Converting to and from Metric to Imperial Units

Converting to and

from Metric to Imperial Units

Lesson 6

MFM2P – Foundations of Mathematics Unit 2 – Lesson 6

Lesson Six Concepts

Explain and use correctly prefixes in the imperial and metric system Convert between imperial and metric units commonly used in everyday applications

Converting To and From Metric and Imperial Units Listed below are the main units of conversion from metric to imperial and then from imperial to metric for length, mass and capacity.

First, find the unit being converted in the far left column.

Second, multiply by the value in the far right column to complete the conversion.

Copyright © 2005, Durham Continuing Education Page 2 of 42


First, find the unit being converted in the far left column.

Second, multiply by the value in the far right column to complete the conversion.

Example 1 Convert 2 yards to metres. Solution

Answer: 2 x .9144 = 1.8288 m

Use the tables above to answer this question.

Example 2 Convert 8.5 L to gallons. Solution

Answer: 8.5 x .2642 = 2.2457 gal



Example 3 Convert 11.45 kg to pounds. Solution

Answer: 11.45 x 2.2046 = 25.24267 lbs

Support Questions 1. Convert each measurement to the unit given.

a. 2 miles to km b. 12 in to cm

c. 27 ft to m d. 144 in to cm

e. 144 m to yds f. 100 cm to in

g. 7040 km to miles h. 3 miles to km

i. 100 mm to in j. 250 yds to m

k. 8 L to gals l. 3 gals to L

m. 12 lbs to kg n. 160 kg to lbs

o. 2 oz to g p. 100 g to oz



Key Question #6 1. Convert each measurement to the unit given. (13 marks)

a. 84 ft to m b. 15 in to cm

c. 36 ft to m d. 48 cm to in

e. 10 miles to km f. 7 yds to m

g. 251 m to yds h. 2 miles to km

i. 3520 km to miles j. 3 in to mm

k. 1/2 lb to kg l. 12 mm to in

m. 30 cm and in

2. Nathan has a piece of string that is 30 feet long and needs a piece of string that is

10 m long. Does he have enough string? Explain with words and calculations. (3 marks)

3. Jill is emptying one 8 L container of apple juice into two 1 gallon jugs. Will the two

gallon jugs hold all of the apple juice? Explain with words and calculations. (3 marks)

4. Brianna can drive 325 miles on a full tank of gas. She has a full tank and needs to

drive 590 km. Does she have enough gas to drive the entire 590 km? (3 marks)


Volume

Lesson 7


Lesson Seven Concepts

Radius and diameter Calculations using pi (π) Solving volume questions using formulas and substitution

Volume Volume is the amount of space occupied by a 3-dimensional object.

Formulas to be used to calculate volume.

hrV 2π=

3r4V

3π=

3hbV

2

=

lwhV =



2

bhlV =

3

hrV2π

=

Example Find the volume of each figure. 10 m

a. 8 m

b. 22 cm 8cm 7cm



c. 8 m 12 m 6m 7.41 m 3 cm 5 cm d. 8cm Solution Find the volume of each figure. 10 m

a. 8 m

3

2

2

m628V8)5)(14.3(V

hrV

=

=

π= 3

111111

m628m628m8m5m5

=

=×× ++

Volume is always measured in cubed units.



b. 22 cm 8cm 7cm

3cm1232V)8)(7)(22(V

lwhV

=

==

c. 8 m 12 m 6m 7.41 m

3m76.266V2

)12)(41.7)(6(V

2bhlV

=

=

=

This value is not the height.

Don’t confuse the height of the end triangle with the side value of the triangle.



3 cm 5 cm d.

8cm

Break this object into two pieces. One is the cylinder on top and the other is the cube on the b tt

Since l, w, and h are all the same value we could use the formula

; 3sV =where s = side.

3)8(V =

3

cylinder

2cylinder

2cylinder

cm52.56V

)8()5.1)(14.3(V

hrV

=

=

π=

3cube

cube

cube

cm512V)8)(8)(8(V

lwhV

=

=

=

Volume Total = Volume of Cylinder + Volume of Cube = 56.52 + 512 3cm 3cm = 568.52 3cm

Support Questions 1. Calculate the volume for each of the following objects. 8.5 cm

a. b.

4.5 cm

6 cm

9 cm 7 cm



Support Questions



Key Question #7

1. Calculate the volume for each of the following objects. (12 marks)



Key Question #7 (continued)

2. Calculate the volume of each solid. (8 marks)

a. 14 cm

18 cm 22 cm 19 cm

b. 6 cm 18 cm 35 cm 26 cm



Key Question #7 (continued)

3. A cone has a height of 8.5 cm and a volume of 400 . What is the radius of the cone? (3 marks)

3cm

4. Look at the formula for the volume of a rectangular prism. How does the volume of

a rectangular prism change in each case? (6 marks)

a. The length is doubled. b. Both the length and width are doubled. c. All the length, width, and height are doubled.

5. The box of a truck has dimensions 2 m by 2 m by 3m. Explain how this truck

was able to carry 13 of sand. (4 marks) 3m

6. A circular swimming pool has a diameter of 9 m and a depth of 3 m. What is the volume of the swimming pool? (3 marks)


Surface Area

Lesson 8


Lesson Eight Concepts

Introduction to surface area Radius and diameter Calculations using pi (π) Solving surface area questions using formulas and substitution

Surface Area

Surface Area is a measure of the area on the surface of a three-dimensional object. Formulas to be used to calculate surface area.

rh2r2.A.Srh2.A.S

r.A.S

r.A.S

2total

side

2base

2top

π+π=

π=

π=

π=

2r4.A.S π=

2total

2total

2base

triangle

bbs2.A.S

b2bs4.A.S

b.A.S2bs.A.S

+=

+⎟⎠⎞

⎜⎝⎛=

=

=

)lhlwwh(2.A.S ++=



lbls22

bh2.A.S ++⎟⎠⎞

⎜⎝⎛=

)rs(r.A.S

r.A.Srs.A.S

total

2base

cone

+π=

π=

π=

Example Find the surface area of each figure.

10 m

a. 8 m

b. 22 cm 8cm 7cm



c. 8 m 12 m 6m 7.41 m

3 cm

5 cm

d. 8cm

Solution Find the surface area of each figure.

10 m

a. 8 m

2

2

2

m2.408A.S2.251157A.S

)8)(5(2)5)(14.3(2A.Srh2r2A.S

=

+=π+=

π+π=

This is still area so the units are squared.



b. 22 cm 8cm 7cm

[ ]( )

2cm772.A.S)386(2.A.S

176154562.A.S)8)(22()7)(22()8)(7(2.A.S

)lhlwwh(2.A.S

=

=++=

++=++=

c.

8 m 12 m 6m 7.41 m

2m46.308.A.S7219246.44.A.S

)6)(12()8)(12(22

)41.7)(6(2.A.S

lbls22

bh2.A.S

=

++=

++⎟⎠⎞

⎜⎝⎛=

++⎟⎠⎞

⎜⎝⎛=

There are two

sbh '2

because of the

triangles on each end.

There are 2 ls’s because two of the rectangles that make the sides of the prism have the same dimensions

3 cm

5 cm

d. 8cm



The top of the cylinder would take the place of the circle missing on the top of the cube. There is no bottom of the cylinder. Therefore all that is needed to be calculated is the surface area of the cube and the side of the cylinder.

2

2

2

cm1.431.A.S1.47384.A.S

)5)(5.1)(14.3(2)8(6.A.Srh2)b(6.A.S

=

+=+=

π+=

Since l = b and w = b and h = b then S.A.=2(wh+lw+lh) = 2(bb+bb+bb)

)b3(2 2= 2b6=

Support Questions 1. Calculate the surface area for each of the following objects.



Support Questions (continued)

3m

Key Question #8

1. Calculate the surface area for each of the following objects. (8 marks) 4 m

a. 9.5 m

b. 18 cm 7 cm 4 cm

c.

6 m 10 m

8 m



Key Question #8 (continued) 3.5 cm

d. 9 cm

2. Determine the minimum amount of packaging needed to completely cover a

triangular prism Oblerone bar with these dimensions: length 22 cm; triangular face has edges 4 cm and height 5 cm. Express the surface area to the nearest square centimetre. (3 marks)

3. Calculate the surface area of the solid below. (4 marks)

12 cm

16 cm 18 cm 20 cm

4. Look at the formula for the volume of a rectangular prism. How does the surface area change in each case? (6 marks)

a. The length is doubled. b. Both the length and the width are doubled. c. All the length, width, and height are doubled.



Key Question #8 (continued) 5. A cooler has a 60-L capacity. Its internal length is 60 cm and its internal width is 35

cm. Determine the internal height and the internal surface area of the cooler. (4 marks)

Hint ml 1cm 1 3 =


Algebra

Lesson 9


Lesson Nine Concepts

Continued introduction to algebra Solving for unknowns Checking solutions to algebraic equations

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 without 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 =−

c. 324x

−=+

d. 32n

21n6 +=+−



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

-4 - 8k = 12

+4 from both sides to keep scale

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

Divide 8 from both sides to keep scale

- 8 = - 8 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 side 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

c. 324x

−=+

324x

−=+

13

12

4x −

=+

⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛

13

12

4x

⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛

134

124

4x4

⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛

134

124

4x4 441

128x −=+ 81288x −−=−+ 20x −=

Checking the solution -20 +2 -3

4 -5 +2 -3 -3 -3

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

Find the (lowest common multiple) LCM of 1, and 4 (the denominators) and multiply each term by the LCM.

Write each term as a fraction

Continue to do the same to both sides keeping the scale balanced

Divide the LCM by the denominator in each term then multiply this quotient by the numerator.



d. 32n

21n6 +=+−

32n

21n6 +=+−

32

1n

21

1n6

+=+−

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ −

326

1n6

216

1n66

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ −

326

1n6

216

1n66 2636

4n63n36 +=+− 34n633n36 −+=−+−

1n6n36 +=− 1n6n6n6n36 +−=−− 1n42 =−

421

42n42

−=

−−

421n −=

Find the (lowest common multiple) LCM of 1, 2 and 3(the denominators) and multiply each term by the LCM.

Divide the LCM by the denominator in each term then multiply this quotient by the numerator.

Continue to do the same to both sides keeping the scale balanced



Support Questions

a. Solve each equation. Check your solution.

a. b. 28w7 = 246x9 =+ c. 17t218 =−

d. 126n= e. 2w49w5 −=+ f. 7r325r10 +=−

g. h. 5x4x27 −=+ 7cc99 −=+ i. )w1(5)3w(3 +=+−

j. k. )3j2()j53(2 −=+ 12h53

= l. 5n2

3n

+=

m. 141

3d

=− n. )x3(9 −−=

b. The formula for the volume of a cone is 3

hrV2π

= . The volume of a cone is 78.5

and its radius 5cm. Using algebra, what is the height of the cone? 3cm

c. The cost of a renting a limousine for 12 hours is $500. Each extra hour using the

limousine costs an additional $47.50.

a. Write and algebraic equation for the total cost of the limousine. b. Calculate how many extra hours the limousine was rented if the cost totalled

$785.00.



Key Question #9

1. Solve each equation. Check your solution. (14 marks)

a. b. 37w3 =− 174n2 =− c. 44w11 =+ d. e. 32c7c3 =− 5h7h39 −=− f. 5.7x2.14x5.2 =++

g. h. x5.37)x4.2(3x5.1 +=−+ 1257

7x3

=+ i. 3r

452 −=−

j. k. 17)1x(5 =+− 12)2x(5 =− l. 18)6t2(3 =+

m. n. )1p(3)4p(2 +=− 159n3 =+

2. The formula for the circumference of a circle is C=2πr. The distance around the

outside of a circular paved driveway is 38.88 m. What is the radius of the driveway? (2 marks)

3. Volcanoes prove that the Earth’s center is hot. The formula T = 15 d +18 is used to

estimate the temperature, T degrees Celsius, at a depth of d kilometres (km). (5 marks)

a. What does each term on the right side of the equation represent? b. Estimate the depth where the temperature is 75°C. c. What is the approximate temperature at a depth of 4 km?

4. The cost, C dollars, to produce a school newspaper is given by the equation C= 725

+2.75n. Where n is the number of magazines printed. (7 marks)

a. What does each term on the right side of the equation represent? b. Suppose $12500 was spent on producing magazines. How many

magazines were printed? c. How many magazines can be produced for $20 000? d. How much would 2500 magazines cost?

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

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

12x5)5x(3 +=−


Converting to and from y-intercept

form and standard form

Lesson 10


Lesson Ten Concepts

Using algebra to convert to y-intercept form from standard form Using algebra to convert to standard form from y-intercept form

Converting to and from Y-intercept to Standard Form

Example 1 Convert into standard form, . 0Ax By C+ + =

3x21y +−=

Solution

( ) ( )12 2 22

2 1 62 62 6

2 6 6 62 6 0

y x

y xy x x xy x

y xx y

−⎛ ⎞= +⎜ ⎟⎝ ⎠

= − ++ = − + ++ =

+ − = −+ − =

3

Multiply each term by 2 to eliminate the fraction.

A=1, B=2 and C=-6



Example 2 Convert into standard form.

5x32y +=



Example 3 Convert into y-intercept form, . y mx b= +

07y4x3 =++

Support Questions 1. Convert the following linear equations from y-intercept form to standard form.

a. b. 1x2y +=92x

73y −−= c.

21x5y += d. 3x4y +−=

2. Convert the following linear equations from standard form to y-intercept form.

a. b. 06y3x5 =+− 010y5x2 =−+ c. 04y2x =+−



Key Question #10 1. Convert the following linear equations from y-intercept form to standard form.

(8 marks)

a. b. 3x2y −=52x

31y −= c.

21x3y +−= d. 7x2y +=

2. Convert the following linear equations from standard form to y-intercept form.

(12 marks) a. b. 03yx5 =+− 014y7x =−+ c. 09y3x6 =+−

d. 7x +2y –3 = 0 e. 2x +5y + 5 = 0 f. 12x –30y +13 =0

3. The cost of renting a banquet hall is given by the linear equation y = 25x +100.

Write this equation in standard form. (2 marks) 4. Brianna sells computers and earns base salary of $300 per week and $50 for every

computer sold. Her salary is modelled by the equation y = 50x +300. Write this linear equation in standard form. (2 marks)

5. The following conversion from y –intercept to standard form is incorrect. Explain

where the error(s) occurs and how to correct the mistake(s). (4 marks)

( )

1 54

14 44

4 1 54 54 5

4 5 5 54 5 0

y x

y x

y xy x x xy x

y xx y

= +

⎛ ⎞= +⎜ ⎟⎝ ⎠

= +− = − +− =

− − = −− + − =

5


MFM2P – Foundations of Mathematics Support Question Answers

Answers to Support Questions Lesson Six 1. a. 2(1.6093) = 3.2186 km b. 12(2.54) = 30.48 cm

c. 27(0.3048) = 8.2296 m d. 144(2.54) = 365.76 cm e. 144(1.0936) = 157.4784 yds f. 100(0.3937) = 39.37 in

g. 7040(0.6214) = 4374.656 miles h. 3(1.6093) = 4.8279 km i. 100(0.03937) = 3.937 in j. 250(0.9144) = 228.6 m k. 8(0.2642) = 2.1136 gal l. 3(3.785) = 11.355 L

m. 12(0.4536) = 5.4432 kg n. 160(2.2046) = 352.736 lbs o. 2(28.35) = 56.7 g p. 100(0.0353) = 3.53 oz

Lesson Seven 1. a. b. c.

3

2

2

cm01.397V

)7()25.8)(14.3(V

hrV

=

=

π=

3cm243V

)9)(5.4)(6(VlwhV

=

==

3

2

2

m44.301V3

)8()6)(14.3(V

3hrV

=

=

π=

d. e. f.

3

2

2

m1.305V3

)3.11()9(V

3hbV

=

=

=

3

3

3

43

4(3.14)(7.2)3

1562.66

rV

V

V m

π=

=

=

3m540V2

)15)(9)(8(V

2bhlV

=

=

=



Lesson Eight 1. a. b.

2

2

2

cm48.571.A.S)6)(7)(14.3(2)7)(14.3(2.A.S

rh2r2.A.S

=

+=

π+π=

2cm258.A.S)129(2.A.S

)543045(2.A.S)6)(9()6)(5()9)(5[(2.A.S

)whlhlw(2.A.S

=

=++=

++=++=

c. d.

( )

( ) ( ) (2

S.A.

3.14 9 15 9

678.24m

r s rπ= +

=

=

)+

)

( ) ( )

( ) ( ) ( )

2 2 2

2 2

2

2

2

3.6 12156.9612.5

S.A. 2

2 7.2 12.5 7.2

231.84m

s a b

s

bs b

= +

= +

==

= +

= +

=e. f.

( ) (

2

2

2

S.A. 4 r

4 3.14 8.7

950.66m

π=

=

=

( )( ) ( ) ( ) ( )( )

2

S.A. 2 223 4

2 2 11 5 12

12 110 33155m

bh ls lb⎛ ⎞= + +⎜ ⎟⎝ ⎠⎛ ⎞

= + +⎜ ⎟⎝ ⎠

= + +

=

1 3

Lesson Nine 1. a. b. c.

7 27 27 7

4

ww

w

=

=

=

88

9 6 249 6 6 24

9 189 189 9

2

xx

xx

x

6+ =

+ − = −=

=

=

18 2 1718 18 2 17 18

2 12 12 2

12

tttt

t

− =− − = −

− = −− −

=− −

=



d. e. f.

126

6 6(16

72

n

n

n

=

⎛ ⎞ =⎜ ⎟⎝ ⎠

=

2)1

5 9 4 25 9 9 4 2 9

5 4 115 4 4 4 1

11

w ww w

w ww w w w

w

+ = −+ − = − −

= −− = − −

= −

10 25 3 710 25 25 3 7 25

10 3 3210 3 3 3 32

7 327 327 7

327

r rr r

r rr r r r

rr

r

− = +− + = + +

= +− = − +

=

=

=

g. h. 7 2 4 5

7 7 2 4 5 72 4 12

2 4 4 4 12 122 122 2

6

x xx xx x

x x x xxx

x

+ = −− + = − −

= −− = − −− = −− −

=− −

=

2

9 9 79 9 9 7 9

9 169 1

8 168 168 8

2

c cc cc c

c c c ccc

c

6

+ = −− + = − −

= −− = − −

= −−

=

= −

i. j.

3( 3) 5(1 )3 9 5 5

3 3 9 5 5 39 5 8

9 5 5 5 814 814 88 8148

w ww w

w w w ww

www

w

− + = +− − = +

− + − = + +− = +

− − = − +− =−

=

−=

2(3 5 ) (2 3)6 10 2 3

6 6 10 2 3 610 2 9

10 2 2 2 98 98 98 8

98

j jj jj jj j

j j j jjj

j

+ = −+ = −

− + = − −= −

− = − −= −−

=

−=



k. l.

3 125

35( ) 5(12)53 603 603 3

20

h

h

hh

h

=

=

=

=

=

23 5

15( ) 15(2) 15( )3 55 30 3

5 3 30 3 32 302 302 2

15

n n

n n

n nn n n n

nn

n

= +

= +

= +− = + −

=

=

=

m. n.

1 13 4

112( ) 12( ) 12(1)3 4

4 3 124 3 3 12

4 154 154 4

154

d

d

dd

dd

d

− =

− =

− =− + = +

=

=

=

3

9 (3 )9 3

9 3 3 312

xx

xx

= − −= − +

+ = − + +=

2.

2

2

2

2

2 2

(3.14)(5)78.53

(3.14)(5)3(78.5) 3( )3

235.5 (3.14)(5)235.5 (3.14)(5)

(3.14)(5) (3.14)(5) 3

h

h

hh

h

=

=

=

=

=



3. a. C= 47.5h+500 b.

785.00 47.5 500785 500 47.5 500 500

285 47.5285 47.547.5 47.5 6

hhhh

h

= +− = + −

=

=

=

Lesson Ten 1. a. b.

2 12 2 22 1

2 1 1 12 1 0

1( 2 1) 02 1 0

y xx y x xx y

x yx y

x yx y

= +− + = − +− + = +

− + − = + −− + − =

− − + − =− + =

1

3 27 9

3 263( ) 63( ) 63( )7 9

63 27 1463 27 27 27 1463 27 14

63 27 14 14 1463 27 14 027 63 14 0

y x

y x

y xy x x xy x

y xy xx y

= − −

−= −

= − −+ = − + −+ = −

+ + = − ++ + =+ + =

c. d.

152

12( ) 2(5 ) 2( )2

2 10 12 10 10 102 10 1

2 10 1 1 12 10 1 010 2 1 0

1( 10 2 1) 010 2 1 0

y x

y x

y xy x x xy x

y xy xx y

x yx y

= +

= +

= +− = − +− = +

− − = + −− − =

− + − =− − + − =

− + =

13

4 34 4 44 3

4 3 3 34 3 0

4 3 0

y xy x x xy x

y xy x

x y

= − ++ = − + ++ =

+ − = −+ − =+ − =



2. a. b.

5 3 6 05 5 3 6 0 5

3 6 53 6 6 5 6

3 53 53 3

5 23

x y

66

x x y xy x

y xy xy x

y x

− + =− − + = −

− + = −− + − = − −

− = − −− −

= −− −

= +

2 5 10 02 2 5 10 0 2

5 10 25 10 10 2 10

5 2 15 2 15 5

2 25

x y

00

x x y xy x

y xy xy x

y x

+ − =− + − = −

− = −− + = − +

= − +− +

=

= − +

c.

2 4 02 4 02 4

2 4 4 42 42 42 2

1 22

x yx x y x

y xy x

y xy x

y x

− + =− − + = −

− + = −− + − = − −

− = − −− − −

=− −

= +


Date post:	24-Jan-2022
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Converting to and from Metric to Imperial Units

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