Solution manual for college algebra enhanced with graphing utilities 7th edition by sullivan

Chapter R Review Section R.1 16 ( A ∩ B) ∪ C = ({1, 3, 4,5, 9} ∩ {2, 4, 6, 7,8} ) ∪ {1,3, 4, 6} rational = {4} ∪ {1, 3, 4, 6} + ⋅ − = + 30 − = 31 = {1,3, 4, 6} Distributive 17 A = {0, 2, 6, 7, 8} ( x + 3) = 18 C = {0, 2, 5, 7, 8, 9} a 19 A ∩ B = {1, 3, 4, 5, 9} ∩ {2, 4, 6, 7, 8} b = {4} = {0, 1, 2, 3, 5, 6, 7, 8, 9} True False; The Zero-Product Property states that if a product equals 0, then at least one of the factors must equal False; is the Greatest Common Factor of 12 and 18 The Least Common Multiple is the smallest value that both numbers will divide evenly The LCM for 12 and 18 is 36 20 B ∪ C = {2, 4, 6, 7, 8} ∪ {1, 3, 4, 6} = {1, 2, 3, 4, 6, 7, 8} = {0, 5, 9} 21 A ∪ B = {0, 2, 6, 7, 8} ∪ {0, 1, 3, 5, 9} = {0, 1, 2, 3, 5, 6, 7, 8, 9} 22 B ∩ C = {0, 1, 3, 5, 9} ∩ {0, 2, 5, 7, 8, 9} 10 True = {0, 5, 9} 11 A ∪ B = {1, 3, 4,5, 9} ∪ {2, 4, 6, 7,8} 23 a = {1, 2,3, 4, 5, 6, 7,8, 9} 12 A ∪ C = {1, 3, 4,5, 9} ∪ {1, 3, 4, 6} = {1, 3, 4, 5, 6, 9} 13 A ∩ B = {1, 3, 4,5, 9} ∩ {2, 4, 6, 7,8} = {4} 14 A ∩ C = {1, 3, 4,5, 9} ∩ {1, 3, 4, 6} = {1, 3, 4} 15 b {−6, 2,5} c { d {π } e { ( A ∪ B) ∩ C = ({1, 3, 4,5, 9} ∪ {2, 4, 6, 7,8} ) ∩ {1,3, 4, 6} = {1, 2,3, 4,5, 6, 7,8,9} ∩ {1,3, 4, 6} = {1, 3, 4, 6} {2,5} 24 a −6, , −1.333 = −1.3, 2,5 } −6, , −1.333 = −1.3, π , 2,5 } {1} b {0,1} c { d { 5} − , 2.060606 = 2.06,1.25, 0,1 Copyright © 2017 Pearson Education, Inc } Chapter R: Review e 25 a − , 2.060606 = 2.06,1.25, 0,1, { {1} b {0,1} c { 1 0,1, , , } d None e 1 0,1, , , { } None b {−1} c {−1.3, −1.2, −1.1, −1} d e 28 a 2, π , + 1, π + 2, π , + 1, π + } } { 36 a 1.001 b 1.000 37 a 0.429 b 0.428 38 a 0.556 b 0.555 39 a 34.733 b 34.733 40 a 16.200 b 16.200 49 x =6 50 =6 x 51 − + = + = 52 − + = + = + 10.3 e 9.998 48 − y = None {− b 47 x − = None d 9.999 46 x = ⋅ None { 35 a 45 y = + b None c 0.053 44 + y = + {−1.3, −1.2, −1.1, −1} { { b 43 x + = ⋅ b None c 0.054 42 ⋅ = 10 d None 27 a 34 a 41 + = 26 a e } } 2, π + 53 −6 + ⋅ = −6 + 12 = } 54 − ⋅ = − = − 2, π + 2, + 10.3 } 55 + − = − = 29 a 18.953 b 18.952 30 a 25.861 b 25.861 31 a 28.653 b 28.653 32 a 99.052 b 99.052 33 a 0.063 b 0.062 56 − − = − = 57 + 12 + 13 = = 3 58 − −1 = = 2 2 Copyright © 2017 Pearson Education, Inc Section R.1: Real Numbers 59 − 3 ⋅ + ⋅ ( − )  = − 15 + ⋅ (1)  = − 17 = −11 60 ⋅ 8 − ⋅ ( + )  − = ⋅ 8 − ⋅ ( )  − = ⋅ [8 − 18] − = ⋅ [ −10] − 70 5⋅3 5⋅3 ⋅ = = = 10 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 71 10 ⋅ ⋅ ⋅ 2 ⋅ ⋅ ⋅ = ⋅ = = 25 27 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 45 72 21 100 ⋅ ⋅ ⋅ 25 ⋅ ⋅ ⋅ 25 = 28 ⋅ = = 25 25 ⋅ 25 ⋅ 73 15 + 23 + = = 20 20 74 + 11 + = = 6 75 25 + 54 79 + = = 30 30 76 15 16 + 135 151 + = = 18 18 77 10 + 13 + = = 18 12 36 36 78 + 40 46 + = = 15 45 45 79 − 35 32 16 − = =− =− 30 18 90 90 45 80 9−4 − = = 14 21 42 42 81 9−8 − = = 20 15 60 60 82 12 − 15 − = =− 35 14 70 70 = −20 − = −23 61 ⋅ ( − ) + ⋅ − = ⋅ ( −2 ) + 16 − = −4 + 16 − = 12 − = 11 62 − ( ⋅ − + ) = − (12 − + ) = − 12 = −11 63 10 − 6 − ⋅ + ( − 3)  ⋅ = 10 − [ − + 5] ⋅ = 10 − [ + 5] ⋅ = 10 − [ ] ⋅ = 10 − 14 = −4 64 − ⋅ − 6 ⋅ ( − )  = − 20 − 6 ⋅ ( −1)  = −18 − [ −6] = −18 + = −12 1 65 ( − 3) = ( ) = 66 ( + ) 13 = ( ) 13 = 67 + 12 = =6 5−3 68 − −2 = = −1 5−3 69 10 ⋅ ⋅ ⋅ ⋅ ⋅ = = = 21 ⋅ ⋅ ⋅ ⋅ 7 5  18  27 ⋅ ⋅ ⋅ ⋅ 15 83   = ⋅ = = =  11  18 11 ⋅ ⋅11 ⋅ ⋅11 22  27     5  21  35 ⋅ ⋅ 5 ⋅ ⋅ 25 = = 84   = ⋅ =   21 ⋅ ⋅ ⋅ ⋅  35    Copyright © 2017 Pearson Education, Inc Chapter R: Review 85 86 7 + 10 ⋅ + = + = = =1 10 10 10 10 10 2⋅2 2⋅ 2 + ⋅ = + = + = + ⋅ ⋅ ⋅ ⋅ 15 10 10 + 12 = ⋅ + = + = = 15 15 15 15 15 4⋅3 4⋅ = = = 5⋅3 5⋅ 98 ( x − )( x + 1) = x + x − x − = x − 3x − 99 ( x − )( x − ) = x − x − x + 16 = x − 10 x + 16 100 ( x − )( x − ) = x − x − x + = x2 − x + 3 3 6 87 ⋅ + = ⋅ + = + = ⋅ + 8 8 12 12 + 15 = + = = 8 8 101 x + x = ⋅ x + ⋅ x 5 3⋅5 ⋅5 88 ⋅ − = ⋅ − = − = − 6 3⋅ 2 ⋅ 2 5 −1 = − = = =2 2 2 102 + ⋅ = + 12 = 14 since multiplication comes before addition in the order of operations for real numbers = ( + 3) ⋅ x = ( 5) ⋅ x = 5x ( + 3) ⋅ = ⋅ = 20 since operations inside parentheses come before multiplication in the order of operations for real numbers 89 ( x + ) = x + 24 90 ( x − 1) = x − 103 ( ⋅ ) = (12 ) = 24 91 x ( x − ) = x − x ( ⋅ 3) ⋅ ( ⋅ ) = ( )(8 ) = 48 92 x ( x + 3) = x + 12 x 104 1 ⋅ 3x 3 − 93  x −  = ⋅ x − ⋅ = 4 2⋅2   ⋅ 3x = − = x −1 ⋅2 2 1 3⋅ 2x 2 + 94  x +  = ⋅ x + ⋅ = 6 3 ⋅2   ⋅ 2x = + = 2x + 3 ⋅2 95 ( x + )( x + ) = x + x + x + = x2 + x + 96 ( x + 5)( x + 1) = x + x + x + = x + 6x + 97 ( x − )( x + 1) = x + x − 2x − 4+3 = = , but 2+5 4 ⋅ + ⋅ 20 + 26 13 + = = = = = 2.6 10 10 10 105 Subtraction is not commutative; for example: − = −1 ≠ = − 106 Subtraction is not associative; for example: ( − ) − = ≠ = − ( − 1) 107 Division is not commutative; for example: ≠ 108 Division is not associative; for example: (12 ÷ ) ÷ = ÷ = , but 12 ÷ ( ÷ ) = 12 ÷ = 12 109 The Symmetric Property implies that if = x, then x = 2 = x −x−2 Copyright © 2017 Pearson Education, Inc Section R.2: Algebra Essentials 110 From the principle of substitution, if x = , then ( x )( x ) = ( 5)( )  x = 25 b True True  x + x = 25 +  x + x = 30 111 There are no real numbers that are both rational and irrational, since an irrational number, by definition, is a number that cannot be expressed as the ratio of two integers; that is, not a rational number Every real number is either a rational number or an irrational number, since the decimal form of a real number either involves an infinitely repeating pattern of digits or an infinite, nonrepeating string of digits 112 The sum of an irrational number and a rational number must be irrational Otherwise, the irrational number would then be the difference of two rational numbers, and therefore would have to be rational 10 False; the absolute value of a real number is nonnegative = which is not a positive number 11 False; a number in scientific notation is expressed as the product of a number, x, ≤ x < 10 or −10 < x ≤ −1 , and a power of 10 12 False; to multiply two expressions with the same base, retain the base and add the exponents 13 0.25 −2.5 14 113 Answers will vary −2 −1.5 114 Since day = 24 hours, we compute 12997 = 541.5416 24 Now we only need to consider the decimal part of the answer in terms of a 24 hour day That is, ( 0.5416 ) ( 24 ) ≈ 13 hours So it must be 13 hours later than 12 noon, which makes the time AM CST 115 Answers will vary 15 >0 16 < 17 −1 > −2 18 −3 < − 19 π > 3.14 Section R.2 20 variable strict base; exponent (or power) 1.2345678 × 10 d > 1.41 21 = 0.5 22 > 0.33 23 < 0.67 24 = 0.25 origin −1 Copyright © 2017 Pearson Education, Inc 2 3 2 Chapter R: Review 25 x > 46 x + y −2+3 1 = = =− x − y − − −5 47 3x + y 3(− 2) + 2(3) − + = = = =0 2+ y 2+3 5 48 x − 2(− 2) − − − = = =− y 3 26 z < 27 x < 28 y > −5 29 x ≤ 30 x ≥ 31 Graph on the number line: x ≥ −2 −2 32 Graph on the number line: x < 4 33 Graph on the number line: x > −1 −1 49 x + y = + (− 2) = = 50 x − y = − (− 2) = = 51 x + y = + −2 = 3+ = 52 x − y = − −2 = 3− =1 53 3 x = = =1 x 3 54 y −2 = = = −1 y −2 −2 55 x − y = 4(3) − 5(− 2) 34 Graph on the number line: x ≤ 7 35 d (C , D ) = d (0,1) = − = = = 12 + 10 = 22 36 d (C , A) = d (0, −3) = − − = − = 37 d ( D, E ) = d (1,3) = − = = 38 d (C , E ) = d (0,3) = − = = = 22 56 57 x + y = 3(3) + 2(− 2) = − = = 4x − y 39 d ( A, E ) = d (−3,3) = − (−3) = = = 12 − − 10 = 12 − 10 40 d ( D, B) = d (1, −1) = − − = − = = 41 x + y = − + ⋅ = − + = =2 42 3x + y = 3(− 2) + = − + = −3 58 x + y = 3 + − = 3⋅3 + ⋅ 43 xy + = 5(− 2)(3) + = −30 + = − 28 = 9+4 = 13 44 − x + xy = − 2(− 2) + (− 2)(3) = − = − 2(− 2) − 4 2x 45 = = = x − y − − −5 = 4(3) − 5(− 2) 59 x2 − x Part (c) must be excluded The value x = must be excluded from the domain because it causes division by Copyright © 2017 Pearson Education, Inc Section R.2: Algebra Essentials 60 x2 + x Part (c) must be excluded The value x = must be excluded from the domain because it causes division by 61 x x = ( 3)( x − x + 3) x −9 Part (a) , x = , must be excluded because it causes the denominator to be 62 x x +9 None of the given values are excluded The domain is all real numbers 63 64 65 69 −6 x+4 x = −4 must be excluded sine it makes the denominator equal Domain = { x x ≠ −4} x x+4 x = −4 must be excluded sine it makes the denominator equal Domain = { x x ≠ −4} x2 x +1 None of the given values are excluded The domain is all real numbers 70 x3 x3 = x − ( x − 1)( x + 1) Parts (b) and (d) must be excluded The values x = 1, and x = −1 must be excluded from the domain because they cause division by x + x − 10 x + x − 10 = x( x − 1)( x + 1) x −x Parts (b), (c), and (d) must be excluded The values x = 0, x = 1, and x = −1 must be excluded from the domain because they cause division by −9 x − x + −9 x − x + = 66 x3 + x x( x + 1) Part (c) must be excluded The value x = must be excluded from the domain because it causes division by 67 68 x −5 x = must be exluded because it makes the denominator equal Domain = { x x ≠ 5} x−2 x−6 x = must be excluded sine it makes the denominator equal Domain = { x x ≠ 6} 5 71 C = ( F − 32) = (32 − 32) = (0) = 0°C 9 5 72 C = ( F − 32) = (212 − 32) = (180) = 100°C 9 5 73 C = ( F − 32) = (77 − 32) = (45) = 25°C 9 5 74 C = ( F − 32) = (− − 32) 9 = (−36) = − 20°C 75 (− 4) = (− 4)(− 4) = 16 76 − 42 = −(4) = −16 77 4−2 = 1 = 42 16 78 − 4−2 = − 1 =− 16 42 79 3−6 ⋅ 34 = 3−6 + = 3−2 = 1 = 32 80 4−2 ⋅ 43 = 4−2 + = 41 = 81 −2 −1 (3 ) −2 −1 = 3( )( ) = 32 = Copyright © 2017 Pearson Education, Inc Chapter R: Review 82 83 −1 −3 −1 −3 = 2( )( ) = 23 = (2 )  3x −1  95  −1   4y  25 = 52 =  x −2  96  −2   6y  84 36 = = 85 ( −4 )2 86 ( −3) 87 (8x ) 88 ( −4 x ) 89 (x 90 ( x y) = ( x ) 93 ( ) = 1 =− 2 −4 x 4x 2 −1 ) = (x ) ⋅( y ) 97 xy −1 = = 64 x −1 = x y −2 = ⋅ y = x −3 y = y3 x3 x −2 y = x −2 −1 y1− = x −3 y −1 = xy x y (− 2)3 x ( y z ) 2 3 xy z = − 8x4 y2 z 9x y z 23 x y x −2 y −1 z −1 8x4 y = x −2− y −1−1 z −1 = x −6 y −2 z −1 = 2x y z = x4 y2  x2  =   5y  = 3 216 x 125 y −3 y −3 ( −1) = = x ( 2) 2 99 x + y = ( ) + ( −1) = + = 2 100 x y = ( ) ( −1) = ⋅1 = 101 ( xy )2 = ( ⋅ ( −1) ) 102 ( x + y )2 = ( + ( −1) ) 103 x2 = x = = 104 − −1 −3 −1 x y z − −1 x y z = x3 z =− 9y x −2 ( y z ) −1 −3 2x ( 2) = = −4 y ( −1) 98 −3x −1 y = = 94  y2  =   6x  42 x 16 x  4x  =  = 2 = y y2  3y  ( ) = (y ) = −3 = = 82 x −2 63 x x2 y3 x 91 = x −1 y 3− = x1 y −1 = y xy 92  3y  =   4x  −1 −1 −3 = −4 = −1 y −2 ( x) 2 = ( −2 ) = 2 = (1) = =x=2 ( )2 + ( −1)2 105 x2 + y = 106 x + y = x + y = + −1 = + = 107 x y = 2−1 = 2 108 y x = ( −1) = Copyright © 2017 Pearson Education, Inc = +1 = Section R.2: Algebra Essentials 109 If x = 2, 121 454.2 = 4.542 × 102 x − x + x − = ⋅ 23 − ⋅ 22 + ⋅ − = 16 − 12 + 10 − = 10 123 0.013 = 1.3 × 10−2 If x = 1, x − x + x − = ⋅13 − ⋅12 + ⋅1 − = 2−3+5− =0 124 0.00421 = 4.21× 10−3 125 32,155 = 3.2155 × 104 126 21, 210 = 2.121× 104 110 If x = 1, x + 3x − x + = ⋅13 + ⋅12 − + = + −1 + 127 0.000423 = 4.23 × 10−4 128 0.0514 = 5.14 × 10−2 =8 129 6.15 × 104 = 61,500 If x = 2, x3 + 3x − x + = ⋅ 23 + ⋅ 22 − + = 32 + 12 − + = 44 111 122 32.14 = 3.214 × 101 (666)  666  =  = = 81 222   (222) 3 1 112 (0.1)3 (20)3 =   ⋅ ( ⋅10 )  10  = ⋅ 23 ⋅103 10 = 23 = 130 9.7 × 103 = 9700 131 1.214 ×10−3 = 0.001214 132 9.88 × 10−4 = 0.000988 133 1.1× 108 = 110, 000, 000 134 4.112 × 102 = 411.2 135 8.1× 10−2 = 0.081 136 6.453 × 10−1 = 0.6453 113 (8.2)6 ≈ 304, 006.671 137 A = lw 114 (3.7)5 ≈ 693.440 138 P = ( l + w ) 115 (6.1) −3 ≈ 0.004 139 C = π d 116 (2.2) −5 ≈ 0.019 140 A = 117 (− 2.8)6 ≈ 481.890 118 − (2.8)6 ≈ − 481.890 119 (− 8.11) −4 ≈ 0.000 120 − (8.11) −4 ≈ −0.000 141 A = bh x 142 P = 3x 143 V = π r 3 144 S = 4π r Copyright © 2017 Pearson Education, Inc Chapter R: Review 145 V = x3 b 209 volts is not acceptable 146 S = x 147 a b If x = 1000, C = 4000 + x = 4000 + 2(1000) = 4000 + 2000 = $6000 The cost of producing 1000 watches is $6000 If x = 2000, C = 4000 + x = 4000 + 2(2000) = 4000 + 4000 = $8000 The cost of producing 2000 watches is $8000 148 210 + 80 − 120 + 25 − 60 − 32 − = $98 His balance at the end of the month was $98 149 We want the difference between x and to be at least units Since we don’t care whether the value for x is larger or smaller than 4, we take the absolute value of the difference We want the inequality to be non-strict since we are dealing with an ‘at least’ situation Thus, we have x−4 ≥ 150 We want the difference between x and to be more than units Since we don’t care whether the value for x is larger or smaller than 2, we take the absolute value of the difference We want the inequality to be strict since we are dealing with a ‘more than’ situation Thus, we have x−2 >5 151 a x − 110 = 108 − 110 = − = ≤ 108 volts is acceptable b x − 220 = 209 − 220 = − 11 = 11 > 153 a x − = 2.999 − = − 0.001 = 0.001 ≤ 0.01 A radius of 2.999 centimeters is acceptable b x − = 2.89 − = − 0.11 = 0.11 ≤/ 0.01 A radius of 2.89 centimeters is not acceptable 154 a x − 98.6 = 97 − 98.6 = − 1.6 = 1.6 ≥ 1.5 97˚F is unhealthy b x − 98.6 = 100 − 98.6 = 1.4 = 1.4 < 1.5 100˚F is not unhealthy 155 The distance from Earth to the Moon is about ×108 = 400, 000, 000 meters 156 The height of Mt Everest is about 8848 = 8.848 × 103 meters 157 The wavelength of visible light is about × 10−7 = 0.0000005 meters 158 The diameter of an atom is about 1× 10−10 = 0.0000000001 meters 159 The smallest commercial copper wire has a diameter of about 0.0005 = × 10−4 inches 160 The smallest motor ever made is less than 0.05 = × 10−2 centimeters wide x − 110 = 104 − 110 = − = > 104 volts is not acceptable 152 a x − 220 = 214 − 220 = − = ≤ 214 volts is acceptable 10 Copyright © 2017 Pearson Education, Inc Section R.7: Rational Expressions 51 x − x + ( x − 3)( x − 2) ( x + 4)( x + 2) − = − x + x − ( x + 2)( x − 2) ( x − 2)( x + 2) ( 2 x − 5x + − x − x − ( x + 2)( x − 2) −(11x + 2) −11x − or = ( x + 2)( x − 2) ( x + 2)( x − 2) = x − x + (2 x − 3)( x + 1) (2 x + 1)( x − 1) − = − ( x − 1)( x + 1) ( x + 1)( x − 1) x −1 x +1 = x − x = x ( x − 1) Therefore, LCM = x ( x + 1)( x − 1) 58 3x − 27 = x − = ( x + 3)( x − 3) ( x − x − 15 = ( x + )( x − 3) Therefore, LCM = ( x + 5)( x − 3)( x + 3) 59 x3 − x + x = x x − x + ( 60 x − x + x = x ( x + 3) x3 − x = x x − = x ( x + 3)( x − 3) ( = 54 ( 61 x3 − x = x x − = x ( x + 1)( x − 1) ( ) x − x + x = x ( x − x + 1) = x ( x − 1) x − = ( x − 1) ( x + x + 1) Therefore, LCM = x ( x + 1)( x − 1) ( x + x + 1) ) x2 − ( ) x ( x − )( x + ) ) ) = x3 − x + x − + x = x + x3 − x + x − ) Therefore, LCM = ( x + )( x − )( x + 1) x − x − 12 = ( x + 3)( x − ) Therefore, LCM = x ( x + ) x x − x − x + x − x − 24 x x = − ( x − 6)( x − 1) ( x − 6)( x + 4) x( x + 4) x( x − 1) = − ( x − 6)( x − 1)( x + 4) ( x − 6)( x + 4)( x − 1) = x − x + 16 = ( x − )( x − ) ( x + )3 63 x − x − = ( x + 1)( x − ) x3 + x = x ( x + ) ) x3 x + ( 62 x + x + = ( x + ) x3 x + ( 2 ( x − 1) x + + x x −1 x + = x3 x2 + x3 x + ( ( 55 x − = ( x + )( x − ) 56 ) Therefore, LCM = x ( x + 3)( x − 3) x2 − x x −4 ( x − 1) Therefore, LCM = x3 ( x − 1) ) 2 x3 x x2 + x2 − + = 53 x −4 x x x2 − = 2x − x = x = ( ) = x ( x − 1)( x − 1) 2x − x − − 2x + x + ( x + 1)( x − 1) −2 = ( x + 1)( x − 1) =− ( x + 1)( x − 1) ) 2 x − x − − (2 x − x − 1) ( x + 1)( x − 1) ) x − x + − ( x + x + 8) ( x + 2)( x − 2) = 52 57 x3 − x = x x − = x ( x + 1)( x − 1) x2 + x − x2 + x 5x = ( x − 6)( x + 4)( x − 1) ( x − 6)( x + 4)( x − 1) Therefore, LCM = ( x + 3)( x − ) 33 Copyright © 2017 Pearson Education, Inc Chapter R: Review 64 x x +1 − x − x + x − 24 x x +1 = − ( x − 3) ( x − 3)( x + 8) x( x + 8) x +1 = − ( x − 3)( x + 8) ( x − 3)( x + 8) 68 = = x2 + 8x − x − x2 + x − = = ( x − 3)( x + 8) ( x − 3)( x + 8) 65 4x − x −4 x + x−6 4x = − ( x − 2)( x + 2) ( x + 3)( x − 2) x( x + 3) 2( x + 2) = − ( x − 2)( x + 2)( x + 3) ( x + 3)( x − 2)( x + 2) = x + 12 x − x − ( x − 2)( x + 2)( x + 3) = x + 10 x − ( x − 2)( x + 2)( x + 3) 69 2(2 x + x − 2) = ( x − 2)( x + 2)( x + 3) 66 67 x−4 x−4 3x 3x − = − x − x − x + ( x − 1) ( x − 1) = x( x − 1) x−4 − ( x − 1)( x − 1) ( x − 1) = 3x − 3x − x + ( x − 1) = 3x − x + ( x − 1) ( x − 1) = = = ( x + 1) + ( x − 1)( x + 1) ( x + 1) + ( x − 1) 2 ( x − 1) − ( x + )( x − 1) 2 ( x − 1) − ( x + ) ( x + )2 ( x − 1)2 x − − x − 12 ( x + )2 ( x − 1)2 −4 x − 14 ( x + )2 ( x − 1)2 −2 ( x + ) = ( x + )2 ( x − 1)2 = ( x + 2) 70 2x + x+4 − x2 − x − x2 + x − 2x + x+4 = − ( x − 2)( x + 1) ( x + 4)( x − 2) ( x + 4)( x + 4) (2 x + 3)( x + 1) = − ( x − 2)( x + 1)( x + 4) ( x + 4)( x − 2)( x + 1) = x + x + 16 − (2 x + x + 3) ( x − 2)( x + 1)( x + 4) = − x + 3x + 13 ( x − 2)( x + 1)( x + 4) 2x − x−2 − x + x + ( x + 1) 2x − x−2 = − ( x + 1)( x + 7) ( x + 1) (2 x − 3)( x + 1) ( x − 2)( x + 7) = − ( x + 1)( x + 7)( x + 1) ( x + 1) ( x + 7) = x − x − − ( x + x − 14) ( x + 1) ( x + 7) = x − x + 11 ( x + 1) ( x + 7) ( x − 1)2 ( x + 1)2 3x + + x − ( x − 1)2 ( x + 1)2 5x + ( x − 1)2 ( x + 1)2 34 Copyright © 2017 Pearson Education, Inc Section R.7: Rational Expressions 71 − + x x + x x3 − x 2 = − + x x ( x + 1) x ( x − 1) = = 72 x ( x + 1)( x − 1) − x ( x − 1) + ( x + 1) x ( x + 1)( x − 1) x x − − x + x + 3x + ( ) x ( x + 1)( x − 1) = x3 − x − x + x + x ( x + 1)( x − 1) = x3 − x + x + x ( x + 1)( x − 1) x ( x − 1) = 74 + x ( x − 1) = = x +1 − x x ( x − 1) = = = 73 x3 + x ( x − 1) − ( x + 1)( x − 1) x ( x − 1) x3 + x x − x + − x − ( ) ( x ( x − 1) ) x3 + x3 − x + x − x + x ( x − 1)  x +   x +1     x =  x x  =  x  = x +1 ⋅ x = x +1 75  x   x −1  x x −1 x −1 1− − x  x x   x  1+ = − x h − h2 hx ( x + h) h( − x − h) hx ( x + h) − 2x − h = x ( x + h) 2x + h =− x ( x + h) 2 x +1 − x x3 − x + 1 1  −   h  ( x + h) x  1( x + h)   1⋅ x2 =  −  h  ( x + h ) x x ( x + h)   x − ( x + xh + h )  =   h x ( x + h)   4x2   x2 +   + 2   x   x2  x2 =  x 76 = 2 −  x −   x −  2 x x   x2   x 4+ x2 + x2 ⋅ 3x − x2 4x +1 = 3x − = 3x3 − x + x + x ( x − 1) 1( x + h)  1 1   1⋅ x − =  −  h  x + h x  h  ( x + h) x x( x + h)  1 x−x−h =  h  x( x + h)  −h = hx( x + h) −1 = x ( x + h) x +1 2x x +1 2x − x −1 − x x x x = = 77 x − ( x + 1) x − 3x + + x − 3+ + x +1 x +1 x +1 x +1 x −1 x −1 x +1 = x = ⋅ 4x + x ( x + 1) x +1 ( x − 1)( x + 1) = x ( x + 1) 2− 35 Copyright © 2017 Pearson Education, Inc Chapter R: Review  x +1 − x    x  x +1 x +1   x +1  =  x + = 78 x −1 x x x 1 − +     2−  x − x   x  x     x = ⋅ x +1 x +1 x = ( x + 1) 1− x + x −3 − 79 x − x + x +1  ( x + 4)( x + 1) ( x − 3)( x − 2)   ( x − 2)( x + 1) − ( x + 1)( x − 2)   = x +1  x + x + − ( x − x + 6)    ( x − 2)( x + 1)  = x +1 10 x − = ⋅ ( x − 2)( x + 1) x + 2(5 x − 1) = ( x − 2)( x + 1) x−2 x − + − x x 80 x+3 x( x + 1)   ( x − 2)( x − 2)  ( x + 1)( x − 2) − ( x − 2)( x + 1)   = x+3  x − x + − ( x + x)    ( x − 2)( x + 1)  = x+3 −5 x + = ⋅ ( x − 2)( x + 1) x + −5 x + = ( x − 2)( x + 1)( x + 3) = x − x −1 + x + x +1 81 2x − x − x +1 x  ( x − 2)( x + 1) ( x − 1)( x + 2)   ( x + 2)( x + 1) + ( x + 1)( x + 2)   =  (2 x − 3)( x + 1)  x2  ( x + 1)( x) − x( x + 1)    x2 − x − + x2 + x −    ( x + 2)( x + 1)   = 2  x − (2 x − x − 3)    x( x + 1)    2x2 −   ( x + 2)( x + 1)   =  −x + x +   x( x + 1)    = x( x + 1) 2( x − 2) ⋅ ( x + 2)( x + 1) −( x − x − 3) = x( x − 2) −( x + 2)( x − x − 3) = −2 x ( x − 2) ( x + 2)( x − x − 3) − (5x − 4) ( x − 2)( x + 1)( x + 3) 36 Copyright © 2017 Pearson Education, Inc Section R.7: Rational Expressions 2x + x − − x x 82 ( x + 1) x2 − x −3 x+3 x( x)   (2 x + 5)( x − 3) −  x( x − 3) x( x − 3)   =  x ( x + 3) ( x − 3)( x + 1)   ( x − 3)( x + 3) − ( x − 3)( x + 3)     x − x − 15 − x    x( x − 3)  =   x + x − ( x3 − x − x − 3)    ( x − 3)( x + 3)    x − x − 15   x( x − 3)   =   x + 5x +   ( x − 3)( x + 3)    = x − x − 15 ( x − 3)( x + 3) ⋅ x( x − 3) x + x + = ( x − x − 15)( x + 3) x(4 x + x + 3) 83 − 1 1− x ( x − 1) 2 +3 + 1 x x x −1 − − = = 85 −1 ( x − 1) 3 ( x − 1) + +2 + x −1 x −1 x −1 + ( x − 1) x −1 = + ( x − 1) x −1 + ( x − 1) x −1 = ⋅ + ( x − 1) x −1 −1 ( x − 1) + = = + ( x − 1) + ( x − 1) = + 3x − 3 + 2x − 3x − 2x + 3( x + 2) 4 −3 − x+2 = x+2 = x+2 86 −1 x + 2) ( 3( x + 2) −1 −1 − x+2 x+2 x+2 − 3( x + 2) x+2 = − ( x + 2) x+2 − 3( x + 2) x+2 = ⋅ x+2 − ( x + 2) −1 ( x + 2) − x −1 x x = 1− x −1 x −1− x = x −1 −1 = x −1 = 1− = = 1 = 1− = 1− 84 − −x 1− x −1 1− 1− x 1− x 1− x 1− x 1− x = 1− = 1+ −x x x +1− x = x = x 87 − 3( x + 2) − ( x + 2) − 3x − 3− x − −3 x − x + = −x +1 x −1 ( x + 3) ⋅ − ( 3x − 5) ⋅ x + − x + 10 = ( x − )2 ( x − )2 = 88 = 19 ( 3x − 5)2 ( x + 1) ⋅ − ( x − ) ⋅ 20 x + − 20 x + = ( x − )2 ( x − )2 37 Copyright © 2017 Pearson Education, Inc = 13 ( x − )2 Chapter R: Review 89 x ⋅ x − x + ⋅1 ( (x ) ) +1 2 x2 − x2 − = (x (x ( (x ) −4 ) 2 ) +1 2 (x = = +1 ) (x −4 ) = x2 + (x −4 ) 95 x +4 ( x + 2) ( x − )2 3x2 + x = f = x3 − 15 x + ⋅ − ( 3x + ) ⋅ x ) +1 = = −2 x + 10 x + 18 +9 (x 96 2 − 5x − +9 ) ) R R + R1 R3 + R1 R2 1 1 = + + = R R1 R2 R3 R1 R2 R3 ⋅ ⋅10 ⋅10 + ⋅10 + ⋅ 200 20 = = ohms 110 11 = ) +1 −3x − x + 2 ( x + 1) − ( x + x − 3) = ( x + 1) 2 ( 3x − 1)( x + 3) (x ) +1 38 Copyright © 2017 Pearson Education, Inc +9 R1 ⋅ R2 (n − 1) ( R2 + R1 ) R1 R2 R3 R2 R3 + R1 R3 + R1 R2 ) 0.1(0.2) (1.5 − 1)(0.2 + 0.1) 0.02 0.02 = = = meters 0.5(0.3) 0.15 15 R= 3x + − x − x =− = (x f = ( x − )2 x ( x − 15 ) = ( x − )2 ) (x  1  = (n − 1)  +  f R R    R + R1  = (n − 1)   f  R1 ⋅ R2  ( x − ) ⋅ 3x − x3 ⋅ x3 − 15 x − x3 = ( x − )2 ( x − )2 ) x + 18 − x + 10 x R1 ⋅ R2 = (n − 1) ( R2 + R1 ) f f = R1 ⋅ R2 (n − 1) ( R2 + R1 ) ( 3x + 1) ⋅ x − x ⋅ x + x − 3x = ( 3x + 1)2 ( 3x + 1)2 (x +9 = (x −2 ( x = (x ( 3x + 1) x ( 3x + ) = ( 3x + 1)2 93 2 x2 − x2 + = 92 ) (x + ⋅ − ( x − 5) ⋅ x 2 91 (x ( x − 1)( x + 1) = x ⋅ x − x − ⋅1 ) +1 x2 − = 90 94 ) Section R.8: nth Roots; Rational Exponents x +1 =  a = 1, b = 1, c = x x 1 x 1+ = 1+ = 1+ +1 x x +   1+  x  x   x + + x 2x + = = x +1 x +1  a = 2, b = 1, c = 97 + 1+ 1+ = 1+ 1 x 1+ x +1 = 1+ 2x +  2x +1   x +1    x + + x + 3x + = 2x +1 2x +1  a = 3, b = 2, c = = 1+ 1+ 1+ 1 1+ x 2x + = 1+ = 1+ 3x +  3x +   2x +1    3x + + x + x + = = 3x + 3x +  a = 5, b = 3, c = If we continue this process, the values of a, b and c produce the following sequences: a :1, 2,3,5,8,13, 21, b :1,1, 2,3,5,8,13, 21, c : 0,1,1, 2,3,5,8,13, 21, In each case we have a Fibonacci Sequence, where the next value in the list is obtained from the sum of the previous values in the list d c c true 10 False; ( −3 ) = −3 = 11 27 = 33 = 12 16 = 24 = 13 −8 = ( −2 )3 = −2 14 −1 = ( −1)3 = −1 15 = 4⋅2 = 2 16 75 = 25 ⋅ = 17 700 = 100 ⋅ = 10 18 45 x3 = x x = x x 19 32 = ⋅ = 20 54 = 27 ⋅ = 3 21 −8 x = −8 x3 ⋅ x = −2 x x 98 Answers will vary 22 192 x5 = 64 x3 ⋅ x = x 3x 99 Answers will vary 23 243 = 81⋅ = 24 48 x5 = 16 x ⋅ 3x = x 3x 25 x12 y = 26 x10 y = x 27 x9 y = x y = x2 y xy Section R.8 9; −9 4; −4 = 4 4 (x ) ( y ) ( ) y5 = x2 y index cube root b 39 Copyright © 2017 Pearson Education, Inc = x3 y Chapter R: Review 28 3xy = 81x y = 27 x3 3 27 x 36 x = x 30 x5 = x ⋅ x = 3x x 4 = 3x y 43 )( ) ( 3) −1 = ( 5−2 )( ) ( 5) = 53 −63 3 =−3 x 20 x3 = 100 x = 10 x ) = ( 5) ( 9) 3 10 = (18 − 3) 3 ) = ( ) ( 10 ) = 15 3 (3 )( 2 ) = 38 (5 )( −3 ) = −15 47 ( ) = ( x) x −1 48 ( x+ 24 = −30 2 +2 ( x )( ) + ( ) = x + 5x + 49 16 x − x = x3 ⋅ x − x = 2x 2x − 2x = ( x − 1) x 39 + = ( + ) = 40 − = ( − ) = 50 32 x + x5 = 16 ⋅ x + x ⋅ x = 2x + x 2x 41 − 18 + = − ⋅ + ⋅ = ( + x ) x or = −3 + = ( −3 + ) 51 ( x + 2) 2x x3 − 50 x = x ⋅ x − 25 ⋅ x = = x x − 15 x = ( x − 15 ) x 42 12 − 27 = ⋅ − ⋅ = ( − 9) − x +1 ) = ( x) 12 = ⋅ = 12 = −9 = x − x +1 = 34 ⋅102 = 3 ⋅100 = 300 3 37 46 24 − 81 = ⋅ 3 − 3 = 18 3 − 3 = ⋅ 92 = 81 = ⋅ 3 = 15 3 = (5 − 6) 5x y 34 −2 +3 −6 = −1 3x 12 x = 36 x ⋅ x = x x 2 = 5+ −6 ( ) y(y ) 539 +3 − −3 45 − 54 = − ⋅ 3 −40 x14 y10 = 5( −2)3 x x ( +3 = 33 ( 3+3 = 3+ −3 2x = −2 x y 36 ( ( ) (y ) 33 35 3x 44 162 x9 y12 = (3) x x 2 34 32 = =2 29 31 52 3x y + 25 y = x y + 20 y = ( x + 20 ) y = −5 40 Copyright © 2017 Pearson Education, Inc Section R.8: nth Roots; Rational Exponents 53 16 x y − 3x xy + −2 xy 62 = x3 ⋅ xy − 3x xy + − y ⋅ xy −1 −1 − = ⋅ +3 +3 −3 = x xy − x xy − y xy = = ( x − x − y ) xy = ( − x − y ) xy or − ( x + y ) xy 63 54 xy − 25 x y + x3 y = xy − xy + xy = −1 = = ( − + ) xy = xy 55 1 2 = ⋅ = 2 2 56 2 3 = ⋅ = 3 3 57 58 5+ 59 = ⋅ 5− 5− 5+ = 60 = −3 5−4 ⋅ 5+4 5−4 5 53 = ⋅ = 2 32 34 66 −2 −2 3 −2 3 = ⋅ = 3 39 33 x+h − x = x+h + x ) ( x + h) − x ( x + h) + x ( x + h) − x ) = x + h − x + xh + x x+h−x = x + h − x + xh h 25 − ( 5+ 23 ( or 3+ 23 −2 ) 7−4 ( −2 x+h − x x+h − x ⋅ x+h + x x+h − x = 2 −2 = ⋅ +2 +2 −2 = 61 ( −3 = 5+4 65 67 = +5 = +5 −1 −3 + 12 −3 + 12 = −11 − 16 − 12 = 11 − − − − − = = ⋅ = = 2 2 2⋅2 5+ +1 ⋅ −1 +1 = − − − 15 = ⋅ = 5 5 64 6−2 −3 +3 9−5 = 12 − ) or 14 − 2 2− 2− 2−3 = ⋅ 2+3 2+3 2−3 − − + 15 − 45 19 − 8 − 19 = = −41 41 = 41 Copyright © 2017 Pearson Education, Inc Chapter R: Review 8 79   9 x+h + x−h x+h − x−h 68 −3/ 9 =  8 x+h + x−h x+h + x−h ⋅ x+h − x−h x+h + x−h = = ( x + h ) + ( x − h )( x + h ) + ( x − h ) = ( x + h) − ( x − h) x + h + x2 − h2 + x − h x+h−x+h = = x + x − h2 2h = x + x − h2 h 69 82 / = ( 8) 70 43/ = ( 4) 71 = 3   80    27  81 = 22 = = 23 = 72 16 3/ 73 16 3/ 74 25 = = = 75 9−3/ = ( 16 ) ( 9 77   8 25 83 = 93/ −3/4 84 −81 = = 64 = = 125 ( 9) = 163/ = ( 16 ) 1 = 33 27 = −2/3 =3− 2/3 = = =− =− 125 − 64 25 16 3/ 1 =− 1000 10 1 =− 25 =− 125 = − 64 1/3 81 27 86 x / x1/ x −1/ = x / +1/ −1/ = x11/12 ( ) 2/3 2  27  3 =3  =  =     1/2 =− 81 27 27 27 = = ⋅ ⋅ 2 16 16 2 1/ = x3 3/ = x4 87 (x y ) 88 (x y ) 89 1/ 1/ ( ) (y ) 1/ 3/ = x3 y 2/3 1/ ( y )1/ ( x )2 / ( y ) x2 / y / 3/ = xy ( ) (y ) ( x y ) ( xy ) 27 = 32  27  78     2/3 85 x3/ x1/ x −1/ = x3 +1/ −1/ = x /12 1 = 64 27 27 = ⋅ 2 16 = 1000 − 25 =−  9   =  =  =   2 2 2 =  27  =    = − 3 ( 2) 3 27 27 ⋅ = 32 16 2 =2 =8 ) ( 16 ) 76 16−3/ = 3/ ( −1000)−1/3 = 64 − 125  9   =  =   8 2 2 33 82 −25−1/ = − ( −27 )1/ = −27 = −3 3/ −2 / 3/ (x ) = = x2 / y2 / x / y1/ x / y / x2 / y2 / = x / + / 3− / y1/ + / 3− / 2  27  3 =3  =  =     42 Copyright © 2017 Pearson Education, Inc = x / y1 = x / y 2/3 Section R.8: nth Roots; Rational Exponents 1/ 90 ( xy )1/ ( x y ) x2 y ( ) = 1/ ( ) (y ) (x ) y = 3/ 1/ x1/ y1/ x 2 3/ 94 3/ x1/ y1/ xy x3/ y 3/ 91 16 x y −1/ ( 3/ ( y1/ x1/ ) 1/ ( xy ) 3/ 163/ x y −1/ 3/ 1/ 2 x x2 + 1/ ) ) 1/ 92 −1 1/ 3/ y ) ( xy )3/ = ) +1 −1/ = 1/ 1/ 2 1/ 2 x3 + x + x3 1/ 2 ( x + 1) x ( 3x + ) = ( x + 1) x5 / y 3/ ⋅ 2x ) 1/ 2 y = 2 ) + 1) x3 + x (x +1 3/ 1/ 3/ ( ) (y ) x3/ y 3/ 96 x −3/ y1/ 1/ ) 1/ 1/ x3/ y 3/ + x ( x + 1) 2/3 ( x + 1)1/ + x 2/3 ( x + 1) / 3+1/ 3 ( x + 1) + x ( x + 1) + x = = 2/3 2/3 ( x + 1) ( x + 1) = 23 x −3/ −3/ y1/ −3/ = = x −3 y −1 = x y 1/ 1/ x + (1 + x ) (1 + x ) x 1/ 93 + (1 + x ) = 1/ (1 + x) (1 + x)1/ = ( x + 1)1/ + x ⋅ 13 ( x + 1)−2 / , x ≠ −1 = ( x + 1) = ( x + 1) 2/3 3x + + x ( x + 1) 2/3 x + (1 + x ) (1 + x)1/ x + + 2x = (1 + x)1/ 3x + = (1 + x)1/ 43 Copyright © 2017 Pearson Education, Inc = 4x + 3 ( x + 1) 2/3 + + x3 43/ x −1 1/ +1/ 2 3/ −1/ / −3/ ( 4) = ( (x 2 (4x + x +1 x3 ) ⋅ ( x + 1) + x = ( x + 1) +x x ( x + 1) 2x ( x = = ( x + 1) (x = 23 x3/ −1/ y −1/ −1/ = + x2 ⋅ = x x2 + ( x1/ y1/ = 8x ) ( ( ) ( = x (y ) ( 16 ) x y = 1/ 1/ 95 x x + = x1/ +1−3/ y1/ +1−3/ = x −1/ y1/ = 1+ x + x + x1/ ⋅ x1/ 1/ x + = x1/ 2 x1/ + x + x 3x + = = 1/ 2 x1/ 2x 1/ Chapter R: Review 97 4x + ⋅ = = = = = 98 x −5 4x + + x − 5 4x + 4x + ⋅ ⋅ 4x + + x − ⋅ ⋅ x − 10 x − x + ( x + 3) + ( x − ) 10 ( x − 5)( x + 3) 20 x + 15 + x − 10 10 ( x − )( x + 3) 22 x + 10 3 1 + x−5⋅ ,x >5 x−5 4x + 3 ( x − )( x + 3) 8x + ( x − )2 + 24 x−2 (8 x + 1)2 , x ≠ 2, x ≠ − = = 8 x + ⋅ ( x + 1) + x − ⋅ ( x − ) 24 3 = ( x − )2 ⋅ (8 x + 1)2 (8 x + 1)3 + ( x − )3 = 64 x + + x − 24 24 = ( x − )2 ⋅ (8 x + 1)2 ( x +1) + x − = 2 24 ( x − ) ( x + 1) 24  2x   x + − x ⋅  x2 +   100 x2 +  x2   x + −  x2 +   = x2 +  x2 + x2   x +1 ⋅  −  x2 + x +   = x2 +  x2 + − x2    1 x2 +   = = ⋅ 2 x +1 x +1 x +1 ( x − )2 (8 x + 1)2 65 x + ( x − )2 (8 x + 1)2    x   1+ x − x ⋅   1+ x −  1+ x   1+ x  = 99  1+ x 1+ x  1+ x 1+ x − x    1+ x  = 1+ x 2(1 + x) − x = ⋅ 2(1 + x)1/ + x 2+ x = 2(1 + x)3/ 101 (x +1 ) 3/ ( x + )1/ − x ( x + )−1/ x+4   1/ 2x  ( x + 4) −  1/   x + ( )  = x+4 1/  2x 1/ ( x + )  ( x + 4) ⋅ − 1/  ( x + 4) ( x + )1/ = x+4  x + − 2x     ( x + )1/    = x+4 −x + = ⋅ 1/ x + ( x + 4) = = −x + ( x + )3/ 4− x ( x + )3/ 44 Copyright © 2017 Pearson Education, Inc     Section R.8: nth Roots; Rational Exponents 1/ 102 (9 − x ) ( ) −1/ , −3 < x < − x2   − x2    1/ ( ) ( ( = 9− x + 1/ ⋅ − x2 ) ( ) (9 − x ) + x2 1/ = ) x2 − 1/ ) − x −1 ( ) 1/ ) ⋅(x = ( x − 1) x − ( x − 1) = ⋅ ( x − 1) x 1/ = x2 − x2 + 1 ⋅ 1/ x x2 − ( ) x2 x2 − ( 1/ ) −1/ x2 − (x +4 +4 x +4 1/ 1/ 1/ 2 ) ⋅ ( x + 4) ( x + 4) x2 + − x2 (x ) − x     +4 1/ ) ⋅ − x2 x2 + ⋅ 1/ 2     1/ ) ( ) ( ) = x +4 x +4 ( 3/ ) )  + x2 − x x x      x   = −1 1/ ) ⋅ )( ) 2      = x2 = + x2 (1 + x ) − ( x )( x x ) ⋅ x + x2 − 4x2 ⋅ x + x2 ( 2 = 1/ ) 1/ ) ( ) ( ( ) ( (x = , x < −1 or x > x2 − x2 − ) 1/ (  x − x − 1/ ⋅ x − 1/  1/   x2 −  = x2 ( ( + x2 − 2x x x ,x > 105 + x2 1/ x2 − x2 x2 + x2 + = (9 − x ) 1/ ) ( ⋅ − x2 3/ x2 +4  x2 + ⋅ x2 +  1/   x2 + = x2 + ⋅ − x2 1/ − x2 ( + x     − x2 1/ (x   x +4  = 2 1/ = ( 1/ ) ⋅ (9 − x ) (9 − x ) − x2 + x2 ( 1/ (9 − x ) 1/ 104      x2 − x2   9− x   = 103 + x2 − x2 106 x − x2 ( ) = − 3x 2 x + x2 ( ) + 23 x (1 − x ) (1 − x ) 1/ 1/ ) + x3 ( 1− x 2/3 2/3 2/3 2/3 45 Copyright © 2017 Pearson Education, Inc ) 2/3 (1 − x )  2x − x ) (1 − x )  (   (1 − x ) = (1 − x ) 1/ , x ≠ −1, x ≠ 2/3 ( ) −2 / 3   2x − x  = 2 (1 + x )     + x3     Chapter R: Review = = = x − x2 ( 1/ 3+ / ) − x2 ( ) + x3 x − x + x3 ( ) − x2 ( ) ( 1− x 4/3 ) − x2 ( ( ( 1− x ) 4/3 ) ( ( = 2x = x1/ ( x + 3)3/ + ( 3x + 5)4 / ( x + 3)1/ 1/ 1/ = ( 3x + ) ( x + 3)  ( x + 3) + ( x + )  1/ ( x + 3)1/ (8 x + 12 + x + 15) 1/ 1/ = ( 3x + ) ( x + 3) (17 x + 27 ) 1/ ( x − 3)3/ + ( x + 1)4 / ( x − 3)1/ 1/ 1/ = ( x + 1) ( x − 3) ( x − 3) + ( x + 1)  114 ( x + 1) 1/ ( x − 3)1/ (10 x − ) 1/ 1/ = ( x + 1) ( x − 3) ( )( x − 1) 1/ 1/ = 12 ( x + 1) ( x − 3) ( x − 1) = ( x + 1) where x ≥ 1/ x ,x > 3 = 1/ + x1/ 2 x 115 3x −1/ + = ⋅ + x1/ ⋅ x1/ + x ( x + ) = 1/ = x1/ 2x x1/ 116 x1/ − x −2 / , x ≠ = x1/ − / x ( 3( x + x) − x − 4) ( 3x − x − ) = x1/ (3 x − 4)( x + 1) = 110 x1/ ( x + 3) + x3/ ⋅ = x1/ ( 3(2 x + 3) + x ) x1/ ⋅ x / − x − 4 ( x − 1) = 2/3 = x2 / x x2 / 117 ≈ 1.41 118 ≈ 2.65 = x1/ (10 x + ) 1/ ( ) + x ⋅ ( x + 4) ⋅ 2x = ( x + ) 3 ( x + ) + x    = ( x + ) 3 x + 12 + x  = ( x + ) (11x + 12 ) 1/ 1/ 1/ 2 where x ≥ − ) 4/3 (5x + 4) = ( 3x + 5) ) 111 x + 1/ 1/ ) 109 x1/ x + x − x3/ − x1/ 1/ ( 3x + ) + x  113 ( x + ) ) ( x + 4) / + x ⋅ ( x + 4)1/ ⋅ x   = ( x + 4)1/  x + + x     11  = ( x + 4)1/  x +  3  1/ = x2 + 11x + 12 ( 1/ + x ⋅ ( 3x + ) 1/ = x ( 3x + ) 4/3 107 ( x + 1)3/ + x ⋅ ( x + 1)1/ 2  1/  = ( x + 1)  x + + x    5  = ( x + 1)1/  x + 1   = ( x + 1)1/ ( x + ) 108 4/3 = x ( 3x + ) 2/3 (1 − x ) x − x2 = 112 x ( 3x + ) x − x3 + x3 = / 3+ / x − x3 ⋅ 2/3 2 46 Copyright © 2017 Pearson Education, Inc Section R.8: nth Roots; Rational Exponents 119 120 121 3 ≈ 1.59 126 a −5 ≈ −1.71 2+ ≈ 4.89 3− v = 64 ⋅ + 02 = 256 = 16 feet per second b v = 64 ⋅16 + 02 = 1024 = 32 feet per second c v = 64 ⋅ + 42 = 144 = 12 feet per second 127 T = 2π 64 = 2π ≈ 8.89 seconds 32 128 T = 2π 16 2π = 2π = 32 2 = π ≈ 4.44 seconds 129 Answers may vary One possibility follows: If a = −5 , then ( −5)2 = 25 = ≠ a Since we use the principal square root, which is always non-negative, a if a ≥ a2 =  − a if a < −2 ≈ 0.04 2+4 122 a2 = which is the definition of a , so 123 35− ≈ 2.15 124 3− ≈ 1.33 125 a b 96 − 0.608 12 ≈ 15, 660.4 gallons V = 40 (12 ) V = 40 (1) 2 96 − 0.608 ≈ 390.7 gallons 47 Copyright © 2017 Pearson Education, Inc a2 = a