Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Chapter F Foundations: A Prelude to Functions (e) Positive y-axis (f) Negative x-axis Section F.1 − ( −3) = = 32 + 42 = 25 = 112 + 602 = 121 + 3600 = 3721 = 612 Since the sum of the squares of two of the sides of the triangle equals the square of the third side, the triangle is a right triangle A = 13 The points will be on a vertical line that is two units to the right of the y-axis bh Angle-Side-Angle, Side-Side-Side, Side-AngleSide x-coordinate, or abscissa; y-coordinate, or ordinate quadrants d = ( x1 − x2 )2 + ( y1 − y2 )2 10 midpoint 11 (a) (b) (c) (d) (e) (f) Quadrant II Positive x-axis Quadrant III Quadrant I Negative y-axis Quadrant IV 12 (a) (b) (c) (d) Quadrant I Quadrant III Quadrant II Quadrant I 14 The points will be on a horizontal line that is three units above the x-axis 15 d ( P1 , P2 ) = (2 − 0) + (1 − 0) = + = 16 d ( P1 , P2 ) = (− − 0)2 + (1 − 0) = + = Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Chapter F Foundations: A Prelude to Functions 29 A = (− 2,5), B = (1,3), C = (−1, 0) 17 d ( P1 , P2 ) = (− − 1) + (2 − 1) = + = 10 d ( A, B ) = 18 d ( P1 , P2 ) = ( − (−1) )2 + (2 − 1)2 = 32 + (− 2) = + = + = 10 = 13 19 d ( P1 , P2 ) = (5 − 3) + ( − ( −4 ) ) = 22 + ( ) d ( B, C ) = ( − ( −1) ) + ( − ) 2 = ( 3) +4 = 13 d ( A, C ) = = + 16 = 25 = 21 d ( P1 , P2 ) = ( −1 − 1)2 + (0 − 3)2 = (− 2) + (− 3) = + = + 64 = 68 = 17 20 d ( P1 , P2 ) = (1 − (− 2) )2 + (3 − 5)2 ( −1 − (− 2) )2 + (0 − 5)2 = 12 + (− 5)2 = + 25 = 26 ( − (−3) )2 + (0 − 2)2 = 92 + (− 2) = 81 + = 85 22 d ( P1 , P2 ) = ( − )2 + ( − (−3) )2 = 2 + 52 = + 25 = 29 23 d ( P1 , P2 ) = (−2 − 4) + ( −5 − (−2) ) = (−6) + (−3) = 36 + = 45 = 24 d ( P1 , P2 ) = ( − (− 4) )2 + ( − (−3) )2 = 102 + 52 = 100 + 25 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: = 125 = 5 [ d ( A, B)]2 + [ d ( B, C )]2 = [ d ( A, C )]2 ( 13 ) + ( 13 ) = ( 25 d ( P1 , P2 ) = ( 2.3 − (− 0.2 ) + (1.1 − 0.3) ) 26 = 26 = 6.89 ≈ 2.62 The area of a triangle is A = problem, A = ⋅ [ d ( A, B) ] ⋅ [ d ( B, C ) ] = ⋅ 13 ⋅ 13 = ⋅13 2 13 = square units ( − 0.3 − 1.2 )2 + (1.1 − 2.3)2 = (−1.5) + (−1.2) = 2.25 + 1.44 = 3.69 ≈ 1.92 27 d ( P1 , P2 ) = (0 − a ) + (0 − b) = a + b 28 d ( P1 , P2 ) = (0 − a) + (0 − a ) = a + a = 2a = a Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ 26 13 + 13 = 26 = (2.5) + (0.8) = 6.25 + 0.64 26 d ( P1 , P2 ) = ⋅ bh In this Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Section F.1: The Distance and Midpoint Formulas 31 A = (− 5,3), B = (6, 0), C = (5,5) 30 A = (− 2, 5), B = (12, 3), C = (10, − 11) d ( A, B) = (12 − (− 2) ) + (3 − 5) d ( A, B) = ( − (− 5) )2 + (0 − 3)2 = 142 + (− 2) = 112 + (− 3) = 121 + = 196 + = 200 = 130 = 10 d ( B, C ) = d ( B, C ) = (10 − 12 )2 + (−11 − 3)2 = (−1) + 52 = + 25 = (− 2) + (−14) = 26 = + 196 = 200 d ( A, C ) = = 10 d ( A, C ) = (10 − (− 2) ) 2 = 12 + (−16) ( − )2 + (5 − 0)2 ( − (− 5) )2 + (5 − 3)2 = 102 + 22 = 100 + + (−11 − 5) = 104 = 26 = 144 + 256 = 400 = 20 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ d ( A, C )]2 + [ d ( B, C )]2 = [ d ( A, B)]2 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: ( [ d ( A, B)]2 + [ d ( B, C )]2 = [ d ( A, C )]2 (10 ) + (10 ) 2 = ( 20 ) 104 ) +( ) =( 130 ) 104 + 26 = 130 130 = 130 The area of a triangle is A = bh In this problem, A = ⋅ [ d ( A, C ) ] ⋅ [ d ( B, C ) ] = ⋅ 104 ⋅ 26 = ⋅ 26 ⋅ 26 = ⋅ ⋅ 26 = 26 square units 200 + 200 = 400 400 = 400 The area of a triangle is A = bh In this problem, A = ⋅ [ d ( A, B) ] ⋅ [ d ( B, C ) ] = ⋅10 ⋅10 2 = ⋅100 ⋅ 2 = 100 square units Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ 26 Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Chapter F Foundations: A Prelude to Functions 33 A = (4, −3), B = (0, −3), C = (4, 2) 32 A = (− 6, 3), B = (3, − 5), C = (−1, 5) + (−5 − 3) d ( A, B ) = (0 − 4) + ( −3 − (−3) ) = 92 + (− 8) = 81 + 64 = (− 4)2 + 02 = 16 + = 145 = 16 d ( A, B) = d ( B, C ) = ( − (− 6) ) =4 ( −1 − 3)2 + (5 − (−5))2 d ( B, C ) = = (− 4) + 102 = 16 + 100 = 41 ( −1 − (− 6) )2 + (5 − 3)2 ( − )2 + ( − (−3) )2 = 42 + 52 = 16 + 25 = 116 = 29 d ( A, C ) = d ( A, C ) = (4 − 4) + ( − (−3) ) = + = 25 + = 02 + 52 = + 25 = 29 = 25 =5 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ d ( A, C )]2 + [ d ( B, C )]2 = [ d ( A, B)]2 ( 29 ) +( 116 ) =( 145 ) Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: 29 + 116 = 145 145 = 145 The area of a triangle is A = bh In this problem, A = ⋅ [ d ( A, C ) ] ⋅ [ d ( B, C ) ] = ⋅ 29 ⋅ 116 = ⋅ 29 ⋅ 29 = ⋅ ⋅ 29 = 29 square units [ d ( A, B)]2 + [ d ( A, C )]2 = [ d ( B, C )]2 + 52 = 41 ) 16 + 25 = 41 41 = 41 The area of a triangle is A = problem, A = ⋅ [ d ( A, B) ] ⋅ [ d ( A, C ) ] = ⋅4⋅5 = 10 square units Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ ( bh In this Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Section F.1: The Distance and Midpoint Formulas 35 The coordinates of the midpoint are: x + x y + y2 ( x, y ) = , 34 A = (4, − 3), B = (4, 1), C = (2, 1) d ( A, B) = (4 − 4) + (1 − (−3) ) = 02 + 42 3+5 −4+ , = 8 0 = , 2 2 = (4, 0) = + 16 = 16 =4 d ( B, C ) = ( − )2 + (1 − 1)2 36 The coordinates of the midpoint are: x + x y + y2 ( x, y ) = , = (− 2) + 02 = + = =2 d ( A, C ) = (2 − 4) + (1 − (−3) ) −2 + + , = 0 4 = , 2 2 = ( 0, ) = (− 2) + 42 = + 16 = 20 =2 37 The coordinates of the midpoint are: x + x y + y2 ( x, y ) = , −3 + + , = 3 2 = , 2 2 3 = ,1 2 38 The coordinates of the midpoint are: x + x y + y2 ( x, y ) = , Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: + −3 + , = −1 = , 2 1 = 3, − 2 [ d ( A, B)]2 + [ d ( B, C )]2 = [ d ( A, C )]2 ( 42 + 22 = ) 16 + = 20 20 = 20 The area of a triangle is A = bh In this problem, A = ⋅ [ d ( A, B) ] ⋅ [ d ( B, C ) ] = ⋅4⋅2 = square units Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Chapter F Foundations: A Prelude to Functions 39 The coordinates of the midpoint are: x + x y + y2 ( x, y ) = , 43 The coordinates of the midpoint are: x + x y + y2 ( x, y ) = , a+0 b+0 = , a b = , 2 + (−2) −2 + (−5) , = 2 −7 = , 2 −7 = 1, 44 The coordinates of the midpoint are: x + x y + y2 ( x, y ) = , 40 The coordinates of the midpoint are: x + x y + y2 ( x, y ) = , a+0 a+0 = , a a = , 2 2 − + −3 + , = − −1 , = 2 45 Consider points of the form ( 2, y ) that are a distance of units from the point ( −2, −1) 1 = −1, − 2 ( x2 − x1 )2 + ( y2 − y1 )2 d= 41 The coordinates of the midpoint are: x + x y + y2 ( x, y ) = , = ( −2 − )2 + ( −1 − y )2 = ( −4 )2 + ( −1 − y )2 = 16 + + y + y − 0.2 + 2.3 0.3 + 1.1 = , 2 2.1 1.4 = , 2 = (1.05, 0.7) y + y + 17 = y + y + 17 5= 52 = 42 The coordinates of the midpoint are: x + x y + y2 ( x, y ) = , ( y + y + 17 ) 25 = y + y + 17 = y2 + y − = ( y + )( y − ) 1.2 + (− 0.3) 2.3 + 1.1 = , 2 0.9 3.4 = , 2 = ( 0.45, 1.7 ) y+4=0 or y − = y = −4 y=2 Thus, the points ( 2, −4 ) and ( 2, ) are a distance of units from the point ( −2, −1) 46 Consider points of the form ( x, −3) that are a distance of 13 units from the point (1, ) Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Section F.1: The Distance and Midpoint Formulas d= = ( x2 − x1 )2 + ( y2 − y1 )2 are a distance of units from the point ( 4, ) (1 − x )2 + ( − ( −3) ) d= = x − x + + (5) 2 = 42 + 16 − y + y = 16 + 16 − y + y = x − x + 26 13 = x − x + 26 ( x − x + 26 ( − )2 + ( − y )2 = = x − x + + 25 132 = ( x2 − x1 )2 + ( y2 − y1 )2 ) = y − y + 32 5= y − y + 32 52 = 169 = x − x + 26 ( y − y + 32 = ( x − 13)( x + 11) x − 13 = or x + 11 = 0 = y2 − y + x = 13 x = −11 Thus, the points (13, −3) and ( −11, −3) are a y − = or = ( y − )( y − 1) y=7 distance of 13 units from the point (1, ) 0+6 0+0 , 49 The midpoint of AB is: D = = ( 3, ) ( x2 − x1 )2 + ( y2 − y1 )2 ( − x )2 + ( −3 − )2 = 16 − x + x + ( −3) 0+4 0+4 , The midpoint of AC is: E = = ( 2, ) = 16 − x + x + 6+4 0+4 , The midpoint of BC is: F = = ( 5, ) = x − x + 25 = x − x + 25 52 = ( x − x + 25 ) y =1 axis and a distance of units from the point ( 4, ) are a distance of units from the point ( 4, −3) y −1 = Thus, the points ( 0, ) and ( 0,1) are on the y- 47 Points on the x-axis have a y-coordinate of Thus, we consider points of the form ( x, ) that = 25 = y − y + 32 = x − x − 143 d= ) d (C , D) = ( − )2 + (3 − 4)2 = (− 4) + (− 1) = 16 + = 17 25 = x − x + 25 = x2 − 8x d ( B, E ) = = x ( x − 8) x = or x − = ( − )2 + (2 − 0)2 = (− 4) + 22 = 16 + = 20 = x=8 Thus, the points ( 0, ) and ( 8, ) are on the x- d ( A, F ) = (2 − 0) + (5 − 0) axis and a distance of units from the point ( 4, −3) = 22 + 52 = + 25 = 29 48 Points on the y-axis have an x-coordinate of Thus, we consider points of the form ( 0, y ) that Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Chapter F Foundations: A Prelude to Functions 52 Let P1 = ( 0, ) , P2 = ( a, ) , and 50 Let P1 = (0, 0), P2 = (0, 4), P = ( x, y ) d ( P1 , P2 ) = (0 − 0) + (4 − 0) a 3a P3 = , To show that these vertices 2 form an equilateral triangle, we need to show that the distance between any pair of points is the same constant value = 16 = d ( P1 , P ) = ( x − 0) + ( y − 0) = x2 + y = d ( P1 , P2 ) = → x + y = 16 d ( P2 , P ) = ( x − 0) + ( y − 4) = = x + ( y − 4) = d ( P2 , P3 ) = = d ( P1 , P3 ) = x = ±2 Two triangles are possible The third vertex is 51 Let P1 = ( 0, ) , P2 = ( 0, s ) , P3 = ( s, ) , and P4 = ( s, s ) Y (s,s) X (0,0) The points P1 and P4 are endpoints of one diagonal and the points P2 and P3 are the endpoints of the other diagonal 0+s 0+s s s M 1,4 = , = , 2 2 0+s s+0 s s M 2,3 = , = , 2 2 The midpoints of the diagonals are the same Therefore, the diagonals of a square intersect at their midpoints Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ 4a = a2 = a a 3a 4a + = = a2 = a 4 Since all three distances have the same constant value, the triangle is an equilateral triangle Now find the midpoints: 0+a 0+0 a D = M P1P2 = , = , 0 2 a a 3a a E = M P2 P3 = a + + = , , a 3a 0+ 0+ a 3a 2 = , F = M P1P3 = , 4 ) (s,0) ( x2 − x1 )2 + ( y2 − y1 )2 = − 3, or 3, (0,s) a 3a + = 4 a 3a = − + − 2 x = 12 ) ( ( x2 − x1 )2 + ( y2 − y1 )2 a 3a = − a + − 2 y = y − y + 16 y = 16 y=2 which gives x + 22 = 16 ( ( a − )2 + ( − )2 = a2 = a → x + ( y − 4) = 16 Therefore, y2 = ( y − 4) ( x2 − x1 )2 + ( y2 − y1 )2 Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Section F.1: The Distance and Midpoint Formulas 3a a a − d ( D, E ) = − + 2 a 3a = + = 54 d ( P1 , P2 ) = ( − (−1) )2 + (2 − 4)2 = + (− 2) 2 = 49 + = 53 a 3a = − + = (− 2)2 + (− 7) 2 = + 49 = 53 = 52 + (− 9) = 25 + 81 3a 3a a a − d ( E , F ) = − + 4 4 = 106 Since [ d ( P1 , P2 ) ] + [ d ( P2 , P3 ) ] = [ d ( P1 , P3 ) ] , the triangle is a right triangle Since d ( P1 , P2 ) = d ( P2 , P3 ) , the triangle is isosceles Therefore, the triangle is an isosceles right triangle a a2 = Since the sides are the same length, the triangle is equilateral = 55 d ( P1 , P2 ) = ( − (− 2) )2 + ( − (−1) )2 = 22 + 82 = + 64 = 68 53 d ( P1 , P2 ) = (− − 2) + (1 − 1) = 17 ( − )2 + (2 − 7)2 d ( P2 , P3 ) = = 36 =6 d ( P2 , P3 ) = 2 a = + 02 2 = (− 6) + ( − (−1) )2 + (−5 − 4)2 d ( P1 , P3 ) = a a 3a + = 16 16 2 ( − )2 + (−5 − 2)2 d ( P2 , P3 ) = a a 3a + = 16 16 2 a a 3a d ( D, F ) = − + − 4 2 = = 32 + (− 5) = + 25 = 34 ( − − (− 4) )2 + (−3 − 1)2 = + (− 4) ( − (−2) )2 + ( − (−1) )2 d ( P1 , P3 ) = = 52 + 32 = 25 + = 16 =4 = 34 Since d ( P2 , P3 ) = d ( P1 , P3 ) , the triangle is isosceles d ( P1 , P3 ) = (− − 2) + (−3 − 1) = (− 6) + (− 4) Since [ d ( P1 , P3 ) ] + [ d ( P2 , P3 ) ] = [ d ( P1 , P2 ) ] , = 36 + 16 the triangle is also a right triangle Therefore, the triangle is an isosceles right triangle = 52 = 13 Since [ d ( P1 , P2 ) ] + [ d ( P2 , P3 ) ] = [ d ( P1 , P3 ) ] , 2 the triangle is a right triangle Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Chapter F Foundations: A Prelude to Functions Y ( − − )2 + ( − )2 56 d ( P1 , P2 ) = (0,90) (90,90) = (−11) + (− 2) = 121 + = 125 =5 ( − (− 4) )2 + (6 − 0)2 d ( P2 , P3 ) = X (0,0) = 82 + 62 = 64 + 36 b Using the distance formula: = 100 = 10 d = (310 − 90) + (15 − 90) ( − )2 + ( − )2 d ( P1 , P3 ) = (90,0) = 2202 + (−75)2 = 54025 = 2161 ≈ 232.43 feet = (−3) + 42 = + 16 = 25 =5 c d = (300 − 0) + (300 − 90)2 Since [ d ( P1 , P3 ) ] + [ d ( P2 , P3 ) ] = [ d ( P1 , P2 ) ] , 2 Using the distance formula: = 3002 + 2102 = 134100 the triangle is a right triangle = 30 149 ≈ 366.20 feet 57 Using the Pythagorean Theorem: 902 + 902 = d 60 a 8100 + 8100 = d First: (60, 0), Second: (60, 60) Third: (0, 60) y 16200 = d (0,60) (60,60) d = 16200 = 90 ≈ 127.28 feet 90 90 d x (0,0) 90 90 (60,0) b Using the distance formula: d = (180 − 60) + (20 − 60) 58 Using the Pythagorean Theorem: 602 + 602 = d 2 3600 + 3600 = d → 7200 = d = 1202 + (− 40) = 16000 = 40 10 ≈ 126.49 feet d = 7200 = 60 ≈ 84.85 feet c d = (220 − 0) + (220 − 60)2 60 60 = 2202 + 1602 = 74000 d 60 59 a Using the distance formula: = 20 185 ≈ 272.03 feet 60 61 The Focus heading east moves a distance 30t after t hours The truck heading south moves a distance 40t after t hours Their distance apart First: (90, 0), Second: (90, 90) Third: (0, 90) 10 Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Chapter F Foundations: A Prelude to Functions b Let y = For 200 kWh, C = 0.0855(200) + 7.00 = $24.10 c d For 500 kWh, C = 0.0855(500) + 7.00 = $49.75 e For each usage increase of kWh, the monthly charge increases by $0.0848 (that is, 8.48 cents) 0=− x = 30 25 25 25 x = ( 30 ) 25 x = 375 The x-intercept is (375, 0) This means that the ramp meets the floor 375 inches (or 31.25 feet) from the base of the platform 121 (°C , ° F ) = (0, 32); (°C , ° F ) = (100, 212) 212 − 32 180 = = 100 − 100 ° F − 32 = (°C − 0) ° F − 32 = (°C ) 5 °C = (° F − 32) If °F = 70 , then 5 °C = (70 − 32) = (38) 9 °C ≈ 21.1° slope = 122 a b 123 a x + 30 25 c No From part (b), the run is 31.25 feet which exceeds the required maximum of 30 feet d First, design requirements state that the maximum slope is a drop of inch for each 12 inches of run This means m ≤ 12 Second, the run is restricted to be no more than 30 feet = 360 inches For a rise of 30 inches, this means the minimum slope is 30 1 = That is, m ≥ Thus, the 12 360 12 only possible slope is m = The 12 diagram indicates that the slope is negative Therefore, the only slope that can be used to obtain the 30-inch rise and still meet design requirements is m = − In words, for 12 every 12 inches of run, the ramp must drop exactly inch K =º C + 273 º C = (º F − 32) K = (° F − 32) + 273 160 K = ºF − + 273 9 2297 K = ºF + 9 K = ( 5° F + 2297 ) 124 a The y-intercept is (0, 30), so b = 30 Since the ramp drops inches for every 25 inches −2 = − Thus, of run, the slope is m = 25 25 the equation is y = − x + 30 25 The year 2000 corresponds to x = 0, and the year 2009 corresponds to x = Therefore, the points (0, 20.6) and (11, 10.3) are on the 20.6 − 10.3 10.3 = = 0.936 line Thus, m = 11 − 11 The y-intercept is 20.6, so b = 20.6 and the equation is y = −0.936 + 20.6 b x-intercept: = −0.936 x + 20.6 x = 22.0 y-intercept: y = −0.936(0) + 20.6 = 20.6 The intercepts are (22.0, 0) and (0, 20.6) c The y-intercept represents the percentage of twelfth graders in 2000 who had reported daily use of cigarettes The x-intercept represents the number of years after 2000 when 0% of twelfth graders will have reported daily use of cigarettes 42 Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Section F.3: Lines 2x − y = − 2x − y = 2x − y = All the lines have the same slope, The lines are parallel d The year 2025 corresponds to x = 25 y = −0.936(25) + 20.6 = −2.8 This prediction is not reasonable since it is negative 125 a Let x = number of boxes to be sold, and A = money, in dollars, spent on advertising We have the points ( x1 , A1 ) = (100, 000, 40, 000); ( x2 , A2 ) = (200, 000, 60, 000) 60, 000 − 40, 000 200, 000 − 100, 000 20, 000 = = 100, 000 A − 40, 000 = ( x − 100, 000 ) A − 40, 000 = x − 20, 000 A = x + 20, 000 slope = 128 Refer to Figure 65 + m12 length of OB = d ( O, B ) = + m2 length of AB = d ( A, B ) = m1 − m2 b If x = 300,000, then A = ( 300, 000 ) + 20, 000 = $80, 000 c length of OA = d ( O, A ) = Now consider the equation ( 1+ m ) + ( 1+ m ) ( 1+ m ) + ( 1+ m ) 2 Each additional box sold requires an additional $0.20 in advertising 2 = ( m1 − m2 ) If this equation is valid, then ΔAOB is a right triangle with right angle at vertex O 126 Find the slope of the line containing ( a, b ) and ( b, a ) : 2 = ( m1 − m2 ) + m12 + + m2 = m12 − 2m1m2 + m2 a −b = −1 b−a The slope of the line y = x is slope = + m12 + m2 = m12 − 2m1m2 + m2 But we are assuming that m1m2 = −1 , so we have Since −1⋅1 = −1 , the line containing the points ( a, b) and (b, a ) is perpendicular to the line y = x + m12 + m2 = m12 − ( −1) + m2 2 + m12 + m2 = m12 + + m2 0=0 Therefore, by the converse of the Pythagorean Theorem, ΔAOB is a right triangle with right angle at vertex O Thus Line is perpendicular to Line The midpoint of ( a, b) and (b, a ) is a+b b+a M = , Since the coordinates are the same, the midpoint lies on the line y = x 129 (b), (c), (e) and (g) The line has positive slope and positive y-intercept a+b b+a = Note: 2 130 (a), (c), and (g) The line has negative slope and positive y-intercept 127 2x − y = C Graph the lines: 43 Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Chapter F Foundations: A Prelude to Functions 131 (c) The equation x − y = −2 has slope and yintercept (0, 2) The equation x − y = has slope and y-intercept (0, −1) Thus, the lines are parallel with positive slopes One line has a positive y-intercept and the other with a negative y-intercept b1 b2 b b = − =− m m m m b b But − = − Line and Line have the m m same x-intercept, which contradicts the original assumption that the lines have unequal x-intercepts Therefore, Line and Line cannot have the same y-intercept b1 = b2 132 (d) The equation y − x = has slope and yintercept (0, 2) The equation x + y = −1 has 141 Yes Two distinct lines with the same y-intercept, but different slopes, can have the same x-intercept if the x-intercept is x = Assume Line has equation y = m1 x + b and Line has equation y = m2 x + b , 1 and y-intercept 0, − The lines 2 1 are perpendicular since − = −1 One line 2 has a positive y-intercept and the other with a negative y-intercept slope − Line has x-intercept − b and y-intercept b m2 Assume also that Line and Line have unequal slopes, that is m1 ≠ m2 If the lines have the same x-intercept, then b b − =− m1 m2 Line has x-intercept − 133 – 135 Answers will vary 136 No, the equation of a vertical line cannot be written in slope-intercept form because the slope is undefined 137 No, a line does not need to have both an xintercept and a y-intercept Vertical and horizontal lines have only one intercept (unless they are a coordinate axis) Every line must have at least one intercept b b =− m1 m2 − m2 b = − m1b − m2 b + m1b = − But − m2b + m1b = b ( m1 − m2 ) = b=0 or m1 − m2 = m1 = m2 Since we are assuming that m1 ≠ m2 , the only way that the two lines can have the same x-intercept is if b = 138 Two lines with equal slopes and equal y-intercepts are coinciding lines (i.e the same) 139 Two lines that have the same x-intercept and yintercept (assuming the x-intercept is not 0) are the same line since a line is uniquely defined by two distinct points 140 No Two lines with the same slope and different xintercepts are distinct parallel lines and have no points in common Assume Line has equation y = mx + b1 and Line has equation y = mx + b2 , 142 Answers will vary 143 m = y2 − y1 − − −6 = = =− x2 − x1 − ( −3) It appears that the student incorrectly found the slope by switching the direction of one of the subtractions b Line has x-intercept − and y-intercept b1 m b2 and y-intercept b2 Line has x-intercept − m Assume also that Line and Line have unequal x-intercepts If the lines have the same y-intercept, then b1 = b2 Section F.4 add; 25 44 Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ b and y-intercept b m1 Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Section F.4: Circles ( x − )2 = General form: x + y − = x−2 = ± x − = ±3 x = 2±3 x = or x = −1 The solution set is {−1, 5} False For example, x + y + x + y + = is not a circle It has no real solutions radius 12 ( x − h) + ( y − k ) = r 2 ( x − h) + ( y − k ) = r ( x − 0) + ( y − 0) = 32 x2 + y = unit General form: x + y − = Center = (2, 1) Radius = distance from (0,1) to (2,1) = (2 − 0) + (1 − 1) = = Equation: ( x − 2) + ( y − 1) = Center = (1, 2) Radius = distance from (1,0) to (1,2) = (1 − 1) + (2 − 0) = = Equation: ( x − 1) + ( y − 2) = 13 ( x − h) + ( y − k ) = r Center = midpoint of (1, 2) and (4, 2) 1+ + = , = 5, 2 2 Radius = distance from , to (4,2) ) ( ) ( ( x − 0) + ( y − 2) = 22 x + ( y − 2) = ( ) General form: x + y − y + = x2 + y2 − y = 5 = − + (2 − 2)2 = = 5 Equation: x − + ( y − 2) = 2 10 Center = midpoint of (0, 1) and (2, 3) + 1+ , = = (1, ) Radius = distance from (1, ) to (2,3) = Equation: ( − 1)2 + (3 − 2)2 ( x − 1) = 14 ( x − h) + ( y − k ) = r + ( y − 2) = ( x − 1) + ( y − 0)2 = 32 ( x − 1) + y = 11 ( x − h) + ( y − k ) = r ( x − 0) + ( y − 0) = 22 x2 + y = 45 Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Chapter F Foundations: A Prelude to Functions General form: x + x + + y − y + = 16 General form: x − x + + y = x + y + x − y − 11 = x2 + y − x − = 15 ( x − h) + ( y − k ) = r 18 ( x − h) + ( y − k ) = r ( x − 4) + ( y − (−3)) = 52 ( x − ( −5 )) + ( y − (−2)) = ( x − 4) + ( y + 3) = 25 General form: x − x + 16 + y + y + = 25 ( x + 5) + ( y + 2) = 49 General form: x + 10 x + 25 + y + y + = 49 x2 + y − 8x + y = x + y + 10 x + y − 20 = 16 ( x − h) + ( y − k ) = r ( x − 2) + ( y − (−3)) = 42 19 ( x − h) + ( y − k ) = r ( x − 2) + ( y + 3) = 16 2 1 1 x − + ( y − 0) = 2 2 General form: x − x + + y + y + = 16 x2 + y2 − x + y − = 2 1 x− + y = 2 1 + y2 = 4 2 x +y −x=0 General form: x − x + 17 ( x − h) + ( y − k ) = r ( x − ( −2 )) + ( y − 1) = 42 ( x + 2) + ( y − 1) = 16 46 Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Section F.4: Circles c x-intercepts: x + ( ) = x2 = x = ± = ±2 y-intercepts: ( ) + y = y2 = y = ± = ±2 The intercepts are ( −2, ) , ( 2, ) , ( 0, −2 ) , and ( 0, ) 20 ( x − h) + ( y − k ) = r = ( x − )2 + y − − 22 x + ( y − 1) = x + ( y − 1) = 12 a 1 x + y+ = Center:(0, 1); Radius = b 1 = 4 x2 + y2 + y = General form: x + y + y + c x-intercepts: x + (0 − 1) = x2 + = x2 = x=± =0 y-intercepts: ( ) + ( y − 1) = ( y − 1) = 21 x + y = y −1 = ± x + y = 22 a y − = ±1 Center: (0, 0); Radius = y = 1±1 y = or y = b The intercepts are ( 0, ) and ( 0, ) 23 ( x − 3) + y = ( x − 3)2 + y = a Center: (3, 0); Radius = 47 Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Chapter F Foundations: A Prelude to Functions b c 2 x-intercepts: ( x + 1) + ( − 1) = ( x + 1)2 + ( −1)2 = ( x + 1)2 + = ( x + 1)2 = x +1 = ± x + = ±1 c 2 ( x − 3) x = −1 ± x = or x = −2 x-intercepts: ( x − 3) + ( ) = y-intercepts: ( + 1) + ( y − 1) = 2 =4 (1)2 + ( y − 1)2 = 2 + ( y − 1) = ( y − 1)2 = x −3 = ± x − = ±2 x = 3± x = or x = y −1 = ± y-intercepts: ( − 3) + y = ( −3) y − = ±1 y = 1±1 y = or y = +y =4 + y2 = The intercepts are ( −2, ) , ( 0, ) , and ( 0, ) y = −5 No real solution The intercepts are (1, ) and ( 5, ) 25 x + y − x − y − = x2 − x + y − y = 24 ( x + 1) + ( y − 1) = ( x − x + 1) + ( y − y + 4) = + + ( x − 1) + ( y − 2) = 32 ( x + 1)2 + ( y − 1)2 = a Center: (–1,1); Radius = a Center: (1, 2); Radius = b b c x-intercepts: ( x − 1) + (0 − 2) = 32 ( x − 1) + (−2) = 32 ( x − 1)2 + = ( x − 1)2 = x −1 = ± x = 1± 48 Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Section F.4: Circles y-intercepts: (0 − 1) + ( y − 2) = 32 27 (−1) + ( y − 2) = 32 + ( y − 2) = ( y − )2 = x2 + y2 + x − y − = x2 + x + y − y = ( x + x + 4) + ( y − y + 4) = + + ( x + 2) + ( y − 2) = 32 Center: (–2, 2); Radius = y b a y−2 = ± y − = ±2 y = 2±2 ( )( ) , and ( 0, + 2 ) (−2, 2) ) The intercepts are − 5, , + 5, , ( 0, − x −5 x + y + x + y − 20 = 26 −5 x + x + y + y = 20 c ( x + x + 4) + ( y + y + 1) = 20 + + x-intercepts: ( x + 2) + (0 − 2) = 32 ( x + 2) + = ( x + 2) + ( y + 1) = 52 ( x + 2) = a Center: (–2,–1); Radius = b x+2= ± x = −2 ± y-intercepts: (0 + 2) + ( y − 2) = 32 + ( y − 2)2 = ( y − 2)2 = y−2 = ± c ( −2 + ( x + 2) + = 25 ( x + 2) = 24 28 x + = ± 24 x + = ±2 x = −2 ± y-intercepts: (0 + 2) + ( y + 1) = 52 + ( y + 1) = 25 b y + = ± 21 y = −1 ± 21 ) The intercepts are −2 − 6, , 21 , and 49 Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ ) Center: (3, –1); Radius = ( y + 1) = 21 ) )( ( ) 5, , 0, − , and 0, + x2 + y2 − x + y + = x − x + y + y = −9 ( x − x + 9) + ( y + y + 1) = −9 + + ( x − 3) + ( y + 1) = 12 a ( −2 + 6, 0) , ( 0, − − ( 0, − + 21 ) ) The intercepts are −2 − 5, , x-intercepts: ( x + 2) + (0 + 1) = 52 ( ( y = 2± Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Chapter F Foundations: A Prelude to Functions c x-intercepts: ( x − 3) + (0 + 1) = 12 1 1 y-intercepts: − + ( y + 1) = 2 2 1 + ( y + 1) = 4 ( y + 1)2 = y +1 = y = −1 ( x − 3) + = ( x − 3)2 = x−3 = x=3 y-intercepts: (0 − 3) + ( y + 1) = 12 + ( y + 1) = The only intercept is ( 0, −1) ( y + 1)2 = −8 No real solution The intercept only intercept is ( 3, ) 30 x2 + y − x + y + = 29 x − x + y + y = −1 1 2 x − x + + ( y + y + 1) = −1 + + 4 2 1 1 x − + ( y + 1) = 2 2 a =0 x2 + x + y + y = 1 1 1 x + x+ + y + y+ = + + 4 4 4 x2 + y2 + x + y − 2 1 1 x + + y + =1 2 2 a 1 Center: , −1 ; Radius = 2 1 Center: − , − ; Radius = 2 b b 2 c 1 1 x-intercepts: x − + (0 + 1) = 2 2 1 x − +1 = c 1 x+ = 2 x+ =± 2 −1 ± x= 2 1 x− = − 2 No real solutions 50 Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ 1 1 x-intercepts: x + + + = 12 2 2 1 x + + =1 2 Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Section F.4: Circles y-intercepts: (0 − 3) + ( y + 2) = 52 1 1 y-intercepts: + + y + = 12 2 2 1 + y + =1 2 + ( y + ) = 25 ( y + )2 = 16 y + = ±4 y = −2 ± y = or y = −6 1 y+ = 2 y+ =± 2 −1 ± y= −1 − −1 + , , , , The intercepts are −1 − −1 + 0, , and 0, 2 ( 0, −6 ) , 32 a x + y − x + y = 12 x − x + y + y = 12 ( x − x + 9) + ( y + y + 4) = 12 + + ( x − 3)2 + ( y + 2) = 52 a ) (3 + and ( 0, ) 2x2 + y2 + 8x + = x + x + y = −7 x2 + x + y2 = − 2 ( x + x + 4) + y = − + 2 ( x + 2) + y = 2 2 2 ( x + 2) + y = x + y − 12 x + y − 24 = 31 ( The intercepts are − 21, , 2 Center: (–2, 0); Radius = Center: (3,–2); Radius = b b c 2 x-intercepts: ( x − 3) + (0 + 2) = c 2 = x-intercepts: ( x + 2) + ( ) = ( x + )2 ( x − 3)2 + = 25 ( x − 3)2 = 21 2 x+2= ± x+2= ± x − = ± 21 x = ± 21 x = −2 ± 51 Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ 2 ) 21, , Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Chapter F Foundations: A Prelude to Functions 34 x + y − 12 y = 2 4+ y = y-intercepts: (0 + 2) + y = x2 + y − y = x2 + y − y + = + x2 + ( y − 2) = No real solutions The intercepts are −2 − , and , −2 + y2 = − 33 a Center: ( 0, ) ; Radius: r=2 b x2 + 8x + y2 = x2 + x + y2 = x2 + x + + y = + a ( x + )2 + y = 22 Center: ( −2, ) ; c Radius: r = 2 x-intercepts: x + ( − ) = x2 + = b x2 = x=0 y-intercepts: + ( y − ) = ( y − )2 = y−2 = ± y − = ±2 c y = 2±2 y = or y = x-intercepts: ( x + ) + ( ) = 22 The intercepts are ( 0, ) and ( 0, ) ( x + 2)2 = ( x + 2) =± 35 Center at (0, 0); containing point (–2, 3) x + = ±2 r= x = −2 ± x = or x = −4 y-intercepts: ( + ) + y = 2 ( −2 − )2 + ( − )2 = + = 13 Equation: ( x − 0)2 + ( y − 0) = ( 13 ) x + y = 13 4+ y = 36 Center at (1, 0); containing point (–3, 2) y2 = y=0 r= The intercepts are ( −4, ) and ( 0, ) ( −3 − 1)2 + ( − )2 = 16 + = 20 = Equation: ( x − 1) + ( y − 0) = ( ( x − 1) + y = 20 52 Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ 20 ) Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Section F.4: Circles 37 Center at (2, 3); tangent to the x-axis r =3 Equation: ( x − 2) + ( y − 3) = 32 47 Let the upper-right corner of the square be the point ( x, y ) The circle and the square are both centered about the origin Because of symmetry, we have that x = y at the upper-right corner of the square Therefore, we get x2 + y = ( x − 2) + ( y − 3) = 38 Center at (–3, 1); tangent to the y-axis r =3 Equation: ( x + 3) + ( y − 1) = 32 x2 + x2 = 2x2 = 9 x2 = ( x + 3) + ( y − 1) = 39 Endpoints of a diameter are (1, 4) and (–3, 2) The center is at the midpoint of that diameter: + (−3) + Center: , = ( −1,3) = 2 The length of one side of the square is 2x Thus, the area is x= Radius: r = (1 − (−1)) + (4 − 3) = + = Equation: ( x − (−1)) + ( y − 3) = ( 5) 2 A = s = ⋅ = 2 2 ( x + 1) + ( y − 3) = 2 x = 36 41 Center at (–1, 3); tangent to the line y = This means that the circle contains the point (–1, 2), so the radius is r = Equation: ( x + 1) + ( y − 3) = (1) x = 18 x=3 The length of one side of the square is 2x Thus, ( ( x + 1) + ( y − 3) = the area of the square is ⋅ 46 (a) ; Center: ( −3,3) ; Radius = Therefore, the area of the shaded region is A = 36π − 72 square units 49 The diameter of the Ferris wheel was 250 feet, so the radius was 125 feet The maximum height was 264 feet, so the center was at a height of 264 − 125 = 139 feet above the ground Since the center of the wheel is on the y-axis, it is the point (0, 90) Thus, an equation for the wheel is: ( x − )2 + ( y − 139 )2 = 1252 x + ( y − 139 ) = 15, 625 53 Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ = 72 square 43 (c); Center: (1, −2 ) ; Radius = ( −1, ) ; Radius = 2 π r = π ( ) = 36π square units ( x − 4) + ( y + 2) = 45 (b) ; Center: ) units From the equation of the circle, we have r = The area of the circle is 42 Center at (4, –2); tangent to the line x = This means that the circle contains the point (1, –2), so the radius is r = Equation: ( x − 4) + ( y + 2) = (3) ( −3,3) ; Radius = = 18 square units x + x = 36 ( x − 2) + ( y − 2) = 44 (d) ; Center: The circle and the square are both centered about the origin Because of symmetry, we have that x = y at the upper-right corner of the square Therefore, we get x + y = 36 Radius: r = (4 − 2) + (3 − 2) = + = ( 5) ) 48 The area of the shaded region is the area of the circle, less the area of the square Let the upperright corner of the square be the point ( x, y ) 40 Endpoints of a diameter are (4, 3) and (0, 1) The center is at the midpoint of that diameter: + +1 Center: , = ( 2, ) Equation: ( x − 2) + ( y − 2) = ( Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Chapter F Foundations: A Prelude to Functions 50 The diameter of the wheel is 520 meters, so the radius is 260 meters The maximum height is 550 meters, so the center of the wheel is at a height of 550 − 260 = 290 meters above the ground Since the center of the wheel is on the y-axis, it is the point (0, 290) Thus, an equation for the wheel is: (1 + m ) x + 2bmx + b − r = − 2bm −bm −bmr −mr = = = b 2(1 + m ) b b2 2 r −mr y = m +b b x= ( x − 0)2 + ( y − 290)2 = 2602 x + ( y − 290) = 67, 600 = 51 x + y + x + y − 4091 = c x + x + y + y − 4091 = x + x + + y + y + = 4091 + ( x + 1)2 + ( y + )2 = 4096 The circle representing Earth has center ( −1, −2 ) and radius = 4096 = 64 So the radius of the satellite’s orbit is 64 + 0.6 = 64.6 units The equation of the orbit is 53 x + y = Center: (0, 0) x + y + x + y − 4168.16 = 2 2 2 x + m x + 2bmx + b = r ) is 2 −0 2 = =2 1− (1 + m ) x + 2bmx + b − r = There is one solution if and only if the discriminant is zero (2bm) − 4(1 + m )(b − r ) = Slope of the tangent line is −1 =− 2 Equation of the tangent line is: y−2 = − ( x − 1) 2 y−2 = − x+ 4 4y −8 = − x + 4b m − 4b + 4r − 4b m + 4m r = − 4b + 4r + 4m r = − b2 + r + m2 r = r (1 + m ) = b b ( Slope from center to 1, 2 x + (mx + b) = r 52 a The slope of the tangent line is m The slope of the line joining the point of tangency and the center is: r2 − 0 b = r ⋅ b =−1 m −mr b − mr − b Therefore, the tangent line is perpendicular to the line containing the center of the circle and the point of tangency ( x + 1)2 + ( y + )2 = ( 64.6 )2 −m2 r −m2 r + b2 r +b = = b b b x + 4y = Using the quadratic formula, the result from part (a), and knowing that the discriminant is zero, we get: 54 x + y − x + y + = ( x − x + 4) + ( y + y + 9) = −4 + + ( x − 2) + ( y + 3) = Center: (2, –3) ( ) Slope from center to 3, 2 − is 2 − − (−3) 2 = =2 3− 54 Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Section F.4: Circles Slope of the tangent line is: −1 2 =− ( 2, −3) ( x + x + 9) + ( y + y + 4) = − + + Equation of the tangent line: ( x − 3) y − 2 −3 = − y−2 +3 = − x+ 4 y − + 12 = − x + ( ( x + 3) + ( y + 2)2 = ) Center: ( −3, −2 ) Find the slope of the line containing the centers: − − (−3) m= =− −3 − Find the equation of the line containing the center: y + = − ( x − 2) 5 y + 15 = − x + x + y = −13 x + y = 11 − 12 55 Let (h, k ) be the center of the circle x − 2y + = 2y = x + y = x+2 The slope of the tangent line is The slope from (h, k ) to (0, 2) is –2 57 Consider the following diagram: (2,2) 2−k = −2 0−h − k = 2h The other tangent line is y = x − , and it has slope The slope from (h, k ) to (3, –1) is − −1 − k =− 3−h 2 + 2k = − h Therefore, the path of the center of the circle has the equation y = 58 6π = 2π r 59 (b), (c), (e) and (g) We need h, k > and ( 0, ) on the graph h = − 2k Solve the two equations in h and k : − k = 2(1 − 2k ) − k = − 4k 3k = k =0 h = − 2(0) = The center of the circle is (1, 0) 60 (b), (e) and (g) We need h < , k = , and h > r 61 Answers will vary 62 The student has the correct radius, but the signs of the coordinates of the center are incorrect The student needs to write the equation in the standard form ( x − h ) + ( y − k ) = r Find the centers of the two circles: x2 + y2 − 4x + y + = ( x + 3) + ( y − ) 2 ( x − 2) + ( y + 3) = = 16 = 42 Thus, ( h, k ) = ( −3, ) and r = Center: 55 Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ ( x − ( −3)) + ( y − 2) ( x − x + 4) + ( y + y + 9) = − + + C = 2π r 6π 2π r = 2π 2π 3=r The radius is units long 2k = − h 56 x2 + y + x + y + = Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Chapter F Foundations: A Prelude to Functions Chapter F Project Internet Based Project 56 Copyright © 2015 Pearson Education, Inc Full file at https://TestbankDirect.eu/ .. .Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Chapter F Foundations: A Prelude to Functions 29 A =... Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Section F.1: The Distance and Midpoint Formulas 3a a a. .. Solution Manual for Precalculus Concepts Through Functions A Unit Circle Approach to Trigonome Full file at https://TestbankDirect.eu/ Chapter F Foundations: A Prelude to Functions 39 The coordinates