Polynomials - WordPress.com · MFM1P – Foundations of Mathematics Unit 2 – Lesson 6 Lesson Six Concepts Overall Expectations ¾ Simplify numerical and polynomial expressions in

Polynomials

Lesson 6

MFM1P – Foundations of Mathematics Unit 2 – Lesson 6

Lesson Six Concepts Overall Expectations

Simplify numerical and polynomial expressions in one variable, and solve simple first-degree equations.

Specific Expectations

Add and subtract polynomials involving the same variable up to degree three; Multiply a polynomial by a monomial involving the same variable to give results up to

degree three; Solve first-degree equations with non-fractional coefficients, using a variety of tools

and strategies; Substitute into algebraic equations and solve for one variable in the first degree.

Polynomials Polynomials are a mathematical expression with one or more terms, in which the exponents are whole numbers and the coefficients are real numbers. Monomial is a polynomial with one term; e.g. -8 and 3x2. Binomial is a polynomial with two terms; e.g. -4x + 7 and y3 – 12. Trinomial is a polynomial with three terms; e.g. 4x2 – 3x + 9 and –w3 + 7w + 2.

-4x is an example of a term.

Term -4x “x” is called the variable or “unknown” “-4” is called the coefficient Adding and Subtracting Polynomials To add and/or subtract polynomials the terms must be “like terms”. Like terms are terms that have the same variables with the same value of exponent for each variable. Example Underline the terms that are like terms.

Copyright © 2005, Durham Continuing Education Page 2 of 51


a) –3x, x, , 2y, 12x 2x7 b) , , 3xy7 , yx5 3− xy3− 2xy5 Solution

Example Simplify. a) ( ) )7x3(1x5 +++

b) ( ) )3a2(4a −−+− c) ( ) ( )5w2w98w6w4 22 ++−−+



Solutions

Support Questions 1. State the like terms in each group.

a) v3 ,w3 ,x ,x ,z5 ,y5 ,x3 2 −b) w4 ,y ,y2 ,z4 ,y3 ,x4 22

2. Simplify.

a) b) 2x43x3 +++− )2n6n53 −−− c) d) )7a4a5()3a2a8( 22 ++−+−+ )x25x4()2x5x6( 22 −+−++−e) f) )nm67()nm23( 22 +−+−− )x37()x62( 22 −−+g) h) )w3()w65( 22 −−− )

)

x5x4()x3x5( 22 +−+− 3. Simplify. Then determine the value of the polynomial when n = 2 and when n = −1.

a) b) )3n2()4n3( −−++− 3n6n2()2n7n3( 22 ++−++−



Multiplying and Dividing Polynomials To multiply and/or divide polynomials the terms do not have to be “like terms”. Example Simplify.

a) b) )x3)(x4( 2 −xy8yx16 33

−

c) ( ) )wy5(yw2 2− d) 232 xyz25yzx100 ÷ Solutions a) = )x3)(x4( 2 − xxx43 ⋅⋅⋅⋅− = 3x12− or = ( )( )( )( )12 xx34 −

Use the multiplication of exponents rule

= 12x12 +− = 3x12−

b) xy8yx16 33

− =

yx8yyyxxx16

⋅⋅−⋅⋅⋅⋅⋅⋅

Divide common factors

= 1

yyxx2 ⋅⋅⋅⋅−

= 22yx2−

or = 1313 yx8

16 −−

−

= 22yx2− c) ( ) )wy5(yw2 2− = yywww10 ⋅⋅⋅⋅− = 23yw10−



d) = 232 xyz25yzx100 ÷ 231112 zyx4 −−−

= xz4

Support Questions

Any base to the zero exponent equals 1

4. Determine the product or quotient.

a) b) )m7)(m3( 2 − )n6()n12( 2 −÷

c) ba3

ab7 2

− d) )y2)(xy4)(x3( −−

e) a7a28 3

−− f) )7)(z8( 3

g) h) )ab2()ba4( 3 ÷− )x4)(x5(

i) ⎟⎠⎞

⎜⎝⎛−⎟⎠⎞

⎜⎝⎛− 22 a

910ab

53

5. Simplify. Then determine the value of the polynomial when a = 2 and when b = −1.

a) b) 2)ab2( )ab5)(ab3( 2−−

Multiplying Polynomials with a Monomial This process requires the use of the distributive law. Example Expand. a) b) )4x3(2 − )5x3(x2 −− c) )p53)(p3( 2 −− Solutions a) = )4x3(2 − 8x6 −

(2)(3x)= 6x and (2)(-4)= -8



b) = )5x3(x2 −− )5(x2)x3(x2 −+− = x10x6 2 −−

c) = -3p 2 (3) - 3p (-5p) )p53)(p3( 2 −− 2

= 32 p15p9 +−

Support Questions 6. Expand.

a) b) )9x3(x − )3n2)(n4( −− c) )1b3b2(b 2 +−d) e) )2x)(x( −− )mm)(m4( 2 −−

Key Question #6 1. State the like terms in each group. (2 marks)

a) b) v3,w3,x,x,z5,w5,w4 22 − w4,y,y2,z4,x3,x4 222 − 2. Simplify. (5 marks)

a) b) 5t72t12 +++− 1r5r46 −−− c) d) 6n2n71n4n4 22 −−−++ x7x31x3x4 22 +−−++−e) 22 n4m15nm52 −+−++

3. Simplify. (3 marks)

a) b) )x64()x73( 22 −++ )w84()w71( 22 +−−−c) )x9x2()x7x6( 22 +−+−

4. Simplify. Then determine the value of the polynomial when n = -3 and when n = 2.

(4 marks) a) b) )1n4n3()4n3n5( 22 −+−+−+ )8n6n()2n5n7( 22 ++−−−−

5. Expand. (5 marks)

a) b) )4w3(w2 + )9n5(n4 −− c) )6c5c7(c 2 −−d) e) )6h)(h( +− )xx)(x6( 2 +−−



Key Question #6 (continued) 6. Determine the product or quotient. (9 marks)

a) b) )x4)(x2( 2 − )m9()m36( 2 −÷

c) ab5ba11

3

33− d) )y)(y2)(x8( 3 −

e) 2

3

w15w75

−− f) )d9)(d3( 2

g) h) )ab3()ba12( 332 ÷− )k)(k7( −

i) ⎟⎠⎞

⎜⎝⎛−⎟⎠⎞

⎜⎝⎛ ab

812ba

64 22

7. Simplify. Then determine the value of the polynomial when a = -1 and when b = 3.

(4 marks) a) b) 3)ab3( 2)ab4(−

8. When the terms of a polynomial in x are arranged from the highest to the lowest

powers of x, the polynomial is in descending order. Simplify the following polynomial in descending order then evaluate for n = 1. (3 marks)

( ) ( ) ( )11n2n46nn5nn26 222 −+−−+−+−

9. When are the sum, difference, product and quotient of two monomials also a

monomial? (3 marks)


Algebra

Lesson 7


Lesson Seven Concepts Overall Expectations






Algebra

Solving Equations When solving algebraic equations we must try to think of a scale always in equilibrium (balanced). It is important to keep the scale balanced at all times. What you do to one side of the equation must also be done to the other side of the equation.

=

+2 +2

You need to get all the terms with the variable to one side and the constants (the ones with out any letters) to the other. It does not matter which side you choose for the isolating of each.



Example Solve each equation algebraically. Check your solution

a) k12k74 +=−− b) 2817x3 =− Solution a) -4 - 7k = 12 + k

This side chosen for the “k’s” -4 - 7k = 12 + k -4 - 7k - k = 12 + k - k

- 1k from both sides to keep scale balanced

-4 - 8k = 12

+4 from both sides to keep scale balanced

-4 + 4 -8k = 12 + 4 - 8k = 16 - 8k = 16 - 8 - 8

Divide 8 from both sides to keep scale balanced

k = - 2 Checking the solution - 4 – 7k 12 + k - 4 – 7(-2) 12 + (-2) - 4 +14 12 – 2 10 10 same

Should now check the answer

Substitute your answer into the original equation.

If both sides equal the same amount then your answer is correct. K = -2



b) 3x – 17 = 28 3x –17 +17 = 28 + 17 3x = 45 3x = 45

3 3 x = 15

Checking the solution 3x – 17 28

3(15) - 17 28 45 - 17 28 28 28

Support Questions

both sides equal the same amount therefore the answer is correct.

1. Solve each equation. Check your solution

a) b) 30w5 = 165x3 =− c) 17t10 =− d) e) 3w212w4 +=− 7r425r12 +=− f) 5x2x65 +=− g) h) 3c2c73 −=+ )w5(3)13w(2 −=+ i) )10j2(7)j21(4 +=− j) )x3(27 −−−=

2. The formula for the area of a parallelogram is bhA = . The area of a parallelogram is

24 and its base is 12 cm. Using algebra, what is the height of the parallelogram?

2cm

3. The cost of a hall rental for a wedding is $500. Each meal at the wedding will cost

and additional $25.00.

a) Write and algebraic equation for the total cost of the wedding. b) Calculate how many people attended the wedding if the wedding cost

totalled $4500.00.



Key Question #7 1. Solve each equation. Check your solution. (11 marks)

a) b) 3 226w2 =− 1n 7− = c) 15 d) 7545w =− c 32c =− e) 6 f) 2.5h2h6 −=+ 3 4 1.2 3.5x x− + = g) 1 41)x4.2(3.3x2. − =+ h) 2 12)2x( =− i) 5 j) 6315)3x( −=− t2( 0) =+ k) )1p(3)1p(2 −=+

2. The formula for the perimeter of a rectangle with length l and width w is P=2l+2w. A

rectangular field is 130 m long and requires 425 m of fencing to enclose it. Determine the width of the field. (3 marks)

3. Volcanoes prove that the Earth’s center is hot. The formula T = 10d +20 is used to

estimate the temperature, T degrees Celsius, at a depth of d kilometres (km). (3 marks)

a) What does each term on the right side of the equation represent? b) Estimate the depth where the temperature is 60°C. c) What is the approximate temperature at a depth of 6 km?

4. The cost, C dollars, to produce a school yearbook is given by the equation

C= 7500 +8n. Where n is the number of yearbooks printed. (4 marks)

a) What does each term on the right side of the equation represent? b) Suppose there is $11 500 to spend on yearbooks. How many yearbooks can be

purchased? c) How many yearbooks can be produced for $20 000? d) How much would 700 yearbooks cost?

5. Suppose you were asked to explain how to solve the equation 3 over the

phone to a friend. Explain in detail the steps that you would tell your friend to solve the equation. (4 marks)

12)2x( =+


Problem Solving Algebraically

Lesson 8


Lesson Eight Concepts Overall Expectations






Solving Problems using Algebraic Modeling Problem Solving Steps

1. Use a variable to represent the unknown quantity. 2. Express any other unknown quantities in terms of this variable. 3. Write an equation and solve. 4. Check your answer to the question. 5. State the answer to the question with a therefore statement.

Example

Members of a Girl Guide troop sold boxes of cookies to raise money for their year end camp. Brianna sold 8 more boxes than her friend Nicola. They sold a total of 46 boxes. How many boxes did each sell?

Solution 1. Use a variable to represent the unknown quantity.

Let b represent the number of boxes sold by Brianna. 2. Express any other unknown quantities in terms of this variable.

If boxes sold by Brianna = b; then boxes sold by Nicola = b − 8. 3. Write an equation and solve.



8 462 8 46

2 8 8 462 542 542 2

27

b bb

bbb

b

+ − =− =

− + = +=

=

=

8

4. Check your answer to the question.

LHS RHS b+ b – 8 46 27 + 27 - 8 46 54 – 8 46 46 46

5. State the answer to the question with a therefore statement.

∴Brianna sold 27 boxes of cookies and Nicola sold 19.

Example

An airplane travels 7 times faster than a train. The difference in their speeds is 420 km/hr. How fast is each vehicle traveling? Solution 1. Use a variable to represent the unknown quantity.

Let v represent the speed the train travels. 2. Express any other unknown quantities in terms of this variable.

If the train travels at v; then the airplane travels 7v.



3. Write an equation and solve. 7 4

6 4206 4206 6

70

v vvv

v

− ==

=

=

20

4. Check your answer to the question.

LHS RHS 7v - v 420 7(70) - 70 420 490 – 70 420 420 420

5. State the answer to the question with a therefore statement.

∴ the train travels at 70 km/hr and the airplane travels at 490 km/hr.

Support Questions

1. Find two consecutive numbers with a sum of 143. 2. A set of golf clubs and bag cost $225. The clubs cost $60 more than the bag. How

much do the clubs cost? 3. Brianna and Noah ran as far as they could in 60 min. Brianna ran 2.5 km less than

Noah. Together they ran 17.5 km in total. How far did each run. 4. Don and his brother Dan went fishing. Don caught 6 times more mass of fish than

his brother. Together they caught 4.2 kg of fish. What mass of fish did each brother catch?



Key Question #8 1. Convert the following phrases into algebraic equations. (5 marks)

a) Three times a number is twenty four. b) Twice a number increased by eight is forty four. c) Five less than four times a number is eleven. d) One more than triple a number is five less than double a number. e) A number decreased by seven is six times a number.

2. Solve for each of the statements in question one. (5 marks) 3. The combined mass of two children is 75 lbs. The first child is four times the mass

of the second child. What are the masses of the two children? (3 marks) 4. A rectangle has a width that is 3cm less than its length. The perimeter of the

rectangle is 22 cm. What is the length and width of the rectangle? (3 marks) 5. The same number of each type of coin has a total of $6.00. There

are nickels, dimes and quarters. How many of each type of coin are there? (4 marks)

6. Are there more ways to solve problems such as the ones given in

questions 1 – 5? Explain. (3 marks)


Slope

Lesson 9


Lesson Nine Concepts Overall Expectations

Apply data-management techniques to investigate relationships between two variables;

Determine the characteristics of linear relations; Demonstrate an understanding of the constant rate of change and its connection to

linear relations; Connect various representations of a linear relation, and solve problems using the

representations. Specific Expectations

Interpret the meanings of points on scatter plots or graphs that represent linear relations, including scatter plots or graphs in more than one quadrant;

Determine through investigation, that the rate of change of a linear relation can be found by choosing any two points on the line that represents the relation, finding vertical change between the points and the horizontal change between the points;

Coordinate Plane

Cartesian Plane uses the x and y axes to plot a point identified by a pair of numbers. If it is known that a point has an x-coordinate of –2, then this point could be located anywhere along the vertical line passing through –2 on the x-axis. If it is known that a point has a y-coordinate of 5, then the point could be located anywhere along the horizontal line passing through 5 on the y-axis. The point of intersection of the lines formed by x= -2 and y= 5 is the location of a point given by the ordered pair (-2, 5). x-coordinate y-coordinate



A

Horizontal axis is the x-axes

Vertical axis is the y-axes

This grid system is the Cartesian plane The

Cartesian plane is divided into four quadrants

Example Plot and label each point: A(3, 5), B(0, -6), C(-7, -4)

Solution



Support Questions 1. Properly draw and label a Cartesian plane then plot and label the following ordered

pairs. A(1, 1), B(1, 5), C(2, -4), D(7, 0), E(0, 8), F(-4, -8), G(-5, 9), H(-3, 0) 2. State the quadrant for each of the ordered pairs plotted in question one. 3. The points (-3, 5) and (3, 5) are two vertices of a square. State all other order pairs

that could be the other two vertices of the square.

Slope Slope describes the steepness of a line or line segment; the ratio of the rise of a line or line segment to its run. Slope is either one of the following four types:

a) positive b) negative c) no slope d) undefined

The formula for determining slope is:runriseslope = .

Example a) Draw a ramp that would lie on the staircase

b) State the slope of the staircase



Solution

+1

runriseslope =

+2

21

21slope =

++

=

Example c) State the slope of the hypotenuse of the triangle given below.

4 12

Solution

−4 +12

runriseslope =

12

4slope −=

Reduced to simplest form.

31slope −=



Support Questions 4. Find the slope of each staircase.

a) b) 5. This drawing represents the side view of road passing through a mountain range.

The road moves from left to right. Calculate the slope of each line segment.

6. The slope of a line segment is –6. What is a possible rise and run? 7. The slope of a line segment is 4. What is the rise if the run is 3? 8. State the slope of

each line segment.



Slope between two points on a Cartesian Plane The slope of line segment can be determined if two ordered pairs of the line segment are known.

The following formula is used: z2

12

xxyy

slope−−

=

Example a) Find the slope of the in line containing the coordinate pairs

A(7,3) and B(-9,7). b) Find the slope of the in line containing the coordinate pairs

A(-3,1) and B(4,6). Solutions )y,x( 11 )y,x( 22

The 1’s in the first ordered pair denotes the x and y values of that term and the 2’s in the second ordered pair denotes the x and y values of that term.

a) (7 , 3) and (-9, 7)

Substitute the values into the equation for slope.

)7()9()3()7(

xxyyslope

12

12

−−−

=−−

=

= 164

−

Fraction needs to be simplified.

= 41

−

Therefore the slope is 41

− .

b) (-3,1) and (4,6)

)3()4(

)1()6(xxyyslope

12

12

−−−

=−−

=

= 75

++

= 75

Therefore the slope is 75 .



Support Questions 9. Find the slope of the in line containing the ordered pairs.

a) A(4, -5), B(-4, 6) b) C(3, 8), D(7, 2) c) E(0, 3), F(2, 1) d) G(5, 4), H(1, -3) e) I(2, 3), J(-6, -2) f) L(3, 7), M(3, -2) g) N(4, -6),P(2, -6)

Key Question #9 1. Properly draw and label a Cartesian plane then plot and label the following ordered

pairs. (5 marks) A(2, 1), B(1, -5), C(3, 4), D(-7, 0), E(0, -8), F(1, -8), G(-4, -9), H(3, 0) 2. State the quadrant for each of the ordered pairs plotted in question one. (2 marks) 3. The points (-4, 2) and (4, 2) are two vertices of a square. State all other order pairs

that could be the other two vertices of the square. (3 marks) 4. This drawing represents the side view of road passing through a mountain range.

The road moves from left to right. Calculate the slope of each line segment. (3 marks)

5. The slope of a line segment is 3. What is a possible rise and run? (2 marks) 6. The slope of a line segment is -5. What is the rise if the run is 4? (2 marks)



Key Question #9 (continued) 7. State the slope of each line segment. (3 marks)

8. Find the slope of the line containing the ordered pairs. (7 marks)

a) A(2, -5), B(-2, 6) b) C(3, 6), D(5, 2) c) E(1, 2), F(2, 1) d) G(-3, 4), H(1, -2) e) I(2, 3), J(-6, 3) f) L(4, 5), M(4, -7) g) N(2, -5),P(3, -8)

9. Decide if the following statements are true, sometimes true or not true. Explain.

(3 marks)

a) An ordered pair with 1 positive coordinate and 1 negative coordinate lies in the 3rd quadrant.

b) An ordered pair with both coordinates positive coordinates lies in the 1st

quadrant.

c) An ordered pair where the x and y coordinate are the same lies in the 1st or 3rd quadrant.


Relationships In Data

Lesson 10


Lesson Ten Concepts Overall Expectations

Apply data-management techniques to investigate relationships between two variables;

Determine the characteristics of linear relations; Demonstrate an understanding of the constant rate of change and its connection to

linear relations; Connect various representations of a linear relation, and solve problems using the

representations. Specific Expectations

Interpret the meanings of points on scatter plots or graphs that represent linear relations, including scatter plots or graphs in more than one quadrant;

Pose problems, identify variables, and formulate hypotheses associated with relationships between two variables;

Describe trends and relationships observed in data, make inferences from data, compare the inferences with hypotheses about the data, and explain any differences between the inferences and the hypotheses;

Construct tables of values and graphs, using a variety of tools; Construct tables of values, scatter plots, and lines or curves of best fit as appropriate

using a variety of tools; Determine values of a linear relation by using a table of values, by using the

equation of the relation, and by interpolating or extrapolating from the graph of the relation.

Relationships in Data

Tables and graphs of data help to show the relationships between quantities. In mathematics the relationship between a pair of quantities is called a relation. Example Use the graph following to answer each question. a) State the percentage of Canadians who enjoy professional wrestling for each of the

following years. Show the popularity in a table. 1960 1970 1980 1990 2000

b) When did the popularity of professional wrestling reach 16% of all Canadians?



Solution

a) State the percentage of Canadians who enjoy professional wrestling for each of the following years? Show the popularity in a table.

1960 1970 1980 1990 2000

Year 1960 1970 1980 1990 2000 Popularity (%)

10 12 14 18 27

b) When did the popularity of professional wrestling reach 16% of Canadians?

Approximately 1985



Example

Which graph below best represents each situation?

a) the height of a person over time b) the height of roller coaster over time c) the amount of hours of sunlight over a year d) the number of D.V.D. players sold compared to selling price

i) ii) iii) iv) Solution

Which graph below best represents each situation? a) the height of a person over time b) the height of roller coaster over time c) the amount of hours of sunlight over a year d) the number of D.V.D. players sold compared to selling price

i) ii) iii) iv) a d c b

Support Questions 1. Use the graph following to answer each question.

a) What year was Canada’s population in the following years?

1955 1965 1975 1985 1995

b) When did Canada’s population reach 25 million?



Canadians Population Since 1950

0

5

10

15

20

25

30

35

1950 1960 1970 1980 1990 2000Year

Popu

latio

n (m

illio

ns)

2. Brianna walks to her grandparents. This graph shows her distance from home during one of her walks. Describe her walk.

3. Noah is riding his bike from his grandparents to his home. Describe Noah’s possible

ride home



Support Questions 4. Refer to the graph given below:

a) How many textbooks make the following heights?

45 cm, 60cm, 75cm

b) Approximately how high is the following number of textbooks?

7 books, 9 books, 4 books, 2 books

5. a) Construct a graph using this data.

Number of stairs climbed

4 6 14 22 28

Height (cm) 75 112.5 262.5 412.5 525

b) Did you join the points? Explain. c) What is the height of 15 stairs? d) How many stair will reach a height of 444 cm.? e) If the number of stairs is doubled will the height double? Explain.



Support Questions 6. a) Construct a graph using this data.

Radius of a circle, (cm)

1 2 3 4 5

Area of a circle. ( ) 2cm

3.14 12.56 28.26 50.24 78.5

b) Did you join the points? Explain. c) What is the approximate radius of circle with an area of ? 2cm100d) What is the approximate area of circle with a radius of 4.5 cm? e) If the radius is doubled will the area double? Explain.

Graphing Relations Relations can be either linear or non-linear. Linear means the relation forms a single straight line and non-linear produces anything that is not a single straight line.

The following formula is used: z2

12

xxyy

slope−−

=

Example Draw a graph of the relation described by the equation. a) 1x3y +−=



Solution a) 1x3y +−= First step is to make a table of values and choose values to place in x column of table.

Second step is to substitute each value into the equation to determine the y value. For x = -2 y = −3x+1 = −3(-2)+1 = 6 + 1 = 7

For x = -1 y = −3x+1 = −3(-1)+1 = 3 + 1 = 4

For x = 0

x y -2 -1 0 1 2

x y -2 7 -1 0 1 2

x y -2 7 -1 4 0 1 2



y = −3x+1 = −3(0)+1 = 0 + 1 = 1

x y -2 7 -1 4 0 1 1 2

For x = 1 y = −3x+1 = −3(1)+1 = -3 + 1 = -2 x

y -2 7 -1 4 0 1 1 -2 2

For x = 2 y = −3x+1 = −3(2)+1 = -6 + 1 = -5 x

y -2 7 -1 4 0 1 1 -2 2 -5

Next, plot the coordinates (x, y) on a grid and join the points with a straight edge and label the equation.



Support Questions

7. Complete the table of values.

a) b) 5x2y −= 6xy +−= c) 3x21y +=

x y x y x y -4 -2 -2

-1 -1 0 2 0 2 5 1 4

8. The cost of D dollars, to print and bind y copies of a yearbook is given by the equation D = 60 +10n.

a) Make a table of values to show the costs for up to 400 Yearbooks. b) Use the table to draw a graph. c) Use the graph to estimate the cost of 325 copies. d) Use the graph to estimate how many copies can be made for $1250.



Support Questions 9. Complete a table of values and graph each relation.

a) b) 4x3y −= 2x41y += c) 5.7w5.5C +=

10. Which of the following ordered pairs satisfy the relation modelled by 7x3y +−= .

Show by substituting the values into the relation.

a) (-3, 6) b) (4, 5) c) (7, 3.5) d) (-4, 19) e) (-4, 6.5)

Key Question #10

1. Which graph below best represents each situation? (4 marks)

a) the height of a person over time b) the height of roller coaster over time c) the amount of hours of sunlight over a year d) the number of D.V.D. players sold compared to selling price

i) ii) iii) iv)

2. Complete a table of values and graph each relation. (8 marks)

a) b) 10n3C += 4x52y −= c) 50.n75.4W −= d) 12l2P +=

3. The time that passes between the time you see lightning and you hear the thunder

depends on your distance from the lightning. With each km from the lightning 3 seconds pass. (4 marks)

a) Make a table of values for distances from 0 to 5 km. b) Graph this relation. State whether it is a linear or non-linear relation. c) Using your graphed relation, how much time passes before you hear lightning

that occurs 4.5 km away? d) Using your graphed relation, how far away are you if you hear the thunder in 1.5

seconds?



Key Question #10 (continued) 4. The amount a taxi driver charges a customer is given by the equation

, where A is the total amount charged and k is the kilometres driven. (5 marks)

25.5k25.1A +=

a) What do the numbers in the equation represent? b) Make a table of values for distances from 0 to 10 km. c) Graph this relation. State whether it is a linear or non-linear relation. d) Using your graphed relation, how much is charged if a person goes 7.5 km? e) Using your graphed relation, how far can a person go in a taxi for $15?

5. Ashlee repairs DVD players. She charges $25 to inspect the problem and $20/h to

repair the device. (5 marks)

a) Write an equation to model this relation. b) Make a table of values. c) Graph this relation. d) Using your graphed relation, how long did it take Ashlee to repair the DVD player

if she charged $55? e) Using your graphed relation, how much should Ashlee charge if it takes her 4.5

hours to repair the DVD player?


MFM1P – Foundations of Mathematics Support Question Answers

Support Question Answers Lesson 6 1. a. b. x,x3 − 22 y,y3 2. a. b.

5x23x4x32x43x3

)2x4()3x3(

+=+++−=+++−=+++−

1n23n6n5

2n6n53)2n6()n53(

+=−++−=

−+−=+−−−

c. d.

4a6a373a4a2a5a87a4a53a2a8

)7a4a5()3a2a8(

2

22

22

22

++=

+−++−=

++−−+=

++−+−+

3x7x1052x2x5x4x6x25x42x5x6

)x25x4()2x5x6(

2

22

22

22

−+−=

−+++−−=

+−−++−=

−+−++−

e. f.

m810nm67nm23

)nm67()nm23(22

22

−=+−+−−=

+−+−−

2

22

22

22

x95x3x672x37x62

)x37()x62(

+−=

++−=

+−+=

−−+

g. h.

2

22

22

22

w52ww635w3w65

)w3()w65(

−=

+−−=

+−−=

−−−

x7x10x4x3x5x5

x5x4x3x5)x5x4()x3x5(

2

22

22

22

−=

−−+=

+−−=

+−+−



3. a. b.

615

1)1(51n51nfor

9110

1)2(51n52nfor

1n534n2n33n24n3

)3n2()4n3(

=+=

+−−=+−−=

−=+−=

+−=+−=+−=

−+−−=−−+−=−−++−

…

…

7511

5)1()1(5nn1nfor

7524

5)2()2(5nn2nfor5nn

32n6n7n2n33n6n22n7n3

)3n6n2()2n7n3(

22

22

2

22

22

22

=++=

+−−−=+−

−==

+−=+−=+−

=+−=

+++−−=

++−+−=

++−++−

…

…

4. a. b. c.

3

2

m21)m7)(m3(

−=

−

n2)n6()n12( 2

−=−÷

b37

ba3ab7 2

−=

−

d. e. f.

22yx24)y2)(xy4)(x3(

=

−−

2

3

a4a7a28

=

−−

3

3

z56)7)(z8(

=

g. h. i.

2

3

a2)ab2()ba4(

−=

÷− 2x20

)x4)(x5(=

23

23

22

ba32

ba4530

a9

10ab53

=

=

⎟⎠⎞

⎜⎝⎛−⎟⎠⎞

⎜⎝⎛−



5. a. b.

16)1)(4(4

)1()2(4ba41b,2afor

ba4)ab2)(ab2(

)ab2(

2222

22

2

==

−=

−===

=

……

60)1)(60(

)1()2(15ba151b,2afor

ba15)ab5)(ab3(

3232

32

2

−=−=

−=

−===

−−

……

6. a. b. x9x3)9x3(x 2 −=− n12n8)3n2)(n4( 2 +−=−− c. d. bb3b2)1b3b2(b 232 +−=+− x2x)2x)(x( 2 +−=−− e. 232 m4m4)mm)(m4( +−=−− Lesson 7 1 a. b. c.

5 35 35 5

6

ww

w

=

=

=

00

3 5 163 5 5 16

3 213 213 3

7

xx

xx

x

5− =

− + = +=

=

=

10 1710 10 17 10

71 71 1

7

tttt

t

− =− − = −

− =−

=− −

= −

d. e.

4 12 2 34 12 12 2 3 12

4 2 154 2 2 2 1

2 152 152 2

152

w ww w

w ww w w w

ww

w

− = +− + = + +

= +− = − +

=

=

=

5

12 25 4 712 25 25 4 7 25

12 4 3212 4 4 4 32

8 328 328 8

4

r rr r

r rr r r r

rr

r

− = +− + = + +

= +− = − +

=

=

=



f. g.

5 6 2 55 6 6 2 6 5

5 4 55 5 4 5 5

0 40

x xx x x x

xx

xx

− = +− − = − +

= − +− = − + −

==

3 7 2 33 3 7 2 3 3

7 2 67 2 2 2

5 65 65 5

65

c cc cc c

c c c ccc

c

6

+ = −− + = − −

= −− = − −

= −−

=

= −

h. i.

2( 13) 3(5 )2 26 15 3

2 26 26 15 26 32 11 3

2 3 11 3 35 115 115 5

115

w ww w

w ww w

w w w www

w

+ = −+ = −

+ − = − −= − −

+ = − − += −−

=

= −

4(1 2 ) 7(2 10)4 8 14 70

4 8 8 14 8 704 22 70

4 70 22 70 7066 2266 2222 22

3

j jj j

j j j jjjjj

j

− = +− = +

− + = + += +

− = + −− =−

=

− =

j.

7 2( 37 6 2

7 6 6 6 21 21 22 212

)xx

xxx

x

= − − −= +

− = − +=

=

=



2.

24 (12)24 1224 1224 1212 12

2

A bhh

hhh

h

====

=

=

Therefore the height of the parallelogram is 2 cm 3. a. C = 25.00m+500 b.

4500.00 25.00 5004500.00 500 25.00 500 500

4000.00 25.004000.00 25.0025.00 25.00

160

mmmm

m

= +− = + −

=

=

=

Therefore 160 people would attend.

Lesson 8 1. Let n be the first consecutive number.

If the first consecutive number is “n” then the next number is “n+1”

1 143

2 1 1432 1 1 143

2 1422 1422 2

71

n nn

nnn

n

+ + =+ =

+ − = −=

=

=

1

∴the consecutive numbers are 71 and 72



2. Let b be the cost of the bag If “b” is the cost of the bag then the clubs are “b+60”

60 225

2 60 2252 60 60 225 60

2 1652 1652 2

82.5

b bb

bbb

b

+ + =+ =

+ − = −=

=

=

∴the bag cost $82.50 and the golf clubs cost $142.50. 3. Let b be the distance ran by Brianna

If “b” is the distance ran by Brianna then Noah ran b+2.5

2.5 17.5

2 2.5 17.52 2.5 2.5 17.5 2.5

2 152 152 2

7.5

b bb

bbb

b

+ + =+ =

+ − = −=

=

=

∴Brianna ran 7.5 km and Noah ran 10 km. 4. Let d be the mass of the fish Dan caught.

If “d” is the mass of the fish Dan caught then Don’s fish’s mass is 6d.

6 4

7 4.7 4.7 7

.6

d ddd

d

+ ==

=

=

.222

∴Don caught 3.6 kg of fish and Dan caught .6 kg of fish.



Lesson 9 1.

2 A. I B. I C. IV D. I and IV E. I and II

F. III G. II H. II and III 3.

4. a. rise = +1 run = +1 111

runrisem ===

b. rise = -2 run = +3 32

32

runrisem −=

+−

==

5.

rise 0slope m 0run 2rise 1 slope mrun 2rise 0slope m 0run 3

rise 1slope m 1run 1

EF

FG

GH

HI

= = = =

= = =

= = = =

−= = = =

rise 2 slope mrun 3rise 1 slope mrun 3rise 0 slope m 0run 4rise 2 slope m 2run 1

AB

BC

DE

= = =

= = =

= = = =

−= = = = −

+ −

CD



6. Answers may vary. 6318

−=− so a possible rise is –18 and a possible run is +3.

7. Therefore the rise is +12.

43

3 33

12

x

x

x

=

⎛ ⎞ =⎜ ⎟⎝ ⎠

=

(4) 8.

11111

runrisemslope f

undefined03

runrisemslope e

133

runrisemslope d

81

runrisemslope c

020

runrisemslope b

23

runrisemslope a

−=−

===

=−

===

−=+−

===

===

====

===

9.

02

0)4()2(

)6()6(xxyymslope g

undefined09

)3()3()7()2(

xxyymslope f

85

85

)2()6()3()2(

xxyymslope e

47

47

)5()1()4()3(

xxyymslope d

122

)0()2()3()1(

xxyymslope c

23

46

)3()7()8()2(

xxyymslope b

811

811

)4()4()5()6(

xxyymslope a

12

12

12

12

12

12

12

12

12

12

12

12

12

12

=−

=−

−−−=

−−

==

=−

=−−−

=−−

==

=−−

=−−−−

=−−

==

=−−

=−−−

=−−

==

−=−

=−−

=−−

==

−=−

=−−

=−−

==

−=−

=−−−−

=−−

==



Lesson 10 1. a. 16 million; 19 million; 23 million; 26 million; 29 million

b. ≈1982

Canadians Population Since 1950

0

5

10

15

20

25

30

35

1950 1960 1970 1980 1990 2000Year

Popu

latio

n (m

illio

ns)

2. Walked at a steady rate, then took a brief rest then walked slightly slower but steadily then reached her grandparents and finally walked quickly back to home.

3. Noah was riding his bike up a hill then once he reached the top rode his bike

quickly down the hill.

4. a. 6,8 and 10 textbooks b. approx 48, 68, 30 and 15 cm.

5. Walk Those Stairs

0

100

200

400

300

500

600

0 5 10 15 20 25 30

Number of Stairs

Hei

ght (

cm)



b. Yes, points are joined to see the possible values between points. c. ≈ 290 cm d. ≈ 23 stairs e. Yes because this is a linear relationship.

6.

Circles

0

20

40

60

80

100

0 1 2 3 4 5

Radius of Circle (cm)

Are

a of

Circ

le

6

b. Yes, points are joined to see the possible values between points. c. ≈ 5.4 cm d. ≈62 2cme. No, this is not a linear relationship.

7. y b. y5x2 −= 6x +−= c. 321wy +=

X 2x -5 Y

-4 2(-4) -5 -13 -1 2(-1) -5 --7 2 2(2) -5 -1 5 2(5) -5 5

X -x + 6 Y -2 -(-2) + 6 8-1 -(-1) + 6 70 -(0) + 6 61 -(1) + 6 5

X w^2 - 1 Y -2 (-2).5 +3 20 (0).5 +3 32 (2).5 +3 44 (4).5 +3 5

8.

X Y 0 60 10 160 50 560 100 1060



b.

c. ≈ $3300 d. ≈ 100 yearbooks

9. a.

X Y -2 -10 4)2(3 −−-1 -7 4)1(3 −− 0 -4 4)0(3 − 1 3 -1 4)1( − 2 2 4)2(3 −

b.

X Y

-8 2)8(21

+− -2

-4 2)4(21

+− 0

0 2)0(21

+ 2

4 2)4(21

+ 4

8 2)8(21

+ 6



c.

X Y -2 5 -3.5 5.7)2(5. +−

+−-1 5 2 5.7)1(5. 0 7.5 5.7)0(5.5 + 1 13 5.7)1(5.5 + 2 18.5 5.7)2(5.5 +

10. (-4, 19)

191971219

7)4(3197x3y

=+=

+−−=+−=


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Polynomials - WordPress.com · MFM1P – Foundations of Mathematics Unit 2 – Lesson 6 Lesson Six Concepts Overall Expectations ¾ Simplify numerical and polynomial expressions in