Math Makes Sense 9 Preparation and Practise Book
p2
1. a) 49, 7 b) 4 2 =16, 16 square units. 16 =4, 4 units c) 82 =64, 64 square units. 64 =8, 8 units
d) 112 =121, 121 square units. 121 =11, 11 units
p3
1. a) 4 b) 144, 12 c) 5, 52 d) 10, 10 2 =100
1. 7, 7, 49, 49  7
3,3,9, 9  3 4,4,16, 16  4 5,5,25, 25  5 6,6,36, 36  6 7,7,49, 49  7 8,8,64, 64 ,8
9,9,81, 81 ,9 10,10,100, 100 ,10 11,11,121, 121 ,11 12,12,144, 144 ,12
p5
3 x3
9 9
3
3
3 x3
9 9
1. a)
=
,
b) x
=
= ,
c) 0.25, 0.25 d) 6.25, 6.25
8 x8
2
2
4 4
64 64
2 x2
1. a) is, 3x3 = 9
is, 7x7 = 49
is
b) is, 5x5 = 25
is not, it is not the product of 2 equal factors
is not
c) is, 8x8 = 64
is, 9x9 = 81
is
3
4
3 x3
4 x4
2. a)
=
b)
=
2
9
2 x2
9 x9
p7
1. b) 2.4, Terminating, Yes c) 0.5, Terminating, Yes d) 1.581 138 830…, Non-repeating, Nonterminating, No
1. a)
1x1 1 1
,
,
4 x 4 16 16
b)
2 2 2 x2 4
4
x 
,
,
7 7 7 x7 49 49
c) 0.6,0.6,0.36, 0.36
d) 1.1, 1.1, 1.21, 1.21
2. b) Yes; 5x5 = 25, Yes; 7x7 = 49, Yes c) Yes; 6x6 = 36, Yes; 11x11 = 121, Yes
d) No, Yes; 5x5 = 25, No e) Yes; 3x3 = 9, Yes; 10x10 = 100, Yes
7
5
3
7 x7
5 x5
1x1 1
3 x3
3. a)
,
b)
,
c)
,
d)
,
10 x10 10
12 x12 12
4 x4 4
20 x20 20
4. a) 2.9 b) 0.26 c) 7.15 d) 2.5
5. a) 1.2, terminates, is b) 5.5, terminates, is
c) 2.915 475 947, does not appear to repeat or terminate, is not
d) 0.16, terminates, is
2
3
3 3 3 9 9
6. a) units,   , x , , square units
5
 5  5 5 25 25
2
6
36
 6  6 6 36
b) units,   , x ,
, The area is
square units
7
49
 7  7 7 49
c)  5.4  , 5.4 x 5.4, 29.16, The area is 29.16 square units d)  2.1 , 2.1 x 2.1, 4.41, The area is 4.41
square units
3 3
5
9
3x3
25
5 x5 5
7. a)
,
,
,
b)
,
, , The side length is units
6
100
10 x10 10 10
36
6 x6 6
c) 0.01 , 0.1, The side length is 0.1 units d) 46.24 , 6.8, The side length is 6.8 units
2
2
p.10
1. a) 5 b) 7 c) 7 d) 6
2. a) 2.5 b) 2.6 c) 2.6 d) 2.7
p.11
1. a) 16 , 25 , 16 , 4, 25 , 5, 4, 5 b) 4, 9, 4 , 9 , 4 , 2, 9 , 3, 2, 3
p.12
1. a) 92 , 122 , 81, 144, 225, 225 , 15, 15 b) 10 2 , 242 , 100, 576, 676, 676 , 26, 26
p.13
1. a) 9, 16, 9 , 16 , 3, 4, 16, 9, 4, 3, 3, 4, 4 b) 49, 64, 49 , 64 , 7, 8, 49, 64, 7, 8, 7, 8, 7
p.14
25
25 5 5
9
9 3 3
1. a) 25, 81,
,
, ,
b) 9, 16,
,
, ,
81
81 9 9
16
16 4 4
p.15
1. a) 49, 64, 49, 64, (answers may vary) 60.5, 60.5 , 7.8, 60.5
2
b) 11 , 121, 122 , 144, 121, 144, (answers may vary) 129.1, 129.1 , 11.4, 129.1
p16
1. a) 36, 49; 6, 7
b) 9, 16; 3, 4 c) 121, 144; 11, 12 d) 81, 100; 9, 10
2. a) 4, 9; 12,3
b) 16, 49; 4,7
c) 4, 25; 2, 5
d) 9, 64; 3, 8
3. a) 1, 4; 1 , 4 ; 1, 2; 1, 4, 1, 2; 1, 2, 1
b) 36, 49; 36 , 49 ; 6, 7; 49, 36, 7, 6; 6, 7, 7
c) 64, 81; 64 , 81 ; 8, 9; 81, 64, 9, 8; 8, 9, 9
4. a) 9, 16;
9 3
;
16 4
1 1
;
9 3
b) 1, 9;
c) 36, 36;
36
,1
36
d) 100, 121;
100 10
;
121 11
5. a) 1, 4; 1, 4; (answers may vary) 1.5; 1.5 , 1.2; 1.5
b) 64, 81; 64, 81; (answers may vary) 70.3; 70.3 , 8.4; 70.3
c) 2.52 , 6.5, 3.52 , 12.25; 6.25, 12.25; (answers may vary) 11.5, 11.5 , 3.4; 11.5
d) 20 2 , 400, 212 , 441; 400, 441; (answers may vary) 431.1; 431.1 , 20.8; 431.1
6. a) 26.01, 39.69; 65.7; 65.7 ; 8.1; 8.1
b) 7.0, 10.5; 49, 110.25; 159.25; 159.25 ; 12.6; 12.6
p. 19
1. a)
4
4
;
49 49
b)
8 8 64
,
,
11 11 121
c) 0.1, 0.1, 0.01; 0.01
d)1.96; 1.96
2. b) Yes; 3 x 3 = 9; Yes; 5 x 5 = 25; Yes
c) Yes; 5 x 5 = 25; No; No
3 3 3
4 4 4
6 6 6
3. a)
b)
c)
4. a) i), iii), and iv) are checked
x ;
x ;
x ;
7 7 7
5 5 5
11 11 11
b) I used a calculator to find the square root of each decimal. If the square root is a repeating or terminating
decimal, the decimal is a perfect square.
5. a)
5 5 2 5 5 25 25
,( ) , x ,
,
9 9
9 9 81 81
6. a) 4, 9;
4,
b)
81
,
100
9 ; 2, 3; 9, 4, 3, 2; 2, 3, 3
9 9 9 9
,
x ,
10 10 10 10
b) 64, 81;
64 ,
81 ; 8, 9; 64, 81, 8, 9; 8, 9, 8
16 9
25 5
;
b) 25, 49,
;
8. a) 4, 9; 4, 9; (answers may vary) 5.2; 5.2 , 2.3; 5.2
49 7
81 4
b) 36, 49; 36, 49; (answers may vary) 48.2; 48.2 , 6.9; 48.2
9. a) 1.12 , 2.2 2 ; 1.21, 4.84; 6.05; 6.05 ; 2.5; 2.5 b) 1.82 , 2.82 . 3.24, 7.84; 11.08; 11.08 ; 3.3; 3.3
p. 23
1. a) Front & Back: 5, 2; 5, 2, 20 Top & Bottom: 5, 3; 5, 3, 30 Right & Left: 3, 2; 3, 2, 12 Total: 62
b) Front & Back: 12, 12; 12, 12, 288 Top & Bottom: 12, 12; 12, 12, 288;
Right & Left: 12, 12; 12, 12, 288 Total: 864
p. 25
1. Front & Back: 3, 6 Top & Bottom: 4, 8 Right & Left: 2, 4 Total: 18
p. 27
1. a) 3,1; 1, 3; 3
b) 40, 70, 3; 104
p.28
2. 6; 6, 6; 6,6, 216; Total: 216; 2, 2; 2, 2, 24; 24; Total: 24; 2, 2; 2, 2, 4; Area of Overlap: 4; SA smaller
cube, 2(Area of Overlap), 216, 24, 4, 232, 232
1. Back, 2(4)=8; Bottom, 2(4)=8; Left, 2(3)=6; Total = 22; 22
p. 29
2. a) 3, 6; 4, 8; 2, 4; Total = 18; 18
b) 5,10; 2(3)=6; 2(3)=6; Total = 22; 22
p.30
3. SA of larger prism: 15, 5, 150; 2(20  15) =600; 2(20  5) =200; Total = 950; 950
SA of smaller prism: 5, 10, 100; 2(5  10) =100; 2(10  10) =200; Total = 400; 400
Area of Overlap: 10, 5, 50; 50; SA large prism, SA smaller prism, 2(Area of overlap); 950, 400, 50, 1250;
1250
p. 31
4. SA of cube: 30, 30; 30, 30, 5400; 5400; 5400
SA of rectangular prism: 20, 10, 400; 2(20  10) =400; 2(10  10) =200; 1000; 1000
Area of Overlap: 10, 10, 100; 100
SA of composite object: SA cube, SA prism, 2(Area of overlap); 5400, 1000, 2(100); 6200; 6200
p. 32
5. No
SA of warehouse to be painted: 20, 20, 800; 30, 20, 1800; 2600; 2600
SA of loading dock to be painted: 30, 10, 600; 30, 20, 600; 20, 10, 400; 1600; 1600
Area of overlap: 20, 10, 200; 200
SA area of composite object to be painted: 2600, 1600, 2(200), 3800; 3800; 3800; 2.50; 3800, 2.50, 9500
p. 34
1. Triangular = 9, 12; 9, 12, 108; Rectangular = 15, 5; 15, 5, 75; 9, 5; 9, 5, 45; 12, 5; 12, 5, 60; Total = 288;
288
p. 35
1. Top Bottom: 4; 2, 4 2 , 100.53
Curved surface: (4), 6; 2, 4, 6, 150.80; Total =251.33; 251
p. 37
1. a) 6, 12; 6, 12, 72 b) 300,720, 2(72), 876, 876 cm 2
1
2. Triangular: 4, 3; , 3, 4, 12 Rectangular: 5, 3; 5, 3, 15; 4, 3; 4, 3, 12; 3, 3; 3, 3, 9; Total= 48; 48
2
p. 38
SA of cube: 3, 3; 3, 3, 54; Total: 54; 54
Area of Overlap: 3, 3; 3, 3, 9; 9
48, 54, 2(9); 84; 84
p. 39
7. a) 16, 81;
1. a) circle of radius 3; 32 ; 28.27
b) prism, cylinder, Area of overlap; 240,56.5, 2(28.27);239.96; 240
p. 40
2. SA of cube: 6, 6; 6, 6, 216; Total =216
SA of cylinder: 1; 2, 12 , 6.28; (1), 4; 2, 1, 4, 25.13; Total
=31.41
2
Area of Overlap: 1 , 3.14
SA composite object: cube, cylinder, Area of Overlap; 216, 31.41,
2(3.14); 241.13; 241
p. 41
1. SA of rectangular prism: 4(24  12), 1152; 2(12  12), 288; Total =1440; 1440
1
2. SA of triangular prism: 2(  3  8) = 24; (8.5  4) = 34; (3  4) = 12; (8  4) = 32; Total = 102; 102
2
Area of overlap: 8, 4, 32; 32
SA of composite object: 1440 + 102 - 2(32); 1478; 1478
p. 42
2. SA of cube: 2, 2, 24; Total = 24; 24
SA of cylinder: 2, 6, 226.19; 2, 6, 4, 150.80; Total = 376.99;
376.99
Area of Overlap: 2, 2, 4; 4
Surface Area of composite object: 24, 376.99, 2(4); 392.99; 393
p. 43
3. SA of smaller cake: 2, 52 , 157.08; 2, 5, 5, 157.08; Total = 314.16; 314.16
SA of larger cake: 2, 10 2 , 628.32; 2, 10, 7, 439.82; Total = 1068.14; 1068.14
Area of Overlap: 52 , 78.54; 78.54
SA of Cake: 314.16, 1068.14, 2(78.54); 1225.22; 1225
p. 44
List all the square root day in a year:
1/1 January 1; 2/4 February 4; 3/9 March 9; 4/16 April 16; 5/25 May 25
List all the square days in a year:
1/1 January 1; 4/2 April 2; 9/3 September 3
List all the square years from 1000 to present
1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936
p. 46
3 3 9
9
1. a) , ,
;
b) 98.01; 98.01
7 7 49 49
2. a) Yes, 5  5 =25; Yes, 9  9=81; Yes
b) No; Yes, 2  2 =4; No
c) Yes, 7  7 =49; No; No
3. a) 2.3; Terminating; Yes
b) 12.5; Terminating; Yes
c) 2.529822128; Appears to be nonterminating and non- repeating; No
5
4. a) b) 7.7
9
5. 14.5 is between 9 and 16; so, 14.5 is between 9 and 16 . That is, 14.5 is between 3 and 4. Since
14.5 is closer to 16 than 9, 14.5 is closer to 4 than 3. So, 14.5 is between 3 and 4, and closer to 4.
p. 47
1
1
9
3
6. a)1; 16;
;
b) 9; 64;
;
64
16
4
8
2
7. a) 4,9; 4, 9; 8.5; 8.5 , 2.9; 8.5
b) 6.5 , 42.25, 7.52 , 56.25; 42.25, 56.25; 47.1; 47.1 , 6.9; 47.1
8. a) 4.2 2 , 7.82 ; 17.64, 60.84; 78.48; 78.84 ; 8.9; 8.9
b) 7.12 , 10.52 ; 50.41, 110.25; 160.66; 160.66 ; 12.7; The length of the hypotenuse is about 12.7cm.
p. 48
9. Front Back; 2(6) =12 Top Bottom; 2(3) =6 Right left; 2(3) =6 Total: 24 Surface area: 24
10. Surface area of Cube
Front/Back, Top/Bottom. Right/Left; 50, 50, 15000
Total: 15000
Surface Area =15000 cm 2
Surface area of rectangular prism
Front/Back, 2(20  10) =400; Top/Bottom, 2(40  10) =800; Right/Left, 2(20  40) =1600; Total: 2800
Surface area: 2800 cm 2
p. 49
10. Area of overlap: 20, 10, 200; Area of overlap: 200
SA cube, SA smaller prism, 2(Area of Overlap); 15000, 2800, 2(200); 17400; Surface Area: 17400
11. Surface area of rectangular prism
Front/Back, 2(8  4) =64; Top/Bottom, 2(9  8) =144; Right/Left, 2(4  9) =72; Total: 280; SA: 280
Surface area of triangular prism
1
2(  3  4) =12; 5  5 =25; 4  5 =20; 3  5 =15; Total =72; SA =72
2
Area of Overlap
4  5=20; The area of overlap is 20 cm 2
SA rectangular prism, SA triangular prism, 2(Area of Overlap); 280, 72, 2(20); 312; SA =312
p.50
12. 16, 8; 4, 2
Surface area of smaller cylinder
2    2 2 =25.13; 2    8  12 =150.80; Total: 175.93; Surface area: 175.93
Surface area of larger cylinder
2    82 =402.12; 2    8  12 =603.19; 1005.31; Surface area: 1005.31
Area of overlap:   2 2 =12.57; The area of overlap is about 12.57
175.93, 1005.31 -2(12.57); 1156.10; 1156
p.52
1. a) Positive b) Negative c) Negative d) Positive
2. a) 28 b) -18 c) -90 d) 45 e) 15 f) -10 g) -16 h) -12
p. 53
1. a) -6 b) 16 c) -16 d) 18
p. 55
1. a) 2 6 b) 54 c) (10)3 d) 4 2 e) (7)8 2. a) 84 b) 7 2 , 7 squared c) 36 d) 23 , 2 cubed
p. 56
1.
Power
Repeated multiplication
Standard Form
23
2 2 2
8
62
6 6
36
34
3 3 3 3
81
10 4
8 squared
10  10  10  10
8 8
10000
63
7 cubed
7 7 7
343
p. 57
1. a) -1; (-1)(-1)(-1); -1 b) 10; -(10  10  10); -1000 c) -7; (-7)(-7); 49 d) -5; -(-5)(-5)(-5)(-5); -625
Practice
1. a) 7, 7; 87 b) 10, 5, 5; 105 c) -2; -2, 3, 3; (2)3 d) 6, -13; -13, 6, 6; (13)6
2. a) 94 b) 56 c) 112 d) (12)5
p. 58
3. a) 3  3 b) 3  3  3  3 c) 2  2  2  2  2  2  2 d) 10  10  10  10  10  10  10  10
4. a) Positive b) Positive c) Negative d) Negative
5. a) (-3)(-3);9 b) 6  6  6; 216 c) (-10)(-10)(-10); -1000 d) –(4  4  4); -64
6. a) (3)3 ; -27 b) (8)2 ; 64 c) 83 ; -512 d) - (1)7 ; 1
7. a) 43 = (4)(4)(4); 64
b) (-2) 9 is negative, because there is an odd number of negative factors.
c) (-9) 2 is positive, because there is an even number of negative factors.
d) 3 2 is not equal to 2 3 , because 3 2 =(3)(3) =9, and 2 3 =(2)(2)(2) =8
e) (10) 2 =(-10)(-10) =100
p. 59
1. a) The input starts at 1, and increases by 1 each time. The output starts at 5, and increases by 5 each
time. You can also multiply the input by 5 to get the output.
b) The input in the last row is 4+1 =5. The output in the last row is 20 + 5=25.
p. 60
2. a) The input starts at 10, and decreases by 1 each time. The output starts at 100, and decreases by 10
each time. You can als0 multiply the input by 10 to get the output.
b) To extend the table 3 more rows, continue to decrease the input by 1 each time. Decrease the
output by 10 each time.
Input
Output
5
50
4
40
3
30
1. a) 7  1000 b) 9  100 c) 4  100 d) 3  10
p. 61
1. a) 1 b) -1 c) 1 d) 1
p.62
1a) 5 b) -7 c) 10 d) 1 1a) 125, 5  5 , 25 , 5, 5 b) 5 c) 1
p.63
2a) 1 b) 1 c) 1 d) -1 e) 10 f) -8 3a) 10 4 b) 106 c) 107 d) 100 e) 109 f) 101 4a) -1 000 000 b) -1
c) -100 000 000 c) -10 5a) 1012 b) 1000 000 000 000, 1014 c) 100  1 000 000 000 000  1014
6a) 50 000 b) (3 10)  (7 1)
c) (2 1000)  (6 100)  (4 10)  (9 1) d) (7 1000)  (8 1)
400+30+7
2000  600  40  9
7000  8
437
2649
7008
p.64
1a) -7 b) 4 c) -3 d) -8 2a) -10, 10 b) -10, -2, -2
p.65
1a) -8 b) 8 c) -17 d) 17
p.66
1a) -2 b) 2 c) -5
p.68
a) (4)(4) + 3 b) (5)(5) – (2)(2) c) 32
d) (1)2
16 + 3
25 – 4
(3)(3)
(-1)(-1)
19
21
9
1
p.69
(32  60 ) 2  21
102  (2  22 ) 2
5  32
82  4
 (9  1) 2  21
2
2
 5  (3)(3)  (8)(8)  4
1. a)
b)
c)  10 2  21
d)  10  8
 5 9
 64  4
 100  64
 100  2
 45
 16
 164
 50
(6  4)  32 10  102  4
 10  9 10  100  4
1.
 10  90  25
 75
p.70
(2  1) 2 (2  1) 2
(4  2) 2 ( 4) 2  2
22  1
22  1
4  22
42  2
2
 2 2 1
 2  2  1  32
 4  2  2  4  4  2  82
 (4)(4)  2
1. a)
b)
c)
d)  1
2. a)
b)
c)
d)
 4 1
 4 1
 4 4
 16  2
 16  2
 3 3
 88
 1 1
5
3
 16
 32
8
9
 64
1
3
3
2
2
2  (1)
3  (1)
23  (1)3
3
2
3
3
(2  1)
(2  1)
 (2)(2)(2)  (1)
 (3)(3)  (1) 2 (3  1)
3
 (2)(2)(2)  (1)
3
2
3
3
2

1

8

(

1)
3. a)
b)
c)  8  (1)(1)(1) d)  3
4. a)  9  (1)
b)  3
 (3)(3)(3)
 3 3
 (1)(1)(1)  8  (1)
 8  (1)(1)(1)
 9  (1)( 1)
 27
9
1
 8  (1)
 9 1
9
7
9
2
2
3
2
3  ( 2)
2  (2)
52  (5)1
 (3)(3)  (2) 2
 (5)(5)  ( 5)1
c)  9  (2)
d)  25  (5)1
 9  (2)(2)
 25  (5)
 9 4
 (5)
2
(2)0  (2)
5. a)  1 (2)
 (2)
 (2)(2)(2)  ( 2) 2
b)  8  (2)
 8  (2)(2)
2
 84
(3  2)0  (3  2)0
c)  1  1
2
d)
(3  52 ) 0
1
 36
2
2
(2)(3)  (4)
70  (6)(6)  60  (6)(6)
3(2  1) 2 (2) 2  (3)(4)
(2)  30  (2)
 (2)(3)  (4)(4)
2
 70  36  60  36
 (2)(2)  (3)(4)  (2)  1 (2)
 3(1)
e)  (2)(3)  16
f)
g)
h)
6.
$4680
 2520  2160
 4  12
 (2)  (2)

3(1)
 6  16
 4680
 16
 4
3
 10
p.72
1. a) Base: 6 Exponent:2 ; There are 2 factors of 6. b) Base: 4 Exponent: 5; There are 5 factors of 4.
c) Base: -3 Exponent: 8; There are 8 factors of (-3). d) Base: 3 Exponent: 8; There are 8 factors of 3.
2. a) 7 6 b) 2 4 c) 51 d) (5)5 3. a) 52  5  5  25 b) 23  2  2  2  8 c) 34  3  3  3  3  81
p.73
73  7  7  7  343
100
10000 1000000 1
4. a) 7 2  7  7  49
b)1 5. a)
b)
c)
d)
6. a)1 b)1 c)12 d)-1
6
2
4
0

10

10

10

10
71  7  7
(1103 )  (3  102 )  (2 101 )  (1100 ) (4 103 )  (2 10 2 )  (3 101)  (6 10 0)
4  103
 (1 1000)  (3  100)  (2  10)  (1 1)  (4 1000)  (2 100)  (3 10)  (6 1)
7. a)  4  1000 b)
c)
 1000  300  20  1
 4000  200  30  6
 4000
 1321
 4236
2
1
0
(8 10 )  (110 )  (9 10 )
 (8 100)  (110)  (9 1)
d)
 800  10  9
 819
p.74
5 2  23
23  (3)3
 5  5  23
32  5
8. a)
 3 3  5
 95
 14
(2  3) 3
 (2)(2)(2)  ( 3)3
5  32
82  4
3
3
b)  25  2
c)  5
d)  8  (3)
9. a)  5  9 b)  64  4
 25  (2)(2)(2)  5  5  5  8  (3)(3)(3)
 45
 16
 125
 25  8
 8  (27)
3
 17
 19
2
3
(10  2)  2
(7  1)  (2  2)
32  42
 (49  1)  (8  2)
 (3)(3)  (4)(4)
 12  2 2
c)
d)
10. a)exponents
b)Square brackets
 50  10
 9  16
 12  4
5
 25
3
3
0
(2)  (3)
0
(6  3)3  (2  2) 2 
 (2)(2)(2)  1
c)Exponents
d)Evaluate the 0 exponent
 8  1
1
 7
p.75
3 3 3
5 5 5 5 5
2 2 2 2 2 2 2 2
3
5 5 5
2 2 2 2 2
3 3
5 5
88888
2 2 2
1. a) 
b)
c) 
d) 
88888
1
1
1
9
 25
8
1
2
p.77
25  2 4
52  55
(3) 2  (3)3
105  10
1. a)  2 (5 4) b)  5(25) c)  (3) (23)
d)  10(51)
 57
 106
 29
p.78
 (3)5
 (3)  2
3
 ( 5)3
 (5)(5)( 5)
 125
(3)9
(3)5
(5)6  (5)3
84  83
98  92
b)  (3)(9 5) c)  8(43) d)  9(8 2)
1. a)  (5)(63)
 (3) 4
 (5)3
 81
 96
p.79
1. a)
4  43  4 2
(3)  (3)  (3)
 4(13)  42
 (3)(11)  (3)
 44  42
 (3)0  (3)
b)
 4(4 2)
 (3)(01)
 42
 (3)1
 16
 3
p.80
76  72
(4)5  (4)3
1.a)  7 (6  2) b)  (4) (53)
 78
105 105
(2)  (2) 3
c)  (2) (13)
 (4)8
7 0  71
(3) 4  (3)5
d)  10(55) e)  7 (0 1) f)  (3) (45)
 (2) 4
 1010
 71
(3)5  (3) 2
56  5 4
2.a)  (3)(5 2)
b)  5(6 4)
 (3)3
 52
 (3)9
103  105  102 (1)9  (1)5  (1)0
23  2 4  2 5
(6)8
58
47
6 4  6 4 (6) 7
 10(35)  102
 2(3 4)  25
 (1)(95)  (1)0
56
44
2
2
3 3
c)  4(7  4) d)  5(86) e)  6 (4  4) f)  (6) (87) 3.a)  27  25
b) 2 2 c)  108  102
d)  (1) 4  (1)0
3 3
 (6)1
 60
 10(8 2)
 43
 52
 2(7 5)
 (1)(4 0)
(2  2)
3
 (2  2)
 106
 212
 (1) 4
3
34
 4
3
 3(4  4)
 30
p.81
99  9 7  9 0
(3)1  (3) 2  2
 (9) (9  7)  90
 (3)(1 2)  2
4.a)  (3)3  2
 (27)  2
 54
b)
 9 2  90
 9(2  0)
9
55
5
54
 5(54)  5
52
50
c)  5(2 0) d)  51  5
2
 81
 52
 5(11)
 25
 52
5.a) 43  45  4(35)  48
 25
b) 2  2  2
 2 not 2 c) (3)  (3)  (3) (6 2)  ( 3) 4 not (3)3 d) 70  7 2  7(0 2)  7 2 not 7 0
e) 62  62  (6)(6)  (6)(6)  36  36  72 not 6 4 which equals to 1296. f) 106  10  10(61)  105 not 106
g) 23  52  (2)(2)(2)  (5)(5)  8  25  200 not 105 which equals to 100,000
p.82
5
5
(55)
10
25
6
2
3 3 3
1 1 1 1 1 1
 
    
4 4 4
2 2 2 2 2 2
2 10  2 10  2
2 5 2 5 2 5 25
3 3 3
1 1 1 1 1 1
1.a)  2  2  2 10 10 b)  2  2  2  2  5  5  5  5 1.a) 
b) 
4 4 4
2 2 2 2 2 2
 23 102
 2 4  54
3
3
16
 3
 6
4
2
p. 84
1. a) (93 )4  93  94  912
b) [(2)5 ]3  (2)53  (2)15 c) (54 )2  (542 )  58
p.85
1. a) 54  7 4 b) 82  22 2. a) (-6) 2 = 36
b) 43 = 64
p. 86
Check
35
13
1. a) 5 b)
2. a)  42 , 16
b) 63 , 216\
3
4
(10)
Practice
50
( 6) 7
32
(1)3
4
4
3
3
5
5
2
3
1. a) 5 2 b) 12 12 c) 3  (2) d) (4)  (5) 2. a) 0 b)
c) 2 d)
8
57
5
(2)3
p. 87
3. a) 5 23 = 56 b) (2)35 = (2)15 c) 441 = 4 4 d) 803 = 80
4. a) (-12) 2 =144 b) (12)2 = -144
c) 4 2 = 16 d) 30 1 = 30
5. a) (32 )3  323 = 36 , not 35 b) (3  2)2  52 = 25, not 32  22 which is equal to 13
4
4
2
28
2
22
c) (53 )3  533 = 59 d) ( )8  8 e) (3  2)2  62 = 36 f) ( ) 2  2 = , not
9
6
3
3
3
3
g) [(3)3 ]0  (3)30 = (3)0 , not (3)3 h) [(2)  (3)] 4 = 64 , not  64
p. 88
1. 54 ( R) 2. 8 (I) 3. 34 (F) 4. 45 (N) 5. -8 (Y) 6. 0 (S) 7. 56 (P) 8. 1 (E) 9. 100000 (A)
10. 6(G) 11. 4 6 (O)
Final Answer: A PERSON FRYING AN EGG
p. 90
1. a) Base: 6 Exponent: 2
b) Base: -3 Exponent: 8 2. a) 43 b) (3)5
3. a) (-2) (-2) (-2) (-2) (-2) = -32 b) 10  10  10  10 = 10000 c) 6 2 = 6  6 = 36 d) 53 = 5  5  5 =125
4. a) 1
b) 1
c) 8
d) -1 5. a) 9  10  10  10 = 9  1000 = 9000
p. 91
b) (1100)  (3 10)  (5 1) = 100  30  5 = 135
c) (2 1000)  (4 100)  (110)  (9 1) = 2000  400 10  9 = 2419
d) (5 10000)  (3 100)  (7 10)  (2 1) = 50000+ 300+ 70+ 2 = 50372
6. a) 3  3+3 = 9 +3 = 12 b) 23 = 2  2  2 = 8 c) 25  52 = 25  25 = 1
d) (64-4)  (36-6) = 60+30 = 2
7. a) 5  9 = 45
b) 10  (9+1) = 10  10 = 100
d) (-3) + 1  3 = (-3)+ (-3) = -6
c) (-8) + (-12) = -20
p. 92
8. a) 6(3 7) = 6 10 b) (-4) (23) = (-4) 5
b) 10(53)  102 c) (6)(82)  (6) 6
10. a) 534  512 b) (-3) 26 = (3)12
11. a) 32  52
b) 25  52
c) (-2) (5 4) = (-2) 9 d) 10 (7 1) = 10 8 9. a) 5 (7 3) = 54
d) 5(106)  54 e) 8(31)  82 f) (-3) (40) = (-3) 4
c) 824 = 88 d) (-5) 54 = (-5) 20
c) (4)3  (5)3
d)
45
35
e) 12 4  10 4
f) (7)6  (9)6
p. 94
14
28
24
4
,  4,  4,
b)  2,  2,
, 3, 3,
20
40
30
5
12
3
10
20
2. a)  4,  4,
b)  3,  3,
, 3, multiply c)
, 5, 5 d)
, 4 , 4 , 4, divide
20
12
15
24
p. 95
1. a)>
b) < c) < d) > 2. a) >
b) <
p. 96
4
3
1. a) 4 , 4 , ,  1 ,  1 , , A common denominator is 8.
8
8
9
10
b) 3, 3 ,
, 2, 2 , , Multiples of 4: 4, 8, 12, Multiples of 6: 6, 12, 18 …Common denominator: 12
12
12
9
10
c) 3, 3 ,
, 5, 5 ,
, Multiples of 5= 5, 10, 15 Multiples of 3= 3, 6, 9, 12, 15
15
15
Common Denominator = 15
4 3
9 12
9 10
, ,<
, ,<
2. a) , , >
b)
c)
12 12
15 15
8 8
p. 97
1
1. a) 0.75
b) 2  3  0.6
c) 5  8  0.625
d) 0.5 e) 1  5 , 0.2, 4.2
f) , 1  3 ,
3
2 5 1
3 5 1
3
9
11
87
5
1
b) , , 4
3. a)
b)
c)
d)
e) 1 or 1
0.3 , 2.3 2. a) , , 2
3 9 3
4 8 5
10
10
100
100
10
2
7
f) 5
10
p. 98
4
1. a) Answers may vary. 1 , -2
b) Answer may vary. 0, 0.3
5
p.99
9
4
1 4
1. a) 2.1, -2, 1.8 , -0.3, 0.7
b) -2, - 1 , - 1 , -1 ,
10
5 5
5
p. 100
1. a) 3, 5, 0.6
b) 5  3 = 1.6
c) 3  5= -0.6
d) -3, 5, -0.6
e) -5, 3, 1.6 f) 3  (-5)=-0.6
3 3 3
Parts c, d, and f match. This shows that  , ,
are all equal.
5 5 5
2
1
5
1
2. a) <
b)  
c) From the number line, -5 <-5 3. b) -1.6, -0.7, -0.1, 0.8
3
3
6
6
4
4. Answers may vary a) -2 and –1.9
b) 4.2 and 4.3
c) Two possible numbers are: -2 and -1
5
1. a)
5. -2, -1.7, -1
1 3 3
, ,
6. -1.5, -1.3, -0.8, 0.9, 2.4
2 4 2
p. 102
5
1. a)
6
P. 103
b)
1. a) 3
b) 2
5
6 1
2. a) , 1
6
5 5
4
2
or 2
6
3
b)
2. a) 3+2+
4 5 9 1
 , ,1
8 8 8 8
2 3
5
5
+ = 5 = 5
7 7
7
7
1 2
1 6
7
7
b) 4+1+  = 5+  = 5+ = 5
9 3
9 9
9
9
p. 104
1. a) -2.2
b) -2
2
3
c) 
3
8
p. 105
1. a) 12,
2
5
2
,  , 2, 2,
12
12
12
b) 15,
10
9
10
1
9
, ( ) , - , 3, 3,
and 5, 5, 
15
15
15
15
15
p. 106
5
3
5
6
1
1
6
)  = (-1)+ 3 + (  ) 
= 2+
=2
Use a denominator of 16. 2, 2,
16 8
16 16
16
16
16
3 1
12 5
17
17
12
5
b) 2+1+  = 2+1+ 
= 3+
=3
Use a denominator of 20. 4, 4,
and 5, 5,
5 4
20 20
20
20
20
20
p. 107
Practice
2
1
10
1. a) -4.5 + (-1.2) = -5.7
b)  + 2 = 1 2. a) -3.3
b) -0.2 3. a) i) 10 ii)10.5 iii)
b) i) -2 ii)
3
3
11
2
2
10
-2.3 iii) c) i) 2 ii) 2.3 iii)
d) i) -10 ii) -10.5 iii) 11
11
11
p. 108
Practice
4. a) 1.2 b) -2.3
c) -6.5
d) -44.6
3 2
5
6
1
9
8
17
 ( ) = 
5. a) - ,
b)
c)   ( ) = 9 9
15
15
24
24
24
15
2
1
4
1
1
1
6. a) (-2 + 6) + (- ) +
= (-2+6) + ( )  = 4 +
=4
5
2
10 2
10
10
1
1
2
3
5
5
b) (-1) + (-3) + (- )  ( ) = (-1) + (03) + ( )  ( ) = (-4) + 
= -4
6
4
12
12
12
12
1
1
7
3
12
10
c) (-3) + (-5) + (- )  (  ) = (-3) + (-5) + ( )  ( ) = (-8) + (- ) = -8
3
7
21
21
21
21
p. 109
1. a) (-1)+3+ (
1 a) 1
3
7
or
4
4
3 11
b) 1 or
8 8
c) 2
5
17
4
31
2
2 8
3
3 23
or
d) 3 or
2. a) 1, 5, 7 b) 2, , 6, ,
c) 5, , 20, ,
6
6
9
9
3
3 3
4
4 4
p. 110
1. a) -
4
11
3
,  , -1
8
8
8
P. 111
b)
4 2 12 10 22
7
,( ),
,( ) ,
,1
5 3 15 15 15 15
6
31
1
1. a) , 18, 18,  , -2
Use 15 as a common denominator
5
15
15
b) 
19 7 19 28 19
28
47
7
, ,  , ,  , (- ),  , -5
8 2
8 8
8
8
8
8
Use 8 as a common denominator
p. 112
1. a) -2.3
b) 5.5
c) -6.9
d) 6.9
2. 2.7 – 9.7  2.7 + (-9.7)  -7
-2.7 – 9.7  -2.7 + ( -9.7)  -12.4
-2.7 – (-9.7)  -2.7+ 9.7  7
2.7 – (-9.7)  2.7 + 9.7  12.4
p. 113
3. a) 2.4
c) i) 9
b) -5.1
ii) 9.4 iii)
c) 34.6
d) -1.4 4. a) i) 3
9
2
, or1
7
7
d) i) -3
iii) -
30
47
7
) = ( ), or  2
20
20
20
c)
9
5
14
+
=
15 15 15
6. a)
13 4 13
4
13
8
5
 =
 ( ) =
 ( ) =
6 3
6
3
6
6
6
b)
iii)
3
7
b) i) -9
ii) -9.4
9
2
iii)  , or  1
7
7
3
7
ii) -3.2
5. a)
b) (
ii) 3.2
9
7
36
35
1
 ( ) =
 ( ) =
5
4
20
20
20
3
7
3 7
9 14
23
 ( ) =  = 
=
2
4
2 3
6 6
6
7. Subtraction Sentence: 42.35 – 24.50 = 17.85
Jenny still owes the cashier $17.85
p. 114
1. Answers may vary. a) -
1
and -1
6
b) -0.2 and 0.1
2 4 7
2. For least to greatest, read the points from left to right; 1 ,  ,
3 5 10
3. a) -
2
=-0.4
5
1
1 =-1.5
2
5
 =-1. 6
3
5
- = -2.5 3. b) -2.5, -1. 6 , -1.5, -0.4
2
p. 115
4. a) 2.3
b) -21.5 5. a)
2 1
,8 8
b)
3 2 5
 =
8 8 8
3
2
5
c) -  ( ) =8
8
8
d)
3
2
1
 ( ) =
8
8
8
6. a)
22
45
23
 ( ) = 33
33
33
1
2
7
8. a) ( ) , (- ), 
12
6
12
5 7
20 21
1
b) (-1+3) + (   ) = 2 + (   ) = 2
7. a) 3.4 b) -9.1
6 8
24 24
24
b) 3
c) 11 d) 3.3
1
18 10
54
70 16
=- +
=+
=
3
7 3
21
21 21
p. 116
9. a) 1.3+5.4=6.7 The temperature rose by 6.7. b) -4.2-2.7 = -4.2+ (-2.7) = -6.9 The temperature fell by
6.9 c) -4.2 – (-5.4) = -4.2 + 5.4 = 1.2 The temperature rose by 1.2 degrees Celsius.
p. 117
2
7
2
2
1. a)
b) 2, 2, , 2
c) 4, 4, , 4
d) 6, 6, , 6
3
10
3
3
p. 118
3 1 3
97
93  71 3 1 3
 , 2 b)
1. a)
CF between 9 and 3 is 3. CF between 7 and 14 is 7.



2  5 10
14  3 142  31 2 1 2
6  3 3 3 9
4  15 2  3 6
12  5 2 1 2






2. a)
b)
c)
7  4 7  2 14
5  14 1 7 7
5 18 1 3 3
p. 119
3  5  4 19
3 7  2
23
112  5 17
1. a)
=
b)
=
c)
=
5
5
7
12
12
7
17  1 17
3 4
3 4
1
1
2. a)
=
b)  =
= 2 Rewrite 1 and 1 as improper fractions.
5 4
20
2 3
23
2
3
p. 120
1 (3)
3
1. a)
=The fractions have different signs, so their product is negative.
5 5
25
p. 121
(9)  ( 7)
( 3)  ( 7)
21
b)
=
=
The fractions have the same sign, So their product is positive.
1112
11 4
44
A common factor of 9 and 12 is 3.
CHECK
5 6 (5)  3
15
1
14 11
( 7)  ( 11)
77
7
  , or  1
, or 7
1. a) ( )  
b) ( )( ) =
=
4 7
27
14
14
5
4
5 2
10
10
p. 122
1. a) The change in value is : 80  (-1.13). The product is negative. To find 80  (-1.13), multiply: 80 
(-113). 80  (-113)=-9040. Estimate: 80  (-1.13) is about 80  (-1) = -80. So, 80  (-1.13) = -90.40.
The shares changes in value by -$90.40 that day.
p. 123
1. a) Negative b) Positive c) the same sign; positive d) different signs; negative
2. a) Yes, since changing the order of the factors does not change the product.
b) No, since the product is positive, not negative. c) No, since the product is positive, not negative.
d) Yes, since the signs and numerical values match.
2  ( 5) 1 ( 5)
5
( 4)  ( 11)
(1)  ( 11)
11
3. a)
=
=b)
=
=
76
73
5  12
5 3
13
21
p. 124
(4) 1
4
1
17
6 (17)  (1) 17

4. a)3,
=  , or  1
b) ( )( ) 
3 1
5
3
6
5
1 5
5
5. a) -128, 1, (-3), -3, -1.28
b) (-303), (-7), 2121, -3, -1, 3, 2.121
6. a) The total change in temperature is: 8  (-2.2). The product is negative. To find 8  (-2.2), multiply
8  -22 = -176. 8  (-2.2) is about 8  (-29 = -16. So, 8  (-2.2) = -17.6. The temperature fell by
17.6 degree Celcius in 8h.
p. 125
1
3 1 3 1
3
3
1. a) 12 b)
b) 
=
c) 4   2   6 \
6
5 3 5 5
2
1
p. 126
2
4
2  ( 4)
8
2
7
( 1)  ( 7)
7
1. a)  (  ) =
=b) (  )  ( ) =
=
5
3
5 3
15
9
4
9 2
18
p. 127
1. Divide integers: (-75)  5 = -15. (-7.5)  5 is about (-5)  5=-1. So, (-7.5)  5 = -1.5
Practice
1. a) Positive b) Different signs the quotient is negative. c) Different signs; the quotient is negative.
d) Same sign; the quotient is positive.
p. 128
2
5
2. a) Yes, since the reciprocal of
is . To divide, you can multiply by the reciprocal.
5
2
b) No, since the quotient is positive, not negative. c) No, since changing the order of the factors changes
the quotient. d) Yes, since the sign and numerical value of the quotient are the same.
2
6
(2)  2
4
15
8
(3)  ( 1)
3
1
3, a) ( ) 
=
= 
b) ( )  ( ) =
= , or1
3
7
1 7
7
16
5
2 1
2
2
8
3
( 8) 1
8
2
2
7
(2)  ( 7) 14
4. a) (- )  =
=  , or  2
b) (- )  ( ) =
=
9
1
3 1
3
3
5
3
5 3
15
5. a) The quotient is negative;(-294)  7=-42; (-3) 1 =-3; -4.2
b) The quotient is positive; (-552)  (-8)=69; (-6)  (-1)=6; 6.9
p. 130
1. a)3.8 + (-4) = -0.2
b) 4.6-9+3.9  (-1.3) = 4.6 – 9 + 3 = -4.4 + (-3) = -7.4
p. 131
3 (1)  (1)
3 1
9 2
7

1. a) 
=  =
=
4
3 2
4 6 12 12 12
1 1
3 ( 1)  5
3
5
3
5
9
14
 ( ) = ( )  ( ) = ( )  ( ) = ( )
b) - )   ( ) 
6 5
2
6 1
2
6
2
6
6
6
1
Divide first. Multiply by the reciprocal of . Add. Use a common denominator of 6.
5
p. 132
1. 32+9  (-12.5)  5 = 32+(-112.5) 5 = 32+ (-22.5) = 9.5
-12.5 digrees Celsius is equivalent to 9.5 degrees Fahrenheit.
Multiply First. Then divide. Then add.
P. 133
1. a) Add. b) Multiply c) Multiply d) Divide
2. a) (-3.6)  1.8 +(-0.3) = -2 + (-0.3) = -2.3
1 3 1
1 2 1
2 1
8 3
5
b) ( )   = (  )   =   = - 
=4 8 4
1 3 4
3 4
12 12
12
3. a) 10  (-2.5) = -4
b) (-4.2) + (-10.2) = -14.4
c) 2.3-3.6  2 = 2.3-7.2 =-4.9
d) 7.5  [-0.7-0.9] = 7.5  (-1.6) = -12
p.134
1 (1)(8)
1
2
3
2
1
1 (1)  82

 ( ) =
=  1
=  ( ) =
5 4 15
5
15
15
15
15
5 4 15
7 3 1
(7)  3 1
21 2
19
3
 =   =  or  2
5. b) ( )   =
4 2 4
4 2
4
8 8
8
8
1 3 1
11 5
1 5
2 15
13
1
 =  =

c)   =
= - , or  1
9 2 4 3 2 4
6 4 12 12
12
12
5. a) Line 2: 4-(-4.1) = 4+(+4.1) = 8.1
1 2  ( 1)
1
2
1
b) Line 1: 
=  ( ) = 3
3 1
3
3
3
6. Substitute h = 3.5, a = 8, and b = 12 in the formula A=h  2 .
A = 3.5  (8+12) 2
= 3.5  20 2
= 70 2
= 35 The trapezoid has area 35 cm 2 .
p. 135
1
4
1
1
1
1. -5.8 2. -7.3 3. -3.8 4. 1.58
5. -1.44
6.
7.
8. 9. 10. 2
5
12
20
6
The winning card is Card A.
p. 137
11
1. a)i) -16 9 = -1. 7
ii) (-7) 3 = -2. 3
iii) - = (-11) (5) = -2.2
5
16
7
b) Sample Answer. Two rational numbers between  and  are: -2 and -2.1
9
3
4 7
2. From least to greatest: -3.9, 3 ,  , 3.3 3. a) 2.7 b) 6.7
c) -10.2
5 2
1
6
7
16 11
5
2 8
6 8
2
2
4. a)   ( ) = 
b) -   
c) (-1+2)+ (  ) = (-1+2)+(-  ) = 1+ = 1
8
8
8
12 12
12
3 9
9 9
9
9
p. 138
7 8
1
3
15
21
75
54
19
 ( ) = - , or  1
5. a)-  
b)  ( ) =
12 12 12
5
7
35
35
35
35
31
8
31
16
47
7
c) +(- ) = -  (  ) =  , or  4
10
5
10
10
10
10
6. a) 5959.1 –(-417.3) = 5959.1 + 417.3 = 6376.4 The difference in elevation is 6376.4m.
b) -410.9 –(-417.3) = -410.9 + 417.3 = 6.4 The difference in elevations is 6.4m.
c) 8849.7 – 5959.1 = 8849.7 + (-5959.1) = 2890.6 The difference in elevations is 2890.6m.
7. a) Positive b) Negative c) Positive d) Negative
p. 139
1 (11)
(1)  (11)
11
(1)  5
5
2
8. a)
=
=
b)
=  , or  1
5 10
5 10
1 3
3
3
50
15 4
( 5) 1
5
1
11
25 (1)  (25)
25
1
, or 8
c) -  =
=  , or  1
d)   ( ) 
=
16 3
4 1
4
4
3
11
3 1
3
3
9. a) 141.6
b) 3.78
c) 978.56
d) 60
10. The distance the diver descends is: -0.8  3.5
The product is negative. Multiply the whole numbers: (-8)  35=-280
Estimate: -0.8  3.5 is about (-1)  4=-4.
The exact answer is -0.8  3.5 = -2.8
The diver descends 2.8m in 3.5 min.
4. a)
p. 140
1
10
1 (2)
2
3
7
( 1)  ( 7)
7
 ( ) =
=b) ( )  ( ) =
=
5
7
1 7
7
5
12
5 4
20
12. a) 1.1- 21.7 = 1.1 + (-21.7) = -20.6
b) -1.8  (-0.3) +[5.1+2.9] = -1.8  (-0.3) +8 = 6+8 =14
( 5)  1 5
5
5
5 10
5
c)
+
=- 
=- 
=
6 4
12
24 12
24 24
24
3 2
9
3 1 ( 3)
3
3
7
3
4
d) 1   ( ) = 1 
= 1  (  ) =  ( ) = , or1
4 3
8
4
1 4
4
4
4
4
4
p. 142
1. a) 3(3) + 5 = 9+5 = 14
b) 6+8(3) = 6+24 = 30 2. a) 8(8)-4 = 64-4 = 60
b) 20-2(8) = 20-16 = 4
p. 143
1.b)i)
Figure Number
Number of Squares
1
2
2
5
3
8
4
11
5
14
ii)
Figure Number
Number of Dots
1
3
2
6
3
9
4
12
5
15
p. 145
1. +1, +1, +1, +4, +4, +4
a) The number of hits is 4 times the number of swings, plus 1.
b) Write an equation to describe the relationship.
h=4s=1
c) Use your equation to find h when s = 10
h=4(10)+1
=40+1
=41
p. 146
1. a) P = 50+2b
b) P=50+2(20) = 50+40 = 90 Marcel got paid $90.
p. 147
1. a) T = 8+6 = 14
b) T=3(6)-2 = 18-2 = 16 c) T=12(6)+9 = 72+9 = 81 d) T = 7(6)+3 = 42+3 = 45
2. b) The expression in part ii: 3f represents the number of dots in terms of the figure number.
3. b) Left side: +1, +1, +1, +1
Right Side: +2. +2. +2. +2
Number of Shaded Tiles, s
Number of White Tiles, w
1
8
2
10
3
12
4
14
5
16
c) w=2s+6
d) w=2(25)+6 = 50+6 = 56 When the number of shaded tiles is 25, there are 56 white tiles.
4. a) C = 400+3n b) C = 400+3(200) = 400+600 = 1000 The total cost is $1000.
p. 149
11. a)
1. A(-5,-4) B(3,5) C) (2,-5)
p. 150
1.b) The points lie on a straight line, so it is a linear relation.
p. 152
1.b) The points lie on a straight line, so it is a linear relation.
c) A building cannot have part of a floor, so the points should not be joined.
p.153
1.
X
Y=4x-2
-1
-6
0
-2
1
4(1)-2 = 2
2
4(2)-2 = 6
p. 154
1. a) Linear b) Not Linear
c) Not Linear
d) Linear
2. a) x increases by 1 each time. y decreases by 1 each time. The relation is linear, because a constant change
in x produces a constant change in y.
b) x increases by 2 each time. y increases by 3 each time. The relation is linear, because a constant change
in x produces a constant change in y.
c) x increases by 1 each time. y does not increase or decrease by a constant value. The relation is not
linear, because the change in y is not constant.
p. 155
3. a) i)
X
Y
2
4
3
6
4
8
5
10
ii)
X
Y
-3
0
-2
1
-1
2
0
3
b) i) when x increases by 1, y increases by 2
ii) When x increases by 1, y increases by 1.
c) i) To get from one point to the next, move 1 unit right and 2 units up.
ii) To get from one point to the next move 1 unit right and 1 unit up.
4. a) y=4x
X
Y
-1
-4
0
0
1
4
2
8
b) y=-3x
X
-1
0
1
2
c) y=1-x
X
0
1
2
3
p. 156
5.
x
-1
0
1
2
Y
3
0
-3
-6
Y
1
0
-1
-2
y = 2x-4
-6
-4
2(1) – 4 = 2 – 4 = -2
2(2) – 4 = 4 – 4 = 0
6. a) No, because the fee is for each hour, not for part of an hour
b)Yes; the relation is linear because the points lie on a straight line
c) 1, 10, 10
p. 157
2x 8
2 x 12
 , x = 4 b) 3 – 3 – 2x = -9 – 3, -2x = -12,

1. a) 3, 3, 2x = 8,
,x=6
2 2
2
2
p. 158
1. vertical, x, 1
p. 159
1. b) horizontal line, y, 3
p. 160
1. a) Horizontal line b) Oblique line c) Vertical line d) Oblique line
p. 161
1. a) y = 4 b) x = 2 2a) 3 b) -2 c) x = -3 3a) i) Vertical ii) Horizontal b) i) -1 ii) y, -4
p. 162
4. a) i) Vertical line ii) Horizontal line iii) Oblique line b) i) x + 3 – 3 = -1 – 3, x = -4
2 y 10
2y 8
 , y = 5; 0, 2y = 8,
 , y = 4;
ii) 1 + y – 1 = 0 – 1, y = -1 c) -2, -2 + 2y + 2 = 8 + 2, 2y = 10,
2
2
2 2
2y 6
 , y = 3,
2, 2 + 2y – 2 = 8 – 2, 2y = 6,
2 2
X
-2
0
2
y
5
4
3
5. a) y + 3 – 3 = 0 – 3, y = -3, horizontal, y, -3
p. 163
1. a) 9, 13, 17 b) 1, 4 c) 4, 3 d) 4, 3
p. 164
2.
Number of Red Buttons, r
2
3
4
5
6
6
Number of Blue Buttons, b
10
13
16
19
22
25
a) 1, 3 b) 3, 4 3b) The points lie on a straight line, so this is a linear relation c) 1, 90
d) 90 units e) 90, 60
p. 165
4. b) 1, 2, 2 units c) 2, 2 5a) Vertical line b) Horizontal line c) Oblique line d) Horizontal line
6. a) y = 3 b) x = -2 c) y = -1
p. 166
1.
X
Y=x+2
0
Y=0+2=2
1
Y=1+2=3
2
Y=2+2=4
X
Y=x-2
0
Y=0-2=-2
1
Y=1-2=-1
2
Y=2-2=0
(0,2), (1,3), and (2,4) do not lie on the graph. (0.-2), (1,-1), and (2,0) lie on the graph.
Equation=Y=x-2.
p. 168
1. Left side= y=1
Right side = y=2(0) +1 = 0+1 = 1
Substitute x=1 and y=3
Left side: y=3
Right side: 2(1) + 1 = 2 + 1 = 3
Both coordinates satisfy the equation. So, the graph has equation y=2x+1
Practice
1.
X
Y=x+2
-2
Y=-2+2=0
-1
Y=-1+2=1
0
Y=0+2=2
(-2,0), (-1,1), and (0,2) lie on the graph.
2.
X
Y=3x
X
Y=-3x
-1
Y=3(-1)=-3
-1
y=-3(-1)=3
0
Y=3(0)=0
0
y=-3(0)=0
1
Y=3(1)=3
1
Y=-3(1)=-3
Y=3x: (-1,-3), (0,0), (1,3); Graph B; Graph B.
Y=-3x: (-1,3), (0,0), (1,-3); Graph A; Graph A
3.
X
Y=x-1
X
Y=1-x
-1
y=(-1)-1=-2
-1
Y=1-(-1)=2
0
y=0-1 = -1
0
Y=1-(0)=1
1
Y=1-1=0
1
Y=1-1=0
Y=1-x: (-1,2), (0,1), (1,0); Graph A; Graph A
Y=x-1: (-1,-2), (0,-1), (1,0); Graph B; Graph B
4. Left side: y=0
Right Side: x-3 = -3-3 = -6
The left side does not equal the ride side. So, y=x-3 does not match Graph A.
Left side: y=-3
Right Side: x-3=0-3=-3 The left side equals the right side.
Left side y=0
Right Side: x-3=3-3=0
The left side equals the right side. Graph B
p. 170
1. a) $14
b)11L
p. 171
1. a) About $27
p. 172
1. a) x=5
b) y=-1
2. a) x=3
b) y=-3
3. a) 500m b) 1.5 days
c) 1500m d) 4 days
p. 173
1. b)
c)
X
Y
X
Y
-4
-4
5
4
0
0
9
0
4
4
13
-4
What do you sea: Sample Answer: A prairie road.
p. 175
1. b)
Figure Number, n Number of Squares, s
1
1
2
4
3
7
4
10
5
13
c) 1;3
d) s=3n-2
e) s=3(10)-2=30-2=28 There are 28 squares in figure 10.
2. a) 8,9
b) n+4
3. a) y= 2,3,4,5
b) y=1,3,5,7
4. b) The points lie on a straight line, so the graph is linear.
c) Hayden only gets a gift on each birthday, not in between, so the points should not be joined.
d) 1, 10; 1 unit right, 10 units up
5. a) x=5
b) y=9
c) y=-10
6. a) Vertical line b) Oblique line c) Horizontal line d) Vertical line
7. a)
X
y=-2x
X
Y=2x
-1
2(-1) = -2
-1
-2(-1)=2
0
2(0)=0
0
-2(0)=0
1
2(1)=2
1
-2(1)=-2
Y=2x: (-1,-2), (0,0), (1,2)
Y=-2x: (-1,2), (0,0), (1,-2)
The graph passes through the points (-1,-2), (0,0), (1,-2) y=2x
8. For A(-2,0)
Left side: -2-0=-2 Right side: 2
The left side does not equal the right side.
So, Graph I does not match equation x-y=2
For C(0,-2)
Left side: x-y=0-(-2) = 2
Right side: 2
The left side equals the right side.
For D(2,0): Left side: x-y=2-0 =2
Right side: 2 The left side equals the right side. Graph ii
9. About 40km.
10. a)i) y=2 ii) y=1
b)i) x=-2
ii) x=4
p. 180
1. a) 3x+1
b) 2x-3
c) -3x+4
d) -2x-2
p. 183
1. The same tiles are used in parts b and c. So, 5-3a 2 -2a and -2c+5-3c 2 represent the same polynomial.
p. 184
2. a) 1 term; monomial
b) 2 terms; binomial
c) 3 terms; trinomial
d) 2 terms; binomial
e) 1 term; monomial
3. a) 2; 2
b) 4b; 1; 1 c) 4d 2 ; 2; 2 d) -4x 0 ; 0
4. a) 2 f 2  f  5 b) 3n 2  2
c) -7p+3
5. Answers may vary. For example, the tiles in question 4 part b can represent the polynomial -2+3p 2 .
6. a) 3, 1 b) 4, 5, 9 c) 9, 4, 5
d) 3, 1, 1
Parts b and c use the same algebra tiles. So, 4r 2  5  9 and 9  4 z 2  5 z both represent the same polynomial
p. 186
1a) +2 b) -1 c) -3
d) 0
p. 188
1. a) 2x-1
b) 2 x 2  3 x  2
c)  x 2  3x  2
p. 189
1. a) 8d+1 5+3=8 and 2+(-1)=1
b) 2a 2  5a 2  3a  7a  7a 2  4a
2+5=7 and -3+7=4
2
2
2
2
c)  x  2 x  4 x  x  5  3  1x  5 x  2  x  5 x  2
d) 2 x 2  2 x 2  6 x  7 x  7  11  0 x 2  13x  4  13x  4
p. 190
1. a) 2 b) 6 c) -3
d) 7
e) 1
f) -1
2
2
2. a) variable z and exponent 2. -z , 2z , -4z 2
b) variable and exponent 1. 4 x.7 x,  x
2
3. a) –x+3 b) 4x-2 c) 2 x  3
4. a) 2; 2c b) 3; 3s
c) -2+7=5, -2 x 2  7 x 2  5 x 2
d) 8+(-8)=0; 8 e2  8e2  0
5. a) 5m -2m +7+1 = 3m+8
b) 7c 2  4c 2  6c  c = 3c 2 5c
c) 11  2  9v  v  v 2  13  10v  v 2  v 2  10v  13 d) 7 f 2  3 f 2  12 f  3 f  2  5  10 f 2  9 f  3
6. a) 3x and 2 are not like terms. They cannot be combined.
b) 3s and 3s are like terms. To combine like terms, add the coefficients and leave the cariable alone. Since
5+3=8, 5s+3s=8s
c) Correct: x 2 and -x 2 are like terms. Since 1-1=0, x 2 -x 2 =0
p. 192
1, a) 4+5=9 b) 6+(-2)=4
c) (-3)+(-5)=-8
d) 3+(-3)=0
e) 5+(-8)=-3
p. 194
1. a) 4p+5 b) 3x 2  x  2
c) e2  3e  3
p. 195
1. a) 10g-7
7+3 =10 and -8+1=-7
2
2
b) 2a  9a  5a  12a  2a 2  5a 2  9a  12a  3a 2  3a 2(-5)=-3 and -9+12=3
c) c 2  11c  3  4c 2  5  1c 2  4c 2  11c  3  5  3c 2  11c +2
p. 196
1. a) 6x+11 b) 3p 2 +3p+5
c) -7b 2 +7b-11
Practice
1. a) (4x-1)+(-x+3)=3x+2 b) (-x 2 +2x-3)+(3x 2 -2x+5)=2x 2 +2
2. a) 2w+5
b) -2 t 2  4t  3
3. a) 5r-4 2+3=5 and -3+(-1)=-4
b) 7h 2  2h  4h 2  9h  4 = 7h 2 -4h 2 -2h+9h-4=3h 2 +7h-4
c) 2 y 2  6 y  1  2 y 2  6 y  5  2 y 2  2 y 2  6 y  6 y  1  5  4
4. a) 11r+4
b) 4a 2  6a  5
c) 4v 2  2v  6
5. a) (s+2)+(2s+5)+(2s+3) = s+2+2s+5+2s+3 = s+2s+2s+2+5+3 = 5s+1p. 198
1. a) 2; 2,8; 8
b) -4; 3+(-4)=-1; -1
c) 5; (-8)+(5) = -3; (-8)-(-5)=-3
d) -4; (-9)+(-4)=-13; (-9)-(4)=-13
p. 200
1. a) (4 p  3)  (2 p  1)  2 p  2 b) (5t  1)  (2t  3)  7t  2 c) (3e2  2e  4)  (4e2  3e  2)  e2  e  2
p. 201
(2  5 g  7 g 2 )  (9 g  4 g 2  2)
(8 f  3)  (7 f  5)
 2  5 g  7 g 2  (9 g  4 g 2  2)
 8 f  3  (7 f  5)
 2  5 g  7 g 2  (9 g  4 g 2  2)
 8 f  3  (7 f  5)
(4 x  3)  (2 x  1)
1) a)
b)  2  5 g  7 g 2  9 g  4 g 2  2
1) a)
 8 f 37 f 5
 2x  2
 2  2  5g  9 g  7 g 2  4g 2
 8 f 7 f 35
 4 g  3g 2
 f 8
 3 g 2  4 g
b)
(3x 2  6 x  5)  (2 x 2  3x  4)
  x  3x  1
p. 202
2
3) a) 9 b) -3r c) 2s 2 d) –t 4) a)
2) a) 5r  1 b) v 2  v  1
(4 p  1)  ( p  10)
(3h 2  5h  4)  (h 2  4h  6)
 4 p  1  ( p  10)
 3h 2  5h  4  (h 2  4h  6)
 4 p  1  ( p  10)
 3h 2  5h  4  (h 2  4h  6)
 4 p  1  p  10
 4 p  p  1  10
 3p 9
b)
 3h 2  5h  4  h 2  4h  6
 3h 2  h 2  5h  4h  4  6
 2h 2  9h  10
(4q 2  3)  (3q  q 2  3)
 4q 2  3  (3q  q 2  3)
c)
 4q 2  3  ( 3q  q 2  3)
 4q 2  3  3q  q 2  3
 4q 2  q 2  3q  3  3
(7 x 2  3 x  7)  (3 x 2  4)
5) a)
 7 x 2  3x  7  3x 2  4
 7 x 2  3x 2  3x  7  4
 4 x 2  3 x  11
(3a 2  2a  4)  (2a 2  3)
b)
 3a 2  2a  4  2a 2  3
 3a 2  2a 2  2a  4  3
 a 2  2a  1
 5q 2  3q
p. 203
1) a) 1 term, monomial b) 2 terms, binomial c) 3 terms, trinomial d) 2 terms, binomial 2) a) 5g – 7
b) 3r 2  4r c) w2  3w  4 3) a) -3x + 4 b) 2 x 2  2
p. 204
4d 2  3d  11  d 2  5d  13
8e  9  5e  4
4) a)  8e  5e  9  4 b)  4d 2  d 2  3d  5d  11  13 5) a) 2v + 3 b) 3u 2  2u 5) a) 4t  7 b) 7 y 2  4 y  11
 3e  5
 3d 2  2d  2
(3s  4)  (4 s  5)  (3s  4)  (4 s  5)
7) Perimeter =
 3s  4  4 s  5  3 s  4  4 s  5
 3s  4 s  3 s  4 s  4  5  4  5
 14 s  2
p.205
8) a) 6n – 3 b) – v – 7 9) a)
(11h  3)  (9h  2)
(7 j 2  11 j  7)  (12 j 2  8 j  3)
 11h  3  (9h  2)
 7 j 2  11 j  7  (12 j 2  8 j  3)
 11h  3  (9h  2)
 7 j 2  11 j  7  (12 j 2  8 j  3)
 11h  3  9h  2
b)
 7 j 2  11 j  7  12 j 2  8 j  3
 11h  9h  3  2
 7 j 2  12 j 2  11 j  8 j  7  3
 2h  5
 5 j 2  3 j  4
p.206
1) a)
Area  length  width
6 15  6  (10  5)
b) 8  35  8  (30  5) 2) a)
 140  7
 147
 (8  40)  (8  3)
b)  320  24
 344
p.207
1) a) Positive b) Negative c) Negative d) Positive 2) a) 30 b) – 40 c) – 21 d) 48 e) – 60 f) 32 3) a) Positive
b) Negative c) Negative d) Positive 4) a) – 7 b) 7 c) 8 d) – 9 e) 9 f) – 6
p.208
1) a) 12p – 9 b) 2 s 2  2 s  6
p.209
(5)(2d 2 )  (5)(3d )  (5)(6)
7(6 y 2 )  7(8 y )  7(9)
2
(4)(5e )  (4)(8e)
1) a) 21s 2  27 b)
c)  10d 2  15d  (30)
d)  42 y 2  (56 y )  63
2
 20e  32e
 10d 2  15d  30
 42 y 2  56 y  63
p. 210
1) a) g2 + 4g b) -2b2 + 3 c) s2-s+2 d) -2t2 + 3t -1
p.211
12
3v 2  v  2
1) a) 12r2, 8,
, 2, 3  r2+2, 3r2+2 b) 18v2, -6v, 12, 18, v2, -6, v, 2, 3  v2  (1)  v  2 ,
4
4e 2 8e 4 2 8
 e   e  2  e2  (4)  e , 2e2  4e
,

2
2
2
2
p. 212
5( w  6)  (5  w)  (5  6)
1) a) m, 4, 4m+16
b)
, 5w 30
c)
2) a) 3(3 x  1)  9 x  3 b) 2( x 2  3x  2)  2 x 2  6 x  4
3) a) 18r 12 b) 4b 2  2b  6
4) a) 4t2, 3, -24t2 + 18 b) (8)(3k 2 )  (8)(2k )  (8)(4) , 24k 2  16k  (32) . 24k 2  16k  32
5) c)
6) a) h 2  5h b) a 2  3a  2
10
7
7 x 2 7 x 21 7
 x2 
 x  (3) ,
,
,
7) a) 10z 2 , 15,
, 3, (2)  z 2  3 , 2 z 2  3 b)
,
7
5
7 7 7 7
(1)  x 2  1 x  3 ,  x 2  x  3
p. 214
1) a) b2 b) -c2 c) f2 d) –g2
2
2) a) 30  r 2 ,30r 2
b) 16  d 2 , 16d 2
c) 4  a  (7)  a, 4  (7)  a  a,(28)  a2 , 28a
d) (5)  (9)  v  v,(45)  v 2 , 45v 2
p. 216
1) a) 8, 8 b) -6m2+12m
p. 217
1) a) 28, 35, 28, 35 b)-3s2+4s, -3, 4 c) (-9r)(4r)+(-9r)(-5) , -36r2+45r
p. 218
12
14c 2 21c 14 c 2 21 c
9 b2 3 b
1) a)
, -2, -2  a, -2a b) 9b2, 3b, , , , , 3  b+1  1, 3b+1 c)

,
   ,
6
7c
7c 7 c 7 c
3 b 3 b
2  c+(-3)  1, 2c-1
p. 218 - 219 (Practice)
1) a) (2 x)(2 x)  4 x 2 , b) (2 x)( x  3)  2 x 2  6 x
2) a) 2s2 + 8s b) -2t2+3t
3) a) 5r, -1, 20, -4, 20r2 – 4r b) (7s)(-3s) + (7s)(6), -21s2 + 42s c) (-6t)(t) + (-6t)(-3) , -6t2 + 18t
18 y 2  12 y
12v 2
2y
4v
18 y 2 12 y
12 v 2
15w2
28 x 2


 
2
y
2y
4 v
3w
7 x
2
2
2
18 y 12 y
v
d) (-8u(-6u)+(-8u)(7), 48u2 – 56u 4) a)  3 
b)  15  w c)  28  x 5) a)    
2 y
2 y
1
3 w
7 x
 3 v
 5  w
 4 x
 9  y  6 1
 3v
 5w
 4x
 9y  6
2
2
32 z  24 z
15n  21n
8 z
3n
2
32 z
24 z
15n 2 21n




8 z
8 z
3n
3n
2
2
32 z
24 z
15 n
21 n
   c) 
  
b) 
8 z 8 z
3 n 3 n
 4  z  (3)  1
 5  n  (7)  1
 4z  3
 5  7
p. 220
Alphabet soup: 1) 2 x 2  6 x 2) 2 x 2  6 x 3) x  2 4) 7 x  4 5) x 2  3 x  5 6) 4 x 2  12 x  4 7) 6 x  3
8) 6 x  3 9) 0 10) 4 x  3 11) 4 x  3 12) 4 x 2  12 x  4 13) x  2 15) x 2  3 x  5 0) 7 x  4
C, D, P, R, T, O, U = Product
p. 222 – 224
7d  8d  4  2
3e2  2e2  8e  11e
1) a) Binomial b) monomial c) trinomial d) binomial 3) a)
b)
 d 2
 5e2  3e
13  9  6h  7 h  2 h 2
c)  4  h  2h
2
 2h 2  h  4
d)
9k 2  2k 2  15k  4k  8  3
 11k  11k  5
2
4) a) 2x 2 and 5x are not like terms. They cannot be
combined. 2 x 2  5 x cannot be simplified. b) Correct: 5s and 7s are like terms. To combine like terms, add
7r  11  2r  3
2
the coefficients. Since 5  7  2 , 5s  7s  2s . 5) a) e  6 b) 2 f  3 f  2 6) a)  7r  2r  11  3
 5r  14
9 s  5s  16 s  9 s  14
2
6v  5  (13v  3)
2
b)  9 s  16 s  5s  9s  14 7) a) 3t  2 b) u  4u  1 8) a)  6v  5  13v  3
2
2
 7 s 2  4 s  14
10 w2  7  (2 w  9 w2  5)
b)
 10 w2  7  2 w  9 w2  5
 10 w  9 w  2 w  7  5
2
2
2
 7v  8
9) a) 4(2 x  4)  8 x  16 b) 2(2 x 2  3x  1)  4 x 2  6 x  2
 w2  2 w  12
27b 2  9b  36
16a  40
9
8
27b 2 9b 36
16a 40





2
8
8
9
9 9
(

9)(

2
z
)

(

9)(

4
z
)

(

9)(5)
6(7 y 2 )  6(1)
16
27 2 9
2

18
z

36
z

(

45)
10) a)
b)
11)
a)
b)


a

(

5)

 b   b  4

9
9
8
 42 y 2  6
 18 z 2  36 z  45
2
 2 a  5
 (3)  b  1 b  4
 2a  5
 3b 2  b  4
21k 2
7k
21 k 2


4 f (5 f  2)
3e(5e  2)
7
k

(

4
f
)(5
f
)

(

4
f
)(2)
 (3e)(5e)  (3e)(2)
k
12) a) 2c 2  10c b) 3d 2  12d 13) a)
b)
14) a)  3 
2
2
 20 f  (8 f )
1
 15e  (6)e
 3  k
 20 f 2  8 f
 15e 2  6e
 3k
2
2
81m  45m
33n  36n
9 m
3n
2
81m 45m
33n 2 36n




3n
3n
9 m 9 m
2
2
33 n 36 n
81 m 45 m
  
b) 


 c) 
3 n 3 n
9 m
9 m
 11 n  ( 12) 1
 9  m  5  1
 11n  12
 9 m  5
p. 226
9  16  8  ( 5)
4  24  (8)
 9  2  ( 5)
17  12
1) a) Multiply b) Divide c) Power d) Divide 2) a)
b)  4  (3)
c)
 11  ( 5)
 5
 (7)
6
p. 227
6(2  y )
3(b  2)
1) a)  3b  3(2) b)  6(2)  6( y )
 12  6 y
 3b  6
p. 229
n  2  10
n  2  2  10  2
1)
n  12
Check :12  2  10
p. 230 – 232
2(t  1)  12
2(t )  2(1)  12
2t  2  12
1) a) First subtract 4, then divide by – 5 b) First subtract 2, then divide by 5 2) 2t  2  2  12  2 Substitute
2t  14
2t 14

2
2
t7
 2(t  1)
 2(7  1)
t = 7 into the equation. Left side
Right side = 12. Since left side = right side, t = 7 is correct.
 2(6)
 12
s  4  4  12  4 6  c  6  2  6
5 2  v 2 2
z  9  9  10  9
Practice: 1) a)
b)
c)
d)
s  8
c  4
7v
z 1
n
n
 3
3 x  15
4
4
3x
3 x 15
n
12
2) a) 4   4(3) Left side = 
Right side = - 3, n = 12 is correct. b)
Left side =  3(5) Right

3
3
4
4
 15
n  12
 3
x5
5k  6  6  24  6
 5k  6
4 x  16
5k  30
 5(6)  6
4 x 16
side = 15, x = 5 is correct. 3)
4) 5k 30
Left side
Right side = 24, k = 6 is

4
4
 30  6

5
5
x4
 24
k 6
3  4 y  3  9  3
 3 4y
4 y  12
 3  4( 3)
correct b) 4 y 12
Left side
Right side = -9, y = - 3 is correct. 5) a) She divided by 3

 3  12
4
4
 9
y  3
3 x  6  15
3 x  6  6  15  6
 3x  6
 3(7)  6
instead of adding 6 first. b) 3 x  21
Left side
Right side = 15, Since left = right, x = 7 is
 21  6
3 x 21

 15
3
3
x7
8  2 w  12
 8  2w
8  2 w  8  12  8
 8  2(2)
correct. 6) a) 4  4  w  w  12 b)
c) Left side
Right side = 12, Since the left
2w  4
 8 4
w2
 12
side equals the right side, w = 2 is correct.
p. 233, p. 235 and p. 236 - 237
4b  2  2b  6
 c  5  2c  4
4 m  6  6  2  6
4b  2  2b  2b  6  2b
 c  5  2c  2c  4  2 c
4 m  8
2b  2  6
3c  5  4
Check: 1) 4m 8

4
4
m  2
Check: 1) 2b  2  2  6  2
2b  4
2b 4

2 2
b2
Check: 1) 3c  5  5  4  5
3c  9
3c 9

3c 3
c3
x
5
4
20
x

Check: 1) a) 4   4(5) Left side
4 Right side = 5; Since the left side equals the right side x = 20 is
4
5
x  20
2 7
x
7
5


4  4  4
4 4
4
4
4
2  7
5
correct. b) x  7  5
Left side 
Right side = ; Since the left side equals the right side,
4
4
x 77  57
5

x  2
4
x = - 2 is correct \
p. 238 – 239
y  4  4  2  y  4
3w  2  2  w  4  2
2  x  3x  2  3 x  3 x
2x  4  6x
y  2 y
3w  w  6
2  2 x  2
2x  4  2x  6x  2x
y y  2 y y
3w  w  w  6  w
2  2 x  2  2  2
1) 4  4 x
2) a) 2 w  6
b) 2 x  4
c) 2 y  2
4 4x
2y 2
2w 6
2 x 4




4 4
2 2
2
2
2
2
x 1
y 1
w3
x  2
2  j  j  8  4 j  j
2  8  5 j
2  8  8  5 j  8
d) 10  5 j
t
22  42
6
t
2
3) a) 6
Left side
t
6   6(2)
6
t  12
t
2
6
12
  2 Right side = 4, t=12 is correct
6
 22

10 5 j

5
5
4
2 j
w
5 5  25
15
5
5
5
w
 3
b) 5
Left side  5  (3) Right side = 2, -15 4a)He forgot to write that -3 – 3 is -6 in the
w
2
5   5( 3)
5
w  15
4c  3  3  c  3  3
4c  c  6
4c  c  c  6  c
second line b) 3c  6
Left side = 4(2) – 3 = 8 – 3 = 5 Right side = 2+3 = 5, 2
3c 6

3 3
c2
p.240
2 x  10
2 x 10
2(5) = 10, so the solution is correct

2
2
x5
b) y – 3 = 15; y – 3 + 3 = 15 + 3; y = 18; 18 – 3 = 15, so the solution is correct
m
4c 20
6 x  6(3)
3. a) x + 7 – 7 = -2 – 7; x = -9 b) 
; c = 5 c) 4 + 2 = y – 2 + 2; 6 = y d) 6
4
4
m  18
2(3)

2(
p
)


4
3q  1  1  17  1
6  2 p  4
3q  18
4. a) 3q 18
3(6) – 1 = 17; the solution is correct b) 6  2 p  6  4  6 2(3 – 5) = -4; so the

2 p  10
3
3
q6
p  5
solution is correct.
p.241
1. a) Divide by 3
b) Add 2
c) Add 3
d) Add 4
2a)
6 x  2  2  3x  5  2
3a  a  4
6 x  3x  3
3a  a  a  4  a
5. 6, 2, 3, 5; 6 x  3 x  3 x  3  3 x
3x  3
x 1
h  3  2h
6a) 2a  4
Left side = 3(-2)– 2 = -8, Right side = -2 – 6 = -8; -2
2 a 4

2
2
a  2
h  2h  3  2h  2h
b) 3h  3
3h 3

3
3
h  1
5a
 6(10)
6
5a  60
Left side = 4 + (-1) = 3; Right side = 1 – 2(-1) = 3; -1
6x
c)
5(12)
Left side = 6 Right side = 10; 12
 10
5a 60

5
5
a  12
p.242
1. a) < b) 
p.243
1. a) right, is b) Sample answers: 1, 2, 3
p.244
1. a) 0, 1, 4 b) -2, 0, 1
p.245
1. a) True b) False; 3 > -10 c) True d) True 2a) yes b) no c) no d) yes e) yes
3. b) i) 4, 5, 6 ii) 1, 0, -1 iii) -5, -4, 0 4a) s  100 b) n  12 c) p  70 d) n  10
5. i) x < 1 ii) x  0 iii) x > 1 iv) x  2 6a) x  6 b) x < -3
p.247
1. a) -3, -3, 4 – 3, p  1 b) -5 – 2 > 2 + a – 2; -7>a, or a < -7
p.248
3z  1  1  2 z  2  1
4  4x  4  6  5x  4
3z  2 z  3
4 x  2  5 x
1. a)
b)
1a) Subtract 1 b) Add 3 c) Add 4 d) Subtract 1
3z  2 z  2 z  3  2 z
4 x  5 x.2  5 x  5 x
z  3
x2
p  6  6  2  6
n  4  4  2  4
u  3  3  4  3
2. a) -5, -5, 5 b) 4, +4, 16 3a)
b)
c)
p  8
n2
u  1
2  y  2  2  2
d)
i) u  1 ii) n > 2 iii) y > -4 iv) p < -8
y  4
p.249
y  3  3  2  3
42  n22
33  t 33
2b2  52
4. a) i)
ii)
iii)
or n  6 iv)
or t  6
y  5
6  n,
6t
b3
b) i) -5, -6, -7 ii) 0, 1, 2 iii) 4, 5, 6 iv) 8, 9, 10 c) i) -4 ii) 3 iii) 7 iv) 5
6 a  2  2  5a   1  2
5. a)
4v  6  6  3v  3  6
6 a  5a  1
b)
6 a  5a  5a  1  5a
a  1
p.252
4v  3v  3
4v  3v  3v  3  3v
v3
1. a) Reverse b) Do not reverse c) Reverse
2 m 8

2a) 2
2
m  4
2m 8

b) 2
2
m4
 y 
(2)    ( 2)3
c)
 2 
y  6
p.253
 2 f 
5
  5(4)
 5 
2 f  20
1.
1a) i) No ii) Yes iii) Yes iv) No b) i) 8 > -24 ii) -1 < 3
2 f 20

2
2
f  10
3. a) i) Divide by 3 ii) Divide by -4 iii) Divide by -3 iv) Multiply by -2
p.254
3 y 15
 4 p 8
 q 
3x 9
(2)    (2)(5)



b)i) No ii) Yes iii) Yes iv) Yes c) i) 3 3 ii) 4
4 iii) 3 3 iv)
 2 
p2
y  5
x3
q  10
5w
11  4 1
8
p
5w
 5 
 2  2  3  2
5
6

6(3)


5
8
3  2r  3  9  3
 6 
p
5w
4. a) 2r  6
b)  5
c)  s  18
d) 8 x
 5(8)
5
8
s  18
r  3
p  25
5w  40
5w 40

5
5
w8
p. 257
g  5  5  2  5
f  66  36
1. a)
f  3
Check : 3  6  3
3
b)
g 3
Check : 3  5  2
3
5h 25

5
5
c) h  5
2 k
6

2 2
d) k  3
Check : 5(5)  25
Check : 2( 3)  6
5
3
4x  2  2  6  2
2  3c  2  7  2
2v  3  3  9  3
4x  8
3c  9
2v  6
2) a) 4 x 8

4 4
x2
 4x  2
Left Side:  4(2)  2
6
Right Side = 6
2 is correct
(2)(2)  (2)( w)  20
4  2 w  20
4  2 w  4  20  4
d) 2 w  16
2 w 16

2
2
w8
 2(2  w)
b) 3c 9

3 3
c3
 2  3c
Left side:  2  3(3)
c) 2v 6

2
2
v  3
 2v  3
Left side:  2(3)  3
 7
 9
Right Side = -7
Right Side = -9
3 is correct
-3 is correct
2a  3  3a  2
3  x  x  3
2a  3a  3  3a  2  3a
3  x  3  x  3  3
5a  3  2
x  x  6
3) a) 5a  3  3  2  3
b)  x  x  x  6  x
5a  5
2 x  6
5a 5
2 x 6


5
5
2 2
x  3
a  1
Left Side:  2(2  8)
 20
Right Side = -20
8 is correct
9  2w  w  w  6  w
9  3w  6
9  3w  9  6  9
4) a) 3w  15
3w 15

3
3
w5
e6e  6ee
3n  1  n  3  n  n
m  2  3m  3m  4  3m
2e  6  6
2n  1  3
2m  2  4
2e  6  6  6  6
2n  1  1  3  1
2m  2  2  4  2
b) 2e  12
c) 2n  2
d)
2e 12
2n 2


3
2
2 2
n 1
e6
2x
4
4  24
3
2(3)
s
 4
2x
6 6  76
3
2
 2
2
 6
3
 42
2
s
2x
1
5) a) 2
, 7
b) 3 
6) a) 0,1,2
 3(2) ,  2
3
s
7
2
2   2(1)
2 x  6
2
2
3
2 x 6
s2

2
2
x  3
2m  6
2m 6

2
2
m  3
b) 0, -1, -2 c) -1,0,1 d) 5,4,3
4 j 1 1  2 j  3 1
7) a  0 b) s > 2 8) a)
d 66  46
d  10
2 f  1  1  3  1
4j 2j4
2 f  4
4j 2j  2j 42j
b) 2 f 4

2
2
f  2
9) a) 2 j  4
2j 4

2 2
j2
k 22  2k 2
k  4k
k k  4k k
b) 2k  4
2k 4

2 2
k2
10) a) Do not reverse b) Reverse
c) Reverse
d) Do not reverse
3b  4  4  5  4
c
3b  9
2 x 4
v
2 z 4


 2    (2)(4) d) 2   2(4) 12) a)
11) a) 2
b)
c)
2 2
2
2
3b 9
 2 

x  2
v  8
z  2
c  8
3 3
b3
5  m  3  3  m  3
x
2   2  1 2
n  2  2  2n  2  2
8  m  m
2
n  2n  4
x
8  m  m  m  m
  1
b) n  2n  2n  4  2n c) 8  2m
d) 2
 n  4
 x
8 2m
(2)     (2)(1)

 2
n4
2
2
x2
4  m
p. 262
1) a) 100, 100cm, 700cm
b) 0.1 cm, 21(0.1)cm, 2.1 cm 2) a) 0.01, 0.01m, 3.46m
b) 0.001m, 1800(0.001m) , 1.8m 3)a) 10, 10mm, 65mm b) 1000mm, 3.8(1000mm), 3800mm
p. 263
33mm
4.5cm
1) a) 3.3, 33,
, 5.5, 5.5 b) 4.5cm, 1.5cm,
, 3, 3
6mm
1.5cm
p. 265 (Check)
a) 15 cm, 15 cm, 60 cm, 10 cm, 10 cm, 40 cm, 40 cm by 60 cm
13
 13  4  3.25 , 15cm, 3.25  10cm, 32.5, 32.5cm by 48.75cm
b)
4
Practice
27 mm
15mm
1. a) 2.7cm, 27mm;
, 3; 3
b) 1.5cm, 15mm;
, 2.5; 2.5
9mm
6mm
p.266
4.5cm
2. 3cm, 4.5cm;
; 1.5; 1.5
3cm
7
 7  4 =1.75; 4cm;1.75  4cm, 7cm; 7cm
4
4. 3cm, 4cm, 5cm; 2.75; 2.75  3cm=8.25cm; 2.75  4cm=11cm; 2.75  5cm=13.75cm
The side lengths of the enlargement are 8.25cm, 11cm, and 13.75cm.
p. 267
1.6cm
1. a) Length=3.2cm; Length=1.6cm;
=0.5;0.5
3.2cm
p. 268
1.5cm
b) 6cm; 1.5cm;
; 0.25; 0.25
6cm
Check
1. 0.05; 104cm; 0.05  104cm=5.2cm; 89cm;0.05  89cm=4.45cm; 5.2cm by 4.45cm
1
2.
=0.02; 10m; 0.02  10m=0.2m; 0.2m=0.2(100cm)=20cm;
50
5m;0.02  5=0.1m;0.1m=0.1(100cm)=10cm; 20cm by 10cm
Practice
0.5cm
1.2cm
1.a) 2.5cm;0.5cm;
;0.2;0.2
b) 4.8cm; 1.2cm;
;0.25;0.25
2.5cm
4.8cm
3
 0.15 ; 36cm; 0.15  36cm, 5.4cm
2.
20
3
3. a) 170cm; 0.04  170cm=6.8cm
b)
=0.06; 4m; 0.06  4m, 0.24m; 0.24m=0.24  100cm=24cm
50
1
4. 3cm, 4cm,5cm;  0.25 ; 0.25  3cm=0.75cm; 0.25  4cm=1cm; 0.25  5cm=1.25cm
4
The side lengths of the reduction are 0.75cm,1cm, and 1.25cm.
p. 271
1. a) Polygon b) Non-polygon c) Non-polygon
d) Polygon
e) Non-Polygon f) Polygon
p. 273
LengthofAB 5.0cm
LengthofBC 2.5cm
1. 90 ; equal; AB, EF, BC, FG;

 0.625 ;

 0.5
LengthofEF 8.0cm
LengthofFG 5.0cm
Are not; Are not; Are not
2. E  135 ; F  45 ; P  G  135 ; Q  H  45 ; are
LengthofMN 2.25cm
LengthofNp 1.5cm
MN, EF, NP, FG;

 0.75 ;

 0.75 ; are; are; are
LengthofEF
3.0cm
LengthofFG 2.0cm
p. 275
3.6cm
 1.5 ; 1.5; DE=1.8cm; 1.5; 1.5  1.8cm=2.7cm; 2.7cm
1. CD=2.4cm and HJ=3.6cm;
2.4cm
Lengt hom Re duction 1.8cm
2. Reduction; WX = 1.8cm and ST =3cm;

 0.6 ; 0.6; UV=2cm; 0.6;
Lengthonoriginal
3cm
0.6  2cm=12.cm; 1.2cm.
p. 276
1,  E =100  ; F  100 ; C  G  60 ; D  H  100 ; are
LengthofAB
2cm
LengthofBC
3cm
EF; FG;

 1.25 ;

 1.25 ; are; are; are
LenfthofEF 1.6cm
LengthofFG 2.4cm
LengthofAB 4.5m
LengthofBC 3.2m
2. 90  ; equal;

 1.6 ;

 1.6 ; are; are; are/
lengthofEF
3m
LengthofFG
2m
3. a) 4cm; 4cm, 10cm; 10cm
b)
LengthofAb 4.8m
LengthofBC 3.2m

 0.8 ;

 1.28 ; are not; are not; are not
LengthofJK
6m
LengthofKL 2.5m
is similar; is not similar; is not similar
2.1cm
 1.5 ; 1.5; 4.5cm; 1.5; 1.5  3.4cm=5.1cm; 5.1cm
3. LM=1.4cm and ST=2.1cm;
1.4cm
p. 278
1. a) 50  -60  =70 
b) 180  -65  -65  =50 
2. 70  =110  ; 110  , 55 
p. 280
1. a) 36  -68  =76  ; P  76 ; Q  68 ; R  36 ; are; are; PQR
LengthofEF 2.8cm
LengthofDE 3.2cm
b) EF, DE, FD; JK, KL, LJ;

 2;

 2;
LengthofJK 1.4cm
LengthofKl 1.6cm
LengthofFJ 5.4cm

 2 ; the same, are smiliar; D and L; E and K; F and J; LKJ
LengthofLJ 2.7cm
p. 282
2cm
 0.2 ; 0.2; 20cm; 0.2;
1. X , V , W ;XVW; XVW , FGH ; FH=10cm and XW=2cm;
10cm
0.2  20cm=4cm; 4cm.
p. 283
1. a) 110  -40  =30  ; U  30 ; T  40 ; S  110 ; are; are; UTS
LengthofJK 3.3cm
LengthofKL 4.8cm
b) JK, KL, LJ; QR, SQ, RS;

 1.5 ;

 1.5 ;
LengthofQR 2.2cm
LengthofSQ 3.2cm
LengthofLJ 5.7cm

 1.5 ; the same; are similar; J and R; K and Q; L and S; RQS
LengthofRS 3.8cm
LengthofQR 6cm
LengthofRP 8cm
2. QR, RP, PQ; CD, DC, BC;

 3;

 2;
LengthofCD 2cm
LengthofDB 4cm
LengthofPQ 12cm

 2.4 ; different, are not similar.
LengthofBC 5cm
2cm
 0.4 ; 0.4; FG; FG, 4cm; 0.4;
3. G, H F ; GHF; CDE , GHF ; DE=2cm and GF=5cm;
5cm
0.4  4cm=1.6cm; 1.6cm
3.6m
 1.8 ; 3m, 1.8; 1.8  3m=5.4m; 5.4m
4. equal; XYZ; enlargement; ZX=3.6m and CA=2m;
2m
5. 1.8, 3.0; A, D, E ; ADE; ADE , ABC ; AD=1.8cm and AB=3.0cm
1.8cm
 0.6 ;0.6;DE and BC; BC, 4.0cm; 0.6; 0.6  4.0cm=2.4cm; 2.4cm
3.0cm
p. 286
15mm
 2.5 ; 2.5
1. 1.5, 15;
6mm
2. 7cm; 7cm, 22.4cm; 5cm; 3.2,5cm,16cm; 16cm by 22.4cm
2.7cm
 0.75 ;0.75
3. 3.6cm; 2.7cm;
3.6cm
p. 287
4. 0.14; 100cm; 0.14, 100cm, 14cm; 14cm
5. Reduction; DE=4.0cm and HJ=2.0cm;
Lengthon Re duction 2.0cm

 0.5 ; 0.5; CD; CD,3.2cm; 0.5;
Lengthonoriginal
4.0cm
0.5  3.2cm=1.6cm; 1.6cm
p. 288
6. KL, MK, LM; QN, PQ, NP;
LengthofKl 3.6cm
LengthofMK 4.0cm

 0.8 ;

 0.8 ;
LengthofQN 4.5cm
LengthofPQ 5.0cm
LengthofLM 6.8cm

 0.8 ; the same, are similar; K, Q; L, N; M, P; QNP
LengthofNP 8.5cm
6.0m
 2.4 ; 2.4; RT; RT, 2.0m, 2.4; 2.4  2.0m=4.8m
7. equal; VUW; VUW; RST; ST=2.5m; UW=6.0m;
2.5m
4.8m
p. 289
1. a) 1 b) 4 c) 2 d) 0
p. 290
1. a) Yes b) No
p. 291
1. a) Yes; Yes; Yes
b) Yes; Yes; No
p. 293
1. left; vertical; is not; below; horizontal; is
p. 294
1. Point E reflects onto itself; Point F’ is 2 squares right of line of reflection; 1 square left of line of
reflection; Point G’ is 1 square right of line of reflection; on the line of reflection; Point H reflects onto
itself.
p. 295
2. a) Hexagon 1: above; horizontal; is
Hexagon 2: above, right; diagonal; is not
Hexagon 3: right; vertical; is
3. a)
Point
Image
A(0,5)
B(2,5)
B(2,5)
B(2,5)
C(3,3)
C’(3,7)
D(2,1)
D’(2,9)
b)
Point
Image
E(3,5)
E(3,5)
F(5,5)
F’(1,5)
G(5,4)
G’(1,4)
H(4,3)
H’(2,3)
J(6,1)
J’(0,1)
K(3,1)
K(3,1)
c)
Point
Image
P(1,3)
P(1,3)
Q(3,3)
Q’(1,1)
R(3,1)
R(3,1)
p. 297
1. a) 180 
b) 90  counterclockwise
p. 299
1. a) 4; 4
p. 300
1. a) 2; 2;
b) 3; 3
360
; 180  ; 180 
2
b) 3; 3;
360
; 120  ; 120 
3
p. 302
1. a) 4; 4 b) 5; 5
2. a) 4; 4; 90  ; 90 
b) 5; 5; 72  ; 72 
3. I have to rotate a tracing of the shape a complete turn before it matches the shape again. So, the shape
does not have rotational symmetry.
p.303
4. 10; 10
p. 304
1. a) 3 squares left and 1 square down
b) 3 squares right and 2 squares up
p. 306
1. a) Yes; horizontal; Yes; No; off; Yes; Yes
b) Yes; No; No; No; off; No; No
p. 307
1.
Point
Image
W(6,5)
W’(6,3)
X(5,4)
X”(5,2)
Y(3,4)
Y’(3,2)
Z(2,5)
Z’(2,3)
The shape has 2 lines of symmetry. The shape has rotational symmetry.
p. 308
2.
Point
Image
P(1,5)
P’(5,5)
Q(3,5)
Q(3,5)
R(3,1)
R(3,1)
S(2,1)
S’(4,1)
T(2,4)
T’(2,4)
U(1,4)
U’(5,4)
The shape has 1 line of symmetry. No, there isn’t a point about which you can turn the tracing. No, the shape
doesn’t have rotational symmetry.
p. 309 - 310
1. Polygons A and B and polygons A and D; Polygons A and B and polygons A and D
2. Rectangle F; Rectangle H; Rectangles E and F and rectangles G and H.
3. a) No, the triangles don’t face opposite ways. No, the triangles aren’t related by a reflection. Yes, the
triangles touch. So, try a point of rotation ON the triangles. Vertex M; After triangle LMN is rotated 90 o
counterclockwise about M. Yes, the triangles are related by a rotation. b) Yes, the triangles face opposite
ways. One triangle is above the other, so try a HORIZONTAL line of reflection. Yes, the triangles are
related by reflection. No, the triangles don’t touch. So, try a point of rotation OFF the triangles. No, the
triangles don’t match. No, the triangles aren’t related by a rotation.
4.
Point
Image
P(1,5)
P’(5,1)
Q(2,5)
Q’(4,1)
R(3,4)
S(3,2)
S(3,2)
R(3,4)
T(1,2)
T’(5,4)
There is no line on which I can place a Mira so one polygon matches the other. So, the shape does not have
line symmetry. Yes, the shape has rotational symmetry.
p. 311
Yes, the logo has rotational symmetry of order 2 about its centre, (6,5).
The logo has 2 lines of symmetry:
- The horizontal line through 5 on the y –axis and the vertical line through 6 on the x-axis
p. 313
1. 3.5; 6cm; 3.5  6cm  21cm ; 4cm; 3.5  4cm  14cm ; 14cm by 21cm. 2. Original=5.0cm; Reduction=
1.5cm
 0.3 ; The scale factor is 0.3 3. Matching angles: A  E  120 ; B  F  60 ;
1.5cm;
5.0cm
C  G  120 ; D  H  60 ; All matching angles are equal; Matching sides: AB and EF, and BC
lengthAB 2.2cm
LengthBC 1.4cm
and FG;

 0.4 ;

 0.4 ; The scale factors ARE equal so, the
LengthEF 5.5cm
LengthFG 3.5cm
parallelograms ARE similar.
p. 314
lengthBC
4cm
lengthCA 6cm
lengthAB 7cm
4. BC, CA, AB; GE, EF, FG;

 1.6 ;

 1.5 ;

 1.4 ; All
lengthGE 2.5cm
lengthEF 4cm
lengthFG 5cm
scale factors are DIFFERENT, so the triangles ARE NOT SIMILAR. 5. JKL is a reduction of EFG ;
12cm
 0.8 ; The scale factor is 0.8; Length of EF=20cm; Scale factor =0.8;
EG=15cm and JL = 12cm;
15cm
Length of JK = 0.8  20cm  16cm so JK has a length of 16cm.
p. 315
7.
Point
Image
P(3,5)
P(3,5)
Q(4,5)
Q’(2,5)
R(4,4)
R’(2,4)
S(5,4)
S’(1,4)
T(5,2)
T’(1,2)
U(3,2)
U(3,2)
8. The shape and its image match 4 times. So the shape has rotational symmetry of order 4. Angle of rotation
360
 90
symmetry is: 360 
4
p. 316
10. Yes, the polygons face opposite ways. Yes, the polygons are related by a reflection. No, the polygons
don’t touch. Try a point of rotation OFF the polygons. Yes, the polygons are related by a rotation.
11. b) The shape has 2 lines of symmetry and rotational symmetry.
p. 318
x 2  5.0 2  3.0 2
1. a) x  180  50  60  70 b) x  5.0 2  3.0 2 x is about 5.8cm
x  5.8
p. 319
OBC  90
1.
x  180  90  65  25
p. 320
OST  90
OT 2  OS 2  ST 2
162  r 2  122
1.
256  r 2  144
r 2  256  144
OS is about 10.6cm long.
r 2  112
r  112
r  10.6
p. 321
1. a) OE, OF, OJ b) CD, HG c) E, F d)  OEC,  OED,  OFG,  OFH 2. a)  OBP = 90 b)  PQO
OTM  90
OTW  90
= 90 ;  PRO = 90 3. a)
b)
x  180  90  25  65
x  180  90  35  55
p. 322
OCT  90
OPQ  90
12 2  x 2  102
x 2  152  152
2
2
2
x  4 6
144  x 2  100
x 2  225  225
2
x  16  36
4. a) x 2  144  100 OC is about 6.6 km. b) 2
OQ is about 7.2 cm c) x 2  450
OP is about
x

52
x 2  44
x  450
x  52
x 21.2
x  44
x 7.2
x 6.6
21.2 cm
p. 324
OC  AB
OC  AB
1. x  90
2. x  90
Since OA=OB, AOB is isosceles and
y  180  90  40  50
y  180  90  55  35
OAB  OBA so z  =55 
p. 326
Check
132  52  a 2
169  25  a 2
1.
a 2  169  25
a  144
2
So, a = 12 cm; EG=FG so b = 12cm
a  144
a  12
Practice
1. a) Radii: OA, OB, OC, OD; Chords: AB, AC, BD, CD; Diameters: AC, BD b) Radii: ON, OP, OQ, OR,
OS; Chords: NS, PQ, RS; Diameters: NS, PQ
p. 327
2. a) AC = CB = 4cm so a = 4cm b) MN = 2  PN = 2  3cm = 6cm so a = 6cm
1
1
c) OL   KL   18cm  9cm so a = 9 cm d) OS = OT = 1.5cm so a = 1.5cm
2
2
p. 328
OPQ  OQP
x  90
3. a)
b) OP = OQ OPQ is isosceles. so, x  35
y  180  45  90  45
y  180  35  35  110
OB 2  OD 2  DB 2
152  12 2  DB 2
4.
225  144  DB 2
DB 2  225  144
So, DB = 9cm; CD = DB = 0cm by chord properties. So chord BC has a length of
DB 2  81
DB  81
OA2  AN 2  ON 2
17 2  152  ON 2
289  225  ON 2
1
1
2  9cm  18cm 5. AN   AB   30cm  15cm ; ON 2  289  225 So, On is 8 cm
2
2
ON 2  64
ON  64
ON  8
p. 329
p  90
OAB  90
1. a) OGM b) OME and ONE 2. a) q  180  75  90 b) s  180  21  90
q  15
s  69
p. 330
a 2  20 2  252
a 2  400  625
3. a) OPQ  90 OQ is the hypotenuse of triangle OPQ a 2  1025
b) OBC  90 OB is a leg of
a  1025
so, a
20  b  15
2
2
32.0cm
2
400  b 2  225
triangle OBC
b 2  400  225
b 2  175
MN  2  ML
4. a) OX = OY; OXY is isosceles so b  35 b)
MN  2  6cm
MN  12cm
so, d  12cm
b  175
so, x 13.2cm
p. 331
x  90
OBA  OAB
5.
OA = OB, so triangle OAB is isosceles.
y  180  65  90  25
so, z   25
OQ 2  QP 2  OP 2
102  82  OP 2
100  64  OP 2
1
1
6. QP   QR   16cm  8cm ; OP 2  100  64
2
2
OP 2  36
The length of OP is 6cm.
OP  36
OP  6
p. 334
1
ACB   AOB
ADB  ACB
2
1. a) x  2  25  50 b)
c)
2. a)
1
x  28
x   62  31
2
QOP  2  QSP
x  2  30  60
QTP  QSP
y  30
1
ACB   AOB
2
1
b) x   70  35
2
ADB  ACB
y  35
p. 336 – 337
1. a) COB b) AOD, ACD c) DAC d) DAC 2. a) Central: AOB , Inscribed: ACB b) Central:
1
TSR   TOR
DEG  DFG
2
QOR , Inscribed: QPR 3. a) x  2  28  56 b)
c)
d) x  90
1
x  72
x   84  42
2
4. a)
x  34
y  34
b)
x  2 15  30
y  15
ACB  90
AOB  2  ACB
5. x  180  90  25  65 6.
OA=OB so triangle
x  2  50  100
y  2  65  130
y  z 
y  y  180  100
OAB is isosceles. In triangle OAB 2 y  80
80
2
y  40, z  40
y 
p. 340
OHG  90
152  x 2  132
1. a)
x  90
y  180  90  20  70
b)
225  x 2  169
ONM  90
x  180  90  75  15
2. a) x 2  225  169 b)
x 2  56
x 2  144  16
x 2  160
x 12.6cm
7.5cm
p. 341
x  90
y  180  25  90
3.
4. x  90 OM=ON so triangle OMN is isosceles.
y  180  115
y  65
1
1
XY   XZ   12cm  6cm
2
2
2
2
OX  OY  XY 2
x 2  12 2  4 2
x  160
x  56
x
OST  90
ONP  OMP
y   27
z   180  27  90
z   63
OX 2  22  62
5. OX 2  4  36
OX 2  40
OX  40
OX
p. 342
6. a)
6.3
x  2 14
x  28
x  y   90
b) x  43 c) x  90 7. a)
w  w  180  90
2w  90
ACD is isocsceles;
90
2
w  45
w 
x  25
y  35
b)
x  2  35  70
y  35
8.
z   180  25  90
z   180  115
z   65
;
p. 344
35
2 2  20 40
3
3 5
15
17
 35% b) 

 40% c)


 15% 2. a)
 0.17
1. a)
100
5 5  20 100
20 20  5 100
100
13
3
27
25
25  25 1
 13  50  0.26 c)  3  8  0.375 3. a) 27% 


b)
b) 25% 
50
8
100
100 100  25 4
70
70  10
7


c) 70% 
100 100  10 10
p. 345
1. a) The outcomes are: red, green, purple, yellow. The favourable outcome is GREEN. There are 2 green
2
1
, or
marbles. Theoretical probability = Number of green marbles  total number of marbles. 20
10
 0.1, or10%
b) The favourable outcomes are: green, purple, and yellow. Number of favourable outcomes: 2 + 5+ 4 = 11
There are 11 favorable outcomes. Theoretical probability = number of favourable outcomes  total number
11
of marbles. 20
 0.55, or 55%
p.346
11
43 7

1. a)
= 0.44, or 44% b) blue or green,
, 0.28, or 28%
25
25
25
p. 347
1. a) Felix made his prediction based on the results of a survey or experiment.
Explain your thinking: Felix’s prediction is based on the results of a survey (experiment) conducted at
school. So, his decision is based on experimental probability.
b) Natalie made her prediction based on theoretical probability.
Explain your thinking: natalie’s decision Is based on the theoretical probability of rolling double ones
1
since this probability is very small, , Natalie predicts she will lose the game.
36
p. 349
1. a) No; Yes; No; People listen to their dentist’s suggestions, The cost of the toothpaste does not matter,
both toothpastes are readily available for purchase.
b) Great; a few; less; most
2. a) Salima is not injured; the opposing team has the same ability as the last 5 teams
b) higher, injured, lower, the opposing team commits a lot of fouls.
p. 350
1. a) Josh made his decision based on theoretical probability. Explain your thinking: Josh knows that 50% of
the marbles in the bag are red. He found this probability without conducting an experiment or a survey. So,
Josh based his decision on theoretical probability.
b) The manufacturer made the decision based on the results of an experiment. Explain your thinking: The
manufacturer based its decision on the results of the experiment done by the quality control officer. So, the
decision is based on experimental probability.
c) Desi made his decision based on personal thoughts or feelings. Explain your thinking: Desi’s decision
is based on his feelings. He chooses the grean envelelope because green is his favourite colour. So, Desi’s
decision is based on subjective judgement.
p. 351
2. a) Lin will prepare for the tes like she did for the other tests, the test will have the same level of difficulty
as the others.
b) all tunnels on the road have the same height as the first tunnel.
c) all team members are healthy, the opposing team is less able than the Tigers.
3. a)i) the same students always buy a drink, the student’s drink preference does not change. The amount of
water sold does not depend on the outside temperature.
ii) students vary the drinks they buy, the outside temperature decreases; more students buy a drink on
any given day, the outside temperature increases
b)i) there are fewer cars on the road, the weather is ideal for driving, there will be no accidents or road
closures.
ii) If there are more cars on the road, or if the weather is bad, or if there is an accident or construction on
his route, Marcel may be late for work.
p. 352
1. a) Yes, is, What is your favourite type of TV program:
Comedy_____Drama______News_____Sports____Other_____?
b) No, is not, not needed
c) Yes, is, what is your favourite dessert: Pie____Cheesecake____Ice Cream____Fruits_____Other____?
p. 354
1. a) No, No, Yes, Explain your thinking: Easter is not celebrated by all cultures and not everybody eats
chocolate. So, the survey does not apply to every student.
How would you avoid the problems: I would add 2 opening questions. “Do you celebrate Easter? Do
you eat chocolate Easter bunnies?” If a student does not answer yes to both questions, the third question
need not be answered.
b) No, Yes, No, Explain your thinking: The timing of the survey question could be a problem. Most
people use sunscreen in the summer. They may not remember much about their sunscreen in January.
How would you avoid the problem? I would ask the question in the summer when most people are
using sunscreen regularly.
p. 355
1. Yes, Yes, No; Explain: The theatre will have to pay for 2 stamps and 2 envelopes for each survey. This
may be beyong what a local theatre can afford. The time that it will take to print, Mail, and collect the
surveys may be too long.
2. Yes, Yes, No; explain: the survey is too long. Most people won’t have time to spend 30 min on a survey.
The timing is wrong. Most people will still be unhappy with the government for raising the taxes.
p.356
1. a) Yes, No, No; Explain your thinking: The use of language is a problem. The question is biased. The
question is worded in such a way that you are to believe that it is cruel no9t to walk a dog daily. This
wording will influence a person’s answer.
b) No, No, Yes; Explain your thinking: The survey is insensitive to those students who don’t eat meat,
either because they are vegetarians or for religious reasons. Also, some students may not buy lunch in the
cafeteria.
c) No, Yes; Explain your thinking: The tutor is asking the students to share personal information. Many
students, especially those struggling in French, might feel uncomfortable or embarrassed sharing their
marks.
p. 357
2. a) I would ask a better question: How often do you think a dog should be walked: Less than once a
day____More than twice a day___Not at all_____?
b) I would add 2 opening questions. “Do you sometimes buy lunch in the cafeteria? Do you eat meat?” If
a student does not answer yes to both questions, the survey need not be completed.
c) I would suggest the tutor survey the teachers to find out which students might need her help. This way
the students will not feel uncomfortable.
3. a) Yes, No, No; Explain: It will take a lot of time and money to survey 10000 people by phone. And the
survey is too long. Most people will not have 25 min to complete the survey.
b) Yes, No, No; Explain: The timing of the survey could be problematic. Since the school’s volleyball
team just won the championship, there is probably a lot of interest and excitement in the wschool for
volleyball. This may influence a student’s answer.
4. a) I would insert a notice in each subsciber’s monthly bill asking them to complete a short on-line survey
on customer satisfaction.l
b) I would poll the students at the beginning of the school year, before sports have started and before any
championships have been won.
p. 358
1. a) The population is those people who eat at the restaurant regularluy. They will have tried many different
entrees.
b) The population is all teachers in the school.
c) The population is those people in Vancouver who drive a vehicle regularly.
p. 359
1. yes, yes, sample
2. no, no, census
3. yes, yes, sample
p. 360
1. All residents of British Columbia, no, a sample, no. no
2. All people boarding planes at the Edmonton International Airport, yes, a census
3. All customers of the café, No, a sample, yes, One in 5, or 20% of the customer were surveyed and the
sample probably include all types of customers. So, the conclusions would probably be valid.
p. 361
1. a) The population is all apartments in the building.
b) The population is all teenagers who live in Grande Prairie.
c) The population is all thermostats manufactured by the company.
2. a) A sample was used because it would take too logn and cost too much to survey every household in
Canada.
b) A sample was used because if every item of clothin was washed, there would be no clothing left to sell.
p. 362
3. a) A census was used because the information is important and it is used to make important decisions on
matters such as healthy care, education, and transportation.
b) a census was used because the number of curren Grade 9 students is not too large, and the information
will be used to determine the number of classes that will be needed to accommodate all interested students.
4. a) No, No, census
b) yes, Yes, sample
5. a) No, No
b) Neil looks at 1 of every 4 vehicles, or 235% of the vehicles. The sample probably includes all types of
vehicles. So, the conclusions would probably be valid.
p. 363
1. Peyton made his decision based on his personal thoughts of feelings. Explain your thinking: Peyton’s
decision is based on his feelings. He chose the 17th of the month because 17 is his lucky number. So,
peyton’s decision is based on subjective judgement.
2. a) the outside temperature is about the same as it was last week, the caretaker does not adjust the
temperature in side the classroom, Lord is as healthy as she was last week
b) the caretake turns on the air conditioner because it is hot outside, Lord is not feeling well; the caretaker
turns up the heat, the outside temperature increases and the air condition is not turned on.
p. 364
3. a) Yes, Yes, Explain your thinking: The timing of the survey could be a problem. Back to school shopping
is usually done in August and September. And Victoria Day is in May. Also, customers are being asked to
share personal information. They may not feel comfortable telling stranger hos much money they spend.
b) The survey could be conducted in late September after families have done their back to school
shopoping. The survey could be anonymous with the possible choicese given in rangers.
4. all residents of West Vancouver, yes. Yes. Sample
5. The coach did not survey the goalies or the forwards. These players may use a different type of stick. So, I
don’t think the conclusions would be valid. It wouild not take very long to survey all the palyers. I think a
census should have been used.
p.366
1. a) no, no, yes convenience sampling
b) yes, no, systematic sampling
c) no, yes, no, self-selected sampling
p. 367
1. yes, no, no, yes, yes, Method B
2. Yes, yes, yes, no, no, Method A
p. 368
1. Self-selected sampling, no, Why? Only people who go to pet stores will respond. Each member of the
population does not have an equal chance of being surveyed. People who go to pet stores have pets or are
animal lovers. The opinions of those people who do not have pets are equally important.
2. Systematic sampling, no, no, no
p. 369
1. a) The population is divided into groups. Every member in one group is surveyed. So, the method is
cluster sampling.
b) Each member of the population has an equal chance of being selected. So, the method is simple random
sampling.
c) Only shoppers in the mall who walk by the researcher are selected. These are the shoppers who are
convenient. So, the method is convenience sampling.
2. a) The librarian divided the school population into groups but she surveyed the wrong group. The books
students would like to read are not necessarily what the English teachers recommend. The students should be
surveyed.
b) Only residents of Regina who listen to that radiop station will complete the survey. The listeners must
also have access to the internet. This sample will not represent all Canadians.
3. a) no, no, yes, no, Method B
b) The residents of the apartment building, no, yes, Method B
4. a) Cluster sampling, no, Why? Students on the honor roll probably have better study habits than many of
the other students in the school. The sample will not be representativeo f all students in the school.
b) Systematic sampling, yes, Why? One of every 4 people or 25% of the people who order toast are
surveyed. The sample probably included people of different ages, genders, and ethnic background. So, the
sample is probably representative of the patrons of the restaurant.
p. 371
Step 1: CBC Radio, The Vibe 98.5FM, The eagle 100.9, AM 770; Other__None__
Write a question: What is your favourite radio station: CBC radio___ The Vibe 98.5 FM _____The
Eagle 100.9____AM770______None_____?
Yes, no, yes
Step 2: All the students in the school, yes, a sample, no, no, yes, yes, method B
Step 3: I will get permission to hand out the survey at the beginning of homeroom. It won‘t take long for
students to complete the survey, so the teacher could collect them at the end of class.
Step 4: The Vibe 98.5FM, It has the most tally marks, The favourite tradio station is represented by the
tallest bar.
p. 376
1. Craig made his prediction based on the results of a survey or experiment. Explain your thinking: Craig’s
prediction is based on past experience. Craig has seen his hamster run on the wheel at 3:00pm on each of the
last four days. So, the prediction is based on experimental probability.
2. The voters are still satisfied with the mayor’s performance; the other candidates have the same popularity
as the candidates in the other elections.
3. a) No, Yes, No, Explain your thinking: Course-selection forms are handed out in February. Students will
nto know in September if they will need help in Febraruy.
b) The survey should be done in Febraruy, when students have received the forms.
4. a) no, Yes, no, The students are being asked to share personal information. This may make the students
uncomfortable.
b) The company could enclose an anonymous survey with each pay cheque that could be dropped into a
box when completed.
5. a) yes, yes, sample
b) yes, yes, yes, census
6. no, no
7. a) The population is divided into groups. Some members from each group are selected. So, the method is
stratified random sampling.
b) Only people who walk in front of the researcher are selected. These are the people who are convenient.
So, the method is convenience sampling.
8. Every 5th person was surveyed. This is systematic sampling. People entering a Ford dealership are more
likely to drive American-made cars. Most people who drive imports would not go to a Ford dealership. The
sample is not representativeo f the population. So, the conclusions would not be valid.
9. yes, Families with young children, yes, no, yes, yes, Method B.