Name: ______________________
Class: _________________
Date: _________
Math 10C Course Review
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. Which referent could you use for 1 cm?
a. The depth of a kitchen sink
b. The length of a public swimming pool
c. The width of your shortest finger
d. The length of a walking stick
____
2. Which referent could you use for 1 yd.?
a. The width of your shortest finger
b. The length of a screwdriver
c. The height of the kitchen counter above the floor
d. The length of a football field
____
3. Which referent could you use for 1 mi.?
a. The length of a salmon
b. The height of a grizzly bear standing on its hind legs
c. The distance equal to 4 laps on an oval running track
d. The thickness of a sheet of loose-leaf paper
____
4. Which expression represents the area of the shaded region?
a.
2r(2r − π)
b.
2
r (1 − π)
c.
2
r (4 − π)
____
5. Which of the following trinomials can be factored?
2
2
a. z + 33z + 9
c. z + 10z + 2
2
2
b. z + 12z + 63
d. z + 10z + 25
____
6. Which of the following trinomials can be factored?
2
2
a. 4c + 33c + 8
c. 4c + 13c + 8
2
2
b. 4c + 21c + 3
d. 4c + 4c + 15
1
d.
r(r − 2 π)
ID: A
Name: ______________________
____
____
ID: A
7. Which polynomial, written in simplified form, represents the area of this rectangle?
2
2
c.
16x + 72xy − 40y
2
2
d.
8x + 36xy − 20y
a.
8x − 36xy − 20y
b.
8x + 22xy − 20y
2
2
2
2
8. Which arrow diagram shows the association “is less than” from a set of numbers to a set of numbers?
a.
c.
b.
d.
2
Name: ______________________
____
ID: A
9. This graph shows the masses of people, m, as a function of age, a. Determine the range of the graph.
a.
b.
R:(4, 5, 8, 12, 14, 17)
R:(3, 5, 8, 10, 14, 17)
c.
d.
R:(15, 25, 45, 55, 80, 85)
R:(20, 25, 45, 65, 80, 85)
____ 10. This graph shows the volume of water remaining in a leaking hot tub as a function of time. Determine
the domain and range.
a.
b.
Domain: t ≤ 129
Range: 0 ≤ V ≤ 1800
Domain: 0 ≤ V ≤ 1800
Range: t ≤ 129
c.
d.
3
Domain: 0 ≤ t ≤ 129
Range: V ≤ 1800
Domain: 0 ≤ t ≤ 129
Range: 0 ≤ V ≤ 1800
Name: ______________________
ID: A
____ 11. Which graph represents the linear function f (x) = −3x + 4?
a.
c.
b.
d.
4
Name: ______________________
ID: A
____ 12. 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.
5
Name: ______________________
ID: A
____ 13. This graph shows the cost of gas. The cost, C dollars, is a function of the volume, V litres, of gas
purchased. What is the volume of gas purchased when the cost is $10.45?
a.
b.
about 11.5 L
about 10.5 L
c.
d.
about 9.5 L
about 9 L
____ 14. A 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
____ 15. The slope of a line is
1
. What is the slope of a line that is parallel to this line?
7
a.
7
c.
2
14
b.
14
2
d.
−7
____ 16. The slope of a line is
a.
−
b.
7
14
2
1
. What is the slope of a line that is perpendicular to this line?
7
c.
d.
6
1
7
2
14
Name: ______________________
ID: A
____ 17. A line passes through D(–5, 3) and N(12, –4). Determine the coordinates of two points on a line parallel
to DN.
a. (6, –10) and (24, –8)
c. (–10, 6) and (24, –8)
b. (–10, 24) and (6, –8)
d. (–10, 6) and (–8, 24)
____ 18. A 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)
2
____ 19. Which graph represents the equation y = − x + 1?
5
a.
c.
b.
d.
7
Name: ______________________
ID: A
____ 20. Write an equation in slope-point form for the line that passes through A(–2, 4) and
B(–9, 6).
2
2
a. y − 6 = − (x + 2)
c. y − 4 = − (x + 2)
7
7
2
2
b. y + 4 = − (x − 2)
d. y + 6 = (x − 2)
7
7
____ 21. Write 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
____ 22. Which equation is written in general form?
a. −4x − 12y + 15 = 0
b.
12x − 4y + 15 = 0
c.
d.
12x = 4y − 15
1
x − 4y − 12 = 0
15
____ 23. 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
d. 3x + 9y + 27 = 0
____ 24. Which linear system has the solution x = 8 and y = 2.5?
a. 2x + 2y = 21
c. 2x + 2y = 8
2x – 2y = 11
x – y = 21
b. x + 2y = 8
d. x + 3y = 22
2x – 4y = 16
2x – y = 10
____ 25. Use the graph to approximate the solution of the linear system:
y = −5x − 2
y = 5x − 4
a.
b.
(–3, 0.2)
(0, –2.8)
c.
d.
8
(0.2, –3)
(–2.8, 0)
Name: ______________________
ID: A
____ 26. Two life insurance companies determine their premiums using different formulas:
Company A: p = 2a + 24
Company B: p = 2.25a + 13, where p represents the annual premium, and a represents the client’s age.
Use the graph to determine the age at which both companies charge the same premium.
a.
62 years
b.
24 years
c.
59 years
____ 27. Express each equation in slope-intercept form.
–2x + 4y = 68
13x + 4y = 284
a.
b.
1
x – 17
2
13
y = − x – 71
4
284
y=−
x + 17
13
13
4
y=− x+
4
13
y=
c.
d.
9
1
x + 17
2
13
y = − x + 71
4
4
284
y=
x−
13
13
1
284
y= x−
2
13
y=
d.
44 years
Name: ______________________
ID: A
____ 28. Use the table of values to determine the solution of this linear system:
4x + y = 3
2x + y = −5
a.
b.
(–13, –13)
(4, –13)
c.
d.
(–13, 4)
(4, 4)
____ 29. For 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
a.
b.
c.
d.
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.
____ 30. Write an equivalent system with integer coefficients.
3
438
x + 3y =
7
7
5
310
x + 5y =
6
3
a.
b.
3x + 21y = 438
5x + 30y = 620
21x + 3y = 438
5x + 30y = 620
c.
d.
3x + 21y = 438
30x + 5y = 620
3x + 21y = 1
5x + 30y = 1
____ 31. The first equation of a linear system is 2x + 3y = 52. Choose a second equation to form a linear system
with infinite solutions.
i) 2x + 3y = –260
i i ) –10x – 15y = –260 iii) –10x + 3y = –260 iv) –10x + 3y = 255
a.
Equation iii
b.
Equation iv
c.
10
Equation i
d.
Equation ii
Name: ______________________
ID: A
____ 32. The first equation of a linear system is –6x + 12y = –42. Choose a second equation to form a linear
system with no solution.
i) –6x + 12y = 126
i i ) 18x – 36y = 126
iii) 18x + 12y = 126 iv) 18x + 36y = 0
a.
Equation iv
b.
Equation ii
c.
Equation iii
d.
Equation i
____ 33. Determine the angle of inclination of the line to the nearest tenth of a degree.
a.
63.3°
b.
24.2°
c.
65.8°
d.
26.7°
____ 34. A guy wire is attached to a tower at a point that is 10 m above the ground. The wire is anchored 21 m
from the base of the tower. What angle, to the nearest degree, does the guy wire make with the ground?
a. 62°
b. 25°
c. 28°
d. 65°
Short Answer
35. A regular tetrahedron with edge length 12.7 mm has a surface area of 229.0 mm2 . Determine the slant
height of the tetrahedron to the nearest millimetre.
.
36. Determine 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.
11
Name: ______________________
ID: A
37. Tan B = 1.3; determine the measure of ∠B to the nearest tenth of a degree.
.
2
38. Suppose you must use 2 x -tiles and twenty-two 1-tiles. Which numbers of x-tiles could you use to form a
rectangle?
.
39. 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)
.
2
2
40. The area of a square is represented by the trinomial 36m + 84mn + 49n . Determine an expression for
the perimeter of the square.
.
41. Evaluate
4
625 .
12
Name: ______________________
ID: A
1
42. This 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.
.
43. Graph the line with y-intercept 3 and slope –2.
44. Determine the slope of the line of this equation: 9x + 5y − 13 = 0
.
13
Name: ______________________
ID: A
Problem
45. Convert 28 yd. to feet. Use unit analysis to verify the conversion.
.
46. Sheila plans to place crown moulding along the top of each wall in her family room. A total of 554 in. of
moulding is required. The moulding costs $1.59/ft. and is sold in 8-ft. lengths. What is the cost of the
crown moulding, before taxes?
.
47. What referents would you use to estimate the length, in both SI units and imperial units, of the Capilano
Suspension Bridge in North Vancouver? Explain how you could measure the length in both units.
.
48. Determine the measures of ∠A and ∠C to the nearest tenth of a degree.
49. For a spherical space station, this formula is used to estimate the number of rotations per minute, N,
required so that the force inside the station simulates the gravity on Earth:
N=
42
5
−
1
2
⋅ r , where r is the radius of the space station, in metres
π
Suppose the radius of the space station is 12.7 m.
Calculate the number of rotations per minute required to simulate the gravity on Earth.
Write the answer to the nearest hundredth.
14
Name: ______________________
ID: A
50. Four litres of latex paint covers approximately 37 m2 and costs $52.
a) Copy and complete this table.
Volume of Paint, p
(L)
Cost, c ($)
Area Covered, A ( m2 )
0
4
8
12
16
0 52
0 37
b) Graph the area covered as a function of the volume of paint.
c) Graph the area covered as a function of the cost.
d) Write the domain and range of the functions in parts b and c.
15
Name: ______________________
ID: A
51. Construction workers are paving a road. The road must drop 4 cm for every 650 cm measured
horizontally.
a) What is the slope of the road?
b) Suppose a section of the road drops 24.5 cm. How long is this section of the road measured
horizontally?
.
52. 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. (Either by hand or graphing calc.)
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?
.
53. a) Model this situation with a linear system:
To rent a car, a person is charged a daily rate and a fee for each kilometre driven. When Chena
rented a car for 15 days and drove 800 km, the charge was $715.00. When she rented the same car
for 25 days and drove 2250 km, the charge was $1512.50.
b) Determine the daily rate and the fee for each kilometre driven. Verify the solution.
16
ID: A
Math 10C Course Review
Answer Section
MULTIPLE CHOICE
1. ANS:
REF:
TOP:
2. ANS:
REF:
TOP:
3. ANS:
REF:
TOP:
4. ANS:
REF:
TOP:
5. ANS:
REF:
TOP:
6. ANS:
REF:
TOP:
7. ANS:
LOC:
8. ANS:
LOC:
9. ANS:
REF:
TOP:
10. ANS:
REF:
TOP:
11. ANS:
REF:
TOP:
12. ANS:
REF:
TOP:
13. ANS:
REF:
TOP:
14. ANS:
LOC:
15. ANS:
REF:
TOP:
C
PTS: 1
DIF: Easy
1.2 Measuring Length and Distance
LOC: 10.M1
Measurement
KEY: Conceptual Understanding
C
PTS: 1
DIF: Easy
1.2 Measuring Length and Distance
LOC: 10.M1
Measurement
KEY: Conceptual Understanding
C
PTS: 1
DIF: Easy
1.2 Measuring Length and Distance
LOC: 10.M1
Measurement
KEY: Conceptual Understanding
C
PTS: 1
DIF: Moderate
3.3 Common Factors of a Polynomial
LOC: 10.AN5
Algebra and Number
KEY: Procedural Knowledge
D
PTS: 1
DIF: Easy
3.4 Modelling Trinomials as Binomial Products
LOC: 10.AN5
Algebra and Number
KEY: Procedural Knowledge
A
PTS: 1
DIF: Easy
3.4 Modelling Trinomials as Binomial Products
LOC: 10.AN5
Algebra and Number
KEY: Procedural Knowledge
D
PTS: 1
DIF: Moderate
REF: 3.7 Multiplying Polynomials
10.AN4
TOP: Algebra and Number
KEY: Procedural Knowledge
D
PTS: 1
DIF: Moderate
REF: 5.1 Representing Relations
10.RF4
TOP: Relations and Functions
KEY: Conceptual Understanding
D
PTS: 1
DIF: Easy
5.5 Graphs of Relations and Functions
LOC: 10.RF1
Relations and Functions
KEY: Conceptual Understanding
D
PTS: 1
DIF: Easy
5.7 Interpreting Graphs of Linear Functions
LOC: 10.RF5
Relations and Functions
KEY: Conceptual Understanding
A
PTS: 1
DIF: Easy
5.7 Interpreting Graphs of Linear Functions
LOC: 10.RF5
Relations and Functions
KEY: Procedural Knowledge
B
PTS: 1
DIF: Easy
5.7 Interpreting Graphs of Linear Functions
LOC: 10.RF5
Relations and Functions
KEY: Procedural Knowledge
C
PTS: 1
DIF: Easy
5.7 Interpreting Graphs of Linear Functions
LOC: 10.RF8
Relations and Functions
KEY: Conceptual Understanding
B
PTS: 1
DIF: Moderate
REF: 6.1 Slope of a Line
10.RF5
TOP: Relations and Functions
KEY: Procedural Knowledge
C
PTS: 1
DIF: Easy
6.2 Slopes of Parallel and Perpendicular Lines
LOC: 10.RF3
Relations and Functions
KEY: Conceptual Understanding
1
ID: A
16. ANS:
REF:
TOP:
17. ANS:
REF:
TOP:
18. ANS:
REF:
TOP:
19. ANS:
REF:
LOC:
20. ANS:
REF:
LOC:
KEY:
21. ANS:
REF:
LOC:
KEY:
22. ANS:
REF:
LOC:
23. ANS:
REF:
LOC:
24. ANS:
REF:
TOP:
25. ANS:
REF:
TOP:
26. ANS:
REF:
TOP:
27. ANS:
REF:
LOC:
28. ANS:
REF:
LOC:
29. ANS:
REF:
LOC:
30. ANS:
REF:
LOC:
A
PTS: 1
DIF: Easy
6.2 Slopes of Parallel and Perpendicular Lines
LOC: 10.RF3
Relations and Functions
KEY: Conceptual Understanding
C
PTS: 1
DIF: Moderate
6.2 Slopes of Parallel and Perpendicular Lines
LOC: 10.RF3
Relations and Functions
KEY: Procedural Knowledge
C
PTS: 1
DIF: Moderate
6.2 Slopes of Parallel and Perpendicular Lines
LOC: 10.RF3
Relations and Functions
KEY: Procedural Knowledge
B
PTS: 1
DIF: Easy
6.4 Slope-Intercept Form of the Equation for a Linear Function
10.RF7
TOP: Relations and Functions
KEY: Conceptual
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
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
B
PTS: 1
DIF: Easy
6.6 General Form of the Equation for a Linear Relation
10.RF6
TOP: Relations and Functions
KEY: Conceptual
C
PTS: 1
DIF: Moderate
6.6 General Form of the Equation for a Linear Relation
10.RF6
TOP: Relations and Functions
KEY: Conceptual
A
PTS: 1
DIF: Easy
7.1 Developing Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
C
PTS: 1
DIF: Easy
7.2 Solving a System of Linear Equations Graphically LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
D
PTS: 1
DIF: Easy
7.2 Solving a System of Linear Equations Graphically LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
C
PTS: 1
DIF: Moderate
7.3 Using Graphing Technology to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Conceptual
B
PTS: 1
DIF: Easy
7.3 Using Graphing Technology to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Conceptual
B
PTS: 1
DIF: Easy
7.4 Using a Substitution Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Conceptual
A
PTS: 1
DIF: Easy
7.4 Using a Substitution Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Conceptual
2
Understanding
Understanding
Understanding
Understanding
Understanding
Understanding
Understanding
ID: A
31. ANS:
REF:
TOP:
32. ANS:
REF:
TOP:
33. ANS:
LOC:
34. ANS:
LOC:
D
PTS: 1
DIF: Moderate
7.6 Properties of Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
D
PTS: 1
DIF: Moderate
7.6 Properties of Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
B
PTS: 1
DIF: Easy
REF: 2.1 The Tangent Ratio
10.M4
TOP: Measurement
KEY: Procedural Knowledge
B
PTS: 1
DIF: Easy
REF: 2.1 The Tangent Ratio
10.M4
TOP: Measurement
KEY: Procedural Knowledge
SHORT ANSWER
35. ANS:
9 mm
PTS: 1
LOC: 10.M3
36. ANS:
370.5 m3
DIF: Moderate
REF: 1.4 Surface Areas of Right Pyramids and Right Cones
TOP: Measurement
KEY: Procedural Knowledge
PTS: 1
LOC: 10.M3
37. ANS:
Ö
∠B = 52.4°
DIF: Easy
REF: 1.7 Solving Problems Involving Objects
TOP: Measurement
KEY: Procedural Knowledge
PTS: 1
LOC: 10.M4
38. ANS:
15, 24, and 45
DIF: Easy
REF: 2.1 The Tangent Ratio
TOP: Measurement
KEY: Procedural Knowledge
PTS: 1
DIF: Moderate
REF: 3.4 Modelling Trinomials as Binomial Products
LOC: 10.AN5
TOP: Algebra and Number
KEY: Procedural Knowledge
39. 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
40. ANS:
4 (6m + 7n )
DIF: Moderate
REF: 3.6 Polynomials of the Form ax^2 + bx + c
TOP: Algebra and Number
KEY: Procedural Knowledge
PTS: 1
LOC: 10.AN5
DIF: Moderate
REF: 3.8 Factoring Special Polynomials
TOP: Algebra and Number
KEY: Procedural Knowledge
3
ID: A
41. ANS:
5
PTS: 1
DIF: Easy
REF: 4.1 Estimating Roots
LOC: 10.AN2
TOP: Algebra and Number
KEY: Conceptual Understanding
42. 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
43. ANS:
DIF: Moderate
REF: 5.5 Graphs of Relations and Functions
TOP: Relations and Functions
KEY: Conceptual Understanding
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
44. ANS:
9
−
5
PTS: 1
LOC: 10.RF6
DIF: Easy
REF: 6.6 General Form of the Equation for a Linear Relation
TOP: Relations and Functions
KEY: Conceptual Understanding
4
ID: A
PROBLEM
45. ANS:
Since 1 yd. = 3 ft., to convert yards to feet, multiply by 3.
28 yd. = 28(3 ft.)
28 yd. = 84 ft.
Write a conversion factor for yards and feet,
with feet in the numerator:
Then, 28 yd. ×
28 yd. 3 ft.
3 ft.
=
×
1 yd.
1
1 yd.
=
28 yd. 3 ft.
×
1
1 yd.
=
84 ft.
1
= 84 ft.
Since the measurements are equal, the conversion is verified.
PTS: 1
LOC: 10.M2
DIF: Moderate
REF: 1.1 Imperial Measures of Length
TOP: Measurement
KEY: Procedural Knowledge
5
ID: A
46. ANS:
To convert inches to feet and inches, divide by 12.
554
554 in. =
ft.
12
2
554 in. = 46
ft.
12
554 in. = 46 ft. 2 in.
Sheila requires approximately 47 ft. of moulding. To find the number of 8-ft. lengths Sheila needs, divide
47 by 8.
47 ft.
7
=5
8 ft.
8
The number of 8-ft. lengths is greater than 5, so Sheila must buy 6 lengths.
The total number of feet in 6 lengths is: 6(8 ft.) = 48 ft.
The cost, C, is:
C = 48($1.59)
C = $76.32
Before taxes, the crown moulding will cost $76.32.
PTS: 1
DIF: Moderate
REF: 1.1 Imperial Measures of Length
LOC: 10.M2
TOP: Measurement
KEY: Problem-Solving Skills
47. ANS:
Sample answer:
I would walk along the bridge, counting the number of large steps I take to go from one end of the bridge
to the other. Each large step is about 1 yd., so the number of steps is the approximate length of the
bridge, in yards.
Since 1 m is slightly longer than 1 yd., the number of metres that represent the length of the bridge is
slightly less than the number of yards.
Students may use different referents to solve this problem.
PTS: 1
DIF: Moderate
REF: 1.2 Measuring Length and Distance
LOC: 10.M1
TOP: Measurement
KEY: Communication | Problem-Solving Skills
6
ID: A
48. ANS:
Determine the measure of ∠C first.
In right ΔABC:
adjacent
cosC =
hypotenuse
cosC =
BC
AC
cosC =
11
23
∠C = 61.4281. . . °
∠A + ∠C = 90°
∠A = 90° − ∠C
So, ∠A = 90° − 61.4281. . . °
∠A = 28.5718. . . °
∠C is approximately 61.4° and ∠A is approximately 28.6°.
PTS: 1
DIF: Moderate
REF: 2.4 The Sine and Cosine Ratios
LOC: 10.M4
TOP: Measurement
KEY: Problem-Solving Skills
49. ANS:
Use the formula. Substitute: r = 12.7
N=
=
42
5
π
42
5
π
−
⋅ 12.7
⋅
1
2
1
12.7
= 8.3884…
So, the space station simulates the gravity on Earth at about 8.39 rotations/min.
PTS: 1
LOC: 10.AN3
DIF: Moderate
REF: 4.5 Negative Exponents and Reciprocals
TOP: Algebra and Number
KEY: Problem-Solving Skills
7
ID: A
50. ANS:
a)
Volume of Paint, p
(L)
Cost, c ($)
0
4
8
12
16
0
52 104 156
208
Area Covered, A ( m2 )
0
37
148
74
111
b)
c)
d) Part b: domain: 0 ≤ p ≤ 16; range: 0 ≤ A ≤ 148
Part c: domain: 0 ≤ c ≤ 208; range: 0 ≤ A ≤ 148
PTS: 1
DIF: Difficult
REF: 5.5 Graphs of Relations and Functions
LOC: 10.RF1
TOP: Relations and Functions
KEY: Communication | Problem-Solving Skills
8
ID: A
51. ANS:
a) Slope =
rise
run
4
650
2
Slope = −
325
Slope = −
The slope of the road is −
2
.
325
b) The distance the road drops is the rise. Since the road drops, the rise is negative. The section of the
road measured horizontally is the run.
rise
Slope =
run
2
−24.5
−
=
325
run
ˆ
ˆ
Ê −2 ˜
Ê −24.5 ˜
˜
˜
run Á
= run Á
˜
˜
˜
˜
325
run
˜
˜
Ë
Ë
˜
˜
˜
˜
¯
¯
−2 (run )
= −24.5
325
ˆ
Ê −2 (run ) ˜
˜
˜
325 Á
= (325)(–24.5)
˜
˜
325
Ë
˜
˜
¯
˜
(−2) run = (325)(–24.5)
(−2) run = −7963
run = 3981.25
The section of the road measured horizontally is 3981.25 cm long.
PTS: 1
DIF: Moderate
REF: 6.1 Slope of a Line
LOC: 10.RF5
TOP: Relations and Functions
KEY: Communication | Problem-Solving Skills
9
ID: A
52. 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.
10
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: 1
DIF: Difficult
REF: 7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
TOP: Relations and Functions
KEY: Communication | Problem-Solving Skills
11
ID: A
53. ANS:
a) Let d dollars represent the daily rate and let k dollars represent the fee for each kilometre driven.
The linear system is:
15d + 800k = 715
25d + 2250k = 1512.5
b) Multiply equation
by 25 and equation
25 × equation : 25(15d + 800k = 715)
375d + 20000k = 17875
15 × equation
by 15, then subtract to eliminate d.
: 15(25d + 2250k = 1512.5)
375d + 33750k = 22687.5
Subtract equation
from equation
375d + 20000k = 17875
−(375d + 33750k = 22687.5)
− 13750k = −4812.5
.
k = 0.35
Substitute k = 0.35 in equation
15d + 800k = 715
.
15d + 800(0.35) = 715
15d + 280 = 715
15d = 435
d = 29
Verify the solution.
In each equation, substitute: k = 0.35 and d = 29
15d + 800k = 715
L.S. = 15d + 800k
25d + 2250k = 1512.5
L.S. = 25d + 2250k
= 15(29) + 800(0.35)
= 25(29) + 2250(0.35)
= 435 + 280
= 725 + 787.5
= 715
= 1512.5
= R.S.
= R.S.
So, the daily rate is $29 and the fee for each kilometre driven is $0.35.
PTS: 1
DIF: Difficult
REF: 7.5 Using an Elimination Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Problem-Solving Skills
12