Unit 5
Answer Key
1
Unit 5 Answer Key
LS-1.
a) Normal: C = 6h, High: C = 5h + 6
b) Use High Use for more than 6 hours, Normal for less than 6 hours.
c) Same for 6 hours. It is the solution to the system.
LS-2.
a) ª 3.5 hours, Cost ª $60 b) Comfort Car = $71
c) Comfort Car = 8 1 hours
3
LS-3.
a) from -7 to 7, inclusive
LS-4.
a) -k £ x £ k
LS-5.
a) 2x - 17 > 9 or 2x - 17 < -9
b) x > 13 or x < 4
LS-6.
a) horizontal line through (0, 3)
b) vertical line through (-2, 0)
c) (-2, 3)
b) above 4 or below -4
b) x < -k or x > k
LS-7.
b) y = 0 c) x = 0
LS-8.
b) There can be no intersection because the lines are parallel. When we assumed
there was an intersection, a false statement was the result.
LS-9.
-2 < 3, 6 > 4, -5 < -1
a) -10 < 15, 30 > 20, -25 < -5, yes
b) no
c) no d) Multiplying inequalities by positive numbers maintains
the inequality, but multiplying by negative numbers changes the relationship.
LS-10.
Reverse it.
LS-11.
a) Any positive number would be a good choice.
b) No; reverse the inequality sign to get y > -3.
c) i: y < -1 ii: y > -1 iii: y > 3
2
d) Multiplying or dividing an inequality by a negative number reverses the order
of the inequality.
LS.13.
c) 1.5 £ x £ 3.5
LS-14.
multiplying or dividing by a negative
© 2000 CPM Educational Program
v. 5.0
Unit 5
Answer Key
2
LS-16.
In order to show that we are not including the line y = 3x - 1.
Point
(Above,Below,or On) e)
A(0, 0)
Above
True
B(3, 4)
Below
False
C(-1, -4)
On
False
D(-2, 3)
Above
True
2
Below
False
E(-3 , -4)
b) A and D
c) Any points above the line y = 3x - 1.
LS-17.
All points where y = 2x - 2, that is, the boundary line itself.
LS-18.
a) y > 3x - 3
LS-20.
a) C = 800 + 60m
b) C = 1200 + 40m
LS-21.
a) x = 9 , y = -3
2
b) x = -6, y = 4
LS-22.
d) D: all reals, R: y ≥ 0; D: x ≥ 0, R: all reals
LS-23.
Solution:
a) B only
b) It shows where both inequalities are true.
b) y < 3
c) y ≥ 3x - 3
2
d) y ≥ x2 - 9
d) 5 years
(-2, 6)
LS-24.
a) 51/2
c) 17x/8
LS-26.
a) rectangle; perpendiculars or slopes
b) (1,4), (-3,-3), (-5, 1), (3,0)
LS-27.
30 square units
LS-28.
a) y = a x - 6 , a > 1
c) y = a x - h + k
LS-29.
y £ 3x + 3, y ≥ 0.5x - 2, and y £ -0.75x + 3
© 2000 CPM Educational Program
b) y = a x - 9, -1 < a < 1
v. 5.0
Unit 5
Answer Key
LS-30.
x = -2, y = 5, z = - 1
2
LS-31.
a) x(1.05)10
d) 0.2 1.08
1.05
( )
10
b) 0.2x
p
=
100
1
2 2
x y
LS-33.
c)
LS-34.
a) -4 < x < 1
e) -1 < x < 4
3
c) 0.2x(1.08)10
10
d) b
a
b) x ≥ 0 or x £ -5 c) -1 < x < 4 d) x ≥ 4 or x £ -1
f) x ≥ 4 or x £ -1
LS-35. a) V shape, shading below
b) V shape, open to right, shading left
LS-36.
b) 6 cars and 2 trucks for a profit of $8
c) 3 cars and 4 trucks for a profit of $11
LS-37.
a) i: 4x + 6y £ 36 ii: 2x + y £ 14 iii: x + 3y £ 15
c) (0, 0), (0, 5), (3, 4), (6, 2), (7, 0)
d) Vertices are located in the first quadrant; profit = x + 2y
e) x + 2y = 8; No, because there are points above the line; The maximum is $11.
f) same slope; y-intercept; slide the graph to find the maximum; Slide it up,
keeping the slope constant, to the highest point at which it still intersects the
feasible region. The point where it intersects the region yields maximum profit.
LS-38.
6 cars and 2 trucks for a profit of $22.
LS-41.
d) It is neither because both the differences and ratios
a) 63
c) n3 - 1
between the terms vary.
LS-42.
x = -7, y = 11
LS-43.
a) 1722
c) 1500(1.047)(n+3) or 1722(1.047)n
LS-45.
b) x(1 + a)
d) x - b
a
LS-46.
a) 0 £ x £ 15; 0 £ y £ 12; 5x + 3y £ 81: 2x + 6y £ 78; 20x + 30y £ 12(37.5)
b) coordinates of the vertices
c) Vertices of the polygon are (0, 0), (0, 12), (3, 12), (6, 11),
(12, 7), (15, 2) and (15, 0). Profit = 500x + 1000y. Maximum Profit of $14,000
when Sand Crabs = 6 and Surf Mobiles = 11.
LS-47.
(-2, 3, -5)
LS-48.
a = 18.5, b = 5.5
© 2000 CPM Educational Program
v. 5.0
Unit 5
Answer Key
LS-49.
a) input x, output x
LS-50.
a) x2 + y2 = 36
c) (x - 4)2 + (y + 5)2= 36
LS-51.
4
b) (x - 2)2 + (y + 3)2 = 36
50 p 4 ÷ 400 ª 9.8%
LS-52. y = 0; x = 0
LS-53.
b) y = -2x + 15
LS-54.
b) the number of parts needed to build 2 cars and 2 trucks
w s g
c) 5A
d) The new matrix is c È 20 10 5 ˘
t ÍÎ 30 5 15˙˚
LS-55.
a)
c t
B = BE [ 20 25]
b) 230 wheels.
w s g
C = BE [ 230 65 95]
e) (# of cars)x(# of seats per car) + (# of trucks)x(# of seats per truck) = total #
of seats needed. (# of cars)x(# of gas tanks per car) + (# of trucks) x (# of gas
tanks per truck) = total number of gas tanks needed.
c) 65 seats and 95 gas tanks
LS-56.
LS-57.
[ 4x + 6y
2x + y
d)
x + 3y ]
c t
w s g
a) B = BE È 20 25 ˘
b) C = BE È 230 65 95 ˘
JC ÍÎ 15 30 ˙˚
JC ÍÎ 240 60 105˙˚
È 20 25 ˘ È 4 2 1 ˘ È 230 65 95 ˘
c) Í
˙ Í 6 1 3˙ = Í 240 60 105˙ ;
15
30
˚ Î
Î
˚ Î
˚
È 20(4) + 25(6) = 230 20(2) + 25(1) = 65 20(1) + 25(3) = 95 ˘
Í 15(4) + 30(6) = 240 15(2) + 30(1) = 60 15(1) + 30(3) = 105˙
Î
˚
LS-58.
Otto will need 105 gas tanks to fill the JC Nickles order.
LS-59.
a) M: 2 ¥ 3, N: 1 ¥ 3, P: 2 ¥ 2, Q: 2 ¥ 1, R: 1 ¥ 1
b) S has no entry in the 2nd row, 2nd column, T is not arranged in rows and
columns, U has no square brackets, and the entries in V are not numbers or
algebraic expressions.
© 2000 CPM Educational Program
v. 5.0
Unit 5
LS-60.
LS-61.
Answer Key
È 0
b) 3 ¥ 4 c) Í 0
Í
ÍÎ 0
e
b) 3C= af È 18
df ÍÎ 9
4 -21˘
0 0 0˘
È -16 -3
d) -F = Í -19 -31 -12 -17 ˙
0 0 0˙
˙
Í
˙
-8 -11˙˚
0 0 0 ˙˚
ÍÎ -25 6
s b
3 15 ˘ = ingredients for 3 of each cake.
4. 5 12 ˙˚
LS-63.
a) square
LS-64.
a) y = -3x + 8 b) y = -x - 1
2
LS-65.
N(315) = 72million, N = 5; there were only 5 bacteria at first.
LS-66.
LS-67.
LS-68.
5
b) (0, -3), (4, 1), (-4, 1), (0, 5)
c t
È 230 65 95 ˘
È 20 25 ˘
BE È 20 25 ˘
4
2
1
È
˘
= Í 240 60 105˙
B=
a) C = Í 15 30 ˙ Í
JC Í 15 30 ˙
˙
˙ Î 6 1 3˙˚ Í
Í
Í
˙
ÍÎ 192 48 84 ˙˚
ÍÎ 12 24 ˙˚
M ÍÎ 12 24 ˙˚
b) 60 seats are needed to complete the JC Nickles order.
c) C = stores x parts matrix. Students should notice that column label of the first
matrix is the same as the row label of the second matrix and is not present in the
product matrix.
$
c È 2.75˘
t ÍÎ 3.10 ˙˚
a) The 1x2 matrix represents: cost x vehicles.
b) vehicles x cost
c) You would use the vehicles x cost matrix for the D
matrix. The matrix product that is possible is BD = (stores x vehicles)(vehicles x
cost) = (stores x cost)
$
BE È 132.50 ˘
d) E =
JC Í 134.25 ˙
˙
Í
M ÍÎ 107.40 ˙˚
e) It will cost Otto $107.40 to complete the order for Marcey's.
c t
and
$ [ 2.75 3.10 ]
a)
BE JC M
% [.86 .74. .93]
© 2000 CPM Educational Program
v. 5.0
Unit 5
Answer Key
6
$
BE JC M BE È 132.50 ˘
b)
˙ = [ 313.177 ]
Í
% [.86 .74 .93] JC Í 134.25 ˙
M ÍÎ 107.40 ˙˚
It's a 1 x 1 matrix that represents his profit for the week; 313.18
È 19 22 ˘
È 23 34 ˘
b)
BA
=
˙
Í 31 46 ˙
Î 43 50 ˚
Î
˚
c) No--we just showed it wasn't.
LS-69.
a) C = Í
LS-70.
each column; second
Bouquet Styles
LS-71.
a) 6
#1 # 2 # 3
c)
A
Workers B
C
È 30 20 35 ˘
Í 20 40 25 ˙
˙
Í
ÍÎ 25 30 30 ˚˙
L R D
A
Bouquet Styles B
C
È 5 4 3˘
Í 4 3 3 ˙ b) bouquet x flowers
˙
Í
ÍÎ 4 6 6 ˙˚
LS-72.
a) Flowers
LS-73.
WB makes sense. It represents the number of flowers each worker will use in a
day. Notice that (workers x bouquets) x (bouquets x flowers) = (workers x
flowers)
Flowers
L
R D
Workers
A
B
C
È 74 78 72 ˘
Í 72 70 66 ˙
˙
Í
ÍÎ 73 74 69 ˙˚
LS-74.
The second one is shifted to the right 4 units and up 2 units and shading is
between and above the intersecting branches.
LS-75.
a) 1
b) -2 c) The product of the slopes is -1, or one is the negative
2
reciprocal of the other.
© 2000 CPM Educational Program
v. 5.0
Unit 5
Answer Key
7
LS-76.
a) y = 1 x - 2
2
LS-77.
b) You get a different matrix that does not have any relationship to the problem.
LS-78.
a) [15 18 18]
b) It represents the total number of each style of bouquet made per day
LS-79.
a) n = p b) m x q
LS-80.
b) 7 and x
LS-83.
b) y = 2x + 2
#1 È 5.10 ˘
b) Bouquet Styles # 2 Í 4.35 ˙
˙
Í
# 3 ÍÎ 7. 50 ˙˚
The cost matrix for each bouquet style (for all flowers).
LS-85.
2¥ 1
LS-86.
a) Í
LS-87.
a) 11
LS-89.
a) y = -2x + 8 b) 63.43° or 116.57°
LS-91.
a) parabola with vertex (2, -5)
c) hyperbola with locator (3, 4)
LS-92.
a) 4x + 4y + 4z = 136, 3x + 2y + 4z = 95, x + 2y + 3z = 56
b) The first equation simplifies to x + y + z = 34.
f) x = 17, y = 12, z = 5
LS-93.
x = -1, y = 3, z = 5
LS-94.
y £ - 3 x +3; y ≥ - 3 x - 3; x £ 3; x ≥ -3
4
4
LS-95.
a) y= -2(x+4)2 + 2
LS-96.
It is not the parent. The second equation does not have a vertical asymptote and
it has a maximum value while y = 1 does not.
x
È 4a + 9b + 2c ˘
˙
Î 6a - 5c ˚
© 2000 CPM Educational Program
È 9 18 1 ˘
b) Í
˙
Î 12 4 - 9 ˚
b) 5 ¥ d
b) y =
b) cubic, locator (-3,0), shade below
1 c) y = -x3 + 3
x- 2
v. 5.0
Unit 5
Answer Key
8
LS-97.
a) x = b
3
b) x = b
5a
LS-98.
a) No. Input equals output only if x ≥ 0.
d) because x 2 = x
LS-99.
Heather is correct and Elissa is wrong because as previouly shown, a 4%
decrease does not cancel out a 4% increase .
b) w = 0, 25
LS-100. a) w = 0, -4
LS-101. a) 4 = -3a + b, -1 = -2a + b; a = -5, b = -11
LS-102. a) 4a + 2b + c = 3, a - b + c = 6, c = 3
LS-103. a) y = 2x2 - 3x + 1
b) y = -5x - 11
b) y = x2 - 2x + 3
b) y = - 1 x + 3
2
LS-104. b) The result is the equation of a line. These three points are collinear, which
yields a = 0. you must have three non-collinear points to get a parabola.
LS-105. y £ -x + 4; y > 1 x
3
LS-106. He should substitue z = 7 into any two of the original equations and solve the
resulting system for x and y: (-1, -3, 7)
LS-108. Both are parabolas that open up. The first has its vertex at (3, 0), and the
second's is at (0, -3).
LS-109. a) 1
4
b)
12 - x
12
LS-110. a) y = 1 x + 5 b) y = 2x + 5
3
È 9 - 3 1˘
LS-111. A = Í 1 1 1˙
˙
Í
ÍÎ 16 4 1˙˚
c) y = - 1 x + 7 1
2
2
È - 7˘
È x˘
X = Í y˙ B = Í - 3 ˙
˙
Í
Í ˙
ÍÎ 21 ˙˚
ÍÎ z ˙˚
È 2 7 5˘
LS-112. AX = B in that order; A = Í 3 1 4 ˙
˙
Í
ÍÎ 6 0 9 ˙˚
© 2000 CPM Educational Program
d) y = 2x
È 15 ˘
È x˘
X = Í y ˙ B = Í 41 ˙
˙
Í
Í ˙
ÍÎ 34 ˙˚
ÍÎ z ˙˚
v. 5.0
Unit 5
Answer Key
9
LS-113. a) Add -149 to both sides to get x by itself.
b) 1; 1/4; 10; reciprocal
È 0 0˘
LS-114. Í
˙
Î 0 0˚
È 1 0˘
LS-116. Í
˙
Î 0 1˚
È 1 0 0˘
LS-117. a) Í 0 1 0 ˙
˙
Í
ÍÎ 0 0 1 ˚˙
b) They are.
È 9 -1 ˘
LS-118. a) Í
˙
Î 4 10 ˚
b) Í
È 4 -5˘
LS-119. b) Í
˙
Î -3 4 ˚
È 20 -5 ˘
˙
Î 35 10 ˚
È m ˘ È -2 ˘
Í n ˙ =Í 9 ˙
Î
˚ Î
˚
È -20 5 ˘
˙
Î -44 -19 ˚
d) Í
È 7 - 3
Í - 2 1
d) Í
Í 4 0
Í 1 0
Î
Pc Pn
LS-120. a)
Juan
Huang
Danusha
È 3 2˘
Í 4 5˙
˙
Í
ÍÎ 6 4 ˙˚
0 2˘
0 - 1˙
˙
1 -2 ˙
0 - 3˙˚
È w ˘ È 41
Í x ˙ Í - 13
Í ˙ =Í
Í y ˙ Í 12
Í z ˙ Í 1
Î ˚ Î
˘
˙
˙
˙
˙
˚
$
Juan È 0.80 ˘
Huang Í 1. 65 ˙
˙
Í
Danusha ÍÎ 1. 60 ˙˚
Huang's pencils and pens are worth $1.65
c)
È ag + bh ˘
È a b˘
g
È ˘
LS-121. Í c d ˙ Í ˙ = Í cg + dh ˙
˙
˙ Î h˚ Í
Í
ÍÎ eg + fh ˙˚
ÍÎ e f ˙˚
LS-122. x-1
LS-123. y = (x + 2)2 - 11
LS-124. a)(5x - 1)(5x + 1)
© 2000 CPM Educational Program
c)(x + 9)(x - 8)
d) x(x - 6)(x + 3)
v. 5.0
Unit 5
Answer Key
È 1 0˘
c) Product in either order is Í
˙
Î 0 1˚
LS-125. a) Identity Matrix
È - 311
Í
LS-126. Í 111
ÍÎ 2 11
21
11
4
11
- 14
11
- 23
- 7
19
10
33 ˘
˙
33 ˙
33 ˙˚
LS-127. Calculator round-off error. These entries should be 0.
LS-128. a) x = 1, y = 3, z = -7
c) X = A-1B
b) left-multiply by A-1
È -2 ˘
È 4 4 -5 ˘
LS-129. A = Í 2 -4 10 ˙ , B = Í 6 ˙ , x = 1 , y = - 3 , z = 1
2
4
˙
Í
˙
Í
5
ÍÎ 0 ˙˚
ÍÎ 1 2 5 ˙˚
LS-131. x = -2, y = 3 , z = 4
3
5
È 4 5˘
LS-133. b) Í
˙
Î 3 4˚
È - 2 ˘ È 37 ˘
Í 9 ˙ = Í 30 ˙ so m = 37, n = 30
Î
˚ Î ˚
È - 7 15 ˘
LS-134. a) Í
˙
Î 12 - 1˚
c) impossible
È 1 0 0˘
LS-135. a) impossible, since M is not square
c) Í 0 1 0 ˙
˙
Í
0
0
1
˙˚
ÍÎ
LS-136. a) -8p + 4q - 2r + s = -22, p + q + r + s = 2, 8p + 4q + 2r + s = -2,
125p + 25q + 5r + s = 118
È -8 4 -2 1˘ È p ˘ È -22 ˘
Í 1
1 1 1˙ Í q ˙ Í 2 ˙
˙
˙ Í ˙ =Í
Í
c) g(x) = 2x3 - 5x2 - 3x + 8
b)
4 2 1˙ Í r ˙ Í -2 ˙
Í 8
Í 125 25 5 1˙ Í s ˙ Í 118 ˙
˚
˚ Î ˚ Î
Î
LS-137. b)
3 d)
x- 4
2
x +1
x - 3x + 2
LS-139. a) (0, 750); (1, 635); (2, 530)
© 2000 CPM Educational Program
b) parabola
v. 5.0
Unit 5
Answer Key
11
c) y = 5x2 - 120x + 750
d) After 10 minutes the ship is 50 thousand
miles away; therefore, the ship burns up.
LS-140. a) (0, 900); (10, 1600); (15, 1875) c) v = -t2 + 80t + 900
d) after 90 days e) 10 days
LS-141. x = 3, y = 1, z = 3
LS-142. a) -7
b) 94 c) 94
d) 124
LS-144. a) y = -x2 + 4x b) y = 5 ± x - 3
LS-145. The first graph opens downward, is stretched, and has its vertex at
(-1, -3). The second is the parent graph.
LS-146. b) exterior of the same circle
LS-147. 384 ft as the crow flies
LS-148. b) (30, 60), (40, 50)
c) 100a + 10b + c = 40, 900a + 30b + c = 60, 1600a + 40b + c = 50
d) y = -0.067x2 + 3.667x + 10 or y = - 1 x2 + 11 x + 10
3
15
e) 56.5 minutes
f) -2.7 min; no, she cannot arrive before she leaves.
g) 8.2 < x < 46.7
h) Leave at or before 7:12 or after 7:42; x ª 12.3, 42.4
i) She would need to leave by 7:19. The other choice is 7:51, but this falls out of
the domain.
LS-149. a) Triangle with vertices (-1, 5), (-1, -1), (1, 1), shaded inside
b) 6 square units
LS-151. a) 24
È 11 11˘
LS-152. a) Í
˙
Î 21 13˚
b)2-3 c)21/2 d) 22/3
b) impossible
LS-153. a) 2a2 - 4
d) 2x2 + 28x + 94
c) [14 12 ]
d) [11 23]
b) 18a2 - 4
c) 2a2 + 4ab + 2b2 - 4
e) 50x2 + 60x +14 f) 10x2 - 17
LS-154. -1, 1 , 2
2
LS-155. a) $20.65
© 2000 CPM Educational Program
b) $2.21
v. 5.0
Unit 5
Answer Key
LS-157. a) y = 2x2 - 4x + 6
LS-158. a) 1
5
© 2000 CPM Educational Program
b) 1
3
12
b) (1, 4)
c) 100
d) x + 60 = 0.4
x + 300
v. 5.0