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2 , y = –10 7-18. a: 0 b: 3 c: 4 d: 64 7-19. y : 3; 4; 5; undefined; 7; 8 a: See graph at right. It is linear. The data does not all connect because f (1) is undefined. b: y = x + 5, f (0.9) = 5.9, f (1.1) = 6.1, no asymptote. c: The complete graph is the line y = x +5 with a hole at (1, 6). 7-20. a: An exponential function b: y = 60000 +12000(0.93)t 7-21. If he drives down the center of the road, the height of the tunnel at the edge of the house
is only approximately 23.56 feet. The house will not fit. 7-22. a: x ≈ 33.752 b: x ≈ 9.663 7-23. x = 18, y = 13, z = 9
3 , y = 1 7-41. y = 3(x +1)2 ! 2 ; See graph at right. 7-42. x ≤ –5 7-43. No real solution. 7-44. C +W + P = 40 , W = C ! 5 , C = 2P ; 18 from California, 13 from Washington, and
7-54. P: (cos 50º , sin 50º ) or (~0.643, ~0.766); Q: (cos110º , sin110º ) or (~ –0.342, ~0.940) 7-55. a: 300 ° b: 12 ,!
32 c: 1
2 ,!!32( )
7-56. a: 30 ° b: 60 ° c: 67° d: 23 ° 7-57. x = 11
5 7-58. a: downward parabola, vertex (2, 3), see graph above right. b: cubic, point of inflection (1, 3), see graph below right. 7-59. Solving graphically: x ≈ –3.2 7-60. a: y = 25d + 0.50m and y = 0.03(2)m!1
b: R vs. T: $55 vs. $15.36, $60 vs. $15,728.64, $100 vs. ~ $1.901!1028 7-61. All of these problems could be solved using the same system of equations.
Lesson 7.1.4 (Day 2) 7-62. 58º, 122º, 238º, or 302º 7-63. a: An angle in the 4th quadrant. b: 270 ° or –90 ° c: An angle in the 3rd quadrant. d: Approximately 160 ° e: No, an angle with sine equal to 0.9 has cosine equal to ±0.4359, so the point (0.8, 0.9)
is not on the unit circle. 7-64. a: (0.3420, 0.9397) b: (cos70° , sin70 ° )
c: (cos 70°)2 + (sin 70°)2 = 0.1170 + 0.8830 = 1 7-65. Graph 2 is sine, while graph 1 is cosine. Answers will vary. 7-66. a: All yes. b: Answers will vary. c: x = (!180° + 360°n) for all integers n 7-67. y-intercept: (0, –17), x-intercepts: (!2 + 21, 0) and (!2 ! 21, 0) 7-68. a: x = –4 b: x = 5± 57
4 c: no solution
d: If a = 3x+2 , then a + 5 ≠ a. Or, solving yields x = –2, but when substituted, –2 gives a
zero denominator. 7-69. 7.07 ' 7-70. Tess is correct: A sequence has no more than one output for each input. A sequence is a
function with domain limited to positive integers.
d: Yes, there are infinitely many, at intervals of 2π. 7-117. a: ! b: y = sin(x +! ) 7-118. a: This may go up and down, but the cycles are probably of differing length. b: This may or may not be periodic. c: This is probably approximately periodic. 7-119. y =100 sin x + !
2( )! 50 or y = 100 cos x ! 50 7-120. Only one needs to be a parent, since y = sin(x + 90°) is the same as y = cos x. 7-121. a: y = 3!6x b: y = !2(0.5)x 7-122. a: x = ± 3
5 = ± 155 b: x = 4, –1 c: x = 4
7-123. a: – 3 b: 3
3 7-124. a = ! 3
3125 = !0.00096 , possible equation: y = ! 33125 (x !125)
3( )!1 7-130. 360 ° is the period of y = cos! , so shifting it 360° left lines up the cycles perfectly. 7-131. Graphing form: y = 2(x !1)2 + 3 ; vertex (1, 3); See graph at right. 7-132. a: x = (0, 0), 5±3 3
2 ,!0( ) and y = (0, 0)
7-133. 17.67 years
7-134. a: y = !2 x + 14( )2 + 1058 , x = all real numbers, y = !" < y < 25
8 ; Yes it is a function.
b: y = !3(x +1)2 +15 , domain: all real numbers, range: !" < y <15 ; Yes it is a function. 7-135. 64.16° , unsafe 7-136. a: 5,000,000 bytes b: ≈ 12.3 minutes c: According to the equation, technically never, but for all practical purposes, after
23 minutes. 7-137. See graph at right. a: The vertex of the graph is at (6, –4) with two rays emanating at slopes of ±1 . b: See graph at right. Flip all parts of the graph that are below the x-axis above the x-axis.
Lesson 7.2.3 7-144. a: Amplitude 3, period 4π b: See graph at right. c: The differences are the period and amplitude, and therefore some of the x-intercepts. They have the same basic shape. 7-145. 1, 2!2! = 1 or 2! (1) = 2! 7-146. Colleen’s calculator was in radian mode, while Jolleen’s calculator was in degree mode.
Colleen’s calculation is wrong. 7-147. y = sin 2(x !1) is correct. To shift the graph one unit to the right, subtract 1 from x
before multiplying by anything. 7-148. They are both wrong. The equation needs to be set equal to zero before the Zero Product
Property can be applied. 2x2 + 5x ! 3= 4 is equivalent to (2x + 7)(x !1) = 0 . x = 1 or x = ! 7
2 7-149. a: 3 b: 1.5 c: 2 d: 12
7-150. a: y = 20 12( )x + 5 b: w = 5.078
7-151. a: Answers vary, if g(x) is linear, tangent lines only. b: Any line y = b such that b < –8.