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Name: ______________________ Class: _________________ Date: _________
ID: A
Gr 10 IAP Year End Review
Multiple Choice
Identify the choice that best completes the statement or answers the question.
1.
(1 point) Convert
A. 21 in.
2.
(1 point) Marie
7 yd. to inches.
B. 252 in.
B. 21
3.
(1 point) Nancy
4.
(1 point) A
5.
(1 point) Baseboards
6.
(1 point) Which
C. 2
D. 10
has 7 yd. of material. She wants to make curtains that are 18 in. wide. How many curtains
can Nancy make?
A. 92
B. 14
C. 4
D. 1
map of British Columbia has a scale of 1:1 723 000. The distance on the map between Prince
11
George and Cache Creek is 8
in. What is this distance to the nearest mile?
16
A. 945 mi.
C. 708 mi.
B. 79 mi.
D. 236 mi.
are sold in 8-ft. lengths. Nelia requires 73 yd. of baseboard. How many 8-ft. lengths
does Nelia need to purchase?
A. 29
B. 28
C. 26
D. 27
A.
B.
C.
D.
The
The
The
The
referent could you use for 1 m?
width of a computer keyboard
length of a dinner fork
length of your stride
width of a classroom in your school
(1 point) Which
A.
B.
C.
D.
8.
D. 84 in.
has 17 yd. of material that she will cut into strips 19 in. wide. How many strips can Marie
make?
A. 32
7.
C. 43 in.
The
The
The
The
referent could you use for 1 cm?
depth of a kitchen sink
length of a public swimming pool
width of your shortest finger
length of a walking stick
(1 point) Which
A. The
B. The
C. The
D. The
referent could you use for 1 km?
1
distance equal to 2 laps on an oval running track
2
length of an iPod
length of a snowboard
length of your arm span
1
9.
(1 point) Which
A.
B.
C.
D.
10.
The
The
The
The
referent could you use for 1 in.?
distance from where you are now to the nearest restaurant
diameter of a bicycle wheel
length of your calculator
width of your largest toe
(1 point) Which
A. Metres
11.
(1 point) Which
A. Miles
SI unit is most appropriate for measuring the length of a soccer field?
B. Millimetres
C. Kilometres
D. Centimetres
imperial unit is most appropriate for measuring the width of a snowboard?
B. Inches
C. Feet
D. Yards
12.
(1 point) Which
13.
(1 point) A
14.
(1 point) Convert
imperial unit is most appropriate for measuring the distance between the nearest lake and
the nearest mountain peak?
A. Feet
B. Inches
C. Miles
D. Yards
penalty box on a soccer field measures 44 yd. by 18 yd. What are these dimensions to the
nearest tenth of a metre?
A. 39.6 m by 16.2 m
C. 39.6 m by 17.6 m
B. 47.7 m by 16.2 m
D. 47.7 m by 17.6 m
165 cm to feet and the nearest inch.
A. 5 ft. 8 in.
B. 6 ft. 6 in.
C. 5 ft. 4 in.
15.
(1 point) A
16.
(1 point) A
17.
(1 point) A
18.
(1 point) A
D. 5 ft. 6 in.
regular tetrahedron has edge length 20.0 m and a slant height of 17.3 m. Calculate the surface
area of the tetrahedron to the nearest square metre.
A. 1384 m2
B. 173 m2
C. 519 m2
D. 692 m2
right rectangular pyramid has base dimensions 8 ft. by 6 ft. and a height of 12 ft. Calculate the
surface area of the pyramid to the nearest square foot.
A. 223 square feet B. 159 square feet C. 271 square feet D. 216 square feet
right pyramid has a square base with side length 12 m and a height of 7 m. Calculate the surface
area of the pyramid to the nearest square metre.
A. 312 m2
B. 443 m2
C. 664 m2
D. 365 m2
right cone has a height of 13 cm and a base diameter of 17 cm. Determine the surface area of
the cone to the nearest square centimetre.
A. 642 cm2
B. 574 cm2
C. 415 cm2
D. 1057 cm2
2
19.
(1 point) Calculate
the volume of this right square pyramid to the nearest cubic foot.
A. 58 cubic feet
20.
(1 point) A
21.
(1 point) This
B. 62 cubic feet
C. 54 cubic feet
D. 163 cubic feet
right square pyramid has a base side length of 11 ft. and a slant height of 19 ft. Calculate the
volume of the pyramid to the nearest cubic foot.
A. 734 cubic feet
B. 2201 cubic feet C. 539 cubic feet
D. 766 cubic feet
regular tetrahedron has a height of 4.7 cm. Calculate its volume to the nearest cubic
centimetre.
A. 58 cm3
22.
(1 point) A
23.
(1 point) A
B. 45 cm3
C. 68 cm3
D. 23 cm3
right cylindrical can has a volume of 263.1 cm3 . What is the volume of a right cone with the
same base and the same height, to the nearest tenth of a centimetre?
A. 131.6 cm
B. 91.7 cm
C. 89.7 cm
D. 87.7 cm
sphere has a surface area of 10.1 m2 . What is the radius of the sphere to the nearest tenth of a
metre?
A. 3.7 m
B. 4.8 m
C. 0.9 m
3
D. 1.8 m
24.
(1 point) Determine
A. 18.4°
25.
(1 point) Calculate
A. 77.3°
26.
27.
C. 70.5°
D. 71.6°
the angle of inclination, to the nearest tenth of a degree, of a road with a grade of 22%.
B. 77.6°
C. 12.4°
D. 12.7°
the length of side l to the nearest tenth of a metre.
B. 27.4 m
C. 11.1 m
D. 5.0 m
(1 point) The
angle of inclination of a solar panel on the roof of a cottage is 57°. Determine the height of
the roof, to the nearest tenth of a metre.
A. 3.8 m
28.
B. 19.5°
(1 point) Determine
A. 5.4 m
the measure of ∠D to the nearest tenth of a degree.
B. 4.9 m
C. 2.1 m
D. 5.9 m
(1 point) A
road has an angle of inclination of 16°. Determine the increase in altitude of the road, to the
nearest metre, for every 150 m of horizontal distance.
A. 523 m
B. 144 m
C. 43 m
4
D. 41 m
29.
(1 point) Determine
sin G and cos G to the nearest hundredth.
A. sin G = 0.99; cos G = 6.54
B. sin G = 0.15; cos G = 0.99
30.
(1 point) A
rectangle is 5.1 cm wide and each diagonal is 9.3 cm long. What is the measure of the angle
between a diagonal and the shorter side of the rectangle to the nearest tenth of a degree?
A. 33.3°
31.
B. 61.3°
(1 point) Determine
A. 8.4 cm
32.
C. sin G = 1.01; cos G = 0.15
D. sin G = 0.99; cos G = 0.15
(1 point) Solve
C. 56.7°
D. 28.7°
the length of XY to the nearest tenth of a centimetre.
B. 15.2 cm
C. 31.4 cm
D. 19.9 cm
this right triangle. Give the measures to the nearest tenth.
A. ∠J = 68.6°; ∠L = 21.4°; JL = 24.7 cm
C. ∠J = 21.4°; ∠L = 68.6°; JL = 24.7 cm
B. ∠J = 68.6°; ∠L = 21.4°; JL = 63.1 cm
D. ∠J = 21.4°; ∠L = 68.6°; JL = 63.1 cm
5
33.
(1 point) Solve
this right triangle. Give the measures to the nearest tenth.
A. ∠G = 40°; GH = 21.8 cm; FH = 18.3
cm
B. ∠G = 40°; GH = 21.8 cm; FH = 44.3
cm
34.
(1 point) The
front of a tent has the shape of an isosceles triangle with equal sides 163 cm long. The
measure of the angle at the peak of the tent is 105°. Calculate the maximum headroom in the tent to the
nearest centimetre.
A. 129 cm
35.
C. ∠G = 40°; GH = 18.3 cm; FH = 21.8
cm
D. ∠G = 50°; GH = 21.8 cm; FH = 18.3
cm
(1 point) Determine
A. 16.5 cm
B. 125 cm
C. 99 cm
D. 231 cm
the perimeter of this rhombus to the nearest tenth of a centimetre.
B. 33.8 cm
C. 25.1 cm
6
D. 7.2 cm
36.
(1 point) Determine
A. 6.7 cm
37.
(1 point) Determine
A. 5.2 cm
38.
(1 point) Determine
A. 85 m
the length of RS to the nearest tenth of a centimetre.
B. 9.3 cm
C. 11.4 cm
D. 8.3 cm
the length of MN to the nearest tenth of a centimetre.
B. 3.4 cm
C. 2.9 cm
D. 4.5 cm
the length of QR to the nearest metre.
B. 170 m
C. 127 m
7
D. 118 m
39.
(1 point) From
the top of an 80-ft. building, the angle of elevation of the top of a taller building is 49° and
the angle of depression of the base of this building is 62°. Determine the height of the taller building to
the nearest foot.
A. 211 ft.
40.
D. 276 ft.
the top of a 25-m lookout tower, a fire ranger observes one fire due east of the tower at an
angle of depression of 7°. She sees another fire due north of the tower at an angle of depression of 3°.
How far apart are the fires to the nearest metre?
(1 point) Determine
A. 14
42.
C. 129 ft.
(1 point) From
A. 205 m
41.
B. 112 ft.
(1 point) Determine
A. 2
43.
(1 point) A
44.
(1 point) Determine
B. 681 m
C. 477 m
D. 519 m
the greatest common factor of 84, 210, and 336.
B. 1680
C. 21
D. 42
the least common multiple of 10 and 22.
B. 55
C. 220
D. 110
cube has volume 15 625 cm3 . What is the surface area of the cube?
A. 132 893.3 cm2
B. 3750 cm2
C. 25 cm2
D. 10 416.7 cm2
A. 301.87 cm
45.
(1 point) A
46.
(1 point) Which
the edge length of this cube.
B. 45 cm
C. 6.71 cm
D. 3375 cm
cube has surface area 3750 square feet. What is its volume?
A. 5625 cubic feet
C. 1448 cubic feet
B. 25 cubic feet
D. 15 625 cubic feet
of the following numbers is not both a perfect square and a perfect cube?
A. 531 441
B. 12 544
C. 117 649
D. 15 625
8
47.
2
(1 point) Factor
the binomial 44a + 99a .
A. a(44 + 99a)
C. 11a(4 + 9a)
2
B. 11(4a + 9a )
48.
(1 point) Factor
D. 22a(2 + 9a)
2
the trinomial 4 − 8n + 12n .
2
2
A. 4(−2n + 3n )
C. 2(2 − 4n + 6n )
2
2
B. 4(1 − 2n + 3n )
49.
(1 point) Factor
D. 4(1 + 2n + 3n )
3
2
B. 8cd(3c + 5cd + 4d )
D. −8cd(3c + 5cd + 4d )
2
(1 point) Identify
C. 3s t
2 2
D. 3s t
2
B. r (1 − π)
C. x + 10x + 3
2
D. x + 13x + 5
D. r(r − 2 π)
2
2
2
−4d − 28d + 240
A. −4(d + 3)(d − 20)
C. −4(d − 3)(d + 20)
B. −4(d + 5)(d − 12)
D. −4(d − 5)(d + 12)
(1 point) Factor:
2 3
of the following trinomials can be represented by a rectangle? Use algebra tiles to check.
B. x + 11x + 30
2
25x + 58x + 16
A. (25x + 4)(x + 4)
C. (5x + 4)(5x + 4)
B. (25x + 8)(x + 2)
55.
2
C. r (4 − π)
2
(1 point) Factor:
4 2
expression represents the area of the shaded region?
A. x + 2x + 14
54.
2
3 2
A. 2r(2r − π)
53.
2
2 3
B. 3s t
(1 point) Which
2
3 4
2 2
(1 point) Which
2
the greatest common factor of the terms in the trinomial 6s t + 12s t − 15s t .
A. 6s t
52.
3
C. 8cd(−3c + 5cd + 4d )
2
51.
2
2
A. −8cd(3c − 5cd − 4d )
50.
2
the trinomial −24c d − 40c d − 32cd .
(1 point) Factor:
180 − 175a + 30a
A. 5(4 + 3a)(9 + 2a)
B. (20 − 15a)(9 − 2a)
D. (5x + 8)(5x + 2)
2
C. 5(4 − 3a)(9 − 2a)
D. 10(18 − 1a)(1 − 3a)
9
56.
(1 point) Expand
and simplify: (f + 5g)(2f − 5g + 7)
2
2
A. 2f + 5fg + 7f + 25g + 35g
2
2
B. 2f − 15fg + 7f − 25g + 35g
57.
(1 point) Which
C. 16x + 72xy − 40y
2
2
2
D. 8x + 36xy − 20y
2
36 − 60r + 25r
C. (6 + 5r)
2
B. (6 − 5r)(6 + 5r)
D. (6 − 5r)
2
(1 point) Identify
7
62.
4
3
30 ,
4
(1 point) Which
(1 point) Write
A.
65.
A. 7
66.
3
A.
6
9604
4
25
100 ,
3
75 ,
D. 8
C.
16
25
D.
16
5
17
3
30
C.
D.
14
3
75
3
D. –1
150
C.
180
D.
C. 14
3
D. 7
C.
686
900
1372 in simplest form.
B. 4
7
C. 1.8
5 as an entire radical.
B.
28
(1 point) Write
8
B. 0
30
(1 point) Write
C. 6
8
4 .
of these numbers is a natural number?
9, 0, –1, 1.8
A. 9
64.
D. 2
which of these numbers is the least.
B.
100
6
C. 7
256
.
625
B.
(1 point) Determine
A.
63.
4
4
5
14 ,
7
the radicand of
B. 4
(1 point) Evaluate
2
2 .
B. 3
(1 point) Identify
A.
3
the index of
2
2
A. (9 − 5r)(4a − 5r)
A. 4
61.
2
2
(1 point) Factor:
A. 2
60.
2
D. 2f − 5fg + 7f − 25g + 35g
2
B. 8x + 22xy − 20y
59.
2
polynomial, written in simplified form, represents the area of this rectangle?
A. 8x − 36xy − 20y
58.
2
C. 2f + 5fg + 7f − 25g + 35g
3
7
7
3
4
14 as an entire radical.
B.
98
10
D.
1372
67.
(1 point) Write
A.
68.
69.
5
4
5
12 as an entire radical.
(1 point) Evaluate
A.
64
1
3
5
995 328
D.
5
12 288
without using a calculator.
B. 4
64
625
ÊÁ 3
ÁÁ
ÁÁ
ÁË 4
ÊÁ 3
ÁÁ
ÁÁ
ÁË 4
(1 point) Write
4
4
25
ÊÁ 3
ÁÁ
ÁÁ
ÁË 4
B.
ˆ˜
˜˜
˜˜
˜¯
9
2
1.25
C.
4
5
C.
ÊÁ 4
ÁÁ
ÁÁ
ÁË 3
D.
16
25
D.
ÊÁ 3
ÁÁ
ÁÁ
ÁË 4
2
ˆ˜ − 9
˜˜
˜˜
˜¯
ˆ˜
˜˜
˜˜
˜¯
C. 0.0256
2
9
D. 0.010 24
as a radical.
4
ÁÊ 5
˜ˆ
7.5 , or ÁÁÁÁ 7.5 ˜˜˜˜
Ë
¯
4
1
5
1
3
5
2
0.16 .
B. 0.1012
7.5
D. 21
without using a calculator.
9
A. 0.4804
A.
1
4
C. –4
ˆ˜ 9
˜˜ as a power.
˜˜
˜¯
ˆ˜ − 2
˜˜
˜˜
˜¯
(1 point) Evaluate
5
˜ˆ˜
˜˜
˜˜
¯
B.
(1 point) Write
A.
73.
C.
square has an area of 12 square inches. Determine the side length of the square as a radical in
simplest form.
A. 4 3 in.
B. 2 6 in.
C. 3 2 in.
D. 2 3 in.
ÁÊ 256
70. (1 point) Evaluate ÁÁÁÁ
ÁË 625
72.
2304
(1 point) A
A. 8
71.
5
B.
192
C.
5
1
ÁÊ 4
˜ˆ 5
B.
7.5 , or ÁÁÁÁ 7.5 ˜˜˜˜
Ë
¯
0.8
ÁÊ 243 ˜ˆ˜
˜˜ .
74. (1 point) Evaluate ÁÁÁÁ −
ÁË 32 ˜˜¯
81
A.
B. does not exist
32
D.
C.
11
4
ÊÁ 15
ÁÁ
ÁÁ
ÁË 2
ÊÁ
ˆ˜ 4
ˆ˜ 4
˜˜ , or ÁÁÁÁ 5 15 ˜˜˜˜
˜˜
ÁÁ
˜¯
2 ˜˜˜
Á
Ë
¯
5
ÁÊ 4
˜ˆ
5
7.5 , or ÁÁÁÁ 7.5 ˜˜˜˜
Ë
¯
81
16
D. −
81
16
75.
(1 point) Arrange
2
3
3
9 ,
A.
B.
76.
3
these numbers in order from greatest to least.
1
2
3
9 ,9 ,
9 ,
3
9 ,9
3
9 ,9
9 ,9
1.2
(1 point) Suppose
1.2
2
3
1.2
2
3
,9 ,9
1
2
,9 ,9 ,
3
1
2
C.
D.
9
3
9 ,9
2
3
1
2
1.2
9 ,9 ,
,
3
3
2
3
1
2
3
1.2
9 ,9 , 9
9 ,
9 ,9
you want $2000 in 3 years. The interest rate for a savings account is 2.8% compounded
annually. The money, P dollars, you must invest now is given by the formula P = 2000 ( 1.028 )
much must you invest now to have $2000 in 3 years?
A. $ 1845.02
B. $ 2172.75
C. $ 1840.99
D. $ 1836.58
−3
ÊÁ 5 −4 7 ˆ˜
77. (1 point) Simplify ÁÁÁÁ a b ˜˜˜˜ .
ÁË 2
˜¯
A.
125b
8a
21
12
B.
8a
12
125b
21
C.
12
125a
8b
12
21
D.
8b
4
125a
7
−3
. How
78.
(1 point) Consider
the relation represented by this graph. Represent the relation as a table.
A.
C.
Shoe Siz e
Stude nt
Stude nt
Shoe Siz e
10
John
John
10
7
Kelly
Kelly
7
11
Natalie
Martin
11
8
Martin
Natalie
8
B.
D.
Shoe Siz e
Stude nt
Stude nt
Shoe Siz e
7
John
John
7
10
Kelly
Kelly
10
8
Martin
Martin
8
11
Natalie
Natalie
11
13
79.
(1 point) Which
arrow diagram shows the association “is less than” from a set of numbers to a set of
numbers?
80.
A.
C.
B.
D.
(1 point) Each
A. E and F
B. C and D
point on this graph represents a person. Which two people are the same age?
C. D and E
D. B and C
14
81.
(1 point) Joshua
went on a bike ride. For part of the ride, Joshua stopped to play in a park with a friend.
Which segment of the graph best describes this part of his bike ride?
A. CD
B. AB
C. OA
15
D. BC
82.
(1 point) A
person in a car drives away from a stop sign, cruises at a constant speed, and then slows down as
she approaches another stop sign. Which graph best represents this situation?
A.
C.
B.
D.
16
83.
(1 point) Gail
leaves the house for her morning jog. She stops for a quick drink, then continues jogging
before stopping again to chat with a friend. She then jogs back home. Which graph best represents Gail’s
run?
17
A.
B.
C.
D.
18
84.
(1 point) Which
graph best represents the cost of renting a kayak as a function of time?
A.
C.
B.
D.
19
85.
(1 point) Which
i)
of these graphs represents a function?
ii)
iii)
A. ii
iv)
B. i
C. iii
20
D. iv
86.
(1 point) This
graph shows the masses of people, m, as a function of age, a. Determine the range of the
graph.
A.
B.
87.
ÏÔ
¸Ô
ÌÔ 4, 5, 8, 12, 14, 17 ˝Ô
Ó
˛
ÏÔ
¸Ô
ÌÔ 3, 5, 8, 10, 14, 17 ˝Ô
Ó
˛
ÏÔ
¸Ô
C. ÌÔ 15, 25, 45, 55, 80, 85 ˝Ô
Ó
˛
ÏÔ
¸Ô
D. ÌÔ 20, 25, 45, 65, 80, 85 ˝Ô
Ó
˛
(1 point) This
graph shows the cost of hosting a dance, c, as a function of the number of students attending,
n. What is a restriction on the domain?
A.
B.
C.
D.
The
The
The
The
domain
domain
domain
domain
can only contain positive numbers.
can only contain whole numbers between 1000 and 3500.
can only contain whole numbers.
can only contain whole numbers that are multiples of 50.
21
88.
(1 point) Determine
the domain of this graph.
A. x > −5
B. x ≥ −5
89.
C. −5 ≤ x ≤ 5
D. y ≥ 2
(1 point) The
relation between x and y is linear. Which number would complete this table?
x
y
3
19
A. –7
90.
11
7
15
19
–5
B. 1
(1 point) This
C. –6
D. 6
table of values represents a linear relation. Determine the rate of change of the relation.
Time (s)
Distance
(m)
A. 5 m/s
91.
7
13
0
0
1
5
2
10
3
15
B. 10 m/s
4
20
C. 2 m/s
D. 1 m/s
(1 point) The
altitude of a plane, a metres, is related to the time, t minutes, that has elapsed since it started
its ascent. Determine the rate of change of this linear relation.
t (min)
a (m)
A. 1500 m/min
0
4000
2
5400
B. 1400 m/min
4
6800
6
8200
8
9600
C. 1200 m/min
22
D. 700 m/min
92.
(1 point) This
graph shows distance, d kilometres, as a function of time, t minutes. Determine the vertical
and horizontal intercepts.
A. Vertical intercept: 80
Horizontal intercept: 96
C. Vertical intercept: 96
Horizontal intercept: 80
B. Vertical intercept: 64
Horizontal intercept: 96
D. Vertical intercept: 80
Horizontal intercept: 64
23
93.
(1 point) This
graph shows the volume of water remaining in a leaking hot tub as a function of time.
Determine the domain and range.
A. Domain: t ≤ 129
Range: 0 ≤ V ≤ 1800
B. Domain: 0 ≤ V ≤ 1800
Range: t ≤ 129
C. Domain: 0 ≤ t ≤ 129
Range: V ≤ 1800
D. Domain: 0 ≤ t ≤ 129
Range: 0 ≤ V ≤ 1800
24
94.
(1 point) Each
graph below shows distance, d metres, as a function of time, t hours. Which graph has a rate
of change of 4 m/h and a vertical intercept of 3 m?
A.
C.
B.
D.
25
95.
(1 point) Each
graph below shows distance, d metres, as a function of time, t hours. Which graph has a rate
of change of 0.75 m/h and a horizontal intercept of 3 m?
A.
C.
B.
D.
26
96.
(1 point) This
graph shows the fuel consumption of a jeep with a full tank of gas at the beginning of a
journey. When the jeep has travelled 150 km, about how much fuel is left in the tank?
A. about 49 L
B. about 12 L
97.
(1 point) Determine
A.
3
2
B. −
98.
C. about 51 L
D. about 11 L
the slope of the line that passes through G(3, –3) and H(–5, 9).
C.
2
3
2
3
D. −
(1 point) A
3
2
straight section of an Olympic downhill ski course is 34 m long. It drops 16 m in height.
Determine the slope of this part of the course.
15
8
A. −
C. −
8
17
8
17
B. −
D. −
15
8
27
99.
(1 point) Determine
A. –
B.
100.
the slope of the line that is parallel to this line segment.
3
7
C.
7
3
(1 point) The
3
7
D. –
slope of a line is
3
2
2
B. –
3
A. –
101.
(1 point) A
102.
(1 point) A
103.
(1 point) Predict
7
3
2
. What is the slope of a line that is perpendicular to this line?
3
3
C.
2
30
D.
20
line passes through J(–10, 10) and K(7, –9). Determine the coordinates of L so that line JL is
perpendicular to line JK.
A. L(27, 9)
C. L(17, –19)
B. L(–19, 17)
D. L(9, 27)
line passes through R(8, 1) and F(–5, –4). Determine the coordinates of two points on a line
perpendicular to RF.
A. (16, –11) and (21, 2)
C. (16, 2) and (21, –11)
B. (2, 16) and (21, –11)
D. (16, 2) and (–11, 21)
what will be common about the graphs of these equations.
i) y = 2x – 6
iii) y = – 5x – 6
ii) y = –3x – 6
iv) y = 5x – 6
A. All the graphs will have the same
y-intercept.
B. All the graphs will have the same
x-intercept.
C. All the graphs will have the same slope.
D. None of the above.
28
104.
(1 point) Which
2
graph represents the equation y = − x + 1?
5
A.
C.
B.
D.
29
105.
(1 point) Write
an equation to describe this graph.
A. d = 2t
B. d = −2t
106.
107.
C. d = −2
D. d = 2
(1 point) Write
an equation for the graph of a linear function that has slope 1 and y-intercept 8.
A. y = x + 8
C. y = 8x + 1
B. y = −8x + −1
D. y = −x − 8
(1 point) Which
equations represent perpendicular lines?
A. y = 6x − 7, y = 6x + 7
B. y = −7x + 11, y =
108.
1
x+6
7
C. y = 11x − 7, y = 11x +
D. y =
(1 point) Write
1
7
1
x + 6, y = 6x + 6
6
an equation for the graph of a linear function that has slope 8 and passes through R(4, −3).
A. y + 3 = −8(x − 4)
B. y + 3 = 8(x − 4)
1
C. y + 3 = (x − 4)
8
D. y − 3 = 8(x + 4)
30
109.
(1 point) Write
an equation in slope-point form for this line.
1
(x − 2)
2
1
B. y + 2 = − (x + 2)
2
A. y − 2 =
110.
(1 point) Distance
111.
(1 point) Write
112.
(1 point) Which
1
C. y − 2 = − (x − 2)
2
1
D. y + 2 = (x + 2)
2
travelled, d, is a linear function of time, t. After 75 min. a bus travelled 50 km. After
165 min. the bus travelled 110 km. Write an equation to represent this function.
2
3
A. d − 50 = (t − 75)
C. d − 50 = (t − 75)
3
2
2
2
B. d − 50 = − (t − 75)
D. d − 75 = (t − 50)
3
3
an equation in slope-point form for the line that passes through A(1, 4) and B(6, 8).
4
4
A. y + 8 = (x − 1)
C. y − 4 = (x − 1)
5
5
4
4
B. y + 4 = − (x − 1)
D. y − 8 = − (x + 1)
5
5
equation is written in general form?
A. −4x − 12y + 15 = 0
C. 12x = 4y − 15
1
B. 12x − 4y + 15 = 0
D.
x − 4y − 12 = 0
15
31
113.
114.
(1 point) Which
A.
C.
B.
D.
(1 point) Determine
A. –
B.
115.
graph represents the equation 4x − 5y − 20 = 0?
7
3
3
7
the slope of the line with this equation: 7x + 3y + 5 = 0
7
C.
3
3
D. –
7
(1 point) A
line has x-intercept –9 and y-intercept 3. Determine the equation of the line in general form.
A. 3x + 9y − 27 = 0
C. 3x − 9y + 27 = 0
B. 3x − 9y − 27 = 0
116.
D. 3x + 9y + 27 = 0
(1 point) Create
a linear system to model this situation:
The perimeter of an isosceles triangle is 36 cm. The base of the triangle is 9 cm longer than each equal
side.
A. s + b = 36
B. 2s + b = 36
C. 2b + s = 36
D. 2s + b = 36
b–9=s
b+9=s
s+9=b
s+9=b
32
117.
(1 point) Create
118.
(1 point) Write
a linear system to model this situation:
Cheri operates a grass-cutting business. She charges $19 for a small lawn and $29 for a large lawn. One
weekend, Cheri made $287 by cutting 13 lawns.
A. s + l = 13
C. s + l = 13
19s + 29l = 287
29s + 19l = 287
B. s + l = 287
D. s + l = 287
19s + 29l = 13
29s + 19l = 13
a linear system to model this situation. Then verify which of the given solutions is correct.
A crate of 32 grapefruit has a total mass of 4.648 kg.
When 9 grapefruit are removed, the total mass is 3.622 kg.
Verify the mass of the crate and the average mass of one grapefruit.
A. c + 32g = 4648
B. c + g = 32
C. c + 6g = 32
D. c + g = 32
c + 23g = 3622
c − 8 = 5g
c − 8g = 3622
c − 5g = 8
i) The crate has a mass of 1 kg, and the mass of one grapefruit is 114 g.
ii) The crate has a mass of 1.2 kg, and the mass of one grapefruit is 114.2 g.
iii)
The crate has a mass of 1 kg, and the mass of one grapefruit is 114.2 g.
iv) The crate has a mass of 1.2 kg, and the mass of one grapefruit is 57 g.
A. Part A-i
B. Part C-ii
C. Part B-iii
D. Part D-iv
33
119.
(1 point) Which
graph represents the solution of the linear system:
y = –2x + 2
y + 6 = 2x
A. Graph B
B. Graph A
C. Graph C
D. Graph D
34
120.
(1 point) Use
the graph to approximate the solution of the linear system:
y = −5x − 2
y = 5x − 4
A. (–3, 0.2)
B. (0, –2.8)
121.
(1 point) Express
C. (0.2, –3)
D. (–2.8, 0)
each equation in slope-intercept form.
6
y = –83
12
12x + 4y = –1772
x+
443
x −166
3
1
y = −3x +
3
B. y = −2x −166
443
y = −2x –
3
A. y =
122.
(1 point) Use
C. y = −2x −166
y = −3x
D. y = −2x −166
y = −3x −443
the table of values to determine the solution of this linear system:
4x + y = 3
2x + y = −5
A. (–13, –13)
B. (4, –13)
C. (–13, 4)
D. (4, 4)
35
123.
(1 point) Use
substitution to solve this linear system.
x = 2y – 56
5x + 13y = 410
A. (4, –30)
124.
B. (–4, 30)
(1 point) Identify
two like terms and state how they are related.
5
7
B. 8x and –96; by a factor of −12
D. 8x and 7x; by a factor of
1
2
7
8
each equation, identify a number you could multiply each term by to ensure that the
coefficients of the variables and the constant term are integers.
5
1
47
(1)
x+ y=
4
6
12
4
6
(2)
x – y = 16
5
7
Multiply
Multiply
Multiply
Multiply
equation
equation
equation
equation
(1)
(1)
(1)
(1)
by
by
by
by
35; multiply equation (2) by 12.
12; multiply equation (2) by 35.
2; multiply equation (2) by 3.
3; multiply equation (2) by 2.
(1 point) Write
3
x + 3y =
7
5
x + 5y =
6
an equivalent system with integer coefficients.
438
7
310
3
A. 3x + 21y = 438
5x + 30y = 620
B. 21x + 3y = 438
5x + 30y = 620
127.
C. 8x and –4y; by a factor of −
(1 point) For
A.
B.
C.
D.
126.
D. (–4, –30)
8x − 4y = −96
7x − 5y = −114
A. 7x and –5y; by a factor of −
125.
C. (4, 30)
C. 3x + 21y = 438
30x + 5y = 620
D. 3x + 21y = 1
5x + 30y = 1
(1 point) Write
an equivalent system with integer coefficients.
3
5x + y = 14
2
5
755
x + 5y =
6
6
A. 10x + 3y = 1
5x + 30y = 1
B. 3x + 10y = 28
5x + 30y = 755
C. 10x + 3y = 28
30x + 5y = 755
D. 10x + 3y = 28
5x + 30y = 755
36
128.
(1 point) Write
A. 8x + 6y = 20 and 8x − 6y = 12
C. 8x + 24y = 20 and 8x − 6y = 12
B. 24x + 8y = 20 and 6x + 8y = 12
D. 12x + 24y = 20 and 12x − 6y = 12
129.
(1 point) Use
130.
(1 point) Model
131.
(1 point) Write
132.
(1 point) Use
133.
an equivalent linear system where both equations have the same x-coefficients.
2x + 6y = 5
8x − 6y = 12
an elimination strategy to solve this linear system.
2
3
m + n = 16
3
4
1
3
− m + n = 18
2
8
A. m = −12 and n = 32
C. m = 12 and n = −32
201
52
32
B. m =
and n =
D. m = −12 and n =
10
15
3
this situation with a linear system:
Frieda has a 13% silver alloy and a 31% silver alloy. Frieda wants to make 26 kg of an alloy that is 47%
silver.
A. s + t = 0.47 and 0.13s + 0.31t = 26
C. s + t = 26 and 0.13s + 0.31t = 0.47
B. s + t = 47 and 0.13s + 0.31t = 26
D. s + t = 26 and 0.13s + 0.31t = 12.22
an equivalent linear system where both equations have the same a-coefficients.
0.3a + 0.2b = 4
0.2a − 0.3b = 1
A. 0.06a + 0.04b = 0.3 and 0.06a − 0.09b = 0.8
B. 0.06a − 0.04b = 0.8 and 0.06a − 0.09b = 0.3
C. 0.06a + 0.04b = 0.3 and 0.06a + 0.09b = 0.8
D. 0.06a + 0.04b = 0.8 and 0.06a − 0.09b = 0.3
an elimination strategy to solve this linear system.
20x − 24y = −52
8x + 32y = 104
A. x = −1 and y = −3
C. x = 1 and y = −3
B. x = 3 and y = 1
D. x = 1 and y = 3
(1 point) Which
linear system is modelled by these balance scales? (Each small square on the right side of
the balance scales represents 2 kg.)
A. 2x + y = 14 and x + 3y = 12
C. 2x + y = 6 and x + 3y = 7
B. 2x + y = 7 and x + 3y = 6
D. x + 2y = 14 and 3x + y = 12
37
134.
(1 point) Without
graphing, determine the equation whose graph intersects the graph of –6x + 3y = 11
exactly once.
i ) –6x + 3y = 13
ii) –24x + 12y = 44
iii) –4x + 3y = 11
A.
135.
B. none
iii
(1 point) Determine
C. ii
D. i
the number of solutions of the linear system:
2x – 5y = 23
–6x + 15y = 21
A. one solution
B. no solution
C. two solutions
D. infinite solutions
Short Answer
136.
(1 point) Convert
137.
(1 point) On
138.
(1 point) A
window is 35 in. high. Convert this height to the nearest centimetre.
139.
(1 point) A
right cone has a slant height of 14 in. and a base diameter of 10 in. Determine the surface area
5 yd. 6 in. to inches.
a map of British Columbia, the distance between Vancouver and Squamish is 52 km. Convert
this distance to the nearest mile.
of the cone to the nearest square inch.
38
140.
(1 point) A
141.
(1 point) In
142.
(1 point) A
143.
(1 point) A
144.
(1 point) Determine
right square pyramid has a height of 15 cm and a slant height of 17 cm. Determine the side
length of the base of the pyramid to the nearest centimetre.
2008, the Queen Sesheshet Pyramid was discovered in Egypt. Archeologists determined that
the original height of this right square pyramid was about 14 m and the original base side length was about
22 m. Determine its original volume to the nearest cubic metre.
regular tetrahedron has base area 98.9 m2 and height 8.6 m. Determine its volume to the
nearest tenth of a cubic metre.
right rectangular pyramid has base dimensions 11 cm by 7 cm and height 9 cm. Determine the
volume of the pyramid to the nearest cubic centimetre.
the surface area of this sphere to the nearest square inch. Determine its volume to the
nearest cubic inch.
39
145.
(1 point) A
146.
(1 point) A
147.
(1 point) Determine
148.
(1 point) A
149.
(1 point) Each
hemisphere has radius 12 m. Determine the volume of the hemisphere to the nearest tenth of a
cubic metre.
spherical globe has diameter 41.3 cm. What is the volume of the globe to the nearest tenth of a
centimetre?
the volume of this composite object, which is a right square prism and a right
rectangular pyramid, to the nearest tenth of a cubic metre.
pencil has a cylindrical body with a cone-shaped end. The cylinder is 5 cm long with a radius of
0.29 cm. The cone has a slant height of 1 cm and has the same radius as the cylinder. Determine the
surface area of the pencil to the nearest tenth of a square centimetre.
layer of a three-layer wedding cake is a cylinder with height 8 cm. The bottom layer has
diameter 24 cm, the middle layer has diameter 19 cm, and the top layer has diameter 14 cm. The cake is
covered in frosting. Determine the area of frosting to the nearest square centimetre.
40
150.
(1 point) Determine
151.
(1 point) Determine
the surface area of this composite object, which is a right triangular prism and a right
cylinder, to the nearest square inch.
the angle of inclination of the line to the nearest tenth of a degree.
41
152.
(1 point) a)
153.
(1 point) A
154.
(1 point) On
For ∠M in the triangle below, label the hypotenuse and the opposite and adjacent sides.
b) Determine tan M to the nearest hundredth.
rectangle has length 19.0 cm. The angle between one shorter side of the rectangle and a
diagonal is 56°. Calculate the width of the rectangle, to the nearest tenth of a centimetre.
a clinometer, how does the acute angle between the thread and the straw relate to the angle of
inclination of the straw?
42
155.
(1 point) A
156.
(1 point) A
157.
(1 point) A
158.
(1 point) A
159.
(1 point) A
160.
(1 point) A
161.
(1 point) Determine
ladder is 7 m long. It leans against a house. The base of the ladder is 2 m from the house. What
is the angle of inclination of the ladder to the nearest tenth of a degree?
ski jump is 116 m long. It has a vertical rise of 54 m. What is the angle of inclination of the
jump to the nearest tenth of a degree?
coast guard patrol boat is at point G, which is 7.8 km south of Smuggler’s Cove, S. A sailboat in
distress is at D, which is due west of Smuggler’s Cove. The patrol boat travels 10.1 km directly to the
sailboat. What is the angle between true north and the patrol boat’s path to the nearest tenth of a
degree?
tree is supported by a guy wire. The wire is anchored to the ground 7.0 m from the base of the
tree. The angle of inclination of the wire is 65°. Calculate the length of the wire to the nearest tenth of a
metre.
diagonal in a rectangle has length 14 in. The angle between a diagonal and the longer side of the
rectangle is 22° . Calculate the width of the rectangle to the nearest inch.
guy wire supports a flagpole. The wire is anchored 8 ft. from the base of the pole and the angle
of inclination of the wire is 47°. Calculate the length of the wire to the nearest foot.
the perfect square whole number closest to 54 362.
43
162.
(1 point) Write
163.
(1 point) Identify
2
an expression for the width of this rectangle.
the greatest common factor of the terms in this set.
2
8x y, 24y , 18xy
164.
(1 point) Suppose
2
you must use 1 x -tile and 10 x-tiles. Which numbers of 1-tiles could you use to form a
rectangle?
165.
(1 point) Expand
and simplify: (12 + q)(2 − q)
166.
(1 point) Factor:
s − 33s + 32
167.
(1 point) Expand
and simplify: (11t + 2)(4t − 3)
2
44
168.
(1 point) Find
and correct the error(s) in this solution of factoring by decomposition.
2
2
90y + 77y − 52 = 90y + 117y − 40y − 52
= 9y(10y + 13) + 4(10y + 13)
= (10y + 13)(9y + 4)
169.
(1 point) Expand
170.
(1 point) Find
2
and simplify: (9z − 2z + 10)(3z + 12)
an integer to replace
2
121x − 308xy +
171.
(1 point) Estimate
172.
(1 point) Which
3.12, –4,
173.
y
so that the trinomial is a perfect square.
2
the value of
35 to one decimal place.
of these numbers are rational numbers, but not integers?
3
4
1 4
5 , , 2.4, −8 , 0, 5 , 16
7
2
2
(1 point) Determine
the side length of a square with area 72 cm .
Write your answer to the nearest tenth of a centimetre.
45
174.
(1 point) Determine
175.
(1 point) Write
176.
(1 point) Arrange
the edge length of a cube with volume 55 cm3 .
Write your answer to the nearest tenth of a centimetre.
9
7
12 ,
6
5
7
28 125 in simplest form.
these numbers in order from least to greatest.
1
9
1
7
12 , 12 , 12 ,
ÊÁ 8
177. (1 point) Evaluate ÁÁÁÁ
ÁË 27
178.
(1 point) Evaluate
179.
(1 point) Simplify
7
12
6
2
ˆ˜ − 3
˜˜
without using a calculator.
˜˜
˜¯
( 0.027 )
−
1
3
without using a calculator.
−3a b c
−3
−7
−6
−6
−3
−3
12a b c
. Write using powers with positive exponents.
46
180.
(1 point) Simplify
ÊÁ 3 −3 −7 −2 ˆ˜ −4
ÁÁ m n p ˜˜ .
ÁÁ
˜˜
ÁË 4
˜¯
47
181.
(1 point) Different
coloured game pieces can be associated with their lengths, in centimetres. Consider the
relation represented by this arrow diagram. Represent the relation as a graph.
48
182.
(1 point) Countries
can be associated with the year they won gold in Olympic Men’s Ice Hockey. Consider
the relation represented by this table. Describe this relation in words.
Country
United States
Soviet Union
Soviet Union
CIS
Sweden
Czech Republic
Canada
Sweden
Canada
183.
(1 point) For
Year
1980
1984
1988
1992
1994
1998
2002
2006
2010
the function f ( x) = 4x − 7, determine f(−7.5).
49
184.
(1 point) The
graph shows the speed of a windsurfer as a function of time.
a) For how long did the windsurfer travel at a speed of 45 km/h?
b) How long did the windsurfer’s ride last?
185.
(1 point) This
table shows the refund, r dollars, for different numbers of pop cans, n. Write the domain and
range.
Number of Pop
Cans, n
9
13
16
24
33
Refund, r ($)
0.45
0.65
0.80
1.20
1.65
50
186.
(1 point) Explain
187.
(1 point) This
why the points on this graph are not joined.
1
is a graph of the function h ( x) = − 2 x + 1.
a) Determine the range value when the domain value is –2.
b) Determine the domain value when the range value is –1.
51
188.
(1 point) This
table represents the approximate relation between a distance in miles and the same distance
in kilometres. Determine the rate of change of the relation..
Miles (mi.)
Kilometres
(km)
189.
9
14.4
18
28.8
(1 point) This
27
43.2
36
57.6
45
72.0
graph shows the volume of gasoline left in a car’s tank, v litres, as a function of the distance
travelled, d in hundreds of kilometres. Determine the domain and range of the graph.
52
190.
(1 point) This
191.
(1 point) Determine
192.
(1 point) The
graph shows the cost, C dollars, of printing an advertising flyer for the school play as a
function of the number of flyers printed, n. What is the cost when 1000 flyers are printed?
the slope of this line segment.
slopes of two lines are
6
6
and
. Are the two lines parallel, perpendicular, or neither?
11
11
53
1
. Are the two lines parallel, perpendicular, or neither?
2
193.
(1 point) The
194.
(1 point) Describe
the graph of the linear function whose equation is y = 3x +
195.
(1 point) Describe
the graph of the linear function whose equation is y =
slopes of two lines are −2 and
54
3
.
2
3
x + 7.
2
196.
(1 point) Graph
the line with y-intercept 3 and slope –2.
55
197.
(1 point) Write
198.
(1 point) Describe
199.
(1 point) Write
200.
(1 point) Desmond
an equation in slope-point form for this line.
the graph of the linear function with this equation: y + 7 = −8(x + 6)
this equation in general form: y − 5 =
3
(x + 5)
5
works as a babysitter for two families. Family A pays $5.75 per hour. Family B pays
$7.5 per hour. Last weekend, Desmond earned $75. Write an equation in general form for the relation.
56
201.
(1 point) Create
a linear system to model this situation:
Two ships start sailing towards each other at the same time from two islands that are 365 km apart. One
ship travels 5 km/h faster than the other. They meet in 5 h. What is the average speed of each ship?
Verify that 34 km/h and 39 km/h represent the solution of the linear system.
57
202.
(1 point) Solve
this linear system by graphing.
–3x – 2y = 16
–x + y = –8
58
59
203.
(1 point) A
submarine cruises underwater at 20 km/h and on the surface at 30 km/h. The submarine travels a
distance of 650 km in 25 h. A linear system that models this situation is:
u + s = 25
20u + 30s = 650
where u represents the time in hours cruising underwater, and s represents the time in hours cruising on
the surface.
a) Graph the linear system above.
b) Use the graph to solve the problem:
How long did the submarine travel underwater, and how long did it travel on the surface?
60
61
204.
(1 point) a)
Write a linear system to model this situation:
Angela is 24 years older than her cousin Zack. In 13 years, she will be double his age.
b) Use a graph to solve this problem:
How old are Angela and Zack now?
62
63
205.
(1 point) a)
Write a linear system to model this situation:
A hockey coach bought 25 pucks for a total cost of $70. The pucks used for practice cost
$2.50 each, and the pucks used for games cost $3.25 each.
b) Use a graph to solve this problem:
How many of each type of puck did the coach purchase?
64
206.
(1 point) Use
graphing technology to solve this linear system.
Where necessary, write the coordinates to the nearest tenth.
3
x + y = –3
2
6
x + 7y = –8
7
65
207.
(1 point) Fill
in the each blank below with the correct integer.
System A
System B
ÊÁ 56x + 48y = −3008 ˆ˜ −____:
7x + 6y = –376
Ë
¯
–4x – 6y = 256
ÊÁ 32x + 48y = −2048 ˆ˜ −____:
Ë
¯
208.
(1 point) Use
209.
(1 point) Model
210.
(1 point) Determine
substitution to solve this linear system:
7
x + y = –34
8
–3x + 4y = –4
this situation with a linear system:
A recycling depot pays 0.06¢ for a small can and 0.23¢ for a large can. Chara took 70 cans to the
recycling depot and her total refund was $22.35.
the number of solutions of this linear system.
15x + 30y = –240
17x + 21y = 53
66
Problem
211.
(1 point) Explain
how to convert a measurement of 20 000 ft. to miles, yards, and feet.
67
212.
(1 point) A
right cylinder has base radius 22.9 cm and height 17.1 cm. Determine the volume of a right
cone with the same base and the same height, to the nearest tenth of a cubic centimetre.
68
213.
(1 point) Determine
the surface area of this composite object, which is a right square prism and a right
square pyramid, to the nearest square foot. Explain your answer.
69
70
214.
(1 point) A
solid sphere just fits inside a cube that has an edge length equal to the diameter of the sphere.
The edge length of the cube is 4.9 cm. What is the volume of air in the cube to the nearest cubic
centimetre?
71
72
215.
(1 point) A
sculpture comprises a right rectangular prism with base dimensions 29 m by 33 m, and height 15
m. A right cylinder with base diameter 7 m and height 14 m sits on top of the prism.
a) Determine the volume of the sculpture to the nearest cubic metre.
b) Determine the surface area of the sculpture to the nearest square metre.
73
74
216.
(1 point) The
approximate latitudes for several cities in western and northern Canada are shown.
Calgary
51.1° Dawson Creek 55.8° Edmonton 53.6°
Fort Nelson
58.8° Inuvik
68.3° Kelowna
49.9°
Prince Rupert 54.3° Regina
50.5° Saskatoon 52.1°
Vancouver
49.3° Victoria
48.4° Whitehorse 60.7°
For which locations might the following roof design be within 1° of the recommended angle for solar
panels? Justify your answer.
75
217.
(1 point) Three
squares with side length 9 mm are placed side-by-side as shown. Thomas says ∠ACB is
approximately 71.6°.
a) Is he correct? Justify your answer.
b) Describe what the value of tan C indicates.
76
77
218.
(1 point) a)
In ΔBCD, identify the side opposite ∠D and the side adjacent to ∠D.
b) Determine sin D to the nearest tenth. Describe what the value of sin D indicates.
c) Determine the measure of ∠D to the nearest tenth of a degree.
78
219.
(1 point) A
cone is formed by cutting out the shape below and joining the straight edges with tape. Calculate
the angle of inclination of the side of the cone to the nearest tenth of a degree.
79
80
220.
(1 point) Determine
the length of the diagonals in this kite to the nearest tenth of a centimetre.
81
82
83
221.
(1 point) A
rectangle is divided into 2 smaller rectangles. The area of the rectangle on the left is 209 square
inches, and the area of the rectangle on the right is 319 square inches. Determine the greatest possible
measure of the side that the two rectangles share.
84
222.
(1 point) A
math textbook has 6 chapters of equal length. The textbook is made by stitching together
40-page booklets. Determine the fewest number of pages the book can have.
85
223.
(1 point) Jordan
wants to cut a rectangular carpet with dimensions 32 cm by 80 cm into squares of equal
size.
a) What is the side length of the largest possible square Jordan can cut?
b) How many squares can she cut from carpet?
86
87
224.
(1 point) A
cube has surface area 2646 m2 . What is its volume?
88
225.
(1 point) a)
Here are a student’s solutions for factoring polynomials. Identify the errors in each
solution. Write a correct solution.
2
3
i) Factor: 15s − 35s + 5s
2
3
2
Solution: 15s − 35s + 5s = 5s(3s − 7s )
2
3
ii) Factor: −22h − 32h + 16h
2
3
2
Solution: −22h − 32h + 16h = −2h(11 + 16h + 8h )
b) What should the student have done to check her work?
89
226.
(1 point) Harish
3
1024 =
=
3
3
simplified
8 ⋅
8 ⋅
= 2⋅5⋅
3
3
3
3
1024 as shown:
128
125 ⋅
3
3
3
3
= 10 ⋅ 3
Identify the error Harish made, then write a correct solution.
90
227.
(1 point) Another
2
5
formula for the approximate surface area, SA square metres, of a person’s body is
1
2
SA = 0.025h m , where h is the person’s height in centimetres, and m is the person’s mass in
kilograms.
a) Calculate the surface area of a newborn with height 48 cm and mass 7.3 kg. Write the answer as a
decimal to the nearest hundredth of a square centimetre.
b) Calculate the surface area of a person with height 170 cm and mass 66 kg. Write the
answer as a decimal to the nearest hundredth of a square centimetre.
91
228.
(1 point) A
2
tree farmer used the formula V = 0.5d h to estimate the volume, V cubic metres, of a tree with
height h metres and mean trunk diameter d metres. The height of a tree is 20 times its mean trunk
3
diameter, and its volume is 230 m . What is the mean trunk diameter of this tree to the nearest metre?
92
229.
(1 point) Identify
any errors in each simplification. Write a correct solution.
ÁÊÁ − 1 ˜ˆ˜
1
−
Ê −6 6 ˜ˆ ÁÁÁ 6 5 ˜˜˜
6
−6
6
5
Á
a) ÁÁÁ x y ˜˜˜ ÁÁÁ x y ˜˜˜ = x ⋅ x ⋅ y ⋅ y
Á
˜
Ë
¯ ÁÁ
˜˜
ÁË
˜¯
1
= x ⋅y
= xy
ÊÁ
1
ÁÁ
ÁÁ
ÁÁ 2m 4
b) ÁÁÁ 4
ÁÁ n
ÁÁ
ÁÁ
Ë
30
30
ˆ˜ −4
˜˜
˜˜
−1
˜˜
˜˜ = − 8m
˜˜
0
˜˜
n
˜˜
˜
¯
= −8m
1
=
8m
−1
93
94
230.
(1 point) Use
ÊÁ 8 ˆ˜ ÁÊÁ 5 3 ˜ˆ˜
exponent laws to simplify ÁÁÁ x ˜˜˜ ÁÁÁ x ˜˜˜ . Explain your strategy.
ÁË
˜¯ Á
˜
Ë
¯
95
231.
(1 point) Consider
the relation represented by this arrow diagram.
a) Represent the relation as a set of ordered pairs.
b) Does the order of the numbers in each ordered pair matter? Explain.
96
232.
(1 point) This
table contains information about a women’s hockey team. Use two columns in this table to
represent a relation.
a) Name two relations that are functions.
b) Name two relations that are not functions.
Justify your answers.
Team Member
Julie
Hayley
Cassie
Jennifer
Marie
Meaghan
Angela
Kim
Age
15
16
16
15
17
15
16
17
Position
Right Wing
Center
Left Wing
Left Defence
Right Wing
Right Defence
Left Wing
Center
97
Points
36
43
38
17
42
19
45
37
98
233.
(1 point) The
equation C = 11g + 250 represents the total cost, C dollars, for a sports banquet when g
people attend.
a) Describe the function.
Write the function in function notation.
b) Determine C(46).
What does this number represent?
c) Determine the value of g when C(g) = 1581.
What does this number represent?
99
100
234.
(1 point) This
graph represents the relation between the distance a vehicle travels and the number of
revolutions of a tire. An equation for the distance travelled, d metres, after r revolutions of the tire is
d = 2.07r.
a) Identify the dependent and independent variables.
b) Does the graph represent a linear relation? How do you know?
c) Describe another strategy you could use to determine whether this relation is linear.
101
235.
(1 point) This
graph shows the length, l metres, of an object’s shadow as a function of the height of the
object, h metres.
a) What is the rate of change? What does it represent?
b) A tree has height 13 m. About how long is its shadow?
c) The length of the shadow of a building is 45 m. About how tall is the building?
102
103
236.
(1 point) a)
Determine the rise, run, and slope of this line segment.
b) Draw a line segment whose slope is:
i) zero
ii) not defined
iii) the same as the slope of the line segment in part a
104
105
237.
(1 point) Four
students determined the slope of the line through S(7, –5) and T(–15, 11). Their answers
11
11 8
8
were:
,− ,
, and − .
8
8 11
11
Which answer is correct? How do you know?
106
238.
(1 point) Given
A(18, 9), B(6, 27), and C(6, 9), determine the coordinates of point D such that CD is
parallel to AB and D is on the:
i) y-axis
ii) x-axis
107
108
239.
(1 point) Describe
the graph of the linear function whose equation is y =
Draw this graph without using technology.
109
2
x + 2.
5
240.
(1 point) Students
at Tahayghen Secondary School sell punch during the school carnival. The number of
cups sold, n, is a linear function of the temperature in degrees Celsius, t. The students sold 458 cups when
the temperature was 25°C. They sold 534 cups when the temperature was 29°C.
a) Write an equation in slope-point form to represent this function.
b) Use the equation in part a to determine the approximate temperature when the students sell 325 cups
of punch.
110
111
241.
(1 point) a)
Write a linear system to model this situation:
The coin box of a vending machine contains $23.75 in quarters and loonies. There are 35 coins in
all.
b) Use a graph to solve this problem:
How many of each coin are there in the coin box?
112
113
242.
(1 point) Sales
clerks at an appliance store have a choice of two methods of payment:
Plan A: $580 every two weeks plus 4.2% commission on all sales
Plan B: $880 every two weeks plus 1.2% commission on all sales
a) Write a linear system to model this situation.
b) Graph the linear system in part a.
c) Use the graph to solve this problem:
What must the sales for a two-week period be for a clerk to receive the same salary with both plans?
114
115
116
243.
(1 point) a)
Write a linear system to model the situation:
For the school play, the cost of one adult ticket is $6 and the cost of one student ticket is $4. Twice
as many student tickets as adult tickets were sold. The total receipts were $2016.
b) Use substitution to solve the related problem:
How many of each type of ticket were sold?
117
118
244.
(1 point) Use
an elimination strategy to solve this linear system. Verify the solution.
20x + 35y = 705
10x − 5y = 195
119
120
245.
(1 point) Determine
the number of solutions of this linear system.
4x + 12y = 28
8x + 24y = 48
121
122
ID: A
Gr 10 IAP Year End Review
Answer Section
MULTIPLE CHOICE
1. ANS: B
PTS: 1
DIF: Easy
REF: 1.1 Imperial Measures of Length
LOC: 10.M2
TOP: Measurement
KEY: Procedural Knowledge
2. ANS: A
PTS: 1
DIF: Moderate
REF: 1.1 Imperial Measures of Length
LOC: 10.M2
TOP: Measurement
KEY: Procedural Knowledge
3. ANS: B
PTS: 1
DIF: Moderate
REF: 1.1 Imperial Measures of Length
LOC: 10.M2
TOP: Measurement
KEY: Procedural Knowledge
4. ANS: D
PTS: 1
DIF: Moderate
REF: 1.1 Imperial Measures of Length
LOC: 10.M2
TOP: Measurement
KEY: Procedural Knowledge
5. ANS: B
PTS: 1
DIF: Moderate
REF: 1.1 Imperial Measures of Length
LOC: 10.M2
TOP: Measurement
KEY: Procedural Knowledge
6. ANS: C
PTS: 1
DIF: Easy
REF: 1.2 Measuring Length and Distance
LOC: 10.M1
TOP: Measurement
KEY: Conceptual Understanding
7. ANS: C
PTS: 1
DIF: Easy
REF: 1.2 Measuring Length and Distance
LOC: 10.M1
TOP: Measurement
KEY: Conceptual Understanding
8. ANS: A
PTS: 1
DIF: Easy
REF: 1.2 Measuring Length and Distance
LOC: 10.M1
TOP: Measurement
KEY: Conceptual Understanding
9. ANS: D
PTS: 1
DIF: Easy
REF: 1.2 Measuring Length and Distance
LOC: 10.M1
TOP: Measurement
KEY: Conceptual Understanding
10. ANS: A
PTS: 1
DIF: Easy
REF: 1.2 Measuring Length and Distance
LOC: 10.M1
TOP: Measurement
KEY: Conceptual Understanding
11. ANS: B
PTS: 1
DIF: Easy
REF: 1.2 Measuring Length and Distance
LOC: 10.M1
TOP: Measurement
KEY: Conceptual Understanding
1
ID: A
12. ANS: C
PTS: 1
DIF: Easy
REF: 1.2 Measuring Length and Distance
LOC: 10.M1
TOP: Measurement
KEY: Conceptual Understanding
13. ANS: A
PTS: 1
DIF: Easy
REF: 1.3 Relating SI and Imperial Units
LOC: 10.M2
TOP: Measurement
KEY: Procedural Knowledge
14. ANS: D
PTS: 1
DIF: Moderate
REF: 1.3 Relating SI and Imperial Units
LOC: 10.M2
TOP: Measurement
KEY: Procedural Knowledge
15. ANS: D
PTS: 1
DIF: Easy
REF: 1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Procedural Knowledge
16. ANS: A
PTS: 1
DIF: Moderate
REF: 1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Procedural Knowledge
17. ANS: D
PTS: 1
DIF: Moderate
REF: 1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Procedural Knowledge
18. ANS: A
PTS: 1
DIF: Moderate
REF: 1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Procedural Knowledge
19. ANS: C
PTS: 1
DIF: Moderate
REF: 1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Procedural Knowledge
20. ANS: A
PTS: 1
DIF: Moderate
REF: 1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Procedural Knowledge
21. ANS: D
PTS: 1
DIF: Moderate
REF: 1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Procedural Knowledge
22. ANS: D
PTS: 1
DIF: Moderate
REF: 1.5 Volumes of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Procedural Knowledge
23. ANS: C
PTS: 1
DIF: Moderate
REF: 1.6 Surface Area and Volume of a Sphere
LOC: 10.M3
TOP: Measurement
KEY: Procedural Knowledge
24. ANS: A
LOC: 10.M4
PTS: 1
DIF: Easy
TOP: Measurement
2
REF: 2.1 The Tangent Ratio
KEY: Procedural Knowledge
ID: A
25. ANS: C
LOC: 10.M4
PTS: 1
DIF: Moderate
TOP: Measurement
REF: 2.1 The Tangent Ratio
KEY: Procedural Knowledge
26. ANS: A
PTS: 1
DIF: Easy
REF: 2.2 Using the Tangent Ratio to Calculate Lengths LOC: 10.M4
TOP: Measurement
KEY: Procedural Knowledge
27. ANS: B
PTS: 1
DIF: Easy
REF: 2.2 Using the Tangent Ratio to Calculate Lengths LOC: 10.M4
TOP: Measurement
KEY: Procedural Knowledge
28. ANS: C
PTS: 1
DIF: Moderate
REF: 2.2 Using the Tangent Ratio to Calculate Lengths LOC: 10.M4
TOP: Measurement
KEY: Procedural Knowledge
29. ANS: D
LOC: 10.M4
PTS: 1
DIF: Easy
TOP: Measurement
REF: 2.4 The Sine and Cosine Ratios
KEY: Procedural Knowledge
30. ANS: C
LOC: 10.M4
PTS: 1
DIF: Moderate
TOP: Measurement
REF: 2.4 The Sine and Cosine Ratios
KEY: Procedural Knowledge
31. ANS: B
PTS: 1
DIF: Easy
REF: 2.5 Using the Sine and Cosine Ratios to Calculate Lengths
LOC: 10.M4
TOP: Measurement
KEY: Procedural Knowledge
32. ANS: C
PTS: 1
DIF: Easy
REF: 2.6 Applying the Trigonometric Ratios
LOC: 10.M4
TOP: Measurement
KEY: Procedural Knowledge
33. ANS: A
PTS: 1
DIF: Easy
REF: 2.6 Applying the Trigonometric Ratios
LOC: 10.M4
TOP: Measurement
KEY: Procedural Knowledge
34. ANS: C
PTS: 1
DIF: Moderate
REF: 2.6 Applying the Trigonometric Ratios
LOC: 10.M4
TOP: Measurement
KEY: Procedural Knowledge
35. ANS: A
PTS: 1
DIF: Moderate
REF: 2.6 Applying the Trigonometric Ratios
LOC: 10.M4
TOP: Measurement
KEY: Procedural Knowledge
36. ANS: D
PTS: 1
DIF: Easy
REF: 2.7 Solving Problems Involving More than One Right Triangle
LOC: 10.M4
TOP: Measurement
KEY: Procedural Knowledge
37. ANS: A
PTS: 1
DIF: Easy
REF: 2.7 Solving Problems Involving More than One Right Triangle
LOC: 10.M4
TOP: Measurement
KEY: Procedural Knowledge
38. ANS: D
PTS: 1
DIF: Easy
REF: 2.7 Solving Problems Involving More than One Right Triangle
LOC: 10.M4
TOP: Measurement
KEY: Procedural Knowledge
3
ID: A
39. ANS: C
PTS: 1
DIF: Moderate
REF: 2.7 Solving Problems Involving More than One Right Triangle
LOC: 10.M4
TOP: Measurement
KEY: Procedural Knowledge
40. ANS: D
PTS: 1
DIF: Moderate
REF: 2.7 Solving Problems Involving More than One Right Triangle
LOC: 10.M4
TOP: Measurement
KEY: Procedural Knowledge
41. ANS: D
PTS: 1
DIF: Moderate
REF: 3.1 Factors and Multiples of Whole Numbers
LOC: 10.AN1
TOP: Algebra and Number
KEY: Procedural Knowledge
42. ANS: D
PTS: 1
DIF: Easy
REF: 3.1 Factors and Multiples of Whole Numbers
LOC: 10.AN1
TOP: Algebra and Number
KEY: Procedural Knowledge
43. ANS: B
PTS: 1
DIF: Moderate
REF: 3.2 Perfect Squares, Perfect Cubes, and Their Roots
LOC: 10.AN1
TOP: Algebra and Number
KEY: Procedural Knowledge
44. ANS: B
PTS: 1
DIF: Easy
REF: 3.2 Perfect Squares, Perfect Cubes, and Their Roots
LOC: 10.AN1
TOP: Algebra and Number
KEY: Procedural Knowledge
45. ANS: D
PTS: 1
DIF: Moderate
REF: 3.2 Perfect Squares, Perfect Cubes, and Their Roots
LOC: 10.AN1
TOP: Algebra and Number
KEY: Procedural Knowledge
46. ANS: B
PTS: 1
DIF: Moderate
REF: 3.2 Perfect Squares, Perfect Cubes, and Their Roots
LOC: 10.AN1
TOP: Algebra and Number
KEY: Procedural Knowledge
47. ANS: C
PTS: 1
DIF: Easy
REF: 3.3 Common Factors of a Polynomial
LOC: 10.AN5
TOP: Algebra and Number
KEY: Procedural Knowledge
48. ANS: B
PTS: 1
DIF: Easy
REF: 3.3 Common Factors of a Polynomial
LOC: 10.AN5
TOP: Algebra and Number
KEY: Procedural Knowledge
49. ANS: D
PTS: 1
DIF: Moderate
REF: 3.3 Common Factors of a Polynomial
LOC: 10.AN5
TOP: Algebra and Number
KEY: Procedural Knowledge
50. ANS: B
PTS: 1
DIF: Easy
REF: 3.3 Common Factors of a Polynomial
LOC: 10.AN5
TOP: Algebra and Number
KEY: Procedural Knowledge
51. ANS: C
PTS: 1
DIF: Moderate
REF: 3.3 Common Factors of a Polynomial
LOC: 10.AN5
TOP: Algebra and Number
KEY: Procedural Knowledge
4
ID: A
52. ANS: B
PTS: 1
DIF: Easy
REF: 3.4 Modelling Trinomials as Binomial Products
LOC: 10.AN5
TOP: Algebra and Number
KEY: Procedural Knowledge
53. ANS: D
PTS: 1
DIF: Moderate
REF: 3.5 Polynomials of the Form x^2 + bx + c
LOC: 10.AN5
TOP: Algebra and Number
KEY: Procedural Knowledge
54. ANS: B
PTS: 1
DIF: Easy
REF: 3.6 Polynomials of the Form ax^2 + bx + c
LOC: 10.AN5
TOP: Algebra and Number
KEY: Procedural Knowledge
55. ANS: C
PTS: 1
DIF: Moderate
REF: 3.6 Polynomials of the Form ax^2 + bx + c
LOC: 10.AN5
TOP: Algebra and Number
KEY: Procedural Knowledge
56. ANS: C
LOC: 10.AN4
PTS: 1
DIF: Easy
TOP: Algebra and Number
REF: 3.7 Multiplying Polynomials
KEY: Procedural Knowledge
57. ANS: D
LOC: 10.AN4
PTS: 1
DIF: Moderate
TOP: Algebra and Number
REF: 3.7 Multiplying Polynomials
KEY: Procedural Knowledge
58. ANS: D
PTS: 1
DIF: Easy
REF: 3.8 Factoring Special Polynomials
LOC: 10.AN5
TOP: Algebra and Number
KEY: Procedural Knowledge
59. ANS: B
LOC: 10.AN2
PTS: 1
DIF: Easy
TOP: Algebra and Number
REF: 4.1 Estimating Roots
KEY: Procedural Knowledge
60. ANS: B
LOC: 10.AN2
PTS: 1
DIF: Easy
TOP: Algebra and Number
REF: 4.1 Estimating Roots
KEY: Procedural Knowledge
61. ANS: A
LOC: 10.AN2
PTS: 1
DIF: Easy
TOP: Algebra and Number
REF: 4.1 Estimating Roots
KEY: Conceptual Understanding
62. ANS: B
LOC: 10.AN2
PTS: 1
DIF: Moderate
TOP: Algebra and Number
REF: 4.2 Irrational Numbers
KEY: Conceptual Understanding
63. ANS: A
LOC: 10.AN2
PTS: 1
DIF: Easy
TOP: Algebra and Number
REF: 4.2 Irrational Numbers
KEY: Procedural Knowledge
64. ANS: C
LOC: 10.AN2
PTS: 1
DIF: Easy
TOP: Algebra and Number
REF: 4.3 Mixed and Entire Radicals
KEY: Conceptual Understanding
65. ANS: D
LOC: 10.AN2
PTS: 1
DIF: Moderate
TOP: Algebra and Number
REF: 4.3 Mixed and Entire Radicals
KEY: Conceptual Understanding
66. ANS: C
LOC: 10.AN2
PTS: 1
DIF: Easy
TOP: Algebra and Number
REF: 4.3 Mixed and Entire Radicals
KEY: Conceptual Understanding
67. ANS: D
LOC: 10.AN2
PTS: 1
DIF: Easy
TOP: Algebra and Number
REF: 4.3 Mixed and Entire Radicals
KEY: Conceptual Understanding
5
ID: A
68. ANS: D
LOC: 10.AN2
PTS: 1
DIF: Easy
TOP: Algebra and Number
REF: 4.3 Mixed and Entire Radicals
KEY: Conceptual Understanding
69. ANS: B
PTS: 1
DIF: Easy
REF: 4.4 Fractional Exponents and Radicals
LOC: 10.AN3
TOP: Algebra and Number
KEY: Conceptual Understanding
70. ANS: C
PTS: 1
DIF: Easy
REF: 4.4 Fractional Exponents and Radicals
LOC: 10.AN3
TOP: Algebra and Number
KEY: Conceptual Understanding
71. ANS: B
PTS: 1
DIF: Easy
REF: 4.4 Fractional Exponents and Radicals
LOC: 10.AN3
TOP: Algebra and Number
KEY: Conceptual Understanding
72. ANS: D
PTS: 1
DIF: Moderate
REF: 4.4 Fractional Exponents and Radicals
LOC: 10.AN3
TOP: Algebra and Number
KEY: Conceptual Understanding
73. ANS: D
PTS: 1
DIF: Moderate
REF: 4.4 Fractional Exponents and Radicals
LOC: 10.AN3
TOP: Algebra and Number
KEY: Conceptual Understanding
74. ANS: C
PTS: 1
DIF: Moderate
REF: 4.4 Fractional Exponents and Radicals
LOC: 10.AN3
TOP: Algebra and Number
KEY: Conceptual Understanding
75. ANS: B
PTS: 1
DIF: Moderate
REF: 4.4 Fractional Exponents and Radicals
LOC: 10.AN3
TOP: Algebra and Number
KEY: Conceptual Understanding
76. ANS: C
PTS: 1
DIF: Moderate
REF: 4.5 Negative Exponents and Reciprocals
LOC: 10.AN3
TOP: Algebra and Number
KEY: Conceptual Understanding
77. ANS: B
PTS: 1
DIF: Moderate
REF: 4.6 Applying the Exponent Laws
LOC: 10.AN3
TOP: Algebra and Number
KEY: Conceptual Understanding
78. ANS: C
LOC: 10.RF4
PTS: 1
DIF: Easy
TOP: Relations and Functions
REF: 5.1 Representing Relations
KEY: Conceptual Understanding
79. ANS: D
LOC: 10.RF4
PTS: 1
DIF: Moderate
TOP: Relations and Functions
REF: 5.1 Representing Relations
KEY: Conceptual Understanding
80. ANS: C
PTS: 1
DIF: Easy
REF: 5.3 Interpreting and Sketching Graphs
LOC: 10.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding
81. ANS: D
PTS: 1
DIF: Easy
REF: 5.3 Interpreting and Sketching Graphs
LOC: 10.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding
6
ID: A
82. ANS: D
PTS: 1
DIF: Moderate
REF: 5.3 Interpreting and Sketching Graphs
LOC: 10.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding
83. ANS: C
PTS: 1
DIF: Easy
REF: 5.3 Interpreting and Sketching Graphs
LOC: 10.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding
84. ANS: A
PTS: 1
DIF: Easy
REF: 5.3 Interpreting and Sketching Graphs
LOC: 10.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding
85. ANS: B
PTS: 1
DIF: Easy
REF: 5.5 Graphs of Relations and Functions
LOC: 10.RF2
TOP: Relations and Functions
KEY: Conceptual Understanding
86. ANS: D
PTS: 1
DIF: Easy
REF: 5.5 Graphs of Relations and Functions
LOC: 10.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding
87. ANS: C
PTS: 1
DIF: Easy
REF: 5.5 Graphs of Relations and Functions
LOC: 10.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding
88. ANS: B
PTS: 1
DIF: Easy
REF: 5.5 Graphs of Relations and Functions
LOC: 10.RF5
TOP: Relations and Functions
KEY: Conceptual Understanding
89. ANS: B
PTS: 1
DIF: Easy
REF: 5.6 Properties of Linear Functions
LOC: 10.RF4
TOP: Relations and Functions
KEY: Procedural Knowledge
90. ANS: A
PTS: 1
DIF: Easy
REF: 5.6 Properties of Linear Functions
LOC: 10.RF3
TOP: Relations and Functions
KEY: Procedural Knowledge
91. ANS: D
PTS: 1
DIF: Easy
REF: 5.6 Properties of Linear Functions
LOC: 10.RF3
TOP: Relations and Functions
KEY: Procedural Knowledge
92. ANS: A
PTS: 1
DIF: Easy
REF: 5.7 Interpreting Graphs of Linear Functions
LOC: 10.RF5
TOP: Relations and Functions
KEY: Conceptual Understanding
93. ANS: D
PTS: 1
DIF: Easy
REF: 5.7 Interpreting Graphs of Linear Functions
LOC: 10.RF5
TOP: Relations and Functions
KEY: Conceptual Understanding
94. ANS: B
PTS: 1
DIF: Easy
REF: 5.7 Interpreting Graphs of Linear Functions
LOC: 10.RF5
TOP: Relations and Functions
KEY: Procedural Knowledge
7
ID: A
95. ANS: B
PTS: 1
DIF: Easy
REF: 5.7 Interpreting Graphs of Linear Functions
LOC: 10.RF5
TOP: Relations and Functions
KEY: Procedural Knowledge
96. ANS: A
PTS: 1
DIF: Easy
REF: 5.7 Interpreting Graphs of Linear Functions
LOC: 10.RF8
TOP: Relations and Functions
KEY: Conceptual Understanding
97. ANS: D
LOC: 10.RF5
PTS: 1
DIF: Easy
TOP: Relations and Functions
REF: 6.1 Slope of a Line
KEY: Procedural Knowledge
98. ANS: B
LOC: 10.RF5
PTS: 1
DIF: Moderate
TOP: Relations and Functions
REF: 6.1 Slope of a Line
KEY: Procedural Knowledge
99. ANS: C
PTS: 1
DIF: Easy
REF: 6.2 Slopes of Parallel and Perpendicular Lines
LOC: 10.RF3
TOP: Relations and Functions
KEY: Procedural Knowledge
100. ANS: A
PTS: 1
DIF: Easy
REF: 6.2 Slopes of Parallel and Perpendicular Lines
LOC: 10.RF3
TOP: Relations and Functions
KEY: Conceptual Understanding
101. ANS: D
PTS: 1
DIF: Moderate
REF: 6.2 Slopes of Parallel and Perpendicular Lines
LOC: 10.RF3
TOP: Relations and Functions
KEY: Procedural Knowledge
102. ANS: C
PTS: 1
DIF: Moderate
REF: 6.2 Slopes of Parallel and Perpendicular Lines
LOC: 10.RF3
TOP: Relations and Functions
KEY: Procedural Knowledge
103. ANS: A
PTS: 1
DIF: Easy
REF: 6.3 Investigating Graphs of Linear Functions
LOC: 10.RF7
TOP: Relations and Functions
KEY: Conceptual Understanding
104. ANS: B
PTS: 1
DIF: Easy
REF: 6.4 Slope-Intercept Form of the Equation for a Linear Function
LOC: 10.RF7
TOP: Relations and Functions
KEY: Conceptual Understanding
105. ANS: A
PTS: 1
DIF: Easy
REF: 6.4 Slope-Intercept Form of the Equation for a Linear Function
LOC: 10.RF7
TOP: Relations and Functions
KEY: Conceptual Understanding
106. ANS: A
PTS: 1
DIF: Easy
REF: 6.4 Slope-Intercept Form of the Equation for a Linear Function
LOC: 10.RF7
TOP: Relations and Functions
KEY: Conceptual Understanding
107. ANS: B
PTS: 1
DIF: Moderate
REF: 6.4 Slope-Intercept Form of the Equation for a Linear Function
LOC: 10.RF7
TOP: Relations and Functions
KEY: Conceptual Understanding
8
ID: A
108. ANS: B
PTS: 1
DIF: Easy
REF: 6.5 Slope-Point Form of the Equation for a Linear Function
LOC: 10.RF6
TOP: Relations and Functions
KEY: Conceptual Understanding
109. ANS: A
PTS: 1
DIF: Easy
REF: 6.5 Slope-Point Form of the Equation for a Linear Function
LOC: 10.RF6
TOP: Relations and Functions
KEY: Conceptual Understanding
110. ANS:
REF:
LOC:
KEY:
A
PTS: 1
DIF: Easy
6.5 Slope-Point Form of the Equation for a Linear Function
10.RF6
TOP: Relations and Functions
Conceptual Understanding | Procedural Knowledge
111. ANS:
REF:
LOC:
KEY:
C
PTS: 1
DIF: Easy
6.5 Slope-Point Form of the Equation for a Linear Function
10.RF6
TOP: Relations and Functions
Conceptual Understanding | Procedural Knowledge
112. ANS: B
PTS: 1
DIF: Easy
REF: 6.6 General Form of the Equation for a Linear Relation
LOC: 10.RF6
TOP: Relations and Functions
KEY: Conceptual Understanding
113. ANS:
REF:
LOC:
KEY:
C
PTS: 1
DIF: Easy
6.6 General Form of the Equation for a Linear Relation
10.RF6
TOP: Relations and Functions
Conceptual Understanding | Procedural Knowledge
114. ANS:
REF:
LOC:
KEY:
A
PTS: 1
DIF: Easy
6.6 General Form of the Equation for a Linear Relation
10.RF6
TOP: Relations and Functions
Conceptual Understanding | Procedural Knowledge
115. ANS: C
PTS: 1
DIF: Moderate
REF: 6.6 General Form of the Equation for a Linear Relation
LOC: 10.RF6
TOP: Relations and Functions
KEY: Conceptual Understanding
116. ANS: D
PTS: 1
DIF: Easy
REF: 7.1 Developing Systems of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
117. ANS: A
PTS: 1
DIF: Easy
REF: 7.1 Developing Systems of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
118. ANS: A
PTS: 1
DIF: Moderate
REF: 7.1 Developing Systems of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
9
ID: A
119. ANS: B
PTS: 1
DIF: Easy
REF: 7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
120. ANS: C
PTS: 1
DIF: Easy
REF: 7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
121. ANS: D
PTS: 1
DIF: Moderate
REF: 7.3 Using Graphing Technology to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
122. ANS: B
PTS: 1
DIF: Easy
REF: 7.3 Using Graphing Technology to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
123. ANS: C
PTS: 1
DIF: Easy
REF: 7.4 Using a Substitution Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
124. ANS: D
PTS: 1
DIF: Easy
REF: 7.4 Using a Substitution Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
125. ANS: B
PTS: 1
DIF: Easy
REF: 7.4 Using a Substitution Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
126. ANS: A
PTS: 1
DIF: Easy
REF: 7.4 Using a Substitution Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
127. ANS: D
PTS: 1
DIF: Easy
REF: 7.4 Using a Substitution Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
128. ANS: C
PTS: 1
DIF: Easy
REF: 7.5 Using an Elimination Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Procedural Knowledge
129. ANS: A
PTS: 1
DIF: Moderate
REF: 7.5 Using an Elimination Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Procedural Knowledge
130. ANS: D
PTS: 1
DIF: Moderate
REF: 7.5 Using an Elimination Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
131. ANS: D
PTS: 1
DIF: Easy
REF: 7.5 Using an Elimination Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Procedural Knowledge
10
ID: A
132. ANS: D
PTS: 1
DIF: Moderate
REF: 7.5 Using an Elimination Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Procedural Knowledge
133. ANS: A
PTS: 1
DIF: Easy
REF: 7.5 Using an Elimination Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Procedural Knowledge
134. ANS: A
PTS: 1
DIF: Easy
REF: 7.6 Properties of Systems of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
135. ANS: B
PTS: 1
DIF: Easy
REF: 7.6 Properties of Systems of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
SHORT ANSWER
136. ANS:
186 in.
PTS: 1
LOC: 10.M2
DIF: Easy
REF: 1.1 Imperial Measures of Length
TOP: Measurement
KEY: Procedural Knowledge
137. ANS:
31 mi.
PTS: 1
LOC: 10.M2
DIF: Easy
REF: 1.3 Relating SI and Imperial Units
TOP: Measurement
KEY: Procedural Knowledge
138. ANS:
88 cm
PTS: 1
LOC: 10.M2
DIF: Easy
REF: 1.3 Relating SI and Imperial Units
TOP: Measurement
KEY: Procedural Knowledge
139. ANS:
298 square inches
PTS: 1
DIF: Moderate
REF: 1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Procedural Knowledge
11
ID: A
140. ANS:
16 cm
PTS: 1
DIF: Moderate
REF: 1.4 Surface Areas of Right Pyramids and Right Cones
LOC: 10.M3
TOP: Measurement
KEY: Procedural Knowledge
141. ANS:
2259 m3
PTS: 1
LOC: 10.M3
DIF: Easy
REF: 1.5 Volumes of Right Pyramids and Right Cones
TOP: Measurement
KEY: Procedural Knowledge
142. ANS:
283.5 m3
PTS: 1
LOC: 10.M3
DIF: Easy
REF: 1.5 Volumes of Right Pyramids and Right Cones
TOP: Measurement
KEY: Procedural Knowledge
143. ANS:
231 cm3
PTS: 1
LOC: 10.M3
DIF: Easy
REF: 1.5 Volumes of Right Pyramids and Right Cones
TOP: Measurement
KEY: Procedural Knowledge
144. ANS:
SA = 380 square inches
V = 697 cubic inches
PTS: 1
LOC: 10.M3
DIF: Moderate
REF: 1.6 Surface Area and Volume of a Sphere
TOP: Measurement
KEY: Procedural Knowledge
145. ANS:
3619.1 m3
PTS: 1
LOC: 10.M3
DIF: Easy
REF: 1.6 Surface Area and Volume of a Sphere
TOP: Measurement
KEY: Procedural Knowledge
146. ANS:
36 884.9 cm3
PTS: 1
LOC: 10.M3
DIF: Moderate
REF: 1.6 Surface Area and Volume of a Sphere
TOP: Measurement
KEY: Procedural Knowledge
12
ID: A
147. ANS:
370.5 m3
PTS: 1
LOC: 10.M3
DIF: Easy
REF: 1.7 Solving Problems Involving Objects
TOP: Measurement
KEY: Procedural Knowledge
148. ANS:
10.3 cm2
PTS: 1
LOC: 10.M3
DIF: Moderate
REF: 1.7 Solving Problems Involving Objects
TOP: Measurement
KEY: Procedural Knowledge
149. ANS:
1885 cm2
PTS: 1
LOC: 10.M3
DIF: Difficult
REF: 1.7 Solving Problems Involving Objects
TOP: Measurement
KEY: Procedural Knowledge
150. ANS:
694 square inches
PTS: 1
LOC: 10.M3
DIF: Moderate
REF: 1.7 Solving Problems Involving Objects
TOP: Measurement
KEY: Problem-Solving Skills
151. ANS:
55.1°
PTS: 1
LOC: 10.M4
DIF: Moderate
REF: 2.1 The Tangent Ratio
TOP: Measurement
KEY: Procedural Knowledge
152. ANS:
a)
b) tan M = 0.25
PTS: 1
DIF: Moderate
REF: 2.1 The Tangent Ratio
LOC: 10.M4
TOP: Measurement
KEY: Conceptual Understanding | Procedural Knowledge
13
ID: A
153. ANS:
12.8 cm
PTS: 1
LOC: 10.M4
DIF: Moderate
REF: 2.2 Using the Tangent Ratio to Calculate Lengths
TOP: Measurement
KEY: Procedural Knowledge
154. ANS:
The angle of inclination of the straw is equal to 90° minus the angle between the thread and the straw.
PTS: 1
DIF: Easy
REF: 2.3 Math Lab: Measuring an Inaccessible Height
LOC: 10.M4
TOP: Measurement
KEY: Conceptual Understanding | Communication
155. ANS:
73.4°
PTS: 1
LOC: 10.M4
DIF: Moderate
REF: 2.4 The Sine and Cosine Ratios
TOP: Measurement
KEY: Procedural Knowledge
156. ANS:
27.7°
PTS: 1
LOC: 10.M4
DIF: Moderate
REF: 2.4 The Sine and Cosine Ratios
TOP: Measurement
KEY: Procedural Knowledge
157. ANS:
39.4°
PTS: 1
LOC: 10.M4
DIF: Moderate
REF: 2.4 The Sine and Cosine Ratios
TOP: Measurement
KEY: Procedural Knowledge
158. ANS:
16.6 m
PTS: 1
DIF: Moderate
REF: 2.5 Using the Sine and Cosine Ratios to Calculate Lengths
LOC: 10.M4
TOP: Measurement
KEY: Procedural Knowledge
159. ANS:
5 in.
PTS: 1
DIF: Moderate
REF: 2.5 Using the Sine and Cosine Ratios to Calculate Lengths
LOC: 10.M4
TOP: Measurement
KEY: Procedural Knowledge
14
ID: A
160. ANS:
12 ft.
PTS: 1
LOC: 10.M4
DIF: Moderate
REF: 2.6 Applying the Trigonometric Ratios
TOP: Measurement
KEY: Procedural Knowledge
161. ANS:
54 289
PTS: 1
DIF: Easy
REF: 3.2 Perfect Squares, Perfect Cubes, and Their Roots
LOC: 10.AN1
TOP: Algebra and Number
KEY: Procedural Knowledge
162. ANS:
a + 6b
PTS: 1
LOC: 10.AN5
DIF: Moderate
REF: 3.3 Common Factors of a Polynomial
TOP: Algebra and Number
KEY: Procedural Knowledge
163. ANS:
2y
PTS: 1
LOC: 10.AN5
DIF: Moderate
REF: 3.3 Common Factors of a Polynomial
TOP: Algebra and Number
KEY: Procedural Knowledge
164. ANS:
9, 16, 21, 24, and 25
PTS: 1
LOC: 10.AN5
DIF: Moderate
REF: 3.4 Modelling Trinomials as Binomial Products
TOP: Algebra and Number
KEY: Procedural Knowledge
165. ANS:
24 − 10q − q
2
PTS: 1
LOC: 10.AN4
DIF: Easy
REF: 3.5 Polynomials of the Form x^2 + bx + c
TOP: Algebra and Number
KEY: Procedural Knowledge
166. ANS:
( s − 32 ) ( s − 1 )
PTS: 1
LOC: 10.AN5
DIF: Easy
REF: 3.5 Polynomials of the Form x^2 + bx + c
TOP: Algebra and Number
KEY: Procedural Knowledge
15
ID: A
167. ANS:
2
44t − 25t − 6
PTS: 1
LOC: 10.AN4
DIF: Easy
REF: 3.6 Polynomials of the Form ax^2 + bx + c
TOP: Algebra and Number
KEY: Procedural Knowledge
168. ANS:
2
2
90y + 77y − 52 = 90y + 117y − 40y − 52
= 9y(10y + 13) − 4(10y + 13)
= (10y + 13)(9y − 4)
PTS: 1
LOC: 10.AN5
DIF: Moderate
REF: 3.6 Polynomials of the Form ax^2 + bx + c
TOP: Algebra and Number
KEY: Procedural Knowledge
169. ANS:
3
2
27z + 102z + 6z + 120
PTS: 1
LOC: 10.AN4
DIF: Easy
REF: 3.7 Multiplying Polynomials
TOP: Algebra and Number
KEY: Procedural Knowledge
170. ANS:
196
PTS: 1
LOC: 10.AN5
DIF: Moderate
REF: 3.8 Factoring Special Polynomials
TOP: Algebra and Number
KEY: Procedural Knowledge
171. ANS:
5.9
PTS: 1
LOC: 10.AN2
DIF: Moderate
REF: 4.1 Estimating Roots
TOP: Algebra and Number
KEY: Conceptual Understanding
172. ANS:
3.12,
4
1
, 2.4, and 5
7
2
PTS: 1
LOC: 10.AN2
DIF: Easy
REF: 4.2 Irrational Numbers
TOP: Algebra and Number
KEY: Conceptual Understanding
173. ANS:
8.5 cm
PTS: 1
LOC: 10.AN1
DIF: Easy
REF: 4.2 Irrational Numbers
TOP: Algebra and Number
KEY: Conceptual Understanding
16
ID: A
174. ANS:
3.8 cm
PTS: 1
LOC: 10.AN1
DIF: Easy
REF: 4.2 Irrational Numbers
TOP: Algebra and Number
KEY: Conceptual Understanding
175. ANS:
5
5
9
PTS: 1
LOC: 10.AN2
DIF: Moderate
REF: 4.3 Mixed and Entire Radicals
TOP: Algebra and Number
KEY: Conceptual Understanding
176. ANS:
1
9
1
7
12 , 12 ,
7
6
12 ,
PTS: 1
LOC: 10.AN3
6
7
12 , 12
9
7
DIF: Moderate
REF: 4.4 Fractional Exponents and Radicals
TOP: Algebra and Number
KEY: Conceptual Understanding
177. ANS:
9
4
PTS: 1
LOC: 10.AN3
DIF: Moderate
REF: 4.5 Negative Exponents and Reciprocals
TOP: Algebra and Number
KEY: Conceptual Understanding
178. ANS:
10
3
PTS: 1
LOC: 10.AN3
DIF: Moderate
REF: 4.5 Negative Exponents and Reciprocals
TOP: Algebra and Number
KEY: Conceptual Understanding
179. ANS:
−
a
3
4
4b c
3
PTS: 1
LOC: 10.AN3
DIF: Easy
REF: 4.6 Applying the Exponent Laws
TOP: Algebra and Number
KEY: Conceptual Understanding
17
ID: A
180. ANS:
256 12 28 8
m n p
81
PTS: 1
LOC: 10.AN3
DIF: Moderate
REF: 4.6 Applying the Exponent Laws
TOP: Algebra and Number
KEY: Conceptual Understanding
181. ANS:
PTS: 1
DIF: Moderate
REF: 5.1 Representing Relations
LOC: 10.RF4
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
182. ANS:
The relation shows the association “won gold in Olympic Men’s Ice Hockey in” from a set of countries
to a set of years.
PTS: 1
DIF: Easy
REF: 5.1 Representing Relations
LOC: 10.RF4
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
183. ANS:
–37
PTS: 1
LOC: 10.RF8
DIF: Moderate
REF: 5.2 Properties of Functions
TOP: Relations and Functions
KEY: Procedural Knowledge
18
ID: A
184. ANS:
a) 4 s
b) 10 s
PTS: 1
LOC: 10.RF1
DIF: Moderate
REF: 5.3 Interpreting and Sketching Graphs
TOP: Relations and Functions
KEY: Conceptual Understanding
185. ANS:
ÏÔ
¸Ô
The domain is: ÌÔ 9, 13, 16, 24, 33, … ˝Ô
Ó
˛
ÏÔ
¸Ô
The range is: ÌÔ 0.45, 0.65, 0.80, 1.20, 1.65, … Ô˝
Ó
˛
PTS: 1
LOC: 10.RF1
DIF: Easy
REF: 5.4 Graphing Data
TOP: Relations and Functions
KEY: Conceptual Understanding
186. ANS:
The points are not joined because the data are only valid for whole numbers of people.
PTS: 1
DIF: Easy
REF: 5.4 Graphing Data
LOC: 10.RF1
TOP: Relations and Functions
KEY: Communication | Conceptual Understanding
187. ANS:
a) When the domain value is –2, the range value is 2.
b) When the range value is –1, the domain value is 4.
PTS: 1
LOC: 10.RF8
DIF: Moderate
REF: 5.5 Graphs of Relations and Functions
TOP: Relations and Functions
KEY: Conceptual Understanding
188. ANS:
approximately 1.6 km/mi.
PTS: 1
LOC: 10.RF3
DIF: Moderate
REF: 5.6 Properties of Linear Functions
TOP: Relations and Functions
KEY: Procedural Knowledge
189. ANS:
Domain: 0 ≤ d ≤ 5
Range: 0 ≤ v ≤ 60
PTS: 1
LOC: 10.RF5
DIF: Moderate
REF: 5.7 Interpreting Graphs of Linear Functions
TOP: Relations and Functions
KEY: Conceptual Understanding
19
ID: A
190. ANS:
$200.00
PTS: 1
LOC: 10.RF8
DIF: Easy
REF: 5.7 Interpreting Graphs of Linear Functions
TOP: Relations and Functions
KEY: Conceptual Understanding
191. ANS:
−1
PTS: 1
LOC: 10.RF5
DIF: Easy
REF: 6.1 Slope of a Line
TOP: Relations and Functions
KEY: Procedural Knowledge
192. ANS:
Parallel
PTS: 1
LOC: 10.RF3
DIF: Easy
REF: 6.2 Slopes of Parallel and Perpendicular Lines
TOP: Relations and Functions
KEY: Conceptual Understanding
193. ANS:
Perpendicular
PTS: 1
LOC: 10.RF3
DIF: Easy
REF: 6.2 Slopes of Parallel and Perpendicular Lines
TOP: Relations and Functions
KEY: Conceptual Understanding
194. ANS:
The graph has a slope of 3 and a y-intercept of
PTS: 1
LOC: 10.RF7
3
.
2
DIF: Easy
REF: 6.3 Investigating Graphs of Linear Functions
TOP: Relations and Functions
KEY: Conceptual Understanding
195. ANS:
The graph has a slope of
PTS: 1
LOC: 10.RF7
3
and a y-intercept of 7.
2
DIF: Moderate
REF: 6.3 Investigating Graphs of Linear Functions
TOP: Relations and Functions
KEY: Conceptual Understanding
20
ID: A
196. ANS:
PTS: 1
DIF: Easy
REF: 6.4 Slope-Intercept Form of the Equation for a Linear Function
LOC: 10.RF6
TOP: Relations and Functions
KEY: Procedural Knowledge
197. ANS:
7
y − 7 = − (x + 4)
2
PTS: 1
DIF: Moderate
REF: 6.5 Slope-Point Form of the Equation for a Linear Function
LOC: 10.RF7
TOP: Relations and Functions
KEY: Conceptual Understanding
198. ANS:
The graph passes through (−6, −7) and has slope –8.
PTS: 1
DIF: Easy
REF: 6.5 Slope-Point Form of the Equation for a Linear Function
LOC: 10.RF7
TOP: Relations and Functions
KEY: Conceptual Understanding
199. ANS:
3x − 5y + 40 = 0
PTS: 1
DIF: Easy
REF: 6.6 General Form of the Equation for a Linear Relation
LOC: 10.RF6
TOP: Relations and Functions
KEY: Conceptual Understanding
21
ID: A
200. ANS:
5.75a + 7.5b − 75 = 0
PTS: 1
DIF: Moderate
REF: 6.6 General Form of the Equation for a Linear Relation
LOC: 10.RF6
TOP: Relations and Functions
KEY: Conceptual Understanding
201. ANS:
Let x represent the speed of the slower ship and y represent the speed of the faster ship.
A linear system is:
x+5=y
5x + 5y = 365
Since x = 34 and y = 39 satisfy each equation, these numbers are the solution of the linear system.
PTS: 1
LOC: 10.RF9
DIF: Difficult
REF: 7.1 Developing Systems of Linear Equations
TOP: Relations and Functions
KEY: Conceptual Understanding
202. ANS:
(0, –8)
PTS: 1
DIF: Easy
REF: 7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
22
ID: A
203. ANS:
a)
b)
(10, 15)
PTS: 1
DIF: Moderate
REF: 7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
23
ID: A
204. ANS:
a)
a = z + 24
a + 13 = 2(z + 13)
b)
Zack is approximately 11 years old and
Angela is approximately 35 years old.
PTS: 1
DIF: Moderate
REF: 7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
24
ID: A
205. ANS:
a)
p + g = 25
2.5p + 3.25g = 70
b)
The team purchased 15 pucks for practice
and
10 pucks for games.
PTS: 1
DIF: Moderate
REF: 7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
206. ANS:
(–1.6, –1)
PTS: 1
DIF: Moderate
REF: 7.3 Using Graphing Technology to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
207. ANS:
System A
ÊÁ 56x + 48y = −3008 ˆ˜ −8:
Ë
¯
System B
7x + 6y = –376
–4x – 6y = 256
ÊÁ 32x + 48y = −2048 ˆ˜ − ( −8 ) :
Ë
¯
PTS: 1
DIF: Easy
REF: 7.4 Using a Substitution Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
25
ID: A
208. ANS:
x = –20; y = –16
PTS: 1
DIF: Moderate
REF: 7.4 Using a Substitution Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
209. ANS:
s + l = 70
0.06s + 0.23l = 22.35
PTS: 1
DIF: Moderate
REF: 7.5 Using an Elimination Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
210. ANS:
One solution
PTS: 1
LOC: 10.RF9
DIF: Easy
REF: 7.6 Properties of Systems of Linear Equations
TOP: Relations and Functions
KEY: Conceptual Understanding
PROBLEM
211. ANS:
Since 5280 ft. = 1 mi., to convert feet to miles, divide by 5280.
20 000
20 000 ft. =
mi.
5280
4160
20 000 ft. = 3
mi.
5280
20 000 ft. = 3 mi. 4160 ft.
Since 3 ft. = 1 yd., to convert feet to yards, divide by 3.
4160
4160 ft. =
yd.
3
2
4160 ft. = 1386 yd.
3
4160 ft. = 1386 yd. 2 ft.
So, 20 000 ft. = 3 mi. 1386 yd. 2 ft.
PTS: 1
DIF: Moderate
REF: 1.1 Imperial Measures of Length
LOC: 10.M2
TOP: Measurement
KEY: Communication | Problem-Solving Skills
26
ID: A
212. ANS:
Use the formula for the volume of a right cylinder.
2
V = πr h
2
V = π(22.9) (17.1)
V = 28 171.9525. . .
The volume of a right cone is
V=
1
the volume of a right cylinder with the same base and the same height.
3
1
(28 171.9525. . . )
3
V = 9390.6508. . .
The volume of the right cone is approximately 9390.7 cm3 .
PTS: 1
LOC: 10.M3
DIF: Easy
REF: 1.5 Volumes of Right Pyramids and Right Cones
TOP: Measurement
KEY: Problem-Solving Skills
27
ID: A
213. ANS:
The surface area of the composite object is: area of the 4 rectangular faces of the prism + area of square
base of the prism + area of 4 triangular faces of the pyramid
The area of the 4 rectangular faces of the prism, in square feet, is:
A = 4(6)(9)
A = 216
The area of the square base of the prism, in square feet, is:
A = (6)(6)
A = 36
To determine the surface area of the triangular faces, calculate the slant height, s.
Sketch a triangle to represent a triangular face.
Use the Pythagorean Theorem in right ΔADB.
2
2
s = AD + BD
2
2
s = 3 +3
2
2
2
s = 9+9
2
s = 18
s=
18
The area of the 4 triangular faces of the pyramid, in square feet, is:
1
A = 4( )(6)( 18 )
2
A = 50.9116. . .
The surface area of the composite object, in square feet, is:
216 + 36 + 50.9116… = 302.9116…
The surface area of the composite object is approximately 303 square feet.
PTS: 1
DIF: Difficult
REF: 1.7 Solving Problems Involving Objects
LOC: 10.M3
TOP: Measurement
KEY: Communication | Problem-Solving Skills
28
ID: A
214. ANS:
Volume of air in the cube = volume of cube – volume of sphere
Use the formula for the volume of a cube.
V = lwh
V = (4.9)(4.9)(4.9)
V = 117.649
Use the formula for the volume of a sphere.
The radius, r, is:
1
r = (4.9 cm)
2
r = 2.45 cm
4 3
V = πr
3
V=
4
3
π(2.45)
3
V = 61.6008. . .
The volume of air in the cube is:
117.649 − 61.6008. . . = 56.0481. . .
The volume of air in the cube is approximately 56 cm3 .
PTS: 1
LOC: 10.M3
DIF: Moderate
REF: 1.7 Solving Problems Involving Objects
TOP: Measurement
KEY: Problem-Solving Skills
29
ID: A
215. ANS:
a) Volume of sculpture = volume of prism + volume of cylinder
Use the formula for the volume of a right rectangular prism.
V = lwh
V = (29)(33)(15)
V = 14 355
Use the formula for the volume of a right cylinder.
The radius, r, is:
1
r = (7 m)
2
r = 3.5 m
2
V = πr h
2
V = π(3.5) (14)
V = 538.7831. . .
The volume of the sculpture is:
14 355 + 538.7831. . . = 14 893.7831. . .
The volume of the sculpture is approximately 14 894 m3 .
b) The surface area of the sculpture is the sum of the areas of the faces of the right rectangular prism
and the curved surface of the cylinder.
The area of the rectangular faces of the prism, in square metres, is:
A = 2(29)(15) + 2(33)(15)
A = 1860
The area of the rectangular bases of the prism, in square metres, is:
A = 2(29)(33)
A = 1914
Use the formula to find the area of the curved surface of the cylinder.
The radius, r, is:
1
r = (7 m)
2
r = 3.5 m
SA = 2πrh
SA = 2π(3.5)(14)
SA = 307.876. . .
30
ID: A
The surface area of the sculpture is:
1860 + 1914 + 307.876… = 4081.876. . .
The surface area of the sculpture is approximately 4082 m2 .
PTS: 1
LOC: 10.M3
DIF: Difficult
REF: 1.7 Solving Problems Involving Objects
TOP: Measurement
KEY: Problem-Solving Skills
216. ANS:
The best angle of inclination for a solar panel is the same as the latitude.
Draw a right triangle to represent the cross-section of the roof and solar panel.
In right ΔABC:
opposite
tan C =
adjacent
tan C =
AB
BC
tan C =
5.0
3.8
∠C =Ö 52.8°
The angle of inclination of the solar panel is about 52.8°. Saskatoon or Edmonton would be an ideal
location for this roof design because the angle of inclination is within 1° of the latitude.
PTS: 1
DIF: Moderate
REF: 2.1 The Tangent Ratio
LOC: 10.M4
TOP: Measurement
KEY: Communication | Problem-Solving Skills
31
ID: A
217. ANS:
a)
AB = 3 × 9 mm = 27 mm
In right ΔABC:
opposite
tan C =
adjacent
tan C =
AB
BC
tan C =
27
9
∠C = tan
−1
ÁÊÁ 27
ÁÁ
ÁÁ 9
Ë
˜ˆ˜
˜˜
˜˜
¯
∠C =Ö 71.6°
Thomas is correct.
b)
tan C =
27
9
tan C = 3
In ΔABC above, tan C indicates that the length of the side opposite ∠C is 3 times the length of the side
adjacent to ∠C.
PTS: 1
DIF: Moderate
REF: 2.1 The Tangent Ratio
LOC: 10.M4
TOP: Measurement
KEY: Communication | Problem-Solving Skills
32
ID: A
218. ANS:
a) In right ΔBCD, BD is the hypotenuse. BC is opposite ∠D and CD is adjacent to ∠D.
opposite
b) sin D =
a
hypotenuse
sin D =
BC
BD
sin D =
11
20
sin D = 0.5500. . .
sin D is approximately 0.6. This means that in any similar right triangle, the length of the side
opposite ∠D is approximately 0.6 times the length of the hypotenuse.
c) sin D = 0.5500...
So, ∠D = 33.3670. . . °
∠D is approximately 33.4°.
PTS: 1
DIF: Moderate
REF: 2.4 The Sine and Cosine Ratios
LOC: 10.M4
TOP: Measurement
KEY: Communication | Conceptual Understanding | Problem-Solving Skills
33
ID: A
219. ANS:
The circumference, C, of the base of the cone is
3
4
3
4
the circumference of a circle with radius 17.9 cm. So,
of the circumference is:
C=
3
4
(2πr)
C=
3
4
(2)(π)(17.9)
C = 84.3517. . .
To determine the angle of inclination of
the side of the cone, we first need to find
the radius, r, of the cone. Use the formula
for the circumference of a circle.
C = 2πr
84.3517… = 2πr
84.3517… 2πr
=
2π
2π
84.3517…
=r
2π
r = 13.425
In right ΔABC, ∠C is the angle of
inclination.
The radius, r, is equal to the length of BC.
Since BC is adjacent to ∠C and AC is the
hypotenuse, use the cosine ratio.
adjacent
cosC =
hypotenuse
cosC =
BC
AC
cosC =
13.425
17.9
cosC = 0.75
∠C = 41.4096. . . °
The angle of inclination of the side of the cone is approximately 41.4°.
PTS: 1
LOC: 10.M4
DIF: Difficult
REF: 2.6 Applying the Trigonometric Ratios
TOP: Measurement
KEY: Problem-Solving Skills
34
ID: A
220. ANS:
We need to find the lengths of the diagonals, BD and AC.
To find the length of BD:
Use right ΔBEC to calculate the length of BE.
Use the cosine ratio in ΔBEC.
adjacent
cosB =
hypotenuse
cosB =
cos45.1° =
BE
BC
BE
17.9
17.9 cos45.1° = BE
BE = 12.6351…
Use right ΔDEC to calculate the length of DE.
Use the cosine ratio in ΔDEC.
adjacent
cosD =
hypotenuse
cosD =
cos25.7° =
DE
CD
DE
29.0
29.0 cos25.7° = DE
DE = 26.1312…
BD = BE + DE
BD = 12.6351… + 26.1312…
BD = 38.7663…
The length of BD is approximately 38.8 cm.
To find the length of AC:
Use right ΔDEA to calculate the length of AE.
Use the cosine ratio in ΔDEA.
35
ID: A
cosA =
adjacent
hypotenuse
cosA =
AE
AD
cos64.3° =
AE
29.0
29.0 cos64.3° = AE
AE = 12.5761…
Use right ΔBEC to calculate the length of CE.
Use the sine ratio in ΔBEC.
opposite
sin B =
hypotenuse
sin B =
sin 45.1° =
CE
BC
CE
17.9
17.9 sin 45.1° = CE
CE = 12.6792…
AC = AE + CE
AC = 12.5761… + 12.6792…
AC = 25.2553…
The length of AC is approximately 25.3 cm.
PTS: 1
DIF: Difficult
REF: 2.7 Solving Problems Involving More than One Right Triangle
LOC: 10.M4
TOP: Measurement
KEY: Problem-Solving Skills
36
ID: A
221. ANS:
The area of a rectangle is the product of its dimensions.
List the factors of 209 and 319. The factors represent possible lengths of a side of each rectangle.
Check to see which factors of 209 are also factors of 319.
The greatest common factor will be the greatest possible measure of the side that the two rectangles
share.
The factors of 209 are: 1, 11, 19, 209
The factors of 319 are: 1, 11, 29, 319
The greatest common factor of 209 and 319 is 11.
Τhe greatest possible measure of the side that the two rectangles share is 11 in.
PTS: 1
LOC: 10.AN1
DIF: Moderate
REF: 3.1 Factors and Multiples of Whole Numbers
TOP: Algebra and Number
KEY: Problem-Solving Skills
222. ANS:
The number of pages is a multiple of 6 and 40.
The fewest number of pages is the least common multiple of 6 and 40.
List the multiples of each number until the same multiple appears in both lists.
Multiples of 6 are: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, 108, 114, 120, . . .
Multiples of 40 are: 40, 80, 120, . . .
The least common multiple of 6 and 40 is 120.
The fewest number of pages the book can have is 120.
PTS: 1
LOC: 10.AN1
DIF: Moderate
REF: 3.1 Factors and Multiples of Whole Numbers
TOP: Algebra and Number
KEY: Problem-Solving Skills
37
ID: A
223. ANS:
a) The shorter side of the carpet measures 32 cm.
So, the side length of the square must be a factor of 32.
The longer side of the carpet measures 80 cm.
So, the side length of the square must be a factor of 80.
So, the side length of the square must be a common factor of 32 and 80.
The side length of the largest square will be the greatest common factor of 32 and 80.
Check to see which factors of 32 are also factors of 80.
The factors of 32 are: 1, 2, 4, 8, 16, 32
The factors of 80 are: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80
The greatest common factor of 32 and 80 is 16.
The side length of the largest possible square is 16 cm.
b) Find the number of squares Jordan can cut from the carpet:
Area, A, of the carpet:
A = lw
A = (32 cm)(80 cm)
A = 2560 cm
2
Area, A, of the largest possible square is:
A = lw
A = (16 cm)(16 cm)
A = 256 cm
2
Divide the area of the carpet by the area of the largest possible square.
2560 cm
256 cm
2
2
= 10
Jordan can cut 10 squares from the carpet.
PTS: 1
LOC: 10.AN1
DIF: Difficult
REF: 3.1 Factors and Multiples of Whole Numbers
TOP: Algebra and Number
KEY: Problem-Solving Skills
38
ID: A
224. ANS:
To calculate the volume, first determine the edge length of the cube.
The surface area of a cube is the sum of the areas of its 6 congruent square faces.
So, the area, A, of one face is:
2646
A=
6
A = 441
The edge length, e, of the cube is the square root of the area of one square face.
e = 441
e = 21
So, the volume, V, of the cube is the cube of its edge length.
3
V = 21
V = 9261
The volume of the cube is 9261 m3 .
PTS: 1
DIF: Easy
REF: 3.2 Perfect Squares, Perfect Cubes, and Their Roots
LOC: 10.AN1
TOP: Algebra and Number
KEY: Problem-Solving Skills
225. ANS:
a) i) Correction:
2
5s(3s − 7s + 1)
The student did not remove the common factor from the third term correctly. When the common
factor is the same as the term, a factor of 1 remains. This must be written as a term in the factored
polynomial.
ii) Correction:
2
−2h(11 + 16h − 8h )
When the student removed the common factor from the third term, she made a sign error. The
sign should be negative, not positive.
b) The student should have expanded her solutions to check that the trinomial was the same as the
original trinomial each time.
PTS: 1
DIF: Moderate
REF: 3.3 Common Factors of a Polynomial
LOC: 10.AN5
TOP: Algebra and Number
KEY: Communication | Problem-Solving Skills
39
ID: A
226. ANS:
Harish made an error in the second line, when he wrote
3
128 as the product of
3
125 and
3
3.
Correct solution:
3
1024 =
=
3
3
8 ⋅
8 ⋅
= 2⋅4⋅
=8
3
3
3
3
128
64 ⋅
3
2
2
2
PTS: 1
DIF: Moderate
REF: 4.3 Mixed and Entire Radicals
LOC: 10.AN2
TOP: Algebra and Number
KEY: Problem-Solving Skills | Communication
227. ANS:
2
5
1
2
Use the formula: SA = 0.025h m
a) Substitute: h = 48 and m = 7.3
SA = 0.025 ⋅ 48
SA = 0.025 ⋅ (
5
2
5
⋅ 7.3
1
2
2
48 ) ⋅ (
7.3 )
Use a calculator.
SA = 0.3177…
2
The surface area of a newborn with height 48 cm and mass 7.3 kg is approximately 0.32 cm .
b) Substitute: h = 170 and m = 66
SA = 0.025 ⋅ 170
SA = 0.025 ⋅ (
5
2
5
⋅ 66
2
1
2
170 ) ⋅ (
66 )
Use a calculator.
SA = 1.5844…
2
The surface area of a person with height 170 cm and mass 66 kg is approximately 1.58 cm .
PTS: 1
LOC: 10.AN3
DIF: Moderate
REF: 4.4 Fractional Exponents and Radicals
TOP: Algebra and Number
KEY: Problem-Solving Skills
40
ID: A
228. ANS:
Use the formula.
Substitute: V = 230 and h = 20d
2
230 = 0.5d ⋅ 20d
3
230
d =
10
3
d = 23
d = 2.8438…
The mean trunk diameter is approximately 3 m.
PTS: 1
LOC: 10.AN3
DIF: Moderate
REF: 4.6 Applying the Exponent Laws
TOP: Algebra and Number
KEY: Problem-Solving Skills
229. ANS:
a) There is an error in the second line. When multiplying powers with the same base, the exponents
should have been added, not multiplied.
A correct solution:
ÁÊÁ − 1 ˜ˆ˜
1
−
Á
˜
ÁÊÁ x −6 y 6 ˜ˆ˜ ÁÁÁ x 6 y 5 ˜˜˜ = x −6 ⋅ x 6 ⋅ y 6 ⋅ y 5
ÁÁ
˜˜ ÁÁ
˜˜
˜˜
Ë
¯ ÁÁÁ
˜˜
ÁË
¯
−
=x
=
y
x
37
6
y
11
11
37
6
b) There are two errors in the first line. The coefficient 2 was incorrectly multiplied by the exponent
–4. And, the exponent of the variable n was added to –4 instead of being multiplied by –4.
A correct solution:
ÊÁ
ˆ˜ −4
1 ˜
ÁÁ
˜˜
ÁÁ
−4
−1
ÁÁ 2m 4 ˜˜˜
2 m
ÁÁ
˜
ÁÁ 4 ˜˜˜ =
−16
ÁÁ n
˜˜
n
ÁÁ
˜˜
ÁË
˜¯
=
n
16
4
2 m
1
16
=
n
16m
PTS: 1
DIF: Moderate
REF: 4.6 Applying the Exponent Laws
LOC: 10.AN3
TOP: Algebra and Number
KEY: Problem-Solving Skills | Communication
41
ID: A
230. ANS:
ÊÁ 8 ˆ˜ ÁÊÁ 5 3 ˜ˆ˜
ÁÁÁ x ˜˜˜ ÁÁÁ x ˜˜˜
ÁË
˜¯ Á
˜
Ë
¯
Write each radical as a power.
1
3
ÁÊÁ 8 ˜ˆ˜ ÊÁÁÁ 5 3 ˆ˜˜˜
8
5
ÁÁ x ˜˜ ÁÁ x ˜˜ = x ⋅ x
ÁË
˜¯ Á
˜
Ë
¯
Use the product of powers law:
1
8
3
5
1
3
+
8
5
x ⋅x = x
Write equivalent fractions with a common denominator, 40.
x
1
3
+
8
5
=x
5
24
+
40
40
29
40
=x
I can write this power as a radical:
x
29
40
40
=
x
29
29
ÁÊ 40 ˜ˆ
or ÁÁÁÁ x ˜˜˜˜
Ë
¯
PTS: 1
DIF: Moderate
REF: 4.6 Applying the Exponent Laws
LOC: 10.AN3
TOP: Algebra and Number
KEY: Problem-Solving Skills | Communication
231. ANS:
a) {(81, 9), (90, 10), (99, 11), (108, 12), (117,13)}
b) Yes, the order of the numbers in each ordered pair does matter. The statement “81 divided by 9 is 9”
is true. However, if the numbers in each ordered pair were reversed, the statement “9 divided by 9 is
81”is not true.
PTS: 1
DIF: Moderate
REF: 5.1 Representing Relations
LOC: 10.RF4
TOP: Relations and Functions
KEY: Communication | Problem-Solving Skills
42
ID: A
232. ANS:
Answers may vary.
a) Functions:
Team Member
Julie
Hayley
Cassie
Jennifer
Marie
Meaghan
Angela
Kim
For example:
Age
15
16
16
15
17
15
16
17
Team Member
Julie
Hayley
Cassie
Jennifer
Marie
Meaghan
Angela
Kim
Points
36
43
38
17
42
19
45
37
Each of these relations is a function because each element in the first column is different.
b) Not functions:
Age Team Member
15
Julie
16
Hayley
16
Cassie
15
Jennifer
17
Marie
15
Meaghan
16
Angela
17
Kim
Position
Right Wing
Center
Left Wing
Left Defence
Right Wing
Right Defence
Left Wing
Center
Points
36
43
38
17
42
19
45
37
Each of these relations is not a function because some of the elements in the first column are the
same.
PTS: 1
DIF: Difficult
REF: 5.2 Properties of Functions
LOC: 10.RF2
TOP: Relations and Functions
KEY: Communication | Problem-Solving Skills
43
ID: A
233. ANS:
a) The total cost of the banquet is a function of the number of people attending. In function notation:
C(g) = 11g + 250
b) To determine C(46), use:
C(g) = 11g + 250
Substitute: g = 46
C(46) = 11(46) + 250
C(46) = 506 + 250
C(46) = 756
C(46) is the value of C when g = 46.
This means that when 46 people attend the banquet, the total cost is $756.
c) To determine the value of g when C(g) = 1581, use:
C(g) = 11g + 250
Substitute: C(g) = 1581
1581 = 11g + 250
1331 = 11g
g = 121
C(121) = 1581 means that when g = 121, C = 1581; that is, when 121 people attend the banquet, the
total cost is $1581.
PTS: 1
DIF: Difficult
REF: 5.2 Properties of Functions
LOC: 10.RF2
TOP: Relations and Functions
KEY: Communication | Problem-Solving Skills
234. ANS:
a) The dependent variable is distance, d. The independent variable is number of tire revolutions, r.
b) The graph represents a linear relation because the graph is a straight line.
c) I could create a table of values for the relation, then calculate the change in each variable. If the
changes in both variables are constant, the relation is linear.
PTS: 1
DIF: Moderate
REF: 5.6 Properties of Linear Functions
LOC: 10.RF4 | 10.RF3 | 10.RF1
TOP: Relations and Functions
KEY: Communication | Problem-Solving Skills
44
ID: A
235. ANS:
a) Choose two points on the line. Calculate the change in each variable from one point to the other.
Change in length of shadow:
60 m − 0 m = 60 m
Change in height of object:
20 m − 0 m = 20 m
Rate of change:
60 m
=3m
20 m
The rate of change is positive so the length of the shadow increases with the height of the object.
For every 1 m of height, the length of the shadow is 3 m.
b) To estimate the length of the shadow, use the graph.
From 13 on the h-axis, draw a vertical line to the graph, then a horizontal line to the l-axis. From
the graph, the length of the shadow will be about 39 m.
c) To estimate the height of the building, use the graph.
From 45 on the l-axis, draw a horizontal line to the graph, then a vertical line to the h-axis. From
the graph, the height of the building will be about 15 m.
PTS: 1
DIF: Moderate
REF: 5.7 Interpreting Graphs of Linear Functions
LOC: 10.RF3 | 10.RF8
TOP: Relations and Functions
KEY: Communication | Problem-Solving Skills
45
ID: A
236. ANS:
a) Point A has coordinates (–3, 4).
Point B has coordinates (1, –4).
From A to B:
The rise is the change in y-coordinates.
Rise = −4 − 4
= –8
The run is the change in x-coordinates.
Run = 1 − (−3)
=4
rise
Slope of AB =
run
8
Slope of AB = −
4
Slope of AB = −2
b) Sample answer:
Any horizontal line segment has slope 0. The slope of QR is zero.
Any vertical line segment has a slope that is not defined. The slope of JK is not defined.
The slope of AB is the same as the slope of the line segment in part a. It has a rise of –8 and a run of
4.
PTS: 1
LOC: 10.RF5
DIF: Moderate
REF: 6.1 Slope of a Line
TOP: Relations and Functions
KEY: Problem-Solving Skills
46
ID: A
237. ANS:
Subtract corresponding coordinates to determine the change in x and in y.
From S to T:
The rise is the change in y-coordinates.
Rise = 11 − (−5)
The run is the change in x-coordinates.
Run = − 15 − 7
11 − (−5)
Slope of ST =
−15 − 7
8
Slope of ST = −
11
The slope of ST is −
8
.
11
The correct answer is −
8
.
11
PTS: 1
DIF: Moderate
REF: 6.1 Slope of a Line
LOC: 10.RF5
TOP: Relations and Functions
KEY: Communication | Problem-Solving Skills
47
ID: A
238. ANS:
Slope of AB =
y2 − y1
x2 − x1
27 − 9
6 − 18
18
Slope of AB =
−12
Slope of AB =
3
The slope of AB is − .
2
Since CD is parallel to AB, the slopes of CD and AB are equal.
3
So, the slope of CD is − .
2
i)
Point D is on the y-axis. So, it has coordinates (0, y).
Use the formula for the slope of a line:
y2 − y1
Slope of CD =
x2 − x1
y−9
3
=
2
0−6
y−9
3
− =
2
−6
ÊÁ y − 9
Á
3
(–6)(− ) = (–6) ÁÁÁÁ
2
ÁË −6
9 = y−9
18 = y
The coordinates of point
−
ˆ˜
˜˜
˜˜
˜˜
¯
D are (0, 18).
ii) Point D is on the x-axis. It has coordinates (x, 0).
Use the formula for the slope of a line:
y2 − y1
Slope of CD =
x2 − x1
3
0−9
=
2
x−6
3
−9
− =
2
x−6
ÊÁ −9
3
(x − 6)(− ) = (x − 6) ÁÁÁÁ
ÁË x − 6
2
−
ˆ˜
˜˜
˜˜
˜¯
−3x + 18
= −9
2
ÊÁ −3x + 18 ˆ˜
˜˜ = (2)(−9)
(2) ÁÁÁÁ
˜˜
ÁË
˜¯
2
48
ID: A
−3x + 18 = –18
–3x = –36
x = 12
The coordinates of point D are (12, 0).
PTS: 1
LOC: 10.RF3
DIF: Difficult
REF: 6.2 Slopes of Parallel and Perpendicular Lines
TOP: Relations and Functions
KEY: Problem-Solving Skills
239. ANS:
The graph has a slope of
2
and a y-intercept of 2.
5
PTS: 1
DIF: Moderate
REF: 6.3 Investigating Graphs of Linear Functions
LOC: 10.RF7
TOP: Relations and Functions
KEY: Communication | Problem-Solving Skills
49
ID: A
240. ANS:
a) n = f(t), so two points on the graph have coordinates C(25, 458) and D(29, 534).
Use this form for the equation of a linear function:
n − n1
n2 − n1
=
t − t1
t2 − t1
Substitute: n 1 = 458, t 1 = 25, n 2 = 534, and t 2 = 29
n − 458
534 − 458
=
t − 25
29 − 25
n − 458
= 19
t − 25
ÊÁ n − 458 ˆ˜
˜˜ = 19 ( t − 25 )
( t − 25 ) ÁÁÁÁ
˜
ÁË t − 25 ˜˜¯
n − 458 = 19 ( t − 25 )
In slope-point form, the equation that represents this function is: n − 458 = 19 ( t − 25 )
b) Use:
n − 458 = 19 ( t − 25 )
Substitute: n = 325
n − 458 = 19 ( t − 25 )
325 − 458 = 19 ( t − 25 )
−133 = 19t − 475
342 = 19t
t = 18
When the students sell 325 cups of punch, the approximate temperature is 18°C.
PTS: 1
DIF: Difficult
REF: 6.5 Slope-Point Form of the Equation for a Linear Function
LOC: 10.RF7
TOP: Relations and Functions
KEY: Problem-Solving Skills
50
ID: A
241. ANS:
a) Let q represent the number of quarters, and l represent the number of loonies.
The value of q quarters is 25q cents, and the value of l loonies is 100l cents.
Then, a system of equations is:
q + l = 35
25q + 100l = 2375
b)
Since the intersection point is at (15, 20), there are 15 quarters and 20 loonies in the coin box.
PTS: 1
DIF: Moderate
REF: 7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
TOP: Relations and Functions
KEY: Problem-Solving Skills
51
ID: A
242. ANS:
a) Let P represent the clerk’s two-week salary, in dollars, and s represent the clerk’s two-week sales, in
dollars.
Then, a linear system is:
P = 580 + 0.042s
P = 880 + 0.012s
b) P = 580 + 0.042s
P = 880 + 0.012s
For each equation, determine the P-intercept and the coordinates of another point on the line.
For equation 1:
P = 580 + 0.042s
Substitute: s = 0
P = 580
Substitute: s = 10 000
P = 580 + 0.042 × 10 000
P = 580 + 420
P = 1000
On a grid, use a scale of 1 square to 200 units on the P-axis, and a scale of 2 squares to
10 000 units on the s-axis. Mark a point at 580 on the P-axis and mark a point at (10 000, 1000).
Join the points with a line.
For equation 2:
P = 880 + 0.012s
Substitute: s = 0
P = 880
Substitute: s = 10 000
P = 880 + 0.012 × 10 000
P = 880 + 120
P = 1000
On the grid, mark a point at 1000 on the P-axis and mark a point at (10 000, 1000).
Join the points with a line.
52
ID: A
c) From the graph, a clerk will receive the same salary, $1000, with both plans when the two-week sales
is $10 000.
Check that this solution satisfies both equations.
Substitute P = 1000 and s = 10 000 in each equation.
For equation 1:
P = 580 + 0.042s
L.S. = P
R.S. = 580 + 0.042s
= 1000
= 580 + 0.042(10 000)
= 580 + 420
= 1000
For equation 2:
P = 880 + 0.012s
L.S. = P
R.S. = 880 + 0.012s
= 1000
= 880 + 0.012(10 000)
= 880 + 120
= 1000
Since the left side is equal to the right side for each equation, the solution is correct.
PTS:
REF:
LOC:
KEY:
1
DIF: Difficult
7.2 Solving a System of Linear Equations Graphically
10.RF9
TOP: Relations and Functions
Communication | Problem-Solving Skills
53
ID: A
243. ANS:
a) Let a represent the number of adult tickets sold, and s represent the number of student tickets sold.
There were twice as many student tickets as adult tickets.
The first equation is:
2a = s
The total receipts were $2016.
The second equation is:
6a + 4s = 2016
The linear system is:
2a = s
(1)
6a + 4s = 2016
(2)
b) Solve for s in equation (1).
2a = s
(1)
s = 2a
Substitute s = 2a in equation (2).
6a + 4s = 2016 (2)
6a + 4(2a) = 2016
6a + 8a = 2016
14a = 2016
2016
a=
14
a = 144
Substitute a = 144 in equation (1).
2a = s (1)
2(144) = s
288 = s
144 adult tickets and 288 student tickets were sold.
PTS:
REF:
LOC:
KEY:
1
DIF: Moderate
7.4 Using a Substitution Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
Problem-Solving Skills | Communication
54
ID: A
244. ANS:
20x + 35y = 705
10x − 5y = 195
Multiply equation
by 7, then add to eliminate y.
7 × equation : 7(10x − 5y = 195)
70x − 35y = 1365
Add:
20x + 35y = 705
+ 70x − 35y = 1365
90x
= 2070
x = 23
Substitute x = 23 in equation
20x + 35y = 705
.
20(23) + 35y = 705
460 + 35y = 705
35y = 245
y=7
Verify the solution.
In each equation, substitute: x = 23 and y = 7
20x + 35y = 705
10x − 5y = 195
L.S. = 20x + 35y
L.S. = 10x − 5y
= 20(23) + 35(7)
= 10(23) − 5(7)
= 460 + 245
= 230 − 35
= 705
= 195
= R.S.
= R.S.
For each equation, the left side is equal to the right side, so the solution is: x = 23 and y = 7
PTS:
REF:
LOC:
KEY:
1
DIF: Moderate
7.5 Using an Elimination Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
Communication | Problem-Solving Skills
55
ID: A
245. ANS:
Write each equation in slope-intercept form to identify the slope and y-intercept of each line.
4x + 12y = 28
8x + 24y = 48
For equation
4x + 12y = 28
:
12y = −4x + 28
y=−
4
28
x+
12
12
1
7
y = − x+
3
3
1
7
The slope is − and the y-intercept is .
3
3
For equation
8x + 24y = 48
:
24y = −8x + 48
y=−
8
48
x+
24
24
1
y = − x+2
3
1
The slope is − and the y-intercept is 2.
3
Because the slopes are equal and the y-intercepts are different, the lines are parallel and the linear system
has no solution.
PTS: 1
DIF: Moderate
REF: 7.6 Properties of Systems of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Communication | Problem-Solving Skills
56
