Solution manual for trigonometry 10th edition by larson

Solution Manual for Trigonometry 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson CHAPTER P Prerequisites Section P.1 Review of Real Numbers and Their Properties Section P.2 Solving Equations Section P.3 The Cartesian Plane and Graphs of Equations 17 Section P.4 Linear Equations in Two Variables .29 Section P.5 Functions .42 Section P.6 Analyzing Graphs of Functions 51 Section P.7 A Library of Parent Functions .61 Section P.8 Transformations of Functions 66 Section P.9 Combinations of Functions: Composite Functions 76 Section P.10 Inverse Functions 85 Review Exercises 98 Problem Solving 112 Practice Test .118 © 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson C H A P T E R Prerequisites P Section P.1 Review of Real Numbers and Their Properties irrational 11 (a) origin 4 −5 −4 −3 −2 −1 −1 terms 12 (a) x −1 (d) Rational numbers: −9, − 72 , 5, 32 , 0, 1, − 4, 2, −11 x 10 11 12 (c) Integers: −9, 5, 0, 1, − 4, 2, −11 (c) x −4.75 (d) 5, − 7, − 73 , 0, 3.14, 54 , − 3, 12, (a) Natural numbers: 12, −8 13 −4 > −8 (c) Integers: −7, 0, − 3, 12, −8 (d) Rational numbers: −7, − 73 , 0, 3.14, 54 , − 3, 12, x −5 −4 −3 −2 −1 (b) Whole numbers: 0, 12, −7 14 < −6 −5 x 16 16 2.01, 0.6, −13, 0.010110111 , 1, − 15 (a) Natural numbers: > x (b) Whole numbers: (c) Integers: −13, 1, − (d) Rational numbers: 2.01, 0.6, −13, 1, − 1π , 16 − 87 < − 73 (b) Whole numbers: 25, − 87 −2 7, −11.1, 13 9, 7, 13 9, 7, 13 9, 7, 13 (d) Rational numbers: x (e) Irrational numbers: 0.010110111 (a) Natural numbers: 25, −4 (e) Irrational numbers: x −7 −6 −5 −4 −3 −2 −1 (e) Irrational numbers: (b) (b) Whole numbers: 0, 5, 1, 25, −17, − 12 , x 8.5 (a) Natural numbers: 5, 1, (c) Integers: 25, −17, −7 −6 −5 −4 −3 −2 −1 2, 0, 1, − 4, 2, −11 9, 3.12, x −5.2 Zero-Factor Property − 12 , − (d) −9, − 72 , 5, 32 , (c) composite 10 25, −17, (b) absolute value x −2 −1 − 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: 12π © 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Section P.1 Review of Real Numbers and Their Properties 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 35 − = 5(5) = 25 (c) The interval is bounded 20 (a) The inequality < x ≤ denotes the set of all real numbers greater than zero and less than or equal to (b) x 21 (a) The interval [4, ∞ ) denotes the set of all real numbers greater than or equal to 4 than 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 + 2) x + = −1 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) = x + 39 −4 = because −4 = and = (−∞, 2) denotes the set of all real numbers less (b) x + 38 If x > 1, then x − is positive (c) The interval is unbounded 22 (a) 37 If x < −2, then x + is negative So, x 36 − − = − 4( 4) = −16 So, (c) The interval is bounded (b) 33 −1 − −2 = − = −1 34 −3 − −3 = −3 − (3) = −6 x −2 31 − = −5 = −( −5) = 32 − = = (c) The interval is unbounded (b) 30 = ( ) ( ) 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file4 at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-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 − x 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 x3 + 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 − 64 2x2 , − Coefficient: − 3( −1) + = + + = (a) −( −1) + 5( −1) − = −1 − − = −10 Coefficients: 6, − Terms: 2 62 − x + x − Terms: x , − x 58 − 3(0) + = 2 59 x − (a) 4( −1) − = −4 − = −10 1+1 = 1−1 Division by zero is undefined −1 + = = −1 − −2 x − x + (a) − = = + (b) −2 − −4 = −2 + Division by zero is undefined (b) 4(0) − = − = −6 65 (h (h + 6) = 1, h ≠ −6 + 6) Multiplicative Inverse Property 66 (x + 3) − ( x + 3) = Additive Inverse Property 67 x(3 y ) = ( x ⋅ 3) y = (3 x ) 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Section P.2 Solving Equations 68 (7 ⋅ 12) = ( 17 ⋅ 7)12 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 Solving Equations equation ax + b = − x = 18 extraneous − 2x 18 = −2 −2 factoring; square roots; completing; square; Quadratic Formula x + 11 = 15 x = −9 x + 11 − 11 = 15 − 11 − x = 25 − − x = 25 − 7 x + = 23 x + − = 23 − x = x = 21 − x = 19 7x 21 = 7 x = − x + x = 19 + x = 19 + x − 19 = 19 + x − 19 −12 = x 3x − = x + 3x − x − = x − x + x−5 = x −5+5 = 7+5 x = 12 © 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file6 at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Chapter P Prerequisites 10 y + − y = − y 4y − 5y + = − 6y 14 −y + = − 6y −y + 6y + = − 6y + 6y 5y + = 5x 1 + = x − 2 5x 1 (4) + (4) = (4) x − (4) 2 5x + = x − x = −4 5y + − = − 5y = 15 5y = 5 y =1 5x − = 5x + 3(5 x − 4) = 2(5 x + 4) 15 x − 12 = 10 x + x = 20 11 x − 3( x + 3) = − x x = x − x − = − 5x −5 x − = − x −5 x + x − = − x + x 16 −9 =/ 10 x + = 5x + 2(10 x + 3) = 1(5 x + 6) 20 x + = x + Because − = is a contradiction, the equation has no 15 x = solution x = 12 x − 10 = x + 2( x − 5) x − 10 = x + x − 10 17 x − 10 = x − 10 Because the equation is an identity, the solution is the set of all real numbers 13 3x 4x − = 3x 4x (24) − (24) = (24)4 x − 32 x = 96 −23 x = 96 x = − 96 23 13 = 4+ x x 10 x − 13 4x + = x x 10 x − 13 = x + 10 − x = 18 x = 18 + = x x −5 = x( x − 5)0 x( x − 5) + x( x − 5) x x −5 x − + 2x = 3x − = 3x = x = 19 x + x + x x x = −2 + + = −2 + =/ − Because = − is a contradiction, the equation has no solution © 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Section P.2 Solving Equations 20 8x − = −4 Multiply each term by ( x + 1)( x − 1) 2x + 2x − 7( x − 1) − x( x + 1) = − 4( x + 1)( x − 1) 14 x − − 16 x − x = −16 x + x = 11 11 x = 21 (x 2 = + x − x − − 4)( x − 2) Multiply each term by ( x − 4)( x − 2) = 1( x − 2) + 2( x − 4) = x − + 2x − = x − 10 12 = x = x A check reveals that x = yields a denominator of zero So, x = is an extraneous solution, and the original equation has no real solution 22 (x − 1)( x + 3) (x 12 = + − 1)( x + 3) x −1 x + (x 12 = ( x − 1)( x + 3) + ( x − 1)( x + 3) − 1)( x + 3) x −1 x + Multiply each term by ( x − 1)( x + 3) 12 = 3( x + 3) + 2( x − 1) 12 = 3x + + x − 12 = x + = 5x x =1 A check reveals that x = yields a denominator of zero So, x = is an extraneous solution, and the original equation has no real solution + x −3 x + x −3 x 23 10 = Multiply each term by ( x + 3)( x − 3) +3 x −9 10 = +3 ( x + 3)( x − 3) 1( x + 3) + 1( x − 3) = 10 x = 10 x = + x −2 x + x −2 x 24 (x = x + x−6 +3 = +3 ( x + 3)( x − 2) Multiply both sides by ( x + 3)( x − 2) + 3) + 3( x − 2) = x + + 3x − = 4x − = 4x = 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file8 at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Chapter P Prerequisites 25 x + 3x = 33 3x( x + 1) = ( 3x = or x + = 26 x + 20)( x + 4) = 3x + 20 = x = or x − = x = or x = x = −4 − x − 16 = x − 16 =  x = 16 x + =  x = −8 x = −5 35 x = 49 x2 − 2x − = x = ±7 − 4)( x + 2) = 36 x = 43 x+2 = x = ± x = −2 43 x ≈ 6.56 + 5x − x = (3 − x)(1 + x) 37 x = 81 = x = 27 − x = or + x = x = or 30 or x − x − 128 = x +5 = 29 x+4 = ( x − 16)( x + 8) = + 5)( x + 5) = x = or or 27 x + 10 x + 25 = x − = or x2 34 (x − 20 x = x( x − 1) = 28 ) + x + 20 = 4(0) x + 32 x + 80 = (3 x 8x2 − 2x = (x + x + 20 = x = − 12 x = or x x = ±3 x = − 12 ≈ ±5.20 x + 12 x + = (2 x 38 x = 36 + 3)( x + 3) = x2 = x + =  x = − 32 31 x = ± 16 x − = (4 x 39 + 3)( x + 3) = 4x + =  x = 4x − =  x = − x + x − 12 = (−1)(− x + x − 12) = ( −1)(0) − 4) = 49 x − = ±7 − 34 − x + x = 12 32 (x x = 4±7 x = 11 or x = − 40 (x + 9) = 24 x +9 = ± − 6)( x − 2) = x−6 =  x = x−2 =  x = 24 x = −9 ± ≈ − 4.10, −13.90 x − x + 12 = (x = ±2 41 ( x − 1) = 18 2 x − = ± 18 2x = ± 1±3 2 ≈ 2.62, −1.62 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Section P.2 Solving Equations 42 (x − 7) = ( x + 3) 2 47 2 x − x + = − + 12 2 ( x − 1) = x − = ± ( x + 3) x2 − 2x = − x − = x + or −7 =/ x − 12 x = − x − = −x − 2x = or x = The only solution of the equation is x = x + x = 32 x + x + 22 = 32 + 22 (x x −1 = ± 43 x + x − 32 = + 2) = 36 x =1± x =1± 2 x + = ±6 x = −2 ± x2 − x = 48 x = or x = − x2 − x = x2 − 2x − = 44  1  1 x2 − x +  −  = + −   2   x − 2x = x − x + ( −1) = + (1) 2 (x 2 2 = ± 2 x− x =1±2 x = x = or x = −1 45 x2 + x + = x = ± 2 1± 49 (x 2x2 + 5x − = + 2) = 2 x2 + 5x = x + = ± x = −2 ± x2 + 5 5 x +  = 4+     4 2 x + x = −14 x + x + 42 = −14 + 16 + 4) = 5 89  x +  = 4 16  x + x + = ± x = 2 x2 + 46 x + x + 14 = (x 2 x2 + 4x = − x + x + 22 = − + 22 1  x −  = 2  − 1) = x −1 = ± x = −4 ± 89 = ± 4 x = − x = ± 89 −5 ± 89 © 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file10at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Chapter P Prerequisites 3x − x − = 50 52 3x − x = x2 − x + 10 x + 25 + 49 = ( x + 5) + 49 x = 3 = x2 − 4 = x + 10 x + 74 x + 10 x + (5) − (5) + 74  2  2 x + −  = + −  3  3   2 2 25  x −  = 3  x − = ± 3 x = ± 3 53 + 2x − x = = = x = −1 or x = = 1 = 51 x − 2x + x − x + 12 + = ( x − 1)2 + 54 12 + x − x = 2 −1( x − x − 12) = −1( x − x + 4) − 16 = −b ± b − 4ac 2a −1 ± 12 − 4( 2)( −1) = = 1± 1+8 1±3 = = 1, − 4 − ( x − 1) −1 x − x + (2) − (2) − 12 = 16 − ( x − 2) x = = −b ± −30 ± b − 4ac 2a 302 − 4(9)( 25) 2(9) 28 x − 49 x = 58 b − 4ac 2a (−1)2 2( 2) −1( x − x + 1) + −30 ± = = − 18 56 x − x − = −( −1) ± −1  x − x + (1) − (1) − 3 2( 2) −b ± 2 −1 ± = = , −1 x = −1( x − x − 3) 57 x + 30 x + 25 = 55 x + x − = x = = − 4( 2)( −1) − 49 x + 28 x − = x = = = −b ± − 28 ± b − 4ac 2a 282 − 4( − 49)( − 4) 2( − 49) −28 ± = − 98 © 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file16at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Chapter P Prerequisites x ( x − 1) 13 102 x 2 x( x − 1)  13 + x( x − 1) 43 x − = 11 103 = x − = 11  x = + 3( x − 1)  =  43 − ( x − 5) = 11  x = − x( x − 1) 2 x + 3( x − 1) = 13 x( x − 1) 13 (5 x 3x + =  x = 2x =  x = − (3x + 2) = x −1 =  x = 5x − =  x = 3x + = 104 − 3) = − 3x − =  x = − 3 105 x + = x − First equation: Second equation: − ( x + 1) = x − x +1 = x −5 x2 − x − = (x − x − = x2 − − 3)( x + 2) = x + x − = x−3 =  x = x = x + =  x = −2 Only x = and x = −1 + 17 −1 − 17 −1 + 17 are solutions of the original equation x = − and x = are extraneous 2 106 x + x = x + 18 First equation: x + x = 3x + 18 x + x − 18 = (x Second equation: − ( x + x) = x + 18 − 3)( x + 6) = x −3 =  x = x + =  x = −6 = x + x + 18 = ( x + 3)( x + 6) = x +  x = −3 x = x +  x = −6 The solutions of the original equation are x = ± and x = − 107 Let y = 18 y = 0.514 x − 14.75 18 = 0.514 x − 14.75 108 Let y = 23 y = 0.532 x − 17.03 23 = 0.532 x − 17.03 32.75 = 0.514 x 40.03 = 0.532 x 32.75 = x 0.514 63.7 = x 40.03 = x 0.532 75.2 = x So, the height of the female is about 63.7 inches The height of the missing man is about 75.2 inches  ft  Because 75.2 in   = 6.27 ft is about feet  12 in  inches, it is possible the femur belongs to the missing man © 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Section P.3 The Cartesian Plane and Graphs of Equations 17 109 False 111 2x + = − + x +1 2x + = − x + + ( x + 1) x − = −4 x − x + 16 = 16( x + 1) x − 24 x = x( x − 24) = x = ≠ − + 1, x = 24 ≠ −2 + x − 10 = x − 10 = x − 10 False The equation is an identity, so every real number is a solution x +1 x − 10 − 112 (a) The formula for volume of the glass cube is V = Length × Width × Height The volume of water in the cube is the length × width × height of the water So, the volume is x ⋅ x ⋅ ( x − 3) = x ( x − 3) (b) Given the equation x ( x − 3) = 320 The 110 2( x − 3) + = x − dimensions of the glass cube can be found by solving for x Then, substitute that value into the expression x3 to find the volume of the cube 2x − + = 2x − 2x − = 2x − False The equation is an identity, so every real number is a solution 113 3x + = x + = 35 and x + = 20 x = 33 x = 11 x = 11 Yes, they are equivalent equations They both have the solution x = 11 Section P.3 The Cartesian Plane and Graphs of Equations Cartesian 12 y Distance Formula (− 2, 4) Midpoint Formula −8 −6 −4 (− 2, − 7) intercepts (1, − 5) −8 14 ( −12, 0) circle; ( h, k ); r A: ( 2, 6), B : ( − 6, − 2), C : ( 4, − 4), D: ( − 3, 2) ( −6 13 ( − 3, 4) y-axis , x −2 −4 graph 10 A: (3, 3) solution or solution point (23, 52) (0, 5) ) ( − , B : (0, − 2), C : − 3, 11 y ) , D : ( − 6, 0) 15 x > and y < in Quadrant IV 16 x < and y < in Quadrant III 17 x = −4 and y > in Quadrant II 18 x < and y = in Quadrant II 19 x + y = 0, x ≠ 0, y ≠ means x = − y or y = − x (2, 4) (− 6, 2) (− 4, 0) This occurs in Quadrant II or IV −8 −6 −4 −2 −4 −6 x (3, − 1) (1.5, − 3.5) 20 ( x, y), xy > means x and y have the same signs This occurs in Quadrant I or III (− 1, − 8) © 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file18at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Chapter P Prerequisites 21 24 d = Year, x Number of Stores, y Number of stores 7276 = (0 − 8)2 2008 7720 = (−8)2 2009 8416 = 64 + 225 2010 8970 = 289 2011 10,130 2012 10,773 2013 10,942 2014 11,453 25 d = x Year y Month, x Temperature, y – 39 – 29 Temperature (in °F) 40 30 20 2 −10 − x1 ) + ( y2 − y1 ) = ( −6 ) = 36 + 25 = 61 units ( x2 + ( −1 − 4) + ( −5) 2 − x1 ) + ( y2 − y1 ) 2 − 9.5) + (8.2 − ( −2.6)) = (−3.9 = (−13.4)2 = 179.56 + 116.64 = 296.2 + (10.8) x 10 12 −20 −30 27 (a) (1, 0), (13, 5) (13 − 1) Distance = + (5 − 0) –5 17 27 (13, 5), (13, 0) 35 Distance = − = = 32 (1, 0), (13, 0) 22 Distance = − 13 = −12 = 12 10 11 – 23 12 – 34 ( x2 − x1 ) + ( y2 − y1 ) = (3 − (−2)) = (5) = 25 + 144 2 −40 Month (1 ↔ January) + ( −12) 122 + 52 = = 169 = 13 (b) 52 + 122 = 25 + 144 = 169 = 132 28 (a) The distance between ( −1, 1) and (9, 1) is 10 The distance between (9, 1) and (9, 4) is + ( −6 − 6) 2 ≈ 17.21 units 10 ( x2 (−5 − 1)2 2007 2008 2009 2010 2011 2012 2013 2014 – 39 + (15) = 26 d = + ( 20 − 5) = 17 units 11,500 11,000 10,500 10,000 9500 9000 8500 8000 7500 7000 23 d = − x1 ) + ( y2 − y1 ) 2007 y 22 ( x2 The distance between ( −1, 1) and (9, 4) is (9 − (−1)) + ( − 1) (b) 102 + 32 = 109 = ( = 109 ) 100 + = 109 = 13 units © 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Section P.3 The Cartesian Plane and Graphs of Equations 19 29 d1 = (4 − 2) + (0 − 1) = 4+1 = d2 = (4 + 1) + (0 + 5) = 25 + 25 = d3 = (2 + 1) + (1 + 5) = + 36 = 2 ( 5) + ( 45 ) = ( 50 ) y 34 (a) (6, 5) 50 45 x 2 −2 (6, −3) −4 (3 − (−1)) 30 d1 = d2 = (5 − 3) d3 = (5 − (−1)) ( 20 ) + ( 20 + (5 − 3) + (1 − 5) ) 2 ( 40 = = + (1 − 3) = ) 16 + = + 16 = = 20 20 36 + = 40 (5 − (−3)) + (6 − 6) = (b) d = 64 =  + + (−3)  (c)  ,  = (6, 1)   y 35 (a) 10 (5, 4) 31 d1 = (1 − 3)2 + ( −3 − 2) d2 = (3 + 2)2 + ( − 4) d3 = (1 + 2)2 + ( −3 − 4) = + 25 = = 29 25 + = = (−1, 2) 29 + 49 = −1 58 x −1 d1 = d (4 32 d1 = − 2) + (9 − 3) d2 = ( −2 d3 = (2 − (−2)) − 4) + (7 − 9) 2 = + 36 = = + (3 − ) 40 36 + = = = 40 16 + 16 = d1 = d 33 (a) (b) d = 32 (5 + 1)2 + ( − 2) 36 + = 10  −1 + +  , (c)   = ( 2, 3)   y 36 (a) y (− 25 , 34 ) 12 ( 12, 1) 10 (9, 7) − −2 − −1 2 −1 2 x (1, 1) −2 (b) d = (9 x 10 − 1) + (7 − 1)  + + 1 , (c)   = (5, 4)   (b) d = = 64 + 36 = 10 = 5 4 1   +  + 1 −  2 3    9+ = 82  −(5 2) + (1 2) ( 3) +   (c)  ,  =  −1, 2    37 d = = 7  6 1202 + 1502 36,900 = 30 41 ≈ 192.09 The plane flies about 192 kilometers © 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file20at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Chapter P Prerequisites 38 d = ( 42 − 18) + (50 − 12) 2 43 (a) ? − 3( 2) + = ? = 242 + 382 = 2020 = = 505 Yes, the point is on the graph 4−6+ = ≈ 45 The pass is about 45 yards  x + x2 y1 + y2  39 midpoint =  ,     2010 + 2014 35,123 + 45,998  , =   2   = ( 2012, 40,560.5) (b)  x + x2 y1 + y2  40 midpoint =  ,     2013 + 2015 1.17 + 3.25  , =   2   = ( 2014, 2.21) (0, 2): = 12 ≠ No, the point is not on the graph 44 (a) ? (−1, 1): = − 2(−1)2 ? = − 2(1) 1=1 Yes, the point is on the graph (b) ? ( − 2, 11): 11 = − 2( − 2)2 ? 11 = − 2( 4) 11 ≠ − No, the point is not on the graph 45 (a) 0+ (3, − 2): (3)2 ? + = 20 13 ≠ 20 Yes, the point is on the graph (5, 3): ? = ? = 5+ ? + ( −2) = 20 = (b) ? − 3( −2) + = ? In 2014, the revenue per share for Twitter, Inc was approximately $2.21 ? (−2, 8): (−2)2 4+ 6+ = In 2012, the sales for the Coca-Cola Company were about $40,560.5 million 41 (a) (2, 0): (2)2 No, the point is not on the graph (b) ( −4, 2): (−4)2 ? + ( 2) = 20 ? 16 + = 20 = 20 = 20 Yes, the point is on the graph Yes, the point is on the graph ? 42 (a) (1, 5) : = − − ? 46 (a) = −1 (6, 0): 2(6) ? ? 72 + = No, the point is not on the graph (6, 0): 72 ≠ ? = 4− 6− No, the point is not on the graph ? = 4− = Yes, the point is on the graph 2(36) + 5(0) = ≠ (b) ? + 5(0) = (b) (0, 4): 2(0) ? + 5( 4) = ? 2(0) + 5(16) = ? + 80 = 80 ≠ No, the point is not on the graph © 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Section P.3 The Cartesian Plane and Graphs of Equations 21 47 y = −2 x + 50 y = − x x −1 x −2 –1 y y (x, y) (−1, 7) (0, 5) (1, 3) (2, 1) ( 52 , 0) x, y (−2, 1) (−1, 4) (0, 5) (1, 4) (2, 1) y y 4 3 2 1 −3 −2 −1 −1 48 y + = x 4 −4 −3 x x −1 −1 −2 51 y = x − x-intercept: = x − x −2 y − 52 –1 − 14 (x, y) (−2, ) (0, −1) (1, ) ( 0) (2, ) − 52 − 14 4, = 5x = x ( 56 , 0) y-intercept: y = 5(0) − = − y (0, − 6) 52 y = − 3x x −4 −3 −2 −1 x-intercept: = − x −2 3x = −3 x = −4 ( 83 , 0) 49 y + x = x x −1 y-intercept: y = − 3(0) = y –2 –2 (0, 8) (x, y) (−1, 4) (0, 0) (1, − 2) (2, − 2) (3, 0) = x + = x + −4 = x −2 x + x-intercept: y −2 −1 −1 53 y = (− 4, 0) x y-intercept: y = 0+ = (0, 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file22at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Chapter P Prerequisites 54 y = 57 y = x3 − x 2x − = x-intercept: 2x − 2x − = x = x-intercept: = x3 − x = x ( x − 2) x = or x = ( 12 , 0) (0, 0), (2, 0) 2(0) − y-intercept: y = = y-intercept: y = 2(0) − 4(0) y = −1 (0, 0) There is no real solution There is no y-intercept 58 y = x − 25 55 y = 3x − x-intercept: = x − x-intercepts: = x − 25 x = 25 = 3x − 7 x = ± 52 = ± = ( 0) , y-intercept: y = 3(0) − = (± 5, ) y-intercept: y = (0) − 25 = − 25 (0, − 25) (0, 7) 59 y = − x 56 y = − x + 10 x-intercept: = − x x = = − x + 10 x-intercept: (6, 0) x + 10 = x = −10 (−10, 0) y-intercepts: y = − y = ± (0, ), (0, − ) y-intercept: y = − + 10 = − 10 = −10 (0, −10) 60 y = x + x-intercept: = x + x = −1 (−1, 0) y-intercepts: y = + y = ±1 (0, 1), (0, −1) 61 x − y = ( − x)2 − y =  x − y =  y -axis symmetry x − ( − y ) =  x + y =  No x-axis symmetry ( − x) − ( − y ) =  x + y =  No origin symmetry © 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Section P.3 The Cartesian Plane and Graphs of Equations 23 62 x − y = ( − x) − y =  − x − y =  No y -axis symmetry x − ( − y ) =  x − y =  x-axis symmetry ( − x) − ( − y ) =  − x − y =  No origin symmetry 63 y = x3 y = ( − x)  y = − x3  No y -axis symmetry − y = x3  y = − x3  No x-axis symmetry − y = ( − x)  − y = − x3  y = x3  Origin symmetry 64 y = x − x + y = ( − x) − ( − x) +  y = x − x +  y -axis symmetry − y = x − x +  y = − x + x −  No x-axis symmetry − y = ( − x) − ( − x) +  y = − x + x −  No origin symmetry x x +1 −x 65 y = y = 2 ( − x) +1  y = −x  No y -axis symmetry x2 + x −x  y =  No x-axis symmetry x2 + x +1 −x −x x −y =  −y =  y =  Origin symmetry x +1 x +1 ( − x) + −y = x2 + 1 66 y = y = ( − x) +1  y =  y -axis symmetry x2 + 1 −1  y =  No x-axis symmetry x2 + x +1 −1 −y =  y =  No origin symmetry +1 x − + x ( ) −y = 67 xy + 10 = (− x) y + 10 =  − xy + 10 =  No y -axis symmetry x( − y ) + 10 =  xy + 10 =  x-axis symmetry (− x)( − y ) + 10 =  − xy + 10 =  No origin symmetry 68 xy = (− x) y =  xy = −4  No y -axis symmetry x( − y ) =  xy = −4  No x-axis symmetry (− x)(− y ) =  xy =  Origin symmetry y 69 −4 −3 −1 x −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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file24at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Chapter P Prerequisites y 70 74 y = x − x-intercept: x −1 ( 32 , 0) y-intercept: (0, − 3) No symmetry −2 −3 y −4 y 71 −3 −2 −1 ( 32 , 0) −1 x −2 −4 −3 −2 (0, − 3) −3 x −2 −3 75 y = x − x −4 72 x-intercepts: (0, 0), ( 2, 0) y y-intercept: (0, 0) No symmetry x −4 −3 −2 −1 −2 x –1 y –1 −3 y −4 73 y = −3 x + x-intercept: ( 13 , 0) (0, 0) y-intercept: (0, 1) −2 −1 No symmetry (2, 0) −1 x −2 y 76 x = y − −4 −3 −2 −1 −1 −2 −3 x-intercept: ( −1, 0) y-intercepts: (0, −1), (0, 1) (0, 1) ,0 (3 ( x x-axis symmetry x –1 y ±1 ±2 y (0, 1) (−1, 0) x −2 (0, −1) −2 −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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Section P.3 The Cartesian Plane and Graphs of Equations 25 77 y = x3 + x-intercept: 80 y = ( −3, ) 1− x x-intercept: (1, 0) y-intercept: (0, 3) y-intercept: (0, 1) No symmetry No symmetry y x –2 –1 y –5 11 y (0, 1) (1, 0) −4 −3 −2 −1 −1 x (0, 3) ( −3, ( 81 y = x − −4 −3 −2 −1 x x-intercept: (6, 0) y-intercept: (0, 6) 78 y = x − No symmetry x-intercept: (1, 0) y-intercept: (0, −1) x –2 10 y 2 No symmetry y y 12 10 −3 −2 (1, 0) −1 −2 x (0, − 1) (6, 0) −3 −2 −4 79 y = (0, 6) −2 10 12 x 82 y = − x x −3 x-intercepts: (1, 0), ( −1, 0) x-intercept: (3, 0) y-intercept: none y-intercept: (0, 1) No symmetry y-axis symmetry y x 12 y 3 (− 1, 0) y −3 −2 (0, 1) (1, 0) x −1 −2 −3 (3, 0) –1 x 83 Center: (0, 0); Radius: (x − 0) + ( y − 0) = 2 x + y = 49 © 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file26at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Chapter P Prerequisites 84 Center: ( − 4, 5); Radius: (x 87 Endpoints of a diameter: (3, 2), ( − 9, − 8) − h) + ( y − k ) = r 2 2 (− − 3)2 + (− − 2)2 = (−12)2 + (−10)2 144 + 100 = 1 244 = ( 2) 61 = 61 = 2 r =  x − ( − 4) + [ y − 5] = (x + 4) + ( y − 5) = 2 85 Center: (3, 8) ; Solution point: ( − 9, 13) (x r = − h) + ( y − k ) = (− = (−12)2 = 144 + 25 = 169 − 3) + (13 − 8) + (5)  + ( − 9) ( h, k ) :  (x (x (x − 3) + ( y − 8) = 132 (x − 3) + ( y − 8) = 169 2 2 − h) + ( y − k ) 2 1 − ( − 2) + −  10 − ( − 6) = (3) = + 16 + ( − 4) 2 = = = 2 − h) + ( y − k ) = r 2 2  x − ( − 2) +  y − ( − 6) = 52 + 2) + ( y + 6) = 25 2 61 ) 2 + 15 − ( − 5) + ( 20) 2 64 + 400 464 =  25 (− 8)2 (h, k ):  = ( + 3) + ( y + 3) = 61 (3 − 11)2 r = = (x 88 Endpoints of a diameter: (11, − 5), (3, 15) 86 Center: ( − 2, − 6) ; Solution point: (1, −10) (x 2 − h) + ( y − k ) = r = − h) + ( y − k ) = r 2  x − ( − 3) +  y − ( − 3) = (x (x + ( − 8)   −6 −6 ⋅  =   2    = ( − 3, − 3) = 13 r =  ⋅ (4) 29 = 29 11 + − + 15   14 10  ⋅  =  ,  = (7, 5) 2  2 2 (x − h) + ( y − k ) = r (x − 7) + ( y − 5) = 29 (x − 7) + ( y − 5) = 116 2 ( 2 ) 2 89 x + y = 25 Center: (0, 0), Radius: y −4 −3−2−1 −2 −3 −4 (0, 0) x −6 © 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Section P.3 The Cartesian Plane and Graphs of Equations 27 90 x + ( y − 1) = 92 Center: (0, 1), Radius: (x − 2) + ( y + 3) = 16 Center: ( 2, − 3), Radius: y y 1 −1 −2 x −1 x −1 (0, 1) −2 2 (2, − 3) −3 −1 −4 91 ( x − 12 ) ( Center: ( 12 , 12 ), 2 + y − ) = Radius: 93 y = 1,200,000 − 80,000t , ≤ t ≤ 10 y y Depreciated value 1,200,000 ( 12 , 12) 1,000,000 800,000 600,000 400,000 200,000 t x −1 3 10 Year 94 y = 9500 − 1000t , ≤ t ≤ Depreciated value y 10,000 9000 8000 7000 6000 5000 4000 3000 2000 1000 t Year 95 (a) x 10 20 30 40 50 60 70 80 90 100 y 414.8 103.7 25.9 11.5 6.5 4.1 2.9 2.1 1.6 1.3 1.0 y Resistance (in ohms) (b) 450 400 350 300 250 200 150 100 50 x 20 40 60 80 100 Diameter of wire (in mils) When x = 85.5, the resistance is about 1.4 ohms (c) When x = 85.5, y = 10,370 = 1.4 ohms (85.5) (d) As the diameter of the copper wire increases, the resistance decreases © 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file28at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Chapter P Prerequisites 96 (a) 100 97 False, you would have to use the Midpoint Formula 15 times 98 True Two sides of the triangle have lengths the third side has a length of 100 The model fits the data well (b) Graphically: The point (50, 74.7) represents a life expectancy of 74.7 years in 1990 Algebraically: y = 63.6 + 0.97(50) + 0.01(50) 112.1 1.5 = 74.7 = So, the life expectancy in 1990 was about 74.7 years (c) Graphically: The point ( 24.2, 70.1) represents a life expectancy of 70.1 years during the year 1964 Algebraically: 63.6 + 0.97t + 0.01t 63.6 + 0.97t 70.1 = + 0.01t 70.1(1 + 0.01t ) = 63.6 + 0.97t y = 70.1 + 0.701t = 63.6 + 0.97t 6.5 = 0.269t t = 24.2 When y = 70.1, t = 24.2 which represents the year 1964 (d) y = 63.6 + 0.97(0) 149 and 18 99 False The line y = x is symmetric with respect to the origin 100 True Depending upon the center and radius, the graph of a circle could intersect one, both, or neither axis 101 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 102 The y-coordinate of a point on the x-axis is The x-coordinates of a point on the y-axis is 103 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   104 (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) + 0.01(0) (d) Because ( x0 , y0 ) lies in Quadrant II, ( − x0 , − y0 ) 63.6 = = 63.6 must lie in Quadrant IV Matches (i) The y-intercept is (0, 63.6) In 1940, the life expectancy of a child (at birth) was 63.6 years (e) Answers will vary 105 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 ) (a) ( x2 , y2 ) = ( xm − x1 , ym − y1 ) = ( ⋅ − 1, 2( −1) − ( −2)) = (7, 0) (b) ( x2 , y2 ) = ( xm − x1 , ym − y1 ) = ( ⋅ − ( −5), ⋅ − 11) = (9, − 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Section P.4 Linear Equations in Two Variables 29 Section P.4 Linear Equations in Two Variables 14 The line appears to go through (0, 7) and (7, 0) linear slope Slope = point-slope y2 − y1 − = = −1 x2 − x1 − 15 y = x + parallel Slope: m = 5 perpendicular y-intercept: (0, 3) rate or rate of change y linear extrapolation general (a) m = (0, 3) Because the slope is positive, the line rises x −4 −3 −2 −1 Matches L2 (b) m is undefined The line is vertical Matches L3 (c) m = −2 The line falls Matches L1 10 (a) m = The line is horizontal Matches L2 16 Slope: m = −1 y -intercept: (0, −10) (b) m = − 34 Because the slope is negative, the line y falls Matches L1 (c) m = Because the slope is positive, the line rises Matches L3 −12 −10 −6 (2, 3) 2 −4 y 11 x −6 −4 −2 −2 m=0 −10 (0, −10) −12 m=1 m = −3 17 y = − 34 x − m=2 Slope: m = − 34 x 12 y -intercept: (0, −1) undefined m = −3 y y m=3 4 m = 21 (− 4, 1) −6 x −2 −2 −6 −4 −2 x −2 (0, −1) −4 −6 −8 13 Two points on the line: (0, 0) and ( 4, 6) Slope = y2 − y1 = = x2 − x1 © 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Solution Manual for Trigonometry 10th Edition by Larson Full file30at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson Chapter P Prerequisites x 18 y = 21 x − = + Slope: m = x = 25 , vertical line y -intercept: (0, 2) Slope: undefined y -intercept: none y y (0, 2) −6 x −2 x −1 −2 −1 −4 −2 19 y − = 22 y + = y = y = −5 Slope: m = y = − 53 y -intercept: (0, 5) Slope: m = y ( y-intercept: 0, − 53 (0, 5) ) y −4 x −2 −2 x −1 −1 −2 −2 20 x + = (0, − 35) −3 x = −4 Slope: undefined ( vertical line) 23 x − y = 30 y -intercept: none − y = − x + 30 y y = Slope: m = −6 x −2 −4 −5 y -intercept: (0, − 5) y −2 x −1 −1 x −2 −3 −4 −5 (0, − 5) −7 © 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 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-10th-Edition-by-Larson ... https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 10th- Edition- by- Larson Solution Manual for Trigonometry 10th Edition by Larson Full file at https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 10th- Edition- by- Larson. .. https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 10th- Edition- by- Larson Solution Manual for Trigonometry 10th Edition by Larson Full file4 at https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 10th- Edition- by- Larson. .. https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 10th- Edition- by- Larson Solution Manual for Trigonometry 10th Edition by Larson Full file at https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 10th- Edition- by- Larson