Solution manual for college algebra 10th edition by larson

Solution Manual for College Algebra 10th Edition by Larson CHAPTER P Prerequisites Section P.1 Review of Real Numbers and Their Properties Section P.2 Exponents and Radicals Section P.3 Polynomials and Special Products Section P.4 Factoring Polynomials 16 Section P.5 Rational Expressions 22 Section P.6 The Rectangular Coordinate System and Graphs 31 Review Exercises 37 Problem Solving 44 Practice Test .48 © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson C H A P T E R Prerequisites P Section P.1 Review of Real Numbers and Their Properties 11 (a) irrational origin −2 −1 4 −5 −4 −3 −2 −1 (b) −1 (c) composite terms − (d) x x −5.2 Zero-Factor Property −9, − 72 , 5, 23 , x absolute value x −7 −6 −5 −4 −3 −2 −1 2, 0, 1, − 4, 2, −11 12 (a) 8.5 (a) Natural numbers: 5, 1, −1 (d) Rational numbers: −9, − 72 , 5, 23 , 0, 1, − 4, 2, −11 10 11 12 x (c) Integers: −9, 5, 0, 1, − 4, 2, −11 (e) Irrational numbers: (b) (b) Whole numbers: 0, 5, 1, 0 (c) x −4.75 x −7 −6 −5 −4 −3 −2 −1 (d) 5, − 7, − 73 , 0, 3.14, 54 , − 3, 12, −8 −5 −4 −3 −2 −1 x (a) Natural numbers: 12, (b) Whole numbers: 0, 12, 13 −4 > −8 (c) Integers: −7, 0, − 3, 12, −8 (d) Rational numbers: −7, − 73 , 0, 3.14, 54 , − 3, 12, (e) Irrational numbers: −7 14 < −6 −5 16 16 2.01, 0.6, −13, 0.010110111 , 1, − 15 (a) Natural numbers: > (d) Rational numbers: 2.01, 0.6, −13, 1, − 9, 3.12, 12 π , 7, −11.1, 13 (a) Natural numbers: 25, (b) Whole numbers: 25, 9, 7, 13 9, 7, 13 9, 7, 13 (d) Rational numbers: 25, −17, x x 16 − 87 < − 73 (e) Irrational numbers: 0.010110111 − 12 , 5 (c) Integers: −13, 1, − (c) Integers: 25, −17, (b) Whole numbers: , 10 25, −17, − 12 x −4 − 87 −2 − 37 −1 x 17 (a) The inequality x ≤ denotes the set of all real numbers less than or equal to (b) x (c) The interval is unbounded 9, 3.12, 7, −11.1, 13 (e) Irrational numbers: 12 π © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Section P.1 18 (a) The inequality x < denotes the set of all real numbers less than zero (b) x −2 −1 19 (a) The inequality −2 < x < denotes the set of all real numbers greater than −2 and less than −1 20 (a) The inequality < x ≤ denotes the set of all real numbers greater than zero and less than or equal to x (c) The interval is bounded numbers greater than or equal to 4 (c) The interval is unbounded 22 (a) than (b) x (c) The interval is unbounded 23 (a) The interval [−5, 2) denotes the set of all real numbers greater than or equal to − and less than (b) x −5 −3 −1 24 (a) The interval ( −1, 2] denotes the set of all real numbers greater than −1 and less than or equal to x −2 −1 (c) The interval is bounded 25 y ≥ 0; [0, ∞ ) 26 y ≤ 25; ( −∞, 25] x + = x + −( x + ) x + = −1 38 If x > 1, then x − is positive x −1 = x −1 x −1 = x −1 40 −5 = − because −5 = −5 41 − −6 < −6 because −6 = and − −6 = −(6) = −6 42 − −2 = − because −2 = −2 43 d (126, 75) = 75 − 126 = 51 44 d ( − 20, 30) = 30 − ( − 20) = 50 = 50 (c) The interval is bounded (b) 37 If x < −2, then x + is negative 39 −4 = because −4 = and = (−∞, 2) denotes the set of all real numbers less 36 − − = − 4( 4) = −16 So, x 33 −1 − −2 = − = −1 So, 21 (a) The interval [4, ∞) denotes the set of all real (b) 31 − = −5 = −( −5) = 35 − = 5(5) = 25 (c) The interval is bounded (b) 30 = 34 −3 − −3 = −3 − (3) = −6 x −2 32 − = = (c) The interval is unbounded (b) Review of Real Numbers and Their Properties ( ) ( ) 45 d − 52 , = − − 52 ( ) = ( ) = − 11 − − 14 46 d − 14 , − 11 4 = − 52 = 47 d ( x, 5) = x − and d ( x, 5) ≤ 3, so x − ≤ 48 d ( x, −10) = x + 10 , and d ( x, −10) ≥ 6, so x + 10 ≥ 27 10 ≤ t ≤ 22; [10, 22] 28 −3 ≤ k < 5; [−3, 5) 29 −10 = −( −10) = 10 © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Chapter P Prerequisites Expenditures, E R − E 49 $2524.0 billion $2982.5 billion 2524.0 − 2982.5 = $458.5 billion 50 $2162.7 billion $3457.1 billion 2162.7 − 3457.1 = $1294.4 billion 51 $2450.0 billion $3537.0 billion 2450.0 − 3537.0 = $1087.0 billion 52 $3021.5 billion $3506.1 billion 3021.5 − 3506.1 = $484.6 billion Receipts, R 53 x + 60 − 7x Terms: x, (a) − 7( −3) = + 21 = 30 Coefficient: (b) − 7(3) = − 21 = −12 54 x − 61 x − 3x + Terms: x, − Coefficient: 55 x − x (a) (0) (b) (−1) (b) −(1) + 5(1) − = −1 + − = 56 x + 0.5 x − Terms: x , 0.5 x, − Coefficients: 4, 0.5 63 x +1 x −1 (a) 57 3 x + Terms: 3 x , (b) Coefficient: 3 2x2 − 2x , − Coefficient: − 3( −1) + = + + = (a) −( −1) + 5( −1) − = −1 − − = −10 Coefficients: 6, − Terms: 2 62 − x + x − Terms: x , − x 58 − 3(0) + = 2 64 59 x − 1+1 = 1−1 Division by zero is undefined −1 + = = −1 − −2 x − x + (a) − = = + (b) −2 − −4 = −2 + (a) 4( −1) − = −4 − = −10 Division by zero is undefined (b) 4(0) − = − = −6 65 (h + 6) = 1, h ≠ −6 ( h + 6) Multiplicative Inverse Property 66 (x + 3) − ( x + 3) = Additive Inverse Property 67 x(3 y ) = ( x ⋅ 3) y = (3x) y Associative Property of Multiplication Commutative Property of Multiplication © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Section P.2 68 (7 ⋅ 12) = ( 17 ⋅ 7)12 Exponents and Radicals Associative Property of Multiplication = ⋅ 12 Multiplicative Inverse Property = 12 Multiplicative Identity Property 69 2x x 8x 3x 5x − = − = 12 12 12 70 3x x 15 x 4x 19 x + = + = 20 20 20 71 3x x x ⋅ = ⋅ = 10 2 72 2x 2x x 7x ÷ = ⋅ = ⋅ = 3 73 False Because zero is nonnegative but not positive, not every nonnegative number is positive 76 (a) Because the price can only be a positive rational number with at most two decimal places, the description matches graph (ii) (b) Because the distance is a positive real number, the description matches graph (i) A range of prices can only include zero and positive numbers with at most two decimal places So, a range of prices can be represented by whole numbers and some noninteger positive fractions A range of lengths can only include positive numbers So, a range of lengths can be represented by positive real numbers 77 (a) 74 False Two numbers with different signs will always have a product less than zero 75 The product of two negative numbers is positive n 0.0001 0.01 100 10,000 5n 50,000 500 0.05 0.0005 (b) (i) As n approaches 0, the value of n increases without bound (approaches infinity) (ii) As n increases without bound (approaches infinity), the value of n approaches Section P.2 Exponents and Radicals exponent; base 11 (a) (23 ⋅ 32 ) = 23 ⋅ ⋅ 32 ⋅ 2 scientific notation square root index; radicand = 26 ⋅ 34 = 64 ⋅ 81 = 5184 (b)  −   5 like radicals conjugates 12 (a) rationalizing power; index (a) ⋅ 53 = 54 = 625 (b) 10 (a) 1 = −2 = = 54 25 (33 ) =1 = ⋅ 34 = 35 = 243 3− (b) 48( −4) 13 (a) (b) −3 = (−2)0 (−4) = − 48 = − 64 =1 14 (a) 3−1 + 2−2 = (b) 48 ⋅ 3−2 16 = ⋅ 22 ⋅ 3−2 − (−1) = ⋅ ⋅ 3−1 = 2−2 ⋅ 3−1 (b) −3 = −9 52 33 5 3− ⋅ 52 −   = ( −1) ⋅ = −1 ⋅ 3 = −3 ⋅ 5−1 = − (3−2 ) = 3−4 = + 34 = = 12 + 12 = 12 81 © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Chapter P Prerequisites 15 When x = 2, 24 (a) −3x3 = −3( 2) = −24 (b) 16 When x = 4, x −2 = ( ) −2 = 7 = 42 16 x2 = x − = x −1 = x3 x 12( x + y ) 3 4 3 43 34 64 ⋅ 81 5184 25 (a)     = ⋅ = = 3+ y y y y y y7     17 When x = 10,  b −2  b   a  b  (b)  −2   =    = 1, a ≠ 0, b ≠  a  a   b  a  x0 = 6(10) = 6(1) = 18 When x = −3, −1 26 (a) ( x y −2 )    x3 = 2( −3) = 2( −27) = −54 −1 −3 x = −3( −2) = ( x −2 y ) x2 y2 = = −3(16) = −48 (b) (5 x z ) (5 x z ) −3 ( 13 ) 21 (a) (5 z )3 ( 271 ) = = 12 = (1)(1)(1) =1 = z = 125z 27 (a) ( ) (b) 5x x = 5x + = 5x6 22 (a) (b) ( x3 ) 28 (a) = 40 ⋅ x = 1, x ≠ 0 ( − z ) (3 z ) ( x + 5) (2 x2 ) −2 = 1, x ≠ −5 = (2x ) 2 ( ) 23 (a) y y (b) (b) (− x) = (− 2) x2 = 4x2 = (53 )( x )( z18 )(5−3 )( x −6 )( z −18 ) = 50 ( x )( z ) 20 When x = − 13 , 12( − x) = 12 −1 = x y −2 19 When x = −2, 4 −1 ( x + y) = ( x + y) 3 = 9( x + y ) (b) = y ( ⋅ 1) = y ( 4) = 24 y 2 2 = ( −1) ( z )3z = −1 ⋅ ⋅ z + = −3z = x4 (4 y −2 )(8 y4 ) = (4)(8)( y − ) = 32 y (z + 2) −3  x −3 y  29 (a)     (z −3 + 2) −1 = ( z + 2) −4 = (z + 2)  x3  125 x =   = y12  y   a −2  b   b  b3  b5 (b)  −2   =    = a  b  a   a  a  30 (a) (b) 3n ⋅ n n + 2n 3n 1 = 3n + = 3n + = 3n − (3n + 2) = − = = 3n ⋅3 3 x2 ⋅ xn x2 + n = + n = x + n − − n = x −1 = n x ⋅x x x 31 10,250.4 = 1.02504 × 104 35 9.46 × 1012 = 9,460,000,000,000 kilometers 32 −0.000125 = −1.25 × 10−4 36 9.0 × 10 − = 0.000009 meter 33 3.14 × 10−4 = 0.000314 37 (a) 34 − 2.058 × 106 = − 2,058,000 (b) (2.0 × 109 )(3.4 × 10−4 ) = 6.8 × 105 (1.2 × 107 )(5.0 × 103 ) = 6.0 ì 104 â 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Section P.2 38 (a) 6.0 × 108 = 2.0 × 1011 3.0 × 10−3 (b) 3 27 = (b) ( ) (b) 36 ( 2) 5 5/5 = 32 x5 = = = (2 x)5 = x = 47 (a) 16 x5 = 20 = (b) 44 (a) (b) 128 = 64 ⋅ = 3 75 = 64 33 52 ⋅ 2 = = 18 z z2 ⋅2 ⋅ x2 75 x y4 25 x ⋅ y4 (5 x ) = = 23 ⋅ = ( x) = = z b 75 x y −4 = (b) 34 x8 16 = 27 4a 4⋅5 = b = 3x 43 (a) ( 4a ) z = x x2 36 = = z ⋅ z 18 = = 6x 182 32a = b2 (b) = (6) = 216 (3 x ) ⋅ 32 ⋅ x ⋅ y 182 = z3 46 (a) 12 ⋅ 42 (a) (b) ( ) 54 xy = (b) 27 = 40 (a) 36 x ⋅ x = 3y2 27 = = 6x 2x = 3 41 (a) 72 x3 = 45 (a) 2.5 × 10−3 = 0.5 × 10−5 = 5.0 × 10−6 (b) 5.0 × 102 39 (a) Exponents and Radicals = 23 48 (a) 3x y = ( y2 ) x y (b) x4 ⋅ y2 = x ⋅ = ⋅3 3y2 x y 160 x8 z = 32 x ⋅ x z = ( x) ⋅ x3 z = x 5 x3 z 49 (a) 20 x + 125 x = 4x2 ⋅ + ( x) = = x = 29 x (b) 147 x − 48 x = ⋅5 +5 + 25 x 25 x ⋅ (5 x) ⋅5 5 49 ⋅ x − 16 ⋅ x = ⋅ 3x − = 56 x − 12 = 44 3x 42 ⋅ 3x 3x © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Full file8 at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-10th-Edition-by-Larso Chapter P Prerequisites 50 (a) 3 54 x3 + 16 x = 3 27 x3 ⋅ + x3 ⋅ = 3 (3 x ) ⋅ + ( x) = 9x = 11x (b) 3 64 x − + 2x 27 x = 64 ⋅ x − = (4) 3 ⋅ x − = ( − x) = 52 ⋅ = ⋅ 22 = 14 − 53 54 + 55 + 56 −3 = 57 = 3 58 x 2 27 x3 ⋅ x (3 x ) ⋅ x x x 3 = 83 83 = = 43 14 + = 14 + ⋅ 14 − = + = + 6 −3 ⋅ ( 14 + ( 14 ) ( ⋅ − − = ⋅ − − = 3 = − )  16  (b)    81  x2 3 ) −2 14 + = = 14 − ( − ( − 3 ) +3 ) ( 14 + x ,x ≠ 32 35 =  81  =    16  ( 34 32 ) = (2) = ) ( = −3 14 + 2 = ) − ( = ) ) +3 ( −1 100 ) −3 = 10 − = 12  4 =   9 32 = 32 = 31 = (b) ( x + 1)4 64 (a) x3 = x3 = x1 = (b) (3 x ) 1000 = = 63 (a) = ( x + 1) 46 = ( x + 1) 23 = ( x + 1)2 x = 3x 32 = (321 ) 12 65 (a) = 321 = (b) − )  81  27  3 =   =   =  16     ) 10 −1 = − ( = 9 (b)   4 a2 −3 ( ) ( 62 (a) 100 − = 61 (a) 32 −3 = +3 −9 = = +3 +3 60 a 0.4 = a = 5−3 52 = 5−6 ( ( − x = x ⋅ x1 59 x − = ) − ( 2) 64 = 641 = x ⋅2 = x − 3x 51 3 2x = ((2 x) ) 14 12 32 = = ( x) 16 ⋅ = 18 = 2x © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Section P.3 ( 12  243( x + 1) =  243( x + 1)  66 (a) ) Polynomials and Special Products 52 69 t = 0.03125 − (12 − h) , ≤ h ≤ 12   12 = ( 243( x + 1)) 14 h (in centimeters) t (in seconds) = 243( x + 1) 0 = ⋅ 81( x + 1) 2.93 5.48 7.67 9.53 11.08 12.32 13.29 14.00 14.50 = ( x − 1) −1 10 14.80 x −1 11 14.93 12 14.96 = 3( x + 1) (b) 10a 7b = ((10a b) ) 13 12 = (10a 7b) 16 = = a 10a ⋅ a ⋅ b 10ab 67 (a) ( x − 1)1/ ( x − 1) / = ( x − 1)3 / = x − (b) ( x − 1)1/ ( x − 1) −4 / = ( x − 1) −3 / = 68 (a) (4 x + 3) 5/ (4 x + 3) −5/3 15/6 = (4 x + 3) (4 x + 3) = (4 x + 3)5/6 , x ≠ − −10/6 70 Package A: x = 500 x = (b) (4 x + 3) −5/ (4 x + 3) 2/3 = (4 x + 3) −15/ (4 x + 3) 4/ 500 ≈ 7.9 in Package B: x = 250 = (4 x + 3) −11/6 x = = (4 x + 3)11 250 ≈ 6.3 in So, x = 2(6.3) = 12.6 in Because 7.9 < 12.6, the length of a side of package A is less than twice the length of a side of package B 71 False When x = 0, the expressions are not equal ( ) 72 False When a power is raised to a power, you multiply the exponents: a n k = a nk 73 False When a sum is raised to a power, you multiply the sum by itself using the Distributive Property (a + b) = a + 2ab + b ≠ a + b 74 False Raising the numerator and denominator to the second power changes the expression To rationalize the denominator, b = multiply the fraction by b Section P.3 Polynomials and Special Products n; an ; a0 (a) Standard form: 7x monomial; binomial; trinomial (b) Degree: Leading coefficient: like terms (c) Monomial First terms; Outer terms; Inner terms; Last terms © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson 10 Chapter P Prerequisites (a) Standard form: 13 (b) Degree: Leading coefficient: (c) Monomial 14 (a) Standard form: − 12 x5 + 14 x 3x + 4 = 3+ = + x −1 is not a polynomial x x because it includes a term with a negative exponent x2 + x − is a polynomial Standard form: (b) Degree: Leading coefficient: − 12 x + x − 2 15 y − y + y is a polynomial (c) Binomial Standard form: − y + y + y (a) Standard form: x + (b) Degree: Leading coefficient: 16 y − y is not a polynomial because it includes a term with a square root (c) Binomial 17 (6 x + 5) − (8 x + 15) = x + − x − 15 (a) Standard form: −4 x + x + = (6 x − x) + (5 − 15) (b) Degree: Leading coefficient: −4 = −2 x − 10 18 ( x + 1) − ( x − x + 1) = x + − x + x − (c) Trinomial = ( x − x ) + x + (1 − 1) 10 (a) Standard form: 25 y − y + (b) Degree: Leading coefficient: 25 = x2 + 2x 19 (t − 1) + (6t − 5t ) = t − + 6t − 5t (c) Trinomial = (t + 6t ) − 5t − 11 x − 3x + is a polynomial = 7t − 5t − Standard form: −3x3 + x + 12 x − x + x −2 is not a polynomial because it includes a term with a negative exponent 20 ( y − 3) + ( − y + 9) = y − − y + = ( y − y ) + ( − + 9) = − 3y2 + 21 (15 x − 6) + ( − 8.3 x − 14.7 x − 17) = 15 x − − 8.3 x − 14.7 x − 17 = − 8.3 x + (15 x − 14.7 x ) + ( − − 17) = − 8.3 x + 0.3 x − 23 22 (15.6 w4 − 14 w − 17.4) + (16.9 w4 − 9.2 w + 13) = 15.6 w4 − 14 w − 17.4 + 16.9w4 − 9.2 w + 13 = (15.6 w4 + 16.9 w4 ) + ( −14 w − 9.2 w) + ( −17.4 + 13) = 32.5w4 − 23.2w − 4.4 23 z − 3z − (10 z + 8) = z − (3z − 10 z − 8) = z − 3z + 10 z + = (5 z − 3z + 10 z ) + = 12 z + © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Full file34at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-10th-Edition-by-Larso Chapter P Prerequisites y 33 (a) y 36 (a) 5 (5, 4) (− 25 , 34 ) 3 (−1, 2) −1 −1 (5 + 1) (b) d = = x + ( − 2) − −2 − −1 2 (b) d = 36 + = 10 = (2, 10) 37 d = = (10, 2) (b) d = = (2 − 10) x 10 + (10 − 2) 38 d = 64 + 64 =  + 10 10 +  (c)  ,  = (6, 6)   35 (a) 15 10 −20 −15 −10 (5.6, 4.9) −5 x −5 (b) d = = (−16.8 − 5.6) 9+ = 82 7  6 120 + 150 36,900 ( 42 − 18) + (50 − 12) = 242 + 382 = 2020  x + x2 y1 + y2  39 midpoint =  ,     2010 + 2014 35,123 + 45,998  =  ,  2   = ( 2012, 40,560.5) 20 5 4 1   +  + 1 −  2 3 2  = 505 ≈ 45 The pass is about 45 yards y (−16.8, 12.3) x = 30 41 ≈ 192.09 The plane flies about 192 kilometers 2  −(5 2) + (1 2) ( 3) +   , (c)   =  −1, 2    y 10 −1 2  −1 + +  (c)  ,  = ( 2, 3)   34 (a) ( 12, 1) In 2012, the sales for the Coca-Cola Company were about $40,560.5 million + (12.3 − 4.9) 501.76 + 54.76 = 556.52  −16.8 + 5.6 12.3 + 4.9  (c)  ,  = ( −5.6, 8.6) 2    x + x2 y1 + y2  40 midpoint =  ,     2013 + 2015 1.17 + 3.25  =  ,  2   = ( 2014, 2.21) In 2014, the revenue per share for Twitter, Inc was approximately $2.21 © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-10th-Edition-by-Larso Section P.6 The Rectangular Coordinate System and Graphs 35 41 ( − + 2, − + 5) = (0, 1) 43 ( −7 + 4, − + 8) = ( −3, 6) (−2 (−2 (−7 (2 + 2, − + 5) = (4, 2) (−1 + 2, −1 + 5) = (1, 4) 42 ( −3 + 6, − 3) = (3, 3) (−5 + 6, − 3) (−3 + 6, − 3) (−1 + 6, − 3) + 4, + 8) = ( 2, 10) + 4, − + 8) = ( 2, 4) + 4, − + 8) = ( −3, 4) 44 (5 − 10, − 6) = ( −5, 2) = (1, 0) (3 − 10, − 6) (7 − 10, − 6) = (3, − 3) = (5, 0) = ( −7, 0) = ( −3, 0) 45 (a) The minimum wage had the greatest increase from 2000 to 2010 (b) Minimum wage in 1985: $3.35 Minimum wage in 2000: $5.15  5.15 − 3.35  Percent increase:   × 100 ≈ 53.7% 3.35   Minimum wage in 2000: $4.25 Minimum wage in 2015: $7.25  7.25 − 5.15  Percent increase:   × 100 ≈ 40.8% 5.15   So, the minimum wage increased 53.7% from 1985 to 2000 and 40.8% from 2000 to 2015 (c) Minimum wage in 2030 = Minimum wage in 2015  Percent  Minimum wage  +   ≈ $7.25 + 0.408($7.25) ≈ $10.21  increase  in 2015  So, the minimum wage will be about $10.21 in the year 2030 (d) Answers will vary Sample answer: Yes, the prediction is reasonable because the percent increase is over an equal time period of 15 years x y 22 53 29 74 35 57 40 66 44 79 48 90 53 76 58 93 65 83 76 99 y Final exam score 46 (a) 47 True Because x < and y > 0, x < and − y < 0, which is located in Quadrant III 100 90 80 70 60 50 40 30 20 10 48 False The Midpoint Formula would be used 15 times 49 True Two sides of the triangle have lengths 10 20 30 40 50 60 70 80 Math entrance test score (b) The point (65, 83) represents an entrance exam score of 65 (c) No There are many variables that will affect the final exam score x the third side has a length of 149 and 18 50 False The polygon could be a rhombus For example, consider the points ( 4, 0), (0, 6), ( − 4, 0), and (0, − 6) 51 Answers will vary Sample answer: When the x-values are much larger or smaller than the y-values, different scales for the coordinate axes should be used 52 The y-coordinate of a point on the x-axis is The x-coordinates of a point on the y-axis is 53 Because xm = x1 + x2 y + y2 and ym = we have: 2 xm = x1 + x2 xm − x1 = x2 ym = y1 + y2 ym − y1 = y2 So, ( x2 , y2 ) = ( xm − x1 , ym − y1 ) © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Full file36at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-10th-Edition-by-Larso Chapter P Prerequisites 54 (a) ( x2 , y ) = ( xm − x1 , ym − y1 ) = ( ⋅ − 1, 2( −1) − ( −2)) = (7, 0) (b) ( x2 , y ) = ( xm − x1 , ym − y1 ) = ( ⋅ − ( −5), ⋅ − 11) = (9, − 3)  x + x2 y1 + y2  55 The midpoint of the given line segment is  ,    x + x2 y + y2   x1 + y1 +  x + x2 y1 + y2    =  3x1 + x2 , y1 + y2  The midpoint between ( x1 , y1 ) and  is , 2  ,    4    2   y + y2  x1 + x2  + x2 + y2   x1 + 3x2 y1 + y2   x + x2 y1 + y2  = , The midpoint between  , 2   and ( x2 , y2 ) is  ,   4    2    x + x2 y1 + y2   x1 + x2 y1 + y2   x1 + x2 y1 + y2  So, the three points are  , , , ,  , and   4 2 4        3x + x2 y1 + y2   ⋅ + 3( −2) −  7 7 , , 56 (a)   =  ,−   =  4 4    4 4   x1 + x2 y1 + y2  1 + − − 1 5 3 , ,   =   =  ,−  2 2      2  x1 + 3x2 y1 + y2   + ⋅ −2 + 3( −1)   13  , ,  =  ,−    =  4 4 4    4   3x + x2 y1 + y2   3( −2) + 3( −3) +    , , (b)   = − , −   =  4 4      4 3  x1 + x2 y1 + y2   −2 + −3 +   , ,   =   =  −1, −     2    x1 + x2 y1 + y2   −2 + −3 +   3 , ,   =   = − , −  4   4    57 Use the Midpoint Formula to prove the diagonals of the parallelogram bisect each other b + a c + 0  a + b c  , ,    =    2  a + b + c + 0  a + b c  , ,    =  2   2  58 (a) Because ( x0 , y0 ) lies in Quadrant II, ( x0 , − y0 ) must lie in Quadrant III Matches (ii) (b) Because ( x0 , y0 ) lies in Quadrant II, ( −2 x0 , y0 ) must lie in Quadrant I Matches (iii) ( ) (c) Because ( x0 , y0 ) lies in Quadrant II, x0 , 12 y0 must lie in Quadrant II Matches (iv) (d) Because ( x0 , y0 ) lies in Quadrant II, ( − x0 , − y0 ) must lie in Quadrant IV Matches (i) © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-10th-Edition-by-Larso Review Exercises for Chapter P 37 59 (a) (b) First Set d ( A, B ) = ( − 2) + (3 − ) = = d ( B, C ) = ( − 6) + (6 − 3) = 16 + = d ( A, C ) = ( − 6) + (3 − 3) = 16 = 2 Because 32 + 42 = 52 , A, B, and C are the vertices of a right triangle −2 (8 − 5) d ( B, C ) = ( − 2) d ( A, C ) = (8 − 2) x First set: Not collinear + ( − 2) = 10 + ( − 1) = 10 + (3 − 1) = 40 2 −2 Second Set d ( A, B ) = y 2 A, B, and C are the vertices of an isosceles triangle or are collinear: 10 + 10 = 10 = 40 Second set: Collinear (c) A set of three points is collinear when the sum of two distances among the points is exactly equal to the third distance 60 (a) The point ( x0 , − y0 ) is reflected in the x-axis (b) The point ( − x0 , − y0 ) is reflected in the x- and y-axes Review Exercises for Chapter P { } 11, −14, − 89 , 52 , (a) x ≥ denotes the set of all real numbers greater than or equal to 6, 0.4 (a) Natural numbers: 11 (b) (d) Rational numbers: 11, −14, − 89 , 52 , 0.4 (e) Irrational numbers: { 15, − 22, − 10 , 0, 5.2, } −4 −2 (c) The set is bounded (a) − ≤ x < denotes the set of all real numbers greater than or equal to − and less than 15 (e) Irrational numbers: (b) x −4 −3 −2 −1 (c) The set is bounded (a) x ≥ denotes the set of all real numbers greater than or equal to 5 > (b) x −1 10 d ( −112, − 6) = −6 − ( −112) = 106 − 25 > − 57 d ( −74, 48) = 48 − ( −74) = 122 − 57 = − 0.714285 −5 (c) The set is unbounded = − 0.36 (a) − 25 −1 x −6 (d) Rational numbers: −22, − 10 , 0, 5.2, (b) (b) (c) Integers: − 22, 8 greater than − and less than (b) Whole numbers: 0 1 8 (a) − < x < denotes the set of all real numbers (a) Natural numbers: none (c) The set is unbounded (c) Integers: 11, −14 x (b) Whole numbers: 11 −9 25 − 0.8 − 0.6 − 0.4 − 0.2 x 11 d ( x, 7) = x − and d ( x, 7) ≥ 4, thus x − ≥ © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Full file38at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-10th-Edition-by-Larso Chapter P Prerequisites 12 d ( x, 25) = x − 25 and d ( x, 25) ≤ 10, thus 26 5( 20 + 7) = 5( 27) = 135 x − 25 ≤ 10 13 12 x − (a) 12(0) − = −7 (b) 12( −1) − = −19 14 x − x + (a) (−2) (b) (2) x 7x 12 x 35 x 47 x + = + = 12 60 60 60 28 x 2x 5x 4x x − = − = 10 10 10 29 3x x x ⋅ = ⋅ = 10 2 30 9 54 ÷ = ⋅ = x x x − 6( −2) + = 21 − 6( 2) + = −3 27 ( ) 31 (a) 3x x3 15 − x + x − (a) −(1) + − = −1 (b) (b) −( −1) + ( −1) − = −3 16 32 (a) x x−3 (a) (b) −3 = −3 − 33 (a) is undefined (b) 3−3 (b) 17 + ( a − 5) = a − Illustrates the Additive Identity Property 18 ⋅ (3 x + 4) = x + Illustrates the Multiplicative Identity Property 19 x + (3x − 10) = ( x + x) − 10 Illustrates the Associative Property of Addition ( ) ( ) Illustrates the Commutative Property of Addition 22 y +4 ⋅ = 1, y +4 (b) 35 (a) 36 (a) (b) y =/ − 37 (a) Illustrates the Multiplicative Inverse Property 23 − + = 24 − ( − 3) = + = (3a) y =/ (6a3 ) = 9a2 (6a3 ) = 54a5 36 x5 ⋅ ⋅ x5 = = 10 x 9x ⋅ x ⋅ x5 (−2 z) (8 y ) = −8z y2 y2 = Illustrates the Distributive Property 21 t + + = + t + y6 y −1 y5 = = , 10 y 2 34 (a) ( x + 2)  = ( x + 2)   (b) 20 4(t + 2) = ⋅ t + ⋅ = 3x (64 x9 ) = 192 x11 (b) 40(b − 3) 75(b − 3) 8 5− (b − 3) = (b − 3) , 15 15 = b =/ a2 = a 2b b −2 (a2b4 )(3ab−2 ) = 3a2 +1b4 − = 3a3b2 62 u 3v −3 36u − (−2)v −3 −1 3u = = 3u 5v − = −2 v 12u v 12 3− m−1n −3 n3 81 = = = −2 −3 mn mmn 81m m (5a)−2 ( 5a ) 4( x −1 ) = ( 5a ) −3 −2 ( x −1 ) −1 = −2 − = ( 5a ) −4 = ( 5a ) = 625a 41−(−2) x = 43 x −1 = 64 x x 25 ( − 8)( − 4) = 32 © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-10th-Edition-by-Larso Review Exercises for Chapter P 39 (x + 38 (a) y −1 ) −1  x −3  x  (b)     y  y  y y 1 1 = = = = 1⋅ = xy xy + 1 x + y −1 xy xy + +1 x + + y y y y = −1   y  =    = , x  x y  x  y =/ 39 274,400,000 = 2.744 × 108 40 0.3048 = 3.048 × 10 46 (a) (b) x3 = 27 −1 64 x6 = 41 4.84 × 10 = 484,000,000 42 2.74 × 10 43 (a) 27 = (b) 44 (a) ( ( 27 493 = ( 64 = 125 (b) 216 ) 12 x + ( x)6 3x = = 0.00274 2 27 x3 − (b) = (7) = 343 ( ( 32 ) ) = 2x 2x x ⋅ 3x + 3x 3x 3x3 = x ⋅ 3x − x ⋅ 3x = 3x x − x x = (3 x − x ) x = x 3x 92 = 102 10 = ( x + 1) 3x = (3) = ) = x 3x + 43 = 5 = 324 = ) 49 81 = 100 (b) 45 (a) 47 (a) −3 x3 x = 33 3 x3 + 48 (a) (b) = (6) = 216 2x = 2x 2x + 18 x5 − x3 = 3x 2 x = ( x + 1) x 2x − 2x 2x = (3 x − x ) x = x x (3 x − 2) = 32 49 These are not like terms Radicals cannot be combined by addition or subtraction unless the index and the radicand are the same ( 242 − 50 A = wh = 51 = 42 12 12 = ⋅ 4 53 1 2+ = ⋅ 2− 2− 2+ 42 = ( ) ( ) 384 = 8 = 64 18 = 64 = 192 in.2 = 3 12 16 = 3 16 = 3 ⋅ = 52 3 3 = = 4(3) ⋅ ) 2+ = − ( 3) −1 = 5−1 = 2+ 2+ = = 2+ 4−3 54 = +1 ⋅ +1 −1 = −1 55 +1 = +1 ⋅ − 12 −1 −1 = = = −1 −1 −1 ( ) ( −1 ) ( ) ( ) −1 = −1 © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Full file40at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-10th-Edition-by-Larso Chapter P Prerequisites − 56 11 57 163 = 58 64−2 = ( )( − = 163 = 642 ( = 11 16 ) ( 64 − 11 = − + 11 11 = 11 ) = (4) = ( + 13 −1 = ( x − 1) 12 − 12 + 11 Leading coefficient: −12 16 64 Standard form: −7 x + 12 x + Degree: ) ( x − 1) ( x − 1) ) = − Degree: 59 3x x1 = x +1 = x9 10 60 11 63 Standard form: −12 x − = ( 4) = 64 + + ⋅ Leading coefficient: −7 ( = ( x − 1) 12 ) 65 − x + x + (1 − x) = −3 x − x + − x = −3 x − x + 61 Standard form: −11x + 66 y − 2 y − (3 y − 8) = y − y + (3 y − 8) Degree: Leading coefficient: −11 = −2 y + 11y − ( 62 Standard form: − x5 + 3x3 + x − ) ( ) 67 x x − x + = ( x) x + ( x)( −5 x) + ( x)(6) Degree: = x − 10 x + 12 x Leading coefficient: −5 ( ) 68 3x3 − 1.5 x + ( −3x) = (3x)( − 3x) − 1.5 x2 ( −3x) + 4( − 3x) = − x + 4.5 x3 − 12 x 69 (3x − 6)(5 x + 1) = 15 x + x − 30 x − 73 ( x − 3) = ( x) − 2( x)(3) + 33 2 = 15 x − 27 x − = x − 12 x + 70 ( x + 2)( x − 2) = x − x + x − 74 = x3 + x − x − (x − 4) = x3 − x ( 4) + x( 4) − 43 2 = x3 − 12 x + 48 x − 64 71 (6 x + 5)(6 x − 5) = (6 x) − 52 = 36 x2 − 25 75 × 72 (3 + x)(3 − x) = (3 5) − (2 x) = 45 − x x4 x4 − − x3 − + x3 + − x3 − x2 x2 2x2 x2 5x2 4x2 + x + − 7x − − x − 10 − 35 x − 37 x − 10 76 (6 x − 20 x − x + 3) − (9 x − 11x + 16) = x − 20 x − x + − x + 11x − 16 = (6 x − x ) + ( −20 x + 11x ) − x + (3 − 16) = −3 x − x − x − 13 ( ) 77 2500(1 + r ) = 2500( r + 1) = 2500 r + 2r + = 2500r + 5000r + 2500 2 © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-10th-Edition-by-Larso Review Exercises for Chapter P 41 78 (a) The surface is the sum of the area of the side, 2π rh, and the areas of the top and bottom which are each π r S = 2π rh + π r + π r = 2π rh + 2π r r h (b) S = 2π (6)(8) + 2π (6) = 96π + 72π = 168π ≈ 527.79 in.2 79 Area = ( x + 12)( x + 16) 89 x3 − x + x − = x ( x − 1) + 2( x − 1) = ( x − 1)( x + 2) = x + 16 x + 12 x + 192 = x + 28 x + 192 square feet 80 90 x3 − x + x − = x ( x − 4) + 2( x − 4) ( x + 5)( x + 3) = x + 3x + x + 15 = ( x + 2)( x − 4) = x + x + 15 This illustrates the Distributive Property ( 91 The domain of ) 81 x3 − x = x x − = x( x + 1)( x − 1) that x ≠ −1 82 x( x − 3) + 4( x − 3) = ( x − 3)( x + 4) is the set of all real numbers x x +6 such that x ≠ − 92 The domain of 83 25 x − 49 = (5 x) − 72 = (5 x + 7)(5 x − 7) 84 36 x − 81 = 9( x − 9) 93 The domain of x + is the set of all real numbers x such that x ≥ − 2 = ( x) − 32    94 The domain of x + is the set of all real numbers x such that x ≥ − = 9( x + 3)( x − 3) ( ) 85 x3 − 64 = x3 − 43 = ( x − 4) x2 + x + 16 ( ) 95 ( x + 8)( x − 8) = x − , x − 64 = 5(3 x + 24) ⋅ 3( x + 8) 15 96 ( x + 3)( x − 3x + 9) x + 27 = x2 + x − ( x + 3)( x − 2) 86 8x3 + 27 = ( x + 3) x − x + 87 x + 21x + 10 = ( x + 1)( x + 10) 88 x + 14 x + = ( x + 4)(3 x + 2) 97 = x − 3x + , x−2 x =/ − x =/ − ( x − 3)( x − 4) ⋅ x − x + 12 x + x + ⋅ = 2 x + x + 16 x − ( x + 3)( x − 3) ( x + 4) = 98 is the set of all real numbers x such x +1 (x x − ,x ≠ + 4)( x + 3) 2( x − 3) 2( x + 2) 2x − 3− x 2x − 2x + 4 ÷ = ⋅ = ⋅ = − , x ≠ − 2, 3x + 2x + 3x + − x 3( x + 2) − ( x − 3) © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Full file42at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-10th-Edition-by-Larso Chapter P Prerequisites 99 6 + = − x −3 3− x x −3 x −3 = − x −3 100 3x 3x − = − x − 4− x x − − ( x − 4) 3x + x − x − 3x + = x − =      3a   3a       a − 1 a − x     x x 3ax  =  x  = 3a ⋅ ,x ≠ ⋅ = 101  x a − x a − x a  a − x a − x)( a − x) ( −     x   x    2x + 2x − −    ( x − 3)( x + 3) ( x − 3)( x + 3)   2x +  2x −   x( x + 3)   x( x + 3) 2x + − 2x + x − 3)( x + 3) x( x + 3) ( 4x = = ⋅ = , 2x + − 2x 2x − (2 x − 3)( x + 3) x( x + 3)   −   2x − 2x +   102 =   −   2x +   2x  1 x − ( x + 6) −   x  x( x + h) −h −1  2( x + h) , 103 = = ⋅ = h h x( x + h ) h x( x + h) x =/ 0, − h =/ x −3 x + h −3 − 3 x x h x − + − − ( )( ) ( 3)( x + h − 3) h x −3− x − h + = ⋅ ( x − 3)( x + h − 3) h 1   −   x + h −3 x − 3  104 = h −h h( x − 3)( x + h − 3) = = (x y 105 (− 3, 6) −1 , − 3)( x + h − 3) h =/ y 106 6 (5, 5) (− 2, 0) −6 −4 −2 −2 −4 x −4 −2 −2 −6 (− 3, − 3) − −8 −6 (− 1, − 7) (0, 6) (8, 1) x (5, − 4) © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-10th-Edition-by-Larso Review Exercises for Chapter P 43 107 x > and y = − in Quadrant IV 112 (a) y 108 xy = means x and y have the same signs This occurs in Quadrants I and III −3 y 109 (a) (1.8, 7.4) −2 −1 −4 x −8 (−3, 8) (−0.6, −14.5) −16 (1, 5) −2 (− − 1) (b) d = = x + (8 − 5) = 16 + = 13   −3 + +   (c) Midpoint:  ,  =  −1,   2   y 110 (a) (−2, 6) −2 (4, −3) −4 (−2 = − 4) + (6 + 3) 36 + 81 = 117 = 13  −2 + − (c) Midpoint:  ,  (5.6, 0) = = ( 2, 0) = (0, − 5) = ( 2, − 5)  x + x2 y1 + y2  115 midpoint =  ,     2013 + 2015 6.8 + 6.1  =  ,  2   = ( 2014, 6.45) 116 (a) (b) d = (6 − 4, − 8) (4 − 4, − 8) (6 − 4, − 8) In 2014, Barnes & Noble had annual sales of $6.45 billion (0, 8.2) −2 485.37 113 ( − 4, − 8) = (0, 0) 3  3  = 1,    2 y 111 (a) 5.76 + 479.61 = (3 − 2, + 3) = (1, 6) (0 − 2, + 3) = (−2, 8) (−3 − 2, + 3) = (−5, 6) x −2 (b) d = 114 (0 − 2, + 3) = ( −2, 4) −4 0.6) + (7.4 + 14.5)  1.8 − 0.6 7.4 − 14.5  (c) Midpoint:  ,  = (0.6, − 3.55) 2   (5.6 − 0)2 x + (0 − 8.2) 31.36 + 67.24 = y Apparent temperature (in °F) −4 (1.8 + (b) d = 150 140 130 120 110 100 90 80 70 65 70 75 80 85 90 95 100 x Actual temperature (in °F) 98.6  + 5.6 8.2 +  (c) Midpoint:  ,  = ( 2.8, 4.1)   Change in apprarent temperature = 150°F − 70°F = 80°F 117 False, ( a + b) = a + 2ab + b ≠ a + b 2 There is also a cross-product term when a binomial sum is squared © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Full file44at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-10th-Edition-by-Larso Chapter P Prerequisites ( )( x n − ) 118 False, x n − y n = ( x − y ) x n −1 + x n − y +  + y n −1 for odd values of n, not for all values of n x − y = (x n n n + y n y n ) for even values of n Problem Solving for Chapter P (a) Men’s Maximum Volume: V = π (65) ≈ 1,150,347 mm3 Minimum Volume: V = π (55) ≈ 696,910 mm3 Women’s Maximum Volume: V = π (55) ≈ 696,910 mm3 Minimum Volume: V =  95  π   ≈ 448,921 mm3 2 (b) Men’s Maximum density: 7.26 ≈ 1.04 × 10−5 kg mm3 696,910 Minimum density: 7.26 ≈ 6.31 × 10−6 kg mm3 1,150,347 Women’s Maximum density: 4.00 ≈ 8.91 × 10−6 kg mm3 448,921 Minimum density: 4.00 ≈ 5.74 × 10−6 kg mm3 696,910 (c) No The weight would be different Cork is much lighter than iron so it would have a much smaller density Let a = and b = −3 Then a − b = − ( −3) = and a − b = − −3 = Thus, a − b > a − b Let a = 11 and b = Then a − b = 11 − = and a − b = 11 − = Thus, a − b = a − b To prove a − b ≥ a − b for all a, b, consider the following cases Case 1: a > and b < a −b = a + b > a − b Case 3: a > and b > (i) a > b Then, a − b = a − b (ii) a < b  a < b Then, a − b = b − a > a − b Case 2: a < and b > a −b = a + b > a − b Case 4: a < and b < (i) a > b  b > a Then, a − b = b − a > a − b (ii) a < b Then, a − b = a − b Therefore, a − b ≥ a − b for all a, b To say that a number has n significant digits means that the number has n digits with the leftmost non-zero digit and ending with the rightmost non-zero digit For example; 28,000, 1.400, 0.00079 each have two significant digits © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-10th-Edition-by-Larso Problem Solving for Chapter P 45 Perimeter: P = 2l + w + π r Amount of glass = Area of window 13.14 = 2l + + π (1) r = ft l = l 13.14 − − π ≈ feet 2 πr 2 = ( 4)( 2) + π (1) ≈ 9.57 square feet A = lw + ft 60 24 365.25 76.4 × × × 1 1 lifetime beats = 2,812,834,080 lifetime 70 beats × Men: So, the number of beats in a lifetime for a man is 2,812,834,080 60 24 365.25 81.2 × × × 1 1 lifetime beats = 2,989,556,640 lifetime 70 beats × Women: So, the number of beats in a lifetime for a woman is 2,989,556,640 y 300 250 Year, x Population, y 1960 179.32 970 203.30 1980 226.54 1990 248.72 2000 281.42 2010 308.75 200 150 100 50 10 00 20 90 20 80 19 19 19 19 70 x 60 U.S census population (in millions) (a) Year The population of the United States has been increasing each year (b) From 1960 to 1970, the population increased by 203.30 – 179.32 = 23.98 million people From 1970 to 1980, the population increased by 226.54 – 203.30 = 23.24 million people From 1980 to 1990, the population increased by 248.72 – 226.54 = 22.18 million people From 1990 to 2000, the population increased by 281.42 – 248.72 = 32.70 million people From 2000 to 2010, the population increased by 308.75 – 281.42 = 27.33 million people (c) The population increased the most from 1990 to 2000 The population increased the least from 1980 to 1990 (d) From 1960 to 1970, the percent increase in population was 23.98 million people ≈ 0.134 = 13.4% 179.3 million people From 1970 to 1980, the percent increase in population was 23.24 million people ≈ 0.114 = 11.4% 203.30 million people From 1980 to 1990, the percent increase in population was 22.18 million people ≈ 0.098 = 9.8% 226.54 million people From 1990 to 2000, the percent increase in population was 32.7 million people ≈ 0.131 = 13.1% 248.72 million people From 2000 to 2010, the percent increase in population was 27.33 million people ≈ 0.097 ≈ 9.7% 281.42 million people (e) The percent increase of population was the greatest from 1960 to 1970 The percent increase of population was the least from 2000 to 2010 © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Full file46at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-10th-Edition-by-Larso Chapter P Prerequisites 14  3225  r = −    12,000  Planet x x y y ≈ 0.280 or 28% Mercury Venus Earth Mars Jupiter 0.387 0.723 1.000 1.524 5.203 0.622 0.850 1.000 1.235 2.281 0.241 0.615 1.000 1.881 11.862 0.622 0.850 1.000 1.234 2.281 The square root of the distance from a planet to the sun is equal to the cube root of the period of the planet V = lwh Volume: x + x − x − = lw( x + 1) 2 x3 + x − x − = lw 2x + x ( x + 1) − 4( x + 1) = lw 2x + ( x − 4)(2 x + 1) 2x + = lw lw = x − = ( x + 2)( x − 2) Let l = x + and w = x − Surface Area: S = 2lw + 2lh + wh = 2(lw + lh + wh) = ( x + 2)( x − 2) + ( x + 2)( x + 1) + ( x − 2)( x + 1) =  x − + x + x + + x − 3x − 2 = 5 x + x − 4 = 10 x + x − When x = inches: S = 10(6) + 4(6) − = 376 cubic inches 10 The distance between ( x1 , y1 ) and x1 + x2 2y1 + y2 , is 3 2 x1 + x2  y1 + y2     x1 −  +  y1 −  3     d = = y1 + y2   x1 − x2     +  y1 −      = 1 ( x1 − x2 )2 + ( y1 − y2 )2  9 = ( x1 − x2 ) + ( y1 − y2 ) which is 2 of the distance between ( x1 , y1 ) and ( x2 , y2 )   x1 + x2    y1 + y2   + x2   + y2   x + x2 y1 + y2  3      =  The second point of trisection is  , ,  3 3     © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-10th-Edition-by-Larso Problem Solving for Chapter P 47 11 (a) (1, − 2) and ( 4, 1) (b) (−2, − 3) and (0, 0) The points of trisection are: The points of trisection are:  2(1) + 2( −2) +  ,   = ( 2, −1) 3    2( −2) + 2( −3) +    ,   =  − , − 2 3      + 2( 4) −2 + 2(1)  ,   = (3, 0) 3    −2 + 2(0) −3 + 2(0)    ,   =  − , −1 3     x2 12 y1 = x − x − 1− x y2 = 2 − 3x − x2 When x = 0, y1 = When x = 0, y2 = Thus, y1 =/ y2 y1 = = = = = = x − x2 − x3 − x2 x − x2 ⋅ 1 − x2 − x2 − x3 − x2 x(1 − x ) − x3 − x2 x − x3 − x3 − x2 x − 3x3 − x2 x( − x ) Let y2 = − x2 x( − x ) − x2 13 One golf ball: Then y1 = y2 1.6 × 107 ≈ 0.101 pound 1.58 × 108 0.101(16) = 1.616 ounces 14 (a) Either graph could be misleading The scales on the vertical axes make it appear that the rise in profits is either dramatic or small, but the total increase is only 2-6 units (b) If the company wanted to gloss over the dip in profits during July, as at a stockholders meeting, the first graph could be used If the company wished to project an image of rapidly increasing profits, as to potential investors, the second graph could be used © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for College Algebra 10th Edition by Larson Full file48at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-10th-Edition-by-Larso Chapter P Prerequisites Practice Test for Chapter P Evaluate Simplify − 42 − 20 15 − − x z − z y The distance between x and is no more than Use absolute value notation to describe this expression Evaluate 10( − x) for x = (161 x) Simplify ( − x )( − x − ) Change 0.0000412 to scientific notation Evaluate 1252 Simplify 64 x y Rationalize the denominator and simplify 12 10 Simplify 80 − 500 ( ) ( ) 11 Simplify x − x + x − − 3x3 + x + ( ) 12 Multiply ( x − 3) x + x − 13 Multiply ( x − 2) − y  14 Factor 16 x − 15 Factor x + x − 16 Factor x3 − 64 17 Combine and simplify − 18 Combine and simplify x + x x + x − x2 − ÷ 4x x2 1 1−   x 19 Simplify 1− 1 1−   x 20 (a) Plot the points ( − 3, 6) and (5, −1), (b) find the distance between the points, and (c) find the midpoint of the line segment joining the points © 2018 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part