Name: ______________________ Class: _________________ Date: _________ ID: A

1

Math 10C Course Review

Multiple ChoiceIdentify 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 sinkb. The length of a public swimming poolc. The width of your shortest fingerd. The length of a walking stick

____ 2. Which referent could you use for 1 yd.?a. The width of your shortest fingerb. The length of a screwdriverc. The height of the kitchen counter above the floord. The length of a football field

____ 3. Which referent could you use for 1 mi.?a. The length of a salmonb. The height of a grizzly bear standing on its hind legsc. The distance equal to 4 laps on an oval running trackd. The thickness of a sheet of loose-leaf paper

____ 4. Which expression represents the area of the shaded region?

a. 2r(2r− π) b. r2 (1 − π) c. r2 (4 − π) d. r(r− 2π)

____ 5. Which of the following trinomials can be factored?a. z2 + 33z+ 9 c. z2 + 10z+ 2b. z2 + 12z+ 63 d. z2 + 10z+ 25

____ 6. Which of the following trinomials can be factored?a. 4c 2 + 33c + 8 c. 4c 2 + 13c + 8b. 4c 2 + 21c + 3 d. 4c 2 + 4c + 15

Name: ______________________ ID: A

2

____ 7. Which polynomial, written in simplified form, represents the area of this rectangle?

a. 8x 2 − 36xy − 20y 2 c. 16x 2 + 72xy − 40y 2

b. 8x 2 + 22xy − 20y 2 d. 8x 2 + 36xy − 20y 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.

Name: ______________________ ID: A

3

____ 9. This graph shows the masses of people, m , as a function of age, a. Determine the range of the graph.

a. R:(4,5,8,12,14,17) c. R:(15,25,45,55,80,85)b. R:(3,5,8,10,14,17) d. 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. Domain: t ≤ 129Range: 0 ≤ V ≤ 1800

c. Domain: 0 ≤ t ≤ 129Range: V ≤ 1800

b. Domain: 0 ≤ V ≤ 1800Range: t ≤ 129

d. Domain: 0 ≤ t ≤ 129Range: 0 ≤ V ≤ 1800

Name: ______________________ ID: A

4

____ 11. Which graph represents the linear function f x( ) = −3x + 4?

a. c.

b. d.

Name: ______________________ ID: A

5

____ 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.

Name: ______________________ ID: A

6

____ 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. about 11.5 L c. about 9.5 Lb. about 10.5 L d. 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.

a. − 158

c. − 817

b. − 815

d. − 178

____ 15. The slope of a line is 17

. What is the slope of a line that is parallel to this line?

a. 7 c. 214

b. 142

d. −7

____ 16. The slope of a line is 17

. What is the slope of a line that is perpendicular to this line?

a. − 142

c. 17

b. 7 d. 214

Name: ______________________ ID: A

7

____ 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)

____ 19. Which graph represents the equation y = − 25

x + 1?

a. c.

b. d.

Name: ______________________ ID: A

8

____ 20. Write an equation in slope-point form for the line that passes through A(–2, 4) and B(–9, 6).

a. y − 6 = − 27

(x + 2) c. y − 4 = − 27

(x + 2)

b. y + 4 = − 27

(x − 2) d. y + 6 = 27

(x − 2)

____ 21. Write an equation in slope-point form for the line that passes through A(1, 4) and B(6, 8).

a. y + 8 = 45

(x − 1) c. y − 4 = 45

(x − 1)

b. y + 4 = − 45

(x − 1) d. y − 8 = − 45

(x + 1)

____ 22. Which equation is written in general form?a. −4x − 12y + 15 = 0 c. 12x = 4y − 15

b. 12x − 4y + 15 = 0 d. 115

x − 4y − 12 = 0

____ 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 = 0b. 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

2x – 2y = 11c. 2x + 2y = 8

x – y = 21b. x + 2y = 8

2x – 4y = 16d. x + 3y = 22

2x – y = 10

____ 25. Use the graph to approximate the solution of the linear system:y = −5x − 2y = 5x − 4

a. (–3, 0.2) c. (0.2, –3) b. (0, –2.8) d. (–2.8, 0)

Name: ______________________ ID: A

9

____ 26. Two life insurance companies determine their premiums using different formulas:Company A: p = 2a + 24Company 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 d. 44 years

____ 27. Express each equation in slope-intercept form.–2x + 4y = 6813x + 4y = 284

a. y = 12

x – 17

y = − 134

x – 71

c. y = 12

x + 17

y = − 134

x + 71

b. y = − 28413

x + 17

y = − 134

x + 413

d. y = 413

x − 28413

y = 12

x − 28413

Name: ______________________ ID: A

10

____ 28. Use the table of values to determine the solution of this linear system:4x + y = 32x + y = −5

a. (–13, –13) c. (–13, 4)b. (4, –13) d. (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.

(1) 54

x + 16

y = 4712

(2) 45

x – 67

y = 16

a. Multiply equation (1) by 35; multiply equation (2) by 12.b. Multiply equation (1) by 12; multiply equation (2) by 35.c. Multiply equation (1) by 2; multiply equation (2) by 3.d. Multiply equation (1) by 3; multiply equation (2) by 2.

____ 30. Write an equivalent system with integer coefficients.37

x + 3y = 4387

56

x + 5y = 3103

a. 3x + 21y = 438 5x + 30y = 620

c. 3x + 21y = 438 30x + 5y = 620

b. 21x + 3y = 438 5x + 30y = 620

d. 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. Equation i d. Equation ii

Name: ______________________ ID: A

11

____ 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.

Name: ______________________ ID: A

12

37. Tan B = 1.3; determine the measure of ∠B to the nearest tenth of a degree.

.

38. Suppose you must use 2 x 2 -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.90y 2 + 77y − 52 = 90y 2 + 117y − 40y − 52 = 9y(10y + 13) + 4(10y + 13) = (10y + 13)(9y + 4)

.

40. The area of a square is represented by the trinomial 36m 2 + 84mn + 49n 2 . Determine an expression for the perimeter of the square.

.

41. Evaluate 6254

.

Name: ______________________ ID: A

13

42. This is a graph of the function h x( ) = − 12 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

.

Name: ______________________ ID: A

14

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 5π

⋅ r−

12 , where r is the radius of the space station, in metres

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.

Name: ______________________ ID: A

15

50. Four litres of latex paint covers approximately 37 m2 and costs $52.a) Copy and complete this table.

Volume of Paint, p (L)

0 4 8 12 16

Cost, c ($) 0 52Area Covered, A (m2) 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.

Name: ______________________ ID: A

16

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 salesPlan 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.

ID: A

1

Math 10C Course ReviewAnswer Section

MULTIPLE CHOICE

1. ANS: C PTS: 1 DIF: Easy REF: 1.2 Measuring Length and Distance LOC: 10.M1TOP: Measurement KEY: Conceptual Understanding

2. ANS: C PTS: 1 DIF: Easy REF: 1.2 Measuring Length and Distance LOC: 10.M1TOP: Measurement KEY: Conceptual Understanding

3. ANS: C PTS: 1 DIF: Easy REF: 1.2 Measuring Length and Distance LOC: 10.M1TOP: Measurement KEY: Conceptual Understanding

4. ANS: C PTS: 1 DIF: Moderate REF: 3.3 Common Factors of a Polynomial LOC: 10.AN5TOP: Algebra and Number KEY: Procedural Knowledge

5. ANS: D PTS: 1 DIF: Easy REF: 3.4 Modelling Trinomials as Binomial Products LOC: 10.AN5TOP: Algebra and Number KEY: Procedural Knowledge

6. ANS: A PTS: 1 DIF: Easy REF: 3.4 Modelling Trinomials as Binomial Products LOC: 10.AN5TOP: Algebra and Number KEY: Procedural Knowledge

7. ANS: D PTS: 1 DIF: Moderate REF: 3.7 Multiplying PolynomialsLOC: 10.AN4 TOP: Algebra and Number KEY: Procedural Knowledge

8. ANS: D PTS: 1 DIF: Moderate REF: 5.1 Representing RelationsLOC: 10.RF4 TOP: Relations and Functions KEY: Conceptual Understanding

9. ANS: D PTS: 1 DIF: Easy REF: 5.5 Graphs of Relations and Functions LOC: 10.RF1TOP: Relations and Functions KEY: Conceptual Understanding

10. ANS: D PTS: 1 DIF: Easy REF: 5.7 Interpreting Graphs of Linear Functions LOC: 10.RF5TOP: Relations and Functions KEY: Conceptual Understanding

11. ANS: A PTS: 1 DIF: Easy REF: 5.7 Interpreting Graphs of Linear Functions LOC: 10.RF5TOP: Relations and Functions KEY: Procedural Knowledge

12. ANS: B PTS: 1 DIF: Easy REF: 5.7 Interpreting Graphs of Linear Functions LOC: 10.RF5TOP: Relations and Functions KEY: Procedural Knowledge

13. ANS: C PTS: 1 DIF: Easy REF: 5.7 Interpreting Graphs of Linear Functions LOC: 10.RF8TOP: Relations and Functions KEY: Conceptual Understanding

14. ANS: B PTS: 1 DIF: Moderate REF: 6.1 Slope of a LineLOC: 10.RF5 TOP: Relations and Functions KEY: Procedural Knowledge

15. ANS: C PTS: 1 DIF: Easy REF: 6.2 Slopes of Parallel and Perpendicular Lines LOC: 10.RF3TOP: Relations and Functions KEY: Conceptual Understanding

ID: A

2

16. ANS: A PTS: 1 DIF: Easy REF: 6.2 Slopes of Parallel and Perpendicular Lines LOC: 10.RF3TOP: Relations and Functions KEY: Conceptual Understanding

17. ANS: C PTS: 1 DIF: Moderate REF: 6.2 Slopes of Parallel and Perpendicular Lines LOC: 10.RF3TOP: Relations and Functions KEY: Procedural Knowledge

18. ANS: C PTS: 1 DIF: Moderate REF: 6.2 Slopes of Parallel and Perpendicular Lines LOC: 10.RF3TOP: Relations and Functions KEY: Procedural Knowledge

19. ANS: B PTS: 1 DIF: Easy REF: 6.4 Slope-Intercept Form of the Equation for a Linear FunctionLOC: 10.RF7 TOP: Relations and Functions KEY: Conceptual Understanding

20. ANS: C 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 | Procedural Knowledge

21. ANS: C 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 | Procedural Knowledge

22. 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

23. 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

24. ANS: A PTS: 1 DIF: Easy REF: 7.1 Developing Systems of Linear Equations LOC: 10.RF9TOP: Relations and Functions KEY: Conceptual Understanding

25. ANS: C PTS: 1 DIF: Easy REF: 7.2 Solving a System of Linear Equations Graphically LOC: 10.RF9TOP: Relations and Functions KEY: Conceptual Understanding

26. ANS: D PTS: 1 DIF: Easy REF: 7.2 Solving a System of Linear Equations Graphically LOC: 10.RF9TOP: Relations and Functions KEY: Conceptual Understanding

27. ANS: C PTS: 1 DIF: Moderate REF: 7.3 Using Graphing Technology to Solve a System of Linear EquationsLOC: 10.RF9 TOP: Relations and Functions KEY: Conceptual Understanding

28. ANS: B PTS: 1 DIF: Easy REF: 7.3 Using Graphing Technology to Solve a System of Linear EquationsLOC: 10.RF9 TOP: Relations and Functions KEY: Conceptual Understanding

29. ANS: B PTS: 1 DIF: Easy REF: 7.4 Using a Substitution Strategy to Solve a System of Linear EquationsLOC: 10.RF9 TOP: Relations and Functions KEY: Conceptual Understanding

30. ANS: A PTS: 1 DIF: Easy REF: 7.4 Using a Substitution Strategy to Solve a System of Linear EquationsLOC: 10.RF9 TOP: Relations and Functions KEY: Conceptual Understanding

ID: A

3

31. ANS: D PTS: 1 DIF: Moderate REF: 7.6 Properties of Systems of Linear Equations LOC: 10.RF9TOP: Relations and Functions KEY: Conceptual Understanding

32. ANS: D PTS: 1 DIF: Moderate REF: 7.6 Properties of Systems of Linear Equations LOC: 10.RF9TOP: Relations and Functions KEY: Conceptual Understanding

33. ANS: B PTS: 1 DIF: Easy REF: 2.1 The Tangent RatioLOC: 10.M4 TOP: Measurement KEY: Procedural Knowledge

34. ANS: B PTS: 1 DIF: Easy REF: 2.1 The Tangent RatioLOC: 10.M4 TOP: Measurement KEY: Procedural Knowledge

SHORT ANSWER

35. ANS: 9 mm

PTS: 1 DIF: Moderate REF: 1.4 Surface Areas of Right Pyramids and Right ConesLOC: 10.M3 TOP: Measurement KEY: Procedural Knowledge

36. ANS: 370.5 m3

PTS: 1 DIF: Easy REF: 1.7 Solving Problems Involving ObjectsLOC: 10.M3 TOP: Measurement KEY: Procedural Knowledge

37. ANS:

∠B =Ö 52.4°

PTS: 1 DIF: Easy REF: 2.1 The Tangent RatioLOC: 10.M4 TOP: Measurement KEY: Procedural Knowledge

38. ANS: 15, 24, and 45

PTS: 1 DIF: Moderate REF: 3.4 Modelling Trinomials as Binomial ProductsLOC: 10.AN5 TOP: Algebra and Number KEY: Procedural Knowledge

39. ANS: 90y 2 + 77y − 52 = 90y 2 + 117y − 40y − 52 = 9y(10y + 13) − 4(10y + 13) = (10y + 13)(9y − 4)

PTS: 1 DIF: Moderate REF: 3.6 Polynomials of the Form ax^2 + bx + cLOC: 10.AN5 TOP: Algebra and Number KEY: Procedural Knowledge

40. ANS: 4 6m + 7n( )

PTS: 1 DIF: Moderate REF: 3.8 Factoring Special PolynomialsLOC: 10.AN5 TOP: Algebra and Number KEY: Procedural Knowledge

ID: A

4

41. ANS: 5

PTS: 1 DIF: Easy REF: 4.1 Estimating RootsLOC: 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 DIF: Moderate REF: 5.5 Graphs of Relations and FunctionsLOC: 10.RF8 TOP: Relations and Functions KEY: Conceptual Understanding

43. ANS:

PTS: 1 DIF: Easy REF: 6.4 Slope-Intercept Form of the Equation for a Linear FunctionLOC: 10.RF6 TOP: Relations and Functions KEY: Procedural Knowledge

44. ANS:

− 95

PTS: 1 DIF: Easy REF: 6.6 General Form of the Equation for a Linear RelationLOC: 10.RF6 TOP: Relations and Functions KEY: Conceptual Understanding

ID: A

5

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. × 3 ft.1 yd.

=28 yd.

1× 3 ft.

1 yd.

=28 yd.

1× 3 ft.

1 yd.

= 84 ft.1

= 84 ft.

Since the measurements are equal, the conversion is verified.

PTS: 1 DIF: Moderate REF: 1.1 Imperial Measures of LengthLOC: 10.M2 TOP: Measurement KEY: Procedural Knowledge

ID: A

6

46. ANS: To convert inches to feet and inches, divide by 12.

554 in. = 55412

ft.

554 in. = 46 212

ft.

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.8 ft.

= 5 78

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 = $76.32Before taxes, the crown moulding will cost $76.32.

PTS: 1 DIF: Moderate REF: 1.1 Imperial Measures of LengthLOC: 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 DistanceLOC: 10.M1 TOP: Measurement KEY: Communication | Problem-Solving Skills

ID: A

7

48. ANS: Determine the measure of ∠C first.In right ΔABC:

cosC =adjacent

hypotenuse

cosC = BCAC

cosC = 1123

∠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 RatiosLOC: 10.M4 TOP: Measurement KEY: Problem-Solving Skills

49. ANS: Use the formula. Substitute: r = 12.7

N =42 5π

⋅ 12.7−

12

= 42 5π

⋅ 112.7

= 8.3884… So, the space station simulates the gravity on Earth at about 8.39 rotations/min.

PTS: 1 DIF: Moderate REF: 4.5 Negative Exponents and ReciprocalsLOC: 10.AN3 TOP: Algebra and Number KEY: Problem-Solving Skills

ID: A

8

50. ANS: a)

Volume of Paint, p (L)

0 4 8 12 16

Cost, c ($) 0 52 104 156 208

Area Covered, A (m2) 0 37 74 111 148

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 FunctionsLOC: 10.RF1 TOP: Relations and Functions KEY: Communication | Problem-Solving Skills

ID: A

9

51. ANS:

a) Slope = riserun

Slope = − 4650

Slope = − 2325

The slope of the road is − 2325

.

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.

Slope = riserun

− 2325

= −24.5run

run −2325

ÊËÁÁ

ˆ

¯

˜̃̃̃̃˜̃̃̃̃˜̃̃̃̃

= run −24.5run

ÊËÁÁ

ˆ

¯

˜̃̃̃̃˜̃̃̃̃˜̃̃̃̃

−2 run( )325

= −24.5

325 −2 run( )325

ÊËÁÁ

ˆ

¯

˜̃̃̃̃˜̃̃̃̃˜̃̃̃̃

= (325)(–24.5)

(−2) run = (325)(–24.5) (−2) run = −7963 run = 3981.25The 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

ID: A

10

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 + 0.042sP = 880 + 0.012s

b) P = 580 + 0.042sP = 880 + 0.012sFor each equation, determine the P-intercept and the coordinates of another point on the line.For equation 1:P = 580 + 0.042sSubstitute: s = 0P = 580Substitute: s = 10 000P = 580 + 0.042 × 10 000P = 580 + 420P = 1000On 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 + 0.012sSubstitute: s = 0P = 880Substitute: s = 10 000P = 880 + 0.012 × 10 000P = 880 + 120P = 1000On 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.

ID: A

11

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 + 0.042sL.S. = P R.S. = 580 + 0.042s = 1000 = 580 + 0.042(10 000) = 580 + 420 = 1000 For equation 2:P = 880 + 0.012sL.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 GraphicallyLOC: 10.RF9 TOP: Relations and Functions KEY: Communication | Problem-Solving Skills

ID: A

12

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 by 15, then subtract to eliminate d.25 × equation : 25(15d + 800k = 715)

375d + 20000k = 17875

15 × equation : 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

= 15(29) + 800(0.35)

= 435 + 280

= 715

= R.S.

25d + 2250k = 1512.5 L.S. = 25d + 2250k

= 25(29) + 2250(0.35)

= 725 + 787.5

= 1512.5

= 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 EquationsLOC: 10.RF9 TOP: Relations and Functions KEY: Problem-Solving Skills