Chapter 1 Working with surds

Exercise 1A Surds

1 a 2 3 b 35 c 5 d 8

2 a 3 b 3 c 4 d 32

3 a 3 10 b 4 2 c 3 7 d 10 3

e 5 6 f 3 30 g 4 6 h 5 5

4 a 10 15 b 18 c 24 d 36

e 15 f 1 g 5 2 h 16

5 a 6 2 b 7 2 c 3 d 7 2

e 3 5 f +7 2 7 3

g −5 6 3 5 h −8 2 6 3

6 a −3 5 10 b −6 2 16 c 24 + 24 2

d − 1 − 3 e 1 − 5 f 8 + 2 2

7 a 77 b 2 5

5 c 2

2

d 24

e 53

f 2 63

g +3 33 h

−3 2 24

8 a 4 7

7 b 2 6

3 c 2

9 a 15 b 2

10 a 2 cm2 b +(2 2 2 3) cm2

Chapter 2 Simplifying expressions using the laws of indices

Exercise 2A Indices

1 a x10 b x9 c x7 d x5

e x7 f x12 g x11 h x13

i x11 j x16 k x8 l x23

2 a y7 b y c y6 d 1

e y16 f y2 g y2 h y12

i y3 j y7 k y24 l y10

3 a 15a6 b 21a5 c 30a6 d 12a9

e −125a8 f 36a12 g −90a13 h 56a14

4 a 6a b 5 c 3a4 d 6a4

e 10a−4 f 8a2 g 5a4 h 12a−2

5 a 35a8b4 b 15a−2b−2 c 5a4b6

d 19a−8b10 e 2a2b−8 f 4ab−5

g 5a6b−3 h 103

1 7a b−

6 a x12 b x−20 c 8x15 d 81x−8

e x19 f x4 g 144x2 h 200x−18

7 a 6x2 b 4x10 c 72 d 4x6

8 a 1

52 b 141 c

1103 d

133

9 a x12 b y

15 c t

5d n

64

e f1

2 3 f h3

4 5

10 a i 9 ii 43

b i 125 ii 1

125

c i 18 ii 1

11 a 1 12 b 17

72

12 a t34 b m

25

13 a 6 b 12 c 5 d 14

e 2 f 5 g 2 h 3

i 3 j 1

14 a 8 b 625 c 27 d 9

e 16 f 216

Exercise 2B Scientifi c notation

1 a 250 b 31.2 c 0.004 32

d 24.3 e 0.020 719 f 5372

g 203 h 1300 i 817 000

j 0.008 35 k 30 000 000 l 0.000 527

Answers

09087_Answers_P001_022.indd 1 1/26/17 10:59 AM

2 a 2 × 102 b 3.05 × 10−1

c 4.07 × 104 d 3.4 × 109

e 2.078 × 1010 f 5.378 × 10−4

g 2.437 × 103 h 1.73 × 10−1

3 a 2.4673 × 107 b 1.5282 × 104

c 6.13 × 1011 d 9.3 × 107, 2.4 × 1013

e 6.5 × 10−13

4 a 1.581 × 106 b 7.68 × 108

c 7.296 × 109 d 2.142 × 10−1

e 4.41 × 1010 f 6.084 × 10−5

5 a 3 × 104 b 3 × 103

c 5 × 106 d 1.4 × 10−1

6 a 2 × 102 b 4 × 102 c 4 × 1010

7 8 × 108

8 3.9 × 105 km

9 5.28 × 109 miles3

10 3.162 24 × 107

11 63 420 years

12 1.25 × 10 = 12.5 min

13 4.5 × 10−2 g

14 129 times

15 −5.3996 × 107

Chapter 3 Working with algebraic expressions involving expansion of brackets

Exercise 3A Expanding brackets

1 a 3m + 21 b 2x − 2y c x2 + 6x

d a2 − ab e 4x2 − 28x f 6y2 + 4ky

2 a 16x + 3 b 6y + 7

c 6x2 − 18x − 2 d 6x − 13

e 15x2 − 28x f 2 + x − 3x2

3 a x2 + 6x + 8 b x2 − 2x − 3

c x2 + 3x − 4 d x2 − 7x + 10

e x2 − 9 f x2 − 6x + 9

g x2 + 7x + 6 h x2 − 7x + 6

4 a i Added instead of multiplied 3 × 2

ii x2 + 5x + 6

b i Ignored the minus sign in front of 7

ii x2 + 4x − 77

c i Got signs incorrect

ii x2 + 7x − 18

d i −2x + −12x should be −14x

ii x2 − 14x + 24

5 a 12x2 + 22x + 8 b 6y2 + 7y + 2

c 12t2 + 30t + 12 d 6t2 + t − 2

e 18m2 − 9m − 2 f 20k2 − 3k − 9

g 12p2 + p − 20 h 18w2 + 27w + 4

i 15a2 − 17a − 4 j 15r2 − 11r + 2

k 12g2 − 11g + 2 l 12d2 − 5d − 2

m 16p2 + 32p + 15 n 6t2 + 19t + 15

o 15p2 + 11p + 2

6 a x2 + 2x + 1 b x2 − 4x + 4

c x2 − 18x + 81 d x2+ 6x + 9

e 4x2 − 36x + 81 f a2 + 2ab + b2

g a2 − 2ab + b2 h m2 − 4mn + 4n2

i x2 + 2xy + y2 j 4a2 + 12ab + 9b2

k 9a2 − 36ab + 36b2

7 a x3 + 9x2 + 26x + 24 b x3 − 7x − 6

c x3 − 5x2 − 9x + 45 d x3 − 5x2 − 8x + 48

e x3 − x2 − 5x − 3 f x3 − 10x2 + 19x + 14

8 a x3 + x2 − x + 15

b 2x3 + 5x2 − 4x − 3

09087_Answers_P001_022.indd 2 1/26/17 10:59 AM

c 6x3 − 13x2 + 21x − 10

d 12x3 − 11x2 − 26x + 7

e 5x2 − 6x − 6

f 6x2 − 27x + 18

g 6x2 − 26x − 12

h −5x2 −32x + 21

i 6x3 −15x2 + 14x + 5

Chapter 4 Factorising an algebraic expression

Exercise 4A Factorising

1 a 2b(4a − 3c) b 4a(a − 2b)

c 2t(4m − 3p) d 4at(5t + 3)

e 2bc(2b − 5) f 2b(2ac + 3ed)

g 2(3a2 + 2a + 5) h 3b(4a + 2c + 3d)

i t(6t + 3 + a) j 3mt(32t − 1 + 23m)

k 2ab(3b + 1 − 2a) l 5pt(t + 3 + p)

2 a (x + 1)(x + 6) b (t + 2)(t + 2)

c (m + 1)(m + 10) d (k + 3)(k + 8)

e (k − 3)(k − 7) f (f − 1)(f − 21)

g (m − 4)(m − 1) h (p − 2)(p − 5)

i (y + 3)(y − 2) j (t + 8)(t − 1)

k (x + 10)(x − 1) l (r + 7)(r − 1)

m (n − 9)(n + 2) n (m − 22)(m + 2)

o (x − 9)(x + 8) p (t − 21)(t + 3)

3 a (3x + 1)(x + 1) b (3x + 1)(x − 1)

c (2x + 1)(2x + 3) d (2x + 1)(x + 3)

e (5x + 1)(3x + 2) f (2x − 1)(2x + 3)

g (3x − 2)(2x − 1) h (4x + 2)(2x − 3)

i (8x + 3)(x − 2) j (6x − 1)(x − 2)

k (5x − 2)(2x + 3) l (6x − 1)(x + 2)

4 a (x + 9)(x − 9) b (t − 6)(t + 6)

c (2 − x)(2 + x) d (9 − t)(9 + t)

e (k − 20)(k + 20) f (8 − y)(8 + y)

g (x − y)(x + y) h (a − 3b)(a + 3b)

i (3x − 5y)(3x + 5y) j (3x − 4)(3x + 4)

k (10t − 2w)(10t + 2w) l (6a − 7b)(6a + 7b)

5 a All the terms in the quadratic have a common factor of 4.

b 4(x + 2)(x − 1) is the most complete factorisation as it has the highest common factor taken out.

6 a 3(x − 2)(x + 3) b 5(x − 2)(x + 2)

c 4x(x − 2y) d 2(2x − 1)(x + 2)

e 3(3x − 2)(3x + 2) f 2x(x − 3)(x − 4)

g 3x(x − 4)(x + 4) h y(2x − 3)(3x − 2)

Chapter 5 Completing the square in a quadratic expression with unitary x2 coefficient

Exercise 5A Completing the square

1 a (x + 8)2 − 64 b (x − 4)2 − 16

c (x + 3)2 − 9 d (x − 6)2 − 36

e (x − 10)2 − 100 f (x + 9)2 − 81

g (x − 15)2 − 225

h (x + 2.5)2 − 6.25 or x +( ) −52

2 254

2 a a = −1, b = −1 b a = 5, b = −25

c a = −7, b = −49 d a = 6, b = −36

e a = −11, b = −121 f a = 40, b = −1600

g a = − 12 , b = − 1

4 h a = 112 , b = − 121

4

3 a (x + 2)2 + 5 b (x + 4)2 + 4

c (x − 1)2 + 6 d (x − 6)2 − 5

e (x + 3)2 − 11 f (x − 5)2 − 29

g (x − 8)2 − 5 h x +( ) +12

2 34

i (x + 2)2 − 6 j (x + 5)2 − 28

k (x − 3)2 + 6 l (x − 6)2 − 39

m (x + 1)2 − 6 n (x − 9)2 − 24

09087_Answers_P001_022.indd 3 1/26/17 10:59 AM

o (x − 52 )2 − 17

4 p (x − 2)2 − 14

q (x + 6)2 + 4 r (x − 5)2 − 12

s (x + 3)2 + 2 t (x − 1)2 − 10

4 a a = −2, b = 3 b a = 5, b = −27

c a = −3, b = −12 d a = −7, b = −12

e a = 10, b = −143 f a = −9, b = 12

g a = −30, b = −715 h a = 32 , b = 11

4

i a = −3, b = −13 j a = 1, b = 8

k a = −7, b = 1 l a = −11, b = 9

m a = −10, b = −71 n a = −5, b = −7

o a = 72 , b = − 1

4 p a = 20, b = 100

q a = −4, b = −33 r a = 3, b = −29

s a = −1, b = −15 t a = − 14 , b = − 13

4

5 a x = 1 + 3 , x = 1 − 3

b x = 10 − 3, x = − 10 − 3

c x = 5 + 7 , x = 5 − 7

d x = 2 + 2 , x = 2 − 2

e x = 5 + 29 , x = 5 − 29

f x = −4 + 3 , x = −4 − 3

g x = 3 + 13 , x = 3 − 13

h x = 1 + 3 , x = 1 − 3

i x = −6 + 6 , x = −6 − 6

6 a x = 3.7, x = 0.3 b x = −3.2, x = −8.8

c x = 5.7, x = 2.3 d x = 6.4, x = 3.6

e x = 0.7, x = −6.7 f x = 0.7, x = 4.3

Chapter 6 Reducing an algebraic fraction to its simplest form

Exercise 6A Simplest form

1 a a23 b 4

7 c cd3 d

ghk

34

e pq

32 f

qt

911 g

ac

54 h 3hm

2 a x5 b

xx

++

22 5 c

−+

xx

2 32

d x4 e )(

−+

xx

3 24 2 1 f )(

−+

xx

5 22 3

3 a x5

b −−

xx

34 1 c

+xx

24

d ++

xx

25 e

−−

xx

2 34 f

−−

xx

2 53 1

g ))

((

−−

x

x

2 33 1 h

−−

xx

25 i

+−

xx

44

j ++

xx

12 5 k −(x + 2) = −x − 2

l )(− +

+ = − −+

xx

xx

62

62

4 a (x − 3)(x + 4) b ++

xx

43

5 a (2x − 5)(2x + 5) b −xx

2 55

6 a (3x − 1)(x + 4) b −−

xx

3 12( 4)

7 a (4x − 1)(x + 2) b −−

xx

4 13 1

Chapter 7 Applying one of the four operations to algebraic fractions

Exercise 7A Operations on algebraic fractions

1 a 13 b

xy25 c

ay

43

d x65

2

e 38 f

−−

xx

4 23 1

g ))

((

+−

x

x x

3 21

h ) )

)( (

(− −

+x x

x x

2 2 13

2 a 23 b

tg2

2 c x59

2

d 58 e 5

4 f 83

g ))

((

−−

x

x x

2 13

h ))

((

+−

y x

x

45 3

3 a x22

15 b +x5 116

c −x16 11

6 d +x

x3 2

2

e +xx

5 122 2 f ) )( (

−+ −

xx x

7 51 2

g )(−−

pp p8 12

2 =

))

((

−−

p

p p

4 2 32

h ) )( (+ −

− +x xx x

4 93 1

2

i ) )( (+ +

− +x x

x x2 2 8

1 3

2 =

)() )( (+ +

− +x x

x x

2 41 3

2

09087_Answers_P001_022.indd 4 1/26/17 10:59 AM

4 a x7

20 b +x3 8

10

c +x8 5

6 d −x

x5 4

2

e −x

x14 3

6 2 f ) )( (−

+ −x

x x3 19

2 3

g )(−−

yy y10

2

h ) )( (− +

− −x xx x

4 10 63 2

2

= )(

) )( (− +

− −x x

x x

2 2 5 33 2

2

i ) )( (−

− +x

x x3 23

3 4

Chapter 8 Determining the gradient of a straight line, given two points

Exercise 8A Gradient of a straight line

1 a 23 b 1

c 2 d −2

e − 12 f −1

2 a 2 b 4 c 1 d 5

e 3 f 0 g −8 h −6

i 0 j −9 k −10 l −6

3 a −3 b 12 c − 2

3 d 13

e 2 f − 32 g 4

3 h −1

i − 819

4 a − 52 b − 5

2

5 y = 11

6 Yes, as the gradient of the ramp is 120 , which

is less than 115 (or 0.05 < 0.0667).

Chapter 9 Calculating the length of an arc or the area of a sector of a circle

Exercise 9A Arc length and sector area

1 a 2.79 cm b 62.8 cm c 22.0 mm

d 5.34 cm e 35.7 cm f 22.6 mm

2 a 6.28 cm2 b 382 mm2 c 82.1 cm2

d 22.3 cm2 e 3880 mm2 f 76.0 mm2

3 a 25.1 cm b 62.8 cm c 22.0 mm

4 a 628 cm2 b 94.2 cm2 c 942 mm2

5 a 88.4 m2 b 45.3 m

6 a 70° b 322°

7 a 12.0 cm b 22.5 mm

8 268 m2

9 26.1 cm

10 707 cm2

11 Unshaded part is 96.6 cm2.

Chapter 10 Calculating the volume of a standard solid

Exercise 10A Volume of a solid

1 a 1436.755 cm3 b 57 905.836 cm3

c 1047.394 cm3 d 24 429.024 cm3

2 a 418.879 cm3 b 20.944 cm3

c 14 241.887 cm3

3 a 90 cm3 b 65.333 cm3

4 a 64 cm3 b 384 cm3

5 a 113.04 cm3 b 113 040 cm3

6 a 94.2 cm3 b 314 cm3 c 1570 cm3

7 a 70 cm3 b 1440 cm3

8 120 m3

9 867 cm3

10 171.5 cm3

11 3 cm

12 4.6 cm

13 Volume of cone is 419 cm3, volume of pyramid is 396 cm3, so cone has bigger volume and will hold more ice cream (419 > 396).

14 a 660 cm3 b 6.8 cm

09087_Answers_P001_022.indd 5 1/26/17 10:59 AM

Chapter 11 Rounding to a given number of significant figures

Exercise 11A Rounding to significant figures

1 a 50 000 b 60 000 c 30 000

d 90 000 e 90 000 f 50

g 90 h 30 i 100

j 200

2 a 6700 b 36 000 c 69 000

d 42 000 e 27 000 f 7000

g 2200 h 960 i 440

j 330

3 a 50 000 b 6200 c 89.7

d 220 e 8 f 1.1

g 730 h 6000 i 67

j 6 k 8 l 9.75

m 26 n 30 o 870

p 40 q 0.085 r 0.0099

s 0.08 t 0.0620

4 a 5.5−6.5 b 33.5−34.5

c 55.5−56.5 d 79.5−80.5

e 3.695−3.705 f 0.85−0.95

g 0.075−0.085 h 895−905

i 0.695−0.705 j 359.5−360.5

k 16.5−17.5 l 195−205

5 Hellaby 850–949; Hook 645–654; Hundleton 1045–1054

6 A; the parking space is between 4.75 and 4.85 metres long and the car is between 4.25 and 4.75 metres long, so the space is big enough.

7 a 15.5 cm b 14.5 cm

c 310 cm d 290 cm

Chapter 12 Determining the equation of a straight line, given the gradient

Exercise 12A Equation of a straight line

1

x

y

–1 1–5

5

O

10152025303540455055

–10–15–20–25–30–35–40–45–50

2 3 4 5 6 7

a

d

b

f

c

e

8 9 10–2–3–4–5–6–7–8–9–10

2 a y = 0.5x + 3 b y = x

c y = 1.5x − 2 d y = −0.5x − 1

e y = −1.5x − 3

3 a gradient = 4, y-intercept = 3

b gradient = 3, y-intercept = −2

c gradient = 2, y-intercept = 1

d gradient = −3, y-intercept = 3

e gradient = 5, y-intercept = 0

f gradient = −2, y-intercept = 3

g gradient = 1, y-intercept =0

h gradient = − 12 , y-intercept = 3

i gradient = 14 , y-intercept = 2

4 a y = 3x + 4 b y = 14 x − 1

c y = −x + 2

5 a y = −4x + 2 b y = 3x − 14

c y = 5x − 3 d y = −2x + 7

e y = 3x + 10 f y = − 13 x + 4

6 a y = −3x + 24 b y = 1312 x − 0.5

09087_Answers_P001_022.indd 6 1/26/17 10:59 AM

c y = 2x + 7 d y = −3x + 5

e y = 12 x − 2 f y = −3x − 8

g y = 23 x − 3 h y = 3

4 x + 34

7 y = − 12 x + 9

8 a i y = 43 x + 2 ii m = 4

3 iii (0, 2)

b i y = − 52 x − 2 ii m = − 5

2 iii (0, −2)

c i y = 12

32x − ii m = 1

2 iii (0, − 32 )

d i y = x ii m = 1 iii (0, 0)

e i y = 3x + 7 ii m = 3 iii (0, 7)

f i y = 6x ii m = 6 iii (0, 0)

g i y = − 32 x + 3 ii m = − 3

2 iii (0, 3)

h i y = − 23 x + 4 ii m = − 2

3 iii (0, 4)

i i y = 45 x − 8 ii m = 4

5 iii (0, −8)

j i y = − x + 6 ii m = −1 iii (0, 6)

k i y = 32 x − 12 ii m = 3

2 iii (0, −12)

l i y = x + 6 ii m = 1 iii (0, 6)

Exercise 12B Functions

1 a i 49 ii 4 iii 14 iv −46

b 25 c −2 2

5

2 a i −10 ii −5 iii −8 12 iv 1

b − 4 c −23

3 a i 10 ii 73 iii 1

iv 46 v 7

b k = 3 and k = −3

4 a i 4 ii −1 iii −28

iv −8 v 1 vi 7.75

b x = 3 and x = −3

5 a i 12 ii 24 iii 24

b x = 1 and x = 2

Chapter 13 Working with linear equations and inequations

Exercise 13A Solving linear equations

1 a x = 2 b y = 4 c a = 7

d t = 3 e p = 4 f k = 5

g m = 2 h s = −2 i w = 0

2 a x = 3 b x = 7 c t = 1

d x = 5 e y = 6 f x = 3

g t = 2 h t = −2

3 a t = 9 b x = −3

c p = 1 d x = −18

4 a b = 3 b c = 2

5 x = 9

6 a Multiplying out the brackets gives 12x + 20 = 12x + 6, giving 20 = 6, which is impossible.

b Multiplying out the brackets gives 10x + 40 = 10x + 40. Both sides of the equation are the same so x could be any number.

7 Amber and Callum both thought of the number −4.

8 20 crime, 28 science fiction, 17 romance

9 Perimeter is 27 cm (x = 4).

10 Put any pair of sides equal, e.g. 3x + 1 = 4x − 1 and solve. Solution x = 2. Put 2 into each expression for the sides: all sides equal 7. So the answer is yes, the triangle is equilateral when x = 2.

11 a g = 18 b m = 28 c h = 64 d h = 6

e t = 12 f x = 12 g x = 1 h t = −2

i w = 18 j y = 15

12 a x = 8 b t = 2 c m = 6 d p = 2

e x = 3 f t = −4 g x = 52 h x = − 3

2

13 a x = − 110 b x = 1

2 c x = − 524 d x = 105

19

e x = −21 f x = − 14 g x = 23

5 h x = −12

09087_Answers_P001_022.indd 7 1/26/17 10:59 AM

Exercise 13B Solving linear inequations

1 a

–1 0 1 2 3 4 –1 0 1 2 3 4 0 1 2 3 4 5–4 –3 –2 –1 0 1

–4 –3 –2 –1 0 1 0 1 2 3 4 5 –2 –1 0 21 3 4 –2 –1 0 21 3 4

x x x x

xxxx

b

–1 0 1 2 3 4 –1 0 1 2 3 4 0 1 2 3 4 5–4 –3 –2 –1 0 1

–4 –3 –2 –1 0 1 0 1 2 3 4 5 –2 –1 0 21 3 4 –2 –1 0 21 3 4

x x x x

xxxx

c

–1 0 1 2 3 4 –1 0 1 2 3 4 0 1 2 3 4 5–4 –3 –2 –1 0 1

–4 –3 –2 –1 0 1 0 1 2 3 4 5 –2 –1 0 21 3 4 –2 –1 0 21 3 4

x x x x

xxxx

d

–1 0 1 2 3 4 –1 0 1 2 3 4 0 1 2 3 4 5–4 –3 –2 –1 0 1

–4 –3 –2 –1 0 1 0 1 2 3 4 5 –2 –1 0 21 3 4 –2 –1 0 21 3 4

x x x x

xxxx

e

–1 0 1 2 3 4 –1 0 1 2 3 4 0 1 2 3 4 5–4 –3 –2 –1 0 1

–4 –3 –2 –1 0 1 0 1 2 3 4 5 –2 –1 0 21 3 4 –2 –1 0 21 3 4

x x x x

xxxx

f

–1 0 1 2 3 4 –1 0 1 2 3 4 0 1 2 3 4 5–4 –3 –2 –1 0 1

–4 –3 –2 –1 0 1 0 1 2 3 4 5 –2 –1 0 21 3 4 –2 –1 0 21 3 4

x x x x

xxxx

g

–1 0 1 2 3 4 –1 0 1 2 3 4 0 1 2 3 4 5–4 –3 –2 –1 0 1

–4 –3 –2 –1 0 1 0 1 2 3 4 5 –2 –1 0 21 3 4 –2 –1 0 21 3 4

x x x x

xxxx

h

–1 0 1 2 3 4 –1 0 1 2 3 4 0 1 2 3 4 5–4 –3 –2 –1 0 1

–4 –3 –2 –1 0 1 0 1 2 3 4 5 –2 –1 0 21 3 4 –2 –1 0 21 3 4

x x x x

xxxx

2 a x 4 b x < −2 c x 5 d x > 3

e x 1.5 f x 4 g x > 7 h x < −1

3 a x < 10 b x < 2 c x > −6 d t < 4

e y < 6 f x > 12 g w < 7

2 h x < 58

4 a x > − 133 b x > 25

12 c x − 15 d x 23

11

e x > 718 f x > 21

10 g x − 315 h x < − 21

4

5 a x > 45

–1 1 4 3 4 5 6 5 6 7 81 2 3 4

0 1 2 3 –3 –2 –1 0 1 –8 –7 –6 –5 –4 –2 –1 0 1 2

x

x

x

x

x

x

x

x

434

45 61

2

12

25

–

b x 3

–1 1 4 3 4 5 6 5 6 7 81 2 3 4

0 1 2 3 –3 –2 –1 0 1 –8 –7 –6 –5 –4 –2 –1 0 1 2

x

x

x

x

x

x

x

x

434

45 61

2

12

25

–

c x 194

–1 1 4 3 4 5 6 5 6 7 81 2 3 4

0 1 2 3 –3 –2 –1 0 1 –8 –7 –6 –5 –4 –2 –1 0 1 2

x

x

x

x

x

x

x

x

434

45 61

2

12

25

–

d x < 132

–1 1 4 3 4 5 6 5 6 7 81 2 3 4

0 1 2 3 –3 –2 –1 0 1 –8 –7 –6 –5 –4 –2 –1 0 1 2

x

x

x

x

x

x

x

x

434

45 61

2

12

25

–

e x 12

–1 1 4 3 4 5 6 5 6 7 81 2 3 4

0 1 2 3 –3 –2 –1 0 1 –8 –7 –6 –5 –4 –2 –1 0 1 2

x

x

x

x

x

x

x

x

434

45 61

2

12

25

–

f x > −2

–1 1 4 3 4 5 6 5 6 7 81 2 3 4

0 1 2 3 –3 –2 –1 0 1 –8 –7 –6 –5 –4 –2 –1 0 1 2

x

x

x

x

x

x

x

x

434

45 61

2

12

25

–

g x −7

–1 1 4 3 4 5 6 5 6 7 81 2 3 4

0 1 2 3 –3 –2 –1 0 1 –8 –7 –6 –5 –4 –2 –1 0 1 2

x

x

x

x

x

x

x

x

434

45 61

2

12

25

–

h x − 25

–1 1 4 3 4 5 6 5 6 7 81 2 3 4

0 1 2 3 –3 –2 –1 0 1 –8 –7 –6 –5 –4 –2 –1 0 1 2

x

x

x

x

x

x

x

x

434

45 61

2

12

25

–

6 a x > 238 b x < 4 c x −2 d x 73

3

Chapter 14 Working with simultaneous equations

Exercise 14A Solving simultaneous equations graphically and algebraically

1 a (1, 2) b (1, 1) c (0, −2) d (1, 1)

e (6, 11) f (12, −18) g (4, 1) h (−1, −4)

i (5, −1) j (3, 7) k (−1, 3)

l Lines are parallel so there is no solution.

2 a x = 5, y = 2 b x = 3, y = 4

c x = 4, y = 2

3 a x = 2, y = 3 b x = 7, y = 3

c x = 2, y = 5 d x = 4, y = 3

4 a x = 3, y = 1 b x = 7, y = 2

c x = 2 12 , y = 3 d x = 7, y = −1

5 a x = 2, y = 5 b x = 6, y = 4

c x = 4, y = 2

09087_Answers_P001_022.indd 8 1/26/17 10:59 AM

6 a x = 1, y = 4 b x = 5, y = 3

c x = 6, y = 2

7 a x = 7, y = 3 b x = 2, y = 4

8 CD £10.50, book £3.50

9 a 2x + 3y = 28.50, 3x + 2y = 31.50

b x = £7.50, y = £4.50

10 12 g in cakes and 13 g in peanuts

11 £2.28

12 £1.21

13 a My age minus 5 was double my son’s age minus 5.

b x = 61 and y = 33

14 4a + 2n = 204, 5a + n = 171 gives a = 23, n = 56.

Total cost for Marcus is £5.40, so he will get £4.60 change.

15 5c + 4p = 340, 3c + 5p = 321, gives c = 32 kg, p = 45 kg

The bags weigh 552 kg, so Carol cannot carry the bags safely on her trailer.

Chapter 15 Changing the subject of a formula

Exercise 15A Changing the subject of a formula

1 a x = k − h b x = p − m

c x = −y cm

d x = −b kb

92 or x = − k

b92 2

e x = Afg

f x = Ny3

2 a x = y34

b x = +p2 53

c x = 3(c + g) d x = bMn

e x = P − hR

f x = +gLH

j or x = +gL HjH

3 a t = y7

b h = cd

− 2t or h = −c dtd2

c w = )( −J

M k7

4 a = −x

y52

b )(

=+

−ab p q

p q

c = +aA

b c2 2 d )(

=+ −

rs t 1 3

2

e = +rr

s3

3 f = +

−gfx t

f A2

5 a = +xb

a c b = − −x

ba b 1

c = +xa

b d2 d = −x

cdc d2

e = −xb

a2

7 f = −x

bcc b2

6 a x = −S y3

2

b x = −B jh

c x = K − 3 d x = G mh

2

e x = )( −H cf

t2

2 f x = kC( )2

+ p

g x = −t L k2

h x = j rp

2

2 + a

Chapter 16 Recognise and determine the equation of a quadratic function from its graph

Exercise 16A Equations and graphs of quadratic functions

1 a k = 2 b k = 5 c k = 3 12

d k = −3 e k = −5 f k = − 12

2 a k = 3 b k = 4 c k = 2

d k = −5 e k = −4 f k = −0.5

3 a a = −3, b = 1 b a = −4, b = 2

c a = 3, b = 2

4 a a = −4, b = −1 b a = 5, b = −2

c a = 4, b = 2

5 a i a = −1, b = 4 ii f(x) = (x − 1)2 + 4

b i a = 5, b = 3 ii f(x) = (x + 5)2 + 3

c i a = 3, b = −1 ii f(x) = −(x + 3)2 − 1

d i a = 1, b = −1 ii f(x) = (x + 1)2 − 1

e i a = −3, b = −1 ii f(x) = −(x − 3)2 − 1

f i a = −3, b = 5 ii f(x) = −(x − 3)2 + 5

6 a = −3, b = 2

09087_Answers_P001_022.indd 9 1/26/17 11:00 AM

Chapter 17 Sketching a quadratic function

Exercise 17A Sketching a quadratic function

1

xO

y

(4, –4)

12

2 6

3–5

–15

xO

y

(–1, –16)

51

5

xO

y

(3, –4)

4–8

–32

xO

y

(–2, –36)

6–4

–24

xO

y

(1, –25)

7–3

–21

xO

y

(2, –25)

4 xO

y

(2, –4)

6–2

12

xO

y(2, 16)

2

3–3

–9

xO

y

1–5

–5

xO

y

(–2, –9)(0, –9)

2–4

–8

xO

y

(–1, –9)

6 xO

y

(3, 9)

–7 –1

7

xO

y

(–4, –9)

(0, 16)

–4 8 xO

y

(2, 36)32

–3 2

–6

xO

y

–4 4 xO

y16

( ), –614

12–

3–3

–9

xO

y

1–5

–5

xO

y

(–2, –9)(0, –9)

2–4

–8

xO

y

(–1, –9)

6 xO

y

(3, 9)

–7 –1

7

xO

y

(–4, –9)

(0, 16)

–4 8 xO

y

(2, 36)32

–3 2

–6

xO

y

–4 4 xO

y16

( ), –614

12–

3

10

xO

y

(–3, 1)

29

xO

y

y y

(–5, 4)

10

xO

y y

(–2, 6)

9

xO

y

(–2, 5)

7

xO

y

(2, 3)

11

xO

(3, 2)

8

xO(1, 7)

14

xO

(4, –2)

a

c d

b

e

g h

f

a b

c

e

g

d

f

h

a b

c d

09087_Answers_P001_022.indd 10 1/26/17 11:00 AM

10

xO

y

(–3, 1)

29

xO

y

y y

(–5, 4)

10

xO

y y

(–2, 6)

9

xO

y

(–2, 5)

7

xO

y

(2, 3)

11

xO

(3, 2)

8

xO(1, 7)

14

xO

(4, –2)

10

xO

y

(–3, 1)

29

xO

y

y y

(–5, 4)

10

xO

y y

(–2, 6)

9

xO

y

(–2, 5)

7

xO

y

(2, 3)

11

xO

(3, 2)

8

xO(1, 7)

14

xO

(4, –2)

4

x xO

y

(–2, –1)

–5

O

y

(3, –2)

–11

xO

y

(5, –6)

–31

xO

y

(–3, –4)

–13

xO

y

(–4, –5)

–21

xO

y

(–1, –4)

–5

xO

y(–5, 3)

–22

xO

y

(2, –3)

–7

5

x

y

(–3, 2)

20

O

xO

y

(2, –4)

–12

x

y

(–1, –3)

–7

O

xO

y

(–2, –5)

–3

xO

y(–1, 3)

9

xO

y

(–1, 4)

xO

y

(3, –1)

–19

x

y

(4, 1)

49

O

Chapter 18 Identifying features of a quadratic function

Exercise 18A Features of a quadratic function

1 a i x = 3 ii (3, 5) iii minimum

b i x = −2 ii (−2, 1) iii minimum

c i x = 7 ii (7, −1) iii maximum

d i x = −5 ii (−5, −8) iii maximum

e i x = 8 ii (8, 7) iii minimum

f i x = −2 ii (−2, 12) iii maximum

g i x = −3 ii (−3, 9) iii minimum

fe

hg

a b

dc

e f

hg

c d

fe

hg

a b

09087_Answers_P001_022.indd 11 1/26/17 11:00 AM

h i x = −6 ii (−6, 3) iii maximum

i i x = −7 ii (−7, −18) iii minimum

2 a i x = 4 ii (4, 8) iii minimum

b i x = −1 ii (−1, 8) iii minimum

c i x = 6 ii (6, −3) iii maximum

d i x = −1 ii (−1, −5) iii maximum

e i x = 8 ii (8, 6) iii minimum

f i x = −5 ii (−5, 12) iii maximum

g i x = −2 ii (−2, 14) iii minimum

h i x = −4 ii (−4, 7) iii maximum

i i x = −3 ii (−3, −13) iii minimum

3 a i y = (x + 2)2 + 3 ii x= −2 iii (−2, 3)

b i y = (x + 3)2 + 2 ii x = −3 iii (−3, 2)

c i y = (x − 5)2 − 27 ii x = 5 iii (5, −27)

d i y = (x − 1)2 − 8 ii x = 1 iii (1, −8)

e i y = (x − 3)2 − 12 ii x = 3 iii (3, −12)

f i y = (x + 6)2 − 1 ii x = −6 iii (−6, −1)

g i y = (x − 2)2 − 6 ii x = 2 iii (2, −6)

h i y = (x − 7)2 + 4 ii x = 7 iii (7, 4)

i i y = x +( )12

2 + 3

4 ii x = − 12 iii −( )1

234,

4 a iv y = (x + 1)2 + 3

b ii y = −(x − 1)2 + 3

c i y = (x − 1)2 + 3

d iii y = 2(x − 1)2 + 3

e v y = −(x + 1)2 + 3

f vi y = 3(x + 1)2 + 3

Exercise 18B Using quadratic functions to solve problems

1 a 36 m b 12 seconds

2 a 42 seconds

b No, as maximum height (at t = 42 ÷ 2 = 21 s) = 441 m, which is less than 450 m.

3 a (10 − x) m b x(10 − x) = 10x − x2

c x = 5 d A = 25 m2

4 a x = £10 b P = £10 000

Chapter 19 Working with quadratic equations

Exercise 19A Solving quadratic equations by factorising

1 a x = −3, x = −2 b t = −4, t = −1

c a = −5, a = −3 d x = −4, x = 1

e x = −2, x = 5 f t = −3, t = 4

g x = 2, x = −1 h x = 1, x = −4

i a = 6, a = −5 j x = 2, x = 5

k x = 2, x = 1 l a = 2, a = 6

2 a x = 0, x = 3 b x = 0, x = −6

c x = −3, x = 3 d x = −4, x = 5

e x = 12 , x = −3 f x = 1

3 , x = − 52

3 a (x + 5)(x + 1) = 0, x = −1, x = −5

b (x + 3)(x + 6) = 0, x = −3, x = −6

c (x − 8)(x + 1) = 0, x = 8, x = −1

d (x − 7)(x + 3) = 0, x = 7, x = −3

e (x + 5)(x − 2) = 0, x = −5, x = 2

f (x + 5)(x − 3) = 0, x = −5, x = 3

g (t − 6)(t + 2) = 0, t = 6, t = −2

h (t − 6)(t + 3) = 0, t = 6, t = −3

i (x + 2)(x − 1) = 0, x = −2, x = 1

j (x − 2)(x − 2) = 0, x = 2

k (m − 5)(m − 5) = 0, m = 5

l (t − 8)(t − 2) = 0, t = 8, t = 2

m (t + 3)(t + 4) = 0, t = −3, t = −4

n (k − 6)(k + 3) = 0, k = 6, k = −3

o (a − 4)(a − 16) = 0, a = 4, m = 16

4 a x = 0, x = 5 b x = 0, x = −8

c x = 0, x = 7 d x = −3, x = 3

e x = −5 , x = 5 f x = −6 , x = 6

09087_Answers_P001_022.indd 12 1/26/17 11:00 AM

5 a (2x + 1)(x + 2) = 0, x = − 12 , x = −2

b (7x + 1)(x + 1) = 0, x = −1, x = − 17

c (4x + 7)(x − 1) = 0, x = − 74 , x = 1

d (3x + 5)(2x + 1) = 0, x = − 53 , x = − 1

2

e (3x + 2)(2x + 1) = 0, x = − 23 , x = − 1

2

f (3x − 4) (4x − 3) = 0, x = 43 , x = 3

4

6 a x = 3, x = −2 b x = − 14 , x = − 3

2

c x = 6, x = −5 d x = 32 , x = −7

e x = − 34 , x = 3 f x = − 3

2 , x = 1

Exercise 19B Solving quadratic equations using the quadratic formula

1 a x = 1.14, x = −1.47

b x = −0.29, x = −1.71

c x = 3.19, x = −2.19

d x = 0.43, x = −0.77

e x = −0.57, x = −1.77

f x = −0.09, x = −5.41

g x = −0.22, x = −2.28

h x = 2.16, x = −4.16

2 a x = −0.697, x = −4.30

b x = 7.74, x = 0.258

c x = 0.531, x = −7.53

d x = −0.293, x = −1.71

e x = 2.37, x = 0.634

f x = 1.47, x = −1.14

g x = −1.83, x = 3.83

h x = 1.85, x = −1.35

i x = 0.558, x = −0.358

3 a discriminant = −11; no real roots

b discriminant = −8; no real roots

c discriminant = 84; two real, distinct roots

d discriminant = 0; two real, equal roots

e discriminant = 81; two real, distinct roots

f discriminant = 144; two real, distinct roots

g discriminant = 41; two real, distinct roots

h discriminant = −11; no real roots

i discriminant = −23; no real roots

j discriminant = 0; two real, equal roots

k discriminant = −136; no real roots

l discriminant = 41; two real, distinct roots

4 3x2 − 4x − 8 = 0

5 Eric gets = ±x

12 018 and June gets

(3x − 2)2 = 0. Both find that there is only one

solution: x = 23 . The x-axis is a tangent to the

curve.

Exercise 19C Using quadratic equations to solve problems

1 a Ella’s brother is (x + 4) years old.

b x(x + 4) = 1020

x2 + 4x = 1020

x2 + 4x − 1020 = 0

c x = 30, so Ella is 30 years old.

2 a (2x + 1)2 = (2x)2 + (x + 1)2 when expanded, and like terms collected, gives the required equation.

b x = 2

c Area = 12 (x + 1)(2x) = 1

2 2 1 2 2 6( ) ( )+ × = cm2

3 a (x + 5) m

b x(x + 5) = 60

x2 − 5x − 60 = 0

c x = 564 cm, so lawn length = 1064 cm

09087_Answers_P001_022.indd 13 1/26/17 11:00 AM

4 a A = (x + 2)2 cm2 or A = (x2 + 4x + 4) cm2

b (2x + 4)(x − 2) = 2x2 − 8

Areas of square and rectangle are the same, so 2x2 − 8 = x2 + 4x + 4, giving x2 − 4x − 12 = 0.

c x = 6, so length of square = 8 cm, length of rectangle = 16 cm, breadth of rectangle = 4 cm.

Chapter 20 Applying Pythagoras’ theorem

Exercise 20A Applying Pythagoras’ theorem

1 Yes, as 92 + 402 = 1681 = 412.

2 No, as 112 + 232 = 650, 262 = 676 and 650 ≠ 676.

3 Yes, as 142 + 482 = 2500 = 502.

4 No, Jenny is not correct as 112 + 32 = 290, 162 = 256 and 290 ≠ 256.

5 No, as 92 + 322 = 1105, 342 = 1156 and 1105 ≠ 1156.

6 a 14.1 m b 14.5 m

7 a DG = 11.2 cm b HA = 7.1 cm

c DB = 11.2 cm d AG = 12.2 cm

8 26 cm

9 42 cm

10 14.1 cm

11 a AC = 9.9 cm b EX = 10.9 cm

c EM = 11.5 cm

12 a 22.3 cm b 1902.9 cm3

13 a 6.7 cm b 10.55 cm

14 Yes, as 52 + 52 + 22 = 54, and 54 > 72 (49).

Chapter 21 Applying the properties of shapes to determine an angle involving at least two steps

Exercise 21A Using angle properties of circles

1 58°

2 76°

3 72°

4 62°

5 32°

6 156°

7 14°

8 58°

Exercise 21B Using Pythagoras’ theorem

1 7.84 cm

2 17.9 cm

3 19.0 cm

4 6 cm

5 a OH = r − 3 b 12.2 cm

6 14.2 m

7 2.30 m

Exercise 21C Angles in polygons

1 a 70° b 120° c 65°

d 70° e 70° f 126°

2 a No, total is 350°.

b Yes, total is 360°.

c Yes, total is 360°.

d No, total is 370°.

e No, total is 350°.

f Yes, total is 360°.

3 a Pentagon divided into 3 triangles, showing 3 × 180° = 540°

b 80°

4 a 112° b 130°

5 a 6 triangles b 1080° c 135°

6 a 10 triangles b 1800° c 150°

7 a 28 triangles b 5040° c 168°

09087_Answers_P001_022.indd 14 1/26/17 11:00 AM

8 a x = 60°, y = 120° b x = 90°, y = 90°

c x = 108°, y = 72° d x = 120°, y = 60°

e x = 135°, y = 45°

9 a 18 b 12 c 20 d 90

10 a 8 b 24 c 36 d 15

11 Angle AED = 108° (interior angle of a regular pentagon), so angle ADE = 36° (angles in an isosceles triangle).

Chapter 22 Using similarity

Exercise 22A Similar shapes

1 a 4.8 cm b 4.88 cm

2 120 cm

3 a 9.6 cm b 1.9 cm

4 a x = 6.9 cm, y = 3.4 cm

b x = 12 cm, y = 12.5 cm

5 2 m

6 No, Suzie is not correct. The corresponding sides are not in the same ratio; for the triangles to be similar, CD would have to be 12.5 cm.

7 BC = 10 cm, CD = 15 cm

8 a 43 cm b 5

3 cm c 6 cm

Exercise 22B Areas and volumes of similar shapes

1 320 cm2

2 a 10 800 cm3 b 50 000 cm3

3 3612 4

3( ) × = 108 litres

4 12.15 m3

5 The large tin holds 2700 ml. He can fill 3 small tins.

6 a 6 m2 b 20 000 cm3

7 a 21% b 33.1%

8 iv 810 cm3

9 a small 13.9 cm; large 25.2 cm

b medium 63 cm2; small 30.1 cm2

10 16.2 cm

11 17.3 cm and 23.1 cm

Chapter 23 Working with the graphs of trigonometric functions

Exercise 23A Graphs of trigonometric functions

1

2 a y = 5cos x° b y = 4sin x°

c y = 3.5cos x° d y = −3sin x°

3

x

y

–1

0

1

360180 x

y

–1

0

1

360180

x

y

–1

0

1

360180 x

y

–1

0

1

180 360

4 a y = cos 2x° b y = sin 34 x°

c y = sin 4x° d y = cos 4x°

x

y

–3

0

3

360180 x

y

–4

0

4

360180

x

y

0360180 x

y

–2

0

2

360180

–23

23

x

y

–3

0

3

360180 x

y

–4

0

4

360180

x

y

0360180 x

y

–2

0

2

360180

–23

23

a b

c d

a

c d

b

09087_Answers_P001_022.indd 15 1/26/17 11:00 AM

5

x

y

–4

0

4

360180 x

y

–5

0

5

360180

x

y

–1.5

0

1.5

360180 x

y

–4

0

4

360180

6 a y = 3cos 2x° b y = 7sin 2x°

c y = 6sin 3x° d y = −5cos 2x°

Exercise 23B Translations of graphs of trigonometric functions

1

x

y

0

5

3

1

3601800

x

y

–4

–2

0360180

x

y

–10

3

360180

x

y

0

6

2

3601800

x

y

0

5

3

1

3601800

x

y

–4

–2

0360180

x

y

–10

3

360180

x

y

0

6

2

3601800

2 a y = 2cos x° + 2 b y = 3sin 2x° − 2

c y = 4cos 2x° + 1 d y = 2sin 3x° + 5

3 a i amplitude = 2 ii period = 360°

b i amplitude = 3 ii period = 180°

c i amplitude = 4 ii period = 180°

d i amplitude = 2 ii period = 120°

4

x

y

–7

–1 36025070

x

y

0

3

1

36020 200

x

y

–1

0

1

300 36030 50120 210

x

y

–2

0

0

2

360230140320

5 a a = 4, b = −40° (or 320°)

b b = 30° (or −330°), c = 3

c a = 5, b = 20° (or −340°)

d b = −30° (or 330°), c = −1

Chapter 24 Working with trigonometric relationships in degrees

Exercise 24A Solving trigonometric equations

1 a x = 26.9°, 153.1° b x = 41.4°, 318.6°

c x = 66.5°, 246.5° d x = 38.3°, 141.7°

e x = 24.2°, 335.8° f x = 23.6°, 156.4°

g x = 59.0°, 239.0° h x = 64.6°, 295.4°

2 a x = 235.5°, 304.5° b x = 112.3°, 247.7°

c x = 122.0°, 302.0° d x = 195.7°, 344.3°

e x = 148.5°, 211.5° f x = 221.8°, 318.2°

g x = 114.0°, 294.0° h x = 112.0°, 248.0°

3 a x = 56.4°, 123.6° b x = 48.2°, 311.8°

c x = 76.0°, 256.0° d x = 30°, 150°

e x = 73.4°, 286.6° f x = 48.6°, 131.4°

g x = 56.3°, 236.3° h x = 79.5°, 280.5°

a

c d

b a

c d

b

a

c d

b

09087_Answers_P001_022.indd 16 1/26/17 11:00 AM

4 a x = 210.0°, 330.0° b x = 104.5°, 255.5°

c x = 116.6°, 296.6° d x = 225.6°, 314.4°

e x = 102.8°, 257.2° f x = 202.0°, 338.0°

g x = 126.9°, 306.9° h x = 107.5°, 252.5°

5 a x = 19.5°, 160.5°

b (221.8, 0) and (318.2, 0)

6 a 14 m b 101.5 s and 258.5 s

7 a 7 m b 10.46 m

c 3.30 p.m. and 5.30 p.m.

Exercise 24B Trigonometric identities

1 a 6sin x b cos x c tan2 x d −4

2 a LHS: 2 + 3cos2 x = 2 + 3(1 − sin2 x ) = 2 + 3 − 3sin2 x = 5 − 3sin2 x

b LHS: 5 53 2sin sin cos

cosx x x

x+

= 5 2 2sin sin cos

cosx x x

x+( )

= 5 sincos

xx = 5tan x

c LHS: tansin

xx

= tan x ÷ sin x

= sincos

sinxx

x÷ 1

= sincos sin

xx x

× 1 = 1

cos x

d LHS: 4sin2 x + 7(1 − sin2 x) = 4sin2 x + 7 − 7sin2 x

= 7 − 3sin2 x

Chapter 25 Calculating the area of a triangle using trigonometry

Exercise 25A Calculating the area of a triangle using trigonometry

1 a 15.45 cm2 b 124.98 cm2 c 309.21 cm2

2 a 45 cm2 b 48 cm2 c 189 cm2

3 a 37.34 cm2 b 9.74cm2

4 127.31 cm2

5 229.10 cm2

6 a 20 m b 18 cm c 4.5 cm

7 a 16 cm b 24 cm

8 a 72° b 21° c 125°

9 43°

Chapter 26 Using the sine and cosine rules to find a side or angle

Exercise 26A Using the sine rule

1 a 4.4 m b 10.0 cm c 29(.7)° d 37(.2)°

2 a 16 cm b 9 cm

3 a 66(.7)° b 113(.3)°

4 127(.4)°

5 a 47° b 88 m c 131.9 m

6 64.95 m

7 54.2 m

8 a 9.24 cm

b Area of quadrilateral = area of triangle ABD (36.83 cm2)

+ area of triangle DCB (11.41 cm2) = 48.24 cm2

Exercise 26B Using the cosine rule

1 a 9.54 m b 53.94 cm

2 a 102.6° b 114.6°

3 Cos B = + −× ×

4 5 72 4 5

2 2 2

= − = −840

15

4 a 11.86 cm b 37.7° c 27.3°

d 5.63 cm e 54.4 cm2

5 1.65 km

6 66.2°

7 29.9°

8 29.7°

9 22.9 cm

10 a 36.3° b 533 cm2

c Yes, she has enough paint, because 0.5 m2 = 5000 cm2, and 5000 cm2 > 533 cm2.

09087_Answers_P001_022.indd 17 1/26/17 11:00 AM

Exercise 26C Choosing the correct formula

1 a 9.2 m b 125.1° c 23.4°

d 8.2 m e 76.8° f 63.4 cm

2 16.4 cm

3 a 66.8° b 9.4 cm

4 7 cm

5 ∠ABC = 87.3° and is the largest angle.

Chapter 27 Using bearings with trigonometry

Exercise 27A Using bearings with trigonometry

1 130°

2 233°

3 32.5 km

4 a 144° b 44.9 km

5 a 83° b 149°

6 20.2 km

7 a 16.16 km b 035°

Chapter 28 Working with 2D vectors

Exercise 28A 2D vectors

1 a 3

4

b −

3

4

c 3

4−

d −−

3

4

2 a AB� ���

=

5

3 b CD

� ���=

−

1

6

c p = −−

7

2 d q = 4

4−

e FE� ���

=−

3

5 f r =

−

5

0

3 a ab

d

ce f

g

h

i

b ab

d

ce f

g

h

i

c ab

d

ce f

g

h

i

d a

b

d

ce f

g

h

i

4 a

–b a ce

–f–d b

–b a ce

–f–d

c

–b a ce

–f–d

5 a b c

a2b

2d

2c 2f3e

a2b

2d

2c 2f3e

a2b

2d

2c 2f3e

Exercise 28B Using vectors to solve problems

1 a i 12 b ii 1

2 a + 12 b iii 3

2 a

b Both are multiples of a.

2 a b − a b −2a

c 2b − a d 2b − a

3 a 34 b + 1

4 a b 58 a + 3

8 b

4 a 5p − 10q b 4p − 8q c 2q + 4p

5 a i 13 (a – b) ii 1

3a + 23 b iii b + 1

2 a

b They lie on a straight line.

6 a i b − a ii 2b − 2a

b They are parallel.

7 a i −3p + 3q ii −3p + 12q

b 2p + 4q

c They lie on a straight line.

Chapter 29 Working with 3D coordinates

Exercise 29A Working with 3D coordinates

1 a A (0, 0, 0) B (8, 0, 0) C (8, 5, 0)

D (0, 5, 0) E (0, 0, 3) F (8, 0, 3)

G (8, 5, 3) H (0, 5, 3)

09087_Answers_P001_022.indd 18 1/26/17 11:00 AM

b J (3, 0, 0) K (7, 0, 0) L (7, 2, 0)

M (3, 2, 0) N (3, 0, 6) P (7, 0, 6)

Q (7, 2, 6) R (3, 2, 6)

c A (0, 0, 0) B (4, 0, 0) C (4, 4, 0)

D (0, 4, 0) E (2, 2, 7)

d A (0, 0, 0) B (6, 0, 0) C (6, 6, 0)

D (0, 6, 0) E (3, 3, 8)

e E (0, 0, 2) F (8, 0, 2) G (8, 6, 2)

H (0, 6, 2) T (4, 3, 9)

f K (0, 0, 0) L (0, 4, 0) M (0, 2, 2)

N (10, 0, 0) P (10, 4, 0) Q (10, 2, 2)

2 a A (0, 0, 0) B (7, 0, 0) C (7, 4, 0)

D (0, 4, 0) E (0, 0, 3) F (4, 0, 3)

G (7, 0, 3) H (7, 4, 3) J (4, 4, 3)

K (0, 4, 3) L (0, 0, 7) M (4, 0, 7)

N (4, 4, 7) P (0, 4, 7)

b A (0, 0, 0) B (10, 0, 0) C (10, 4, 0)

D (0, 4, 0) E (0, 0, 6) F (10, 0, 6)

G (10, 4, 6) H (0, 4, 6) J (0, 2, 9)

K (10, 2, 9)

c C (4, 0, 5) D (4, 4, 5) E (2, 2, 10)

3 a i (2.5, 0, 10) ii (2.5, 2, 10) iii (5, 4, 5)

b 11.9 units

4 a A (0, 0, 0) B (8, 0, 0) C (8, 8, 0)

D (0, 8, 0) E (0, 0, 8) F (8, 0, 8)

G (8, 8, 8) H (0, 8, 8) J (4, 4, 11)

b 6.4 units

Chapter 30 Using vector components

Exercise 30A Using vector components

1 a −

4

13 b

−−

12

8 c

10

21

d 3

3

e −

3

2 f

12

4−

g 8

26−

h 9

12−

2 a −

−

2

23

28

b 6

6

3

−

c 1

46

45−

d 11

4

2−

e 5

15

13−

f 20

23

28

−

g 18

3

15

h 7

9

1

3 a a = −1, b = 4

b x = −2, y = −5, z = −2

c a = 2, b = −1

Exercise 30B Calculating the magnitude of vectors

1 a 5.39 b 3.16 c 7.21 d 12.04

2 a 2 5 b 3 5 c 2 10 d 3 10

3 a 5.39 b 11.58 c 11.87 d 15.17

4 a 3 5 b 3 2 c 2 11 d 2 41

5 a 3.16 b 9.85 c 18.36 d 26.68

6 a 7.14 b 8.12 c 28.18 d 14.04

Chapter 31 Working with percentages

Exercise 31A Percentage increase and decrease

1 £2519.42

2 a 5.5 cm b 6.05 cm c 7.32 cm d 9.74 cm

3 a £291.60 b £314.93 c £367.33

4 a 1725 b 1984 c 2624

5 £20 240.75

6 Veronika £174.47, Amelia £241.94 , Scarlett £308.46. Scarlett’s phone is worth the most.

7 £3795.96

8 a 87.55 g b 98.54 g c 114.23 g

9 a £32 413.50 b 7 years

10 a i 2012 ii 2015 iii 2020 iv 2030

b 2022

11 £3176.76

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12 After 11 years, the sycamore is 93.27 cm tall and the conifer is 93.04 cm tall. After 12 years, the sycamore is 100.72 cm tall and the conifer is 107 cm tall.

13 2 years

14 4 weeks

Exercise 31B Reverse use of percentages

1 £24

2 51 400

3 3 hours 45 minutes

4 23 612 800

5 2100 cal

6 £150

7 £440

8 220

9 £15

10 £45 000

11 £180

12 a £22 454 b 6.8%

13 £1800

14 100% (still twice as many)

Chapter 32 Working with fractions

Exercise 32A Adding and subtracting fractions

1 a 710 b 5

6 c 1330 d 17

24

e 1920 f 11

15 g 3940 h 9

10

2 a 18 b 3

10 c 715 d 7

20

3 a 169240 b 199

360 c 301468

4 a 12 1720 b 10 1

9 c 9 920 d 12 81

200

e 10 6180 f 12 5

16 g 1 1330 h 1 1

3

i 2 1996 j 1 169

240 k 1 199360 l 1 301

468

5 a 112 b 36

6 110

7 13 125

8 a 4 14 miles b 1 1

4 miles

9 To make a 2 metre pipe, use two 34 metre

pipes and one 12 metre pipe.

10 24

Exercise 32B Multiplying and dividing fractions

1 a 13 b 3

10 c 310 d 2

7

e 59 f 1

5 g 715 h 3

20

i 16 j 3

20 k 716 l 1

2

2 a 35 b 1 3

5 c 1 15 d 9

14

e 34 f 4 g 1 h 1

3 a 3 b 2 13 c 2 d 2 1

6

e 5 15 f 4 2

3 g 4 112 h 12

i 3 1118 j 11 k 8 l 1

4 a 1 13 b 1 1

51 c 1 79 d 88

95

e 1 1125 f 1 37

80 g 4547 h 7

8

i 1819 j 24

25 k 2 23 l 1 4

11

5 2 14 km

6 25

7 23 of 4 2

5 (= 2 1415 ) is smaller than 3

4 of 5 13 (= 4).

8 Yes, as 66 litres were bought.

9 15 sections

10 80 hops

11 215

12 4 tins

Exercise 32C Combinations of operations with fractions

1 a 7 4748 b 2 289

560 c 6 911

d 2265 e 7 61

792 f 38 67234

09087_Answers_P001_022.indd 20 1/26/17 11:00 AM

2 a 2263132 or 17 19

132 b 32 469980 or 33 129

980

c 38275 or 5 7

75 d 10 142

e 1 1320 f 12 7

20

3 a 1124 b 1

14 c 23240

d 2 114 e 2 14

15 f 49144

Chapter 33 Comparing data using statistics

Exercise 33A Mean and standard deviation

1 a mean = 11, s = 4.52

b mean = 19, s = 3.65

c mean = 134, s = 8.15

d mean = 2154, s = 53.47

2 a mean = 5, s = 582

b mean = 3, s = 2 7

3

c mean = 5, s = 8

d mean = 3, s = 343

3 a mean = 70 kg, s = 6.66 kg

b On average, players in Beeton Academy’s team are lighter, and their mass varies less.

4 a mean = 55, s = 14.96

b On average, Jez’s team performed better, but Rebecca’s team was more consistent.

5 a mean = 6, s = 1.4

b mean = 6, s = 3.1

c On average, their scores were the same, but Connie’s scores varied more.

6 a mean = 2, s = 1.63

b On average, Henrietta produced more eggs in the first week, but her results were also more varied in the first week.

7 a mean = 12, s = 3.63

b On average, the pupils’ performance had improved in the December test and their results were more consistent.

8 a mean = £861.50, s = £32.50

b The addition of a £5.50 booking fee increased the mean by £5.50 but did not affect the standard deviation, because the price of each holiday increased by the same amount.

Exercise 33B Median and semi-interquartile range

1 a i median = 6.5

ii Q1 = 4, Q3 = 8, SIQR = 4, SIQR = 2

b i median = 18

ii Q1 = 14.5, Q3 = 22, IQR = 7.5, SIQR = 3.75

c i median = 45

ii Q1 = 27, Q3 = 54, IQR = 27, SIQR = 13.5

2 a median = 51 years, IQR = 5.5 years, SIQR = 2.75 years

b median = 34.5 years, IQR = 10 years, SIQR = 5 years

c On average, the ages of teachers in the music department are higher but less varied.

3 a median = £912 000, IQR = £1 255 000, SIQR = £627 500

b On average, the actors earned more but their salaries were more varied.

4 a median = 3 min 33 s, SIQR = 37.5 s

b On average, the songs were shorter on the first album, but the lengths were more varied.

5 a median = £7.68, SIQR = £1.21

b On average, the quotes for New York were more expensive and more varied.

6 a median = 71, SIQR = 5

b On average, the rugby team completed more press-ups per minute and their results were less varied.

7 a median = 41 min 30 s, SIQR = 6 min 45 s

b On average, the students took longer to complete the history tasks, but their times were more consistent.

09087_Answers_P001_022.indd 21 1/26/17 11:00 AM

8 a mean = 66%, s = 10.8%

b median = 69%, SIQR = 5%

c Using the median and SIQR is better because her result for Art is an outlier.

Chapter 34 Forming a linear model from a given set of data

Exercise 34A Drawing and using a best-fitting line from given data

1 a, b

x40Maths exam

Mus

ic e

xam

0

20

40

60

60200

y

80

100

80 100

c Using (40, 40) and (70, 60): y = 23 x + 40

3

d 43 (to nearest whole number)

e 97

2 a, b

T40Time (min)

Dis

tanc

e (k

m)

0

10

20

30

60200

D

40

50

80 100 120

c Using (0, 0) and (110, 40), D = 411 T

d approx. 16 km

e approx. 82.5 min

3 a D = 39T

b 117 miles

c approx. 4 hours 29 min

4 a P = 34 M + 8 b 32 c 20

5 a H = 3F + 71 b 155 cm c 31 cm

6 a L = −0.5R + 37.5 b 26.5 hours

c 3 hours

7 a N = −3T + 34 b 16 pairs of gloves

09087_Answers_P001_022.indd 22 1/26/17 11:00 AM