x\right)$(x)$Q$Q$\left(x\right)=108$(x)=108. Or we may be asked to find missing terms in a polynomial (we'll see how in example video 3 below). Remember! Every term in one pair of brackets has to be multiplied by every other term in all the other pairs of brackets. #### Examples ##### Question 1 Expand $\left(a+2\right)\left(5a^2-2a+2\right)$(a+2)(5a22a+2). ##### Question 2 Given that $P($P($4$4$)$) $=$= $10$10 and $Q($Q($4$4$)$) $=$= $7$7, evaluate $P($P($4$4$)$)$Q($Q($4$4$)$). ##### Question 3 Consider the expansion of $\left(4x^3-4x^2+ax+b\right)\left(4x^2-2x+9\right)$(4x34x2+ax+b)(4x22x+9). 1. In the expansion, the coefficient of $x^3$x3 is $12$12. Form an equation and solve for the value of $a$a. 2. In the expansion, the coefficient of $x^2$x2 is $4$4. Form an equation and solve for the value of $b$b.
# Lesson 14 Problemas de comparación de fracciones ## Warm-up: Conversación numérica: Múltiplos de diez (10 minutes) ### Narrative The purpose of this Number Talk is to elicit strategies and understandings students have for adding and subtracting multi-digit numbers. These understandings help students develop fluency and will be helpful in later units as students will need to be able to add and subtract multi-digit numbers fluently using the standard algorithm. When students make adjustments and create multiples of ten for mental addition they are looking for and making use of the base ten structure of numbers (MP7). ### Launch • Display one expression. • “Hagan una señal cuando tengan una respuesta y puedan explicar cómo la obtuvieron” // “Give me a signal when you have an answer and can explain how you got it.” • 1 minute: quiet think time ### Activity • Keep expressions and work displayed. • Repeat with each expression. ### Student Facing Encuentra mentalmente el valor de cada expresión. • $$119 + 119$$ • $$139 + 139$$ • $$159 + 159$$ • $$199 + 199$$ ### Activity Synthesis How did you use multiples of ten, for example 20, 40, and 60 to help add these numbers mentally? (I changed the addends by adding one more to each addend and the subtracting the extra two from the final sum.) • “¿Alguien puede expresar el razonamiento de _____ de otra forma?” // “Who can restate _____’s reasoning in a different way?” • “¿Alguien usó la misma estrategia, pero la explicaría de otra forma?” // “Did anyone have the same strategy but would explain it differently?” • “¿Alguien pensó en la expresión de otra forma?” // “Did anyone approach the expression in a different way?” • “¿Alguien quiere agregar algo a la estrategia de ______?” // “Does anyone want to add on to _____’s strategy?” ## Activity 1: Fracciones desconocidas (20 minutes) ### Narrative In this activity, students are given several sets of fractions and some clues about the size of a particular fraction in each set. To identify a fraction that meets certain size requirements or falls within a specified range, students need to use multiple comparison strategies they have learned. For example, they can use comparisons to benchmarks such as $$\frac{1}{2}$$ and 1 to eliminate some fractions, and then use equivalent fractions to compare the remaining ones. ### Launch • Groups of 3–4 • “Hay seis conjuntos de fracciones en esta actividad. Cada conjunto viene con unas pistas. Su tarea en cada conjunto es encontrar una fracción que cumpla con las tres pistas” // “There are six sets of fractions in the activity. Each set comes with some clues. Your task is to find one fraction that meets all three clues in each set.” ### Activity • “Trabajen en grupo para encontrar las fracciones desconocidas” // “Work with your group to find the mystery fractions.” • “Cada miembro del grupo debe empezar con un conjunto de fracciones distinto y debe encontrar las fracción desconocida de al menos dos conjuntos antes de discutir sus respuestas” // “Each group member should start with a different set, and should find at least two mystery fractions before discussing their responses with the group.“ • 5–6 minutes: independent work time • 7–8 minutes: group work time ### Student Facing A cada uno de seis amigos se les dio una lista con 5 fracciones. Cada uno eligió en secreto una fracción y escribió pistas sobre su elección. Usa sus pistas para identificar las fracciones que eligió cada uno. Andre: $$\ \frac{8}{12} \quad \frac{3}{6} \quad \frac{3}{4} \quad \frac{3}{2} \quad \frac{2}{12}$$ menor que 1 mayor que $$\frac{1}{3}$$ menor que $$\frac{2}{3}$$ Tyler: $$\ \ \frac{2}{6} \quad \frac{2}{2} \quad \frac{2}{4} \quad \frac{2}{3} \quad \frac{2}{5} \quad$$ mayor que $$\frac{1}{3}$$ menor que 1 menor que $$\frac{1}{2}$$ Clare: $$\ \ \frac{4}{3} \quad \frac{4}{2} \quad \frac{3}{4} \quad \frac{1}{4} \quad \frac{2}{10} \ \$$ mayor que $$\frac{2}{8}$$ menor que $$\frac{11}{6}$$ mayor que 1 Diego: $$\ \frac{2}{8} \quad \frac{6}{12} \quad \frac{6}{8} \quad \frac{12}{10} \quad \frac{11}{12}$$ mayor que $$\frac{1}{2}$$ menor que 1 mayor que $$\frac{3}{4}$$ Elena: $$\ \frac{2}{12} \quad \frac{50}{100} \quad \frac{4}{10} \quad \frac{3}{5} \quad \frac{7}{5}$$ mayor que $$\frac{2}{10}$$ menor que 1 mayor que $$\frac{3}{6}$$ Noah: $$\ \frac{18}{10} \quad \frac{7}{8} \quad \frac{2}{5} \quad \frac{18}{5} \quad \frac{150}{100}$$ mayor que $$\frac{1}{2}$$ menor que $$\frac{25}{10}$$ mayor que $$\frac{8}{5}$$ ### Activity Synthesis • “¿Cuáles pistas les ayudaron a eliminar fracciones más rápido?” // “Which clues helped you eliminate fractions the fastest?” (clues about size relative to 1) • “¿Qué estrategias usaron para comparar fracciones?” // “What strategies did you use to compare fractions?” (compare fractions to $$\frac{1}{2}$$, 1, or another benchmark, write equivalent fractions to compare two fractions, compare fractions with the same numerator or denominator) • “¿En algún caso tuvieron que usar más de una estrategia para comparar fracciones?” // “Did you ever have to use more than one strategy to compare fractions?” (Yes, two or three were often needed to find the mystery fraction.) ## Activity 2: Distancias a pie (10 minutes) ### Narrative This activity has two purposes: to give students an opportunity to solve fraction comparison problems in context, and to reinforce the idea that two fractions can be compared only if they refer to the same whole. To serve the former, students compare fractional distance measurements. To serve the latter, they investigate fractional measurements in two different units of distance: Chinese “li” and kilometer. When comparing the distances in the first question, students can rely on a number of familiar strategies. Two of the fractional values are close to 2. Some students are likely to use that benchmark for efficient comparison. For example, they may note that the school and the market are both a little over 1 li from home, the library is more than 2 li, and the badminton club is a little under 2 li. Focus the synthesis on the last two questions about interpreting fractional measurements in two different units. MLR8 Discussion Supports. Synthesis: Display sentence frames to support whole-class discussion: “Estoy de acuerdo porque . . .” // “I agree because . . .” and “No estoy de acuerdo porque . . .” // “I disagree because . . . .” Engagement: Provide Access by Recruiting Interest. Optimize meaning and value. Share information about the unit “li” and help students understand the size of one li by referencing a common point of interest. Invite students to share this knowledge with family members and other teachers. Supports accessibility for: Visual-Spatial Processing, Social-Emotional Functioning ### Launch • Groups of 3–4 • “¿Cuáles son algunas unidades que usamos para medir distancia? Digamos todas las que se nos ocurran” // “What are some units that we use for measuring distance? Let’s name as many as we can think of.” (Sample responses: inches, feet, miles, meters, kilometers) • Share and record responses. • “Hoy vamos a pensar en distancias medidas en ‘li’, una unidad que se usa con frecuencia en China” // “Today we’ll look at distances measured in ‘li,’ a unit commonly used in China.” ### Activity • “Tómense unos minutos para trabajar en silencio en las preguntas. Prepárense para explicar su razonamiento” // “Take a few quiet minutes to work on the questions. Be prepared to explain your reasoning.” • “Después, discutan sus respuestas con su grupo y trabajen juntos para terminar la actividad” // “Afterward, discuss your responses with your group and work together to complete the activity.” • 4 minutes: independent work time • 4 minutes: group work time • Monitor for students who: • attend to the location of 1 whole when representing $$\frac{4}{5}$$ and $$\frac{7}{5}$$ on the two number lines (rather than partitioning both number lines into the same 5 parts) • recognize that $$\frac{4}{5}$$ and $$\frac{7}{5}$$ in the two cases refer to different wholes and can articulate their reasoning ### Student Facing En China y en algunos países del este de Asia se usa la unidad “li” para medir distancias. Estas son las distancias que camina un estudiante en China entre su casa y algunos de los lugares que visita con frecuencia. • escuela: $$\frac{7}{5}$$ li • biblioteca: $$\frac{23}{10}$$ li • mercado: $$\frac{7}{4}$$ li • club de bádminton: $$\frac{23}{12}$$ li 1. Cuál queda a menor distancia desde la casa del estudiante: 1. ¿Su escuela o la biblioteca? 2. ¿El mercado o el club de bádminton? 3. ¿La biblioteca o el mercado? 2. Un estudiante en los Estados Unidos camina $$\frac{4}{5}$$ kilómetros (km) de la casa a la escuela. Estas rectas numéricas muestran cómo se relaciona 1 kilómetro con 1 li. ¿Cuál estudiante camina una mayor distancia a la escuela? Usa las rectas numéricas para mostrar tu razonamiento. 3. Explica por qué no podemos simplemente comparar las fracciones $$\frac{4}{5}$$ y $$\frac{7}{5}$$ para ver cuál estudiante camina una mayor distancia. ### Activity Synthesis • Ask previously selected students to share their responses to the last two questions. Display their number lines, or display the number lines from the activity for them to annotate while they explain. • Emphasize that just as 1 km is not the same distance as 1 li, $$\frac{4}{5}$$ km is not the same distance as $$\frac{4}{5}$$ li. We can’t compare two fractions that refer to different wholes. ## Lesson Synthesis ### Lesson Synthesis “Hoy combinamos varias estrategias que nos ayudaron a comparar fracciones. También resolvimos problemas de comparación de fracciones en una situación sobre distancias” // “Today we used a combination of strategies to help us compare fractions. We also solved fraction comparison problems in a situation about distance.” Keep students in groups of 3–4. Give tools for creating a visual display to each group. Assign each group one set of fractions in Activity 1 or the first set of questions in Activity 2. • For the former: “Hagan una presentación visual que explique cómo encontraron la fracción desconocida para el conjunto de fracciones que les asignaron en la actividad 1. Su presentación debe incluir las cinco fracciones, las tres pistas y una explicación de cómo la fracción elegida cumple con todas las pistas” // “Create a visual display that explains how you found the mystery fraction for your assigned set of fractions from Activity 1. Your display should list the five fractions, the three clues, and how the chosen fraction satisfies all the clues.” • For the latter: “Hagan un presentación visual que muestre sus respuestas al primer grupo de preguntas de la actividad 2. Su presentación debe mostrar las cuatro distancias y cómo las compararon” // “Create a visual display that shows your responses to the first set of questions in Activity 2. Your display should show the four walking distances and how you compared them.” “Incluyan diagramas, notas y cualquier descripción que pueda ayudarle a los demás a entender cómo pensaron” // “Include diagrams, notes, and any descriptions that might help others understand your thinking.” Ask students to display their work around the room. “Visiten las presentaciones de al menos otros 2 grupos” // “Visit the display of at least 2 other groups.” “En cada presentación, revisen si las estrategias de razonamiento tienen sentido para ustedes. Piensen en qué se parece y en qué es diferente el razonamiento en las presentaciones” // “At each display, check to see if the reasoning strategies make sense to you. Think about how the reasoning in different displays is alike and how it is different.” “¿En qué se parecen los diagramas, explicaciones o cálculos que vieron? ¿En qué son diferentes?” // “How are the diagrams, explanations, or calculations that you saw alike? How are they different?” Share and record responses. Reference the displays that students created to show similarities and differences in their reasoning strategies.
# Remainder Theorem Tough Questions for Competitive Exams \| Aptitude Questions ## Remainder Theorem Aptitude Examples with Answers \| Remainder Theorem Tutorial Remainder theorem basic rules were given in the following link. Here provides some examples with shortcut methods on remainder theorem aptitude. Remainder Theorem for Number System Basic rules #### Application of the remainder theorem: Finding the last digit of an expression purpose simply find the remainder of that expression divided by 10. In the same way for finding the last two digits of an expression purpose find the remainder of that expression divided by 100. #### Remainder Theorem Sums Example – 1 : Find the last two digits of the expression of 120 x 2587 x 247 x 952 x 854 Solution: Now the above expression divided by 100 and find the remainder then it will equal to last two digits of given sum. For reducing the calculation purpose cancelled the both numerator and denominator by 20 then Now find the remainders of the each number Remainder of the above expression is 2 Now reminder of the initial expression is 40 ( Multiplying the reminder with canceled number i.e ) Example – 2 : Find the remainder when 1! + 2! + 3! +4! + 5! + ————– 1000! is divided by 8 Solution: Here remember about the factorial function Note : 2! is usually pronounced “2 factorial The factorial function says to multiply descending natural numbers series. For example 4! = 4 x 3 x 2 x 1 = 24 Now calculate remainder for each number in the given series The remainder of the remaining terms is also zero. So Example – 3 : Find the remainder when 1! + 2! + 3! +4! + 5! + ————– 1000! is divided by 14 Solution: Now calculate remainder for each number in the given series 5! / 14 8 The remainder of the remaining terms is also zero. NOW = 1 + 2 +6 – 4 + 8 + 6 /14 = 19 / 14 5 Remainder of the given sum is 5 Example – 4 : Find the remainder when 51203 is divided by 7 Solution: Find Remainder of the expression 51 / 7 51 / 7 2 So Example – 5 : Find the remainder when 21875 is divided by 17 Solution: Find Remainder of the expression 21 / 7 Here our aim is obtained number as 24 = 16 ( 16 = 17 – 1) so according to that rewrite the equation as follow as Final remainder is 17 – 4 = 13 Example – 6 : Find the remainder when 2787 x 2345 x 1992 is divided by 23 Solution: While observing the above question middle term is exactly divisible by 23, So the hole expression is also exactly divisible by 23 The final remainder is Zero Example – 7 : Find the remainder when 341 +782 is divided by 52 Solution: Hint : If an +bn and n = odd number then (a +b) is exactly divisible of that number an +bn The given expression can be written as 341 +782 = 341 +4941 From the above information the given expression is exactly divisible by 52 So remainder is Zero Example – 8 : Find the remainder when 53 + 173 +183 +19 is divided by 70 Solution: Hint: If An +Bn + Cn + Dn and n = odd number then (A+B+C+D) is exactly divisible of that number From the above information n = 3 is odd number and 16 + 17 + 18+ 19 = 70 so 70 is exactly divisible by 163 + 173 +183 +193 Remainder of the sum is zero Example – 9 : Find the last two digits of the expression of 12899 x 96 x 997 Solution: Now the above expression divided by 100 and find the remainder then it will equal to last two digits of given example Final remainder = -12 +100 = 88 Example – 10 : Find the remainder when 1 2 3 4 5 – – – – – – 41 digits is divided by 4 Solution: Here first identifying that 1 to 9 numbers having 1 digit after that each number having 2 digits. So 41 digits means – 41 – 9 = 32 / 2 = 16 then last number is 9+ 16 = 25 and given number is 1 2 3 4 5 – – – – – – – – – – 2 4 2 5 Now according to divisibility rules we can find easily ( i. e find remainder while divided last two digits of that number) = 25/4 1 Example – 11 : Find the remainder when 8 8 8 8 8 8 8 8 – – – – – 32 times is divided by 37 Solution: Hint: If any 3-digit which is formed by repeating a digit 3-times then this number is divisible by 3 & 37. The above expression can be written as 10 pairs of 8 8 8 30 times remaining number is 88 So 88 / 37 14 Example – 12 : Find the remainder when 7 7 7 7 7 7 7 – – – – – 43 times is divided by 13 Solution : Hint : If any 6-digit which is formed by repeating a digit 6-times then this number is divisible by 3, 7, 11, 13 and 37. The above expression can be written as 7 pairs of 7 7 7 7 7 7 7 42 times remaining number is 7 So 7 / 13 7 Example – 13 : Find the remainder when 101 + 102 +103 + 10 + – – – – – – + 10100 is divided by 6 Solution: Here find remainder of the each number in the given expression 101/ 6 +4 102/ 6 +4 103/ 6 +4 104/ 6 +4 So final remainder = 100 x 4 / 6 +4 Hint : The above sum simple we identifying that every three terms the remainder is zero ( 4 + 4+ 4 = 12 / 6 0 ) so upto 99 terms the remainder is zero. Example -14: Find the remainder when 2469 + 3268 is divided by 22 Solution: Here find remainder of the each number individually. Final remainder of this term = 3 x 2 /11 6 Final remainder of the give expression = 6 +5 /22 = 11 /22 11 Example – 15 : Find the remainder when 7 7 + 7 77 + 7 777 + 7 7777 + – – – – – – + 7 777777777 is divided by 6 Solution: Here find remainder of the each number individually. = 7 7 / 6 = (6+1) 7 /6 1 So remaining terms remainders are also 1 and total terms in given expression is 9 = 9/ 6 3 Example – 16 : Find the remainder when 2310 – 1024 is divided by 7 Solution: Here 1024 can be written as 2310 The expression is 2310 – 210 Hint: If the given expression like ( an – bn ) and “n” is even number then (a-b) and (a+b) are exactly divides that expression. From the above hint factors of the given expressions = 21 and 25 Factors of 21 = 1 , 3 , 7, 21 Factors of 25 = 1 , 5 , 25 So the numbers 1, 3 , 5 , 21 and 25 are exactly divisible by the given expression 2310 – 1024 Remainder = 0 Example – 17 : Find the remainder when 341 + 782 is divided by 26 Solution: Here 782 can be written as 4941 The expression is 341 + 4941 Hint: If the given expression like ( an + bn ) and “n” is odd number then (a+b) is exactly divides that expression. From the above hint factors of the given expressions = 52 Factors of 52 = 1 , 2, 4, 13 , 26, 52 So the numbers 1, 2, 4 , 13 , 26 and 52 are exactly divisible by the given expression 341 + 782 Remainder = 0 Example – 18 : Find the remainder when 2723 + 1923 is divided by 2 Solution: Hint: If the given expression like ( an – bn ) and “n” is odd number then (a+b) is exactly divides that expression. From the above hint factors of the given expressions = 27 + 19 = 46 Factors of 46 = 1 , 2, 23 , 46 So the numbers 1, 2, 23 and 46 are exactly divisible by the given expression 2723 + 1923 Remainder = 0 Example – 19 : Find the remainder when 52P is divided by 26 where P = (1 !)2 + (2 !)2 + (3 !)2 + – – – – – + (10 !)2 Solution: Here 52p = 25P = (26 – 1)P So the reminder depend upon value of P. If P = even number then remainder has 1 & If P = odd number then remainder has 25 Now find the last digit of the expression (1 !)2 + (2 !)2 + (3 !)2 + – – – – – + (10 !)2 For that purpose Find the remainder when P is divided by 10 (1 !)2/ 10 1 ( 2 !)2/ 10 4 (3 !)2/ 10 6 ( 4 !)2/ 10 = 576 / 10 6 ( 5 !)2/ 10 = 0 Remainder is 1 + 4 + 6 + 6 = 17 /10 7 So P is odd number and Example – 20 : Find the remainder when 22225555 + 55552222 is divided by 7 Solution: Here find remainder of the each number individually. 22225555 / 7 = 35555 = (3)( 3 x 1851 + 2 ) / 7 = (27) 1851 x 9 / 7 5 55552222 / 7 = 45555 = (4)( 3 x 740 + 2 ) / 7 = (64) 740 x 16 / 7 2 Final remainder is = 5 +2 /7 0 Example – 21 : Find last digit of the expression 2017 2017 Solution: The last digit of an expression purpose simply find the remainder of that expression divided by 10. So last digit of the number 2017 2017 is 7 Related Topics : Number Categories Rules for Divisibility of numbers Formulas for Sum of n Consecutive numbers Methods to find HCF & LCM GCD and LCM Problems & Solutions Topics in Quantitative aptitude math for all types of exams Shortcut Math Tricks for helpful to improve speed in all calculations Hi friends Thanks for reading. I Hope you liked it. Give feed back, comments and please don’t forget to share it. ## 6 thoughts on “Remainder Theorem Tough Questions for Competitive Exams \| Aptitude Questions” #### Jitendra kumar dagur (February 23, 2019 - 7:22 pm) Thank you very much sir for this valuable information on remainder theorem . I really appreciate this effot of yours. I will be very thankful to you if you kindly give me the source of these valuable questions and if you please mail me the question bank on this topic and other topics of mathematics . My E-mail id is intelect.jkd1@gmail.com and my contact number is 9035930724. 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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) Published by Pearson # Chapter 5 - Polynomials and Factoring - 5.3 Factoring Trinomials of the Type ax2+bx+c - 5.3 Exercise Set: 99 #### Answer $[(t-3)^{n}+1][7(t-3)^n-2]$ #### Work Step by Step Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 7(t-3)^{2n}+5(t-3)^n-2 \end{array} has $ac= 7(-2)=-14$ and $b= 5 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 7,-2 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 7(t-3)^{2n}+7(t-3)^n-2(t-3)^n-2 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} [7(t-3)^{2n}+7(t-3)^n]-[2(t-3)^n+2] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 7(t-3)^n[(t-3)^{n}+1]-2[(t-3)^n+1] .\end{array} Factoring the $GCF= (a^{3n}-1)$ of the entire expression above results to \begin{array}{l}\require{cancel} [(t-3)^{n}+1][7(t-3)^n-2] .\end{array} After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
# Formula's For Syllogisms 18 May 2020 ##### Formula's For Syllogisms Types of Statements The four basic statements in syllogism are, • All As are B (Eg. All Mobiles are cameras) • Some As are B (Eg. Some cats are dogs) • No A is B (Eg. No rabbit is a pig) • Some As are not B (Eg. Some cats are not humans) These statements can be classified into two categories as shown in below table. Alternate Words Basic Diagrams For (i) statement, i.e. All As are B, Circle A should be inside B or A and B can be equal. But circle A should not exceed B. For (ii) statement, i.e. Some As are B, Circle A and B should be connected always. It should not separate. For (iii) statement, i.e. No A is B, We should not connect circle A and circle B. For (iv) statement, i.e. Some As are not B, We can connect circle A and circle B. Complementary pair In the Complementary pair, subject and predicate should be same in both the conclusions. If one conclusion is true, definitely the other conclusion will be false and vice versa. There are two complementary pairs in syllogism. Pair I : All As are B & Some As are not B If “All As are B” is true, definitely “Some As are not B” is false. If “Some As are not B” is true, definitely “All As are B” is false. Pair II: No A is B & Some As are B If “No A is B” is true, definitely “Some As are B” is false. If “Some As are B” is true, definitely “No A is B” is false. Procedure: Step 1: Draw the basic diagram for the given statements. Step 2a: If all are positive conclusions, Check those conclusions in basic diagram and decide which one is true or false. Don’t draw any other diagram if all are positive in conclusion. Step 2b: If there is negative conclusion and it is true in basic diagram, try to make it false by drawing its complementary pair. While drawing alternate diagram, it should not violate any other given statements. If you are able to draw alternate diagram, without violating any statement. Then the negative statement is false. Examples: (Positive Conclusions) Q.1. Statements 1. All grapes are apples 2. All apples are mangoes Conclusions 1. All grapes are mangoes 2. All mangoes are grapes 3. Some grapes are mangoes Answer: (1) and (3) are true Examples (Negative Conclusion) Q.1. Statements 1. All months are weeks 2. Some week are days Conclusions 1. No month is day 2. Some weeks are months To test yourself, kindly practice questions on Syllogism-https://bit.ly/3fWBP1L Clear all your Aptitude Concepts from Experts with Aptitude Cracker Course!-https://bit.ly/2YFn39h To test yourself, click on the FREE TEST SERIES provided by Talent Battle - https://bit.ly/3dmvTNa Related Articles
Upcoming SlideShare × # Applications Section 1.3 3,576 views Published on Linear Algebra Published in: Technology 0 Likes Statistics Notes • Full Name Comment goes here. Are you sure you want to Yes No • Be the first to comment • Be the first to like this Views Total views 3,576 On SlideShare 0 From Embeds 0 Number of Embeds 21 Actions Shares 0 20 0 Likes 0 Embeds 0 No embeds No notes for slide ### Applications Section 1.3 1. 1. Section 1.3 Applications Polynomial Curve Fitting Section 1.3.1 Suppose a collection of data is represented by n points ( x 1 , y 1 ), ( x 2 , y 2 ), ( x 3 , y 3 ) . . . , ( x n , y n ) in the xy -plane. We can fit a polynomial function to this data of degree n – 1 that passes through each of these points. This polynomial will have the form y To solve for the n coefficients of p ( x ) we substitute each of the n points into the polynomial and obtain n linear equations in n variables a 0 , a 1 , a 2 , . . . , a n -1 : ( x 1 , y 1 ) ( x 2 , y 2 ) ( x 3 , y 3 ) ( x n , y n ) x 2. 2. Section 1.3 Applications Polynomial Curve Fitting Section 1.3.2 Example 1 : The graph of a cubic polynomial has horizontal tangents at (1, – 2) and (– 1, 2). Find the equation for the cubic and sketch its graph. Solution : The general form for a cubic polynomial is The derivative of this function is We know the derivative is zero at the given points so we obtain the equations 3. 3. Section 1.3 Applications Polynomial Curve Fitting Section 1.3.3 Example 1 : These substitutions lead to the following system of equations Which in turn leads to the augmented matrix We now solve this system: 4. 4. Section 1.3 Applications Polynomial Curve Fitting Section 1.3.4 Example 1 : From this matrix we have a 2 = 0 and a 1 = – 3 a 3 . Now using the given points and the solutions from the above system we substitute into equation (1) to obtain: or This is a new system of equations with augmented matrix: 5. 5. Section 1.3 Applications Polynomial Curve Fitting Section 1.3.5 Example 1 : We now solve this system: From the last matrix we see that a 0 = 0 and a 3 = 1, and consequently a 1 = – 3. The equation of the polynomial is p ( x ) = – 3 x + x 3 . We can now sketch the graph of the third degree polynomial that goes through these points and has horizontal tangents at these points. 6. 6. Section 1.3 Applications Polynomial Curve Fitting Example 1 : We now graph the polynomial p ( x ) = – 3 x + x 3 : ( – 1, 2) ( 1, – 2) Section 1.3.6 7. 7. Section 1.3 Applications Network Analysis Section 1.3.7 Networks composed of branches and junctions are used as models in fields as diverse as economics, traffic analysis, and electrical engineering. It is assumed in such models that the total flow into a junction is equal to the total flow out of the junction. For example, because the junction shown below has 30 units flowing into it, there must be 30 units flowing out of it. We show this diagrammatically as and can be represented by the linear system Since we can represent each junction in a network gives rise to a linear equation we can analyze the flow through a network composed of several junctions by solving a system of linear equations. 30 x 1 x 2 8. 8. Section 1.3 Applications Network Analysis Section 1.3.8 <ul><li>Example 2 : The flow of traffic (in vehicles per hour) through a network of streets is shown in the figure below. </li></ul><ul><li>Solve the system for x i , i = 1, 2, 3, 4. </li></ul><ul><li>Find the traffic flow when x 4 = 0. </li></ul><ul><li>Find the traffic flow when x 4 = 100. </li></ul>200 100 100 200 x 1 x 3 x 2 x 4 9. 9. Section 1.3 Applications Network Analysis Section 1.3.9 Example 2 : We label each junction appropriately as shown. We then establish the equations: Junction 1 Junction 2 Junction 3 Junction 4 These equations lead to the following system of linear equations: 200 100 100 200 x 1 x 3 x 2 x 4 1 2 3 4 10. 10. Section 1.3 Applications Network Analysis Section 1.3.10 Example 2 : This system of linear equations leads to the following matrix representation: We now use the Gauss-Jordan elimination technique to solve this system. 11. 11. Section 1.3 Applications Network Analysis Section 1.3.11 Example 2 : From the fourth row we see that x 4 can be any real number, so letting x 4 = t we have the following general solution: x 4 = t , x 3 = 200 + t , x 2 = t – 100 , x 1 = 100 + t where t is a real number. Thus this system has an infinite number of solutions. This is the solution to part (a) of the question. 12. 12. Section 1.3 Applications Network Analysis Section 1.3.12 Example 2 : <ul><li>Given x 4 = t , x 3 = 200 + t , x 2 = t – 100 , x 1 = 100 + t where t is a real </li></ul><ul><li>number. </li></ul><ul><li>(b) We now find the traffic flow when x 4 = 0 (i.e. when t = 0). </li></ul><ul><li>x 1 = 100, x 2 = – 100, x 3 = 200, x 4 = 0 . </li></ul><ul><li>We now find the traffic flow when x 4 = 100 (i.e. when t = 100). </li></ul><ul><li>x 1 = 200, x 2 = 0, x 3 = 300, x 4 = 100 . </li></ul> 13. 13. Section 1.3 Applications Network Analysis Section 1.3.13 <ul><li>Another type of network is an electrical network. An analysis of such a system uses two properties of electrical networks known as Kirchhoff's Laws . </li></ul><ul><li>All the current flowing into a junction must flow out of it </li></ul><ul><li>The sum of the products IR ( I is current and R is resistance) around a closed </li></ul><ul><li>path is equal to the total voltage ( V ) in the path. </li></ul><ul><li>In an electrical network, current is measured in amps, resistance in ohms, and the product of current and resistance in volts. Batteries are represented by the symbol </li></ul><ul><li>, where the current flows out of the terminal denoted by the larger vertical bar. Resistance is denoted by the symbol . The direction of the current is indicated by the arrow in the branch. </li></ul> 14. 14. Section 1.3 Applications Analysis of an Electrical Network Section 1.3.14 Example 3 : Determine the currents I 1 , I 2 , and I 3 or the electrical network shown below. 1 2 Path 1 Path 2 8 Volts 7 Volts R 3 = 4 R 2 = 2 R 1 = 3 I 1 I 2 I 3
Showing posts with label CAT 2014. Show all posts Showing posts with label CAT 2014. Show all posts ## Sunday, June 29, 2014 ### Probability Basics : Lesson 1 Today we explore the basic concepts of Probability. Probability is the measure of the likelihood of an "event" to occur. For example, when we toss a fair coin, the chances of a head turning up are 1/2. When we roll a dice the chance of 2 appearing is 1/6. ### Important Terms: 1. Sample Space : The collection of all results is called the sample space of the event. For example, when a coin is tossed, the Sample space(S) is the set {Head, Tails}. When a dice is rolled, S = {1,2,3,4,5,6} 2. Event: A subset of the S is called an event. For example, when a coin is tossed, {Head}, {Tail} are events. Note that an event set can contain multiple items. For example when 2 coins are tossed, {Head,Head} is considered an event. 3. Equally Probable Events: When 2 events have the same likelihood of occuring, they are known as Equally Probable events. For example the chance of head and tails turning up on the toss of a coin are Equally Probable events. ### Probability The Probability of an event occurring is defined as the number of cases favorable for the event divided by total number of events in the sample space. If the event be called as 'A', the probability is represented as P(A) Example: Probability that head shows up on tossing a coin Favorable Cases : Head (1) Sample Space: Head, Tail (2) P(A) = Favorable Cases/Sample Space = 1/2 P(A') is used to represent the probability of event A not occurring. Note: P(A) + P(A') = 1 If we have 2 events A,B as the overall sample space, then: P(A) = A / (A+B) and P(B) = B / (A + B) ### Independent Events and Mutually Exclusive Events A regular confusion among students is the difference between Independent and Mutually exclusive Events. Let A, B be 2 events. P(A and B) = P (A ∩ B ) is defined as the probability that both A and B occur together. For 2 independent events A, B, P(A ∩ B ) = P(A) * P(B) For Mutually Exclusive events, P(A ∩ B ) = 0 This is best understood with an example. Consider a fair coin and a fair six-sided die. Let event A be obtaining tails, and event B be rolling a 3. Then we can safely say that events A and B are independent, because the outcome of one does not affect the outcome of the other. Here, P(A ∩ B ) = 1/2 * 1/6 = 1/12. Here A and B are Independent Events (Not mutually exclusive). Consider a fair six-sided dice, where even-numbered faces are colored red, and the odd-numbered faces are colored green. Let event A be rolling a green face, and event B be rolling a 6. P(A) = 3/6 = 1/2 P(B) = 1/6 But note that A&B cannot occur simultaneously since 6 is always going to turn up on a red face. Here, P(A ∩ B ) =0 Here A and B are Mutually exclusive events(Not independent). In our next post we will go deeper into complex probability theory and solve a few problems and provide video solutions for the same. ## Monday, June 2, 2014 ### CAT 2014: Registration, Syllabus and Format The Common Admission Test 2014 (CAT 2014) is India's premier MBA entrance exam for admission into the 13 Indian Institutes of Managements. The CAT exam is conducted by the IIMs in a Computer Based Test (CBT) format once every year between October and November. Over 2 lakh candidates write the test from all over India. Apart from the IIMs, a number of other Indian institutes also consider CAT scores for admission. ### Who will be conducting it? CAT 2014 will be conducted by a new test partner of the IIM's - TCS. The news was announced that TCS will be conducting CAT from 2014 to 2018. ### Has the registrations for CAT 2014 started? Not Yet. Expect the advertisement in Mid July. Of course we will update this blog as and when the registrations begin. ### What is the Syllabus for CAT 2014? The IIMs have never defined a syllabus for CAT but since it has always been an aptitutude test, the following is a good guide for the CAT syllabus. • Quantitative ability • Number Systems • HCF, LCM • Averages • Percentages, Profit & Loss • Simple/compound Interest • Equations, Inequalities • Complex numbers • Functions • Probability • Permutation & Combination • Mensuration • Geometry • Co-ordinate Geometry • Trigonometry • Set Theory • Speed & Distance • Work & Time • Series & Progressions • Surds & Indices • Logarithms • Ratio & Proportion • Base Systems • Data Interpretation • Data Tables • Pie Charts • Line Graphs • Bar graphs • Other Chart types • Data Sufficiency • Verbal Ability • Parajumbles • Fill in the blanks • Vocabulary • Sentence Correction • Analogies • Meaning-Usage match • Critical reasoning • Facts/Inference/Judgment • Logical reasoning • Cubes & Dice • Blood Relations • Number & Letter Series • Seating Arrangements • Venn Diagrams • Clocks & Calendars • Analytical Reasoning • Syllogism ### Format of the CAT 2014 Computer Based Test. The CAT 2014 exam will contain 2 sections of 50 questions each. Section1: Quantitative Techniques & Data Interpretation : 50 questions Section2: Verbal Ability & Logical Reasoning : 50 questions Total Time: 170 Minutes. Scoring Scheme: +3 for correct answer, -1 for wrong answer ### How to get Started? Check out our blog post on how to get started for CAT 2014. Oliveboard offers complete prep for CAT 2014 exam. Register now. Get access to Video Lessons for the complete syllabus. Also you can take Mock tests, Practice Tests and do a lot more! ## Thursday, May 29, 2014 ### CAT 2014 : Number Systems - Solutions to Questions that have appeared in Previous CAT Papers. In our previous blog post, we had picked 3 questions from previous CAT paper. Today we will look at solving these questions. In the process, we will be doing a deeper dive into a couple of important concepts. Question 1: If x = -0.5, which of the following has the smallest value? 21/x , 1/x, 1/x2, 2x, 1/√-x Solution: In solving this problem, we will first need to understand a couple of important concepts on Powers of numbers for CAT 2014. The video below explains all the concepts and basics related to it. Now that we have the concept clear, we can look at the solution to this particular problem. Question 2: Which among the following is the largest? 21/2, 31/3, 41/4, 61/6, 121/12 Solution: The solution relies on another basic concept of Number systems to compare different numbers with different exponents . The concept and the solution are given below. Question 3: How may 2 digit numbers increase by 18 when the digits are reversed? Solution: Note that 2 digit numbers can be expressed as ab, where a,b are digits. So the original number is 10a + b, reversed number 10b + a. The complete solution along with a few more basics for CAT 2014 can be viewed below. We will be returning with our next post in a couple of days time! Happy Learning! ## Sunday, May 25, 2014 ### CAT 2014 : Number Systems - Questions that have appeared in Previous CAT Papers. In Today's post, we try to look at 3 problems that have appeared in past CAT papers, and provide clues to solve them. The complete solution will be available in our next post. 1. If x = -0.5, which of the following has the smallest value? 21/x , 1/x, 1/x2, 2x, 1/√-x 2. Which among the following is the largest? 21/2, 31/3, 41/4, 61/6, 121/12 3. How many 2 digit numbers increase by 18 when the digits are reversed? Hints: 1. A positive number raised to any real number will always be positive. For eg, 4-1/10 will be greater than 0. 2. Convert all numbers to have the same exponent. Then compare the numbers. 3. Let the number have digits a,b. Original number is 10a+b, reversed is 10b+a. Take the difference and try to solve it :) ## Saturday, May 24, 2014 ### Solved Examples : Number Systems for CAT and other MBA Exams. In the previous post, we asked a few questions to help us understand the various concepts in number systems by examples. Here are Video Solutions to all these problems :). 1. What happens when you multiply 3 even numbers? 2 even numbers and an odd number? 2 odd numbers and an even number? and 3 odd numbers? 2. What happens when you multiply 'N' even numbers? and 'N' odd numbers 3. Can product of 4 consecutive numbers be Odd? The solution to all these 3 problems are explained in the video below. We then asked the question: Whats the smallest number should be added to 156789 to make it divisible by 11? To solve this problem, we need to understand divisibility tests by 11. This and the solution to the problem is explained in the video below. Our Next question was: Whats the smallest number that should be added to 677 to make it divisible by 4, 5, and 11? To solve this, its important to understand concepts of LCM. The question is solved with concepts below. Our Next question was related to power cycles: What is the units digit of 72999? Detailed solution with basic power cycles concept is below. Our Last question was in advanced Power cycle: What will be the last 2 digits when 25625 is multiplied by 375? Concept explanation with detailed solution is given below. Do let us know what you think about these solutions. We will be back with more a more detailed Number Systems Lesson in the next week. In other news: TCS will be the official test partner for CAT the next 5 years. This was announced a few days back. ## Friday, May 16, 2014 ### Number Systems for CAT and Other MBA Exams. Number systems is an important topic for CAT and all other MBA Exams. The various topics under Number systems include : 1. Classification of Numbers : Natural numbers, Whole numbers, Fractions, Real Numbers, Complex numbers etc. 2. Rules on Number Operations : For eg, sum of 2 odd numbers is even etc. 3. Divisibility Rules. 4. Properties of Prime and Composite numbers, factorisation. 5. Power Cycle of Numbers Today we will start by asking questions for you to ponder over, and answer each of them in the next set of blog posts with detailed video concepts. 1. What happens when you multiply 3 even numbers? 2 even numbers and an odd number? 2 odd numbers and an even number? and 3 odd numbers? 2. What happens when you multiply 'N' even numbers? and 'N' odd numbers 3. Can product of 4 consecutive numbers be Odd? 4. Whats the smallest number should be added to 156789 to make it divisible by 11? 5. Whats the smallest number that should be added to 677 to make it divisible by 4, 5, and 11? 6. What is the units digit of 72999 7. What will be the last 2 digits when 25625 is multiplied by 375? Through these problems, we hope to cover the complete Number systems concepts, and help you prepare better for CAT and other leading MBA Exams. ## Thursday, May 15, 2014 ### CAT 2014: Getting Started With less than 6 months to go for CAT, Oliveboard will be starting its concept blog today. What to Expect: 1. Each topic with basic concepts. 2. Worked out examples for these topics. 3. Exercises 4. Video solutions to selected questions. We are sure this will help you ace CAT 2014 and other MBA exams! Register at Oliveboard for more. Have a doubt? Leave a comment and we will get back to you at the earliest. Or you can email us at : support [at] oliveboard [dot] in
# Trigonometry: Evaluating Angles In these lessons, we will learn • how to find the trigonometric functions of special angles 30°, 45° and 60°. • how to use the calculator to evaluate the trigonometric functions of any angle. ### Special Angles We will first look into the trigonometric functions of the angles 30°, 45° and 60°. Let us consider 30° and 60°. These two angles form a 30°-60°-90° right triangle as shown. The ratio of the sides of the triangle is 1 : √3 : 2 From the triangle we get the ratios as follows: Next, we consider the 45˚ angle that forms a 45°-45°-90° right triangle as shown. The ratio of the sides of the triangle is Combining the two tables we get: Example: Evaluate the following without using a calculator: a) 2 sin 30˚ + 3 cos 60˚ – 3 tan 45˚ b) 3(cos 30˚)2 + 2 (sin 30˚ )2 Solution: a) 2 sin 30˚ + 3 cos 60˚ – 3 tan 45˚ b) 3(cos 30˚)2 + 2 (sin 30˚)2 How to find the trig ratios of the special angles? Using a 45-45-90 triangle and a 30-60-90 triangle find sine, cosine and tangent values of 0, 30, 45, 60 and 90 degrees Find exact values of expressions involving sine, cosine and tangent values of 0, 30, 45, 60 and 90 degrees Example: Determine the exact values of each of the following: a) sin30°tan45° + tan30°sin60° b) cos30°sin45° + sin30°tan30° ### How To Use A Calculator To Find Trig Ratios And Angles? We could make use of a scientific calculator to obtain the trigonometric value of an angle. (Your calculator may work in a slightly different way. Please check your manual.) Example: Find the value of cos 6.35˚. Solution: Press <cos 6.35˚ = 0.9939 (correct to 4 decimal places) Example: Find the value of sin 40˚ 32’. Solution: sin 40˚ 32’ = 0.6499 (correct to 4 decimal places) Example: Find the value of x for the following triangle. (Give your answer correct to 4 decimal places) Solution: x = 6.21 × sin 31.3˚ = 3.2262 Finding trig ratios and angles using your calculator Examples: 1. Use a calculator to find the function value. Use the correct number of significant digits. a) cos 369.18° b) tan 426,62° c) sin 46.6° d) cot 17.9° 2. Determine θ in degrees. Use the correct number of significant digits. a) sin θ = 0.42 b) cos θ = 0.29 c) tan θ = 0.91 3. Determine θ in decimal degrees, 0° ≤ θ ≤ 90°. Use the correct number of significant digits. a) csc θ = 3.6 b) cot θ = 2.1 c) csc θ = 1.63 d) sec θ = 7.25 Determining Trigonometric Function Values on the Calculator Using the TI 84 to find function values for sine, cosine, tangent, cosecant, secant, and cotangent. Examples: 1. sin 30° 2. cos 45° 3. tan(-264°) 4. sec(102.5°) 5. csc(432°) 6. cot(-23.45°) Inverse Trigonometric Functions We can use inverse trigonometric functions to find an angle with a given trigonometric value. We can also inverse trigonometric functions to solve a right triangle. Examples: 1. Use the calculator to find an angle θ in the interval [0, 90] that satisfies the equation. a) sin θ = 0.7523 b) tan θ = 3.54 2. Solve the given right triangle if a = 44.3 cm and b = 55.9 cm 3. Find each angle in a 3, 4, 5 right triangle. Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
# How to calculate the angles of a triangle All triangles have three angles . If the angles have the same size they will measure all 60 degrees and it will be an equilateral triangle, while a right triangle has an angle of 90 degrees that forms an "L". An angle size can be calculated by measuring the other two angles. Keep reading this article and you will discover how to calculate the angles of a triangle. Steps to follow: #### one One of the options you can find is to know two angles of the triangle and have to find the measure of the third angle. Let's take an example: two of the angles of a triangle could be 55 degrees and 25 degrees. #### two In this way, to be able to calculate the angles of a triangle you will have to add the measure of the two angles that you know. For example, following the case that we have proposed previously, we would have to do this operation: • 55 + 25 = 80 degrees, the total of the two angles measured #### 3 Now you must subtract the total sum of the two known angles to 180 degrees. Following the previous example, the operation to be performed would be the following: 180-80 = 100 degrees. In this way, we find as a result that 100º is the measure of the third angle that was the unknown angle. #### 4 So that you can understand perfectly how to calculate the angle of a triangle, we will give you another example: If we know the two angles of a triangle like the one in the image that are 125 and 30 degrees, to find the third angle of the triangle we have to add the known values: 125 + 30 = 155 And then subtract the previous result to 180; therefore the third angle is: 180-155 = 25 degrees #### 5 In case it is an isosceles triangle, that is, with two sides and equal angles and the third different one, you can find yourself in two situations: • If you only know the different angle, you will have to subtract the measure of this angle at 180º and then divide it by two. For example: 180º - 40º = 140º and 140º / 2 = 70º so that both equal angles measure 70º • If you know the size of the equal angles, you must add them and subtract that amount to 180º. Example: 35º + 35º = 70º and 180º - 70º = 110º so that the third angle will measure 110º. #### 6 These other articles of: • How to measure the angles of the triangles • How to calculate the center of a triangle Tips • The total measure of the three angles of a triangle equals 180 degrees.
# Hyperbola Formulas Want to solve all Hyperbola equations in an easy way? Then, this page helps you a lot. Because we have curated simple hyperbola formulas to simplify the problems quickly and effortlessly. So, try to remember the hyperbola formulae for all concepts by using the Hyperbola Formula Cheat Sheet existed here. Also, you can make your calculations so easy at any time if you memorize the formulas of hyperbola. ## Hyperbola Formulas List \| Cheatsheet of Hyperbola Formulae Make the most out of these Hyperbola Formulas and solve the calculations within given time. The list of Hyperbola formulae that exist here helps you to do your homework or math assignments at a faster pace. 1. Standard equation of Hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1 • Length of transverse axis → 2a • Length of conjugate axis → 2b • Directrix: x = a/e and x = – a/e • Focus: S (ae, 0) and S’ (- ae, 0) • Length of Latus Rectum is given by $$\frac{2 b^{2}}{a}$$ 2. Eccentricity (A) For the hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1, b2 = a2 (e2 – 1) (B) Equation of vertical hyperbola is $$\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}$$ = 1 Length of L.R. = $$\frac{2 a^{2}}{b}$$ also a2 = b2 (e2 – 1) 3. The equation ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 will represent an hyperbola if h2 – ab > 0 & Δ = abc + 2fgh – af2 – bg2 – ch2 ≠ 0. 4. Conjugate Hyperbola (i) The equation of the conjugate hyperbola of $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1 is – $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$$ = 1 (ii) If e1 and e2 are the eccentricities of the hyperbola and its conjugate then $$\frac{1}{e_{1}^{2}}+\frac{1}{e_{21}^{2}}$$ 5. The equation of hyperbola in the parametric form will be given by x = a sec Φ, y = b tan Φ 6. Condition of tangency The line y = mx + c touches the hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1, if c = ± $$\sqrt{a^{2} m^{2}-b^{2}}$$ and point of tangency is $$\left(-\frac{\mathrm{a}^{2} \mathrm{m}}{\mathrm{c}},-\frac{\mathrm{b}^{2}}{\mathrm{c}}\right)$$ 7. Equation of tangent (i) The equation of the tangent at any point (x1, y1) on the hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \text { is } \frac{x x_{1}}{a^{2}}-\frac{y y_{1}}{b^{2}}=1$$ (ii) Equation of tangent to the hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1 at the point (a sec θ, b tan θ) is $$\frac{x}{a}$$ sec θ – $$\frac{y}{b}$$ tan θ = 1. (iii) Slope form: y = mx ± $$\sqrt{a^{2} m^{2}-b^{2}}$$ and the point of contacts is $$\left(\pm \frac{\mathrm{a}^{2} \mathrm{m}}{\sqrt{\mathrm{a}^{2} \mathrm{m}^{2}-\mathrm{b}^{2}}}, \pm \frac{\mathrm{b}^{2}}{\sqrt{\mathrm{a}^{2} \mathrm{m}^{2}-\mathrm{b}^{2}}}\right)$$ 8. Equation of the normal (i) The equation of normal to tbe hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1 at (x1, y1) is $$\frac{a^{2} x}{x_{1}}+\frac{b^{2} y}{y_{1}}$$ = a2 + b2 = a2e2. (ii) The equation of normal at (a sec θ, b tan θ) to the hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1 is ax cos θ + by cot θ = a2 + b2. (iii) Slope form: y = mx – $$\frac{m\left(a^{2}+b^{2}\right)}{\sqrt{a^{2}-b^{2} m^{2}}}$$ 9. Pair of Tangents SS1 = T2 10. Chord of Contact T = 0 at (x1, y1) 11. Equation of the chord whose middle point is given T = S1 12. Director Circle The equation of the director circle is x2 + y2 = a2 – b2 13. Diameter If y = mx + c represent a system of parallel chords of the hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$ then the equation of the diameter is y = $$\frac{b^{2}}{a^{2} m}$$x. 14. Asymptotes Equation of the asymptotes of hyperbolas $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1 and –$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$$ = 1 are y = ±$$\frac{b}{a}$$x. 15. Equation of rectangular hyperbola Hyperbola whose eccentricity is $$\sqrt{2}$$, equation is x2 – y2 = a2. Equation of hyperbola referred asymptotes as axes is xy = c2 where c2 = $$\frac{a^{2}+b^{2}}{4}$$. Point on xy = c2 may be taken (ct, c/t) 16. Equation of chord joining points t1 and t2 on the hyperbola xy = c2 is x + yt1t2 – c(t1 + t2) = 0 17. Tangent at the point “t” to the: x + yt2 – 2ct = 0 If we call the point t i.e. (ct, c/t) as (x1, y1) then above tangent can be written as $$\frac{x}{x_{1}}+\frac{y}{y_{1}}=2$$ 18. Normal to the Hyperbola at point “t” xt3 – yt – ct4 + c = 0 in another form as (ct, c/t) = (x1, y1) xx1 – yy1 = x12 – y12 19. Equation of Auxiliary circle if hyperbola is $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1 is x2 + y2 = a2 20. If e1 and e2 be the eccentricities of a hyperbola and its conjugate then $$\frac{1}{e_{1}^{2}}+\frac{1}{e_{2}^{2}}=1$$
# Common Core: 7th Grade Math : Understand Distances Between Numbers on a Number Line: CCSS.Math.Content.7.NS.A.1b ## Example Questions ### Example Question #1 : Understand Distances Between Numbers On A Number Line: Ccss.Math.Content.7.Ns.A.1b Explanation: Simply the signs before solving. A positive sign multiplied with a negative sign will convert the sign to a negative, and a negative multiplied with a negative will convert the sign to a positive. ### Example Question #1 : Understand Distances Between Numbers On A Number Line: Ccss.Math.Content.7.Ns.A.1b is equal to which of the following? Explanation: This is a straightforward problem. Remember that when adding a negative number, you are actually subtracting: Be sure to remember that the first number is also negative, meaning we are subtracting a number from a negative number: The answer is -6.25. ### Example Question #2 : Understand Distances Between Numbers On A Number Line: Ccss.Math.Content.7.Ns.A.1b Evaluate: Explanation: The sum of two numbers of unlike sign is the difference of their absolute values, with the sign of the "dominant" number (the positive number here) affixed: Subtract vertically by aligning the decimal points, making sure you append the 3.2 with a placeholder zero: This is the correct choice. ### Example Question #3 : Understand Distances Between Numbers On A Number Line: Ccss.Math.Content.7.Ns.A.1b Explanation: When we add a negative number, the sign turns negative. Since we are adding two negative numbers, we treat this as an addition problem and add a minus sign in the end. ### Example Question #2 : Understand Distances Between Numbers On A Number Line: Ccss.Math.Content.7.Ns.A.1b Explanation: When dealing with negative numbers, let's see which number is greater. is greater than and is negative so the answer is negative. We treat this as a subtraction problem. . Because our answer should be negative, the correct answer is ### Example Question #1 : Understand Distances Between Numbers On A Number Line: Ccss.Math.Content.7.Ns.A.1b Solve: Explanation: Remember that adding a negative number is the same as subtracting that same number if it were positive. ### Example Question #252 : Operations And Properties Explanation: Since is greater than and is positive, our answer is positive. We treat as a subtraction problem. Answer is . ### Example Question #33 : Negative Numbers Explanation: When a plus and minus sign meet, the sign is negative. The difference is . ### Example Question #34 : Negative Numbers Explanation: When a plus and a minus meet, the sign is negative. When adding two negatie numbers, we treat as addition and add the minus sign in the end. Answer is .
# How do you solve 3x+4y=22 and x-5y=-37? Sep 9, 2015 The solution for the system of equations is color(blue)(x=-2, y=7 #### Explanation: $\textcolor{b l u e}{3 x} + 4 y = 22$...........equation $\left(1\right)$ $x - 5 y = - 37$, multiplying this equation by $3$ $\textcolor{b l u e}{3 x} - 15 y = - 111$....equation $\left(2\right)$ Now we can solve using the method of elimination . Subtracting equation $2$ from $1$ eliminates color(blue)(3x $\textcolor{b l u e}{3 x} + 4 y = 22$ $- \cancel{\textcolor{b l u e}{3 x}} + 15 y = + 111$ $19 y = 133$ color(blue)(y=7 Now we find $x$ by substituting $y$ in equation $1$ $3 x + 4 y = 22$ $3 x = 22 - 4 y$ $3 x = 22 - 4 \cdot 7$ $3 x = 22 - 28$ $3 x = - 6$ color(blue)(x=-2
1. ## Quadratic Equation Part 1 How to do these? 1. Find the real roots of $2x^2+3x-1=0$. Give your answer in surd form. 2. Find the range of values of q if $x^2+(q+1)x+q+1=0$ has complex roots. 3. Find the range of values of m if $x^2+x+1=m(x+2)$ has real and distinct roots. 4. Find the range of values of k if $x^2+(k-3)x+k=0$ has roots with same sign. 1. $\frac{1}{2}(-3 \pm \sqrt{17})$ 2. -1<q<3 3. $m<-3-2\sqrt{3},m>-3+2\sqrt{3}$ 4. k>0 2. For (4), you got part of it correct (namely the product of the roots is positive). But you also want your roots to be real, since it doesn’t make any sense to talk about two conjugate complex numbers having the same sign. 3. 1. Find the real roots of $2x^2+3x-1=0$. Give your answer in surd form. \begin{aligned} x &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} \\ &= \frac{-3 \pm \sqrt{9 + 8}}{4} \\ &= \frac{-3 \pm \sqrt{17}}{4} \\ &= \frac{1}{4}(-3 \pm \sqrt{17}) \end{aligned} Hmm, I'm getting a different answer. 2. Find the range of values of q if $x^2+(q+1)x+q+1=0$ has complex roots. A quadratic equation in the form $ax^2 + bx + c = 0$ has complex roots if the discriminant $b^2 - 4ac < 0$. So a = 1 b = q + 1 c = q + 1 \begin{aligned} b^2 - 4ac &< 0 \\ (q + 1)^2 - 4(1)(q + 1) &< 0 \\ q^2 + 2q + 1 - 4q - 4 &< 0 \\ q^2 - 2q - 3 &< 0 \\ (q - 3)(q + 1) &< 0 \end{aligned} Make a sign chart. Draw a number line and the critical points -1 and 3 (you get these by setting each factor equal to 0). Code: ----+----+----+----+----+----+----+---- -1 3 You have 3 intervals to test: (-∞, -1) (-1, 3) (3, ∞) From each interval, pick a number inside it to test into the inequality (q - 3)(q + 1) < 0, in order to determine the sign. The signs of the polynomial for each interval is as follows: Code: ----+----+----+----+----+----+----+---- pos -1 neg 3 pos So the answer is -1 < q < 3. 01 4. Originally Posted by yeongil \begin{aligned} x &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} \\ &= \frac{-3 \pm \sqrt{9 + 8}}{4} \\ &= \frac{-3 \pm \sqrt{17}}{4} \\ &= \frac{1}{4}(-3 \pm \sqrt{17}) \end{aligned} Hmm, I'm getting a different answer. I find this too... 5. 3. Find the range of values of m if $x^2+x+1=m(x+2)$ has real and distinct roots. This is the same as #2, but this time, since we want real and distinct roots, the discriminant has to be positive, or $b^2 - 4ac > 0$. So \begin{aligned} x^2 + x + 1 &= m(x + 2) \\ x^2 + x + 1 &= mx + 2m \\ x^2 + x - mx + 1 - 2m &= 0 \\ x^2 + (1 - m)x + (1 - 2m) &= 0 \\ \end{aligned} a = 1 b = 1 - m c = 1 - 2m \begin{aligned} b^2 - 4ac &> 0 \\ (1 - m)^2 - 4(1)(1 - 2m) &> 0 \\ 1 - 2m + m^2 - 4 + 8m &> 0 \\ m^2 + 6m - 3 &> 0 \end{aligned} I want to find the zeros of the polynomial $m^2 + 6m - 3$. Let's pretend for a moment that it equals 0 and solve for m: \begin{aligned} m^2 + 6m - 3 &= 0 \\ m^2 + 6m &= 3 \\ m^2 + 6m + 9 &= 3 + 9 \\ (m + 3)^2 &= 12 \\ m + 3 &= \pm\sqrt{12} \\ m &= -3 \pm 2\sqrt{3} \\ \end{aligned} Make a sign chart. Draw a number line and the critical points (the zeros we found earlier): Code: ---------+-------------------+--------- -6.46 0.46 ( $0.46 \approx -3 + 2\sqrt{3}$ and $-6.46 \approx -3 - 2\sqrt{3}$) You have 3 intervals to test: $(-\infty,\;-3 - 2\sqrt{3})$ $(-3 - 2\sqrt{3},\;-3 + 2\sqrt{3})$ $(-3 + 2\sqrt{3}, \infty)$ From each interval, pick a number inside it to test into the inequality $m^2 + 6m - 3 > 0$, in order to determine the sign. The signs of the polynomial for each interval is as follows: Code: ---------+-------------------+--------- pos -6.46 neg 0.46 pos So the answer is $m<-3-2\sqrt{3},m>-3+2\sqrt{3}$. 01 6. Originally Posted by yeongil \begin{aligned} x &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} \\ &= \frac{-3 \pm \sqrt{9 + 8}}{4} \\ &= \frac{-3 \pm \sqrt{17}}{4} \\ &= \frac{1}{4}(-3 \pm \sqrt{17}) \end{aligned} Hmm, I'm getting a different answer. ! I got that answer too! But the answer in my book is $\frac{1}{2}$... Maybe there is a mistake in my book? 7. Yeah. Looks like a mistake. I got $\frac{1}{4}(-3\pm\sqrt{17})$ too
# Common Core Standards: Math ### Geometry 6.G.A.2 2. Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge lengths, and show that the volume is the same as would be found by multiplying the edge lengths of the prism. Apply the formulas V = l w h and V = b h to find volumes of right rectangular prisms with fractional edge lengths in the context of solving real-world and mathematical problems. Some mathematical language sneaks its way out of the math classrooms and into common conversations. For example, people throw around the word, "rate," "exponentially," and "similar" all the time. Other mathematical language stays firmly inside math classrooms. We're talking words like, "integer," "polynomial," and yes, "fractional edge length." So what exactly does fractional edge length mean, and what is this standard talking about? In fifth grade, students were introduced to the concept of volume and the unit cube as a measure of volume. To measure the volume of a solid in cm3, for example, they'd count how many 1 cm × 1 cm × 1 cm cubes can fit inside of the solid. So far, so good. Students also learned that to find the volume of a right rectangular prism with dimensions <em>l</em>, <em>w</em>, and <em>h</em>, the number of unit cubes they can fit inside the prism is given by the formula <em>V</em> = <em>lwh</em>. They then used this formula to find the volume of right rectangular prisms with whole-number edge lengths. Now, we expect them to do the same thing—only this time, the edge lengths of the prism will be fractions or decimals instead of whole numbers. Psych! For instance, how can we get the students to find and understand the volume of a rectangular prism whose dimensions are <img src="http://media1.shmoop.com/images/common-core/comcore6_math_g_latek_1.png"> foot, <img src="http://media1.shmoop.com/images/common-core/comcore6_math_g_latek_2.png"> foot, and <img src="http://media1.shmoop.com/images/common-core/comcore6_math_g_latek_3.png"> foot? It's 1 over the least common multiple of the denominators. In our example, the unit fraction should be <img src="http://media1.shmoop.com/images/common-core/comcore6_math_g_latek_4.png">. Our unit cubes will have edge lengths of <img src="http://media1.shmoop.com/images/common-core/comcore6_math_g_latek_5.png"> foot. To figure out the volume of each of these unit cubes, we can reason that since their edge lengths are <img src="http://media1.shmoop.com/images/common-core/comcore6_math_g_latek_6.png"> foot, we can fit 6 × 6 × 6, or 216, of them into a rectangular solid whose volume is 1 ft3. This tells us that each of our unit cubes has a teeny tiny volume of <img src="http://media1.shmoop.com/images/common-core/comcore6_math_g_latek_7.png"> ft3. We can now fill up our rectangular prism these unit cubes. We should be able to fit 24 in our prism. The volume of rectangular prism, then, is <img src="http://media1.shmoop.com/images/common-core/comcore6_math_g_latek_8.png"> ft3, or <img src="http://media1.shmoop.com/images/common-core/comcore6_math_g_latek_9.png"> ft3. Lo and behold, if we multiply our length, width, and height together, we'll get the same answer. <img src="http://media1.shmoop.com/images/common-core/comcore6_math_g_latek_10.png"> This way, students can understand that (gasp!) volumes of solids with fractional edge lengths work the exact same way as ones with whole-number edge lengths. And we're accomplishing this <em>visually</em> (i.e., by drawing out the fractional unit cubes) and by using ideas that students already know and feel comfortable with (i.e., the unit cube and counting cubes to find volume) to boot. Good goin', teach! These types of activities and calculations will allow them to understand what fractional volumes and fractional unit cubes mean and why the formula <em>V</em> = <em>lwh</em> works for fractions as well as whole numbers. (Hint: because whole numbers and fractions are both, well, numbers.) As always, we also want students to be able to apply this understanding to real-world situations. So long as students working with rectangular prisms with non-whole-number dimensions, they're good to go. Assuming, of course, that they're using the right units and interpreting their answers properly.
Refer to ourÂ Texas Go Math Grade 2 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 2 Lesson 13.4 Answer Key Addition Situations. Unlock the Problem Kendra had 35 crayons. Her dad gave her some more crayons. Then she had 49 crayons. How many crayons did Kendraâ€™s dad give her? The number of crayons did Kendraâ€™s dad give her is 14 crayons. Explanation: Given that Kendra had 35 crayons and her dad gave her some more crayons, then she had 49 crayons. So the number of crayons did Kendraâ€™s dad give her is 49 – 35 = 14 crayons. What information am I given? he gave her some more crayons, the had ______ crayons. he gave her some more crayons, the had 49 crayons. Plan What Â¡s my plan or strategy? I can ___________ to solve the problem. I can subtract to solve the problem. Solve Show how you solve the problem. ___________ crayons 35 + 14 = 49 crayons. Explanation: As Kendra’s had 35 crayons and her dad gave her 14 crayons, so the total number of crayons does Kendra had is 35 + 14 which 49 crayons. HOME CONNECTION â€¢ Your child used a model and a number sentence to represent the problem. These help show what the missing amount is Â¡n order to solve the problem. Try Another Problem Label the model. Write a number sentence with a for the unknown number. Solve. Mr. Kane has 104- red pens. He bought 19 blue pens. How many pens does he have now? ___________ pens The number of pens does he have now is 123 pens. Explanation: Given Mr.Kane has 104- red pens and he bought 19 blue pens. So the number of pens does he have now is 104 + 19 which is 123 pens. Question 2. Jim has some pencils. Hannah has 13 pencils. They have 31 pencils altogether. How many pencils does Jim have? ___________ pencils Jim has 18 pencils. Explanation: Given that Jim has some pencils and Hannah has 13 pencils. They have 31 pencils altogether, so the number of pencils does Jim has is 31 – 13 = 18 pencils. Math Talk Mathematical Processes Explain how you know if amount is a part or the whole in a problem. Share and Show Label the model. Write a number sentence with a for the unknown number. Solve. Question 3. Aimee and Matthew catch 17 crickets in all. Aimee catches 9 crickets. How many crickets does Matthew catch? __________ crickets The number of crickets does Matthew caught is 26 crickets. Explanation: Given that Aimee and Matthew catch 17 crickets in all and Aimee catches 9 crickets. So the number of crickets does Matthew caught is 17 + 9 which is 26 crickets. Problem Solving Solve. Write or draw to show your work. Question 4. H.O.T. Multi-Step There are three groups of owls. There are 17 owls in each of the first two groups. There are 53 owls in all. How many owls are in the third group? __________ owls The number of owls in the third group will be 19 owls. Explanation: Given that there are 17 owls in each of the first two groups, so the owls in the first two groups are 17 + 17 which is 34. And there are 53 owls in all. So the number of owls in the third group will be 53 – 34 which is 19 owls. H.O.T. Percy sees a grasshopper jumping in the grass. He counts 16 jumps. 9 jumps, and then 15 more jumps. How many jumps does he count? _________ jumps The number of jumps does Percy counts is 40 jumps. Explanation: Given that Percy sees a grasshopper jumping in the grass and he counts 16 jumps, 9 jumps, and then 15 more jumps. So the number of jumps does he count is 16 + 9 + 15 which is 40 jumps. Question 6. Analyze The first book has 79 pages. The second book has 67 pages. How many pages are in these books? (A) 146 (B) 136 (C) 149 A. Explanation: Given that the first book has 79 pages and the second book has 67 pages. So the number of pages are in these books is 79 + 67 which is 146. Question 7. Mr. Green saw 60 books on a shelf and some books on a table. He saw 140 books in all. How many books did Mr. Green see on the table? Write a number sentence with a for the unknown number. Then solve. ______________ books The number of books did Mr. Green saw is 80 books. Explanation: Given that Mr. Green saw 60 books on a shelf and some books on a table and He saw 140 books in all. So the number of books did Mr. Green saw on the table is 140 – 60 which is 80 books. Question 8. To solve for the missing number, describe a strategy that you could use. The missing number is 18. Texas Test Prep There are 23 books in a box. There are 29 books on a shelf. How many books are there? (A) 52 (B) 56 (C) 42 52 books. Explanation: Given that there are 23 books in a box and there are 29 books on a shelf. So the number of books are 23 + 29 which is 52 books. TAKE HOME ACTIVITY â€¢ Ask your child to explain how to solve one of the problems in this lesson. ### Texas Go Math Grade 2 Lesson 13.4 Homework and Practice Answer Key Label the model. Write a number sentence with a for the unknown number. Solve. Question 1. June and Mike catch 15 fish in all. June catches 8 fish. How many fish does Mike catch? _________ fish The number of fish did Mike caught is 7 fish. Explanation: Given that June and Mike catch 15 fish in all and June catches 8 fish. So the number of fishes did Mike caught is 15 – 8 which is 7 fish. Problem Solving Solve. Write or draw to show your work. Question 2. Pam sees 12 frogs jumping near a pond. She also sees, 7 turtles, and then 17 rabbits. How many animals does she see? __________ animals The animals did she saw is 36 animals. Explanation: Given that Pam sees 12 frogs jumping near a pond and she also sees, 7 turtles, and then 17 rabbits. So the number of animals she saw is 12 + 7 + 17 which is 36 animals. Multi-Step There are a total of 41 chickens on the farm. There are 13 chickens in each of 2 cages in the barn. The rest of the chickens are outside. How many chickens are outside? ___________ chickens The number of chickens that are outside is 15 chickens. Explanation: Given that there are a total of 41 chickens on the farm and there are 13 chickens in each of 2 cages in the barn which is 13Ã—2= 26 chickens. So the number of chickens are outside is 41 – 26 which is 15 chickens. Lesson Check Question 4. There are 25 toy cars in a box. There are 37 toy cars on a shelf. How many toy cars are there in all? (A) 52 (B) 61 (C) 62 C. Explanation: Given that there are 25 toy cars in a box and there are 37 toy cars on a shelf. So the total number of cars are there in all is 25 + 37 which is 62 toy cars. Question 5. Dennis finds two boxes in the attic. The first box has 53 hats. The second box has 29 hats. How many hats are in the two boxes? (A) 86 (B) 72 (C) 82 The number of hats is 82 hats. Explanation: Given that Dennis finds two boxes in the attic. The first box has 53 hats and the second box has 29 hats, so the total number of hats in the two boxes is 53 + 29Â which is 82 hats. Go Math 2nd Grade Practice and Homework Lesson 13.4 Answer Key Question 6. Mr. and Mrs. Gray went apple picking. Mr. Gray picked some green apples. Mrs. Gray picked 70 red apples. They picked 160 apples in all. How many green apples did Mr. Gray pick? (A) 230 (B) 90 (C) 100 Mr. Gray picked 90 green apples. Explanation: Given that Mr. Gray picked some green apples and Mrs. Gray picked 70 red apples. They picked 160 apples in all. So the number of green apples did Mr. Gray picked is 160 – 70 which is 90 green apples. Question 7. Marcy bought two packages of paper. One package has 88 sheets of paper. The other has 64 sheets. How many sheets of paper did Marcy buy? (A) 152 (B) 142 (C) 154
# Differentiation rules This is a summary of differentiation rules, that is, rules for computing the derivative of a function in calculus. ## Elementary rules of differentiation Unless otherwise stated, all functions are functions of real numbers (R) that return real values; although more generally, the formulae below apply wherever they are well defined12—including complex numbers (C).3 ### Differentiation is linear For any functions f and g and any real numbers a and b the derivative of the function h(x) = af(x) + bg(x) with respect to x is $h'(x) = a f'(x) + b g'(x).\,$ In Leibniz's notation this is written as: $\frac{d(af+bg)}{dx} = a\frac{df}{dx} +b\frac{dg}{dx}.$ Special cases include: $(af)' = af' \,$ $(f + g)' = f' + g'\,$ • The subtraction rule $(f - g)' = f' - g'.\,$ ### The product rule (Leibniz rule) For the functions f and g, the derivative of the function h(x) = f(x) g(x) with respect to x is $h'(x) = f'(x) g(x) + f(x) g'(x).\,$ In Leibniz's notation this is written $\frac{d(fg)}{dx} = \frac{df}{dx} g + f \frac{dg}{dx}.$ ### The chain rule The derivative of the function of a function h(x) = f(g(x)) with respect to x is $h'(x) = f'(g(x)) g'(x).\,$ In Leibniz's notation this is written as: $\frac{dh}{dx} = \frac{df(g(x))}{dg(x)} \frac{dg(x)}{dx}.\,$ However, by relaxing the interpretation of h as a function, this is often simply written $\frac{dh}{dx} = \frac{dh}{dg} \frac{dg}{dx}.\,$ ### The inverse function rule If the function f has an inverse function g, meaning that g(f(x)) = x and f(g(y)) = y, then $g' = \frac{1}{f'\circ g}.\,$ In Leibniz notation, this is written as $\frac{dx}{dy} = \frac{1}{dy/dx}.$ ## Power laws, polynomials, quotients, and reciprocals ### The polynomial or elementary power rule If $f(x) = x^n$, for any integer n then $f'(x) = nx^{n-1}.\,$ Special cases include: • Constant rule: if f is the constant function f(x) = c, for any number c, then for all x, f′(x) = 0. • if f(x) = x, then f′(x) = 1. This special case may be generalized to: The derivative of an affine function is constant: if f(x) = ax + b, then f′(x) = a. Combining this rule with the linearity of the derivative and the addition rule permits the computation of the derivative of any polynomial. ### The reciprocal rule The derivative of h(x) = 1/f(x) for any (nonvanishing) function f is: $h'(x) = -\frac{f'(x)}{[f(x)]^2}.\$ In Leibniz's notation, this is written $\frac{d(1/f)}{dx} = -\frac{1}{f^2}\frac{df}{dx}.\,$ The reciprocal rule can be derived from the chain rule and the power rule. ### The quotient rule If f and g are functions, then: $\left(\frac{f}{g}\right)' = \frac{f'g - g'f}{g^2}\quad$ wherever g is nonzero. This can be derived from reciprocal rule and the product rule. Conversely (using the constant rule) the reciprocal rule may be derived from the special case f(x) = 1. ### Generalized power rule The elementary power rule generalizes considerably. The most general power rule is the functional power rule: for any functions f and g, $(f^g)' = \left(e^{g\ln f}\right)' = f^g\left(f'{g \over f} + g'\ln f\right),\quad$ wherever both sides are well defined. Special cases: • If f(x) = xa, f′(x) = axa − 1 when a is any real number and x is positive. • The reciprocal rule may be derived as the special case where g(x) = −1. ## Derivatives of exponential and logarithmic functions $\frac{d}{dx}\left(c^{ax}\right) = {c^{ax} \ln c \cdot a} ,\qquad c > 0$ note that the equation above is true for all c, but the derivative for c < 0 yields a complex number. $\frac{d}{dx}\left(e^x\right) = e^x$ $\frac{d}{dx}\left( \log_c x\right) = {1 \over x \ln c} , \qquad c > 0, c \ne 1$ the equation above is also true for all c but yields a complex number. $\frac{d}{dx}\left( \ln x\right) = {1 \over x} ,\qquad x \ne 0$ $\frac{d}{dx}\left( \ln \|x\|\right) = {\|x\| \over x^2}$ $\frac{d}{dx}\left( x^x \right) = x^x(1+\ln x).$ ### Logarithmic derivatives The logarithmic derivative is another way of stating the rule for differentiating the logarithm of a function (using the chain rule): $(\ln f)'= \frac{f'}{f} \quad$ wherever f is positive. ## Derivatives of trigonometric functions $(\sin x)' = \cos x \,$ $(\arcsin x)' = { 1 \over \sqrt{1 - x^2}} \,$ $(\cos x)' = -\sin x \,$ $(\arccos x)' = -{1 \over \sqrt{1 - x^2}} \,$ $(\tan x)' = \sec^2 x = { 1 \over \cos^2 x} = 1 + \tan^2 x \,$ $(\arctan x)' = { 1 \over 1 + x^2} \,$ $(\sec x)' = \sec x \tan x \,$ $(\operatorname{arcsec} x)' = { 1 \over \|x\|\sqrt{x^2 - 1}} \,$ $(\csc x)' = -\csc x \cot x \,$ $(\operatorname{arccsc} x)' = -{1 \over \|x\|\sqrt{x^2 - 1}} \,$ $(\cot x)' = -\csc^2 x = { -1 \over \sin^2 x} = -(1 + \cot^2 x)\,$ $(\operatorname{arccot} x)' = -{1 \over 1 + x^2} \,$ ## Derivatives of hyperbolic functions $( \sinh x )'= \cosh x = \frac{e^x + e^{-x}}{2}$ $(\operatorname{arsinh}\,x)' = { 1 \over \sqrt{x^2 + 1}}$ $(\cosh x )'= \sinh x = \frac{e^x - e^{-x}}{2}$ $(\operatorname{arcosh}\,x)' = {\frac {1}{\sqrt{x^2-1}}}$ $(\tanh x )'= {\operatorname{sech}^2\,x}$ $(\operatorname{artanh}\,x)' = { 1 \over 1 - x^2}$ $(\operatorname{sech}\,x)' = - \tanh x\,\operatorname{sech}\,x$ $(\operatorname{arsech}\,x)' = -{1 \over x\sqrt{1 - x^2}}$ $(\operatorname{csch}\,x)' = -\,\operatorname{coth}\,x\,\operatorname{csch}\,x$ $(\operatorname{arcsch}\,x)' = -{1 \over \|x\|\sqrt{1 + x^2}}$ $(\operatorname{coth}\,x )' = -\,\operatorname{csch}^2\,x$ $(\operatorname{arcoth}\,x)' = -{ 1 \over 1 - x^2}$ ## Derivatives of special functions Gamma function $\Gamma'(x) = \int_0^\infty t^{x-1} e^{-t} \ln t\,dt$ $= \Gamma(x) \left(\sum_{n=1}^\infty \left(\ln\left(1 + \dfrac{1}{n}\right) - \dfrac{1}{x + n}\right) - \dfrac{1}{x}\right) = \Gamma(x) \psi(x)$ Riemann Zeta function $\zeta'(x) = -\sum_{n=1}^\infty \frac{\ln n}{n^x} = -\frac{\ln 2}{2^x} - \frac{\ln 3}{3^x} - \frac{\ln 4}{4^x} - \cdots \!$ $= -\sum_{p \text{ prime}} \frac{p^{-x} \ln p}{(1-p^{-x})^2}\prod_{q \text{ prime}, q \neq p} \frac{1}{1-q^{-x}} \!$ ## Derivatives of integrals Suppose that it is required to differentiate with respect to x the function $F(x)=\int_{a(x)}^{b(x)}f(x,t)\,dt,$ where the functions $f(x,t)\,$ and $\frac{\partial}{\partial x}\,f(x,t)\,$ are both continuous in both $t\,$ and $x\,$ in some region of the $(t,x)\,$ plane, including $a(x)\leq t\leq b(x),$ $x_0\leq x\leq x_1\,$, and the functions $a(x)\,$ and $b(x)\,$ are both continuous and both have continuous derivatives for $x_0\leq x\leq x_1\,$. Then for $\,x_0\leq x\leq x_1\,\,$: $F'(x) = f(x,b(x))\,b'(x) - f(x,a(x))\,a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x}\, f(x,t)\; dt\,.$ This formula is the general form of the Leibniz integral rule and can be derived using the fundamental theorem of calculus. ## Derivatives to nth order Some rules exist for computing the nth derivative of functions, where n is a positive integer. These include: ### Faà di Bruno's formula If f and g are n times differentiable, then $\frac{d^n}{d x^n} [f(g(x))]= n! \sum_{\{k_m\}}^{} f^{(r)}(g(x)) \prod_{m=1}^n \frac{1}{k_m!} \left(g^{(m)}(x) \right)^{k_m}$ where $r = \sum_{m=1}^{n-1} k_m$ and the set $\{k_m\}$ consists of all non-negative integer solutions of the Diophantine equation $\sum_{m=1}^{n} m k_m = n$. ### General Leibniz rule If f and g are n times differentiable, then $\frac{d^n}{dx^n}[f(x)g(x)] = \sum_{k=0}^{n} \binom{n}{k} \frac{d^{n-k}}{d x^{n-k}} f(x) \frac{d^k}{d x^k} g(x)$ ## References 1. ^ Calculus (5th edition), F. Ayres, E. Mendelson, Schuam's Outline Series, 2009, ISBN 978-0-07-150861-2. 2. ^ Advanced Calculus (3rd edition), R. Wrede, M.R. Spiegel, Schuam's Outline Series, 2010, ISBN 978-0-07-162366-7. 3. ^ Complex Variables, M.R. Speigel, S. Lipschutz, J.J. Schiller, D. Spellman, Schaum's Outlines Series, McGraw Hill (USA), 2009, ISBN 978-0-07-161569-3 ## Sources and further reading These rules are given in many books, both on elementary and advanced calculus, in pure and applied mathematics. Those in this article (in addition to the above references) can be found in: • Mathematical Handbook of Formulas and Tables (3rd edition), S. Lipschutz, M.R. Spiegel, J. Liu, Schuam's Outline Series, 2009, ISBN 978-0-07-154855-7. • The Cambridge Handbook of Physics Formulas, G. Woan, Cambridge University Press, 2010, ISBN 978-0-521-57507-2. • Mathematical methods for physics and engineering, K.F. Riley, M.P. Hobson, S.J. Bence, Cambridge University Press, 2010, ISBN 978-0-521-86153-3 • NIST Handbook of Mathematical Functions, F. W. J. Olver, D. W. Lozier, R. F. Boisvert, C. W. Clark, Cambridge University Press, 2010, ISBN 978-0-521-19225-5. HPTS - Area Progetti - Edu-Soft - JavaEdu - N.Saperi - Ass.Scuola.. - TS BCTV - TS VideoRes - TSODP - TRTWE TSE-Wiki - Blog Lavoro - InterAzioni- NormaScuola - Editoriali - Job Search - DownFree ! TerritorioScuola. Some rights reserved. Informazioni d'uso ☞
# Lesson 14 Compare with Addition and Subtraction ## Warm-up: True or False: Equal Sign (10 minutes) ### Narrative The purpose of this True or False is to continue to develop and deepen student understanding of the equal sign. These understandings will be helpful when students work with equations with expressions on both sides of the equal sign in a later unit. This is the second time students will do this instructional routine. The teacher should read aloud each equation. When students are more familiar with equations, they do not need to be read by the teacher. ### Launch • Display one statement. • “Give me a signal when you know whether the statement is true and can explain how you know.” • 1 minute: quiet think time ### Activity • Share and record answers and strategy. • Repeat with each equation. ### Student Facing Decide if each statement is true or false. Be prepared to explain your reasoning. • $$7 + 3 = 10$$ • $$10 = 7 + 3$$ • $$10 = 3 + 6$$ ### Activity Synthesis • “How is 10 the same amount as $$7 + 3$$?” (When I add 7 and 3, the total is 10.) • “If $$7 + 3 = 10$$, and $$10 = 7 + 3$$, would $$10 = 3 + 7$$?” (Yes, because 7 + 3 is the same amount as 10, and 10 is the same amount as $$3 + 7$$.) ## Activity 1: Is It Addition or Subtraction? (15 minutes) ### Narrative The purpose of this activity is for students to explore the relationship between addition and subtraction through a Compare, Difference Unknown story problems. They analyze two equations, one addition and one subtraction, that match the same problem and discuss the relationship between the two equations and the story problem. Students should notice that both equations can be used to describe how they solve the problem. This also helps them relate addition and subtraction and see that often either operation can be used to solve a problem (MP7). ### Launch • Groups of 2 • Give students access to connecting cubes or two-color counters. • Display the image in the student book. • 1 minute: quiet think time • 2 minutes: partner discussion • Share responses. ### Activity • 5 minutes: partner work time • Encourage students to use the representation to make sense of both equations. • Monitor for a group who uses the representation to explain the addition equation and one who explains the subtraction equation. ### Student Facing There are 8 glue sticks and 3 scissors at the art station. How many fewer scissors are there than glue sticks? Mai created a picture. She is not sure which equation she should use to find the difference. $$8 - 3 = \boxed{5}$$ $$3 + \boxed{5} = 8$$ Help her decide. Show your thinking using drawings, numbers, or words. ### Advancing Student Thinking If students find the difference using only one operation, consider asking: • "Can you explain how you found the difference between the number of glue sticks and scissors?" • "Where in your drawing (objects) do you see the difference? How could you find that part of the representation by adding (subtracting)?" ### Activity Synthesis • “What are we trying to find out in this story problem?” (How many fewer scissors there are than glue sticks. The difference between the number of scissors and glue sticks.) • Invite previously identified groups to share. • “What is the same? What is different?” (3, 5, and 8 are in each equation. The numbers represent the same things in both. The 5 is boxed in both. One uses addition and the other uses subtraction. The boxed number is in a different place.) ## Activity 2: Which Equation? (10 minutes) ### Narrative The purpose of this activity is for students to identify addition and subtraction equations that match Compare, Difference Unknown story problems. Students may not initially choose more than one equation for each problem, so this is the emphasis of the activity synthesis. Students continue to build their language of Compare problems and solidify the relationship between addition and subtraction. MLR2 Collect and Display. Collect the language students use to explain their thinking. Display words and phrases such as: more, fewer, equation, drawings, words. During the synthesis, invite students to suggest ways to update the display: “What are some other words or phrases we should include?”, etc. Invite students to borrow language from the display as needed. Engagement: Internalize Self-Regulation. Synthesis: Provide students an opportunity to self-assess and reflect on their own progress. For example, ask students to check over their work to make sure they used drawings, numbers, or words to show their thinking, and also included at least one equation to show how they solved the problem. Supports accessibility for: Conceptual Processing, Attention ### Launch • Groups of 2 • Give students access to connecting cubes or two-color counters. ### Activity • 3 minutes: independent work time • 2 minutes: partner discussion • Monitor for a student who can show and explain how they chose which equation matches. ### Student Facing 1. There are 5 red pillows and 3 blue pillows on the reading rug. How many more red pillows are there than blue pillows? Show your thinking using drawings, numbers, or words. Circle the equation that matches the problem. $$5 + 3 = \boxed{\phantom{8}}$$ $$5 - 3 = \boxed{\phantom{2}}$$ $$5 + \boxed{\phantom{2}} = 3$$ $$3 + \boxed{\phantom{2}} = 5$$ 2. There are 7 calculators on the table. There are 8 math books. How many more math books are there than calculators? Show your thinking using drawings, numbers, or words. Circle the equation that matches the problem. $$7 + \boxed{\phantom{1}} = 8$$ $$8 - 7 = \boxed{\phantom{1}}$$ $$7 - 8 = \boxed{\phantom{0}}$$ $$8 + \boxed{\phantom{1}} = 7$$ 3. In Mr. Green’s class, 3 students have purple backpacks and 7 students have black backpacks. How many more students have black backpacks than purple backpacks? Show your thinking using drawings, numbers, or words. Circle the equation that matches the problem. $$3 + 7 = \boxed{\phantom{10}}$$ $$3 + \boxed{\phantom{4}} = 7$$ $$7 - \boxed{\phantom{4}} = 3$$ $$7 + \boxed{\phantom{1}} = 3$$ ### Activity Synthesis • “How did you know which equation matched the problem about backpacks?" (I made a tower of 3 cubes to show the purple backpacks, and a tower of 7 cubes to show the black backpacks. Then I broke off the difference, which is 4. This matches 7 - __ = 3 because the 4 shows the difference.) • “Are there any other equations that represent this problem?” (Yes, 3 + ___ = 7 also matches. I started with 3 cubes, and added cubes until I got to 7. The amount I added was 4.) • "There were two equations that match this problem. Check the other problems to see if there are any other equations that match." ## Activity 3: Centers: Choice Time (15 minutes) ### Narrative The purpose of this activity is for students to choose from activities that offer practice telling and solving story problems and adding and subtracting within 10. Students choose from any stage of previously introduced centers. • Capture Squares • Math Stories • Shake and Spill ### Required Materials Materials to Gather ### Required Preparation • Gather materials from previous centers: • Capture Squares, Stage 1 • Math Stories, Stage 4 • Shake and Spill, Stages 3 and 4 ### Launch • Groups of 2 • “Now you are going to choose from centers we have already learned.” • Display the center choices in the student book. • “Think about what you would like to do.” • 30 seconds: quiet think time ### Activity • Invite students to work at the center of their choice. • 10 minutes: center work time Choose a center. Capture Squares Math Stories Shake and Spill ### Activity Synthesis • “Tell your partner one thing they did that helped you during center time today.” ## Lesson Synthesis ### Lesson Synthesis Display: In Mr. Green's class, 3 students have purple backpacks and 7 students have black backpacks. How many more students have black backpacks than purple backpacks? $$3 + \boxed{4} = 7$$ and $$7 - 3 = \boxed{4}$$. “Today we explained how different equations can match the same story. These are the equations that match this story. Why does the position of the answer change?” (Because one equation is addition and one is subtraction, because they used different methods to solve the problem, so the answer came in different places.) “What does each number represent?”
# Recursive Sequence Formulas via Diagonalization In this post, we introduce an interesting application of matrix diagonalization: constructing closed-form expressions for recursive series. A recursive series is defined according to one or more initial terms and an update rule for obtaining the next term after some number of previous terms. For example, the series $(a_n)_{n\geq 0}$ and $(b_n)_{n\geq 0}$ given below are arithmetic and geometric series given in recursive form. For both of these series, it is straightforward to write a closed-form expression for the Nth term: For other sequences, however, this is not so straightforward. For example, consider the Fibonacci sequence, whose first term is $0$, whose second term is $1$, and whose successive terms are obtained by adding the previous two terms together. \begin{align} 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, \ldots \end{align} The recursive rule for the Fibonacci sequence is as follows: \begin{align} a_0 &= 0 \\ a_1 &= 1 \\ a_{n+2} &= a_n + a_{n+1} \end{align} Notice that we can express the recursive update rule using matrices. \begin{align} \begin{pmatrix} a_{n+1} \\ a_{n+2} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} a_n \\ a_{n+1} \end{pmatrix} \end{align} Repeatedly multiplying by this matrix, we can write a closed-form expression for the Nth and (N+1)st terms. \begin{align} \begin{pmatrix} a_{n} \\ a_{n+1} \end{pmatrix} &= \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^n \begin{pmatrix} a_0 \\ a_{1} \end{pmatrix} \\ &= \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^n \begin{pmatrix} 0 \\ 1 \end{pmatrix} \end{align} We can simplify this expression even further by diagonalizing the matrix. First, we solve for the eigenvalues. \begin{align} \det \left( \begin{pmatrix}0 & 1 \\ 1 & 1 \end{pmatrix} − \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) &= 0 \\ \det \begin{pmatrix} −\lambda & 1 \\ 1 & 1−\lambda \end{pmatrix} &= 0 \\ −\lambda(1−\lambda) − (1)(1) &= 0 \\ \lambda^2 − \lambda − 1 &= 0 \\ \lambda &= \frac{1 \pm \sqrt{5}}{2} \end{align} Now, we find the eigenvectors $v_1$ and $v_2$ that correspond to the eigenvalues $\lambda_1 = \frac{1+\sqrt{5} }{2}$ and $\lambda_2 = \frac{1- \sqrt{5} }{2}$. \begin{align} \left( \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} − \left( \frac{1+\sqrt{5}}{2} \right) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) v_1 &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \begin{pmatrix} −\frac{1+\sqrt{5}}{2} & 1 \\ 1 & \frac{1−\sqrt{5}}{2} \end{pmatrix} v_1 &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \begin{pmatrix} 2 & 0 \\ 0 & \frac{1+\sqrt{5}}{2} \end{pmatrix} \begin{pmatrix} −\frac{1+\sqrt{5}}{2} & 1 \\ 1 & \frac{1−\sqrt{5}}{2} \end{pmatrix} v_1 &= \begin{pmatrix} 2 & 0 \\ 0 & \frac{1+\sqrt{5}}{2} \end{pmatrix} \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \begin{pmatrix}1+\sqrt{5} & −2 \\ 1+\sqrt{5} & −2 \end{pmatrix} v_1 &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ v_1 &= \begin{pmatrix} 2 \\ 1+\sqrt{5} \end{pmatrix} \end{align} \begin{align} \left( \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} − \left( \frac{1−\sqrt{5}}{2} \right) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) v_2 &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \begin{pmatrix} \frac{−1+\sqrt{5}}{2} & 1 \\ 1 & \frac{1+\sqrt{5}}{2} \end{pmatrix} v_2 &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \begin{pmatrix} 2 & 0 \\ 0 & \frac{1−\sqrt{5}}{2} \end{pmatrix} \begin{pmatrix} \frac{−1+\sqrt{5}}{2} & 1 \\ 1 & \frac{1+\sqrt{5}}{2} \end{pmatrix} v_2 &= \begin{pmatrix} 2 & 0 \\ 0 & \frac{1−\sqrt{5}}{2} \end{pmatrix} \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \begin{pmatrix}−1+\sqrt{5} & 2 \\ −1+\sqrt{5} & 2 \end{pmatrix} v_2 &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ v_2 &= \begin{pmatrix} 2 \\ 1−\sqrt{5} \end{pmatrix} \end{align} Finally, we can diagonalize the matrix. \begin{align} \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} &= \begin{pmatrix} 2 & 2 \\ 1+\sqrt{5} & 1−\sqrt{5} \end{pmatrix} \begin{pmatrix} \frac{1+\sqrt{5}}{2} & 0 \\ 0 & \frac{1−\sqrt{5}}{2} \end{pmatrix} \begin{pmatrix} 2 & 2 \\ 1+\sqrt{5} & 1−\sqrt{5} \end{pmatrix}^{−1} \end{align} Substituting the diagonalized matrix into the original formula, we are able to simplify so much that we find a closed-form, non-matrix formula for the Nth term of the sequence. \begin{align} \begin{pmatrix} a_{n} \\ a_{n+1} \end{pmatrix} &= \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^n \begin{pmatrix} 0 \\ 1 \end{pmatrix} \\ \begin{pmatrix} a_n \\ a_{n+1} \end{pmatrix} &= \left[ \begin{pmatrix} 2 & 2 \\ 1+\sqrt{5} & 1−\sqrt{5} \end{pmatrix} \begin{pmatrix} \frac{1+\sqrt{5}}{2} & 0 \\ 0 & \frac{1−\sqrt{5}}{2} \end{pmatrix} \begin{pmatrix} 2 & 2 \\ 1+\sqrt{5} & 1−\sqrt{5} \end{pmatrix}^{−1} \right]^n \begin{pmatrix} 0 \\ 1 \end{pmatrix} \\ \begin{pmatrix} a_n \\ a_{n+1} \end{pmatrix} &= \begin{pmatrix} 2 & 2 \\ 1+\sqrt{5} & 1−\sqrt{5} \end{pmatrix} \begin{pmatrix} \left( \frac{1+\sqrt{5}}{2} \right)^n & 0 \\ 0 & \left( \frac{1−\sqrt{5}}{2} \right)^n \end{pmatrix} \begin{pmatrix} 2 & 2 \\ 1+\sqrt{5} & 1−\sqrt{5} \end{pmatrix}^{−1}\begin{pmatrix} 0 \\ 1 \end{pmatrix} \\ \begin{pmatrix} a_n \\ a_{n+1} \end{pmatrix} &= \begin{pmatrix} 2 & 2 \\ 1+\sqrt{5} & 1−\sqrt{5} \end{pmatrix} \begin{pmatrix} \left( \frac{1+\sqrt{5}}{2} \right)^n & 0 \\ 0 & \left( \frac{1−\sqrt{5}}{2} \right)^n \end{pmatrix} \left[ \frac{1}{−4\sqrt{5}} \begin{pmatrix} 1−\sqrt{5} & −2 \\ −1−\sqrt{5} & 2 \end{pmatrix} \right] \begin{pmatrix} 0 \\ 1 \end{pmatrix} \\ \begin{pmatrix} a_n \\ a_{n+1} \end{pmatrix} &= \frac{1}{2\sqrt{5}} \begin{pmatrix} 2 & 2 \\ 1+\sqrt{5} & 1−\sqrt{5} \end{pmatrix} \begin{pmatrix} \left( \frac{1+\sqrt{5}}{2} \right)^n & 0 \\ 0 & \left( \frac{1−\sqrt{5}}{2} \right)^n \end{pmatrix} \begin{pmatrix} 1 \\ −1 \end{pmatrix} \\ \begin{pmatrix} a_n \\ a_{n+1} \end{pmatrix} &= \frac{1}{\sqrt{5}} \begin{pmatrix} 1 & 1 \\ \frac{1+\sqrt{5}}{2} & \frac{1−\sqrt{5}}{2} \end{pmatrix} \begin{pmatrix} \left( \frac{1+\sqrt{5}}{2} \right)^n \\ − \left( \frac{1−\sqrt{5}}{2} \right)^n \end{pmatrix} \\ \begin{pmatrix} a_n \\ a_{n+1} \end{pmatrix} &= \begin{pmatrix} \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right)^n − \left( \frac{1−\sqrt{5}}{2} \right)^n \right] \\ \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right)^{n+1} − \left( \frac{1−\sqrt{5}}{2} \right)^{n+1} \right] \end{pmatrix} \\ a_n &= \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right)^n − \left( \frac{1−\sqrt{5}}{2} \right)^n \right] \end{align} This formula might be a little surprising – the Fibonacci sequence consists only of whole numbers, yet $\sqrt{5}$ appears often in the formula! However, the formula is indeed correct. We verify the formula for $a_0$, $a_1$, and $a_2$ – and it will work on all the other terms as well. \begin{align} a_0 &= \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right)^0 − \left( \frac{1−\sqrt{5}}{2} \right)^0 \right] \\ &= \frac{1}{\sqrt{5}} \left[ 1−1 \right] \\ &= 0 \end{align} \begin{align} a_1 &= \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right)^1 − \left( \frac{1−\sqrt{5}}{2} \right)^1 \right] \\ &= \frac{1}{\sqrt{5}} \left[ \sqrt{5} \right] \\ &= 1 \end{align} \begin{align} a_2 &= \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right)^2 − \left( \frac{1−\sqrt{5}}{2} \right)^2 \right] \\ &= \frac{1}{\sqrt{5}} \left[ \frac{1+2\sqrt{5}+5}{4} − \frac{1−2\sqrt{5}+5}{4} \right] \\ &= \frac{1}{\sqrt{5}} \left[ \sqrt{5} \right] \\ &= 1 \end{align} This same method applies for any recursive series, though we may need to diagonalize a higher-dimensional matrix and numerically approximate the eigenvalues. For example, consider the following spin-off of the Fibonacci sequence: \begin{align} a_0 &= 0 \\ a_1 &= 1 \\ a_2 &= 1 \\ a_{n+3} &= 2a_n + a_{n+2} \end{align} First, we express the recursive update rule using matrices, and write a closed-form expression involving an exponentiated matrix multiplying the first few terms. \begin{align} \begin{pmatrix} a_{n+1} \\ a_{n+2} \\ a_{n+3} \end{pmatrix} &= \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 2 & 0 & 1 \end{pmatrix} \begin{pmatrix} a_n \\ a_{n+1} \\ a_{n+2} \end{pmatrix} \\ &= \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 2 & 0 & 1 \end{pmatrix}^n \begin{pmatrix} a_0 \\ a_1 \\ a_2 \end{pmatrix} \\ &= \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 2 & 0 & 1 \end{pmatrix}^n \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \end{align} We omit the steps of diagonalizing the matrix since they should be routine by now – but it is worthwhile to discuss the method of approximating the eigenvalues. When solving for the eigenvalues, we reach the following equation: \begin{align} \lambda^3 - \lambda^2 - 2 = 0 \end{align} This cubic cannot be factored manually – not even using synthetic division – since it has no rational roots. Hence, we turn to a graphing utility to approximate a root $\lambda_1 \approx 1.696$. Then, we can perform synthetic division with that root to factor the polynomial into \begin{align} (\lambda - 1.696)(\lambda^2 + 0.696 \lambda + 1.180) = 0. \end{align} We can use the quadratic equation to solve $\lambda^2 + 0.696\lambda + 1.180 = 0$ for the other two roots, which we find as $\lambda_2 \approx -0.348-1.029i$ and $\lambda_3 \approx -0.348+1.029i$. Then, with a bit of grunt work, we can use these approximations to solve for the eigenvectors, substitute the diagonalization into the original equation, and multiply to find the formula for the Nth term. The result, with each term rounded to 3 decimal places, is \begin{align} a_n \approx \hspace{.1cm} &(−0.162+0.164i)(−0.348−1.029i)^n \\ &− (0.162+0.164i)(−0.348+1.029i)^n \\ &+ 0.324(1.696)^n \, . \end{align} Lastly, we can verify that the first several terms match up with the actual sequence $0, 1, 1, 1, 3, 5, 7, 13$. Our estimates are slightly off due to compounded rounding error, but they could be made more accurate by using greater precision in the decimals that occur in the formula for $a_n$. \begin{align} \begin{matrix} a_0 = 0 & a_1 \approx 1.000 & a_2 \approx 1.001 & a_3 \approx 1.001 \\ a_4 \approx 3.003 & a_5 \approx 5.006 & a_6 \approx 7.011 & a_7 \approx 13.022 \end{matrix} \end{align} Practice Problems Use diagonalization to compute a closed-form expression for the recursive sequence $a_n$. (You can view the solution by clicking on the problem.) \begin{align} 1) \hspace{.75cm} a_0 &= 0 \\ a_1 &= 1 \\ a_{n+2} &= 2a_n + a_{n+1} \end{align} Solution: \begin{align} a_n = \frac{2^n−(−1)^n}{3} \end{align} \begin{align} 2) \hspace{.75cm} a_0 &= 0 \\ a_1 &= 1 \\ a_{n+2} &= a_n + 2a_{n+1} \end{align} Solution: \begin{align} a_n = \frac{1}{2\sqrt{2}} \left[ \left( 1+\sqrt{2} \right)^n − \left( 1−\sqrt{2} \right)^n \right] \end{align} \begin{align} 3) \hspace{.75cm} a_0 &= 0 \\ a_1 &= 1 \\ a_2 &= 1 \\ a_{n+3} &= a_n + a_{n+2} \end{align} Solution: \begin{align} a_n \approx \hspace{.1cm} &(−0.209+0.184i)(−0.233−0.793i)^n \\ &− (0.209+0.184i)(−0.233+0.793i)^n \\ &+ 0.417(1.466)^n \end{align} \begin{align} 4) \hspace{.75cm} a_0 &= 0 \\ a_1 &= 1 \\ a_2 &= 1 \\ a_{n+3} &= a_n + a_{n+1}+ a_{n+2} \end{align} Solution: \begin{align} a_n \approx \hspace{.1cm} &(−0.168+0.198i)(−0.420−0.606i)^n \\ &− (0.168+0.198i)(−0.420+0.606i)^n \\ &+ 0.336(1.839)^n \end{align} Tags:
In this post we’ll deal only with finite lists. Questions regarding partial and infinite lists are left as an exercise for the reader. Sometimes we have to answer questions like: I have something that looks like a functional equation, but I miss a function – how do I find it? To be more concrete, given the functions f,g,h, how do I find the function r in the equation: f.g = h.r? The purpose of this post is to show you how to find a function in a specific example and, after you found it, to proove that the functional equation holds. Problem: Given the following equation: reverse . concat = concat . reverse . h, find the function h. Part 1. First of all, we’ll start with the definitions of the functions concat and reverse. concat is a function that takes a list of lists and concatenates the innermost lists into a single one. Example: concat [ [1,2] , [3,4] ] = [1,2,3,4]. Definition of concat: concat :: [[a]] -> [a] concat [] = [] concat (xs:xss) = xs ++ concat xss reverse is a function that takes a list and returns a list with the original elements read from end to beginning. Example: reverse [1,2,3,4] = [4,3,2,1] Definition of reverse: reverse :: [a] -> [a] reverse [] = [] reverse (x:xs) = reverse xs ++ [x] Exercise: Proove that reverse (reverse xs) = xs, for all finite lists xs (in other words, we say that reverse is an involution). Back to our problem, we want to find the function h that satisfies the functional equation reverse . concat = concat . reverse . h. Because reverse is an involution, we can simplify our equation by a standard technique – compose it on the left with reverse: reverse . (reverse . concat) = reverse . (concat . reverse . h) Because composing is associative, we have that: (reverse . reverse) . concat = reverse . (concat . reverse . h) Because reverse . reverse = Id (reverse is an involution), we derive that: concat = reverse . (concat . reverse . h) (E1), which is much more amenable to manipulations than the initial equation, because the left-hand side has been reduced to a single function – concat. Let’s simplify the problem a little and say we’re dealing with a finite list with 2 elements: L = [ [a1,a2] , [a3,a4] ]. Applying the equation E1 on L, we have: concat L = reverse(concat(reverse( h L ))) (E2) E2 is a very important equation, because it gives us a lot of information about the type of h. Remark 1. h needs to work on L, which is a list of lists, so the domain of h is [[a]], for a fixed type ‘a’. Remark 2. reverse(h L) in equation E2 says the function reverse needs to work on the values of function h. But reverse is a function that works on lists, so the value (h L) must be a list! In that way, we found that (h L) :: [b], for a fixed type b. So, the codomain of the function h is [b], for a fixed type b. Conclusion 1: From remarks 1 and 2, we conclude that h :: [[a]] -> [b], for fixed types a and b. As we said, h(L) :: [b]. We can write h(L) = [b1, b2, …, bm]. Equation E2 becomes: concat L = reverse(concat(reverse [b1,…,bm])) = {definition of reverse} reverse(concat [bm,…,b1] ) (1) But concat L = concat [ [a1,a2] , [a3, a4]] = [a1,a2,a3,a4]. Using the derived equation (1), we deduce that: [a1,a2,a3,a4] = reverse(concat [bm,…,b1]) (E3) Remark 3. In the right-hand side of E3, we have the expression concat [bm,…,b1], and knowing that concat works on lists of lists, we deduce that bm,…,b1 are also lists. Remark 3 helps us to write bm,…,b1 as lists in the following way: bm = [cm_1,cm_2,…., cm_im], … , b1 = [c1_1, c1_2, … , c1_i1]. In this way, concat [bm,…,b1] = [cm_1,cm_2, … , cm_im, … , c1_1, c1_2, … , c1_i1] (2) Coming back to the equation E2, we’ll have (together with (2)): [a1,a2,a3,a4] = reverse [cm_1,cm_2, … , cm_im, … , c1_1, c1_2, … , c1_i1] = {definition of reverse} [c1_i1, … , c1_1, …, cm_im, …, cm_1] So, we have now that: [a1,a2,a3,a4] = [c1_i1, … c1_2, c1_1, …, cm_im, …cm_2, cm_1] (E4) Becuase in the left-hand side of the equation E4 we have only 4 elements, it’s clear that in the right-hand side we also have only 4 elements. We’ll pick two elements from b1 and two elements from b2, and see what happens next: [a1,a2,a3,a4] = [c1_2,c1_1,c2_2,c2_1] It’s clear now how to pick: c1_2 = a1; c1_1 = a2; c2_2 = a3; c2_1 = a4. So we’re left only with b1 = [c1_1,c1_2] = [a2,a1] and b2 = [c2_1,c2_2] = [a4,a3]. We first introduced h(L) = [b1,…,bm], which becomes now: h(L) = h [ [a1,a2], [a3,a4] ] = [b1,b2] = [ [a2,a1], [a4,a3] ] = [reverse [a1,a2], reverse [a3,a4]] = map reverse [[a1,a2], [a3,a4]] = map reverse L This derivation says that h = map reverse, and we’ve found the function needed. The equation in the problem becomes: reverse . concat = concat . reverse . map reverse But we’ve been overspecific by now, working on a list of lists of 2 elements each. Is the equation found holding on all finite lists? The answer is yes, and we’ll proove: Proposition (P): reverse . concat = concat . reverse . map reverse, for all finite lists.1 End of Part 1. Part 2. Prove that: Proposition (P): reverse . concat = concat . reverse . map reverse, for all finite lists.1 We can proove this equation using induction, but the scope of this post is to introduce some advanced techniques that you’ll need in your future projects. For that, we’ll start with a little theory. First of all, all the functions used in the proposition P can be rewritten using foldr. We’ll let the proofs as an exercise. concat = foldr (++) [] Also, we need to know a very important theorem, which is demonstrated in [1]. The fusion theorem. Given a function f and a function h with the following conditions: 1. f is a strict function. 2. f(g x y) = h x (f y), for all x,y of appropriate types. Let b = f a. Then f . foldr g a = foldr h b. End of fusion theorem. The fusion theorem says that, in certain conditions, a function composed with a foldr is also a foldr. This theorem will be used to simplify a lot of equations in our exercise. Now we’re ready to give the proof of our proposition, which is reminded below: reverse . concat = concat . reverse . map reverse, for all finite lists We’ll prove that in the following way: 1. Express the left-hand side as a foldr. 2. Express the right-hand side as a foldr. 3. Compare the definitions to see that are equal. 1. Express the left-hand side as a foldr. reverse . concat = { because, as we said, concat = foldr (++) [] } reverse . foldr (++) [] We have a composition of reverse with a foldr function, so the only thing we can think of is to try applying the fusion theorem: reverse . foldr (++) [] = foldr h b. 1. First of all, the first condition of the theorem is satisfied – reverse is a strict function. 2. reverse ((++) xs ys) = h xs (reverse ys) reverse ((++) xs ys) = { infix notation for (++) } reverse (xs ++ ys) = { this can be easily proved } reverse ys ++ reverse xs We have in this way that: reverse ys ++ reverse xs = h xs (reverse ys) We define h to work generally like this: h xs zs = zs ++ reverse xs, and we’re done, we’ve found the desired function h for the fusion theorem. The technique is simple: we just generalized reverse ys (present in left side and right side of the equation) to a fresh variable named zs. The theorem states also that b = reverse [] = []. So b = []. We’ve found our equation: reverse . concat = reverse . foldr (++) [] = foldr h [], where h xs zs = zs ++ reverse xs. (R1) End – express the left-hand side as a foldr. 2. Express the righ-hand side as a foldr (concat . reverse . map reverse) We’ll use the following result, which is left also as an exercise to the reader: map f = foldr ((:) . f) [] (where ‘:’ is the lists operator). Using the map rewriting as a foldr, we’ll have: concat . reverse . map reverse = { function composition is associative } concat . (reverse . map reverse) = { map f function as a foldr } concat . (reverse . foldr ((:) . reverse) [] ) We apply now the fusion theorem for the composition: reverse . foldr ((:) . reverse) [] = foldr h b. It’s clear that reverse is a strict function. Also, b = reverse [] = []. So b = []. The second condition of the theorem becomes (instantiated in our case): reverse ((:) . reverse xs yss) = h xs (reverse yss) (1) reverse ((:). reverse xs yss) = { function composition } reverse ((:)(reverse xs) (yss) ) = { infix notation of (:) } reverse(reverse xs : yss) = { reverse definition, equation 2 } reverse yss ++ [reverse xs] This derivation, together with (1), gives us: reverse yss ++ [reverse xs] = h xs (reverse yss) Generalizing reverse yss (found in both sides of the equation) to the fresh variable zss, we have found our function h: h xs zss = zss ++ [reverse xs]. Applying the fusion theorem, we have that: reverse . foldr ((:) . reverse) [] = foldr h [], where the function h is: h xs zss = zss ++ [reverse xs] I’ll rewrite the function h above as h1, in order to not confuse notations in the following argument. We have so the original equation, which became: concat . reverse . map reverse = concat . foldr h1 [], where h1 xs zss = zss ++ [reverse xs]. We’ll try to apply again the fusion theorem and to write: concat . foldr h1 [] = foldr h b. We know that concat is a strict function, so the first condition of the theorem applies. Let b = concat [] = []. So b = []. concat (h1 xs zss) = h xs (concat zss) (2) concat (h1 xs zss) = { definition of h1} concat(zss ++ [reverse xs]) = {simple to proove: concat (xss ++ zss) = concat xss ++ concat zss} concat zss ++ concat [reverse xs] = { it’s trivial to see that concat [xs] = xs} concat zss ++ reverse xs From this derivation and equation (2), we conclude that: h xs (concat zss) = concat zss ++ reverse xs Generalizing concat zss into the fresh variable ys, we have that: h xs ys = ys ++ reverse xs In this way, we have that concat . foldr h1 [] = foldr h [], where h xs ys = ys ++ reverse xs. We have completed, in this way, the right-hand side of the original equation: concat . reverse . map reverse = foldr h [], where h xs ys = ys ++ reverse xs. (R2) But we also have the result obtained for the left-hand side: reverse . concat = foldr h [], where h xs zs = zs ++ reverse xs (R3) 3. Comparing R2 and R3, it’s clear that are equal. So the proposition holds: reverse . concat = concat . reverse . map reverse End of Part 2. I know the exercise is a bit long and technical, but in this way we’ve shown some powerful techniques for finding functions with some properties given and for proving equations using the fusion theorem of foldr. This are mathematical tools that you can use for optimizing equations, for prooving and even for constructing new functions as folds (see more about it in [2]). If you see some error in my calculations, I’m more than happy to listen your opinion. References [1]. Bird,R. (2015) Thinking Functionally with Haskell. Cambridge University Press. [2]. Hutton,G. (1999) A tutorial on the universality and expressiveness of fold. J. Functional Programming. Constantin Liculescu
550.50K Категория: Математика # Basics of functions and their graphs ## 1. Chapter 2 Functions and Graphs 2.1 Basics of Functions and Their Graphs 1 ## 2. Objectives: Find the domain and range of a relation. Determine whether a relation is a function. Determine whether an equation represents a function. Evaluate a function. Graph functions by plotting points. Use the vertical line test to identify functions. Obtain information about a function from its graph. Identify the domain and range of a function from its graph. Identify intercepts from a function’s graph. 2 ## 3. Definition of a Relation A relation is any set of ordered pairs. The set of all first components of the ordered pairs is called the domain of the relation and the set of all second components is called the range of the relation. 3 ## 4. Example: Finding the Domain and Range of a Relation Find the domain and range of the relation: {(0, 9.1), (10, 6.7), (20, 10.7), (30, 13.2), (40, 21.2)} domain: {0, 10, 20, 30, 40} range: {9.1, 6.7, 10.7, 13.2, 21.2} 4 ## 5. Definition of a Function A function is a correspondence from a first set, called the domain, to a second set, called the range, such that each element in the domain corresponds to exactly one element in the range. 5 ## 6. Example: Determining Whether a Relation is a Function Determine whether the relation is a function: {(1, 2), (3, 4), (6, 5), (8, 5)} No two ordered pairs in the given relation have the same first component and different second components. Thus, the relation is a function. 6 ## 7. Functions as Equations If an equation is solved for y and more than one value of y can be obtained for a given x, then the equation does not define y as a function of x. 7 ## 8. Example: Determining Whether an Equation Represents a Function Determine whether the equation defines y as a function of x. x2 + y 2 = 1 y2 = 1 - x2 y = ± 1 - x2 The ± shows that for certain values of x, there are two values of y. For this reason, the equation does not define y as a function of x. 8 ## 9. Function Notation The special notation f(x), read “f of x” or “f at x”, represents the value of the function at the number x. 9 ## 10. Example: Evaluating a Function If f ( x) = x - 2 x + 7, evaluate f (-5). 2 f ( x) = x 2 - 2 x + 7, f (-5) = (-5) 2 - 2( -5) + 7 = 25 + 10 + 7 = 42 Thus, f (-5) = 42. 10 ## 11. Graphs of Functions The graph of a function is the graph of its ordered pairs. 11 ## 12. Example: Graphing Functions Graph the functions f(x) = 2x and g(x) = 2x – 3 in the same rectangular coordinate system. Select integers for x, starting with –2 and ending with 2. 12 ## 13. Example: Graphing Functions (continued) We set up a partial table of coordinates for each function. We then plot the points and connect them. f ( x) = 2 x g ( x) = 2 x - 3 x y = f (x) x y = f (x) –2 –4 –2 –7 –1 –2 –1 –5 0 0 0 –3 1 2 1 –1 2 4 2 1 13 ## 14. The Vertical Line Test for Functions If any vertical line intersects a graph in more than one point, the graph does not define y as a function of x. 14 ## 15. Example: Using the Vertical Line Test Use the vertical line test to identify graphs in which y is a function of x. not a function function 15 ## 16. Example: Analyzing the Graph of a Function Use the graph to find f(5) f(5)=400 For what value of x is f(x) = 100? f(9) = 125, so x = 9. 16 ## 17. Identifying Domain and Range from a Function’s Graph To find the domain of a function from it’s graph, look for all the inputs on the x-axis that correspond to points on the graph. To find the range of a function from it’s graph, look for all the outputs on the y-axis that correspond to points on the graph. 17 ## 18. Example: Identifying the Domain and Range of a Function from Its Graph Use the graph of the function to identify its domain and its range. Domain { x -2 £ x £ 1} [-2,1] Range { y 0 £ y £ 3} [0,3] 18 ## 19. Example: Identifying the Domain and Range of a Function from Its Graph Use the graph of the function to identify its domain and its range. Domain { x -2 < x £ 1} (-2,1] Range { y -1 £ y < 2} [-1,2) 19 ## 20. Identifying Intercepts from a Function’s Graph To find the x-intercepts, look for the points at which the graph crosses the x-axis. To find the y-intercept, look for the point at which the graph crosses the y-axis. A function can have more than one x-intercept but at most one y-intercept. 20 ## 21. Example: Identifying Intercepts from a Function’s Graph Identify the x- and y-intercepts for the graph of f(x). · · · · The x-intercepts are (–3, 0) (–1, 0) and (2, 0) The y-intercept is (0, –6)
Lesson Explainer: Logarithmic Differentiation \| Nagwa Lesson Explainer: Logarithmic Differentiation \| Nagwa # Lesson Explainer: Logarithmic Differentiation Mathematics • Third Year of Secondary School ## Join Nagwa Classes In this explainer, we will learn how to find the derivatives of positive functions by taking the natural logarithm of both sides before differentiating. We sometimes encounter functions we would like to differentiate but on which our usual methods, such as the chain, product, or quotient rules, are difficult to or cannot be directly applied. For example, in each of the functions below, it is not clear how we might differentiate each function with respect to : Each of these functions has a complicated exponent involving the variable , which makes applying our usual rules of differentiation problematic. In such cases, one method we can use to differentiate is logarithmic differentiation. ### Definition: Logarithmic Differentiation Logarithmic differentiation is a four-step process used to differentiate awkward or complicated functions that do not lend themselves easily, if at all, to the usual methods of differentiation. For a differentiable function the steps are as follows: 1. Apply the natural logarithm to both sides where the natural logarithm . 2. Use the laws of logarithms to simplify or distribute the right-hand side. 3. Differentiate both sides with respect to . 4. Solve for . Note that while we have specified that these steps apply when , if we want to apply logarithmic differentiation for all values of , we take the natural logarithm of the absolute value of : . This then covers both positive and negative values of our function . Aside from taking the natural logarithm, the key step that makes this process so useful is the second step: using the laws of logarithms to simplify the right-hand side. This is because, • if involves a product, then involves a sum, since by the product rule for logarithms • if involves a quotient, then involves a difference, since by the quotient rule for logarithms • if involves an exponent or power, then involves a product, since by the law of exponents for logarithms Once we have rewritten our function in a more manageable form, we can then differentiate the resulting expressions using the usual methods and/or known results. On the left-hand side of our equation, , and since we are now differentiating a function of where is itself a function of , we use implicit differentiation. This involves the chain rule for differentiating a composite function where . Applying this to our left-hand side, differentiating with respect to gives us Then, using the known result , we have Now that we know the steps to follow for logarithmic differentiation, let us try them out on some quite complicated functions. ### Example 1: Finding the First Derivative of a Function Having a Variable in Its Base and Exponent Using Logarithmic Differentiation Find if . Let us begin by isolating on the left-hand side. Dividing both sides by 6, we have Since our function is an exponential function whose base and exponent are variables, we use the process of logarithmic differentiation to evaluate its derivative. Our first step in logarithmic differentiation is to take the natural logarithms of both sides. This gives us In taking the natural logarithm, we have specified that since the logarithm is only defined for positive, nonzero input values. Our second step in logarithmic differentiation is to use the laws of logarithms to simplify our right-hand side. In this case, we note that our function consists of the variable with the exponent , all multiplied by the constant . We first use the product rule for logarithms, to separate the constant. Thus, We may then use the power rule for logarithms, to expand the second term on the right-hand side and bring our exponent down. This gives us Our function is now in a form amenable to the usual rules of differentiation, so we can apply step three of logarithmic differentiation. That is, we can differentiate both sides with respect to . Since the derivative of a sum of two functions is equal to the sum of their derivatives, we have Applying implicit differentiation to our left-hand side, where is a function of , gives us , and since is a constant, its derivative is zero. We therefore have On the right-hand side, we are differentiating a product of two functions and can therefore use the product rule for differentiation. We know that the derivative of with respect to is and that the derivative of with respect to is . Hence, by the product rule for differentiation, we have Our final step in logarithmic differentiation is to isolate on the left-hand side. To do this, we multiply both sides by . And since , we have Combining exponents of , we then have Let us now consider an example of how to use logarithmic differentiation to differentiate a function involving a trigonometric ratio with a variable exponent. ### Example 2: Differentiating Functions Involving Trigonometric Ratios Using Logarithmic Differentiation Use logarithmic differentiation to find the derivative of the function . Logarithmic differentiation is useful when a function we wish to differentiate does not lend itself to the usual rules of differentiation. This is the case in the given function, where both the argument of the cosine and the exponent are variable. Our first step in logarithmic differentiation is to apply the natural logarithm to both sides of our function so that In taking the natural logarithm, we have specified that since the logarithm is only defined for positive, nonzero input values. We then use the laws of logarithms to expand the right-hand side into something we can easily differentiate. To separate the constant, 2, from , we use the product rule for logarithms: Applying this rule gives us Our second term on the right-hand side involves an exponent , so we can apply the power rule for logarithms, to bring our exponent down: As hoped, on our right-hand side we now have a collection of expressions that we are able to easily differentiate. Thus, we can apply the third step in logarithmic differentiation. That is, we differentiate both sides with respect to . And since the derivative of a sum of two functions is equal to the sum of their derivatives, we have Applying implicit differentiation to our left-hand side, where is a function of , gives us , and since is a constant so that its derivative is zero, we have, On our right-hand side, we are now differentiating a product with factors and and can therefore apply the product rule for differentiation. In order to do this, however, we will need to know the derivative of with respect to . Since this is a function of a function, in other words, a composite function, we can apply the chain rule. Letting , where , then and And since , we have We are now able to apply the product rule for differentiation to our right-hand side above. Letting and , so that and , by the product rule, Having taken the natural logarithm on both sides and then differentiated with respect to , we then have Our final step in logarithmic differentiation is to isolate on the left-hand side. We do this by multiplying both sides by to obtain Now, recalling that our function , we have In our next example, we use logarithmic differentiation to differentiate a composite function with variable exponent. ### Example 3: Finding the First Derivative of a Function Using Logarithmic Differentiation Find if . If the exponent outside the parentheses in our function were a constant, we would simply use the chain rule to differentiate. However, the exponent is itself a function of and the easiest way to tackle differentiating a function of this type is to use logarithmic differentiation. The first step is to apply the natural logarithm to both sides of our function so that In taking the natural logarithm, we have specified that , since the logarithm is only defined for positive, nonzero input values. We then use the laws of logarithms to expand the right-hand side into something we can easily differentiate. Our right-hand side involves an exponent so we can apply the power rule for logarithms, to bring our exponent down: On our right-hand side we now have a function that is the product of two differentiable functions that we know how to differentiate. Thus, we can apply the third step in logarithmic differentiation. That is, we differentiate both sides with respect to so that Applying implicit differentiation to our left-hand side, where is a function of , gives us so that On our right-hand side, we are now differentiating a product with factors and . This means we can apply the product rule for differentiation. To do this, however, we need to know the derivative of with respect to . Since this is a function of a function, we can apply the chain rule. Letting , where , then and Since so that , we then have and we can now apply the product rule for differentiation to the right-hand side above. Letting and so that by the product rule, Rearranging and taking the common factor of 8 outside the parentheses, we now have Our final step in logarithmic differentiation is to isolate on the left-hand side. We do this by multiplying both sides by to obtain We may want the derivative purely in terms of , so with , we have Let us now consider an example where we use logarithmic differentiation to differentiate a function which is a constant raised to a complicated exponent involving exponential and trigonometric functions. ### Example 4: Differentiating a Composition of Exponential and Trigonometric Functions Using Logarithmic Differentiation Given that , determine . Considering that the exponent in our function is a complicated function of , the easiest method we have to differentiate this function is by using logarithmic differentiation. Our first step in this process is to take the natural logarithms of both sides, giving us In taking natural logarithms, we have specified that since the logarithm is only defined for positive, nonzero input values. Although this looks possibly even more complicated, our next step will help us simplify things so that we are able to differentiate quite easily. The second step of logarithmic differentiation is to use the laws of logarithms to change our right-hand side into something more manageable. In our case, since we have an exponent, we use the power rule, or law of exponents for logarithms, Using this to bring down our exponent gives us and on our right-hand side now we have the constant, , multiplied by the expression . We know how to differentiate this, so we apply step three of logarithmic differentiation. That is, we differentiate both sides with respect to , giving (taking the constant, , outside of the derivative). Applying implicit differentiation to the left-hand side, where is a function of , we have Now recalling that and , differentiating the right-hand side gives us Our final step in logarithmic differentiation is to isolate on the left-hand side. We can do this by multiplying both sides by so that Finally recalling that and rearranging our right-hand side, we have In our final example, we will use logarithmic differentiation to differentiate a slightly strange function with repeated exponents of . ### Example 5: Finding the First Derivative of a Function in the Form 𝑥𝑥𝑥 Using Logarithmic Differentiation Given , find . The exponents in the given function are variables and the easiest way to differentiate a function of this type is using logarithmic differentiation. To do this, we first take the natural logarithms on both sides. Applying this to our function , we have In taking natural logarithms, we have specified that , since the logarithm is only defined for positive, nonzero input values. Our next step in logarithmic differentiation is to use the laws of logarithms to expand our right-hand side. This should give us something that we can easily differentiate, although with this particular function we will see that we must apply these first two steps twice to reach the point where we can differentiate. Since our function involves exponents, we can use the power or exponent law for logarithms to expand our right-hand side. This says that and applying it to our function so that we bring down our exponent, we have This appears a little more manageable since we now have a product on our right-hand side. However, one of the factors, , is still not something we can easily differentiate. To remedy this, we return to step one and again take natural logarithms on both sides. By doing this, we will then be able to again apply the laws of logarithms to expand our right-hand side. So, taking the natural logarithms once more, we have Now, on our right-hand side, we can use the product rule for logarithms; that is, Applying this to our right-hand side, we have We can use the power rule for logarithms once more to expand the first term on the right-hand side so that we have and now we have something we can actually differentiate using the usual rules of differentiation. This brings us to our third step of logarithmic differentiation, which is to differentiate both sides with respect to : (noting that on the right-hand side we have used the fact that the derivative of a sum is equal to the sum of the derivatives). Let us begin our differentiation with the first term on the right-hand side. We have a product with factors and . Using the product rule for differentiation and the fact that , then Considering now the second term on our right-hand side, we may observe that the technique we will use applies similarly to the expression on the left-hand side. We have an expression, . This is of the form, , where is a differentiable function of . So, we have a function of a function where, in our case, . We know that and that (for ). We therefore have We now have the derivatives of both terms of our right-hand side, so all that remains is to evaluate the derivative on the left-hand side of our equation. That is, . As noted, we can apply the same technique as used for the second term on our right-hand side. Now we wish to differentiate the expression , where . This time we must take into account the fact that is a function of , so We can now assemble the results of our differentiation as follows: In the final step of logarithmic differentiation, we isolate on the left-hand side by multiplying both sides by and rearranging; we have We conclude our discussion of logarithmic differentiation by noting some key points. ### Key Points • For complicated functions where the usual methods of differentiation are either difficult or impossible to apply, we use the method of logarithmic differentiation by taking the natural logarithm, where the natural logarithm “” is . This is particularly useful when applied to functions involving a variable exponent. • For a differentiable function of , we • apply the natural logarithm to both sides • use the laws of logarithms to simplify or expand the right-hand side; • differentiate both sides with respect to ; • solve for . • In taking natural logarithms on both sides, we specify that since the logarithm is only defined for positive, nonzero input values. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions
Courses Courses for Kids Free study material Offline Centres More Store # Let $A={{R}_{0}}\times R$ where ${{R}_{0}}$ denotes the set of all non-zero numbers. A binary operation $'O'$ is defined on $A$ as follows: $\left( a,b \right)O\left( c,d \right)=\left( ac,bc+d \right)$ for all $\left( a,b \right),\left( c,d \right)\in {{R}_{0}}\times R$. Find the identity element in $A$. Last updated date: 18th Jul 2024 Total views: 450.3k Views today: 13.50k Verified 450.3k+ views Hint: Take any general element of ${{R}_{0}}$ and find the identity element of $R$ by forming the equations using properties of the identity element and then solve them. Similarly take any general element of $R$ and find the identity of ${{R}_{0}}$ by forming equations using properties of the identity element. We have the set $A={{R}_{0}}\times R$ where ${{R}_{0}}$ denotes the set of all non-zero numbers. We have a binary relation on $A$ defined as $\left( a,b \right)O\left( c,d \right)=\left( ac,bc+d \right)$ for all $\left( a,b \right),\left( c,d \right)\in {{R}_{0}}\times R$. We have to find the identity element in $A$. Let’s assume that the identity element of $A$ is of the form $\left( x,y \right)$. We know that any identity element has the property that any element operated with identity element returns the element itself, i.e., for any $\left( a,b \right)\in A$, we have $\left( a,b \right)O\left( x,y \right)=\left( x,y \right)O\left( a,b \right)=\left( a,b \right)$. We know that for all $\left( a,b \right),\left( c,d \right)\in {{R}_{0}}\times R$, we have $\left( a,b \right)O\left( c,d \right)=\left( ac,bc+d \right)$. Thus, we have $\left( a,b \right)O\left( x,y \right)=\left( ax,bx+y \right)=\left( a,b \right)$. Comparing the terms on both sides of the equation, we have $ax=a,bx+y=b$. We can clearly see that the solution of the above equations is $x=1,y=0$. Hence, we have $\left( x,y \right)=\left( 1,0 \right)$ as the identity of the given set $A={{R}_{0}}\times R$ where ${{R}_{0}}$ denotes the set of all non-zero numbers .
# CogAT® Question Type: Number Puzzle Sample & Tips In this post, we start with an overview of "Number Puzzles" question types. We also provide a sample question, accompanied by tips and strategies that every student can use to perform successfully on "Number Puzzles" questions during the official CogAT® exam. Number Puzzle questions are part of the quantitative battery on the CogAT. Number Series and Number Analogies are the other two subtests in this battery. ## Number Puzzles Questions: What To Expect? This subtest requires students to solve simple mathematical equations in order to make the amounts on either side of the equal sign the same. K-2 students are provided with pictures that represent math problems. ## How Many Questions To Expect on The Test? The amount of number puzzles questions on the test depends on the age of the student and the test level, as shown by the table below. Number Puzzles Level Number of Questions Grade Level Approximate Age Level 5/6 10 Kindergarten 5/6 Years Level 7 12 1st Grade 7 Years Level 8 14 2nd Grade 8 Years Level 9 16 3rd Grade 9 Years Level 10 16 4th Grade 10 Years Level 11 16 5th Grade 11 Years Level 12 18 6th Grade 12 Years Level 13-18 16 7th-12th Grade 13-18 Years # 2 + 10 = ? X 4 A. 12 B. 3 C. 2 D. 8 E. 5 Correct Answer: B Explanation: The first two numbers add up to 12 (10+2). Therefore, we need to find a number (to insert instead of the question mark) that multiplies with 4 to create a total of 12. The only number that can be multiplied by 4 to make 12 is 3. Therefore, B is the correct answer. ### CogAT Level 5/6 Number Puzzle Question Correct answer: A In the truck on the top row, there are 10 balls. In the truck on the bottom row, there are four balls in one compartment and two balls in another. Your child needs to figure out how many balls are needed in the compartment with the question mark so that both trucks (top and bottom row), have the same amount of balls. We can see that four balls are needed in order to make a total of 10 balls in the truck on the bottom row, because 4 balls from one compartment plus 2 balls from the other compartment results in a total of 6 balls (4+2=6), and we need four more balls in the train to have 10 balls in total (6+4=10). The only answer that shows 4 balls is option A. ## Number Puzzles:Tips and Strategies This question type requires your child to solve basic math equations, so practice with numbers and problem solving is essential. Make sure your child understands the meaning of “equal,” since the object is to supply the missing piece of information that will make two provided equations equal to one another. Buy a workbook that focuses on basic mathematical skills like addition, subtraction, multiplication, and division. Practicing multiplication facts could help improve your child’s accuracy and speed. You can also teach your child to approach the question by “plugging in” the answer choices and solving to see if the result is equal to the other equation in the question. Of course, it is also important to work with practice questions in order to familiarize your child with the content and structure of the actual test. Ask her to explain her reasoning for each question that she answers. This will allow you to praise and reinforce strong problem solving skills for correct answers and address misconceptions for incorrect answers. ## What's Next? Help your child become familiar with the questions on this challenging exam by downloading a free CogAT practice test. Learn more about other quantitative question types on the CoGAT, including Number Series, and Number Analogies. Also see critical teaching tips that cover other areas of the CogAT, including the nonverbal battery and the verbal battery. If you think you need more information and guidance about the CogAT, check out our ultimate guide on the test, and our in-depth article on CogAT scores. Also, learn everything you need to know about other tests that measure a child’s potential to learn in school, like the NNAT or the OLSAT.
Nikoismusic.com Blog What are algebra graphs called? # What are algebra graphs called? ## What are algebra graphs called? The eight most commonly used graphs are linear, power, quadratic, polynomial, rational, exponential, logarithmic, and sinusoidal. Each has a unique graph that is easy to visually differentiate from the rest. How do you plot a function on a graph? Steps for Sketching the Graph of the Function 1. Determine, whether function is obtained by transforming a simpler function, and perform necessary steps for this simpler function. 2. Determine, whether function is even, odd or periodic. 3. Find y-intercept (point f(0)). 4. Find x-intercepts (points where f(x)=0). What are the basic functions of algebra? In mathematics, an algebraic function is a function that can be defined as the root of a polynomial equation. Quite often algebraic functions are algebraic expressions using a finite number of terms, involving only the algebraic operations addition, subtraction, multiplication, division, and raising to a fractional power. ### How do you determine the function of a graph? Use the vertical line test to determine whether or not a graph represents a function. If a vertical line is moved across the graph and, at any time, touches the graph at only one point, then the graph is a function. If the vertical line touches the graph at more than one point, then the graph is not a function. What are the different types of algebraic graphs? Graphs come in all sorts of shapes and sizes. In algebra, there are 3 basic types of graphs you’ll see most often: linear, quadratic, and exponential. What is an example of a function in a graph? Line graphs can also be used to show how functions change. A function is just an equation that gives you a unique output for every input. For example, y=- 4/5x + 3 is a function because you’ll get a unique value for y when you put in any number for x.
# Factors of 504: Prime Factorization, Methods, and Examples The factors of 504Â are numbers that, when divided by 504, leave zero as the remainder. This means the numbers that ultimately divide the given number are named as their factors.Â The given number’s factors can be positive and negative, provided that the given number is achieved upon multiplication of two-factor integers. ### Factors of 504 Here are the factors of numberÂ 504. Factors of 504: 1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 18, 21, 24, 28, 36, 42, 56, 63, 72, 84, 126, 168, 252, and 504Â ### Negative Factors of 504 The negative factors of 504Â are similar to its positive aspects, just with a negative sign. Negative Factors of 504: –1, -2, -3, -4, -6, -7, -8, -9, -12, -14, -18, -21, -24, -28, -36, -42, -56, -63, -72, -84, -126, -168, -252, and -504Â ### Prime Factorization of 504 The prime factorization of 504Â is the way of expressing its prime factors in the product form. Prime Factorization: 2$^3$ x 3$^2$ x 7 In this article, we will learn about the factors of 504Â and how to find them using various techniques such as upside-down division, prime factorization, and factor tree. ## What Are the Factors of 504? The factors of 504 areÂ 1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 18, 21, 24, 28, 36, 42, 56, 63, 72, 84, 126, 168, 252, and 504.Â These numbers are the factors as they do not leave any remainder when divided by 504. The factors of 504Â are classified as prime numbers and composite numbers. The prime factors of the number 504 can be determined using the prime factorization technique. ## How To Find the Factors of 504? You can find the factors of 504Â by using the rules of divisibility. The divisibility rule states that any number, when divided by any other natural number, is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero. To find the factors of 504, create a list containing the numbers that are exactly divisible by 504 with zero remainders. One important thing to note is that 1 and 504 are 504’s factors as every natural number has 1 and the number itself as its factor. 1 is also called the universal factor of every number. The factors of 504 are determined as follows: $\dfrac{504}{1} = 504$ $\dfrac{504}{2} = 252$ $\dfrac{504}{3} = 168$ $\dfrac{504}{4} = 126$ $\dfrac{504}{6} = 84$ $\dfrac{504}{7} = 72$ $\dfrac{504}{8} = 63$ $\dfrac{504}{9} = 56$ $\dfrac{504}{12} = 42$ $\dfrac{504}{14} = 36$ $\dfrac{504}{18} = 28$ $\dfrac{504}{21} = 24$ Therefore,Â 1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 18, 21, 24, 28, 36, 42, 56, 63, 72, 84, 126, 168, 252, and 504Â are the factors of 504. ### Total Number of Factors of 504 For 504, there are twenty-fourÂ positive factors andÂ twenty-fourÂ negative ones. So in total, there are forty-eight factors of 504.Â To find the total number of factors of the given number, follow the procedure mentioned below: 1. Find the factorization/prime factorization of the given number. 2. Demonstrate the prime factorization of the number in the form of exponent form. 3. Add 1 to each of the exponents of the prime factor. 4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number. By following this procedure, the total number of factors of 504 is given as: Factorization of 504 isÂ 1 xÂ 2$^3$ x 3$^2$ x 7. The exponent of 1 and 7 is 1. The exponent of 2 is 3 whereas that of 3 is 2. Adding 1 to each and multiplying them together results in 48. Therefore, the total number of factors of 504 is 48. Twenty-four are positive, and twenty-four factors are negative. ### Important Notes Here are some essential points that must be considered while finding the factors of any given number: • The factor of any given number must be a whole number. • The factors of the number cannot be in the form of decimals or fractions. • Factors can be positive as well as negative. • Negative factors are the additive inverse of the positive factors of a given number. • The factor of a number cannot be greater than that number. • Every even number has 2 as its prime factor, the smallest prime factor. ## Factors of 504 by Prime Factorization The number 504Â is a composite. Prime factorization is a valuable technique for finding the number’s prime factors and expressing the number as the product of its prime factors. Before finding the factors of 504 using prime factorization, let us find out what prime factors are. Prime factors are the factors of any given number that are only divisible by 1 and themselves. To start the prime factorization of 504, start dividing by its most minor prime factor. First, determine that the given number is either even or odd. If it is an even number, then 2 will be the smallest prime factor. Continue splitting the quotient obtained until 1 is received as the quotient. The prime factorization of 504Â can be expressed as: 504 =Â 2$^3$ x 3$^2$ x 7 ## Factors of 504 in Pairs The factor pairs are the duplet of numbers that, when multiplied together, result in the factorized number. Factor pairs can be more than one depending on the total number of factors given. For 504, the factor pairs can be found as: 1 x 504 = 504 2 x 252 = 504Â 3 x 168 = 504 4 x 126 = 504 6 x 84 = 504 7 x 72 = 504 8 x 63 = 504 9 x 56 = 504 12 x 42 = 504 14 x 36 = 504 18 x 28 = 504 Â Â Â Â Â 21 x 24 = 504Â Â Â Â Â Â Â Â Â Â The possible factor pairs of 504Â are given asÂ (1, 504), (2, 252), (3, 168), (4, 126), (6, 84), (7, 72), (8, 63), (9, 56), (12, 42), (14, 36), (18, 28),Â and (21, 24). All these numbers in pairs, when multiplied, give 504 as the product. The negative factor pairs of 504 are given as: -1 x -504 = 504 -2 x -252 = 504Â -3 x -168 = 504 -4 x -126 = 504 -6 x -84 = 504 -7 x -72 = 504 -8 x -63 = 504 -9 x -56 = 504 -12 x -42 = 504 -14 x -36 = 504 -18 x -28 = 504 Â -21 x -24 = 504Â Â It is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign, due to which the resulting product is the original positive number. Therefore,Â 1, -2, -3, -4, -6, -7, -8, -9, -12, -14, -18, -21, -24, -28, -36, -42, -56, -63, -72, -84, -126, -168, -252, and -504Â Â are called negative factors of 504. The list of all the factors of 504, including positive as well as negative numbers, is given below. Factor list of 504:Â 1, –1, 2, -2, 3, -3, 4, -4, 6, -6, 7, -7, 8, -8, 9, -9, 12, -12, 14, -14, 18, -18, 21, -21, 24, -24, 28, -28, 36, -36, 42, -42, 56, -56, 63, -63, 72, -72, 84, -84, 126, -126, 168, -168, 252, -252, 504, and -504Â ## Factors of 504 Solved Examples To better understand the concept of factors, let’s solve some examples. ### Example 1 How many factors of 504 are there? ### Solution The total number of Factors of 504 is 24. Factors of 504 areÂ 1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 18, 21, 24, 28, 36, 42, 56, 63, 72, 84, 126, 168, 252, and 504. ### Example 2 Find the factors of 504 using prime factorization. ### Solution The prime factorization of 504 is given as: 504 $\div$ 2 = 252Â 252 $\div$ 2 = 1256 126 $\div$ 2 = 63 63 $\div$ 3 = 21 21 $\div$ 3 = 7Â Â Â Â 7 $\div$ 7 = 1Â So the prime factorization of 504 can be written as: 2$^3$ x 3$^2$ x 7Â = 504
Module 12: Linear Regression and Correlation # Testing the Significance of the Correlation Coefficient Barbara Illowsky & OpenStax et al. The correlation coefficient, r, tells us about the strength and direction of the linear relationship between x and y. However, the reliability of the linear model also depends on how many observed data points are in the sample. We need to look at both the value of the correlation coefficient r and the sample size n, together. We perform a hypothesis test of the “significance of the correlation coefficient” to decide whether the linear relationship in the sample data is strong enough to use to model the relationship in the population. The sample data are used to compute r, the correlation coefficient for the sample. If we had data for the entire population, we could find the population correlation coefficient. But because we have only have sample data, we cannot calculate the population correlation coefficient. The sample correlation coefficient, r, is our estimate of the unknown population correlation coefficient. • The symbol for the population correlation coefficient is ρ, the Greek letter “rho.” • ρ = population correlation coefficient (unknown) • r = sample correlation coefficient (known; calculated from sample data) The hypothesis test lets us decide whether the value of the population correlation coefficient ρ is “close to zero” or “significantly different from zero”. We decide this based on the sample correlation coefficient r and the sample size n. If the test concludes that the correlation coefficient is significantly different from zero, we say that the correlation coefficient is “significant.” Conclusion: There is sufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is significantly different from zero. What the conclusion means: There is a significant linear relationship between x and y. We can use the regression line to model the linear relationship between x and y in the population. If the test concludes that the correlation coefficient is not significantly different from zero (it is close to zero), we say that correlation coefficient is “not significant.” Conclusion: “There is insufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is not significantly different from zero.” What the conclusion means: There is not a significant linear relationship between x and y. Therefore, we CANNOT use the regression line to model a linear relationship between x and y in the population. #### Note • If r is significant and the scatter plot shows a linear trend, the line can be used to predict the value of y for values of x that are within the domain of observed x values. • If r is not significant OR if the scatter plot does not show a linear trend, the line should not be used for prediction. • If r is significant and if the scatter plot shows a linear trend, the line may NOT be appropriate or reliable for prediction OUTSIDE the domain of observed x values in the data. ## Performing the Hypothesis Test • Null Hypothesis: H0: ρ = 0 • Alternate Hypothesis: Ha: ρ ≠ 0 ### What the Hypotheses Mean in Words • Null Hypothesis H0: The population correlation coefficient IS NOT significantly different from zero. There IS NOT a significant linear relationship(correlation) between x and y in the population. • Alternate Hypothesis Ha: The population correlation coefficient IS significantly DIFFERENT FROM zero. There IS A SIGNIFICANT LINEAR RELATIONSHIP (correlation) between x and y in the population. ### Drawing a Conclusion There are two methods of making the decision. The two methods are equivalent and give the same result. • Method 1: Using the p-value • Method 2: Using a table of critical values In this chapter of this textbook, we will always use a significance level of 5%, α = 0.05 #### Note Using the p-value method, you could choose any appropriate significance level you want; you are not limited to using α = 0.05. But the table of critical values provided in this textbook assumes that we are using a significance level of 5%, α = 0.05. (If we wanted to use a different significance level than 5% with the critical value method, we would need different tables of critical values that are not provided in this textbook.) ## Method 1: Using a p-value to make a decision To calculate the p-value using LinRegTTEST: • On the LinRegTTEST input screen, on the line prompt for β or ρ, highlight “≠ 0” • The output screen shows the p-value on the line that reads “p =”. • (Most computer statistical software can calculate thep-value.) If the p-value is less than the significance level (α = 0.05) • Decision: Reject the null hypothesis. • Conclusion: “There is sufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is significantly different from zero.” If the p-value is NOT less than the significance level (α = 0.05) • Decision: DO NOT REJECT the null hypothesis. • Conclusion: “There is insufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is NOT significantly different from zero.” Calculation Notes: • You will use technology to calculate the p-value. The following describes the calculations to compute the test statistics and the p-value: • The p-value is calculated using a t-distribution with n – 2 degrees of freedom. • The formula for the test statistic is $displaystyle{t}=frac{{{r}sqrt{{{n}-{2}}}}}{sqrt{{{1}-{r}^{{2}}}}}$. The value of the test statistic, t, is shown in the computer or calculator output along with the p-value. The test statistic t has the same sign as the correlation coefficient r. • The p-value is the combined area in both tails. An alternative way to calculate the p-value (p) given by LinRegTTest is the command 2*tcdf(abs(t),10^99, n-2) in 2nd DISTR. ## Method 2: Using a table of Critical Values to make a decision The 95% Critical Values of the Sample Correlation Coefficient Table can be used to give you a good idea of whether the computed value of is significant or not. Compare r to the appropriate critical value in the table. If r is not between the positive and negative critical values, then the correlation coefficient is significant. Ifr is significant, then you may want to use the line for prediction. ### Example Suppose you computed r = 0.801 using n = 10 data points.df = n – 2 = 10 – 2 = 8. The critical values associated with df = 8 are -0.632 and + 0.632. If r < negative critical value or r > positive critical value, then r is significant. Since r = 0.801 and 0.801 > 0.632, r is significant and the line may be used for prediction. If you view this example on a number line, it will help you. r is not significant between -0.632 and +0.632. r = 0.801 > +0.632. Therefore, r is significant. ### try it For a given line of best fit, you computed that r = 0.6501 using n = 12 data points and the critical value is 0.576. Can the line be used for prediction? Why or why not? If the scatter plot looks linear then, yes, the line can be used for prediction, because r > the positive critical value. ### Example Suppose you computed r = –0.624 with 14 data points. df = 14 – 2 = 12. The critical values are –0.532 and 0.532. Since –0.624 < –0.532, r is significant and the line can be used for prediction r = –0.624-0.532. Therefore, r is significant. ### try it For a given line of best fit, you compute that r = 0.5204 using n = 9 data points, and the critical value is 0.666. Can the line be used for prediction? Why or why not? No, the line cannot be used for prediction, because r < the positive critical value. ### Example 3 Suppose you computed r = 0.776 and n = 6. df = 6 – 2 = 4. The critical values are –0.811 and 0.811. Since –0.811 < 0.776 < 0.811, r is not significant, and the line should not be used for prediction. –0.811 < r = 0.776 < 0.811. Therefore, r is not significant. ### Try it For a given line of best fit, you compute that r = –0.7204 using n = 8 data points, and the critical value is = 0.707. Can the line be used for prediction? Why or why not? Yes, the line can be used for prediction, because r < the negative critical value. ### Example Suppose you computed the following correlation coefficients. Using the table at the end of the chapter, determine if r is significant and the line of best fit associated with each r can be used to predict a y value. If it helps, draw a number line. 1. r = –0.567 and the sample size, n, is 19. The df = n – 2 = 17. The critical value is –0.456. –0.567 < –0.456 so r is significant. 2. r = 0.708 and the sample size, n, is nine. The df = n – 2 = 7. The critical value is 0.666. 0.708 > 0.666 so r is significant. 3. r = 0.134 and the sample size, n, is 14. The df = 14 – 2 = 12. The critical value is 0.532. 0.134 is between –0.532 and 0.532 so r is not significant. 4. r = 0 and the sample size, n, is five. No matter what the dfs are, r = 0 is between the two critical values so r is not significant. ### try it For a given line of best fit, you compute that r = 0 using n = 100 data points. Can the line be used for prediction? Why or why not? No, the line cannot be used for prediction no matter what the sample size is. ## Assumptions in Testing the Significance of the Correlation Coefficient Testing the significance of the correlation coefficient requires that certain assumptions about the data are satisfied. The premise of this test is that the data are a sample of observed points taken from a larger population. We have not examined the entire population because it is not possible or feasible to do so. We are examining the sample to draw a conclusion about whether the linear relationship that we see between x and y in the sample data provides strong enough evidence so that we can conclude that there is a linear relationship between x and y in the population. The regression line equation that we calculate from the sample data gives the best-fit line for our particular sample. We want to use this best-fit line for the sample as an estimate of the best-fit line for the population. Examining the scatterplot and testing the significance of the correlation coefficient helps us determine if it is appropriate to do this. The assumptions underlying the test of significance are: • There is a linear relationship in the population that models the average value of y for varying values of x. In other words, the expected value of y for each particular value lies on a straight line in the population. (We do not know the equation for the line for the population. Our regression line from the sample is our best estimate of this line in the population.) • The y values for any particular x value are normally distributed about the line. This implies that there are more y values scattered closer to the line than are scattered farther away. Assumption (1) implies that these normal distributions are centered on the line: the means of these normal distributions of y values lie on the line. • The standard deviations of the population y values about the line are equal for each value of x. In other words, each of these normal distributions of yvalues has the same shape and spread about the line. • The residual errors are mutually independent (no pattern). • The data are produced from a well-designed, random sample or randomized experiment. The y values for each x value are normally distributed about the line with the same standard deviation. For each x value, the mean of the y values lies on the regression line. More y values lie near the line than are scattered further away from the line. ## Concept Review Linear regression is a procedure for fitting a straight line of the form $displaystylehat{{y}}={a}+{b}{x}$ to data. The conditions for regression are: • Linear: In the population, there is a linear relationship that models the average value of y for different values of x. • Independent: The residuals are assumed to be independent. • Normal: The y values are distributed normally for any value of x. • Equal variance: The standard deviation of the y values is equal for each x value. • Random: The data are produced from a well-designed random sample or randomized experiment. The slope b and intercept a of the least-squares line estimate the slope β and intercept α of the population (true) regression line. To estimate the population standard deviation of y, σ, use the standard deviation of the residuals, s. $displaystyle{s}=sqrt{{frac{{{S}{S}{E}}}{{{n}-{2}}}}}$ The variable ρ (rho) is the population correlation coefficient. To test the null hypothesis H0: ρ = hypothesized value, use a linear regression t-test. The most common null hypothesis is H0: ρ = 0 which indicates there is no linear relationship between x and y in the population. The TI-83, 83+, 84, 84+ calculator function LinRegTTest can perform this test (STATS TESTS LinRegTTest). ## Formula Review Least Squares Line or Line of Best Fit: $displaystylehat{{y}}={a}+{b}{x}$ where a = y-intercept, b = slope Standard deviation of the residuals: $displaystyle{s}=sqrt{{frac{{{S}{S}{E}}}{{{n}-{2}}}}}$where SSE = sum of squared errors n = the number of data points ## License Testing the Significance of the Correlation Coefficient Copyright © 2018 by Barbara Illowsky & OpenStax et al. is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.
How do you solve 4x^3 + 3x^2 + x + 2 = 0? Apr 29, 2017 $x = - 1 \text{ }$ or $\text{ } x = \frac{1}{8} \pm \frac{\sqrt{31}}{8} i$ Explanation: Given: $4 {x}^{3} + 3 {x}^{2} + x + 2 = 0$ Note that: $- 4 + 3 - 1 + 2 = 0$ So we can deduce that $x = - 1$ is a solution and $\left(x + 1\right)$ a factor: $0 = 4 {x}^{3} + 3 {x}^{2} + x + 2$ $\textcolor{w h i t e}{0} = \left(x + 1\right) \left(4 {x}^{2} - x + 2\right)$ The remaining quadratic is in the form $a {x}^{2} + b x + c$ with $a = 4$, $b = - 1$ and $c = 2$ This has discriminant $\Delta$ given by the formula: $\Delta = {b}^{2} - 4 a c = {\left(\textcolor{b l u e}{- 1}\right)}^{2} - 4 \left(\textcolor{b l u e}{4}\right) \left(\textcolor{b l u e}{2}\right) = 1 - 32 = - 31$ Since $\Delta < 0$ this quadratic has no real zeros. We can find Complex zeros for it using the quadratic formula: $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ $\textcolor{w h i t e}{x} = \frac{- b \pm \sqrt{\Delta}}{2 a}$ $\textcolor{w h i t e}{x} = \frac{1 \pm \sqrt{- 31}}{8}$ $\textcolor{w h i t e}{x} = \frac{1}{8} \pm \frac{\sqrt{31}}{8} i$ where $i$ is the imaginary unit.
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. ## Get ready for Algebra 1 ### Course: Get ready for Algebra 1>Unit 1 Lesson 3: One-step multiplication & division equations # One-step multiplication & division equations Learn to solve equations like "4x = 20" or "y/3 = 7". Based on our understanding of the balance beam model, we know that to keep a true equation, we always have to do the same thing to both sides of an equation. But how do we know what to do to both sides of the equation? ## Multiplication and division are inverse operations Here's an example of how division is the inverse operation of multiplication: If we start with 7, multiply by 3, then divide by 3, we get back to 7: 7, dot, 3, divided by, 3, equals, 7 Here's an example of how multiplication is the inverse operation of division: If we start with 8, divide by 4, then multiply by 4, we get back to 8: 8, divided by, 4, dot, 4, equals, 8 ## Solving a multiplication equation using inverse operations Let's think about how we can solve for t in the following equation: 6, t, equals, 54 We want to get t by itself on the left hand side of the equation. So, what can we do to undo multiplying by 6? We should divide by 6 because the inverse operation of multiplication is division! Here's how dividing by 6 on each side looks: \begin{aligned} 6t &= 54 \\\\ \dfrac{6t}{\blueD{6}} &= \dfrac{54}{\blueD{ 6}}~~~~~~~~~~\small\gray{\text{Divide each side by six.}} \\\\ t &= \greenD{9}~~~~~~~~~~\small\gray{\text{Simplify.}} \end{aligned} ### Let's check our work. It's always a good idea to check our solution in the original equation to make sure we didn't make any mistakes: $\qquad$ \begin{aligned} 6t &= 54 \\ 6 \cdot \greenD9 &\stackrel{\large?}{=} 54\\ 54 &= 54 \end{aligned} Yes, t, equals, start color #1fab54, 9, end color #1fab54 is a solution! ## Solving a division equation using inverse operations Now, let's try to solve a slightly different type of equation: start fraction, x, divided by, 5, end fraction, equals, 7 We want to get x by itself on the left hand side of the equation. So, what can we do to cancel out dividing by 5? We can multiply by 5 because the inverse operation of division is multiplication! Here's how multiplying by 5 on each side looks: \begin{aligned} \dfrac x5 &= 7 \\\\ \dfrac x5 \cdot \blueD{5} &= 7 \cdot \blueD{5}~~~~~~~~~~\small\gray{\text{Multiply each side by five.}} \\\\ x &= \greenD{35}~~~~~~~~~~\small\gray{\text{Simplify.}} \end{aligned} ### Let's check our work. $\qquad$ \begin{aligned} \dfrac x5 &= 7 \\\\ \dfrac{\greenD{35}}{5} &\stackrel{\large?}{=} 7\\\\ 7 &= 7 \end{aligned} Yes, x, equals, start color #1fab54, 35, end color #1fab54 is a solution! ## Summary of how to solve multiplication and division equations Awesome! We just solved a multiplication equation and a division equation. Let's summarize what we did: Type of equationExampleFirst step Multiplication equation6, t, equals, 54Divide each side by six. Division equationstart fraction, x, divided by, 5, end fraction, equals, 7Multiply each side by five. ## Let's try solving equations. Equation A • Current Which operation would help solve for w? 8, w, equals, 72 w, equals ## Want to join the conversation? • So even if you have a fraction it just means to divide but you should multiply? • Yes exactly so in the last problem you would cancel out the 4 by multiplying it by 4 and what ever you do on one side of the equal sign you have to do on the other so you would multiply 18 by 4 to get 72. • i dont get this, it says "what you do to one side, you do to the other" so if you have x/5=7 you have to multiply each side by 5, so is it 5x5=25 and 7x5=35? if so what happened to the 25? • When x/5 is multiplied by 5, the 5's are not multiplied together to give 25. Rather, multiplication by 5 undoes (or cancels out) the division by 5 that occurs in the expression x/5. So there's no 25 here. The final result is x = 35. • the only one I don't understand is the first one,any tips? • You divide by 8 on each sides so 72/8 =9 • Is there any way i cant get taught better for inverse operation • If you are looking for another video or some reference to use for inverse operation, I would recommend using Math Antics or Mathceraptops (YouTube channel).This are the places i personally do to for second opinions. I am not sure if Math antics has a video about inverse operation but Mathceraptops sure do. • If a letter is close to a number how do you solve the equation? • I'm assuming your referring to something such as 7x = 14 in this case, you would use 7x to say that 7 of that variable is equal to 14 in which case x = 2. This works because 7x is used to refer to multiplication easily during algebra since if you tried 7 x X = 14, it can get a bit confusing. While you can also use 7 ⋅ X = 14 (⋅ being another form of multiplication symbol) but it's much less common. TLDR; when a number is by a variable it means to multiply the variable by the number. • Thank you, Khan! You made math easier for me! When get rich, I will donate to Khan Academy to help many people, students, and children that do not have easy access to education. • You post this in questions not tips and thanks. Did you know that? • About the "what you do to one side, you do to the other". To understand the logic behind this, first let's imagine a beam balance in front of you, it has two blocks on each side each weighing 200 grams, thus the two sides are equal. However if you add another 100g block on one side, the balance becomes uneven, as in 200g < 300g . To solve this we add another 100g block to the other side, so the beam balance becomes even again, as in 300g = 300g . That's why what you do on one side you also do to the other, when solving an equation. TL:DR; We do that to avoid a false equation.
## Tension question A 0.16 kg meter stick is held perpendicular to a vertical wall by a 2.6 m string going from the wall to the far end of the stick. A.Find the tension in the string. B.Find the tension in a 2.0 {\rm m} string. • A 0.10 kg meter stick is held perpendicular to a vertical wall by a 2.6 m string going from the wall to the far end of the stick. Find the tension in the string. (How do you do this?) If a shorter string is used, will its tension be greater than, less than, or the same as that found in part (a)? Find the tension in a 1.9 string. You need geometry to give you the angle that the string makes w/r/t the stick. T = arccos(1m/2.6m) = 67.4º Now let's look at the stick and sum the moments about the wall-end: If the tension in the string is T, then 0.1kg•9.8m/s²•0.5m = T•1m•sin67.4º (That is, the vertical component at 1m must balance the stick weight at 0.5m.) T = 0.53 N • Anonymous commented • T = 0.169.8 =1.568 N B) torque = 21.568 =3.136 • Anonymous commented • We must find first the angle of elevation of the string with respect to the meter stick. Know that the string (hypotenuse), the meter stick (adjacent side) and the wall (opposite side) form a right triangle, so your typical trigonometry will work. cos? = a/h ? = arccos(a/h) ? = arccos(1 m / 2.6 m) ? = 67.38° It's now time to look at the forces in the system. The weight of the meter stick is balanced by the vertical component of the tension in the string. So we can say that the vertical component of the tensile force is: (0.16 kg)(9.8 m/s^2) = 1.56 N We can now find the actual tension in the string. The vertical component of the tension is the opposite side, while the tension is the hypotenuse. sin? = Ty / T T = Ty / sin? T = (1.56 N) / (sin 67.38°)
# Newton's Method Calculator for a System of two Equations An interactive calculator using Newton's method [1] to approximate the solutions of a system of two equations in two variables is presented. It approximates the solutions, if there is any, and gives a table of all values of iterations for educational purposes. ## Newton's Method for Systems of Equations Newton's method is numerical method used to find the roots of an equation by iterations starting from an approximate initial solution. When dealing with a system of equations, the method extends naturally by considering the Jacobian matrix and its determinant. ## One Variable Newton's Method Suppose that we need to solve the following equation $f(x) = 0$ Taylor expansion of $$f(x+\Delta x)$$ is given by $f(x+\Delta x) \approx f(x) + \Delta x f'(x)$ We now solve $$f(x+\Delta x) = 0$$ which gives $f(x) + \Delta x f'(x) = 0$ which gives $\Delta x \approx - \dfrac{f(x)}{f'(x)}$ Suppose we know an approximate value $$x_n$$ of the root of the equation, the approximate root $$x_{n+1}$$ defined by $\Delta x = x_{n+1} - x_n$ is given by $x_{n+1} \approx x_{n} - \dfrac{f(x)}{f'(x)}$ ## System of Equations and the Jacobian Matrix Consider a system of two equations in two variables $$x$$ and $$y$$: \begin{align} f(x, y) &= 0 \\ g(x, y) &= 0 \end{align} The Jacobian matrix $$J$$ of the system is given by: $J = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \end{bmatrix}$ ## Newton's Method Update The update formulas for Newton's method for a system of equations are given by: \begin{aligned} \Delta x &= \frac{-f \cdot g_y + g \cdot f_y}{\text{D}} \\\\ \Delta y &= \frac{-g \cdot f_x + f \cdot g_x}{\text{D}} \end{aligned} hence \begin{aligned} x_{n+1} &\approx x_n + \frac{-f \cdot g_y + g \cdot f_y}{\text{D}} \\\\ y_{n+1} &\approx y_n + \frac{-g \cdot f_x + f \cdot g_x}{\text{D}} \end{aligned} Where $$f$$ and $$g$$ are the functions evaluated at the current $$(x_n, y_n)$$ values. $$f_x, f_y, g_x, g_y$$ are the partial derivatives of $$f$$ and $$g$$ with respect to $$x$$ and $$y$$, respectively. $$\text{D} = f_x \cdot g_y - f_y \cdot g_x$$ is the determinant of the Jacobian matrix. ## Iterative Process Newton's method iteratively updates the variables $$x$$ and $$y$$ using the above formulas until a stopping criterion is met. The common stopping criteria include: Convergence tolerance: Stop when the difference between consecutive iterations is below a certain threshold. Maximum iterations: Stop after a maximum number of iterations is reached. 1. Initialization: Start with initial guesses for $$x$$ and $$y$$: One way to obtain close initial guesses for the solution to the system is to graph $$f(x,y)$$ and $$g(x,y)$$, approximate their points of interscetion use them as initial guesses. 2. Evaluate Functions and Derivatives: Compute $$f(x, y)$$, $$g(x, y)$$, and their partial derivatives at the current $$(x, y)$$ values. 3. Calculate Determinant: Compute the determinant of the Jacobian matrix. 4. Update Variables: Use the Newton's method update formulas to compute $$\Delta x$$ and $$\Delta y$$. 5. Iterate: Update $$x$$ and $$y$$ using $$\Delta x$$ and $$\Delta y$$, and repeat until convergence or maximum iterations are reached. 6. The tolerance $$\epsilon$$ is used to test the absolute value of $$f(x,y)$$ and $$g(x,y)$$ as follows: When $$\|f(x,y)\| \lt \epsilon$$ and $$\|g(x,y)\| \lt \epsilon$$, the iteration process stops. 7. The calculator approximate one solution at the time. Newton's method provides a robust and efficient way to approximate the solutions of a system of equations in two variables, given that the initial guesses are close enough to the actual solutions and the functions are differentiable in the neighborhood of the solutions. ## Results Table including the iteration values of $$x, y, f(x,y)$$ and $$g(x,y)$$. Iteration $$x$$ $$y$$ $$f(x, y)$$ $$g(x, y)$$
# Derivative of Cos he derivative of the cosine function (cos(x)) is the negative sine function (-sin(x)). In mathematical notation, it can be represented as: d/dx(cos(x)) = -sin(x) This means that if you have a function y = cos(x), the rate at which y changes with respect to x (its derivative) is given by -sin(x). In simpler terms, when you differentiate the cosine function, you get the negative sine function. It tells you how the height (y-value) of the cos curve changes as you move along the x-axis. The derivative, -sin(x), represents the slope or rate of change of the cos function at any given point. Here are three examples that demonstrate the application of the derivative of the cosine function: ## Example 1: Consider the function y = cos(x). To find the derivative, we differentiate the cosine function with respect to x: dy/dx = d/dx(cos(x)) = -sin(x) So, the derivative of y = cos(x) is dy/dx = -sin(x). ## Example 2: Let’s find the derivative of y = 3cos(x). Again, we differentiate the cosine function and multiply by the constant factor: dy/dx = 3 * d/dx(cos(x)) = 3(-sin(x)) Therefore, the derivative of y = 3cos(x) is dy/dx = -3sin(x). ## Example 3: Consider the function y = cos(2x). To find the derivative, we differentiate the cosine function and apply the chain rule: dy/dx = d/dx(cos(2x)) = -sin(2x) 2 Thus, the derivative of y = cos(2x) is dy/dx = -2sin(2x). In each example, we differentiate the cosine function with respect to x, and the resulting derivative is the negative sine function. We can apply additional factors or the chain rule when necessary. The derivative gives us information about the rate of change or slope of the cosine function at different points.
# Area Model Algebra_面積模型與代數下載嵌入關閉嵌入模擬教學執行複本利用 HTML 來嵌入模擬教學執行複本。您可以在 HTML 中改變已嵌入模擬教學的寬度及高度。嵌入啓動模擬教學之圖樣 • Polynomials • Factors • Products ### 學習目標 • Develop and justify a method to use the area model to determine the product of a monomial and a binomial or the product of two binomials. • Factor an expression, including expressions containing a variable. • Recognize that area represents the product of two numbers and is additive. • Represent a multiplication problem as the area of a rectangle, proportionally or using generic area. • Develop and justify a strategy to determine the product of two multi-digit numbers by representing the product as an area or the sum of areas. ### 標準對齊 #### 共用核心 - 數學 3.MD.C.7 Relate area to the operations of multiplication and addition. 3.MD.C.7c Use tiling to show in a concrete case that the area of a rectangle with whole-number side lengths a and b + c is the sum of a × b and a × c. Use area models to represent the distributive property in mathematical reasoning. 3.MD.C.7d Recognize area as additive. Find areas of rectilinear figures by decomposing them into non-overlapping rectangles and adding the areas of the non-overlapping parts, applying this technique to solve real world problems. 4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. 6.EE.A.3 Apply the properties of operations to generate equivalent expressions. For example, apply the distributive property to the expression 3 (2 + x) to produce the equivalent expression 6 + 3x; apply the distributive property to the expression 24x + 18y to produce the equivalent expression 6 (4x + 3y); apply properties of operations to y + y + y to produce the equivalent expression 3y. 6.EE.A.4 Identify when two expressions are equivalent (i.e., when the two expressions name the same number regardless of which value is substituted into them). For example, the expressions y + y + y and 3y are equivalent because they name the same number regardless of which number y stands for.. 6.NS.B.4 Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1-100 with a common factor as a multiple of a sum of two whole numbers with no common factor. For example, express 36 + 8 as 4 (9 + 2).. 7.EE.A.1 Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients. 7.NS.A.2c Apply properties of operations as strategies to multiply and divide rational numbers. 8.EE.C.7b Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms. ### 教學提示模擬教學的控制概觀、模型簡化與引導學生思考 ( PDF ). ### 老師提供的活動 Expandiendo y Factorizando Expresiones Algebraicas Amanda McGarry (Traducción de Mayra Tavares) 國中指引數學 Games Remote Lesson ideas Trish Loeblein 國中 K-5 Remote Expanding and Factoring Algebraic Expressions Amanda McGarry 國中指引數學 Algebraic Expressions Amanda McGarry 國中指引數學 Expresiones Algebraicas Amanda McGarry (Traducción por Mayra Tavares) 國中指引數學 Exploring Factoring as Undoing the Distributive Property Sarah Hampton 高中 SECUNDARIA: Alineación PhET con programas de la SEP México (2011 y 2017) Diana López 國中 PREPARATORIA: Alineación de PhET con programas de la DGB México (2017) Diana López 高中 Middle School Math Sim Alignment Amanda McGarry 國中其它數學 Using the area model with expressions Jennifer Knudsen, Teresa Lara-Meloy, Amanda McGarry 國中討論數學 Usando el Modelo de Áreas con Expresiones Jennifer Knudsen, Teresa Lara-Meloy, Amanda McGarry (Traducción Mayra Tavares) 國中討論數學 Modelo de Áreas, Cambio de Expresiones Andrés Gualberto Palomo Cuevas 高中 Remote projectile motion alya qasem 大學-簡介實驗室物理 Would You Rather ... 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# Lesson Rationalizing The Denominator Containing a Radical Algebra -> Algebra -> Radicals -> Lesson Rationalizing The Denominator Containing a Radical Log On Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Algebra: Radicals -- complicated equations involving roots Solvers Lessons Answers archive Quiz In Depth This Lesson (Rationalizing The Denominator Containing a Radical) was created by by solver91311(17077) : View Source, ShowAbout solver91311: Logistics Engineer, Trainer, Public Speaker, US Navy - retired. E-mail me at reecejo@gmail.com, $20/hr,$2-5 per detailed solution depending on complexity. PayPal Rationalizing Denominators Containing Radicals Numbers that have rational denominators are considered to be in simpler form than numbers with irrational denominators. For example: is considered to be simpler form than even though, as we shall see, these two numbers are equal. Hence, the goal of the process of rationalizing the denominator is to change the irrational denominator to a rational one without changing the value of the overall fraction. There are two forms that we will deal with in this lesson but both of them depend on the idea of the multiplicative identity, one of the first concepts we learn in algebra. For any number a we know that: : Multiplicative Identity From which it follows that: Which is to say, you don't change the value of anything if you multiply it by 1, and anything divided by itself is 1. The first form we will deal with is, like the example provided above, an expression in the form of a fraction that contains single term radical in the denominator. For example, consider the fraction: In order to eliminate the radical in the denominator we will make use of the following definition of a radical: From which we can see that it would be helpful if we could multiply the denominator of the example fraction by . However, simply multiplying the denominator by something other than 1 would change the value of the fraction and, as previously stated, that is something we want to avoid. But using the idea of the multiplicative identity, we can multiply the entire fraction by 1 without changing its value, and we can use the idea that anything divided by itself is equal to 1 to develop an appropriate multiplier that is equal to 1. For this first example, we will use: as the multiplier, thus: Now just perform the multiplication operation just like multiplying any other fractions, i.e. numerator times numerator and denominator times denominator: Which result is fraction with a rational denominator that is numerically equivalent to the expression we started with. That's fine for square roots, but what about higher order radical indexes -- cube, 4th, or 5th roots for example? Let's look at: Here we need to realize that From which we can see that the desired multiplier for the denominator in the example is: and the multiplier for the entire fraction is then: and then we have: *********************************************************************** The second form we will deal with is when we have a binomial expression in the denominator where one or both of the terms is a radical. For example: Multiplying the denominator by itself will not work in this case because: And we are no better off than when we started. Fortunately there is a simple technique that eliminates this problem. What we need to do is multiply by the conjugate. Please recall the Difference of Two Squares factorization: The two factors shown above are called conjugates. You obtain the conjugate of a binomial by changing the sign in the middle. This is important to us here because, as you can see from the Difference of Two Squares factorization, multiplying a binomial by its conjugate results in the difference of two squares. Here's how the technique is applied. Again, we must multiply the entire fraction by 1, but this time in the form of the conjugate of the denominator divided by itself, thus: So just multiply: These techniques should allow you to rationalize denominators containing radicals for any problems you are likely to encounter in Algebra II through Calculus. If you have questions on this lesson, feel free to write: reecejo@gmail.com John This lesson has been accessed 20676 times.
Share Explore BrainMass Difference Between Linear Equations and Linear Inequalities Consider first a linear equation of the form 2x - 5y = 8. Now choose a linear inequality of the form 5y < 2x - 8 or 5y > 2x - 8. What are the major differences between the linear equation graph and the linear inequality graph? Solution Preview Consider first a linear equation of the form 2x - 5y = 8 The equation 2x - 5y = 8 is the equation of a line. You can rearrange it so that it's in slope-intercept form: 2x - 5y = 8 -5y = -2x + 8 y = (2/5)x - (8/5) This line has a y-intercept of -8/5 and a slope of 2/5. The graph is below. It is made up of all the points (x, y) for which the equation holds. For example, the point (1, -6/5) is a point on the line because 2(1) - 5(-6/5) = 8 is true. The point (0, 8) is not on the line because 2(0) - 7(8) = 8 is not true. Now choose a linear inequality of the form 5 < 2x - 8 or 5 > 2x - 8 Let's look at the first one. You can write this as y < ... Solution Summary The solution uses an example of a linear equation and an example of a linear inequality to demonstrate how to graph each kind of relation and the differences between the graphs. \$2.19
# IB DP Maths Topic 6.2 Derivatives of xn , sinx , cosx , tanx HL Paper 2 ## Question The diagram shows the plan of an art gallery a metres wide. [AB] represents a doorway, leading to an exit corridor b metres wide. In order to remove a painting from the art gallery, CD (denoted by L ) is measured for various values of $$\alpha$$ , as represented in the diagram. If $$\alpha$$ is the angle between [CD] and the wall, show that $$L = \frac{a }{{\sin \alpha }} + \frac{b}{{\cos \alpha }}{\text{, }}0 < \alpha < \frac{\pi }{2}$$. [3] a. If a = 5 and b = 1, find the maximum length of a painting that can be removed through this doorway. [4] b. Let a = 3k and b = k . Find $$\frac{{{\text{d}}L}}{{{\text{d}}\alpha }}$$. [3] c. Let a = 3k and b = k Find, in terms of k , the maximum length of a painting that can be removed from the gallery through this doorway. [6] d. Let a = 3k and b = k Find the minimum value of k if a painting 8 metres long is to be removed through this doorway. [2] e. ## Markscheme $$L = {\text{CA}} + {\text{AD}}$$ M1 $${\text{sin}}\alpha {\text{ = }}\frac{a}{{{\text{CA}}}} \Rightarrow {\text{CA}} = \frac{a}{{\sin \alpha }}$$ A1 $$\cos \alpha = \frac{b}{{{\text{AD}}}} \Rightarrow {\text{AD}} = \frac{b}{{\cos \alpha }}$$ A1 $$L = \frac{a}{{\sin \alpha }} + \frac{b}{{\cos \alpha }}$$ AG [2 marks] a. $$a = 5{\text{ and }}b = 1 \Rightarrow L = \frac{5}{{\sin \alpha }} + \frac{1}{{\cos \alpha }}$$ METHOD 1 (M1) minimum from graph $$\Rightarrow L = 7.77$$ (M1)A1 minimum of gives the max length of the painting R1 [4 marks] METHOD 2 $$\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = \frac{{ – 5\cos \alpha }}{{{{\sin }^2}\alpha }} + \frac{{\sin \alpha }}{{{{\cos }^2}\alpha }}$$ (M1) $$\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = 0 \Rightarrow \frac{{{{\sin }^3}\alpha }}{{{{\cos }^3}\alpha }} = 5 \Rightarrow \tan \alpha = \sqrt[{3{\text{ }}}]{5}{\text{ }}(\alpha = 1.0416…)$$ (M1) minimum of gives the max length of the painting R1 maximum length = 7.77 A1 [4 marks] b. $$\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = \frac{{ – 3k\cos \alpha }}{{{{\sin }^2}\alpha }} + \frac{{k\sin \alpha }}{{{{\cos }^2}\alpha }}\,\,\,\,\,{\text{(or equivalent)}}$$ M1A1A1 [3 marks] c. $$\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = \frac{{ – 3k{{\cos }^3}\alpha + k{{\sin }^3}\alpha }}{{{{\sin }^2}\alpha {{\cos }^2}\alpha }}$$ (A1) $$\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = 0 \Rightarrow \frac{{{{\sin }^3}\alpha }}{{{{\cos }^3}\alpha }} = \frac{{3k}}{k} \Rightarrow \tan \alpha = \sqrt[3]{3}\,\,\,\,\,(\alpha = 0.96454…)$$ M1A1 $$\tan \alpha = \sqrt[3]{3} \Rightarrow \frac{1}{{\cos \alpha }} = \sqrt {1 + \sqrt[3]{9}} \,\,\,\,\,(1.755…)$$ (A1) $${\text{and }}\frac{1}{{\sin \alpha }} = \frac{{\sqrt {1 + \sqrt[3]{9}} }}{{\sqrt[3]{3}}}\,\,\,\,\,(1.216…)$$ (A1) $$L = 3k\left( {\frac{{\sqrt {1 + \sqrt[3]{9}} }}{{\sqrt[3]{3}}}} \right) + k\sqrt {1 + \sqrt[3]{9}} \,\,\,\,\,(L = 5.405598…k)$$ A1 N4 [6 marks] d. $$L \leqslant 8 \Rightarrow k \geqslant 1.48$$ M1A1 the minimum value is 1.48 [2 marks] e. ## Examiners report Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC. In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for $$\alpha$$ and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question. a. Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC. In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for $$\alpha$$ and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question. b. Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC. In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for $$\alpha$$ and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question. c. Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC. In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for $$\alpha$$ and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question. d. Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC. In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for $$\alpha$$ and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question. e. ## Question The function f has inverse $${f^{ – 1}}$$ and derivative $$f'(x)$$ for all $$x \in \mathbb{R}$$. For all functions with these properties you are given the result that for $$a \in \mathbb{R}$$ with $$b = f(a)$$ and $$f'(a) \ne 0$$ $({f^{ – 1}})'(b) = \frac{1}{{f'(a)}}.$ Verify that this is true for $$f(x) = {x^3} + 1$$ at x = 2. [6] a. Given that $$g(x) = x{{\text{e}}^{{x^2}}}$$, show that $$g'(x) > 0$$ for all values of x. [3] b. Using the result given at the start of the question, find the value of the gradient function of $$y = {g^{ – 1}}(x)$$ at x = 2. [4] c. (i) With f and g as defined in parts (a) and (b), solve $$g \circ f(x) = 2$$. (ii) Let $$h(x) = {(g \circ f)^{ – 1}}(x)$$. Find $$h'(2)$$. [6] d. ## Markscheme $$f(2) = 9$$ (A1) $${f^{ – 1}}(x) = {(x – 1)^{\frac{1}{3}}}$$ A1 $$({f^{ – 1}})'(x) = \frac{1}{3}{(x – 1)^{ – \frac{2}{3}}}$$ (M1) $$({f^{ – 1}})'(9) = \frac{1}{{12}}$$ A1 $$f'(x) = 3{x^2}$$ (M1) $$\frac{1}{{f'(2)}} = \frac{1}{{3 \times 4}} = \frac{1}{{12}}$$ A1 Note: The last M1 and A1 are independent of previous marks. [6 marks] a. $$g'(x) = {{\text{e}}^{{x^2}}} + 2{x^2}{{\text{e}}^{{x^2}}}$$ M1A1 $$g'(x) > 0$$ as each part is positive R1 [3 marks] b. to find the x-coordinate on $$y = g(x)$$ solve $$2 = x{{\text{e}}^{{x^2}}}$$ (M1) $$x = 0.89605022078 \ldots$$ (A1) gradient $$= ({g^{ – 1}})'(2) = \frac{1}{{g'(0.896 \ldots )}}$$ (M1) $$= \frac{1}{{{{\text{e}}^{{{(0.896 \ldots )}^2}}}\left( {1 + 2 \times {{(0.896 \ldots )}^2}} \right)}} = 0.172$$ to 3sf A1 (using the $$\frac{{{\text{d}}y}}{{{\text{d}}x}}$$ function on gdc $$g'(0.896 \ldots ) = 5.7716028 \ldots$$ $$\frac{1}{{g'(0.896 \ldots )}} = 0.173$$ [4 marks] c. (i) $$({x^3} + 1){{\text{e}}^{{{({x^3} + 1)}^2}}} = 2$$ A1 $$x = – 0.470191 \ldots$$ A1 (ii) METHOD 1 $$(g \circ f)'(x) = 3{x^2}{{\text{e}}^{{{({x^3} + 1)}^2}}}\left( {2{{({x^3} + 1)}^2} + 1} \right)$$ (M1)(A1) $$(g \circ f)'( – 0.470191 \ldots ) = 3.85755 \ldots$$ (A1) $$h'(2) = \frac{1}{{3.85755 \ldots }} = 0.259{\text{ }}(232 \ldots )$$ A1 Note: The solution can be found without the student obtaining the explicit form of the composite function. METHOD 2 $$h(x) = ({f^{ – 1}} \circ {g^{ – 1}})(x)$$ A1 $$h'(x) = ({f^{ – 1}})’\left( {{g^{ – 1}}(x)} \right) \times ({g^{ – 1}})'(x)$$ M1 $$= \frac{1}{3}{\left( {{g^{ – 1}}(x) – 1} \right)^{ – \frac{2}{3}}} \times ({g^{ – 1}})'(x)$$ M1 $$h'(2) = \frac{1}{3}{\left( {{g^{ – 1}}(2) – 1} \right)^{ – \frac{2}{3}}} \times ({g^{ – 1}})'(2)$$ $$= \frac{1}{3}{(0.89605 \ldots – 1)^{ – \frac{2}{3}}} \times 0.171933 \ldots$$ $$= 0.259{\text{ }}(232 \ldots )$$ A1 N4 [6 marks] d. ## Examiners report There were many good attempts at parts (a) and (b), although in (b) many were unable to give a thorough justification. a. There were many good attempts at parts (a) and (b), although in (b) many were unable to give a thorough justification. b. Few good solutions to parts (c) and (d)(ii) were seen although many were able to answer (d)(i) correctly. c. Few good solutions to parts (c) and (d)(ii) were seen although many were able to answer (d)(i) correctly. d. ## Question A straight street of width 20 metres is bounded on its parallel sides by two vertical walls, one of height 13 metres, the other of height 8 metres. The intensity of light at point P at ground level on the street is proportional to the angle $$\theta$$ where $$\theta = {\rm{A\hat PB}}$$, as shown in the diagram. Find an expression for $$\theta$$ in terms of x, where x is the distance of P from the base of the wall of height 8 m. [2] a. (i) Calculate the value of $$\theta$$ when x = 0. (ii) Calculate the value of $$\theta$$ when x = 20. [2] b. Sketch the graph of $$\theta$$, for $$0 \leqslant x \leqslant 20$$. [2] c. Show that $$\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = \frac{{5(744 – 64x – {x^2})}}{{({x^2} + 64)({x^2} – 40x + 569)}}$$. [6] d. Using the result in part (d), or otherwise, determine the value of x corresponding to the maximum light intensity at P. Give your answer to four significant figures. [3] e. The point P moves across the street with speed $$0.5{\text{ m}}{{\text{s}}^{ – 1}}$$. Determine the rate of change of $$\theta$$ with respect to time when P is at the midpoint of the street. [4] f. ## Markscheme EITHER $$\theta = \pi – \arctan \left( {\frac{8}{x}} \right) – \arctan \left( {\frac{{13}}{{20 – x}}} \right)$$ (or equivalent) M1A1 Note: Accept $$\theta = 180^\circ – \arctan \left( {\frac{8}{x}} \right) – \arctan \left( {\frac{{13}}{{20 – x}}} \right)$$ (or equivalent). OR $$\theta = \arctan \left( {\frac{x}{8}} \right) + \arctan \left( {\frac{{20 – x}}{{13}}} \right)$$ (or equivalent) M1A1 [2 marks] a. (i) $$\theta = 0.994{\text{ }}\left( { = \arctan \frac{{20}}{{13}}} \right)$$ A1 (ii) $$\theta = 1.19{\text{ }}\left( { = \arctan \frac{5}{2}} \right)$$ A1 [2 marks] b. correct shape. A1 correct domain indicated. A1 [2 marks] c. attempting to differentiate one $$\arctan \left( {f(x)} \right)$$ term M1 EITHER $$\theta = \pi – \arctan \left( {\frac{8}{x}} \right) – \arctan \left( {\frac{{13}}{{20 – x}}} \right)$$ $$\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = \frac{8}{{{x^2}}} \times \frac{1}{{1 + {{\left( {\frac{8}{x}} \right)}^2}}} – \frac{{13}}{{{{(20 – x)}^2}}} \times \frac{1}{{1 + {{\left( {\frac{{13}}{{20 – x}}} \right)}^2}}}$$ A1A1 OR $$\theta = \arctan \left( {\frac{x}{8}} \right) + \arctan \left( {\frac{{20 – x}}{{13}}} \right)$$ $$\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = \frac{{\frac{1}{8}}}{{1 + {{\left( {\frac{x}{8}} \right)}^2}}} + \frac{{ – \frac{1}{{13}}}}{{1 + {{\left( {\frac{{20 – x}}{{13}}} \right)}^2}}}$$ A1A1 THEN $$= \frac{8}{{{x^2} + 64}} – \frac{{13}}{{569 – 40x + {x^2}}}$$ A1 $$= \frac{{8(569 – 40x + {x^2}) – 13({x^{2}} + 64)}}{{({x^2} + 64)({x^2} – 40x + 569)}}$$ M1A1 $$= \frac{{5(744 – 64x – {x^2})}}{{({x^2} + 64)({x^2} – 40x + 569)}}$$ AG [6 marks] d. Maximum light intensity at P occurs when $$\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = 0$$. (M1) either attempting to solve $$\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = 0$$ for x or using the graph of either $$\theta$$ or $$\frac{{{\text{d}}\theta }}{{{\text{d}}x}}$$ (M1) x = 10.05 (m) A1 [3 marks] e. $$\frac{{{\text{d}}x}}{{{\text{d}}t}} = 0.5$$ (A1) At x = 10, $$\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = 0.000453{\text{ }}\left( { = \frac{5}{{11029}}} \right)$$. (A1) use of $$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{{\text{d}}\theta }}{{{\text{d}}x}} \times \frac{{{\text{d}}x}}{{{\text{d}}t}}$$ M1 $$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = 0.000227{\text{ }}\left( { = \frac{5}{{22058}}} \right){\text{ (rad }}{{\text{s}}^{ – 1}})$$ A1 Note: Award (A1) for $$\frac{{{\text{d}}x}}{{{\text{d}}t}} = – 0.5$$ and A1 for $$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = – 0.000227{\text{ }}\left( { = – \frac{5}{{22058}}} \right){\text{ }}$$. Note: Implicit differentiation can be used to find $$\frac{{{\text{d}}\theta }}{{{\text{d}}t}}$$. Award as above. [4 marks] f. ## Examiners report Part (a) was reasonably well done. While many candidates exhibited sound trigonometric knowledge to correctly express θ in terms of x, many other candidates were not able to use elementary trigonometry to formulate the required expression for θ. a. In part (b), a large number of candidates did not realize that θ could only be acute and gave obtuse angle values for θ. Many candidates also demonstrated a lack of insight when substituting endpoint x-values into θ. b. In part (c), many candidates sketched either inaccurate or implausible graphs. c. In part (d), a large number of candidates started their differentiation incorrectly by failing to use the chain rule correctly. d. For a question part situated at the end of the paper, part (e) was reasonably well done. A large number of candidates demonstrated a sound knowledge of finding where the maximum value of θ occurred and rejected solutions that were not physically feasible. e. In part (f), many candidates were able to link the required rates, however only a few candidates were able to successfully apply the chain rule in a related rates context. f. ## Question Consider the triangle $${\text{PQR}}$$ where $${\rm{Q\hat PR = 30^\circ }}$$, $${\text{PQ}} = (x + 2){\text{ cm}}$$ and $${\text{PR}} = {(5 – x)^2}{\text{ cm}}$$, where $$– 2 < x < 5$$. Show that the area, $$A\;{\text{c}}{{\text{m}}^2}$$, of the triangle is given by $$A = \frac{1}{4}({x^3} – 8{x^2} + 5x + 50)$$. [2] a. (i) State $$\frac{{{\text{d}}A}}{{{\text{d}}x}}$$. (ii) Verify that $$\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0$$ when $$x = \frac{1}{3}$$. [3] b. (i) Find $$\frac{{{{\text{d}}^2}A}}{{{\text{d}}{x^2}}}$$ and hence justify that $$x = \frac{1}{3}$$ gives the maximum area of triangle $$PQR$$. (ii) State the maximum area of triangle $$PQR$$. (iii) Find $$QR$$ when the area of triangle $$PQR$$ is a maximum. [7] c. ## Markscheme use of $$A = \frac{1}{2}qr\sin \theta$$ to obtain $$A = \frac{1}{2}(x + 2){(5 – x)^2}\sin 30^\circ$$ M1 $$= \frac{1}{4}(x + 2)(25 – 10x + {x^2})$$ A1 $$A = \frac{1}{4}({x^3} – 8{x^2} + 5x + 50)$$ AG [2 marks] a. (i) $$\frac{{{\text{d}}A}}{{{\text{d}}x}} = \frac{1}{4}(3{x^2} – 16x + 5) = \frac{1}{4}(3x – 1)(x – 5)$$ A1 (ii) METHOD 1 EITHER $$\frac{{{\text{d}}A}}{{{\text{d}}x}} = \frac{1}{4}\left( {3{{\left( {\frac{1}{3}} \right)}^2} – 16\left( {\frac{1}{3}} \right) + 5} \right) = 0$$ M1A1 OR $$\frac{{{\text{d}}A}}{{{\text{d}}x}} = \frac{1}{4}\left( {3\left( {\frac{1}{3}} \right) – 1} \right)\left( {\left( {\frac{1}{3}} \right) – 5} \right) = 0$$ M1A1 THEN so $$\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0$$ when $$x = \frac{1}{3}$$ AG METHOD 2 solving $$\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0$$ for $$x$$ M1 $$– 2 < x < 5 \Rightarrow x = \frac{1}{3}$$ A1 so $$\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0$$ when $$x = \frac{1}{3}$$ AG METHOD 3 a correct graph of $$\frac{{{\text{d}}A}}{{{\text{d}}x}}$$ versus $$x$$ M1 the graph clearly showing that $$\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0$$ when $$x = \frac{1}{3}$$ A1 so $$\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0$$ when $$x = \frac{1}{3}$$ AG [3 marks] b. (i) $$\frac{{{{\text{d}}^2}A}}{{{\text{d}}{x^2}}} = \frac{1}{2}(3x – 8)$$ A1 for $$x = \frac{1}{3},{\text{ }}\frac{{{{\text{d}}^2}A}}{{{\text{d}}{x^2}}} = – 3.5{\text{ }}( < 0)$$ R1 so $$x = \frac{1}{3}$$ gives the maximum area of triangle $$PQR$$ AG (ii) $${A_{\max }} = \frac{{343}}{{27}}{\text{ }}( = 12.7){\text{ (c}}{{\text{m}}^2}{\text{)}}$$ A1 (iii) $${\text{PQ}} = \frac{7}{3}{\text{ (cm)}}$$ and $${\text{PR}} = {\left( {\frac{{14}}{3}} \right)^2}{\text{ (cm)}}$$ (A1) $${\text{Q}}{{\text{R}}^2} = {\left( {\frac{7}{3}} \right)^2} + {\left( {\frac{{14}}{3}} \right)^4} – 2\left( {\frac{7}{3}} \right){\left( {\frac{{14}}{3}} \right)^2}\cos 30^\circ$$ (M1)(A1) $$= 391.702 \ldots$$ $${\text{QR = 19.8 (cm)}}$$ A1 [7 marks] Total [12 marks] c. ## Examiners report This question was generally well done. Parts (a) and (b) were straightforward and well answered. a. This question was generally well done. Parts (a) and (b) were straightforward and well answered. b. This question was generally well done. Parts (c) (i) and (ii) were also well answered with most candidates correctly applying the second derivative test and displaying sound reasoning skills. Part (c) (iii) required the use of the cosine rule and was reasonably well done. The most common error committed by candidates in attempting to find the value of $$QR$$ was to use $${\text{PR}} = \frac{{14}}{3}{\text{ (cm)}}$$ rather than $${\text{PR}} = {\left( {\frac{{14}}{3}} \right)^2}{\text{ (cm)}}$$. The occasional candidate used $$\cos 30^\circ = \frac{1}{2}$$. c. ## Question A function $$f$$ is defined by $$f(x) = {x^3} + {{\text{e}}^x} + 1,{\text{ }}x \in \mathbb{R}$$. By considering $$f'(x)$$ determine whether $$f$$ is a one-to-one or a many-to-one function. ## Markscheme $$f'(x) = 3{x^2} + {{\text{e}}^x}$$ A1 Note: Accept labelled diagram showing the graph $$y = f'(x)$$ above the x-axis; do not accept unlabelled graphs nor graph of $$y = f(x)$$. EITHER this is always $$> 0$$ R1 so the function is (strictly) increasing R1 and thus $$1 – 1$$ A1 OR this is always $$> 0\;\;\;{\text{(accept }} \ne 0{\text{)}}$$ R1 so there are no turning points R1 and thus $$1 – 1$$ A1 Note: A1 is dependent on the first R1. [4 marks] ## Examiners report The differentiation was normally completed correctly, but then a large number did not realise what was required to determine the type of the original function. Most candidates scored 1/4 and wrote explanations that showed little or no understanding of the relation between first derivative and the given function. For example, it was common to see comments about horizontal and vertical line tests but applied to the incorrect function.In term of mathematical language, it was noted that candidates used many terms incorrectly showing no knowledge of the meaning of terms like ‘parabola’, ‘even’ or ‘odd’ ( or no idea about these concepts). ## Question The function f is defined on the domain [0, 2] by $$f(x) = \ln (x + 1)\sin (\pi x)$$ . Obtain an expression for $$f'(x)$$ . [3] a. Sketch the graphs of f and $$f’$$ on the same axes, showing clearly all x-intercepts. [4] b. Find the x-coordinates of the two points of inflexion on the graph of f . [2] c. Find the equation of the normal to the graph of f where x = 0.75 , giving your answer in the form y = mx + c . [3] d. Consider the points $${\text{A}}\left( {a{\text{ }},{\text{ }}f(a)} \right)$$ , $${\text{B}}\left( {b{\text{ }},{\text{ }}f(b)} \right)$$ and $${\text{C}}\left( {c{\text{ }},{\text{ }}f(c)} \right)$$ where a , b and c $$(a < b < c)$$ are the solutions of the equation $$f(x) = f'(x)$$ . Find the area of the triangle ABC. [6] e. ## Markscheme $$f'(x) = \frac{1}{{x + 1}}\sin (\pi x) + \pi \ln (x + 1)\cos (\pi x)$$ M1A1A1 [3 marks] a. A4 Note: Award A1A1 for graphs, A1A1 for intercepts. [4 marks] b. 0.310, 1.12 A1A1 [2 marks] c. $$f'(0.75) = – 0.839092$$ A1 so equation of normal is $$y – 0.39570812 = \frac{1}{{0.839092}}(x – 0.75)$$ M1 $$y = 1.19x – 0.498$$ A1 [3 marks] d. $${\text{A}}(0,{\text{ }}0)$$ $${\text{B(}}\overbrace {0.548 \ldots }^c,\overbrace {0.432 \ldots }^d)$$ A1 $${\text{C(}}\overbrace {1.44 \ldots }^e,\overbrace { – 0.881 \ldots }^f)$$ A1 Note: Accept coordinates for B and C rounded to 3 significant figures. area $$\Delta {\text{ABC}} = \frac{1}{2}\|$$(ci + dj) $$\times$$ (ei + fj)$$\|$$ M1A1 $$= \frac{1}{2}(de – cf)$$ A1 $$= 0.554$$ A1 [6 marks] e. [N/A] a. [N/A] b. [N/A] c. [N/A] d. [N/A] e. Scroll to Top
Instasolv IIT-JEE NEET CBSE NCERT Q&A 4.5/5 # RS Aggarwal Class 9 Chapter 2 Solutions (Polynomials) RS Aggarwal Class 9 Maths Solutions Chapter 2: Polynomials is an extremely helpful practise book for you to prepare for the CBSE exams. RS Aggarwal Maths Solutions for Class 9 Chapter 2 gives you all the details regarding polynomials. It consists of major topics like algebraic expressions, terms of algebraic expressions, polynomials in one variable, various degrees of polynomials, number of terms in a polynomial, about zeros of polynomials, division algorithm, and various theorems like remainder and factor theorem, etc. Chapter 2 of RS Aggarwal’s textbook has 4 exercises. Exercise 2A, 2B, 2C, 2D has 6, 9, 16, and 26 questions respectively. All the questions cover the whole chapter extensively. To understand every question in detail, we suggest you follow our RS Aggarwal Class 9 Maths Solutions. These solutions are based on the latest syllabus and NCERT Textbook. Our RS Aggarwal Solutions is a useful resource for you to develop Maths problem-solving skills. Because it provides you with the solutions in the easiest method. Not only it provides easy solutions but also it helps in enhancing your logical thinking over the concepts. It is also true, Mathematics has lots of formulas and theorems. It’s not easy for Class 9 students to remember it. That’s why our experts provided some tips and shortcuts which can help you in stressful situations. ## Important topics for RS Aggarwal Class 9 Maths Solutions Chapter 2: Polynomials Let’s have a look at some beautiful concepts regarding this chapter: Polynomials You may be familiar with the word “expressions”. So, what are expressions? Expressions are the equations which have several terms which include constants, variables, exponents, etc. However, in the polynomial case, as the word “poly” means many and ” nomials” means terms suggest that it is the algebraic expression that contains many terms. These terms are constants, variables, and exponents, etc. For Example 4x3 + 3x2 + x +3 is a polynomial in variable 4x2 + 3x-1 – 4 is not a polynomial as it has negative powers 3x3/2 + 2x – 3 is not a polynomial. Polynomial in one Variable When an algebraic expression has only one variable, they are called polynomials in one variable. For Example- x – 4 is a polynomial in one variable x. Degree of Polynomials Degrees of polynomials are the highest value of the power of variables of an algebraic expression. For Example– x3 + x – 4, has degree 3. Types of polynomials Based on degree, polynomials are divided into three types of polynomials • Constant Polynomial Polynomials that have degree 0 are termed as constant polynomials. For Example- 1,4,-6, are constant polynomials. • Linear polynomials- Polynomials that have degree one, are termed as linear polynomial For Example– 2x – 4, is a linear polynomial. • Quadratic Polynomials– Polynomials that have degree 2, are termed as quadratic polynomials. For Example– 3x2 + x +8 , is quadratic polynomial • Cubic Polynomials– Polynomials that have degree three, are termed as a cubic polynomial For Example x3 + x – 4, is a cubic polynomial. • Zeros of Polynomials- Zeros of the polynomial is the number at which the value of the polynomial is equal to vanishes. Like P(x) is a polynomial and a number ‘a’ at which the polynomial p(a)= 0 is termed as zeros of the polynomial. For Example X – 5 is a polynomial. For finding the zeros of the polynomial, we to equate p(x)= 0 Let’s see- X – 5 = 0 X = 5 is a zero of the polynomial. This means that for finding the root of a polynomial, we have to find the value of variable x. • Factorization of Polynomials Factorization of the polynomials can be done in three ways: 1. By Common-factor method 2. By grouping Method 3. By splitting the middle term. • Remainder Theorem As we know the property of division- Dividend = (Divisor × Quotient) + Remainder This property also follows in polynomials. If p(a) and g(a) are two polynomials which follows the conditions: degree of p(a) ≥ degree of g(x) and g(a) ≠ 0 are given then we can find the q(a) and r(a) so that: P(x) = g(a) q(a) + r(a), where r(a) = 0 or degree of r(a) < degree of g(a). It states that when p(a) divided by g(a), it gives q(a) as quotient and r(a) as remainder. ### Exercise Discussion for RS Aggarwal Class 9 Maths Solutions Chapter 2: Polynomials • RS Aggarwal Class 9 Maths Solutions Chapter 2: Polynomial contains 4 exercises 2A, 2B, 2C, and 2D respectively. The questions on these exercises cover all the concepts of this chapter extensively. • Exercise 2 A of RS Aggarwal’s Textbook contains 6 questions based on the concepts: types of polynomials like( constant, linear, quadratic, and cubic), degrees of polynomials. These questions are simple to solve at which you need to just identify the types of polynomials and to find the degrees of polynomials. • Exercise 2 B has 9 RS Aggarwal’s problems based on the concept of Zeros of polynomials. There are the questions at which you need to find the zeros of the given polynomial, and in some of which you have to verify that the expression has the correct zero polynomial or not. • Exercise 2 C contains 16 questions that will test your knowledge on the concepts like remainder Theorem, division algorithm, etc. these questions are quite lengthy to solve and important from your exam point of view. • Exercise 2 D of RS Aggarwal’s solutions for chapter 2 has 26 questions in total. Questions generally use the concepts of the remainder and factor theorem. These questions are quite tricky, you need to have a better understanding of the remainder theorem and factor theorem for solving it. Benefits of RS Aggarwal’s solutions for class 9 maths chapter 2 by Instasolv • Instasolv’s RS Aggarwal Solutions are the ultimate source for you to overcome the fear issues regarding solving maths problems. • The best part of these solutions is that it provides you with a brief understanding of all the aspects of the chapter, as it is designed in a layman’s language with a detailed explanation • Our team of experts has made their efforts and provided diagrammatic explanations of the questions also, that helps you in the visualization of the questions. • Solving RS Aggarwal’s solutions for class 9 maths chapter 2 by Instasolv gives you an edge in your upcoming exams and more importantly it helps in managing time and trouble also so that you can outperform in your CBSE Class 9 examination. More Chapters from Class 9
Chapter 11 Class 9 Constructions Class 9 Important Questions for Exam - Class 9 ### Transcript Question 2 Construct an angle of 45° at the initial point of a given ray and justify the construction . Steps of construction Draw a ray OA. Taking O as center and any radius, draw an arc cutting OA at B. 3. Now, taking B as center and with the same radius as before, draw an arc intersecting the previously drawn arc at point C. 4. With C as center and the same radius, draw an arc cutting the arc at D 5. With C and D as centers and radius more than 1/2 CD, draw two arcs intersecting at P. 6. Join OP. Thus, ∠ AOP = 90° Now we draw bisector of ∠ AOP 7. Let OP intersect the original arc at point Q 8. Now, taking B and Q as centers, and radius greater than 1/2 BQ, draw two arcs intersecting at R. 9. Join OR. Thus, ∠ AOR = 45° Justification We need to prove ∠ AOR = 45° Join OC & OB Thus, OB = BC = OC ∴ Δ OCB is an equilateral triangle ∴ ∠ BOC = 60° Join OD, OC and CD Thus, OD = OC = DC ∴ Δ DOC is an equilateral triangle ∴ ∠ DOC = 60° Join PD and PC Now, In Δ ODP and Δ OCP OD = OC DP = CP OP = OP ∴ Δ ODP ≅ Δ OCP ∴ ∠ DOP = ∠ COP So, we can say that ∠ DOP = ∠ COP = 1/2 ∠ DOC ∠ DOP = ∠ COP = 1/2 × 60° = 30° (Radius of same arcs) (Arc of same radii) (Common) (SSS Congruency) (CPCT) (We proved earlier that ∠ DOC= 60° ) Now, ∠ AOP = ∠ BOC + ∠ COP ∠ AOP = 60° + 30° = 90° Now, Join QR and BR In Δ OQR and Δ OBR OQ = OB QR = BR OR = OR ∴ Δ OQR ≅ Δ OBR ∴ ∠ QOR = ∠ BOR (Radius of same arcs) (Arc of same radii) (Common) (SSS Congruency) ∠ QOR = ∠ BOR = 1/2 ∠ AOP ∠ DOP = ∠ COP = 1/2 × 90° = 45° Thus, ∠ AOR = 45° Hence justified
## calyne 3 years ago simplify 1/x + 1/2(x+1) + 1/2(x-1) 1. Rose1285 (1)/(x)+(1)/(2)(x+1)+(1)/(2)(x-1) Multiply the rational expressions to get ((x+1))/(2). (1)/(x)+(x+1)/(2)+(1)/(2)(x-1) Multiply the rational expressions to get ((x-1))/(2). (1)/(x)+(x+1)/(2)+(x-1)/(2) Multiply each term by a factor of 1 that will equate all the denominators. In this case, all terms need a denominator of 2x. The (1)/(x) expression needs to be multiplied by ((2))/((2)) to make the denominator 2x. The ((x+1))/(2) expression needs to be multiplied by ((x))/((x)) to make the denominator 2x. The ((x-1))/(2) expression needs to be multiplied by ((x))/((x)) to make the denominator 2x. (1)/(x)(2)/(2)+(x+1)/(2)(x)/(x)+(x-1)/(2)(x)/(x) Multiply the expression by a factor of 1 to create the least common denominator (LCD) of 2x. (1(2))/(2x)+(x+1)/(2)(x)/(x)+(x-1)/(2)(x)/(x) Remove the parentheses around the expression 2. (2)/(2x)+(x+1)/(2)(x)/(x)+(x-1)/(2)(x)/(x) Multiply the expression by a factor of 1 to create the least common denominator (LCD) of 2x. (2)/(2x)+((x+1)(x))/(2x)+(x-1)/(2)(x)/(x) Multiply the expression by a factor of 1 to create the least common denominator (LCD) of 2x. (2)/(2x)+((x+1)(x))/(2x)+((x-1)(x))/(2x) The numerators of expressions that have equal denominators can be combined. In this case, ((2))/(2x) and ((x+1)(x))/(2x) have the same denominator of 2x, so the numerators can be combined. ((2)+(x+1)(x)+((x-1)(x)))/(2x) Remove the extra parentheses around the factors. ((2)+(x+1)(x)+(x-1)(x))/(2x) Remove the parentheses around the expression 2. (2+(x+1)(x)+(x-1)(x))/(2x) Multiply each term in the first polynomial by each term in the second polynomial. (2+(xx+1x)+(x-1)(x))/(2x) Multiply x by x to get x^(2). (2+(x^(2)+1x)+(x-1)(x))/(2x) Multiply 1 by x to get x. (2+(x^(2)+x)+(x-1)(x))/(2x) Multiply each term in the first polynomial by each term in the second polynomial. (2+(x^(2)+x)+(xx-1x))/(2x) Multiply x by x to get x^(2). (2+(x^(2)+x)+(x^(2)-1*x))/(2x) Multiply -1 by x to get -x. (2+(x^(2)+x)+(x^(2)-x))/(2x) Remove the parentheses around the expression x^(2)+x. (2+x^(2)+x+(x^(2)-x))/(2x) Remove the parentheses around the expression x^(2)-x. (2+x^(2)+x+x^(2)-x)/(2x) According to the distributive property, for any numbers a, b, and c, a(b+c)=ab+ac and (b+c)a=ba+ca. Here, x^(2) is a factor of both x^(2) and x^(2). (2+(1+1)x^(2)+x-x)/(2x) Add 1 to 1 to get 2. (2+(2)x^(2)+x-x)/(2x) Remove the parentheses. (2+2x^(2)+x-x)/(2x) According to the distributive property, for any numbers a, b, and c, a(b+c)=ab+ac and (b+c)a=ba+ca. Here, x is a factor of both x and -x. (2+2x^(2)+(1-1)x)/(2x) To add integers with different signs, subtract their absolute values and give the result the same sign as the integer with the greater absolute value. In this example, subtract the absolute values of 1 and -1 and give the result the same sign as the integer with the greater absolute value. (2+2x^(2)+(0)x)/(2x) Remove the parentheses. (2+2x^(2))/(2x) Reorder the polynomial 2+2x^(2) alphabetically from left to right, starting with the highest order term. (2x^(2)+2)/(2x) Factor out the GCF of 2 from the expression 2x^(2). (2(x^(2))+2)/(2x) Factor out the GCF of 2 from the expression 2. (2(x^(2))+2(1))/(2x) Factor out the GCF of 2 from 2x^(2)+2. (2(x^(2)+1))/(2x) Cancel the common factor of 2 the expression (2(x^(2)+1))/(2x). (<X>2<x>(x^(2)+1))/(<X>2<x>x) Remove the common factors that were cancelled out. (x^(2)+1)/(x) LOL,@Rose1285 :U googled it or what !? 3. calyne no i meant 1/x + 1/[2(x+1)] + 1/[2(x-1)]

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