. Hence r = (1 +ilk)&-- 1. 1.2 SYSTEM OF REAL NUMBERS The system of real numbers as we know it today is a result of gradual progress, as the following indicates. ways of arranging 7 persons in a circle, the required number of ways = 2!6! . . (c) 4 x + 9 y 2 = 3 6 , y 2 = - (9-4, 3 9 Note that if x is greater than 9, y is imaginary. . . 11.1 11.2 11.3 11.4 11.5 Chapter 12 58 58 58 58 59 60 SIMPLE OPERATIONS WITH COMPLEX NUMBERS . 188 SYSTEMS OF EQUATIONS INVOLVING QUADRATICS CHAP. . . . . SOLUTION Choosing the basic column indicated, multiply its elements by -2, -5, 3 and add respectively to the corresponding elements of the lst, 3rd, 4th colums to obtain 7 -2 14 -4 0 1 0 0 -9 5 -27 17 1 -2 6 -5 ' Expand according to the cofactors of the elements in the second row and obtain (O)(its cofactor) + (l)(its cofactor) + (O)(its cofactor) { I1 = (l)(its cofactor) = 1 + -9 -27 l: + (O)(its cofactor) I] 17 . . . . . 25.57 How many two-digit numbers can be formed with the digits 0,3,5,7if no repetition in any of the numbers is allowed? . . Even numbers = 9.10.10.10.5 = 45 O00 numbers. . Assume the statement is true €or n = k. So k3 + 1 L k2 + k is true. CHAP. . . The first place may be filled by any one of the 8 digits, i.e., by the remaining 4 odd digits and the even digits, 2,4,6,8. . . . 22.2 Arithmetic Sequences . For n = k + 1, ( k + 1)3+ 1 = k 3 + 3 k 2 + 3 k + 1 + 1 = k 3 + 3 k 2 + 3 k + 2 = k3 + 2k2 + k2 + 3k + 2 = (k3 + 2k2) (k l)(k + 2) = (k3 + 2k2) + (k + l)[(k + 1) + 11 = (k3 + 2k2) + [(k + 1)2 + (k + l)] + + We know n L 1, so k 2 1 and k3 + 2 k 2 z 3. . . Robert E. 11. The amounts are equal if P(l + r) = P(1.03)2, 1 + r = (1.03)*, r = 0.0609 or 6.09%. . . . 17-21 (g) hyperbola, Fig. . . 19.4 Prove that a* + b2 > 2ab if a and b are real and unequal numbers. . . The values of nPrand nCrcan often be computed on the calculator when n! . . . . . 24.43 During the Black Plague the world population declined by about 1 million from 4.7 million to about 3.7 million during the 50-year period from 1350 to 1400. These terms are: + alb2c3 0 inversions, sign al b 3 ~ 2 1 inversion, sign a2b1c3 1 inversion, sign - a2b3cl 2 inversions, sign + u3b2c1 3 inversions, sign a361c2 2 inversions, sign +. (n - 2)!2! . . (1) (2) (3) (4) (5) (6) a > b means “a is greater than 6” (or a - b is a positive number). . . 25 * 24 - 23 = 2300 1.2.3 Note that in each case the numerator and denominator have the same number of factors. Laws for Radicals . . . . SOLUTION The simple interest on $200 for 1 year at 5% is I = 200(0.05)(1) = $10. . . 7! . . . . . Therefore, a2 = b2 + c2 yields 16 = b2 + 9 and b2 = 7. These last two constraints are called natural or implied constraints, because these conditions are true as a matter of fact and need not be stated in the problem. . . SOLUTION Multiply (2) by 3 and subtract from (1) to obtain 2r2 - 6xy + 4y2 = 0, x2 - 3xy + 2y2 = 0, ( x - y)(x - 2y) = 0 and x = y, x = 2y. Schaum’s Outline of College Algebra pdf Schaum’s Outline of College Algebra pdf : Pages 464 By Murray R. Spiegel and Robert E. Moyer Publisher: McGraw-Hill Education, Year: 2018 ISBN: 1260120767, 9781260120769 Search in Amazon.com Description: Tough Test Questions? . . . . . . . . . . . . (i) The rule of (i) is applied here. How should the company place its order so that costs are minimized? . . & (a) 248 (d) 0.162 (b) 2.48 (c) 0.024 (e) O.OOO6 (f) 18.36 k) 1-06 ( h ) 6OOo (9 4 (j) 40.60 (m)7000000 (n) 0.000007 (k) 237.63 (f) 146.203 23.33 Find the common logarithm of each number. 25.72 How many Morse code characters could be made by using three dots and two dashes in each character? . 295 PERMUTATIONS AND COMBINATIONS CHAP. 2. 25.87 How many different sums of money can be formed from the coins of Problem 25.86? . . . . O! QA157S725 1997 5 12.9’076-dc21 97-31223 CIP McGraw-Hi22 A Division of TheMcGmw-HiUCompanies a In revising this book, the strengths of the first edition were retained while reflecting the changes in the study of algebra since the first edition was written. . . To establish the truth of such a statement, we could prove it for each positive integer of interest separately. . . are then associated with points on the line at distances 1, 2, 3, . PERMUTATIONS AND COMBINATIONS 298 [CHAP. . Algebra - Schaum -murray R. Spiegel.pdf [klzz0vojgylg]. . Horizontal Asymptotes . _ - 1- --- 1 - 1 4!3! . 24 Chapter 25 Permutations and Combinations 25.1 FUNDAMENTAL COUNTING PRINCIPLE If one thing can be done in m different ways and, when it is done in any one of these ways, a second thing can be done in n different ways, then the two things in succession can be done in mn different ways. . . Since n # 0 , l we may divide by n(n - 1) to obtain 7(n - 2) = 6(n + l), n = 20. 18.20 The diagonal of a rectangle is 85 ft. Q 3x2 - 1 is a rational fraction. . The general entry for the matrix is denoted by aii. . . . . + (-4) Upon evaluating each of the 3rd order determinants the result -53 is obtained. . . = 1036 800. If the following two conditions are satisfied: (1) P(1) is true. This illustrates the commutative law for addition. . . PERMUTATIONS AND COMBINATIONS 294 [CHAP. . 3!7! . . . 25.81 If 5*,,P3= 24.,c4, find n. 25.83 How many straight lines are determined by (a) 6, ( 6 ) n points, no three of which lie in the same straight line? . . SOLUTION Let the numbers be p, q. . . . 17-14(6)). . . . . . . . x+2y+3z= 6 - z = 0 is The augmented matrix associated with the system X x- y- z=-4 1 A=[l :I 2 3 0 -1 1 -1 -1 -4 MATRICES 358 [CHAP 30 EXAMPLE 30.9. Hence ( a - 6) + ( c - d ) > 0, (a c ) - (6 + d ) > 0 and ( a + c) > (6 + d ) . SOLUTION A = AOerf 5.0 = 2.5e437) 2 = e37r In2 = 37r lne 0.6931 = 37r(l) 0.6931 = 37r 0.01873 = r r = 0.0187 r = 1.87% 24.21 In Nigeria the rate of deforestation is 5.25% per year. . . . . . . - n(n - l)(n - 2) . 8.1 8.2 8.3 8.4 8.5 Chapter 9 Chapter 10 67 67 67 68 EQUATIONS IN GENERAL . . . . . Subtraction may be defined in terms of addition. Fortunately, there’s Schaum’s. Hence the number of ways n women can be seated in a row if 2 particular women may never sit together = .P, - 2(n-1Pn-1) = n! Combinations of n different things taken r at a time ,c, = nf'r = r! 4+(-2)]=[: -1+2 + :] The matrix -A is called the opposite of matrix A and each entry in -A is the opposite of the corresponding entry in A. . . SOLUTION (a) Number of words = arrangements of 8 different letters taken 5 at a time = = 8 . 25.43 A has 3 maps and B has 9 maps. . 7 n2 - n - 20 = 0, I=- 30(n - 4) 120 ' n=5 n = 8. . . . + 4) < 0. There are 4 inversions. We graph the related equations 2x y = 3 andx - 2y = - 1on the same set of axes. When letters are used, as in algebra, the notation p x q is usually avoided since X may be confused with a letter representing a number. . . . . . Schaum's reinforces the main concepts required in your course and offers hundreds of practice questions to help you suceed. Thus 27 + (48 12) = (27 + 48) + 12. . . . . The first equation is y = f(x) - 7. . . . . . . Amount A = principal P + interest I = $424. . . . . . 25 Chapter 26 The Binomial Theorem 26.1 COMBINATORIAL NOTATION The number of combinations of n objects selected r at a time, nCr, can be written in the form which is called combinatorial notation. . 2 - 1 Number of arrangements = -= 302400. 171 CONIC SECTIONS 177 ( a ) The foci are on a line parallel to the x axis, so the form is ( x - h)2 ---a2 (y - k)2 62 -I The center is half-way between the foci, so c = 3 and the center is at C(-1,5). For more information about the companion Electronic Tutor, including system requirements, please see the back cover. 1= 120 orders. 25.66 How many integers greater than 300 and less than 1O00 can be made with the digits 1,2,3,4,5 if no digit is repeated in any number? . 1 0 +3 . To do this, we choose a point on the line to represent the real number zero and call this point the origin. . . (a) vertices are (0, +2) and foci are (0, f 3 ) (6) foci (1,2) and (- 11,2) and the transverse axis has length 4 SOLUTION (a) Since the vertices are (0, +2), the center is at (0, 0), and since they are on a vertical line the standard form is The vertices are at (0, +a) so a = 2 and the foci are at (0,+3) so c = 3. . . . . . . If the objective is a linear function and the constraints are linear inequalities, the values, if any, that maximize o r minimize the objective occur at the corners of the region determined by the constraints. . . . . . . . . . . Hence x2 - 7x 12 < 0 is satisfied when 3 < x < 4 . . Rationalizing Binomial Denominators . . . 14.1 14.2 14.3 14.4 14.5 14.6 127 Slope of a Line . . 167 Circles . . . 3000-solved problems in physics by schaums.pdf. . 3 . . Since the vertices are on the x axis and the center is at (0, 0), the form of the ellipse is (a) -x2 + - =y2I a2 b2 From a vertex at (5,O) and the center at ( O , O ) , we get a = 5. . . . . . . (2) x - 4 < 0 and x + 4 > 0 simultaneously. . . . 2 -1 -3 1 :I [-; I:; I ;] [ 4 -;-jl [ [:;;: 1 0 2 4 0 (b) 3 3 5 2 0 -1 (4 -3 -1 7 1 1 1 5 0 (4 0 -1 2 -1 1 1 3 -1 0 -2 1 -3 1 2 -1 -4 3 3 2 -2 -3 -1. EXAMPLE 30.10. Hence the required number of orders = 5 . . . 1 When no sign is placed before a number, a plus sign is understood. ways. (2) Assume that the theorem or formula is true for n = k . . . . . . . A = Aoert APPLICATIONS OF LOGARITHMS AND EXPONENTS 278 [CHAP. . . . . 8 , 8.6 (h) (32- 15) (e) 4(7.6), (4.7)6 SOLUTION + 23 = 65, 23 + 42 = 65. . . . Removing factors ( x - 1) and 0,- 1) from 1st and 2nd columns respectively. . and y - - I - y--5 4 Simplify each APPENDIX C] SAMPLE SCREENS 393 Check Using Study Works: Remember extraneous roots may be introduced when taking t h e root of both sides.