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BLM 1–9 Chapter 1 Test
1. B 2. C 3. B 4. D 5. B
6. There are 33 lockers.
7. Brittany travelled 1752 m.
8. 20 358
9. a) d = 6 b) t1 = 4 c) t100 = 598
10. a) r = 9 or –9 b) tn = 5(9)n – 1 or tn = 5(–9)n –
1
11. a) –3, –12, –48
b) The sequence is geometric. tn = –3(4)n – 1
12. a) Example, for the series
2 + 10 + 50 + ..., S10 = 4 882 812.
b) Answers will vary. Students need to change
the sign of the first term, while leaving the
common ratio unchanged. In the example above,
the series becomes –2 – 10 – 50 – ..., S10 = – 4
882 812.
c) Answers will vary. Correct answers must
have positive first term and negative common
ratio.
For example, 2 – 10 + 50 – ….
13. a) 6979, 7537, 8140 b) tn = 6462(1.08)n – 1
c) 27 888
d) Answers will vary. For example, we assume
that population continues to grow at the same
rate.
14. Answers will vary, however will all be in the
form k, k, k, …, where k is a real number.
a) Note that d = 0, so tn = k.
b) Note that r = 1, so tn = k.
c) There are infinitely many such sequences, but
all sequences will have the same form.
BLM 2–9 Chapter 2 Test
1. B 2. B 3. C 4. D 5. A 6. 9.5 yd 7. 21.9
cm
8. 14 cm 9. a)  3 b)
– 3
2
c)
2
2
BLM 3–7 Chapter 3 Test
1. B 2. A 3. C 4. A 5. D
6. a) x b) y c) y
7. a) r and s b) t
8. a) f (x) =
b) f (x) = –
1
(x +
3
x2
3
4)2 + 2
+5
9. a) y = (x – 2)2 + 8; vertex (2, 8)
1
4
b) y = – (x + 8)2 – 2; vertex(–8, –2)
10. vertex: (6, –21); axis of symmetry: x = 6;
direction of opening: upward; domain: x  R;
range: y ≥ –21; x-intercepts: (6  42, 0);
y-intercept: (0, –3)
11. a) R = –50x2 + 400x + 12 000 b) $80
c) $12 800 d) 40 seats
BLM 4–9 Chapter 4 Test
1. A 2. B 3. D 4. B 5. A
6.
5 5
2
or 5.59 s
7. a) In line 2, –4 should be in brackets.
2  10
.
2
b) In step 3, each term should have been divided
by 15.
3  39
15
.
8. a) x = 2 or 8; Example: Factoring, because the
equation is easily factored to (x – 2)(x – 8).
2
3
b) x = –7 or x  ; Example: Quadratic formula,
because the equation is not readily factored.
c) x  3  2; Example: Completing the square,
because it is easy to find the perfect square.
d) x = 1 or 5; Example: Determining square
roots, because it is easy to find the roots for (x –
3)2 = 4
10. a) 20° b) 20°, 200°, 340°
9. x2 + 5x – 10 = 0;
11. a)
10. k 
b) 29.9°, 56.3°, 93.8°
5
2
11. 11.3 m by 9.3 m
12.
2
3
or
3
2
13. 2.57 s
12. 12.5 ft and 6.4 ft
Copyright © 2011, McGraw-Hill Ryerson, ISBN: 9780070738836
5  65
2
c) 2 is not a solution because it is a non-
BLM 5–7 Chapter 5 Test
1. B 2. D 3. A 4. C 5. B
permissible value. d)  1 .
2
6. 6 2, 5 3, 2 19, 4 5, 9
7. False. 16  9  7
8. 7 6 units
9. 5 5
BLM 7–8 Chapter 7 Test
1. C 2. A 3. B 4. C 5. B 6. C
7. | 5 |
10. r  7, r   1
4

3 x  4, if x  3
8. y  
3 x  4, if x  4

3
3
11. 6 5  13
12. a) No. Valerie’s error is in taking the square
root of each term of the radicand. The equation
v0  v 2  20h cannot be further simplified.
b) 60 m
13. a) x  3 b) 3 is not a root of the equation.
2
2
9. Example: | x  5 |  3
10. x  3
11. x  3 and x  2
12. a)
14. a) y  3, y  7 b) y  5
3
BLM 6–9 Chapter 6 Test
1. D 2. D 3. B 4. B 5. B 6. C 7. D
2
( x  3)
8. x  32 ,  3; (2 x  3) 9. 2 x  3x  3
( x  6)( x  3)
11. Example:
x 1
x3
10. 8
does not belong because it
13. a) | A –16 | 
is not a rational expression.
12. a) x  5, 4, 0, 6 b)
13. a) 15  x b)
15  x
x
( x  5)( x  6)
x( x  5)
c)
15  x
 1.25;
x
60
buckets
14. No. Example: Restrictions on a variable in
an expression come from the denominator of the
factored form before any of the common factors
in the numerator and denominator are simplified.
15. Example: Mary cannot cancel 3y from the
denominator and from one term in the
numerator. However, she can divide each term
by 3y, which results in
5y 1
,
1
b) (5, 0)
c) domain: {x | x  R}; range: {y | y  0, y  R}
3
4
b) 15 years and 3 months, or 16 years and 9
months
c) Alain could be older or younger than the
average age.
14. a) x  4 and x  1.5
b)
or 5y  1.
16. 2; Example: I first found the LCD on each
side of the equation and then performed the
required operations: addition on the left side and
subtraction on the right. I then used
multiplication to eliminate the denominator on
each side of the equation. Finally, I isolated the
variable, and solved.
17. a) x  1, 2, 3
1
2
b)  , 2
BLM 8–6 Chapter 8 Test
Copyright © 2011, McGraw-Hill Ryerson, ISBN: 9780070738836
1. A 2. B 3. B 4. D 5. A 6.{(5, 8), (0, 3)}
7. Example: An object is released from a
launcher on the ground, and a person standing
on a platform throws a ball, trying to hit the
object with the ball.
8. Example: ay  a(x2  6x  5), a  R
9. a) LS  2x2  x  7 RS  y

 2(3)2  3  7
 14

 14
LS  RS
LS  3x  y  23 RS  0
  3(3)  14  23
 0
LS  RS
b) (5, 38)

solution half-plane.
  
10.  52 ,  2 , 12 , 1
11. a) m  5, k  2 b) k  8, m  2
12. a) two b) k  4 or k  0
c) (5.43, 1.08) or (1.43, 1.08)
13. a)
{(4.3, 6.8), (21.7, 177.3)}
b) The coordinates represent where the two
streams of water meet. However, only the (4.3,
6.8) solution makes sense because the distance
cannot be negative in this context.
14. a) perimeter: 2y  4x  26;
area: 3y  9  x2  13x  36
b) x  7 and y  1, or x  12 and y  11
c) Substituting 7 results in a negative dimension,
so x must be 12. The dimensions are 8 units and
3 units.
d) perimeter: 22 units; area: 24 square units
BLM 9–7 Chapter 9 Test
1. B 2. D 3. D 4. C 5. A
6. Example: When the point is substituted into
the inequality, it makes a true statement. Shade
the
half-plane containing the point.
7. Example: The point (0, 0) lies on the
boundary. It cannot be used to determine the
Copyright © 2011, McGraw-Hill Ryerson, ISBN: 9780070738836
10. {x   2  3  x  2  3, x  R}
11.
14. {P  30  P  70, P  R}
Example: Test point (0, 3):
1
2
3   (0)2  3(0)  1
31
The test point satisfies the inequality, so the
region above the parabola should be shaded.
12. a) 13.95x  9.50y  150, where x is the
number of tickets for adults and y is the number
of tickets for children
b)
c) Example: The number of tickets must be a
whole number. The number of tickets can be
various combinations of 0 to 10 adult tickets and
0 to 15 children’s tickets, where the total cost
does not go over $150.
13. a) Example: x2  x  2  0
b) Example: x2  x  12  0
Copyright © 2011, McGraw-Hill Ryerson, ISBN: 9780070738836