Answers.pdf

Answers Working with unfamiliar problems: Part 1 1

x6

+

6x5 y

2 99 3 i x = 15 1 4 , $56 8 5 24 cm

+

15x4 y2

+

20x3 y3

ii x = 450

+

15x2 y4

+

6xy5

+

iii x = 6

6 a i 11.8 seconds 1 1 1 b , , 4 2 4 7 XY = 5.6 cm

2 ii 6.5 seconds

3 4 5 6 7

8 35 9 72

( )n 4 → ∞ as 3 n → ∞. The area approaches a finite value as area change √ ( )n 3 2 3 4 → 0 as n → ∞. x × 4 4 9 77 cm, 181 cm √ 2k(2 3 - 3) 22° 1 y=3 4 x = 0.9, y = 3.3 √ 16 10 cm b The perimeter increases indefinitely as 3x

y6

8 20 students; 6 with 100%, 7 with 75%, 7 with 76%, mean = 82.85%. 1 9 9 units 8 10 a 11

10 Charlie 23 years, Bob 68 years D 11 2 1 12 b = 1 3 13 k = 11 14 V = 27 cm3 , TSA = 54 cm2 15 4 cm < third side < 20 cm. Its length is between the addition and subtraction of the other two sides. 16 785

b 28 c 20 3 d 2 11 640 12 P to R : 145°, 1606 m; R to S : 295°, 789 m; S to Q : 51°, 1542 m; Q to P : 270°, 1400 m

17 n + 1

1 1 13 y = x2 - x + 1 or y = (x - 4)2 - 1 8 8 y

18 10 19 3 : 5

x=4

Working with unfamiliar problems: Part 2 ( )5 4 1 a i P = 3 × 45 × x5 or P = 3x ; 3 3 x P = 3 × 4n × n 3 √ √ ( )2 3 2 3 x ii A = x +3× + 4 4 3 ( ) √ √ ( )2 2 3 x 3 x 2 3×4× +3×4 × 4 32 4 33 √ ( ) 3 x 2 Area change = 3 × 4n-1 × 4 3n

(4, 1)

1 O

1.2

(4, −1)

x

6.8 y = −3

774 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Hours

1st and 6th 2nd and 5th 3rd and 4th

% change from equation

6.7%

% change from ‘rule of thumb’

8.3%

18.3%

25%

16.7%

25%

The percentage change per hour for the ‘rule of thumb’ is 1.6 points higher for the 1st and 6th hours, 1.6 points lower in the 2nd and 5th hours and the same in the middle two hours. Overall, this is quite an accurate ‘rule of thumb’. b The proportion of tide height change ] 1[ = cos(30t1 ) - cos(30t2 ) 2 √ 1+ 5 , 1.618034 15 2

9 a 5x + 23 e 10

b 10a + 26 c 21y + 3 d 15m + 6 f 11t - 1 g 3x2 + 15x h 15z - 7

i -11d3 10 a 3(x - 3)

j 9q4 - 9q3

Answers

14 a

b 4(x - 2)

c 10(y + 2) e x(x + 7)

d 6(y + 5) f 2a(a + 4)

g 5x(x - 1) i xy(1 - y)

h 9y(y - 7) j x2 y(1 - 4y)

k 8a2 (b + 5) m -5t(t + 1)

l ab(7a + 1) n -6mn(1 + 3n)

o -y(y + 8z) 11 a -32 b 7 13 1 f e 2 5

c 61 7 g 5

12 a 2x2 + 6x

d 12 h 1

b x2 - 5x x2

13 a P = 4x - 4, A = - 2x - 4 b P = 4x + 2, A = 3x - 1 c P = 4x + 14, A = 7x + 12

Chapter 1

14 a (-2)(-2) = 4, negative signs cancel b a2 > 0 Â - a2 < 0 c (-2)3 = (-2)(-2)(-2) = -8

Exercise 1A

15 a True

1 C 2 D 3 a 7

b 1

1 5 4 a yes 5 a 9

2 7 b yes b -8

6 a 10a e 4ab

f -

i -3m2 n m 3a + 7b

e 30ht i 8a2 b4 n 3ab r -3x 8 a 5x + 5

d -9

d -9

b 15d f 9t

c 0 g 9b

d 5xy h -st2

j -0.7a2 b n 8jk - 7j

k 2gh + 5 l 12xy - 3y o ab2 + 10a2 b

h -

g True

y x+y or x + 2 2 b It could refer to either of the above, depending on c

17 a b

r 3x3 y4 + 2xy2

b 25ab

c -6ad

d -10hm

f 30bl j 24p3 q a o 3 y s 2

g 12s2 t k -18h5 i5 ab p 4 a t 2

h -21b2 d5 l 63m2 pr

b 2x + 8

c True

16 a

7 3 c no c -8

g

p 2mn - m2 nq 5st - s2 t 7 a 12ab

c -4

1 e 2

b False, 1 - 2 ¢ 2 - 1 1 2 d False, ¢ e True 2 1 f False, 3 - (2 - 1) ¢ (3 - 2) - 1 h False, 8 ÷ (4 ÷ 2) ¢ (8 ÷ 4) ÷ 2

q 2b

c 3x - 15

c mx

interpretation. ‘Half ( of the sum ) of a and b’(or ‘a plus ) b all divided by 2’. π π 2 P= 4+ x + 2, A = 1 + x +x 2 4 ) ( ) ( π 2 π x - 6, A = 3 x - 3x P= 6+ 2 4 ( ) π 2 P = 2πx, A = 1 + x 2

1A

Exercise 1B 1 a 1 2 a

2 3

3 a 5x

5 6 3 b 7a b

c 2 c -

d 7t 4xy

a 4

b 4x

c

h -2y

i -

3 4

b2 c 8x2 a 1 d e 5x 3a 4 3x j k 9st y d -

d -20 - 5b e -2y + 6

f -7a - 7c

g 6m + 18 i -2p + 6q + 4

h 4m - 12n + 20 j 2x2 + 10x k 6a2 - 24a

l -12x2 + 16xy n 36g - 18g2 - 45gh

m 15y2 + 3yz - 24y o -8ab + 14a2 - 20a

4 a x+2

b a-5

c 3x - 9

d 1 - 3y

p 14y2 - 14y3 - 28y r -5t4 - 6t3 - 2t

q -6a3 + 3a2 + 3a s 6m4 - 2m3 + 10m2

e 1 + 6b

f 1 - 3x

g 3-t

i x+2

j 3 - 2x

k a-1

h x-4 1 + 2a l 3

t x4 - x

u 3s4 - 6st

g -9b

1 2p

f -2x l

6b 7

775 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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x-1 2x 4 d 9

b

Answers

5 a

g 2 6 a 3 d

3 4

5 3 7 a x+1

d

f

h 15

i

b 3

c

h b

8 a 3x

b

4 x

e

11 12

13

4 3 2 5 2 4 3a 4 7x 3x 1-x 10x 3 35x2 (2 - x)(x - 1) 3(x + 2) -5 1-x

10 b x-3 3(x + 2) c d 2 x-1 e f 2x a x-1 b 4 e d x+2 Factorise and cancel to 1. a 1 - x = -(x - 1) -12 7 -7 ii iii b i 3 x 2 x+2 a b x 2 3 d e 4(a + 1) 2(x + 2) 1 x-y f g (a + 1)(a - 3) xy

9 a

10

c -4

e 5

e

g -

x+4 5x

f i c c f

6 a

5a 2 1 2 18 5 1 25 1 3 4 x+3 5 2b2 b-1

c 2(x - 3)

c

(y + 2) x

d 2x

c e g

3 - 15b 10 2a f 15 c

3a + 8 4a 16 - 3b d 4b 27 - 14y f 18y -27 - 2x h 6x

17 6

6x + 5 12 3x - 23 d 14 18x - 9 6x - 3 f = 6 2 -14x - 7 h 15

x+5 6 -2x + 38 15 14x - 8 21 x + 14 30 -3x + 10 4 7x + 22 (x + 1)(x + 4) 3x - 1 (x - 3)(x + 5) -21 (2x - 1)(x - 4) 41 - 7x (2x - 1)(x + 7) 14 - 17x (3x - 2)(1 - x)

b

7x - 13 (x - 7)(x + 2) x - 18 d (x + 3)(x - 4) 14x - 26 f (x - 5)(3x - 4) 3x + 17 h (x - 3)(3x + 4) b

ii x2

2a - 3 a2

a2 + a - 4 a2

3x + 14 4x2 x-2 10 The 2 in the second numerator needs to be subtracted, . 6 11 a -(3 - 2x) = -3 + 2x (-1 × -2x = 2x) 2 2x x+3 b i ii iii x-1 3-x 7-x b i

h

7x + 11 12 4x + 9 d 9 8x + 3 f 10 5x - 1 h 5 b

9 a i a2

c 14

b

c

i

1 3

b 6

5 a

e

g

3 a 12

g

c

e

d

d

7 a

f 4(x - 1)

c -6x + 12 5 c 14

4a + 3 8 1 - 6a e 9 11b h 14

i

8 a

b -x - 6 17 b 15

b

g

i

1 a 2x - 4 5 2 a 6 3a + 14 21 4x + 6 15 3x 20 2a + 15 3a 7a - 27 9a 4b - 21 14b -12 - 2x 3x

e

g

Exercise 1C

4 a

c

9x + 23 20 3x + 1 4 8x - 1 6 7x + 2 24 x+1 14

ii

5a + 2 a2 3x - x2 d (x - 2)2 yz - xz - xy f xyz

12 a -1

b

3x + 5 (x + 1)2 21x - 9x2 e 14(x - 3)2 c

13 a 2

iii

b 1

Exercise 1D 1 a no

b no

c yes

2 a true 3 a false

b false b true

c false c true

4 a 5 5 e 2

f

b 8 11 4

g -

c -3 1 3

d yes

d 4 h -

11 6

776 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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k -

9 2

4 3 11 p 9

m -2

n 7

o -2

b 9

23 c 2

5 d 6

5 a 1

g 1

h 2

i 4 m 19

2 3 j 7 n 23

k -9 o 1

l 5

6 a 10 e -5

b 13 f 6

c -22 g 16

d 4 h 4

i -9 m 20

j 8 n 15

k 6 o -9

l -7 p 5

e -

9 11

f

7 a x + 3 = 7, x = 4 c x - 4 = 5, x = 9 e 2x + 5 = 13, x = 4 g 3x + 8 = 23, x = 5 8 a 1 27 23 94 i 11 e

9 a 1

b 0 f

28 5

Exercise 1E

l -

j

1 a 3, 6, 10 (Answers may vary.) b -4, -3, -2 (Answers may vary.) c 5, 6, 7 (Answers may vary.) d -8.5, -8.4, -8.3 (Answers may vary.) 2 a B

13 14

c 2

d x > -9

b xÅ5

x

5

c xÅ4

7 d 2 2 h 5

x

4

d x Ä 10

x

4

x

10

e xÄ2 b 6

c xÄ4

5 a x<4

d 25 f x>3

11 17 and 18 12 24 km 13 a $214 b $582

x

3

g x>6 ii 10.5 iii 21 b 90 s = 1 min 30 s

x

2

10 17 cm

c i 1 14 a 41 L

c A

e -2 < x Ä 1 f 8 < x Ä 11 g -9 < x < -7 h 1.5 Ä x Ä 2.5 i -1 Ä x < 1

5 f 2(x - 5) = -15, x = 2 h 2x - 5 = x - 3, x = 2

g

b C

3 11, 12 or 13 rabbits 4 a xÅ1 b x<7

b x + 8 = 5, x = -3 d 15 - x = 22, x = -7

c -17

h xÄ6

b 4 g 1

c -15 h -26

d 20

i x < -18

i -10 16 x = 9. Method 2 is better, expanding the brackets is unnecessary, given 2 is a factor of 8. a 5 17 a 5 - a b c 6 a 2a + 1 3a + 1 c-b d e f a a a c b 18 a a = b a= b+1 b+1 1 b c a= d a= c-b b-1 bc e a = -b f a= b-c ab abc 19 a 6a b c a+b b-a

x

6

e 3

1E x

6

c 250 s = 4 min 10 s 15 a 6 f 6

Answers

3 2

i -4

x

−18 j x > 32

x

32 k xÄ

10 9

x

10 9

l x<-

3 8

−0.375 2 5 e x < -8

6 a xÅ-

x

b x<2

c x Ä -5

d x Ä -7

f xÅ4

g x Å -10

h x < -21

777 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

7 a x>6

b xÄ2

c x<

5 2

d x Å 10

11 4 x 5 b 4 - Å -2, x Ä 12 8 a 2x + 7 < 12, x < 2 2 1 c 3(x + 1) Å 2, x Å 3 d x + (x + 2) Ä 24, x Ä 10 since x must be even e xÄ

1 16

f x<

e (x - 6) + (x - 4) + (x - 2) + x Ä 148, x Ä 40 9 a i C < $1.30 ii C > $2.30 b i less than 9 min 10 a x < -5 e xÅ

27 29

11 4 1 f x< 2

b xÅ

ii 16 min or more 11 14 d xÄ 29 5

c xÅ

11 a An infinite number of whole numbers (all the ones greater than 8). b 1, 3 is the only whole number. a+3 12 a x Å b x < 2 - 4a 10 7 a-7 c x < 1 - or x < if a > 0. a a Reverse the inequality if a < 0. 13 a -4 Ä x < 5 b -9.5 < x Ä -7

Exercise 1F 1 a y = -2x + 5, m = -2, c = 5 b y = 2x - 3, m = 2, c = -3 c y = x - 7, m = 1, c = -7 2 3 2x 3 d y = - - ,m = - ,c = 5 5 5 5 2 a i 3

ii 6

iii

21 2

b i 2

ii 6

iii

8 3

3 a i d iii

b iv e v

c ii f vi

4 a yes d no

b yes e yes

c no f no

5 a m = 5, c = -3

y y = 5x −3 (1, 2) x

O −3

c x = 10

14 a 3 Ä x Ä 9

3 b -7 Ä x Ä 3

9

x

b m = 2, c = 3

y

−7 c -9 Ä x < -7

3

−9

−7

x (1, 5)

y = 2x + 3 3

x

x

O

3 d - ÄxÄ2 2

−1.5 7 e -3 Ä x Ä 3

2

7 3

−4 g 11 Ä x Ä 12

−2

11 15 h 1Äx< 4

12

1 b x<

c m = -2, c = -1

y

−3 f -4 Ä x Ä -2

15 a x Å 23

x

19 5

15 4

x y = −2x −1 x

−1 O

x (1, −3)

x

x

c xÄ1

778 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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y

y y = −x + 2

2

Answers

7 h m = - ,c = 6 2

d m = -1, c = 2

(1, 1)

x

O

6

y = − 72 x + 6

O

(2, −1)

x

i m = 0.5, c = -0.5

e m = 1, c = -4

y

y

y = 0.5x − 0.5 y=x−4 O

x

O

(2, 0.5) −0.5

x

(1, −3)

−4

j m = -1, c = 1

3 f m = - ,c = 1 2

y y

y=1−x y=

1 O

− 32

x+1 1 O

1

x

1F

x (2,−2) 2 k m = ,c = 3 3

4 g m = , c = -2 3

y y= y

O −2

+3

(3 , 5)

3 y = 43 x −2 (3, 2) x

2 3x

x

O

l m = -0.2, c = 0.4

y y = 0.4 − 0.2x (0.4) O

(1, 0.2)

x

779 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Answers

6 a m = -3, c = 12

f m = -1, -c = -

y

y

y = −x −

12 (1, 9)

y = −3x + 12

x

O

b m = -5, c =

5 2

1 3

− 13

O

1 3

x

(1, − ) 4 3

g m = -4, c = -8

y

y

2.5

x

O x

O

(1, −2.5)

y = −4x − 8 −8

5 2

y = −5x +

(1, −12) c m = 1, c = -7

y

1 1 h m = - ,c = 2 4

y x

O

0.25

(1, −6)

−7 d m = 1, c = -2

y = x −7 O

7 a x = 2, y = -6

y=x−2

(1, −1) −2

y x x

−6

b x = -2, y = 4

y y=

4 3x

y

−3

(3, 1) O −3

2

O y = 3x − 6

4 e m = , c = -3 3

x

y = − 12 x + 14

y

O

(2, −0.75)

y = 2x + 4 4

x −2

O

x

780 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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y

y

y = − 32 x + 6

y = 4x + 10

6

10 −2.5

Answers

g x = 4, y = 6

c x = -2.5, y = 10

x

O

4

O

x

h x = 5, y = 2

y 4 d x = , y = -4 3

y = − 25 x + 2 y 2

y = 3x − 4 O O

x

5

x

4 3

i x = -8, y = 6

y

−4

y = 34 x + 6

6 e x = 3.5, y = 7

y

x

O

−8

y = −2x + 7

1F

7 O

x

3.5

j x = 5, y = 2.5

y y = − 12 x +

f x = 8, y = 4

5 2

2.5

y O y = − 12 x + 4 4

O

x

5

7 7 k x = ,y = 3 4

8

x

y y = − 34 x + 74 1.75 O

7 3

x

781 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Answers

l x = -6, y =

12 5

y

e

y y=0 y=

2.4

2 5x

+

x

O

−6

y

f

x=0

y

8 a

x

O

12 5

x

O

y = −4 x

O

y

g

(1, 4)

−4

y = 4x

y

b

x

O

y=1 1 x

O

y

h

y = −3x y

c

x=2

O

2

x

(1, −3)

y

i

y

d

y = −13x

x = − 52 −2.5

x

O O

x

O

(3, −1)

x

y

j

(2, 5) y = −52 x O

x

782 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

3 7 15 a gradient = , y-intercept = a a

y

k

b gradient = a, y-intercept = -b

O

(1, −1)

a 3 c gradient = - , y-intercept = b b d d a 16 a b c a b b

x

17 a 12 sq. units

y

l

d

4

y 35 30 25 20 15 10 5

(10, 30) C = 2n + 10

1 2 3 4 5 6 7 8 9 10 c i $28 ii 23.5 kg 10 a V = 90 - 1.5t b y

c 11 a 12 a 13 a c 14 a b c d

x

2 a 2

b 3

e -3 1 3 a 4

f Undefined

c y = -x - 7

c 0 5 2

b 2

c

e 0

f 0

g -1

i -1

j Undefined

3 2 b y=x-2 k

c y = 3x + 6 e y=4

d y = -3x + 4 f y = -7x - 10

5 a y = 2x + 4 c y=x-4

b y = 4x - 5 d y = -2x + 12

e y = -3x - 4 6 a y = 3x + 5 1 3 c y= x2 2 7 a A = 500t + 15 000

f y = -3x - 2 b y = -2x + 4

d -4 d 3 h

5 2

l -

3 2

1G

d y = -2x - 2

b $15 000 c 4 years more, i.e 10 years from investment

V = 90 − 1.5t

x 5 10 15 20 25 30 35 40 45 50 55 60 i 82.5 L ii 60 hours $7 per hour b P = 7t $0.05/km b C = 0.05k c C = 1200 + 0.05k m = 25 b The cyclist started 30 km from home. (0, 30) 1 y = x + , gradient = 1 2 y = 0.5x + 1.5, y-intercept = 1.5 y = -3x + 7, gradient = -3 1 1 y = x - 2, gradient = 2 2 0

b y = -x + 3

4 a y=x+3

0

90 80 70 60 50 40 30 20 10

121 sq. units 4

121 32 sq. units e sq. units 5 3

1 a y = 4x - 10 1 11 d y= x+ 2 2

9 a C = 2n + 10 b

c

Exercise 1G

x

O

b 9 sq. units

d $21 250 8 a y

90 80 70 60 50 40 30 20 10

C = 10t + 20

x 0 1 2 3 4 5 6 7 8 9 b C = 10t + 20 c i $10 per hour ii $20 up-front fee

783 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Answers

9 a i V = 4t iii V = t + 1

ii V = 3t iv V = 1.5t + 2

11 a = -4, 0

y

b 1 L, 2 L c Initially the flask contains b litres and it is losing 1 litre 10 a c 11 a b c d

per minute. 5 -5 = -1 b m= = -1 m= 5 -5 It doesn’t matter in which pair of points is (x1 , y1 ) and which is (x2 , y2 ). 4 3 4x 13 y=- + 3 3 4x 13 y=- + 3 3 The results from parts b and c are the same (when simplified). So it doesn’t matter which point on the line is

used in the formula y - y1 = m(x - x1 ). 1 2 12 a i = 0.02 ii = 0.04 50 50 b i y = 0.02x + 1.5 ii y = 0.04x + 1.5 c The archer needs m to be between 0.02 and 0.04 to hit the target.

Exercise 1H 1 a 4

b 5

2 a 4

b 4

√ 41 d √ √ c 32 = 4 2

c

d (0, -3) √ √ 3 a d = 20 = 2 5, M = (2, 5) √ b d = 97, M = (2, 3.5) √ c d = 41, M = (-1, 1.5) √ d d = 37, M = (-1, -1.5) √ √ √ 4 a 29 b 58 c 37 √ √ e 37 f 15 g 101 √ i 37 5 a (1, 6.5) e (1, -1.5)

b (1.5, 2.5) f (-3.5, 3)

( 3,

) 9 2

c (-0.5, 1) d (-1, 4.5) g (-3, -0.5) h (2, 2.5)

8 a 3, 7 9 a 1478 m

b -1, 3 b 739 m

d -6, 0

10 a (-0.5, 1) b (-0.5, 1) c These are the same. The order of the points doesn’t matter since addition is commutative (x1 + x2 …) (x1 + x2 = x2 + x1 ). d 5

x

(−2, −1)

( 12 a d 13 a b c

) ( ) 1 1 4 ,2 b - , (2 ) ( 3 3) 16 3 2, e - ,1 5 4 √ 2 2 (x - 7) + y √ (x - 7)2 + (x + 3)2 i 721 m ii 707 m iii 721 m

(

) 4 8 , (3 3) 8 f 0, 5 c

iv 762 m

d x=2 e The distance will be a minimum when the dotted line joining Sarah to the fence is perpendicular to the fence (when it has gradient -1). The closest point is (2, 5).

Progress quiz 1 a 9a2 b + 2ab + 8b

3 a

b -12x3 y 7-x c 2

b a-4

9a 2

f

5 2

6+m 8

b

4x - 15 6x

4 a x=5

b k=-

3 2

c

14 - 3a 24

c m = 30

c 13m + 14 m+3 d 3m

d

3m - 13 (m - 1)(m - 3)

d a=-

9 2

5 a a>3

0 1 2 3 4 5 b x Å -4

b a = -4, b = 5 d a = 11, b = 2 c -1, 9

d = √20 O

e

i (-7, 10.5) 6 B and C are both 5 units away from (2, 3). 7 a a = 3, b = 5 c a = -2, b = 2

(−4, 3)

2 a 4k

√ d 65 √ h 193

y=3

3

e 5

f The order of the points doesn’t matter (x - y)2 = (y - x)2 , as (-3)2 = (3)2 .

−5 −4 −3 −2 −1 0 c m<4 0 1 2 3 4 5 d a Å -2 −4 −3 −2 −1 0 1 6 a (-3, 2) is not on the line. b (-3, 2) is on the line.

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4 a parallel d neither g parallel j perpendicular

4 3 2 1 x

1 2 3 4 (2, −2)

O −4 −3 −2 −1−1 −2 −3 −4

O −4 −3 −2 −1−1 −2 d x = −2 −3 −4

d y = 8.4 f y = -3 4 h x=11 5 54 b y=- x+ 7 7

2 7 a y= x+5 3 2 16 c y= x+ 3 3 3 8 a y=- x+5 2 3 c y=- x+1 2

x

d y = 7x + 20

7 28 b y= x+ 5 5 1 10 d y=- x7 7 2 5 9 The second line has equation y = - x- . It cuts the 3 3 5 x-axis at x = - . 2 9 10 a 14 b -2 c 5 d 7

y 4 3 2 1

b x=0

c y = 11 e y=3 2 g x= 3

1 2 3 4

c, d, e

cy=3

1 2 3 4

x

11 a m

3x 4

12 a y = 2x + b - 2a

ey=− (4, −3)

b -

a b

c -

1 m

d

1I

b a

b y = mx + b - ma 1 a d y=- x+b+ m m

c y=x+b-a 13 a i 1

8 a y = 2x + 3 3 b y=- x+8 2 c x=5

ii -1

iii 1

iv -1

b AB is parallel to CD, BC is parallel to DA, AB and CD are perpendicular to BC and DA; i.e. opposite sides are parallel and adjacent sides are perpendicular.

√ √ 9 midpoint (4, -1); length = 80 (or 4 5)

c rectangle. 4 14 a i 3

10 a = 11 or a = -1

Exercise 1I

3 a 5

b y = -x - 6 2 d y= x-6 3 1 f y=- x+6 2 3 h y=- x+5 2 7 j y = x + 31 2

6 a x=6

4 3 2 1

b -7 b

i perpendicular

4 e y=- x+7 5 1 g y= x-2 4 3 i y=- x-5 4

y

1 3

h parallel

c y = -4x - 1

b x-intercept = -3, y-intercept = -2

2 a -

c neither f perpendicular

5 a y=x+4

O −4 −3 −2 −1−1 −2 −3 −4

1 a 4

b parallel e perpendicular

Answers

3 7 a gradient = - , y-intercept = 1 2 y

1 2

b 4

3 4 8 c 7

8 7 9 d 4

c -

d

ii -

3 4

iii 0

b Right-angled triangle ( AB is perpendicular to BC ). c 20 1 15 y = - x + 4, x-intercept = 8 2

c y = 5x + 4

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Answers

Exercise 1J

y

e (1, 4)

1 a yes

b yes

c no

d no g yes ) ( 7 ,3 2 a 2

e no h no

f no

(1, 4)

y=x+3

y

y = − 23 x + 14 3

y = 2x − 4 y

f (1, -4)

(3.5, 3)

y=3

b

1 , -2 2

y = 3x − 7

x

O (

x

O

y = 2x − 6

)

O

x

(1, −4)

y y = 2x − 3

g No intersection (lines are parallel).

x

O

y

y = −2

y = 3x − 3 y = 3x + 9

y

c (2, 4)

O

x

x=2 y=4 (2, 4)

h No intersection (lines are parallel).

x

O

y y=x

y=x+4

x

O

y

d (-1, 0)

x = −1 y=0 (−1, 0)

O

x

( i

) 1 17 - , 5 5

y

(−0.2, 3.4) x

O y = 3x + 4

y = −2x + 3

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k x = 2, y = 1 5 a x = 1, y = 1

iii Joe’s: C = 0.2k + 60, Paul’s: C = 0.1k + 150 iv 900 km

c x = 2, y = 1 1 e x = ,y = 1 2

b Joe’s Car Rental c Paul’s Motor Mart 4 a x = 2, y = 7 c x = 3, y = 1

b x = 2, y = 5 d x = 2, y = 1

e x = 1, y = 1 g x = 5, y = 1

f x = 1, y = 1 h x = 10, y = 4

i x = 1, y = 2 5 a x = 2, y = 10

j x = 9, y = 2 b x = 1, y = -5

c x = -3, y = 3 e x = 3, y = 1

d x = 13, y = -2 f x = 2, y = 1

f yes

ii V = 40 000 - 3000t

ii $7000

9 197 600 m2 10 a no b no g yes 3 b ) 2

11 a -4 ( k 2k 12 a , 3 3 ( ) k k b ,2 2 c (-1 ( - k, -2 - k) ) -2k - 1 -2k - 4 d , 3 3 b2 b 13 a x = ,y = a-b a-b a -a c x= ,y = 1+b 1+b b2 b d x= ,y = b-a b-a c(1 - b) 2c f x= ,y = a(b + 1) b+1

c yes

d yes

h no

d x = 4, y = -3 1 1 f x = - ,y = 2 2

6 a x = 4, y = -3

b x = 1, y = 1

c x = 3, y = 4 1 e x = , y = -1 2 7 799 and 834 8 $0.60

d x = 2, y = 2 f x = -3, y =

1 3

-( 2y - (-2y) solution ) is (1, -1). ) = 0. ( The correct ) ( 1 13 1 2 2 11 a , -1 b , c - , (a ) 3 3b ( a b ) a+b a-b c c d , e , 2a 2b a+b a+b

b t = 9, E = 180 c i 9 hours ii $180

c i 11 years 8 18 years

l x = -1, y = 2 b x = 4, y = 2

9 A = $15, C = $11 10 Should have been (1) - (2), to eliminate y :

g x = 1, y = 4 h x = 1, y = 3 6 a i E = 20t ii E = 15t + 45

7 a i V = 62 000 - 5000t b t = 11, V = 7000

h x = 2, y = 2 j x = 2, y = 1

Answers

g x = 2, y = -1 i x = 1, y = 2

3 a i Joe’s: $60, Paul’s: $150 ii Joe’s: $0.20 per km, Paul’s: $0.10 per km

e no

12 The two lines are parallel, they have the same gradient. 2 2 2 1 13 a b x-1 x+1 2x - 3 x + 2 3 2 3 2 c d + 3x + 1 2x - 1 3x - 1 x + 2 1 1 1 3 e + f x+3 x-4 7(2x - 1) 7(4 - x)

c 12

Exercise 1L 1 a x + y = 16, x - y = 2; 7 and 9 b x + y = 30, x - y = 10; 10 and 20

1K

c x + y = 7, 2x + y = 12; 5 and 2 d 2x + 3y = 11, 4x - 3y = 13; 4 and 1

b x=

-b a ,y = a+b a+b

2 7 cm × 21 cm 3 Nikki is 16, Travis is 8. 4 Cam is 33, Lara is 30. 5 Bolts cost $0.10, washers cost $0.30. 6 There were 2500 adults and 2500 children.

1 (a - b) e x= ,y = a - 2b a - 2b ab a2 b g x= 2 ,y = 2 a +b a +b

14 Answers will vary.

7 Thickshakes cost $5, juices cost $3. 8 There are 36 ducks and 6 sheep. 9 43 10 $6.15 (mangoes cost $1.10, apples cost $0.65) 11 70 12 1 hour and 40 minutes 1 13 of an hour 7

Exercise 1K 1 a 0

b 0

c 0

d 0

2 a subtract e add

b subtract f add

c add g subtract

d add h subtract

3 a 4x - 6y = 8 c 8x - 12y = 16

b 6x - 9y = 12 d 20x - 30y = 40

4 a x = 2, y = 5 c x = 4, y = 2

b x = 2, y = 3 d x = 2, y = 2

e x = 1, y = 1

f x = 2, y = 1

14 200 m 15

4 L 17

16

210 L 19

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Answers

Exercise 1M

d y Ä 3x - 5

1 a no

b yes

c yes

d no

f no 2 a B

g no b C

h yes c A

i yes

y

e no

y ≤ 3x − 5

3 x Å -1, y Ä 4

y

O x≥1

y≤4 (−1, 0)

5 3

x

−5

(0, 4) x

O

e y < -4x + 2

y y < − 4x + 2

4 a yÅx+4

2

y

O 4 −4

x

1 2

y≥x+4 x

O

f y Ä 2x + 7

y 7

b y < 3x - 6

y

y ≤ 2x + 7 −72

y < 3x − 6 O

x

O

x

2

−6

g y < 4x

y

c y > 2x - 8

y y > 2x − 8

O

4

4 x

O

1

x

y < 4x −8

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y

Answers

l yÅ2

h y > -3x + 6

y

6 y≥2 2 O

x

2

x

O

y > 6 − 3x

y

5 a

i y Ä -x

y 3 O

x

9

x

O y ≤ −x

y

b j x>3

y x

O

x>3

1M

−3 O

3

x y

c

4

k x < -2

y

O

2

x

x < −2

−2 O

x

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y

Answers

d

y

h

9 O

x

x

O

−9

−4 −6 y

e

6 a yes

(0, 5) −2.5 O

x

b no

c no

7 a no b yes c no 8 a yÄx+3 b y Å -2x + 2 2 3 d y> x-2 c y<- x-3 2 5 9 a y

d yes d no

4 2 O

y

f

4

x

1.5 O

−3

x

b

y

(0, 3)

−5

O

4

O 1

y

g

x

c

x

y

−2

O

2

3

x

(0, −2)

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y

10 a

Answers

y

d

y≥0 (0, 1) O 1

(0, 0) O (2, 0)

x

5

y

b

y

y ≥ 2x − 4 (0, 0) O

4 2 O

3

4

x

y

y≥

1 4x

(0, 15)

+4

(4, 5)

(0, 4)

10 y≤

(5, 5)

− 54

O

x

10

x + 15

x≥0

O

x

y

d

2.5

1M

y > − 25 x + 2

(−2.5, 3)

(2, 3) (2, 1.2)

x<2 O

y

g

x≥0

y≤0 c

y

−5

(2, 0)

(0, −4) x

(6, −2)

f

x

y ≤ − 12 x + 1

(0, −3)

e

x≥0

y<3

x

9 y

e

(4, 3) 2 O

−8

x≤0 y<x+7

(−5.4, 1.6)

y

h

7

x

6

−7

−3 −2O

x 2x + 3y ≥ −6

6 2.5 −5

O

(1, 3) 2

x

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Answers

Short-answer questions

y

f

b 12a2 b

1 a 5xy + 6x

y ≥ 12 x + 3

e -2m2 + 12m

(−5.5, 14.5)

O

x y ≥ −3x − 2

6 - 7a 14

b

4 a x = -3

b

5 a x<1

b

6 a x>5

b x Å 10

b 4

c 22

d

c c c c c

7 a V = 2 - 0.4t

11 a y Å 0, y < 2x + 4, y Ä -x + 7 1 b y > - x + 6, y Ä x + 3, x < 8 2 12 a 1

b

3 a

(−1.43, 2.29)

f 2 x+2 5a + 18 6a 3 x=4 x Å -4

2 a 3x - 1

(4, 5)

y ≤ −x + 9

c

b

3 x d 3b + 21 2 x+2 3 4 7x + 26 11 - x d 30 (x + 1)(x - 3) 1 x= d x=2 5 -1 < x Ä 3 2 x > -3 d xÄ 7 1.4 L c 5 minutes

d Ä 3.5 minutes 8 a y

81 20

(3, 0)

x

O

115 578 13 a i ii 6 15 b Answers may vary; e.g. x > 0, x < 3, y > 0, y < 2

y = 3x − 9

Problems and challenges

(0, −9)

1 0.75 km 6 2 8 3 a The gradient from (2, 12) to (-2, 0) = the gradient from (-2, 0) to (-5, -9) = -3.

b

b The gradient from (a, 2b) to (2a, b) = the gradient from b (2a, b) to (-a, 4b) = - . a 3 5 4 The gradient of AC is and the gradient of AB is - . So 5 3 △ABC is a right-angled triangle, as AC is perpendicular to AB. Can also show that side lengths satisfy Pythagoras’ theorem.

y

(0, 5) (2.5, 0) O

c

4840 9680 km/h and km/h. 9 9 6 The distance between the two points and (2, 5) is 5 units.

x

y = 5 − 2x

y

5 The missiles are travelling at

diagonals. It is not a square since the angles at the corners ( are not 90°. ) In particular, AB is not perpendicular to -1 BC mAB ¢ . mBC 8 x = 2, y = -3, z = -1

y=3

3

7 The diagonals have equations x = 0 and y = 3. These lines are perpendicular and intersect at the midpoint (0, 3) of the

x

O

d

y

units2

9 24 10 24, 15 years

Multiple-choice questions 1 E

2D

3B

4C

5D

6B 11 E

7 C 12B

8A 13A

9C 14A

10D 15D

O

x=5

x

16C

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y

17 a

y = 2x O

(1, 2)

(0, 0)

x

O

−4

y

b

y

f

x

4 3

Answers

y

e

y = −5x 8 3

(0, 0) O

x

x

O

−4

(1, −5) 18 The point of intersection is (4, 0)

y

g

y

4

y = 4 − –x 2

O

8

2

x

O 6

y

h

3

x

4

3x y = 3 − –– 8

O

8

Extended-response questions x

1 a i h = 4t + 25 b 16 cm

ii h = 6t + 16

c Shrub B because its gradient is greater. d y 1 5 3 15 9 a y= x+3 b y= x+5 c y=- x+ 2 2 2 2 d y = 2x - 3 3 3 34 10 a m = b y=- x+ 5 5 √ √5 11 a m = (4, = 52 = 2 13 ( 8), d ) √ 11 b M= , 1 , d = 61 (2 ) √ √ 1 5 c M= , - , d = 18 = 3 2 2 2 1 12 a y = 3x - 2 b y = -1 c y=- x+5 2 d y = 3x - 1 13 a a = 7 14 a (-3, -1)

b b = -8 c c = 0 or 4 b (-8, -21)

90 80 70 60 50 40 30 20 10

(12, 88) (12, 73)

x 2 4 6 8 10 12 e after 4.5 months f i 1.24 m ii 26.25 months iii between 8.75 and 11.25 months 0

15 a (-3, -1) b (0, 2) 16 A regular popcorn costs $4 and a small drink costs $2.50.

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j 121 (vertically opposite to cointerior angle in ∥ lines) k 71 (isosceles, cointerior angles in ∥ lines)

Answers

2 a A(0, 0), B(8, 6), C(20, 0)

y

l 60 (isosceles, cointerior angles in ∥ lines) 6 a 50 (angle sum in a quadrilateral)

B (8, 6)

b 95 (angle sum in a quadrilateral) c 125 (angle sum in a pentagon)

D (14, 3) A (0, 0) O

C (20, 0)

d 30 (angle sum in a pentagon) e 45 (angle sum in a hexagon)

x

f 15 (angle sum in a quadrilateral) 7 a 108° b 135° c 144° 8 a 95 (alternate + cointerior) c 85 (alternate + cointerior)

b 43.4 km c The drink station is at (14, 3). 3 d i y= x 4

1 ii y = - x + 10 2

iii y = 0

b 113 (2 × alternate) d 106 (cointerior)

e 147, (cointerior, angles in a revolution) f 292, (angles in a revolution, alternate + cointerior)

3 1 e y Å 0, y Ä x, y Ä - x + 10 4 2

9 a 176.4° 10 a 12

4 80 f y=- x+ 3 3

Exercise 2A

11 x = 36, y = 144 12 115, equilateral and isocles triangle 60 + 55 S + 360 13 a Expand the brackets. b n= 180 360 S 180(n - 2) c I= = d E = 180 - I = n n n 14 a ÒBCA = 180° - a° - b° (angles in a triangle) b c° = 180° - ÒBCA = a° + b° (angles at a line)

1 triangle, quadrilateral, pentagon, hexagon, heptagon, octagon, nonagon, decagon

15 a alternate angles (BA ∥ CD) b ÒABC + ÒBCD = 180° (cointerior), so

Chapter 2

2 a false e false

b true f false

c true g true

d true h false

i true 3 a a = 110 (angles on a line), b = 70 (vertically opposite) b a = 140 (angles in a revolution) c a = 19 (complementary) d a = 113 (cointerior angles in ∥ lines), b = 67 (alternate angles in ∥ lines), c = 67 (vertically opposite to b) e a = 81 (isosceles triangle), b = 18 (angles in a triangle) f a = 17 (angles in a triangle), b = 102 (angles at a line) g a = 106 (cointerior angles in ∥ lines), b = 74 (opposite angles in a parallelogram) h a = 17 (angles in a triangle), b = 17 (complementary) i a = 90 (vertically opposite), b = 60 (angles in a triangle)

b 3.6° b 20

a + b + c = 180. c Angle sum of a triangle is 180°. 16 ÒACB = ÒDCE (vertically opposite), so ÒCAB = ÒCBA = ÒCDE = ÒCED (isosceles) since ÒCAB = ÒCED (alternate) AB ∥ DE. 17 Answers may vary. 18 a 15 (alternate angles in parallel lines) b 315 (angle sum in an octagon) 19 Let M be the midpoint of AC. Then ÒAMB = 60° (△ABM is equilateral). ÒBMC = 120° (supplementary). Therefore, ÒMBC = 30° (△MBC is isosceles). So ÒABC = ÒABM + ÒMBC = 60° + 30° = 90°. 20 Let ÒAOB = x and ÒCOD = y. 2x + 2y = 180° (angles at a line). So ÒBOD = x + y = 90°.

4 a 72 b 60

Exercise 2B

c 56 5 a 60 (equilateral triangle)

1 a SAS

b SSS

e RHS

f RHS

b 60 (exterior angle theorem) c 110 (isosceles, angles in a triangle)

2 a 5 f 2

d 80 (angles in a triangle) e 10 (exterior angle theorem)

3 a AB = DE (given) S

f 20 (isosceles, angles in a triangle) g 109 (angles on a line) h 28 (diagonals meet at a right angle in a rhombus) i 23 (angles in a triangle)

c 48

b 4

c AAS

d SAS

c 3

d 5

e 2

ÒABC = ÒDEF (given) A BC = EF (given) S Â △ABC Ã △DEF (SAS)

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b ÒFED = ÒCBA = 90° (given) R FD = CA (given) H

ÒBCA = ÒDCE (vertically opposite) A

FE = CB (given) S

AC = EC (given) S

 △FED à △CBA (RHS)

 △ABC à △EDC (SAS) b AB = DE (corresponding sides in congruent triangles) c ÒABC = ÒCDE (corresponding sides in congruent

c AC = DF (given) S BC = EF (given) S

triangles). ÒABC and ÒCDE are alternate angles. Â AB ∥ DE.

AB = DE (given) S Â △ACB Ã △DFE (SSS) d ÒEDF = ÒBAC (given) A

d 5 cm 8 a AB = CD (given) S

ÒDFE = ÒACB (given) A

AD = CB (given) S

EF = BC (given) S

BD is common; S

 △EDF à △BAC (AAS) 4 a x = 7.3, y = 5.2 b x = 12, y = 11 c a = 2.6, b = 2.4 5 a AD = CB (given) S

d x = 16, y = 9

 △ABD à △CDB (SSS) b ÒDBC = ÒBDA (corresponding angles in congruent triangles) c ÒDBC and ÒBDA are alternate angles (and equal).  AD ∥ BC.

DC = BA (given) S

9 a CB = CD (given) S

AC is common; S

ÒBCA = ÒDCE (vertically opposite) A

 △ADC à △CBA (AAS)

CA = CE (given) S

b ÒADB = ÒCBD (given) A

Â△BCA Ã △DCE (SAS)

ÒABD = ÒCDB (given) A

ÒBAC = ÒDEC (corresponding angles in congruent

BD is common; S

triangles)

 △ADB à △CBD (AAS) c ÒBAC = ÒDEC (alternate, AB ∥ DE) A ÒCBA = ÒCDE (alternate, AB ∥ DE) A

 Alternate angles are equal, so AB || DE. b ÒOBC = ÒOBA = 90° (given) R OA = OC (radii) H

BC = DC (given) S

2B

OB is common; S

 △BAC à △DEC (AAS)

Â△OAB Ã △OCB (RHS)

d DA = DC (given) S ÒADB = ÒCDB (given) A

AB = BC (corresponding sides in congruent triangles)

DB is common; S

ÂOB bisects AC. c AB = CD (given) S

 △ADB à △CDB (SAS)

AC is common; S

e OA = OC (radii) S OB = OD (radii) S

AD = CB (given) S

AB = CD (given) S

Â△ACD Ã △CAB (SSS) ÒDAC = ÒBCA (corresponding angles in congruent

 △OAB à △OCD (SSS) f

Answers

7 a BC = DC (given) S

triangles)

ÒADC = ÒABC = 90° (given) R

ÂAlternate angles are equal, so AD || BC.

AC is common; H

d AB = AE (given) S

DC = BC (given) S

ÒABC = ÒAED (△ABE is isosceles) A

 △ADC à △ABC(RHS)

ED = BC (given) S

6 a OA = OC (radii) S ÒAOB = ÒCOB (given) A

Â△ABC Ã △AED (SAS)

OB is common; S

AD = AC (corresponding sides in congruent triangles)

 △AOB à △COB (SAS) b AB = BC (corresponding sides in congruent triangles) c 10 mm

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4 a ÒBAC = ÒDCA (alternate angles)

Answers

e OD = OC (given) S ÒAOD = ÒBOC (vertically opposite) A

ÒBCA = ÒDAC (alternate angles)

OA = OB (given) S

AC is common.

Â△AOD Ã △BOC (SAS) ÒOAD = ÒOBC (corresponding angles in congruent triangles) f

 △ABC à △CDA (AAS) b As △ABC à △CDA, AD = CB, AB = CD (corresponding sides). 5 a ÒABE = ÒCDE (alternate angles)

AD = AB (given) S

ÒBAE = ÒDCE (alternate angles)

ÒDAC = ÒBAC (given) A

AB = CD (opposite sides of parallelogram)

AC is common; S

 △ABE à △CDE (AAS) b AE = CE (corresponding sides), BE = DE

Â△ADC Ã △ABC (SAS) ÒACD = ÒACB (corresponding angles in congruent triangles)

(corresponding sides). 6 a AB = CB (given)

ÒACD = ÒACB are supplementary.

AD = CD (given)

ÂÒACD = ÒACB = 90°

BD is common. Â △ABD Ã △CDB (SSS) b ÒABD = ÒADB = ÒCBD = ÒCDB (equal angles in

ÂAC ⊥ BD 10 a OA = OB (radii) S

congruent isosceles triangles). Therefore, BD bisects ÒABC and ÒCDA.

OM is common; S

7 a AE = CE (given)

AM = BM (M is midpoint) S Â△OAM Ã △OBM (SSS)

BE = DE (given)

ÒOMA = ÒOMB (corresponding angles in congruent

ÒAEB = ÒCED (vertically opposite angles)

triangles) ÒOMA and Ò OMB are supplementary.

Â△ABE Ã △CDE (SAS) b ÒABE = ÒCDE (corresponding

ÂÒOMA = ÒOMB = 90° ÂOM ⊥ AB

8 a AD = CB (given)

OC is common; S

ÒDAC = ÒBCA (alternate angles)

Â△OAC Ã OBC (SSS)

AC is common.

ÒAOC = ÒBOC (corresponding angles in congruent triangles) c ÒCAB = ÒCBA = x (△ABC is isosceles) ÒEAB = ÒDBA =

ÒBAE =

(corresponding angles), ÒDAE = ÒBCE (corresponding angles). Therefore, AD ∥ BC (alternate angles are equal).

b OA = OB (radii of same circle) S CA = CB (radii of same circle) S

angles),

ÒDCE (corresponding angles). Therefore, AB ∥ DC (alternate angles are equal). ÒADE = ÒCBE

 △ABC à △CDA (SAS) b ÒBAC = ÒDCA (corresponding angles), therefore AB ∥ DC (alternate angles are equal). 9 a △ABE à △CBE à △ADE à △CDE (SAS) b ÒABE = ÒCDE (corresponding angles), ÒBAE = ÒDCE (corresponding angles), therefore AB ∥ CD.

x 2

ÒADE = ÒCBE (corresponding angles), ÒDAE = ÒBCE (corresponding angles), therefore AD ∥ CB.

Â△AFB is isosceles, so AF = BF.

Exercise 2C 1 a rectangle b parallelogram c square d rhombus 2 a rectangle, square b rectangle, square c parallelogram, rhombus, rectangle, square d rhombus, square e rhombus, square 3 a A trapezium does not have both pairs of opposite sides parallel. b A kite does not have two pairs of opposite sides parallel.

Also, AB = AD = CB = CD (corresponding sides). Therefore, ABCD is a rhombus. 10 a

C

D E

B A ÒCAB = ÒACD and ÒCAD = ÒACB (alternate angles).

796 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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4 a ABCDE ||| FGHIJ d

DC = BC (given) EC is common.

d 12 m

b From part a, ÒECD = ÒECB.

C

A B As ABCD is a parallelogram, ÒBDC = ÒDBA (alternate angles) and ÒDBC = ÒBDA (alternate angles). BD is common. So △CBD Ã △ADB (AAS). So ÒBAD = ÒDCB = 90°. Similarly, ÒADC = Ò180° - ÒBAD (cointerior angles) = 90° and similarly for ÒABC. 12 D

C E

A B First, prove △AED Ã △BEC (SAS). Hence, corresponding angles in the isosceles triangles are

equal. So ÒADC = ÒDCB = ÒCBA = ÒBAC, which sum to 360°. Therefore, all angles are 90° and ABCD is a rectangle.

G

C

F

AB DE = FG IJ

c

3 2

b

EF GH = AB CD

c

4 3

4 cm 3

e 10.5 m

6 a 1.2 e 11.5

b 12.5 f 14.5

7 1.7 m 8 a 1.6

b 62.5 cm

9 a 2 10 a BC

b 1 c 1.875 b △ABC ||| △EDC

d 4.3 c 1

c 4.8

d 3.75

b true f false

d false h false

d 4.5 11 a true e false

c false g false

i true j true 12 Yes, the missing angle in the first triangle is 20° and the missing angle in the second triangle is 75°, so all three angles are equal. 3 13 a 2 b i 4

ii 9

c i 8 d Cube Small

ii 27

Large

equal and △CED Ã △BEA (SAS). Hence, corresponding angles in the isosceles triangles are

13

e

5 a ABCD ||| EFGH

 △CDE à △CBE (SAS) So ÒCED = ÒCEB = 90°. 11 D

3 cm 2

b

Answers

So ÒECB = ÒECD since △ABC and △ADC are isosceles.

Scale factor (fraction)

Length 2

Area 4

Volume 8

3 3 2

9 9 4

27 27 8

e Scale factor for area = (scale factor for length)2 ; Scale factor for volume = (scale factor for length)3 .

2D

b2 b3 ii 3 a2 a 14 Answers will vary f i

Exercise 2E D

B

H

1 a E b ÒC d △ABC ||| △DEF 2 a ÒD (alternate angles) b ÒA (alternate angles)

E

A

First, prove all four corner triangles are congruent (SAS). So EF = FG = GH = HE, so EFGH is a rhombus.

c ÒECD d CA e △ABC ||| △EDC 3 a SAS b AAA

Exercise 2D 1 a Yes, both squares have all angles 90° and all sides of equal length. b 3 c 15 cm 2 a 2

8 5

3 a A b ÒC d △ABC ||| △EFD

c

4 3

c FD

d

3 2

c SAS

d SSS

4 a ÒABC = ÒDEF = 65° ÒBAC = ÒEDF = 70° Â △ABC ||| △DEF (AAA). b

b

c AB

DE 2 = =2 AB 1 EF 6 = = 2 (ratio of corresponding sides) BC 3 ÒABC = ÒDEF = 120° Â △ABC ||| △DEF (SAS).

797 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Answers

c

DF 10 = =2 CA 5 DE 8 = = 2 (ratio of corresponding sides) CB 4 ÒABC = ÒFED = 90° Â△ABC ||| △FED (RHS).

AB 28 d = =4 DE 7 BC 16 = =4 EF 4 AC 32 = = 4 (ratio of corresponding sides) DF 8 Â△ABC ||| △DEF (SSS). 3 b 19.5 c 2.2 d a = 4, b = 15 5 a 2 e x = 0.16, y = 0.325 f a = 43.2, b = 18 6 a ÒABC = ÒEDC (alternate angles) ÒBAC = ÒDEC (alternate angles) ÒACB = ÒECD (vertically opposite angles) Â△ABC ||| △EDC (AAA). b ÒABE = ÒACD (corresponding angles) ÒAEB = ÒADC (corresponding angles) ÒBAE = ÒCAD (common) Â△ABE ||| △ACD (AAA). c ÒDBC = ÒAEC (given) ÒBCD = ÒECA (common) Â△BCD ||| △ECA (AAA). d

AB 3 = = 0.4 CB 7.5 EB 2 = = 0.4 (ratio of corresponding sides) DB 5 ÒABE = ÒCBD (vertically opposite angles) Â△AEB ||| △CDB (SAS).

13 a ÒAOD = ÒBOC (common) ÒOAD = ÒOBC (corresponding angles) ÒODA = ÒOCB (corresponding angles) So △OAD ||| △OBC (AAA). OB OC 3 = = 3 (ratio of corresponding sides), therefore = OD 1 OA 3 OB = 3OA b ÒABC = ÒEDC (alternate angles) ÒBAC = ÒDEC (alternate angles) ÒACB = ÒECD (vertically opposite) So △ABC ||| △EDC (AAA). AC + CE 1 + 2 7 CE CD 2 = = , therefore = = . AC BC 5 AC 5 5 7 AE 7 = and AE = AC. AC 5 5 14 a ÒBAD = ÒBCA = 90° But AC + CE = AE, so

ÒABD = ÒCBA (common) So △ABD ||| △CBA (AAA). AB BD Therefore, = . CB AB AB2 = CB × BD b ÒBAD = ÒACD = 90° ÒADB = ÒCDA (common) So △ABD ||| △CAD (AAA). AD BD Therefore, = . CD AD AD2 = CD × BD c Adding the two equations: AB2 + AD2 = CB × BD + CD × BD = BD(CB + CD)

7 a ÒEDC = ÒADB (common) ÒCED = ÒBAD = 90°

= BD × BD = BD2

 △EDC ||| △ADB (AAA). 4 cm 3 8 a ÒACB = ÒDCE (common) ÒBAC = ÒEDC = 90°

Progress quiz

b

1 a x = 78 (exterior angle of a triangle) b w = 89 (angle sum of a quadrilateral) c x = 120 (interior angle of a regular hexagon)

 △BAC ||| △EDC (AAA). b 1.25 m

d x = 35 (alternate angles in parallel lines) e x = 97 (cointerior angles in parallel lines, vertically

9 1.90 m 10 4.5 m

opposite angles equal) f w = 47 (angle sum of an isosceles triangle)

11 a Yes, AAA for both. b 20 m c 20 m d Less working required for May’s triangles. 12 The missing angle in the smaller triangle is 47°, and the missing angle in the larger triangle is 91°. Therefore the two triangles are similar (AAA).

2 a AB = QB (given) ÒABC = ÒQBP (vertically opposite) ÒCAB = ÒPQB (alternate angles AC ∥ PQ) Â △ABC Ã △QBP (AAS) b CB = PB corresponding sides of congruent triangles and B is the midpoint of CP. 3 Let ABCD be any rhombus with diagonals intersecting at P. AB = BC (sides of a rhombus equal)

798 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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 △ABP à △CBP (SAS) and ÒAPB = ÒBPC (corresponding angles of congruent tri-

Exercise 2F 1 a–e

chord

angles). And ÒAPB + ÒBPC = 180° (straight line)

minor sector

centre

 diagonal AC ⊥ diagonal DB.

A

radius

B

major sector 2 a 55° b 90°

P D

A

4

Answers

ÒABP = ÒCBP (diagonals of a rhombus bisect the interior angles through which they cross)

c 75°

d 140°

3 a 85° each b ÒAOB = ÒCOD (chord theorem 1)

C

B

c 0.9 cm each d OE = OF (chord theorem 2) 4 a 1 cm each b 52° each c AM = BM and ÒAOM = ÒBOM (chord theorem 3)

D

C

Let ABCD be any parallelogram with opposite sides parallel. AC is common. ÒBAC = ÒACD (alternate angles AB ∥ CD) ÒBCA = ÒDAC (alternate angles AD ∥ CB) Â △ABC Ã △CDA (AAS) and AB = DC as well as AD = BC (corresponding sides in congruent triangles). 5 a △ABE ||| △ACD (all angles equal) b 2.5 c x = 7.5

5 a ÒDOC = 70° (chord theorem 1) b OE = 7.2 cm (chord theorem 2) c XZ = 4 cm and ÒXOZ = 51° (chord theorem 3) 6 The perpendicular bisectors of two different chords of a circle intersect at the centre of the circle. 7 a 3.5 m b 9m c 90° 8 a 140 e 30

c 19

d 90° d 72

9 6m √ √ 10 3 + 128 mm = 3 + 8 2 mm

2F

11 a Triangles are congruent (SSS), so angles at the centre of the circle are corresponding, and therefore equal.

6 a ÒCAB = ÒFDE (given) AC AB 1 = = DF DE 3 △CAB ||| △FDE (SAS) b ÒBAO = ÒCDO (alternate angles AB ∥ DC)

b Triangles are congruent (SAS), so chords are corresponding sides, and therefore equal. 12 a Triangles are congruent (SSS), so the angles formed by the chord and radius are corresponding, and therefore equal. Since these angles are also supplementary, they must be 90°.

ÒAOB = ÒDOC (vertically opposite) Â △ABO ||| △DCO (AAA)

b Triangles are congruent (SAS), so the angles formed by the chord and radius are corresponding, and therefore equal.

7 a ÒD is common ÒABD = ÒECD (corresponding angles equal since

Since these angles are also supplementary, they must be 90°.

AB ∥ EC) △ABD ||| △ECD (AAA)

A

13

b 3 cm 8 ÒA is common, as Q and P are both midpoints.

O

AP 1 AQ 1 = and = AB 2 AC 2 Â △AQP ||| △ACB (SAS) and

b 40 f 54

B

QP 1 = (corresponding sides in the same ratio). CB 2

C

First, prove △OAB Ã △OAC (AAS), which are isosceles. So AB = AC, corresponding sides in congruent triangles.

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Answers

c ÒOBC = x° + y° (△OCB is isosceles) ÒCOB = 180° - 2(x + y)°

14 a AD = BD (radii of same circle) AC = BC (radii of same circle)

ÒAOB = 180° - 2x° - (180° - 2(x + y)°) = 2y°

CD is common. Â△ACD Ã △BCD (SSS). b AC = BC ÒACE = ÒBCE (corresponding angles in congruent

15 ÒAOB = 180° - 2x° (△AOB is isosceles) ÒBOC = 180° - 2y° (△BOC is isosceles) ÒAOB + ÒBOC = 180° (supplementary angles), therefore (180 - 2x) + (180 - 2y) = 180

triangles) CE is common. Â△ACE Ã △BCE (SAS). c Using the converse of chord theorem 3 since

360 - 2x - 2y = 180 2x + 2y = 180

ÒACE = ÒBCE, CD ⊥ AB.

2(x + y) = 180

Exercise 2G

x + y = 90

1 a ÒADC e ÒAEC

b ÒADC f ÒAEC

c ÒADC

d ÒAFC

Exercise 2H

2 a ÒAOB 3 a 180°

b ÒACB b 90°

c 80° c 60°

d 61° d 7°

1 a ÒACD

4 a 50 e 250

b 40 f 112.5

c 80 g 38

d 60 h 120

i 18 5 a 70

b 25

c 10

b ÒACD

2 a ÒABD and ÒACD c ÒBAC and ÒBDC

c ÒACD b 85° d 17°

3 a Supplementary angles sum to 180°. b 117° c 109° d Yes, 117° + 109° + 63° + 71° = 360° 4 a x = 37 e x = 22.5

b x = 20 f x = 55

c x = 110

d x = 40

5 a x = 60 b x = 90 e x = 72, y = 108

c x = 30 f x = 123

d x = 88

7 a ÒADC = 75°, ÒABC = 75° b ÒABC = 57.5°, ÒADC = 57.5°

6 a 72 e 52

c 69 g 30

d 57 h 47

c ÒAOD = 170°, ÒABD = 85° 8 a 100° b 94.5° c 100°

i 108 7 a a = 30, b = 100

6 a ÒABC = 72°, ÒABD = 22° b ÒABC = 70°, ÒABD = 45° c ÒABC = 72°, ÒABD = 35°

e 70° 9 a 58°

f 66° b 53°

e 19° 10 a 70°

f 21° b 90°

c 51°

d 119° d 45°

b i false 12 a 2x°

ii true b 360 - 2x

b a = 54, b = 90

c a = 105, b = 105, c = 75 d a = 55, b = 70 e a = 118, b = 21 f a = 45, b = 35 8 a 80° b 71° c ÒCBE + ÒABE = 180° (supplementary angles)

c The angle in a semicircle is 90°. d Theorem 2 is the specific case of theorem 1 when the angle at the centre is 180°. 11 a i false ii true

b 43 f 48

iii true

iv false

iii true

iv false

13 a ÒAOC = 180° - 2x° (△AOC is isosceles) b ÒBOC = 180° - 2y° (△BOC is isosceles) c ÒAOB = 360° - ÒAOC - ÒBOC = 2x° + 2y° d ÒAOB = 2(x° + y°) = 2ÒACB 14 a ÒBOC = 180° - 2x° (△BOC is isosceles). ÒAOB = 180° - ÒBOC = 180° - (180° - 2x°) = 2x° b ÒAOC = 180° - 2x° (△AOC is isosceles) ÒBOC = 180° - 2y° (△BOC is isosceles) Reflex ÒAOB = 360° - ÒAOC - ÒBOC

ÒCBE + ÒCDE = 180° (circle theorem 4) Â ÒCBE + ÒABE = ÒCBE + ÒCDE. Â ÒABE = ÒCDE 9 a ÒACD = ÒABD = x° and ÒDAC = ÒDBC = y° (circle theorem 3) b Using angle sum of △ACD, ÒADC = 180° - (x° + y°). c ÒABC and ÒADC are supplementary. 10 a i 80° ii 100° iii 80° b ÒBAF + ÒDCB = 180°, therefore AF ∥ CD (cointerior angles are supplementary). 11 a ÒPCB = 90° (circle theorem 2) b ÒA = ÒP (circle theorem 3) a c sin P = 2r a a d As ÒA = ÒP, sin A = , therefore 2r = . 2r sin A

= 360° - (180° - 2x°) - (180° - 2y°) = 2x° + 2y° = 2(x + y)° = 2ÒACB.

800 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Exercise 2I b 90°

c 5 cm

2 a ÒBAP 3 a 180°

b ÒBPX b 360°

c ÒABP

d ÒAPY

4 a a = 19 5 a a = 50

b a = 62 b a = 28

c a = 70 c a = 25

d a = 63

6 a 50° b 59° 7 a a = 73, b = 42, c = 65

b a = 26, b = 83, c = 71

c a = 69, b = 65, c = 46 8 a 5 cm b 11.2 cm 9 a a = 115 e a = 26

b a = 163 f a = 26

c a = 33 g a = 36

d a = 28 h a = 26

i a = 30 10 a a = 70

b a = 50

c a = 73

d a = 40

f a = 54

Answers

1 a Once

e a = 19 11 4 cm

32 16 35 b c 2 √3 √3 7 a 65 b 77 64 209 81 74 8 a b c d 7 10 7 7 √ 153 e f 65 - 1 20 9 a x(x + 5) = 7 × 8, x2 + 5x = 56, x2 + 5x - 56 = 0 b x(x + 11) = 10 × 22, x2 + 11x = 220, 6 a

x2 + 11x - 220 = 0 c x(x + 23) = 312 , x2 + 23x = 961, x2 + 23x - 961 = 0 10 For this diagram, the third secant rule states: AP2 = DP × CP and BP2 = DP × CP, so BP = AP. 11 AP × BP = DP × CP AP × BP = AP × CP since AP = DP. BP = CP 12 a ÒA = ÒD and ÒB = ÒC (circle theorem 3) b ÒP is the same for both triangles (vertically opposite), so △ABP ||| △DCP (AAA).

12 a OA and OB are radii of the circle. b ÒOAP = ÒOBP = 90°

c

c ÒOAP = ÒOBP = 90° OP is common

AP BP = DP CP

AP BP = , cross-multiplying gives AP × CP = BP × DP. DP CP 13 a ÒB = ÒC (circle theorem 3) b △PBD ||| △PCA (AAA) d

OA = OB Â △OAP Ã △OBP (RHS) d AP and BP are corresponding sides in congruent triangles. 13 a ÒOPB = 90° - x°, tangent meets radii at right angles b ÒBOP = 2x°, using angle sum in an isocles triangle c ÒBAP = x°, circle theorem 1 14 ÒBAP = ÒBPY (alternate segment theorem) ÒBPY = ÒDPX (vertically opposite angles) ÒDPX = ÒDCP (alternate segment theorem) ÂÒBAP = ÒDCP, so AB ∥ DC (alternate angles are equal). 15 AP = TP and TP = BP, hence AP = BP. 16 a Let ÒACB = x°, therefore ÒABC = 90° - x°. Construct OP. OP ⊥ PM (tangent). ÒOPC = x° (△OPC is isosceles). Construct OM. △OAM Ã △OPM (RHS), therefore AM = PM. ÒBPM = 180° - 90° - x° = 90° - x°. Therefore, △BPM is isosceles with PM = BM. Therefore, AM = BM. b Answers may vary

c

AP CP = , so AP × BP = DP × CP. DP BP

14 a yes b alternate segment theorem c △BPC ||| △CPA (AAA)

2I

BP CP = , so CP2 = AP × BP. CP AP √ √ 15 d = (4r1 r2 ) = 2 r1 r2 d

Problems and challenges 1 21 units2 2 BD = 5 cm, CE = 19 cm 3 ÒADE = ÒABE, ÒEFD = ÒBFA, ÒDEB = ÒDAB, ÒDFB = ÒEFA, ÒCDB = ÒCAE, ÒDAE = ÒDBE, ÒADB = ÒAEB, ÒABD = ÒAED = ÒCBD = ÒCEA 4 42.5% 5 Check with your teacher.

Exercise 2J 1 a 3 b 6 5 21 b 2 a 2 2 3 a AP × CP = BP × DP

c 7 d 8 33 27 c d 7 7 b AP × BP = DP × CP

c AP × BP = CP2 4 a 5 5 a

143 8

b 10 b

178 9

112 15 161 c 9

c

6 a ÒFDE = ÒDFC = ÒABC (alternate and corresponding angles in parallel lines) ÒFED = ÒEFB = ÒACB (alternate and corresponding angles in parallel lines) ÒDFE = ÒBAC (angle sum of a triangle) △ABC ||| △FDE (AAA) b i 4:1 c 4n-1 : 1

ii 16 : 1

801 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Answers

Multiple-choice questions 1 C

2B

3B

4C

5B

6A

7 E

8C

9D

10 B

b 120

c 6 (chord theorem 3) 7 a a = 25 b a = 50, b = 40

c x = 62, y = 118

2 a 148 b 112 3 a AB = DE (given) ÒABC = ÒDEF (given)

b a = 65, b = 130, c = 50, d = 8 c t = 63 40 9 a 5 b 6 c 3

Extended-response questions

ÒBAC = ÒEDF (given)

1 a ÒBAC = ÒBDE = 90° ÒB is common.

 △ABC à △DEF (AAS).

Â△ABC ||| △DBE (AAA). b 1.2 km AC 3 = c i DE 2

b AB = AD (given) ÒBAC = ÒDAC (given) AC is common. Â △ABC Ã △ADC (SAS).

Â

c AB = CD (given) AD = CB (given) Â △ABD Ã △CDB (SSS). 4 a AB = CD (given)

 2(x + 1) = 3x ii 2

ÒBAC = ÒDCA (alternate angles) AC is common. Â△ABC Ã △CDA (SAS). b ÒBCA = ÒDAC (alternate angles), therefore AD ∥ BC (alternate angles are equal). DE 10.5 = = 1.5 5 a AB 7 EF 14.7 = = 1.5 (ratio of corresponding sides) BC 9.8 ÒABC = ÒDEF (given)

d 44.4% 2 a i 106.26°

x = 19.5 b ÒEAB = ÒDAC (common) ÒEBA = ÒDCA (corresponding with EB ∥ DC) Â △ABE ||| △ACD (AAA)

d 70 cm

Chapter 3 Exercise 3A b root b 9 f 16

c 25 g 36

d 4 h 36

3 a irrational e rational

b rational f irrational

c rational g irrational

d rational h irrational

−1.24 −2

c ÒBAC = ÒEDC (given)

x = 8.82 d ÒABD = ÒDBC (given) ÒDAB = ÒCDB = 35° (angle sum of triangle)

√5

0.18

−1

2

0

1 57

2 5

−√2

ÒACB = ÒDCE (vertically opposite) Â △ABC ||| △DEC (AAA)

c non recurring

2 a 4 e 16

x = 6.25

100 7

ii 73.74°

b 12 cm c 25 cm

1 a irrational d rational

 △ABC ||| △DEF (SAS)

 △ABD ||| △DBC (AAA)

AB 3 = (ratio of corresponding sides in similar DB 2 triangles)

x+1 3 = 2 x

BD is common.

x=

c a = 70

d b = 54 e a = 115 f a = 30, b = 120 8 a x = 26, y = 58, z = 64

Short-answer questions 1 a 65 d 46

6 a 65 (chord theorem 1) b 7 (chord theorem 2)

2√3 3 π

4 a yes

b yes

c no

d no

f yes √ 5 a 2 3 √ e 5 3 √ i 8 2 √ 6 a 6 2 √ e√ 21 2 6 i 2

g yes √ b 3 5 √ f 10 5 √ j 6 10 √ b 6 5 √ f √ 20 5 6 j 4

h no √ c 2 6 √ g 7 2 √ k 9 2 √ c 16 3 √ g √5 5 k 5

√ d 4 3 √ h 3 10 √ l 4 5 √ d 6 7 √ h √7 11 l 6

4 e no

802 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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8

9 10 11

12 13 14

15 a Draw triangle with shorter sides length 1 cm and 3 cm. b Draw triangle with shorter sides length 2 cm and 5 cm. c d 2 1

√22

√6

1

1

1

2

4

16 Check with your teacher

Exercise 3B 1 a yes

b no

c no

d yes g yes

e yes h yes

f no

2 a 6x d -2b

b 7y e 17a

c -5x f -5x

g t √ 3 a 4 3 √ b i 5 3 √ 4 a 10 2 √ b i 5 2 √ 5 a 6 5 √ d 3 2 √ g 6 10 √ j -6 11

h -2y √ ii -3 3

b e h k

√ ii -16 2 √ 3 3 √ 11 5 √ 5 2 √ - 13

√ √ √ √ √ √ 6 a 3+5 2 b 3 6 + 7 11 c 4 5-7 2 √ √ √ e 4 3 f 0 d -2 2 + 5 √ √ √ √ g -3 2 - 3 10 h -2 5 + 3 15 √ √ √ √ b 5 2 c 4 3 d 5 7 a 2 √ √ √ √ e 7 2 f 12 3 g 8 11 h 3 2 √ √ √ √ j 5 k 32 2 l 20 2 i 5 6 √ √ √ √ 8 a 13 2 b 9 6 c 2 5- 7 √ √ √ √ √ √ √ f 2 3 + 11 5 - 5 2 d 5 5+6 7 e 6-3 2 √ √ √ √ g 9 3 + 2 2 h 11 - 9 3 i 9 + 18 2 √ √ j -9 2 + 15 5 √ √ √ √ 2 7 7 5 5 3 b c d 9 a 6 12 30 6 √ √ √ √ - 2 13 3 13 5 -7 3 e f g h 10 √ 14 18 30 -11 10 i √ √ √24 √ √ 10 a 4 3 + 2 5 cm b 14 2 cm c 10 + 3 2 cm √ √ √ √ d 2 10 + 4 5 cm e 4 3 + 30 cm √ f 12 3 cm √ √ 11 a 20 = 2 5 √ √ √ √ b 3 72 = 18 2, 338 = 13 2 √ √ √ 12 a 5 3 - 6 3 + 3 = 0 √ √ √ b 6+2 6-3 6=0 √ √ √ c 6 2-8 2+2 2=0 √ √ √ d 2 2-3 2+ 2=0 √ √ √ e 4 5-7 5+3 5=0 √ √ √ √ √ f 3 2-6 3-5 2+6 3+2 2=0 √ √ 13 a 6 3 - 3 2, unlike surds √ √ b 8 2 + 2 5, unlike surds √ √ c 5 2 - 6 5, unlike surds √ √ d 10 10 + 10 3, unlike surds √ √ e 20 2 + 30 3, unlike surds √ √ f 4√ 5 - 6 6, unlike √ √ surds √ 2 3 5 -3 2 7 2 b c d 14 a 3√ 12 4 √15 √ √ 3 -7 7 29 6 e f g - 2 h i 0 2 15 28 √ √ √ 6 6 29 5 j 8 3 k l 35 42

Answers

7

√ √ √ √ 2 17 n 2 2 o 2 2 p m 3 11 √ √ √7 √ 3 3 2 3 3 3 q r 4 6 s t 2 3 √ √ √2 √ 2 2 2 3 3 2 11 a b c d 3 7 5 5 √ √ √ √ 10 21 13 14 f g h e 12 4 5 √ √3 √ √ 3 3 5 i j k √5 l √14 3 2 2 2 19 √ √ √ √ a 12 b 32 c 50 d 27 √ √ √ √ e 45 f 108 g 128 h 700 √ √ √ √ j 125 k 245 l 363 i 810 √ √ √ √ a 15 3 b 13 7 c 19 5 d 31 3 √ √ √ a 4 2m b 2 30 cm c 4 15 mm √ √ a radius = 2 6 cm, diameter = 4 6 cm √ √ b radius = 3 6 m, diameter = 6 6 m √ √ c radius = 8 2 m, diameter = 16 2 m √ √ √ √ a 2 5 cm b 3 5 m c 145 mm d 11 m √ √ e 11 mm f 2 21 cm √ √ 72 = 36 × 2 (i.e. 36 is highest square factor of 72) √ =6 2 √ a 9, 25, 225 b 15 2

3B

Exercise 3C √

√ iii 17 3 iii 0 √ c 4 2 √ f 3 √ i -2 21 √ l -7 30

√ √ 15 √ 42 √ √ = 5b = 6 c 6 × 5 = 30 3 7 √ √ d 11 × 2 = 22 √ √ √ √ 2 a 15 b 21 c 26 d 35 √ √ √ √ f - 30 g 66 h 6 e - 30 √ i 70 √ √ √ √ 3 a 10 b 6 c - 3 d 5 √ √ √ √ f 10 g 5 h - 13 e 3 √ i - 5 1 a

803 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

√ 4 a 21 d 3 √ g 7 2 √ j 5 2 √ 5 a 10 3 √ d -50 3 √ g 42 6 √ j 42 2 √ 6 a 2 2

√ b 10 e 5 √ h 2 11 √ k 4 6 √ b 21 2 √ e -18 3 √ h -60 10 √ k 24 30 √ b 3 6

√ c 30 f 9 √ i 3 6 l 10 √ c 12 14 √ f 15 5 √ i -20 10 √ l 216 7 √ c 5 2 √ 2 5 4 1 d√ e √ f 3 13 3 7 √ √ √ √ √ √ 7 a 6 + 15 b 14 - 10 c - 55 - 65 √ √ √ √ √ d -2 15 - 2 21 e 6 26 - 3 22 f 20 - 20 2 √ √ √ √ √ g 30 2 + 15 30 h -12 3 + 12 2 i 42 + 63 2 √ √ √ √ j 90 3 - 24 10 k -16 + 24 10 l 42 2 + 30 b 18 c -75 √ √ √ e 3 3+4 f - 10 + 5 √ √ 2 h 8 2 i 2-6 √ √ 2 6 b 30 c 6 √ 3 cm2 b 2 6 cm 4√ √ √ √ 6 × 6 = 6 × 6 = 36 = 6 √ √ √ √ - 8 × 8 = - 8 × 8 = - 64 = -8 √ √ √ √ - 5 × - 5 = + 5 × 5 = 25 = 5 Simplify each surd before multiplying.

8 a 28 √ √ d 2- 6 g 9 a 10 a 11 a b c 12 a

b Allows for the multiplication of smaller surds, which is simpler. √ √ √ c i 3 2×3 3=9 6 √ √ √ ii 2 6 × 2 5 = 4 30 √ √ √ iii 5 2 × 3 5 = 15 10 √ √ √ iv 3 6 × 5 3 = 45 2 √ √ √ v 6 2 × 4 3 = 24 6 √ √ √ vi 6 3 × -10 5 = -60 15 √ √ √ vii -12 3 × -2 7 = 24 21 √ √ √ viii 7 2 × 10 3 = 70 6 √ √ √ ix 12 2 × 12 5 = 144 10 13 a 3 1 5√ 14 a 54 2 d 25 √ g -120 5 √ j 14 7 √ m 100√ 3 81 3 p 25 d -

b 2 2 e 5 √ b 375 3 e 9 √ h -108 √ 2 27 2 k 2 n 144 5 q √ 3 3

c -9 f 3 √ c 162 3 √ f 128 2 i 720 l 81 √ o -96 √ 15 9 6 r 2

Exercise 3D 1 a

√ 21

e 13 i 162 2 a 0 √ e 2 3

√ b - 10

√ c 6 6

d 11

f 12

g 125

h 147

b 0 √ f 6 3

c 0

d 0

3 a x2 + 5x + 6 b x2 - 4x - 5 c x2 + x - 12 d 2x2 - 9x - 5 e 6x2 - 11x - 10 f 6x2 - 17x - 28 h 4x2 - 9 i 25x2 - 36 k 4x2 - 4x + 1 √ 4 a - 2-4 √ d 7+3 3 √ g 6 5 - 13 √ 5 a 27 + 12 2 √ d 2 6 - 118 √ g 23 3 - 46 √ 6 a 14 - 6 5 √ d 15 + 4 11 √ g 9 + 2 14 √ j 32 + 2 247 7 a 7 e 4 i -4

l 9x2 - 42x + 49 √ b 2 5-3 c √ e 5+ 7 f √ h 30 - 9 10 i √ b 21 + 2 3 c √ e 35 - 13 10 f √ h 43 - 19 2 i √ b 10 - 4 6 c √ e 28 + 10 3 f √ h 16 - 2 55 i √ k 40 + 2 391 l b 19 c f -6

√ 8 a 118 + 28 10 √ c 195 + 30 30 √ e 207 - 36 33 √ g 66 + 36 2 √ i 107 - 40 6 9 a 97 e 26

g x2 - 16 j x2 + 4x + 4

b 17 f 10

√ 4+ 6

√ -13 - 2 2 √ 23 - 8 7 √ 25 + 32 5 √ 18 7 - 65 √ 24 5 - 89 √ 23 + 8 7 √ 54 - 14 5 √ 13 - 2 30 √ 60 - 2 899 13 d 6

g 3

h 6

√ b 139 + 24 21 √ d 176 - 64 6 √ f 87 - 12 42 √ h 140 + 60 5 c 41 g 0

d 163 h -33

i -40 j -90 √ 10 a 7 + 4 3 cm2 b 2 m2 √ √ √ c 15 6 - 5 2 - 18 + 2 3 mm2 √ 2 d 5√ + 6 5√ m √ 7 6-7 2- 3+1 2 e cm √2 2 f 81 - 30 2 mm 11 a -5 b 7 c 128 √ √ √ √ 12 a 11 21 - 26 3 b 2 5 + 30 √ √ √ √ c 5 35 + 31 7 d 19 7 + 2 √ √ √ √ e 2-2 6 f 10 + 3 5 √ √ 13 Yes. Possible example a = 12, b = 3 √ 14 a 19 - 2 6 b 16 √ √ c 2 15 - 85 d 10 3 - 37 √ f 0 e 30 - 10 2 √ √ √ g 4 3 - 14 h 47 2 - 10 30 + 11

804 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

√

1 a 1 e -2 √ 2 a 3 √ e 3 √ i √13 13 3 a 0.377… 4

5

6

7

8

9

10

11

12 a b 1 f -9 √ b 5 √ f 7

b 2.886…

1 2 h -1 √ d √5 h √7 7

1 2 g 6 c√ 10 g √3 3

d

d -

c

g j 13 a b

c 16.31…

All√pairs of numbers √ are equal. √ √ 2 7 3 11 4 5 b c d a 2√ 7 √5 √11 √ 15 14 5 3 e f 4 2 g h 3 3 7 √ √ √ √ 6 35 66 10 a b c d √7 √3 √11 √5 21 42 30 34 f g h e 3 7 3 2 √ √ √ √ 5 6 3 10 3 42 4 14 b c d a 3 2 7 √7 √ 7 30 2 105 e f 10 √ √ √ 15 √ 2 2 4 21 6 35 b c d a 15 3 3 5 √ √ √ 2 5 10 9 2 3 7 e f g h 9 2 √15 √ √ 2√ 3+ 6 3 7 + 35 a b 7 √ √ √3 √ 2 5 - 15 6 - 10 d c √ 5√ √ 2√ 35 + 14 30 - 21 e f √ 3 √ √ 7√ 2 3 + 42 5 2+2 5 g h 6 √ √ √ 10 √ 30 - 5 2 8 3 - 15 2 i j √ 5 √ √ 6 √ 3 2+2 5 6 5+5 6 k l 2 √ 2 5 3 2 a cm2 b m2 3 √3 √ 10 + 15 2 c mm √ √ 10 √ √ 2 3+3 2 6 5+5 2 a b √ 6 √ √ 10 √ 9 7 - 14 3 5 3-2 2 c d √ 21 √ √ 6 √ 2 2+5 3 9 5+4 3 e f √12 √ 30 √ -2 14 6 30 + 4 6 g h 9 √ √15 3 10 - 2 42 i √ 9 x As √ is equal to 1. x

c d

14 a d g

√ √ √ √ √ 2 3 + 6a 21 + 7a 30 + 5a b c √7 √6 √5 1√ - 3a √ e√ 1 - 5a f 1 √ √ 7a √ 4 10a + 5 2 6a + 2 2 14a + 7 2 h i 2 14 √ √ 10 3 5 - 30a 15 i 14 ii 2 iii 47 Each question is a difference of perfect squares, and each

answer √ is an integer. 4+ 2 √ 4+ 2 √ √ 12 + 3 2 -3 3-3 i ii 14 √ 2 -(6 + 2 30) iv 7 √ √ 5 3-5 b 2 3+2 2 √ √ -3 - 3 3 -4 - 4 2 e 2 √ √ -12 - 4 10 h -14 - 7 5

√ √ j 2 5-2 2

k

√ m6+ 6 5 √ √ b a-b b p a√- b a - ab s a-b

√

√ 7- 3

√ 14 + 2 2 √ √ a a+a b q √a - b√ a b+b a t a-b n

√

√ √ iii 2 2 + 6

√ c 3 5+6 √ 42 + 7 7 f √ 29 √ 2 11 + 2 2 i 9 √ √ 14 2 l 6 √ o 10 - 4 5 √ a + b - 2 ab r a-b

3E

Exercise 3F 1 a 34 e

b 76

15y4

f

c 83

Answers

Exercise 3E

d 6x3

8a3 b2

2 x 4 3 2 1 0 2x 24 = 16 23 = 8 22 = 4 21 = 2 20 = 1 3 a 22 × 23 = 2 × 2 × 2 × 2 × 2 = 25 x5 x × x × x × x × x b 3 = x x×x×x = x2 2 3

c (a ) = a × a × a × a × a × a = a6 0

d (2x) × 2x0 = 1 × 2 =2 4 a a9 d 14m5 p3 g 5 j 6x3 y3 m 150x5 y6

b x5 e 6s7 c7 h 6 k 15a3 b6

c b6 f 2t16 9 2 i s 25 l 18v9 w2

n 12r7 s6

o 20m8 n10

805 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

5 a x3 f d5

b a g j

c q3 h m6

d b4 i 2xy3

k 2p2 1 o v2 4

l 2m4 x 1 p a 2

m 5b3 x q 3

i

b t6

c 4a6

d 5y15

m 53 3s

e 64t6

f 4u4

g 27r9

h 81p16

9 j x12 y

k

j 3r2 s n 4st y2 r 2 6 a x10

x4 y6 z8

16 8 27f 6 m l u 8w 125g3 v 8 12 256p q p 81r 4 c 1 d 1

4 2 n 9a2 b6 4p q 7 a 8

3 9 o a t12 27g b 3

f 3 8 a x8

g -5 b x2 y2 9x8 y2 g 2

h 3 c x6 n8

d xy2

h 2y4

i 2a2 b2

k -45a8 b5

l

n 21y3 z2 9 a -27

o 1 b -27

p c

10 a x12

b a105

c

11 a 13

b 18

c

f r4 s7 j 27m7 n14

16 3 f 3 -6m2 n7 81 a30 b15 81

e y5

6 a x e 62 a

4 i a6 b

2x 3

7 a 164 x e 98 t y4 i 16

e 5 8 a e m

1 xy

e a3 b p i 2 q r

m 2m6 n3

m

d -81

f5 g5

b a3 f

12 x

2 l 3b 4

n 42 3f

o 12 2d

p 52 6t

c 221 x

d 46 d

g 81 x20

h -8 x15

k 7j8

l 2t6

b

1 64m6

f 54 x 12 j h 81 4y2 b 3 a q f 5 p

c

2 h m n

j x 2y

k 4m3 7n

3 7 l 4r s 3

n 61 2 r s

5 o wx 2

5 4 p 5c d 4

16p4 9q2 11 e 324r s b

e 1

g a2 b18

14 h m8 n

1 25 -5 d 81

1 64 1 e 9

g 98

h -48

10 a c 162

b 5

c 2

d -18 7 2 16 a x = 2, y = 4 or x = 4, y = 2 or x = 16, y = 1 b x = 8, y = 2 or x = 4, y = 3 or x = 64, y = 1, d

or x = 2, y = 6 c x = 9, y = 2, or x = 3, y = 4 or x = 81, y = 1 d x = 1, y = any positive integer

Exercise 3G x-1 , x-2 , x-3

1 a c 3x-1 , 3x-2 , 3x-3 2 a 12 b 12 3 5 1 3 a b b ab a

b

c 52 or 54 4 2

d -23 3

4 a 15 x

c 24 m 10y5 z g x2

d 37 y

2 7 k 3u w 8v6 c 4m7 4 3 g 4b a 3

2b3 5c5 d2 d 3b5 3 3 h 5h g 2

5 a x2 e 2b4 d3

b

j

k

ii

7 5

c 54x7 y10 f

2y14 x3

i

27x 2y

c

2 49

f 1 i

1 16

9 4

l 100 iii y 2x

b a The negative index should only be applied to x not to 2: 2 2x-2 = 2 x 5 5 1 7 a b c d 6 18 3 12 71 106 e f 48 ( ) 9 x 1 Proof: = (2-1 )x = 2-1×x = 2-x 2 a -2 b -5 c -3 d -1 e -2 f -3 g -3 h -4 b

2-2 , 2-3 , 2-4

2 e 3a3 b q3 r i 3p2

-64 125 11 0.0041 cm 3 12 a i 2

4 d 18b2 a

6 a5 b2

2 g a2 b

i.e. has only applied it afterwards. Need (-2)4 not -24 . 13 a 3 b 4 c 1 d 3

15 a ± 2

h -18 a10

1 2

d 4a8 b3

f 1 b 2

g -10 m6 k

f 1 g 9 h 8 12 He has not included the minus sign inside the brackets,

e 4 14 a 9

d 33 y

2 j 7d 10

9 a a7 b2 d 64

c 24 b

b 14 a 3 f 4m3 n d2 f5 j 5e4

b 2y3 f 3m2 n4

3 h 3z 4 x y2

l

13

14

15 16

i 0 m -2

j 0 n 1

k 1 o -2

l 2 p 2

806 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

1 a 3

b 4

c 3

f 2 2 a 103

g 1 b 107

h 3 c 10-6

104

3 a 4.3 × b 7.12 × d 1.001 × 104 g 3 × 10-5 4 a 3120

105

d 2

e 3

d 10-3

c 9.012 × e 7.8 × 10-4 f 1.01 × 10-3

j -55 000 5 a 0.0045 d 0.00783 g 0.0001002

e 59 500 h 9 990 000

f -800 200 i 210 500 000

d 4.24 × 105 g 7.25 × 104 j 9.09 × 105 7 a 2.42 × 10-3

Progress quiz 1 a 3.16227766 irrational c 3.141592654 irrational

22 7 π

e -0.000092 f 0.265 h -0.000006235 i 0.98 2 3

e -1.01 × 104 f 3.50 × 107 h 3.56 × 105 i 1.10 × 108

4

k -4.56 × 106 l 9.83 × 109 b -1.88 × 10-2

5

c 1.25 × 10-4 e -7.08 × 10-4

d 7.87 × 10-3 f 1.14 × 10-1

6 7

g 6.40 × 10-6 i 1.30 × 10-4

h 7.89 × 10-5 j 7.01 × 10-7

8

k 9.89 × 10-9

l -5.00 × 10-4

8 a -2.4 × 104 c 7.0 × 108

b 5.71 × 106 d 4.88 × 103

e 1.9 × 10-3 g 9.8 × 10-6

f -7.05 × 10-4 h -3.571 × 10-1

i 5.00 × 10-5 9 a 7.7 × 106 km2

b 2.5 × 106

c 7.4 × 109 km e 1.675 × 10-27 kg

d 1 × 10-2 cm f 9.5 × 10-13 g

10 a 2.85 × 10-3 c 4.41 × 10-8

b 1.55 × 10-3 d 6.38 × 10-3

e 8.00 × 107 g 1.80 × 10-3

f 3.63 × 108 h 3.42 × 1015

i 8.31 × 10-2 11 328 minutes 12 38 is larger than 10. 13 a 2.1 × 104 b 3.94 × 109 d 1.79 × 10-4 g 1 × 107 j 3.1 × 10-14 14 a 9 × 104 d 1.44 × 10-8 g 2.25 × 10-6

e 2 × 103 h 6 × 106

15 3 ×

f 7 × 10-1 i 4 × 10-3

k 2.103 × 10-4 l 9.164 × 10-21 b 8 × 109 c 6.4 × 109 e 4 × 104 h 1.25 × 107

j 1.275 × 10-4 k 1.8 × 10-1 m 8 × 10-1 n 4 × 10-14 10-4

c 6.004 × 101

= 3 ÷ 10000

f 6.25 × 10-12 i 1 × 10-5 l 2 × 102 o 2.5 × 104

ii 4.22 × 10-1 kg iv 1.89 × 10-19 kg

c 5.4 × 1041 J

k 2 350 000 000 l 1 237 000 000 000 b 0.0272 c 0.0003085

j -0.000000000545 k 0.000000000003285 l 0.000000875 6 a 6.24 × 103 b -5.73 × 105 c 3.02 × 104

ii 2.34 × 1021 J iii 2.7 × 1015 J

b i 1.11 × 108 kg iii 9.69 × 10-13 kg

105

h 3.00401 × 10-2 b 54 293 c 710 500

d 8 213 000 g -10 120

16 a i 9 × 1017 J iv 9 × 1011 J

Answers

Exercise 3H

b 3.142857 rational d 3.15 rational

315%

√10

3.14 3.15 √ √ a 7 2 b 10 3 c √ 192 √ √ √ a 3 3 b 6 2+2 5 √ √ d 17 5 - 6 3 √ √ a - 15 b 35 6 c √ 66 2 √ √ a 26√- 18 3b 5√ -2 6 c 3 7 2 15 a b c 7 5

9 a a5

3.16 √ 5 2 2

3.17 √ 5 5 d 4 √ c 16 3

2 √ 3 5 -4 √ √ 2 3 - 3 10 2 1 6 3 m n 2

b 12x3 y4

c h4

e a6

f 9m10

g

10 a 13 x

4 b 2b a2 c3

c 7m2

d 42 5d

6 e 168 f a g 12m k 8 a5 11 a i 7 012 000 ii 0.009206

5 h 4d2 3c

d

4p8 q6 49r2 t4

h 6

3H

b i 3.52 × 108 ii 2.10 × 10-4 4 10 9 12 a a 5 b 9x 13d 5b 8c

Exercise 3I 1 a 3, 2 e 4, 4

b 2, 3 f 5, 5

c 3, 5

d 5, 5

2 a 3 f 3

b 5 g 5

c 11 h 4

d 25 i 2

e 2 j 3

k 2 l 10 3 a 1.91, 1.91 b 1.58, 1.58 c 1.43, 1.43 1

1

4 a 29 2 1

7

5

b 6n 3 2

3

e 2a 3 b 3 3

5

f 2g 4 h 4 1

h 7 2 or 343 2

d b4 1 2

f 43 t3

5 a 7x 2

3

c x5

1 7

1

e 22 a2

4

2

b 35 3

1

1

g 10 5 t 5

h 88 m2

c 3y3

d 5p 3 r 3

2

3

1

4

1

1

g 5 2 or 125 2 i 4 3 or 256 3

807 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

√ 5 6 a 2 √ 3 e 9

√ 7 b 8 √ 3 f 49

√ 3 c 6 √ 5 g 8

√ 10 d 11 √ 7 h 81

7 a 6

b 3

e 2

f 5

c 4 1 g 3 1 k 20

d 7 1 h 2 1 l 10

c 216 1 g 16 1 k 81

d 32 1 h 125 1 l 100

i

1 3

j

8 a 4 1 e 8 i 125 9 a

a2 6

e s7 10 a 5s2 e x

1 10

b 8 1 f 9 1 j 16

c x

d

1 b2

f y9

g 1

h

a2 b

b 3t2 f b4

c 2t2 g t3

d 5t4 h m2

k 2x2 y3 2 o 2 x

l 7r3 t2

b

m3 1

i 4ab4 j 6m2 n 5 2x m n 7 3 11 a method B b i 32 ii 216 iv 81

1

v 625

p 10x

b i no iii yes

ii yes iv no

A = 1200 × 0.9n f A = cell area at any time, n = number of minutes elapsed A = 0.01 × 2n g A = Size of oil spill at any time, n = number of minutes elapsed A = 2 × 1.05n h A = mass of substance at any time, n = number of hours elapsed A = 30 × 0.92n 5 a 1.1 b i $665 500 c 7.3 years

ii $1 296 871.23 iii $3 363 749.97

6 a 300 000 b i $216 750

ii $96 173.13

iii $42 672.53

c 3.1 years 7 a V = 15 000 × 0.94t ii 9727 L

d 55.0 hours 8 a V = 50 000 × 1.11n

32 81 viii 3125 10000 12 It equals 2 since 26 = 64. ii -10 iv -3

A = 172 500 × 1.15n e A = litres in tank at any time, n = number of hours elapsed

b i 12 459 L c 769.53 L

iii 128 1 vi 27

vii

13 a i -3 iii -2

d A = population at any time, n = number of years since initial census

b i $75 903.52

ii $403 115.58

c 6.64 years 9 a 3000 b i 7800 ii 20 280 c 10 hours 11 minutes

c y is a real number when n is odd, for x < 0.

iii 243 220

10 a D = 10 × 0.875t , where t = number of 10 000 km travelled b 90 000 c yes 11 a T = 90 × 0.92t

Exercise 3J 1 a $50 b $1050 c $52.50 e $1276.28 2 2 a 4.9 kg b , 0.98 c 4.52 kg 100 3 a growth b growth c decay e growth f decay

d $55.13

d decay

4 a A = amount of money at any time, n = number of years of investment 1.17n

A = 200 000 × b A = house value at any time, n = number of years since initial valuation A = 530 000 × 0.95n

12 a i $1610.51 ii $2143.59 iii $4177.25 b i $1645.31 ii $2218.18 iii $4453.92 13 a $2805.10 14 a i 90 g

b $2835.25 c $2837.47 ii 72.9 g iii 53.1 g

b 66 years 15 a 60 L b 22.8 minutes 16 0.7%

Exercise 3K

c A = car value at any time, n = number of years since purchase A = 14 200 × 0.97n

b i 79.4°C ii 76.2°C c 3.2 minutes = 3 minutes 12 seconds

1 a $50 b $550 c $55 d $605 e $605 2 a $1102.50 b $1102.50 c $1157.63 d $1157.63 3 a 700(1.08)2 4 a 6, 3% d 14, 2.625%

b 1000(1.15)6 b 60, 1%

c 850(1.06)4 c 52, 0.173%

e 32, 3.75%

f 120, 0.8%

808 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

7 a

2

4200

210

4410

3 4

4410 4630.50

220.50 231.53

4630.50 4862.03

$7000 5% $7000 10%

5 years 5 years

$1750 $3500

$8750 $10 500

5

4862.03

243.10

5105.13

$3300 10% $8000 10%

3 years 3 years

$990 $2400

$4290 $10 400

2 years 2 years

$1440 $2880

$10 440 $20 880

6 a $5105.13 c $13 652.22

b $11 946.33 d $9550.63

7 a $106 000 d $133 822.56

b $112 360 e $179 084.77

c $119 101.60 f $239 655.82

8 a $2254.32 d $789.84

b $87 960.39 e $591.63

c $1461.53 f $1407.76

9 $11 651.92 10 a $5075

$9000 $18 000

8% 8%

b i interest is doubled

b $5228.39

c $5386.42

ii $3348.15 v $3465.96

iii $3446.15

b $226.54 12 a P = 300, n = 12, r = 7%, R = 14%, t = 6 years b P = 5000, n = 24, r = 2.5%, R = 30%, t = 2 years c P = 1000, n = 65, r = 0.036%, R = 0.936%, t = 2.5 years d P = 3500, n = 30, r = 0.0053%, R = 1.9345%, t = 30 days e P = 10 000, n = 10, r = 7.8%, R = 7.8%, t = 10 years

8

Overall time

Interest

$7000 4.56% annually 5 years $7000 8.45% annually 5 years

$1750 $3500

$8750 $10 500

$9000 8% fornightly 2 years $1559.00 $18 000 8% fornightly 2 years $2880

$10 559.00 $21 118.01

13 5.3% compounded bi-annually 14 a i approx. 6 years ii approx. 12 years iv approx. 5 years vi approx. 4 years

b Same answer as part a c yes

Exercise 3L 1 B 2 P = 750, r = 7.5, n = 5 3 I = 225P = 300 r = 3

t = 25

ii $6955.64 v $7916.37

iii $6858.57

b $6000 at 5.7% p.a., for 5 years 5 a i $7080 ii $7080 iii $7428

Period

Amount

9 a 8.45% b 8.19% c 8.12% The more than interest is calculated, the lower the required rate. 10 a i 4.2%

ii 8.7%

b it increases by more than the factor.

Exercise 3M

f P = $6000, n = 91, r = 0.22%, R = 5.72%, t = 1.75 years

iii approx. 9 years v approx. 7 years

ii no change

iii interest is doubled Principal Rate

11 a i $3239.42 iv $3461.88

4 a i $7146.10 iv $7260

Principal Rate Overall time Interest Amount

Answers

5

1 a i 4

ii 8

iii 16

iv 32

b i 3 ii 5 iii 6 2 a 16, 32, 64, 128, 256, 512 b 81, 243, 729, 2187, 6561 c 64, 256, 1024, 4096 e 216, 1296 32

53

3L

d 125, 625, 3125 35

27

36

3 a 4 a 3

b b 3

c c 2

d d 2

e e 3

f 3 k 4

g 4 l 3

h 3

i 4

j 5

5 a -2 e -5 3 6 a 2 1 e 2

b -2 f 3 4 b 3 1 f 3

i -2

j -4

c -2 g 2 3 c 2 1 g 5 3 k 2

d -4 h 6 3 d 2 1 h 4 3 l 2

i 3

7 a 1 iv $7200

v $7710 b 6000 at 5.7% p.a., for 5 years 6 a i I $240, $240 III $1200, $1352.90

II $480, $494.40 IV $2400, $3163.39

ii I $240, $243.60 III $1200, $1375.67

II $480, $502.04 IV $2400, $3224.44

iii I $240, $246.71 III $1200, $1395.40

II $480, $508.64 IV $2400, $3277.59

b i 2 c i 3 min 1 8 a 2 3 e 4 15 i 4

ii 32 ii 8 min

iii 260 iii 10 min

iv 21440

b 1

c 3

d 1

f 2

g 9

h

k 4

l -

j -

11 2

6 7 3 2

b compound interest c compound interest

809 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

9 1 cent doubled every second for 30 seconds. Receive 230 cents, which is more than 1 million dollars. 10 a i 1 ii 1 iii 1 b No solutions. If a = 1, then ax = 1 for all values of x. 3 5 2 d e 11 a 2 b 1 c 3 4 4 3 1 1 g h f 3 10 2 12 a i 0.25 ii 0.125 iii 0.001 iv 0.0016 5-2

b i 13 a -4

e -1.5 14 a 1 e i

5 2

1 5

ii

2-4

iii 2-1

b -6

c -5

d

1

3 2

d -

f -2

g 3

h

j 2

k 0

l -2

b

c 20x7 y10

3

2 3

5

2

b x3

c m3

d a5

f

g 72 1 c 2

h

4 43

d

1 7

1 1 2 3 a3 b 3

b 4

b i 3.08 × 10-4 iii 5.68 × 106

3

ii 4 024 000 iv 0.0000981 ii 7.18 × 10-6 iv 1.20 × 108

9 a V = 800 × 1.07t b V = 3000 × 0.82t 10 a $1215.51 b $3519.60 c $5673.46 11 a 3

b 2

e -2

f -3

i 4

j 3

c 1 3 g 2 k -4

d 6 2 3 l 0

h

Extended-response questions √ √ √ √ 1 a 36 15 + 3 45 = 36 15 + 9 5 cm2 √ √ √ b 360 3 + 144 15 + 90 + 36 5 cm3 √ c 4 3+1

xy(x - y)

d i 10 000 cm2 2 a V = 10 000 × 1.065n b i $11 342.25 c 11.0 years

ii 1.6% ii $13 700.87

d i $14 591 ii V = 14 591 × 0.97t iii $12 917; profit of $2917

Multiple-choice questions 1 C

2D

3B

4E

5A

6D 11 C

7 D 12D

8C

9B

10D

Chapter 4 Exercise 4A

Short-answer questions √ √ 1 a 2 6 b 6 2 c √ √ 2 2 f g 5 7 h 3 √ √ √ 2 a 4+7 3 b 2 5+2 7 √ √ d 4 3+2 2 √ e √ 7 g 2 5 h i √ 3 √ 3 a 2 6+4 2 b √ d 6 5 e 6 f √ h 56 - 16 6

3x2 y4

d

27 4b8

1 1 f 10 5 8 a i 3210 iii 0.00759

( )x 4 9

√

b 6

e

1 5 4 a -8 b 22-a √ 5 length = 10 2 cm, width = 10 cm √ √ -3 - 2 + 7 3 6 7 x-y xy √ 8 12 + 8 2 9 x = 3.5

√ √ c 3 6 d 2 14 √ √ 10 + 2 2 g 2

1

7 a 5 c 8

√ b 5 2 √ 5 2 f 8

f

e 10 2 x 2

3

7 a

3y4 2x3

6 a 21 2

b -1

b

e

1

1 2

Problems and challenges 2 a 5

5 a 25y6

iv 5-4

3 f 4

1 3n

√ 6 6√ 3 2 e 4 √ 4 6-3 h 3

4 a

√ 30 2 √ 2 5 5 √ 2 30

√ d 12 6

√ c 5 2 √ f -12 5

0 √ √ 12 5 - 6 c 12 3 - 4 √ -9 g 16 + 6 7

e

2 7

1 a 0.799

b 0.951

c 1.192

d 0.931

e 0.274 2 a sin h

f 11.664 b cos h

g 0.201 c tan h

h 0.999

3 a 1.80 e 22.33

b 2.94 f 12.47

c 3.42

d 2.38

4 a 1.15 e 2.25

b 3.86 f 2.79

c 13.74 g 1.97

d 5.07 h 13.52

i 37.02 5 a 8.55

j 9.30 b 4.26

k 10.17 c 13.06

l 13.11 d 10.04

f 1.52 j 12.12

g 22.38 k 9.80

h 6.28 l 15.20

e 5.55 i 0.06

810 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

c x = 0.20 m, y = 0.11 m 7 a 125 m b 327 m 8 1.85 m 9 22.3 m 10 7.54 m 11 28.5 m 12 26.4 cm 13 a 4.5 cm

1 1866.03 m

14 The student rounded tan 65° too early. 15 a 3.7 b 6.5 c 7.7

2 39 m 3 28.31 m

a iii tan h = b

ii b = c cos h

4 4.3° 5 320 m

c sin h sin h = c cos h cos h v Answers may vary iv tan h =

iii c2

b i a = c sin h ii b = c cos h iv c2 = (c sin h)2 + (c cos h)2 c2

c2 (sin h)2

=

a2

+

b2

10 299 m 11 a 1.45°

Exercise 4B

2 a 23.58°

1 2 b 60°

e 25.84° 3 a tangent 4 a 60° e 52.12°

b

c 0.75 d 5.74°

f 45° b cosine

g 14.48° c sine

h 31.79°

b 45° f 32.74°

c 48.59°

d 30°

5 a a = 60°, h = 30° c a = 53.1°, h = 36.9°

b a = 45°, h = 45° d a = 22.6°, h = 67.4°

e a = 28.1°, h = 61.9° 6 a 44.4°, 45.6°

f a = 53.1°, h = 36.9° b 74.7°, 15.3° d 23.9°, 66.1° f 42.4°, 47.6°

7 70.02° 8 31.1° 9 47.1° 10 a 66.4° 11 a i 45°

b 114.1° ii 33.7°

c 32.0°

b 11.3° 12 a Once one angle is known, the other can be determined by subtracting the known angle from 90°. b a = 63.4°, b = 26.6° x 13 a b tan 45° = = 1 x 45°

14 a i 8.7 cm b i 17.3 cm

√

b 3.44°

c 1.99°

b i 0.77 m c 3.85 m

45°

d sin 45° = √x = √1 cos 45° also equals √1 . 2x 2 2

ii 72°

iii 36°

iv 54°

ii 2.38 m

iii 2.02 m

iv 1.47 m

4B

d 4.05 m e Proof

Exercise 4D 1 a 0° e 180°

b 45° f 225°

c 90° g 270°

d 135° h 315°

2 a 050° e 227°

b 060° f 289°

c 139°

d 162°

3 a 200° 4 a 220°T

b 082° b 305°T

c 335° c 075°T

d 164° d 150°T

5 a 1.7 km 6 a 121°

b 3.6 km b 301°

7 a 3.83 km b 6.21 km 8 a 14.77 cm b 2.6 cm 9 a 217° b 37° 10 a 1.414 km b 1.414 km c 2.914 km 12 10.032 km 13 a i 045°

2x

ii 5 cm ii 20 cm

c Answers may vary. 15 321.1 km/h 16 a i 18°

11 a 1.62 km

c

b 1.50 m

12 Yes 13 89.12 m

c 11.31°

c 58.3°, 31.7° e 82.9°, 7.1°

6 1509.53 m 7 32° 8 a 1.17 m 9 8.69 cm

c2 (cos h)2

= + Â 1 = (sin h)2 + (cos h)2

1 a 60°

√ 3 2

Exercise 4C

b 8.5 mm

16 a i a = c sin h

iv

Answers

14 a h = 30° b a = 60° √ c 3 √ 1 1 3 d i ii iii 2 2 2 √ √ 3 v vi 3 3 √ (√ ) 1 3 3+1 e AB = x + x= x 2 2 2

6 a x = 2.5 cm, y = 4.33 cm b x = 12.26 cm, y = 6.11 cm

b i 296.6°

b 5.92 km

c 2.16 km

ii 236.3°

iii 26.6°

iv 315°

ii 116.6°

iii 101.3°

iv 246.8°

811 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

8 131.0 m 9 a ÒABC = 80°, ÒACB = 40°

b 122 km

c i 45.2° ii 7.6 km 15 a 229.7°, 18.2 km b 55.1°, 12.3 km

10 a ÒABC = 80° 11 a 147.5° b 102.8°

c 53.9 km d 100.5°

16 a 212.98 m b i 99.32 m

e 123.9° f 137.7° 12 Impossible to find h as such a triangle does not exist.

14 a i 2.5 km b i 4.33 km

ii 2.82 km ii 1.03 km

iii 5.32 km iii 5.36 km

ii 69.20 m

iii 30.11 m

13 37.6° or 142.4° 14 a 59.4° or 120.6°

17 a 38.30 km b 57.86 km c 33.50° 18 a 4.34 km b 2.07 km c 4.81 km

C 2√2

3

2

B

B

b

Exercise 4E 1 a

3

35°

A

b

B D

2√3

A

c 35.3°

2

C

d A triangle can only have one obtuse angle. C

e

2

2√2

A

31.3°

10 120° A

C

35°

120.6°

59.4° C 2

c 31.3°

2

A

b 61.3 km c 126.1°

6

B

Exercise 4G

d 45° 2 61.4°

1 a c2 = 32 + 42 - 2 × 3 × 4 × cos 105°

3 a 37.609 m b 45.47° 4 a 57.409 m b 57.91°

b 72 = 52 + 92 - 2 × 5 × 9 × cos h 2 a 9.6 b 1.5 c 100.3°

5 a i 26.57° b 10.14°

3 a 16.07 cm b 8.85 m e 2.86 km f 8.14 m

c 14.78 cm d 4.56 m

4 a 81.79° e 92.20°

c 64.62°

6 a 7.31 m 7 138.56 m 8 a i 2.25 m b 40.97°

ii 11.18 cm b 6.87 m ii 2.59 m

d 61.20°

5 310 m 6 32.2°, 49.6°, 98.2° 7 a 145.9° b 208.2°

c 3.43 m 9 a i 1.331 km ii 1.677 km b 0.346 km

8 383 km 9 7.76 m

10 a camera C b 609.07 m 11 a 5.5 m b 34.5° c 34.7° 12 a 45° 13 a i 1.55

b 104.48° f 46.83°

d 36.2°

b 1.41 units c 35.26° ii 1.27 iii 2.82

d 0.2° d 1.73 units

b 34.34° 14 22°

c 2.5

3 a 50.3°

b 39.5°

c 29.2°

4 a 7.9 e 8.4

b 16.5 f 22.7

c 19.1

5 a 38.0° e 47.5°

b 51.5° f 48.1°

c 28.8°

1

6 a 1.367 km b 74° 7 27.0°

c 2.089 km

d cosine rule

11 Obtuse, as cos of an obtuse angle gives a negative result. a2 + b2 - c2 12 a cos c = b 121.9° 2ab 13 a AP = b - x b a2 = x2 + h2 c c2 = h2 + (b - x)2 d c2 = a2 - x2 + (b - x)2 = a2 + b2 - 2bx x e cos C = a f x = a cos C substitute into part d.

Exercise 4F a b c = = sin A sin B sin C 2 a 1.9 b 3.6

10 a cosine rule b sine rule c sine rule e sine rule f cosine rule

Progress quiz d 9.2 d 44.3°

1 a 12.58

b 38.14

2 a 39° b 58° √ √ 3 a i 80 or 4 5 √ 4 5 iii √ or 5 80 b 26.6°

c 15.44

d 6.59

c 52°

√ 8 2 5 ii √ or 5 80

812 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

6 A 060° B 150° 7 a 13.65 km b 048.4°

C 288°

4 a 0.139 e -0.799

b 0.995 f -0.259

c -0.530 g 0.777

d -0.574 h -0.087

i 0.900 5 a 140°

j -1.036 b 115°

k 0.900 c 155°

l -0.424 d 99°

f 172° b 86°

c 70°

d 9°

8 a 36.77 m √ 9 a 8 cm

b 61° b 35°

e 143° 6 a 30°

10 a 8.7 11 9.5

b 66.7°

e 21° f 37° 7 a quadrant 2, sin h positive, cos h negative, tan h negative b quadrant 4, sin h negative, cos h positive, tan h negative c quadrant 3, sin h negative, cos h negative, tan h positive

Exercise 4H c 48.0

d quadrant 1, sin h positive, cos h positive, tan h positive e quadrant 4, sin h negative, cos h positive, tan h negative

2 a a b h 3 a 56.44° or 123.56°

c b b 45.58° or 134.42°

f quadrant 2, sin h positive, cos h negative, tan h negative g quadrant 3, sin h negative, cos h negative, tan h positive

c 58.05° or 121.95° 4 a 4.4 cm2 b 26.4 m2

c 0.9 km2

h quadrant 3, sin h negative, cos h negative, tan h positive i quadrant 3, sin h negative, cos h negative, tan h positive

1 a 3.7

b 28.8

d 13.7 m2

e 318.4 m2 f 76.2 cm2 5 a 11.9 cm2 b 105.6 m2 c 1.6 km2 6 a x = 5.7 e x = 10.6

b x = 7.9 f x = 1.3

7 a 59.09 cm2 8 a 35.03 cm2

c x = 9.1

j quadrant 1, sin h positive, cos h positive, tan h positive k quadrant 4, sin h negative, cos h positive, tan h negative d x = 18.2

b 1.56 mm2 b 51.68 m2

c 361.25 km2 c 6.37 km2

9 a 965.88 m2 b 214.66 m2 c 0.72 km2 10 a 17.3 m2 b 47.2 cm2 c 151.4 km2 √ 3 2 1 11 a Area = ab sin h b Area = a2 sin 60° = a 4 2 1 1 c Area = a2 sin(180° - 2h) = a2 sin 2h 2 2 12 a i 129.9 cm2 ii 129.9 cm2 b They are equal because sin 60° and sin 120° are equal. c Same side lengths with included angle 140°. 13 a 65.4°, 114.6° b B

65.4° 8m

14 a i 540° iv 8.09 cm

11 m

114.6°

C

A

ii 108° v 72°, 36°

C 8m iii 11.89 cm2 vi 19.24 cm2 vii 43.0 cm2

b 65.0 cm2 c Answers may vary.

Exercise 4I 1 a quadrant 1 b quadrant 3 c quadrant 4 d quadrant 2 2 a quadrants 1 and 2 c quadrants 2 and 3

b quadrants 2 and 4 d quadrants 1 and 4

e quadrants 1 and 3

f quadrants 3 and 4

3

h 0° sin h 0 cos h tan h

1 0

l quadrant 2, sin h positive, cos h negative, tan h negative 8 a -sin 80° b cos 60° c tan 40° d sin 40° e -cos 55° i tan 47° 9 a 30° e 71°

90° 1

180° 0

270° -1

360° 0

0 undefined

-1 0

0 undefined

1 0

f -tan 45° j sin 68°

g -sin 15° k cos 66°

h -cos 58° l -tan 57°

b 60° f 76°

c 24° g 50°

d 40° h 25°

b 47° f 62°

c 34° g 14°

d 9° h 58°

i 82° 10 a 42° e 33°

11 a 0 < h < 90° c 270° < h < 360°

b 90° < h < 180° d 180° < h < 270°

4H

12 h 150° 315° 350° 195° 235° 140° 100° 35° 55° 2 13 a quadrant 4

B

11 m A

Answers

4 63° 5 23.84 m

b quadrant 1

c quadrant 2

d quadrant 2 e quadrant 1 f quadrant 3 sin h 14 As tan h = and both sin h and cos h are negative over cos h this range, tan h is positive in the third quadrant. sin h 15 As tan h = and cos h = 0 at 90° and 270°, the value of cos h sin h and, hence, tan h is undefined at these values. cos h 16 a true b true c false d true e true i false

f false j true

17 a i 0.17

ii 0.17

g true k true iii 0.59

h false l false iv 0.59

v 0.99

vi 0.99 vii 0.37 viii 0.37 b sin a = cos b when a + b = 90°. c i 90° - h d i 70° e i 90° - h √ 2 5 f 5

ii 90° - h ii 5° b ii c

iii 19° b iii c

iv 38°

813 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

18 a i Proof ii True for these values. b i sin 60° = cos 30° = 0.866,

9

sin 80° = cos 10° = 0.985, sin 110° = cos(-20°) = 0.940, sin 195° = cos(-105°) = -0.259 ii Their values are the same. iii They add to 90°.

10

iv sin h = cos(90° - h) v True for these values. 11

c Answers may vary

Exercise 4J

12

1

Degrees 360° 180° 90° 60° 45° 30° 15° πc πc πc πc π c Radians 2π c πc 2 3 4 6 12 π 180 2 a b 180 π √ 1 1 1 3 a√ b√ c 1 d 3 e 2 2 2 2 √ √ 1 f √1 g 3 h i 3 2 3 2 4 h° 0 30 45 60 90 180 270 360 π π π 6 √4 √3 1 2 3 sin h 0 √2 √2 2 3 2 1 cos h 1 2 2 √2 √ 3 tan h 0 1 3 3 hc

c π 3 g√ 300° 3 6 a 2√ 3 e √2 3 i 2 m -1

c 5π 6 h 165° √

5 a

b

2 2 √ f - 3 √ 3 j 3 n 1

2 2

c 30° d i -

1 2

e 60° √ 3 f i 2 √ 2 8 a 2 e -1 √ 2 i 2

5π 4

b

7 a 45° √ b i -

c

√ 2 ii 2 √ 3 ii 2 ii -

1 2

b 0 √ 3 f 2 j -1

3π 2

π 2

π

1

0

-1

0

0

-1

0

1

undefined

0

undefined

0

0

c

11π 6

d √ 3 √ 2 g 2 1 k 2 o 0 c

c e 135° √

f 210°

2 2√ 2 h 2 1 l 2 p undefined

2π

13

m0 n undefined o 1 p -1 c c c c c c π π π π π π a b c d e f 3 4 6 3 3 6 c c π π h g 12 12√ √ √ 3 2 1 2 1 1 a b c d e f 2√ 2 2 2 2 2 √ 3 g h 3 3 √ √ √ 20 3 a 3 2 b 3 2 c d 14 3 √ e 5 3 f 3 a 45° b 30° c 60° 5π π is the reference angle and is in quadrant 2 with sin h a 6 6 positive. π 2π b is the reference angle and is in quadrant 2 with cos h 3 3 negative. π 11π c is the reference angle and is in quadrant 4 with cos 6 6 h positive. 3π π is in quadrant 2 with tan h d is the reference angle and 4 4 negative. π 3π e is the reference angle and is in quadrant 2 with tan h 4 4 negative π 4π f is the reference angle and is in quadrant 3 with sin h 3 3 negative.

14 a 13 b i 15 a i v b i

d

v

5 13 60°, 120° 45°, 225° c c π 2π , 3 3 c c π 7π , 6 6

12 5 iii 13 12 ii 45°, 135° iii 60°, 300° iv 150°, 210° vi 120°, 300° c c c c c c 5π 7π 3π 5π π 11π ii , iii , iv , 4 4 4 4 6 6 c c 3π 7π vi , 4 4 ii

Exercise 4K 1 a h sinh

0° 0

30° 0.5

60° 0.87

90° 1

120° 0.87

150° 0.5

iii 1 h √ 3 iii 3 √ iii - 3 √ 3 c 2 1 g 2 √ 3 k 3

sinh b

180° 210° 240° 270° 300° 330° 360° 0

-0.5 -0.87 -1 -0.87 -0.5

0

sin θ 1

√ 3 d 2 √ h - 3 l

1 2

O

90°

180°

270°

360°

θ

−1

814 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

h

0° 30° 60° 90° 120°

cos h h

1

0.87 0.5

150°

-0.5 -0.87

0

180° 210° 240° 270° 300° 330° 360°

cos h

-1 -0.87 -0.5

0

0.5 0.87

1

14 a Graph is reflected in the x-axis. b Graph is reflected in the x-axis.

cos θ

b

1

c Graph is dilated and constricted from the x-axis. d Graph is dilated and constricted from the y-axis.

O

90°

180°

270°

360°

θ

1 a 120°, 60° b 8.7 cm

b i maximum = 1, minimum = -1 ii 90°, 270° c i 90° < h < 270° ii 180° < h < 360° ii -0.98 vi 0.26

4 a i 0.82 v -0.17

iv -0.77 viii 0.57

iii 0.87 vii -0.42

b i 37°, 323° ii 53°, 307° iii 73°, 287° iv 84°, 276° v 114°, 246° vi 102°, 258° vii 143°, 217° viii 127°, 233° 5 a i 0.42 ii 0.91

iv -0.77

v 204°, 336° vi 233°, 307° vii 224°, 316° c false

d true

g true k true

h true l true

7 a 110° e 27°

b 60° f 326°

c 350° g 233°

d 260° h 357°

8 a 280° e 136°

b 350° f 213°

c 195° g 24°

d 75° h 161°

9 a 30° e 55°

b 60° f 80°

c 15° g 55°

d 70° h 25°

d 36.9°, 323.1° f 233.1°, 306.9°

g 113.6°, 246.4° i 28.7°, 151.3°

h 49.5°, 310.5°

12 a sin h

0

cos h

1

1 2 1 v 2

1 ii 2 √ 3 vi 2

Essential Mathematics for the Victorian Curriculum Year 10 & 10A

45° √ 2 2 √ 2 2

60° √ 3 2 1 2 √ 2 iii 2 vii0

1 D 6C

2B 7 A

3E 8D

4D 9C

5A 10C

90° 1

b 13.17 b 64.8°

c x = 11.55, y = 5.42

3 6.1 m 4 a A = 115°, B = 315°, C = 250°, D = 030° b i 295° 5 a 98.3 km

ii 070° b 228.8 km c 336.8°

6 a 15.43 m b 52° 7 a i 15.5 cm ii 135.0 cm2

9 a 52.6° 10 a 12.5

11 a 0, the maximum value of sin h is 1. b 0, the minimum value of cos h is -1. 30° 1 2 √ 3 2

7 4.33 cm

b i 14.9 cm 8 28.1 m

k 63° l 14° b 44.4°, 135.6°

c 53.1°, 306.9° e 191.5°, 348.5°

0°

b 308°

6 17.93°

1 a 14.74 2 a 45.6°

f true j false

h

4 514 m 5 a 2 hours 9 minutes

Short-answer questions

e false i true

i 36° j 72° 10 a 17.5°, 162.5°

2 225° 3 Use the cosine rule.

Multiple-choice questions iii -0.64

v 0.34 vi -0.82 vii -0.64 viii 0.94 b i 37°, 143° ii 12°, 168° iii 17°, 163° iv 64°, 116° viii 186°, 354° 6 a true b false

e Graph is translated up and down from the x-axis. f Graph is translated left and right from the y-axis.

Problems and Challenges

−1 3 a i maximum = 1, minimum = -1 ii 0°, 180°, 360°

b i

√ 1 1 2 x xi xii2 √2 √ √2 3 3 3 xiv xv xvi 2 2 2 60°, 120° c 30°, 150° d 210°, 330° 150°, 210°

Answers

√ 3 ix √2 2 xiii 2 13 a 45°, 315° b e 120°, 240° f

2 a

ii 111.3 cm2 b 105.4° b 42.8°

11 a i √ sin 60° 3 b i 2 c i negative c π 12 a 3 13 a 4

ii -cos √ 30° 3 ii 2 ii positive c 5π b 4√ b 5 3

14 a i 0.77 b i 53°, 127

ii -0.97 ii 197°, 343° iii no value

c i true

ii true

iii -tan 45° iii -1 iii negative c 150°

iv -sin √ 45° 2 iv 2 iv positive d 300°

iii false

0 iv 0 viii

√ 2 2

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815

Answers

Extended-response questions

8 a 6x2 + 13x + 5 c 10x2 + 41x + 21

waterfall

1 a

N

3 km

entrance 325° b 2.9 km west d i 21.9 m 2 a 33.646°

c 7.7 km ii 38.0°

b 3177.54 m2

e i 65.66°, 114.34°

c 41.00 m

d 61.60 m

ii 80.1 m, 43.1 m

Chapter 5

1 a x2 + 2x 2 a

+ ab

d (y + x)2 3 a 6x x e 2 i -18x

f 6x2 + 5x - 25 i 25x2 - 49

j 14x2 - 34x + 12 l 56x2 - 30x + 4

k 25x2 - 45x + 18 m 4x2 + 20x + 25 d 8

b x2 + 4x + 3

c x2 + 8x + 16 c a2 - b2

b 2x - 5 a-b e 2 b -20x c 2x2 x f g -4x 3 j 7x k 5x

4 a 2x + 10 d -4x + 8 g -10x + 6 j 3x2 - x m -6x2 - 4x p -4x + 16x2 1 3 14 v -2x 9 5 a 2x2 + 3x s -2x -

d -4x2 h -3x

b 3x2 + 27x + 42 d -4x2 - 44x - 72

e 5x2 + 5x - 60 g -3a2 + 15a + 42

f 3x2 + 6x - 45 h -5a2 + 30a + 80

i 4a2 - 36a + 72 k -2y2 + 22y - 48

j 3y2 - 27y + 60 l -6y2 + 42y - 72

m 12x2 + 48x + 45 o -6x2 - 10x + 56

n 18x2 + 12x - 48 p 2x2 + 12x + 18

q 4m2 + 40m + 100 s -3y2 + 30y - 75

r 2a2 - 28a + 98 t 12b2 - 12b + 3

u -12y2 + 72y - 108 11 a 2x2 + 10x + 11

c -5x - 15

e 6x - 3 h -20x - 15

f 12x + 4 i 2x2 + 5x

k 2x - 2x2 n -18x2 + 6x 8 q 4x + 5 3 t -2x + 2 9 2 w x + 6x 4 b 6x2 - 3x c

l 6x - 3x2 o -10x + 10x2 15 r 6x 4 3 u -9x + 8 14 6 x x - x2 5 5 2x2 + 7x

d 8x2 + 7x e 2x2 - 2x f 25x - 12x2 2 2 6 a x + 10x + 16 b x + 7x + 12 c x2 + 12x + 35 d x2 + 5x - 24 2 2 e x + x - 30f x + x - 6 g x2 - 4x - 21 x2

h - 10x + 24 7 a x2 + 10x + 25 x2

b 2x2 + 20x + 44

c - 4y + 5 e -24a - 45

d 2y2 - y - 43 f b2 + 54b + 5

g x2 + 10x + 18 i -4x2 + 36x - 78

h x2 - 14x + 40 j -25x2 - 30x + 5

12 a x2 - 12x + 36 cm2

b x2 + 10x - 200 cm2

a2

13 a (a + b)(a - b) = - ab + ba - b2 = a2 - b2 b (a + b)2 = (a + b)(a + b) = a2 + ab + ba + b2 = a2 + 2ab + b2 2

c (a - b) = (a - b)(a - b) = a2 - ab - ba + b2

l -13x

b 3x - 12

o 49x2 - 14x + 1 e 1 f 2

10 a 2x2 + 14x + 24 c -2x2 - 20x - 32

2y2

Exercise 5A a2

e 20x2 + 2x - 6 g 16x2 - 25 h 4x2 - 81

n 25x2 + 60x + 36 9 a 3 b 3 c 3

5 km

b 12x2 + 23x + 10 d 9x2 - 9x - 10

= a2 - 2ab + b2 2

d (a + b) - (a - b)2 = a2 + ab + ba + b2 - (a2 - ab - ba + b2 ) = 2ab + 2ab = 4ab 14 a 618 b 220 e 1386 f 891 15 a -x2 + 7x

2x2

c + - 15x - 36 e 2x3 - x2 - 63x + 90

d 1664 h 3480 c 4x2 + 12x + 9

b 10a - 28

d 4x + 8 16 a x3 + 6x2 + 11x + 6 x3

c 567 g 3960

b x3 + 11x2 + 38x + 40 d 2x3 - 13x2 + 17x + 12 f 6x3 - 35x2 - 47x - 12

17 a 2ab b (a + b)2 - c2 2 c (a + b) - c2 = 2ab c2 = a2 + 2ab + b2 - 2ab c2 = a2 + b2

x2

i - 13x + 40 b x2 + 14x + 49 x2

c + 12x + 36 e x2 - 16x + 64

d - 6x + 9 f x2 - 20x + 100

g x2 - 16 i 4x2 - 9

h x2 - 81 j 9x2 - 16

k 16x2 - 25

l 64x2 - 49

Exercise 5B 1 a 7 e 2a

b 6 f 3a

c 8 g -5a

d -5 h -3xy

816 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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b 4(x + 5) e -5(x + 6)

c 7(a + b) f -2(2y + 1)

g -3(4a + 1) j x(5x - 2)

h -b(2a + c) k 6b(b - 3)

i x(4x + 1) l 7a(2a - 3)

m 5a(2 - a) p -4y(1 + 2y)

n 6x(2 - 5x) q ab(b - a)

o -x(2 + x) r 2xy(xz - 2)

s -12mn(m + n) 3 a (x - 1)(5 - a)

t 3z2 (2xy - 1) b (x + 2)(b + 3)

c (x + 5)(a - 4) e (x - 4)(x - 2)

d (x + 2)(x + 5) f (x + 1)(3 - x)

g (x + 3)(a + 1) i (x - 6)(1 - x)

h (x - 2)(x - 1)

4 a (x + 3)(x - 3) c (y + 7)(y - 7)

b (x + 5)(x - 5) d (y + 1)(y - 1)

√ √ f (x + 4 + 3 3)(x + 4 - 3 3) √ √ g (x + 1 + 5 3)(x + 1 - 5 3) √ √ h (x - 7 + 2 10)(x - 7 - 2 10) √ √ i ( 3x + 2)( 3x - 2) √ √ j ( 5x + 3)( 5x - 3) √ √ √ √ k ( 7x + 5)( 7x - 5) √ √ √ √ l ( 6x + 11)( 6x - 11) √ √ m ( 2x + 3)( 2x - 3) √ √ n ( 5x + 4)( 5x - 4) √ √ √ √ o ( 3x + 10)( 3x - 10) √ √ √ √ p ( 13x + 7)( 13x - 7) 9 a (x + 2)(y - 3) b (a - 4)(x + 3)

e 3(x + 5y)(x - 5y) g 3(2x + 3y)(2x - 3y)

f 3(a + 10b)(a - 10b) h 7(3a + 4b)(3a - 4b)

c (a + 5)(x - 2) d (y - 4)(x - 3) e (a - 3)(2x - 1) f (2a - 5)(x + 4) √ √ 10 a 5(x + 2 6)(x - 2 6) √ √ b 3(x + 3 6)(x - 3 6) √ √ c 7(x + 3 2)(x - 3 2) √ √ d 2(x + 4 3)(x - 4 3) √ √ e 2(x + 3 + 5)(x + 3 - 5) √ √ f 3(x - 1 + 7)(x - 1 - 7) √ √ g 4(x - 4 + 2 3)(x - 4 - 2 3) √ √ h 5(x + 6 + 3 2)(x + 6 - 3 2)

i (x + 9)(x + 1) k (a + 5)(a - 11)

j (x - 7)(x - 1) l (a - 8)(a - 6)

11 a 60 e 64

m (4x + 5)(2x + 5)

n (y + 7)(3y + 7)

12 a 4 - (x + 2)2 = (2 - (x + 2))

e (2x - 3)(2x + 3) g (1 + 9y)(1 - 9y)

f (6a - 5)(6a + 5) h (10 - 3x)(10 + 3x)

i (5x - 2y)(5x + 2y) k (3a + 7b)(3a - 7b)

j (8x - 5y)(8x + 5y) l (12a - 7b)(12a + 7b)

5 a 2(x + 4)(x - 4) c 6(y + 2)(y - 2)

b 5(x + 3)(x - 3) d 3(y + 4)(y - 4)

o (3x + 11)(7x + 11) p 3x(3x - 10y) √ √ √ √ b (x + 5)(x - 5) 6 a (x + 7)(x - 7) √ √ √ √ c (x + 19)(x - 19) d (x + 21)(x - 21) √ √ √ √ e (x + 14)(x - 14) f (x + 30)(x - 30) √ √ √ √ g (x + 15)(x - 15) h (x + 11)(x - 11) √ √ √ √ i (x + 2 2)(x - 2 2) j (x + 3 2)(x - 3 2) √ √ √ √ k (x + 3 5)(x - 3 5) l (x + 2 5)(x - 2 5) √ √ √ √ m (x + 4 2)(x - 4 2) n (x + 4 3)(x - 4 3) √ √ √ √ o (x + 5 2)(x - 5 2) p (x + 10 2)(x - 10 2) √ √ q (x + 2 + 6)(x + 2 - 6) √ √ r (x + 5 + 10)(x + 5 - 10) √ √ s (x - 3 + 11)(x - 3 - 11) √ √ t (x - 1 + 7)(x - 1 - 7) √ √ u (x - 6 + 15)(x - 6 - 15) √ √ v (x + 4 + 21)(x + 4 - 21) √ √ w (x + 1 + 19)(x + 1 - 19) √ √ x (x - 7 + 26)(x - 7 - 26) 7 a (x + 4)(x + a) b (x + 7)(x + b) c (x - 3)(x + a) e (x + 5)(x - b)

d (x + 2)(x - a) f (x + 3)(x - 4a)

g (x - a)(x - 4) h (x - 2b)(x - 5) i (x 2a)(3x 7) ( √ )( √ ) ( √ )( √ ) 2 2 3 3 8 a x+ xb x+ x3 3 2 2 ( √ )( √ ) ( √ )( √ ) 7 7 5 5 c x+ xd x+ x4 √ 4 6 6 √ e (x - 2 + 2 5)(x - 2 - 2 5)

b 35 f 40

c 69 g 153

Answers

2 a 3(x - 6) d 3(3a - 5)

d 104 h 1260

(2 + (x + 2)) = -x(x + 4) b i -x(x + 6) ii -x(x + 8) iii x(10 - x)

iv (3 - x)(7 + x)

v (8 - x)(6 + x) vi (6 - x)(14 + x) 13 a (x + a)2 = x2 + 2ax + a2 ¢ x2 + a2

5B

b If x = 0, then (x + a)2 = x2 + a2 . Or if a = 0, then (x + a)2 = x2 + a2 is true for all real values of x. 4 1 1 14 - = (9x2 - 4) = (3x + 2)(3x - 2) 9 9 ( 9 )( ) 2 2 4 xor: x2 - = x + 9 3 3 x2

1 1 = (3x + 2) (3x - 2) 3 3 1 = (3x + 2)(3x - 2) 9 15 a -(2x + 5) b -11(2y - 3) c 16(a - 1) e -12s

d 20b f -28y

g (5w + 7x)(-w - x) i 6f(2f + 6j)

h (4d + 3e)(-2d + 7e) j 0

16 a x2 + 5y - y2 + 5x = x2 - y2 + 5x + 5y = (x - y)(x + y) + 5(x + y) = (x + y)(x - y + 5) b i (x + y)(x - y + 7) iii (2x + 3y)(2x - 3y + 2)

ii (x + y)(x - y - 2) iv (5y + 2x)(5y - 2x + 3)

817 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Answers

Exercise 5C 1 a 9, 2

b 10, 2

c 5, -3

d 4, -3

e -8, 3 2 a b c d 3 a e i 4 a

f -10, 3 g -2, -5 h -12, -3 x - 10 =1 Possible answer: x - 10 3(x - 7) Possible answer: =3 x-7 -5(x + 3) Possible answer: = -5 x+3 1 x+4 = Possible answer: 3(x + 4) 3 x x 1 b c 3 d 2 3 5 1 1 2 f g 5 h 3 4 3 x-3 x+1 j x-2 k l 1 - 2x 2 (x + 6)(x + 1) b (x + 3)(x + 2)

c (x + 3)2 e (x + 4)(x + 3)

d (x + 5)(x + 2) f (x + 9)(x + 2)

g (x - 1)(x + 6) i (x + 4)(x - 2)

h (x + 3)(x - 2) j (x - 1)(x + 4)

k (x + 10)(x - 3) m (x - 2)(x - 5)

l (x + 11)(x - 2) n (x - 4)(x - 2)

o (x - 4)(x - 3) r (x - 2)(x - 9)

p (x - 1)2 q (x - 6)(x - 3) s (x - 6)(x + 2)

t (x - 5)(x + 4)

u (x - 7)(x + 2)

v (x - 4)(x + 3) x (x - 5)(x + 2)

w (x + 8)(x - 4)

5 a 2(x + 5)(x + 2) c 2(x + 9)(x + 2)

f 3(x - 5)(x + 2) h -3(x - 2)(x - 1)

i -2(x - 7)(x + 2) k -5(x + 3)(x + 1)

j -4(x - 2)(x + 1) l -7(x - 6)(x - 1)

g 2(x + 11)2 j -3(x - 6)2 7 a x+6 1 d x+7 2 g x-8 5 8 a x+6 4 d x+5 x+2 g x-1

=1 b Answers will vary. a-b 14 a a 2 (a c + b)2 (a - b)

b (x + 3)2 e (x - 9)2

c (x + 6)2 f (x - 10)2

h 3(x - 4)2 k -2(x - 7)2

i 5(x - 5)2 l -4(x + 9)2

b x-3 1 e x-5 x+4 h 3 x-3 b 3 4 e x+7 x-4 h x+6

c x-3 1 f x-6 x-7 i 5 2(x - 1) c x+5 6 f x-2

√ c x-2 3 √ √ f 7x - 5 √ i x-6- 6 3 c x-3 x+3 f x-1

c x-8 x-7 f 5

b 1 - b) d (a + b)(a a2

15 a

3x - 8 (x + 3)(x - 4)

b

7x - 36 (x + 2)(x - 9)

c

x - 12 (x + 4)(x - 4)

d

3x - 23 (x + 3)(x - 3)(x - 5)

e

x - 14 (x - 3)(x + 2)(x - 6)

f

14x + 9 (x + 3)(x + 4)(x - 8)

g

9 - 3x (x + 5)(x - 5)(x - 1)

h

4x + 11 (x - 1)2 (x + 4)

b 3(x + 4)(x + 3) d 5(x - 2)(x + 1)

e 4(x - 5)(x + 1) g -2(x + 4)(x + 3)

6 a (x - 2)2 d (x - 7)2

√ √ 9 a x- 7 b x + 10 d√ 1 e√ 1 5x - 3 3x + 4 √ √ g x+1- 2 h x-3+ 5 2(x + 3) x-3 10 a b 3(x - 5) 4 3 x-2 d e 2 x+3 2 2 t - 5t - 24 t - 49 × = 11 5t - 40 2t2 - 8t - 42 ( t - 7 )(t + 7) ( t - 8 )( t + 3) t + 7 × = 5( t - 8) 2( t - 7 )( t + 3) 10 12 a x - 3 b x+1 4 6 e d x-2 x+5 a2 - ab a2 + 2ab + b2 × 2 13 a 2 a + ab a - b2  2   (a + b)  a (a - b) = ×    a (a + b)  (a + b)  (a - b)

Exercise 5D 1

Two numbers which multiply ax 2 + bx + c a × c to give a × c and add to give b 6x2 + 13x + 6 8x2 + 18x + 4

36 32

9 and 4 16 and 2

12x2 + x - 6 10x2 - 11x - 6

-72 -60

-8 and 9 -15 and 4

21x2 - 20x + 4 15x2 - 13x + 2

84 30

-6 and -14 -3 and -10

2 a (x + 2)(x + 5)

b (x + 4)(x + 6)

c (x + 3)(x + 7) e (x - 3)(x - 4)

d (x - 7)(x - 2) f (x - 5)(x + 3)

g (3x - 4)(2x + 1) i (2x - 1)(4x + 3)

h (x - 4)(3x + 2) j (x + 4)(5x - 2)

k (5x + 6)(2x - 3)

l (2x - 1)(6x - 5)

818 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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b (2x + 1)(x + 1) d (3x - 2)(x - 1)

e (2x - 1)(x - 5) g (3x + 1)(x - 4)

f (5x - 3)(x + 1) h (3x + 1)(x - 1)

i (7x - 5)(x + 1) k (3x - 4)(x + 2)

j (2x - 7)(x - 1) l (2x - 3)(x + 4)

m (2x + 1)(x - 5) o (5x - 2)(x - 4)

n (13x + 6)(x - 1) p (4x - 5)(2x - 1)

q (3x - 4)(2x + 3) s (3x + 2)(2x + 3)

r (5x - 2)(2x + 3) t (4x - 1)(x - 1)

u (4x - 5)(2x - 1) w (3x - 2)(2x - 3)

v (2x - 5)(4x - 3) x (3x - 2)(3x + 5)

4 a (6x + 5)(3x + 2) c (7x - 2)(3x + 4)

b (4x + 3)(5x + 6) d (5x - 2)(6x + 5)

e (8x + 3)(5x - 2) g (6x - 5)(4x - 3)

f (7x + 2)(4x - 3) h (9x - 2)(5x - 4)

i (5x - 2)(5x - 8) 5 a 2(3x + 4)(x + 5)

b 3(2x + 3)(x - 4)

c 3(8x + 1)(2x - 1) e 8(2x - 1)(x - 1)

d 4(4x - 5)(2x - 3) f 10(3x - 2)(3x + 5)

g -5(5x + 4)(2x + 3) 6 a 2x - 5 b 2 e d 3x + 2 x+4 g h 3x + 1 3x - 2 k j 5x - 2

3)2

h 3(2x 4x - 1 2 7x - 2 3x - 1 2x + 3 2x + 3 7x + 1

i 5(4x - 1)(x - 1) c 3x - 2 4 f 2x - 3 5x + 4 i 7x - 2 2x - 3 l 4x - 5

11 - 3x (3x + 5)(3x - 5)(3x - 2) 12x + 3 h (5x - 2)(2x - 3)(2x + 7)

f

g

Exercise 5E 1 a 9 e 16 i

b 36 f 25

c 1 25 g 4

d 4 9 h 4

81 4

2 a (x + 2)2

b (x + 4)2

d (x - 6)2 e (x - 3)2 √ √ 3 a (x + 1 + 5)(x + 1 - 5) √ √ b (x + 2 + 7)(x + 2 - 7) √ √ c (x + 4 + 10)(x + 4 - 10) √ √ d (x - 3 + 11)(x - 3 - 11) √ √ e (x - 6 + 22)(x - 6 - 22) √ √ f (x - 5 + 3)(x - 5 - 3) 4 a 9, (x + 3)2 c 4, (x + 2)2 e 25, (x - 5)2

c (x + 5)2 f (x - 9)2

b 36, (x + 6)2 d 16, (x + 4)2 f 1, (x - 1)2

= -(12x2 + 5x - 3)

g 16, (x - 4)2 h 36, (x - 6)2 ( )2 ( )2 25 5 81 9 i , x+ j , x+ 4 2 4 2 ( )2 ( )2 49 7 121 11 k , x+ l , x+ 4 2 4 2 ( )2 ( )2 9 3 49 7 m , xn , x4 2 4 2 ( )2 ( )2 1 1 81 9 o , xp , x4 2 4 2 √ √ 5 a (x + 2 + 3)(x + 2 - 3) √ √ b (x + 3 + 7)(x + 3 - 7) √ √ c (x + 1 + 5)(x + 1 - 5) √ √ d (x + 5 + 29)(x + 5 - 29) √ √ e (x - 4 + 3)(x - 4 - 3) √ √ f (x - 6 + 26)(x - 6 - 26) √ √ g (x - 2 + 7)(x - 2 - 7) √ √ h (x - 4 + 21)(x - 4 - 21)

= -(3x - 1)(4x + 3)

6 a not possible

7 a (3x - 4)(x - 5) b -10 m; the cable is 10 m below the water. 4 c x = or x = 5 3 3x + 2 1-x 3x + 4 b c 8 a x-3 4 3 4x - 3 x+2 d e 125 f 5x + 1 5 2 (4x 5) g 1 h (x - 3)2 9 -12x2 - 5x + 3

= (1 - 3x)(4x + 3) a (3 - 2x)(4x + 5)

b (5 - 2x)(3x + 2)

c (4 - 3x)(4x + 1) e (2 - 7x)(2x - 5)

d (3 - 4x)(2x - 3) f (3 - 5x)(3x + 2)

10 Answers will vary. 9x + 2 11 a (2x - 3)(4x + 1) 16x2 + 5x c (2x - 5)(4x + 1) 8x - 5 e (2x + 1)(2x - 1)(3x - 2)

5x + 15 (3x - 1)(2x + 5) 7x - 12x2 d (3x - 2)(4x - 1) b

2 (2x - 5)(3x - 2)

Answers

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

5E

b not possible √ √ c (x + 4 + 15)(x + 4 - 15) √ √ d (x + 2 + 2)(x + 2 - 2) √ √ e (x + 5 + 22)(x + 5 - 22) √ √ f (x + 2 + 10)(x + 2 - 10) g not possible √ √ h (x - 3 + 3)(x - 3 - 3) √ √ i (x - 6 + 34)(x - 6 - 34) j not possible √ √ k (x - 4 + 17)(x - 4 - 17) l not possible

819 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

√ ) √ )( 3- 5 3+ 5 x+ 2√ ) ( 2√ ) ( 7 + 41 7 - 41 x+ x+ 2 )( 2 ) √ √ ( 5 + 33 5 - 33 x+ x+ 2 )( 2 ) √ √ ( 9 + 93 9 - 93 x+ x+ 2 )( √ √2 ) ( 3+ 7 3- 7 xx2√ ) ( 2√ ) ( 5 + 23 5 - 23 xx2 )( 2 ) √ √ ( 5 + 31 5 - 31 xx2 )( 2 ) √ √ ( 9 + 91 9 - 91 xx2 2 √ √ 2(x + 3 + 5)(x + 3 - 5) √ √ 3(x + 2 + 5)(x + 2 - 5) √ √ 4(x - 1 + 5)(x - 1 - 5) √ √ 3(x - 4 + 14)(x - 4 - 14) √ √ -2(x + 1 + 6)(x + 1 - 6) √ √ -3(x + 5 + 2 6)(x + 5 - 2 6) √ √ -4(x + 2 + 7)(x + 2 - 7) √ √ -2(x - 4 + 3 2)(x - 4 - 3 2) √ √ -3(x - 4 + 11)(x - 4 - 11) √ )( √ ) ( 3+ 5 3- 5 3 x+ x+ 2√ ) ( 2√ ) ( 3 - 37 3 + 37 x+ 5 x+ 2 )( 2 ) √ √ ( 5 + 17 5 - 17 2 xx2 )( 2 ) √ √ ( 7 + 37 7 - 37 4 xx2√ ) ( 2√ ) ( 7 - 57 7 + 57 x+ -3 x + 2 )( 2 ) √ √ ( 7 + 65 7 - 65 -2 x + x+ 2 )( 2 ) √ √ ( 3 + 29 3 - 29 -4 x x2 )( 2 ) √ √ ( 3 - 17 3 + 17 x-3 x 2 )( 2 ) √ √ ( 5 + 41 5 - 41 -2 x x2 2

(

Answers

7 a b c d e f g h 8 a b c d e f g h i 9 a b c d e f g h i

x+

10 a x2 - 2x - 24 = x2 - 2x + (-1)2 - (-1)2 - 24 = (x - 1)2 - 25 = (x - 1 + 5)(x - 1 - 5) = (x + 4)(x - 6) b Using a quadratic trinomial and finding two numbers that multiply to -24 and add to -2. 11 a If the difference of perfect squares is taken, it involves the square root of a negative number.

b i yes v no

ii yes vi yes

iii no vii yes

iv no viii no

c i mÄ4 ii m Ä 9 iii m Ä 25 ( ) 3 12 a 2(x + 4) x 2 √ )( √ ) ( 2 + 13 2 - 13 b 3 x+ x+ 3 3 √ )( √ ) ( 7 + 305 7 - 305 c 4 xx8 8 d Unable to be √ factorised. √ ( )( ) 3 + 41 3 - 41 e -2 x + x+ 4 4 √ )( √ ) ( 7 + 13 7 - 13 f -3 x + x+ 6 6 g Unable to be factorised. √ ) √ )( ( 3 - 41 3 + 41 xh -2 x 4 ( 4 ) 7 i 2(x - 1) x + √2 )( √ ) ( 2 + 19 2 - 19 j 3 x+ x+ 3 ( )3 5 k -2 x + (x - 1) ( 2) 4 l -3 x + (x + 1) 3

Exercise 5F 1 a 0, -1

b 0, 5

e -5, 4 f 1, -1 √ √ i 2 2, -2 2 3 3 l - ,8 4 2 a x2 + 2x - 3 = 0 c x2 - 5x + 6 = 0

c 0, 4 √ √ g 3, - 3 1 7 j ,2 3

d 3, -2 √ √ h 5, - 5 5 2 k ,4 5

b x2 - 3x - 10 = 0 d 5x2 - 2x - 7 = 0

e 3x2 - 14x + 8 = 0 g x2 + x - 4 = 0

f 4x2 + 4x - 3 = 0 h 2x2 - 6x - 5 = 0

i -x2 - 4x + 12 = 0 k 3x2 + 2x + 4 = 0

j x2 - 3x - 2 = 0 l x2 + 3x - 6 = 0

3 a 2 f 2

b 2 g 1

4 a x = 0, 4 d x = 0, 4 √ √ g x = 7, - 7

c 1 h 1

e 2

m x = 2, -2

n x = 3, -3

c x = 0, -2 f x = 0, -2 √ √ i x = 5, - 5 1 l x = 0, 7 o x = 6, -6

5 a x = -2, -1 d x = 5, 2

b x = -3, -2 e x = -6, 2

c x = 2, 4 f x = -5, 3

g x = 5, -4 j x = -2

h x = 8, -3 k x = -5

i x = 4, 8 l x=4

mx = 7

n x = 12

o x = -9

j x = 0, 2

b x = 0, 3 e x = 0, 5 √ √ h x = 11, - 11

d 1 i 1

k x = 0, -5

820 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

d x=2 8 a x = 6, -4 d x = -2, -5

1 7 b x = - ,2 2 5 e x = - ,3 3 3 h x = , -4 7 b x = -1, 11 3 e x = , -2 2 b x = 8, -4 e x = 5, 3

g x = 4, -4 j x = -5

h x = -1, -9 k x=8 2 m x = 3, -1 n x = - , -4 3 9 a x = 12, -7 b x = -5, 14 4 5 e x = - ,2 d x = , -4 2 5 1 g x = -3, 1 h x = 1, 2 10 a i x = 1, -2 ii x = 1, -2

7 2 3 f x = - ,2 5 5 2 i x= , 4 5 c x=3 2 5 f x= , 3 2 c x=3 c x = 5,

f x = 3, -3 i x = 5, -1 l x = 8, -8 1 3 o x = - ,4 2 c x = -9, 2 5 f x = 2, 6 i x = 3, -2

b no difference c 3x2 - 15x - 18 = 3(x2 - 5x - 6) and, as seen in part a, the coefficient of 3 makes no difference when solving. 11 This is a perfect square (x + 8)2 , which only has 1 solution; i.e. x = -8. 12 The student has applied the null factor law incorrectly; i.e. when the product does not equal zero. Correct solution is: x2 - 2x - 8 = 7 x2 - 2x - 15 = 0 (x - 5)(x + 3) = 0 x = 5 or x = -3

d x = 8, -6

e x = -6, -2

g x = 8, -3

h x = 5, -3

1 c x = ,5 2 3 f x = , -4 2 i x=2

j x = 4, -3

k x = 5, -2

l x = -5, 3

13 a x = -2, -1

b x=1

Progress quiz 10x 3 d k2 - 6k + 9 g 5x2 - 35x + 60

1 a -8x2 +

4 a x-3

b √

√

5 a (x + 4 + 13)(x + 4 - 13) √ √ 6 - 10) √ ) c not possible b (x √ )-( ( - 6 + 10)(x 5+3 3 5-3 3 d x+ x+ 2 2 √ √ 6 a x = 0 or x = 5 b x = 10 or x = - 10 c x = 4 or x = -4 e x = -3

d x = 3 or x = 4 f x = -5 or x = 11

g x = 4 or x = -8 i x = 2 or x = 5

h x = 2 or x = -2

7 a (3a + 2)(2a + 5) c (3x - 2)(5x - 4) 2x + 5 8 2x - 3 7 5 9 a x = or x = 2 3 4 1 c x = or x = 5 6

c m2 + 7m + 10

e 9m2 - 4 h 19p + 4

f 8h2 - 6h - 35

2 a 4(a - 5) c (x + 5)(4 - x) e (4a - 11b)(4a + 11b) g (k - 5)(k + 9) √ √ i (x - 2 5)(x + 2 5) k (x + 5)(x + a) 3 a (x - 4)(x + 5) c 3(k - 9)(k + 2)

b -6m(2m - 3) d (a - 9)(a + 9)

b (2m - 3)(4m + 3) d (2k - 7)(3k + 5)

b x = -5 or x =

7 2

Exercise 5G 1 b x+5 c x(x + 5) = 24 d x2 + 5x - 24 = 0, x = -8, 3 e width = 3 m, length = 8 m 2 a width = 6 m, length = 10 m b width = 9 m, length = 7 m c width = 14 mm, length = 11 mm 3 height = 8 cm, base = 6 cm 4 height = 2 m, base = 7 m

5G

5 8 and 9 or -9 and -8 6 12 and 14 7 15 m 8 a 6

b 13

c 14

9 1m 10 father 64, son 8 11 5 cm 12 a 55 b i 7

b 4a2 - 7a

x+5 2

Answers

3 6 a x = - , -4 2 1 d x = , 11 2 4 5 g x = ,3 2 7 a x = -2, -6

ii 13

iii 23

13 a 3.75 m b t = 1 second, 3 seconds c The ball will reach this height both on the way up and on the way down. d t = 0 seconds, 4 seconds e t = 2 seconds f The ball reaches a maximum height of 4 m. g No, 4 metres is the maximum height. When h = 5, there is no

f 5(m - 5)(m + 5) h (x - 3)(x + 1) √ √ j (h + 3 - 7)(h + 3 + 7) l (x - 2m)(4x - 5)

solution. 14 a x = 0, 100

b (a - 3)(a - 7) d (m - 6)2

c x = 2 m or 98 m 15 5 m × 45 m

b The ball starts at the tee (i.e. at ground level) and hits the ground again 100 metres from the tee.

16 150 m × 200 m

821 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

Exercise 5H 1 a 1

b 9 c 100 9 25 h g 4 √ √4 (x + 3)(x - 3) = 0 √ √ (x + 7)(x - 7) = 0 √ √ (x + 10)(x - 10) = 0 √ √ (x + 1 + 5)(x + 1 - 5) = 0 √ √ (x + 3 + 11)(x + 3 - 11) = 0 √ √ (x - 1 + 2)(x - 1 - 2) = 0 √ √ x = 2, - 2 √ √ x = 7, - 7 √ √ x = 10, - 10 √ √ x = 3 - 5, 3 + 5 √ √ x = 4 - 6, 4 + 6 √ √ x = -5 - 14, -5 + 14 √ √ x = -3 - 6, -3 + 6 √ √ x = -2 - 2, -2 + 2 √ √ x = -5 - 10, -5 + 10 √ √ x = -2 - 6, -2 + 6 √ √ x = -4 - 19, -4 + 19 √ √ x = -3 - 14, -3 + 14 √ √ x = 4 - 17, 4 + 17 √ √ x = 6 - 39, 6 + 39 √ √ x = 1 - 17, 1 + 17 √ √ x = 5 - 7, 5 + 7 √ √ x = 3 - 5, 3 + 5 √ √ x = 4 - 7, 4 + 7 √ √ x = -3 - 13, -3 + 13 √ √ x = -10 - 87, -10 + 87 √ √ x = 7 - 55, 7 + 55 √ √ x = -4 - 2 3, -4 + 2 3 √ √ x = -3 - 2 2, -3 + 2 2 √ √ x = 5 - 2 5, 5 + 2 5 √ √ x = 2 - 3 2, 2 + 3 2 √ √ x = 5 - 2 7, 5 + 2 7 √ √ x = -4 - 2 6, -4 + 2 6 √ √ x = 1 - 4 2, 1 + 4 2 √ √ x = -6 - 3 6, -6 + 3 6 √ √ x = -3 - 5 2, -3 + 5 2

d 625

e 4

f 25 2 a b c d e f 3 a b c d e f 4 a b c d e f g h i j k l m n o 5 a b c d e f g h i

6 a 2 f 2

b 2 g 2

k 2 7 a b c d

√ √ 1 + 13 1 - 13 , 2√ 2 √ -5 + 33 -5 - 33 x= , 2 √ √2 7 + 41 7 - 41 x= , 2 2 √ √ 9 + 61 9 - 61 x= , 2√ 2 √ -1 + 17 -1 - 17 , x= 2 √ 2 √ -9 + 3 5 -9 - 3 5 x= , 2 2 √ 3 3 √ x = + 3, - 3 2 2 5 √ 5 √ x = - + 5, - - 5 2 2

e x=

c 0 h 0

l 0 √ √ -5 + 17 -5 - 17 x= , 2√ 2 √ -3 + 5 -3 - 5 x= , 2√ 2 √ -7 + 29 -7 - 29 x= , 2 √ √2 3 + 17 3 - 17 x= , 2 2

f g h i j k l

√ -5 ± 17 8 a No real solution. b x= 2√ √ -9 ± 69 5 ± 17 d x= c x= 2√ 2 √ -5 ± 21 f x=3± 5 e x= 2√ √ -3 ± 29 -5 ± 61 9 a x= b x= 2 2 √ c No real solutions. d x=4± 5 √ e x = -5 ± 2 5 f No real solutions. √ √ -3 + 89 3 + 89 10 width = cm, length = cm 2 2 11 a i 1.5 km b i 0 km or 400 km √ c 200 ± 100 2 km

ii 1.5 km ii 200 km

12 a x2 + 4x + 5 = 0 (x + 2)2 + 1 = 0, no real solutions ( )2 3 3 b x+ = 0, no real solutions 2 4 13 Factorise by quadratic trinomial; i.e. (x + 6)(x - 5) = 0, 6 × (-5) = -30, and 6 + (-5) = -1. Therefore, x = -6, 5. √ √ 14 a x = 3 ± 7 b x = -4 ± 10 √ √ c x = -2 ± 11 d x=1± 6 √ √ e x=4±2 3 f x = -5 ± 2 6 15 a Use the dimensions of rectangle BCDE and ACDF d 0 i 0

e 0 j 2

b 16 a c e

and the √ corresponding side lengths in similar rectangles. 1+ 5 a= 2√ √ 6 x = -1 ± b x = -1 ± 5 2 √ √ 3± 5 x = 4 ± 11 d x= 2√ √ -5 ± 17 -1 ± 13 x= f x= 2 2

822 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

1 a a = 3, b = 2, c = 1 c a = 5, b = 3, c = -2 e a = 2, b = -1, c = -5

c i -2 < k < 2 d i no values

b a = 2, b = 1, c = 4 d a = 4, b = -3, c = 2 f a = -3, b = 4, c = -5

b -31 f -44

c 49

d -23

3 a 1 4 a 2

b 0 b 0

c 2 c 1

d 2

e 2

g 0 √l 2 -3 ± 17 x= 2 √ 7 ± 29 x= 2 x = -1, -4 √ -7 ± 65 x= 8 √ 2 ± 22 x= 3 4 x = - ,1 3 √ x = -2 ± 3 √ x = -3 ± 11

h 0

i 2

j 1

5 a c e g i k 6 a c

7 8

9 10 11

√ e x=2±2 2 √ 1± 7 g x= 2√ 4 ± 31 i x= √ 5 -5 + 105 2 √ 3±2 3 a x= 3√ -5 ± 57 c x= 8√ -2 ± 13 e x= 3 √ 1 ± 11 g x= 5√ 5 ± 19 i x= 6 √ √ 3 + 53 -3 + 53 , 2 √2 6 2 + 10 units 63 cm

b ±3 b x = 1, -2

5 144 cm2

8 x2 - 2x + 2 = (x - 1)2 + 1, as (x - 1)2 Å 0, (x - 1)2 + 1 > 0 (x - y)2 > 0 for all 9 Square area - rectangle area = 4 x and y; hence, square area is greater than rectangle area.

f x = -1, -7 √ -5 ± 37 h x= 6 √ 5 ± 65 j x= 4√ -3 ± 19 l x= 5 √ b x = 3 ± 5√ -3 ± 3 5 d x= 2 √ 4 ± 10 f x= 3√ 3±2 3 h x= 3

10 w : p = 1 : 3; t : q = 1 : 9

Multiple-choice questions 1 D

2B

3C

4A

5B

6D 11 A

7 C 12B

8C

9E

10C

Short-answer questions 1 a -2x + 26

b 3x2 + 11x - 20

25x2

d x2 - 12x + 36 f 12x2 - 23x + 10

c -4 e 7x + 22 √ -2 ± 10 2 √ 5 ± 17 d x= 4 √ f x=1± 6 √ 3 ± 41 h x= 4 b x=

single solution.

ii k = 4 9 ii k = 8

3 a ± 2, ± 1 4 a x = 0, 1 6 25 km/h 7 1.6

d x=4

13 Answers will vary. 14 k = 6 or -6 15 a i k > 4 9 b i k> 8

1 b = -4, c = 1 2 47

√ -7 ± 65 b x= 2

12 When b2 - 4ac = 0, the solution reduces to x =

iii k < 4 9 iii k < 8

iii k > 2, k < -2 iii All values of k.

Problems and challenges

2 a -8 e 41

f 2 k 0

ii ± 2 ii no values

Answers

Exercise 5I

2 a x2 + 4x + 4 c x2 + 3x + 21

b 4x2 + 18x

3 a (x + 7)(x - 7) c (2x + 1)(2x - 1)

b (3x + 4)(3x - 4) d 3(x + 5)(x - 5) √ √ f (x + 11)(x - 11) h (x + 5)(x - 3)

5I

e 2(x + 3)(x - 3) √ √ g -2(x + 2 5)(x - 2 5) √ √ i (x - 3 + 10)(x - 3 - 10) 4 a (x - 6)(x - 2) b (x + 12)(x - 2) c -3(x - 6)(x - 1) 5 a (3x + 2)(x + 5)

-b ; i.e. a 2a

b (2x - 3)(2x + 5)

c (6x + 1)(2x - 3) d (3x - 2)(4x - 5) 2x x-4 6 a b x+3 √ 4 √ √ √ 7 a (x + 4 + 6)(x + 4 - 6) b (x + 5 + 29)(x + 5 - 29) √ √ c (x -3-2 √ 3) ) ( - 3 + 2√ 3)(x )( 3 + 17 3 - 17 d x+ x+ 2 )( 2 ) √ √ ( 5 + 13 5 - 13 e x+ x+ 2 )( 2 ) √ √ ( 7 + 31 7 - 31 f x+ x+ 2 2

823 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

8 a x = 0, -4 c x = 5, -5

b x = 0, 3 d x = 3, 7

e x=4

y

f x = -9, 4 2 5 h x = ,3 2

1 2 1 3 i x = ,9 2 9 a x = 3, -3 c x = 4, -7 g x = -2,

y (1, 1)

1 O

b x = 5, -1 d x = -3, 6

10 length = 8 m, width = 6 m √ 11 a x = -2 ±√ 7 3 ± 17 c x= 2 12 a 1 solution c 0 solutions√ -3 ± 33 13 a x = 2 √ 2 ± 14 c x= 2

b d b d b d

4 ii m = - , c = 2 3

3 a i m = 3, c = -2

2

x

1

−2

O

(3,−2)

−2

√ x=3±2 √ 2 -5 ± 3 5 x= 2 2 solutions 2 solutions √ x=1± 5 √ 1 ± 37 x= 6

b

y

i

y

ii

O

x

3

3 O

−6

Extended-response questions

y

iii

c trench = 4x2 + 54x m2 d Minimum width is 1 m.

O

(1, −2)

−2

8 9 4 a y = -x + 3 b y = x 5 5 5 a a = -3 b a = -4 c a = 1 or a = 7

iii r = 4.97 m; i.e. πr2 + 4πr - 140 = 0.

d a = -4 6 a x = -3, y = -7

Semester review 1

y

8 a

Multiple-choice questions 2C

b x = -2, y = -4

c x = -1, y = 4 d x = 3, y = -5 7 A hot dog costs $3.50 and a soft drink $2.

Linear relations 3D

4E

5C

1 a 3 - 2x 2 a i x = -4 b i x Ä 6,

b 20 ii x = 2

3a - 8 9x - 2 c d 4a (x + 2)(x - 3) iii x = 13 iv x = 2

3 4 5 6 7

3

O

0 1 2 3 4

O

3 2

3

x

−4.5

x

y

c

x x

y

b

Short-answer questions

ii x < 3,

x

1

m2

c i 420 = 3πr2 + 12πr ii 3πr2 + 12πr - 420 = 0

1 A

O

x

3

x

5

y

iv

1 a i 15 + 2x m ii 12 + 2x m b area = 4x2 + 54x + 180 m2

2 a S = 63π b 0.46 m

x

3

O

x

−3

824 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

a CD = 6 cm, chord theorem 2

a x = 3, y = 4

y

b, c

b OA = OD (radii of circle) OB = OC (radii of circle)

intersecting region y = 4x − 8

8.5

AB = DC (chord theorem 2) Â△OAB Ã △OCD (SSS). c OM = 4 cm, area = 12 cm2 d 30.6%

(3, 4) O

Answers

Extended-response question

Extended-response question

2

e ÒBOD = 106.2°

x

17 3

Indices and surds

3x + 2y = 17

Multiple-choice questions −8

1 B

Multiple-choice questions 3C

4E

5C

Short-answer questions 1 a AB = DE (given) AC = DF (given) ÒBAC = 60° = ÒEDF (given). Â△ABC Ã △DEF (SAS). a = 35 (corresponding angles in congruent triangles) b BC = DC (given) AC is common.

4E

5C

e 21 √ 10 2 i √7 √ 2 a 7 5- 7 b 0 √ √ 3 a 2 15 - 4 3 √ d 59√+ 24 6 √ 3 2 2 b 4 a 2 5

√ d √10 5 h 3

√ c - 2-4 √ b 11 5 - 62c 4

√ 2 5-5 5 3b2 10 2 2 2 5 a 24x y b 3a b c 5 a 6 a i 37 200 ii 0.0000049 c

d

2x2 5y3

SR1

3

x = 3 (corresponding sides in congruent triangles)

A

√ c 3 6 √ g 3

b i 7.30 × 10-5 ii 4.73 × 109 3 1 1 3 2 2 7 a i 10 ii 7 x , when x > 0 iii 4x 5

ÒABC = 90° = ÒADC (given) Â△ABC Ã △ADC (RHS). 2

√ b 20 3 √ f 48 3

√ 1 a 3 6

Geometry 2B

3E

Short-answer questions

d 167 500 m2

1 D

2D

D

iv 15 2 √ b i 6 1 8 a 5

√ 5 ii 20 1 b 16

9 a x=3

b x=2

√ 4

√ 4 73 or ( 7)3 1 c 3 d 2 3 1 c x= d 2 2 iii

10 a $2382.03 b $7658.36

3 4 5 6

B C ÒDBC = ÒBDA (alternate angles in parallel lines) ÒBDC = ÒDBA (alternate angles in parallel lines) BD is common. Â△BAD Ã △DCB (AAS). Using congruence, BC = AD and AB = DC, corresponding sides in congruent triangles. a x = 6.75 b x = 2 a x=8 b x=5 c a = 32, b = 65 d x = 40 e a = 55 f a = 90, b = 60, c = 70 a x = 20 b x=8 c a = 63, b = 55 47 29 39 a x= b x= c x= 5 3 5

Extended-response question a V = 80 000(1.08)n b i $86 400 ii $108 839 c 11.91 years d 6% per year

Trigonometry Multiple-choice questions 1 E

2B

3A

4D

5C

Short-answer questions 1 a x = 19.5 2 a i 150°T b i 310°T

b h = 43.8, y = 9.4 ii 330°T ii 130°T

825 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

3 a 32.174 m b 52.2° 4 a x = 9.8 b h = 125.3

Chapter 6

5 95.1° c 6 a 5π 18 7 a tan h

Exercise 6A

b 135°

1 a 40 mm d 0.3 km

b i h = 155° ii √ h = 35° iii h√ = 42° 2 3 1 c i ii iii 2 2 3 8 a ¥ 0.34 b h ¥ 233°, 307 c yes

g 1.27 cm

a 104.3 m

b

8 km 108°

20°

c 1m f 0.297 km

h 510.2 cm

i 3.2 m

1 4 1 d 9 3 a 810 m 4 a 36.6 cm

b

1 2 7 e 24 b 9.4 km b 5.1 cm

c

5 a 21.8 m 6 a 43.98 cm

b 3.2 m b 7.54 m

c x = P - 16.8 c 89.22 mm

2 a

Extended-response question

b 9.6 cm e 8m

3 4 11 f 12 c 180 cm c x = P - 28.6

d 3.46 km

13 km 52°

b r=

7 a 75.40 m

Start

C 2π

c 5.57 m

8 a i 8 + 4π m

ii 20.6 m ii 7.1 m

Short-answer questions

b i 4+π m π c i 2 + km 3 d i 12 + 10π cm 70π e i 10 + mm 9 31π f i 6+ cm 12 9 a 3 b 8.8

1 a 9x2 - 1 b 4x2 - 20x + 25 2 c -x + 30x - 5

f 2.4

2 a (2x - y)(2x + y) c 3(x - 4)(x + 4)

√ √ b (x + 2 + 7)(x + 2 - 7) d (x - 2)(x + 7)

e 3.87 10 57.6 m 11 a 12.25

b 53.03

e 19.77

f 61.70

e (x - 5)2 3 a (3x + 4)(x - 2)

f 2(x - 6)(x - 2) b (3x - 1)(2x + 3)

c 17.242 km d 206° T

Quadratic equations Multiple-choice questions 1 C

2B

3D

4B

5D

c (5x - 4)(2x - 3) 4 a d g 5 a 6 a

b

7 a

Extended-response question a 4x2 + 40x c x=3

b 44 m2 d x = 2.2

ii 43.4 cm ii 34.4 mm ii 14.1 cm

b 10 +

12 a 6π m 13 a i 201 cm

1 x = 0, 3 b x = -4, c x = 0, -5 √ 2√ x = 4, -4 e x = 7, - 7 f x=2 1 x = 8, -3 h x = -2, 3 x = -8, 5 b x = 3, 7 c x = -4, 5 √ √ i (x - 3 + 5)(x - 3 - 5) 2 3, does not factorise further ii (x √ ) ( + 2) +√ )( 3 5 3 5 iii x + x+ + 2 √2 2 2 i x=3± √ 5 ii no solutions -3 ± 5 iii x = √2 √ -3 ± 57 x= b x = 2 ± 10 4

ii 3.0 km

c 0.009

d 2.65

c 1.37

d 62.83

5π m 2

c π + 1 km

ii 1005 cm

b 4974 1 000 000 c πd 2n 14 r = √π 15 π 2x P - 2w 1 16 a l = or P - w b l = 5 - w c 0 < w < 5 2 2 d 0
b 1080°

c 540°

d 1440°

Exercise 6B √ b 11 √ √ e 8=2 2

√ 1 a 55 √ d 2 2 a x2 + y2 = z2 c

2x2

=

√ c 77 √ √ f 50 = 5 2

b a2 + d2 = b2

c2

826 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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b 11.18 m e 0.43 km

c 16.55 km f 77.10 cm

c

e 21π km2 , 65.97 km2

4 a 4.58 m b 7.94 m c 0.63 m d 1.11 cm e 14.60 cm f 0.09 cm √ ii 6.16 5 a i 34 √ √ ii 16.61 b i 80(or 4 5) √ c i 10 ii 7.68 √ ii 13.04 d i 89 6 a no d no

b yes e yes

c no f yes

7 a 2.86 m d 4.59 cm

b 2.11 cm e 0.58 km

c 26.38 m f 1.81 km

26 π m2 , 9.08 m2 9 7 f π mm2 , 2.75 mm2 8 d

6 43.2 m2 7 a c d e f

8 8.3 cm 9 a 13.19 mm

b 13.62 m

c 4.53 cm

d 2.61 m √ 10 a 2 13

e 12.27 km √ b 4 2

f 5.23 cm √ c 181

11 a i 22.4 cm ii 24.5 cm b Investigation required. √ 5 12 cm, using Pythagoras’ theorem given that an angle in a 2 semicircle is 90°. √ √ √ √ 13 a √ 4 5 cm by 2 5√ cm b 3 10 cm by 10 cm 100 100 c cm by 10 cm 101 101 14 a i 5.41 m ii 4.61 m iii 5.70 m iv 8.70 m v 8.91 m b 7.91 m

49 π m2 , 51.31 m2 3

Answers

3 a 5 cm d 1.81 mm

25 π + 25 cm2 , 34.82 cm2 8 104 2 289 π+ m , 8.70 m2 200 25 (3969 - 441π) mm2 , 103.34 mm2 25 81π + 324 km2 , 578.47 km2 99 2 49 πm , 0.52 m2 200 400

8 a 66 m2

b 49 m2

b 27 bags

9 a 100 ha b 200 000 m2 d 2.5 acres 2A 10 a a = -b h 1 b i 3 ii 4.7 iii 0 3 c a triangle

c 0.4 ha

11 a

vi 6.44 m

h

15 Research

1 a πr2

b

h × πr2 360°

c l2

d l×w

1 1 xy, where x and y are the diagonals. f (b + l)h 2 2 1 1 h xy i bh g bh 2 2 1 1 j πr2 k πr2 2 4 2 a 30 cm2 b 2.98 m2 e

c 0.205 km2 e 5 000 000 m2

d 5000 cm2 f 100 m2

g 230 cm2 i 2700 m2

h 53 700 mm2 j 10 000 000 mm2

k 2 200 000 cm2 3 a 25 cm2 d 0.03 mm2 g 1472 m2 j 2.36 km2 4 a 2.88 d 1.05 g 1.26

6C

b Let x be the base of each triangle. 1 1 A = (b - x) × h + xh + xh 2 2

Exercise 6C

(i.e. rectangle and two triangles) A = bh - xh + xh A = bh

C

b

D

E

B

l 0.000 145 km2 b 54.60 m2 c 1.82 km2 e 153.94 m2 h 0.05 mm2

f 75 cm2 i 0.17 km2

k 1.12 m2 b 14.35

l 3.97 cm2 c 1.44

e 1.91 h 0.52

f 8.89 i 5753.63

5 a 9π cm2 , 28.27 cm2

b

A Let x = AC and y = BD. 1 AC bisects BD, hence DE = EB = y. 2

25 π m2 , 39.27 m2 2

827 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Answers

A=

1 1 1 1 ×x× y+ ×x× y 2 2 2 2

b

2 cm

(i.e. area of △ACD plus area of △ABC) 1 1 A = xy + xy 4 4 1 A = xy 2 c Consider the following trapezium.

b 1 2

c

a A = Area 1 + Area 2 A=

1 1 ×a×h+ ×a×h 2 2

1 A = (a + b)h 2 12 a 63.7%

b 78.5%

c 50%

3 cm

d 53.9%

Exercise 6D

1 cm

1 a

b

3 a 90 cm2 d 920 m2

2 cm b 47.82 mm2 e 502.91 m2

c 111.3 cm2 f 168.89 m2

4 a 8.64 cm2 d 688 mm2

b 96 mm2 e 4.74 cm2

c 836.6 m2 f 43.99 m2

5 24.03 m2 6 3880 cm2 7 a 121.3 cm2 c 236.5 8 a 66.2

c

b 10.2 m2

m2 b 17.9

e 2308.7 9 a 144.5 cm2 c 1192.7 10 33.5 m2

d 2437.8 cm2 c 243.1

d 207.3

f 65.0 b 851.3 m2

cm2

d 4170.8 m2

11 a 6x2 b 2(ab + ac + bc) ( )2 1 1 1 1 c π d + πdh + dh d πr2 + 2rh + πrh 2 2 2 2 1 12 a 6π b 5 π 2 13 a 0.79 m b 7.71 m 14 1 cm

2 a

15 a 4πr2

4 cm

c 2rh + πr(h + r)

10 cm

b 2x(x + 2y) h d 2rh + πr(h + r) 180°

Exercise 6E 1 bh 2 √ 2 a 29 cm

1 a

b πr2 √ b 221 m

c πrs √ c 109 cm

828 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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2 cm

e 260.53 cm2

d 20.199 cm

Answers

14 Substitute equation given in Question 13. √ h = r into the √ πr(r + r2 + h2 ) = πr(r + r2 + r2 ) √ = πr(r + 2r2 ) √ = πr(r + 2r) √ = πr2 (1 + 2) as required 15 182.3 cm2 √ √ 16 a 4 26 cm b 306.57 cm2 c 4 2 cm

3 a

f 85%

Progress quiz 1 a 36 cm 2 a i 5

b

b 26.85 cm c 30 cm ii 30 cm

iii 30 cm2

ii 90 cm ii 31.40 cm

iii 180 cm2 iii 50 cm2

b i 41 √ c i 41

2 cm 4 cm

3 a C = 25.13 mm b C = 55.29 mm

A = 50.27 mm2 A = 243.28 mm2

4 a 17.45 cm2 5 a 450

b 29.32 cm2 b 0.00045

6 a 3.86 cm2 7 a 158 cm2

b 42.06 cm2 c 28.54 cm2 2 b 2.12 m c 434.29 cm2

d 175.18 cm2 8 a 58.90 cm2

b 2.5 cm

c 7.07 cm

9 13.75 cm c

Exercise 6F 1 a 80 cm3 2 a 2000 mm3

3 cm

4 a 593.76 mm2 5 a 64 m2

b 0.82 m2 b 105 cm2

c 435.90 km2 c 0.16 m2

6 a 62.83 m2 7 a 10.44 cm

b 5.18 cm2 b 126.7 cm2

c 1960.35 mm2

8 a 25.5 cm 9 a 18.9 cm

b 25.0 cm b 17.8 cm

10 a 6.3 m 11 hat B

b 66.6 m2

12 a 105 cm2 c 163.3 cm2

b 63 cm2 d 299.4 m2

e 502.8 mm2 f 76.6 m2 √ 2 13 Slant height, s = r + h2 , √ so πr(r + s) = πr(r + r2 + h2 )

b 32 m3 b 200 000 cm3

c 108 mm3 c 15 000 000 m3

d 5.7 cm3 g 130 000 cm3

e 0.0283 km3 h 1000 m3

f 0.762 m3 i 2094 mm3

j 2700 mL m 5720 kL

k 0.342 ML n 74.25 L

l 0.035 kL o 18 440 L

3 a 40 cm3 4 a 785.40 m3

b 10 500 m3 b 18.85 cm3

c 259.7 mm3 c 1583.36 m3

5 a 12 cm3 6 a 30 km3

b 1570.8 m3 b 196 cm3

c 2.448 mm3 c 30 m3

e 0.002 m3 h 1357.168 cm3

f 4752.51 cm3 i 24 m3

b 223.17 m3 e 142.36 cm3 √ b 3 3 m3

c 6.81 m3 f 42.85 cm3

d 10 cm3 g 0.157 m3

6F

7 1000 8 480 L 9 a 379.33 cm3 d 716.46 mm3 10 a 27 cm3 11 0.5 cm

12 He needs to use the perpendicular height of the oblique prism instead of 5. 13 V =

h πr2 h 360°

14 yes; 69.3 m3 15 a √1 m b 5.8 m3 2

829 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Answers

Exercise 6G

c 688.13 m2 , 1697.40 m3 d 15.71 mm2 , 5.85 mm3

1 4 cm3 2 15 m3 3 a 10 m3

8 cm3 3 b 585 m3

1 mm3 3 c 50 km3

e 8 cm3

f 0.336 mm3

b 9.38 mm3 e 0.12 m3

c 25 132.74 m3 f 523.60 cm3

b 276 cm3

c 48 m3

e 10.35 m3

f 70.79 m3

c 58

b

4 a 4 cm3 8 d cm3 3 5 a 0.82 m3 d 25.13 m3 6 47 mL 7 a 282.74 m3 d 56.88 mm3 8 4.76 cm

2 3 Wood wasted = volume of cylinder - volume of cone 1 Wood wasted = πr2 h - πr2 h 3 2 2 Wood wasted = πr h 3 2 Wood wasted = of the volume of cylinder 3 1 1 10 a i V = x2 h ii V = πx2 h 3 12 π b 4 11 a 3.7 cm √ 3V b i h = 3V2 ii r = πh πr 12 a Similar triangles are formed so corresponding sides are in the same ratio. 1 b π(r21 h1 - r22 h2 ) 3 c i 18.3 cm3 ii 14.7 cm3

9

Exercise 6H 1 a 314.16 e 91.95 √ 3 2 r= π √ 6 3 r= 3 π

b 60.82 f 1436.76

ii 3.50 cm ii 3.09 cm

8 a 113.10 cm3 c 21 345.1 cm3

iii 0.50 km iii 0.18 mm

b 5654.9 cm3

9 11.5 cm 10 52% 11 a 32.72 cm3 12 1570.8 cm2

b 67.02 cm3

c 0.52 m3

13 a 4 m 14 a 235.62 m2

b 234.6 m3 b 5.94 cm2

c 138.23 mm2

d 94.25 m2 15 a 5.24 m3

e 27.14 m2 b 942.48 m3

f 26.85 cm2 c 10.09 cm3

d 1273.39 cm3 16 a i 523.60 cm3

e 4.76 m3 ii 4188.79 cm3

f 0.74 cm3

iii 14 137.17 cm3 b 61.2 cm √ b 5 5 cm √ 3V b r= 3 4π b 8 times

17 a 5 cm √

c 332.7 cm2

S 4π 19 a 4 times 4 20 V = × πr3 3 d Substitute into r, giving: 2 ( )3 4 d V= ×π 3 2 3 πd 4 1 πd3 V= × = × 8 3 3 2 18 a r =

1 V = πd3 6 4 21 h = r 3√ 3 22 a i 3 4π iv 6 units2

ii

√ 3

iii 1

36π

v 80.6% √ ii x =

3

( )2 4π 3 2 iii 6 r 3

4π r 3

c Proof required. Example: 1 c 4

m3

( 6

b 3.14 c 18145.84 mm2 , 229 847.30 mm3 d 1017.88 cm2 , 3053.63 cm3 e 2.66 km2 , 0.41 km3 m2 , 1.32

d 33.51

7 a i 1.53 cm b i 0.89 m

b i 4πr2

1 1 4 a b 2 8 5 a 50.27 cm2 , 33.51 cm3 m2 , 0.52

c 3.14

e 21.99 m2 , 9.70 m3 f 15.21 km2 , 5.58 km3

m3

f 5.81 6 a 113.10 cm2 , 113.10 cm3 b 201.06 m2 , 268.08 m3

4πr2 )2 4π 3

=

3 r2

2π 1 2 3 3 (4π)3

√

1

=

2π 3 1 83

×

1 63

=

3

π , as 6

required. d They are the same.

Exercise 6I 1 Some examples are 3.35, 3.37, 3.40 and 3.42. 2 a 347 cm 3 6.65

b 3m

830 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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ii 44.5 cm to 45.5 cm ii 6.75 mm to 6.85 mm

c i 1m d i 0.1 kg

ii 11.5 m to 12.5 m ii 15.55 kg to 15.65 kg

e i 0.1 g f i 1m

ii 56.75 g to 56.85 g ii 9.5 m to 10.5 m

g i 1h h i 0.01 m

ii 672.5 h to 673.5 h ii 9.835 m to 9.845 m

i i 0.01 km j i 0.001 km

ii 12.335 km to 12.345 km ii 0.9865 km to 0.9875 km

k i 0.01 L l i 0.01 mL

ii 1.645L to 1.655 L ii 9.025 mL to 9.035 mL

5 a 4.5 m to 5.5 m c 77.5 mm to 78.5 mm

b 7.5 cm to 8.5 cm d 4.5 mL to 5.5 mL

Problems and challenges 1 6 2 1.3 m 3 a As the sphere touches the top, bottom and curved surface, the height of the cylinder is 2r, and the radius of the base is r. So the curved surface area = 2 × π × r × h and h = 2r, therefore this equals 4πr2 , which is equal to the surface area of the sphere. b 67% 4 h = 4r 5 (4 - π)r2 √ 6 2:1

e 1.5 km to 2.5 km g 3.85 kg to 3.95 kg

f 34.15 cm to 34.25 cm h 19.35 kg to 19.45 kg

Multiple-choice questions

i 457.85 L to 457.95 L k 7.875 km to 7.885 km

j 18.645 m to 18.655 m l 5.045 s to 5.055 s

1 D 7 B

6 a $4450 to $4550 b $4495 to $4505 c 4.6 km

d 9.0 km e 990 g 8 a 149.5 cm to 150.5 cm

f 990 g (nearest whole) b 145 cm to 155 cm

c 149.95 cm to 150.05 cm 9 a 24.5 cm to 25.5 cm b 245 cm c 255 cm 10 a 9.15 cm b 9.25 cm c 36.6 cm to 37 cm d 83.7225 11 a 9.195 cm

to 85.5625

cm2

b 9.205 cm c 36.78 cm to 36.82 cm cm2

cm2

d 84.548025 to 84.732025 e Increasing the level of accuracy lowers the difference between the upper and lower limits of any subsequent working. 12 a Different rounding (level of accuracy being used) b Cody used to the nearest kg, Jacinta used to the nearest 100 g and Luke used to the nearest 10 g. c yes 13 a Distances on rural outback properties, distances between towns, length of wires and pipes along roadways b building plans, measuring carpet and wood c giving medicine at home to children, paint mixtures, chemical mixtures by students d buying paint, filling a pool, recording water use 14 a ± 1.8% d ± 0.056% g ± 0.12%

2E 8E

3A 9E

4D 10C

5C 11 D

6A 12E

Short-answer questions

c $4499.50 to $4500.50 7 a 30 m b 15 g

cm2

Answers

4 a i 1 cm b i 0.1 mm

b ± 5.6% e ± 0.28% h ± 0.071%

c ± 0.56% f ± 0.056%

1 a 23 cm

b 2.7 cm2

c 2 600 000 cm3

d 8372 mL 2 a 32 m 7 3 a m π √ 4 a 65

e 0.63825 m2 b 28.6 m

f 3 000 000 cm2 c 20.4 cm

b 15.60 m2

5 a 16.12 m2 d 78.54 cm2

b 216 m2 e 100.43 m2

6 a 4.8 m 7 a i 236 m2

b 25.48 m ii 240 m3

b 8.31 c 38.5 m2 f 46.69 m2

6I

b i 184 cm2 ii 120 cm3 2 c i 1407.43 cm ii 4021.24 cm3 d i 360 cm2 e i 201.06 m2

ii 400 cm3 ii 268.08 m3

f i 282.74 cm2 175 8 a cm 3π 9 a 18 cm 10 12 m

ii 314.16 cm3

11 a i 414.25 cm2 b i 124 m2

ii 535.62 cm3 ii 88 m3

c i 19.67 mm2 12 a i 117.27 cm2

ii 6.11 mm3 ii 84.94 cm3

b i 104 cm2 c i 25.73 cm2

b 17.6 cm √ b 3 61 cm

c 2305.8 cm2

ii 75 cm3 ii 9.67 cm3

13 a 4950π cm3 14 a 7.5 m to 8.5 m

b 1035π cm2 b 10.25 kg to 10.35 kg

c 4.745 L to 4.755 L

831 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Answers

Extended-response questions

y

m3

10 9 8 7 6 5 4 3 2 1

1 a 72 √ b 37 m c 138.7 m2 d 6 L, $120 2 a 100 m √ b 50 2 m c 5000 m2 d 36% e athlete A, 0.01 seconds

−4 −3 −2 −1 O

x

1 2 3 4

Chapter 7 Exercise 7A 1

x

-3

-2

-1

0

1

2

3

y

9

4

1

0

1

4

9

4 a b c d e f

5

2 a maximum d -5, 1

b (-2, 4) e x = -2

3 a i (2, -5), min

ii x = 2

iii -1, 5

b i (2, 0), max c i (2, 5), min

ii x = 2 ii x = 2

iii 2 iv -1 iii no x-intercept iv 7

d i (-3, 0), min e i (2, -2), min

ii x = -3 ii x = 2

iii -3 iii 1, 3

iv 4 iv 6

f i (0, 3), max

ii x = 0

iii -3, 3

iv 3

min

Reflected in the x-axis (yes/no) no

Turning point (0, 0)

min

no

(0, 0)

min

no

(0, 0)

max

yes

(0, 0)

max

yes

(0, 0)

max

yes

(0, 0)

Formula

Max or min

y = 3x2 y = 1 x2 2 y = 2x2 y = -4x2 y = - 1 x2 3 y = -2x2

c 2

y value when x = 1 y=3 y=1 2 y=2

Wider or narrower than y = x 2 narrower wider narrower

y = -4 y = -1 3 y = -2

narrower wider narrower

a

Formula y = (x + 3)2

Turning point (-3, 0)

Axis of symmetry x = -3

y value 9

x value -3

b c

y = (x - 1)2 y = (x - 2)2

(1, 0) (2, 0)

x=1 x=2

1 4

1 2

d

y = (x + 4)2

(-4, 0)

x = -4

16

-4

a

Formula y = x2 + 3

Turning point (0, 3)

y value 3

y value when x = 1 y=4

b c

y = x2 - 1 y = x2 + 2

(0, -1) (0, 2)

-1 2

y=0 y=3

d

y = x2 - 4

(0, -4)

-4

y = -3

6

iv -3

832 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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b x=0 e x=0

c x=0 f x=2

h x = -3 k x=0

i x=0 l x = -4

b (0, 7) e (0, -4)

c (0, 0) f (2, 0)

h (-3, 0) k (0, -16)

i (0, -3) l (-4, 0)

b 7 e -4

c 0 f 4

h -9 k -16

i -3 l -16

10 a viii d iv

b iii e i

c vii f v

g ii 11 a i

h vi

g x = -1 j x=0 8 a (0, 0) d (0, 0) g (-1, 0) j (0, 2) 9 a 0 d 0 g 1 j 2

y

22 20 18 16 14 12 10 8 6 4 2 −5 −4 −3 −2 −1−2O

1 2 3 4 5

−5 −4 −3 −2 −1−2O

−5 −4 −3 −2 −1−2O

1 2 3 4 5

x

1 2 3 4 5

x

7A

y

v

22 20 18 16 14 12 10 8 6 4 2

x

22 20 18 16 14 12 10 8 6 4 2

x

y

ii

1 2 3 4 5

y

iv

22 20 18 16 14 12 10 8 6 4 2 −5 −4 −3 −2 −1−2O

y

iii

Answers

7 a x=0 d x=0

22 20 18 16 14 12 10 8 6 4 2 −5 −4 −3 −2 −1−2O

1 2 3 4 5

x

833 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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y

Answers

vi

y

13 a i

22 20 18 16 14 12 10 8 6 4 2

11 10 9 8 7 6 5 4 3 2 1

−5 −4 −3 −2 −1−2O

1 2 3 4 5

x

b The constant a determines the narrowness of the graph.

−5 −4 −3 −2 −1−1O

8 7 6 5 4 3 2 1

20 18 16 14 12 10 8 6 4 2 −5 −4 −3 −2 −1−2O

x

y

ii

y

12 a i

1 2 3 4 5

1 2 3 4 5

−5 −4 −3 −2 −1−1O −2 −3 −4 −5

x

1 2 3 4 5

x

y

ii

b The constant k determines whether the graph moves up or down from y = x2 .

20 18 16 14 12 10 8 6 4 2 −5 −4 −3 −2 −1−2O

14 Answers could be: a y = x2 - 4 b y = (x - 5)2

c y = x2 + 3

15 a y = x2 + 2 d y = (x - 2)2

b y = -x2 + 2 c y = (x + 1)2 e y = 2x2 f y = -3x2 1 2 2 g y = (x + 1) + 2 h y = (x - 4) - 2 8 16 parabola on its side

y 1 2 3 4 5

x

b The constant h determines whether the graph moves left or right from y = x2 .

4 3 2 1 O −1 −2 −3 −4

1 2 3 4 5 6 7 8 9

x

834 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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y

d

1 a (0, 0)

b (0, 0)

c (0, 3)

d (0, -3)

e (0, 7) i (0, 0)

f (2, 0)

g (-5, 0)

h (0, 0)

2 a 3 f 4

b -3 g -16

c -3 h -25

d -1 i -2

k 6 3 a up

l -63 b right

c left

d down

e down 4 a

f left

g right

h up

1 O −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9 −10

e 1 j 2

y 7 6 5 4 3 2 1

−4 −3 −2 −1−1O −2

1 2 3 4

10 9 8 7 6 5 4 3 2 1

x

y

b

1 O −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7

1 2 3 4

x

10 9 8 7 6 5 4 3 2 1 O −4 −3 −2 −1−1

x

7B

8 7 6 5 4 3 2 1

y

c

1 2 3 4 y

f

−8 −9

−4 −3 −2 −1−1O −2 −3 −4

(1, 1 ) 2 1 2 3 4

x

(1, 3)

O −4 −3 −2 −1−1

(1, 3)

1 2 3 4 5

y

e

(1, 2)

(1, − 1) 3

Answers

Exercise 7B

1 2 3 4

x

(1, −3)

x

835 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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y

Answers

g

1 O

(1, 0)

−4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9 −10

10 9 8 7 6 5 4 3 2 1

x

1 2 3 4

O −3 −2 −1−1

y 1

1 O −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9 −10

x

1 2 3 4 5

k

y

h

y

j

x

1 2 3 4

−6 −5 −4 −3 −2 −1−1O −2 −3 −4 −5 −6 −7 −8 −9 −10

(1, −4)

1

10 9 8 7 6 5 4 3 2 1 O −6 −5 −4 −3 −2 −1−1

x

y

l

y

i

1

1

x

−2 −1−1O −2 −3 −4 −5 −6 −7 −8 −9

1 2 3 4 5 6

x

5 a (-3, 1) d (4, -2)

b (-2, -4) e (3, -5)

c (1, 3) f (2, 2)

g (3, 3) j (2, -5)

h (2, 6) k (-1, -1)

i (-1, 4) l (4, -10)

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10 9 8 7 6 5 4 3 2 1 (−1, 1) −4 −3 −2 −1 O

1 2 3 4

−3 −2 −1−1O

1 2

18 16 14 12 10 8 6 4 2

x

−1−2O

x

1 2 3 4 5 6 7 8 y

7B

7 6 5 4 3 2 1

11 10 9 8 7 6 5 4 3 2 1

−6 −5 −4 −3 −2 −1−1O

x

(4, 1)

f

(−3, 2)

1 2 3 4

y

e

9 8 7 6 5 4 3 2 1 −6 −5 −4 −3 −2 −1−1O (−2, −1) y

11 10 9 8 7 6 5 4 3 2 (1, 2) 1

x

y

b

c

y

d

y

Answers

6 a

1

x

−3 −2 −1−1O −2 −3 −4

1 2 3 4 5

x

(1, − 4)

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y

Answers

g

3 2 1 −3 −2 −1−1O −2 −3 −4 −5 −6 −7

−3 −2 −1−1O −2 −3 −4 −5 −6 −7 −8 −9

x

1 2 3 4 5

x

−1−2O −4 −6 −8 −10 −12 −14 −16 −18 −20

y

i

1 2 3 4 5

1

x

x

1 2 3 4 5 6 7 8 (4, −2)

y

l

1 −7 −6 −5 −4 −3 −2 −1−1O (−3, −2) −2 −3 −4 −5 −6 −7 −8 −9 −10 −11

(2, 1)

y

k

(2, 1)

1 −3 −2 −1−1O −2 −3 −4 −5 −6 −7 −8 −9

1

1 2 3 4 5

y

h

y

j

(1, 3)

(−2, 2)

x

2

−6 −5 −4 −3 −2 −1−2 O 1 2 −4 −6 −8 −10 −12 −14

x

7 a y = -x2 b y = (x + 2)2 c y = x2 - 5 d y = x2 + 4 e y = (x - 1)2 f y = -x2 + 2 g y = -(x + 3)2 h y = (x + 5)2 - 3 i y = (x - 6)2 + 1 8 a y = 6x2 b y = x2 + 4 c y = (x - 3)2 d y = -(x + 2)2 1 e y = x2 2

838 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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y

b

24 22 20 18 17 16 14 12 10 8 6 (−2, 5) 4 2

h y = (x - 1)2 i y = -7x2 9 a maximum b (5, 25) c 0 d 25 m e i 21 m 10 a (1, 0) d (0, -4) g (-4, -1)

ii 21 m iii 0 m b (-2, 0)

c (-3, 0)

e (0, -2) h (-2, 3)

f (0, 5) i (5, 4)

j (-2, 3) k (-3, -5) 11 a translate 3 units right

l (3, -3)

−5 −4 −3 −2 −1−2O

b translate 2 units left c translate 3 units down

4 2

f translate 2 units left and 4 units down g translate 5 units right and 8 units up

−1−2O −4 −6 −8 −10 −12 −14 −16 −18

h reflect in x-axis, translate 3 units left i reflect in x-axis, translate 6 units up

24 22 20 18 16 14 12 10 8 6 4 2 −1−2O

b ah2 + k

y

x

1

y

c

d translate 7 units up e reflect in x-axis

12 a (h, k) 13 a

Answers

f y = -x2 + 2 g y = x2 - 1

(3, 4)

1 2 3 4 5 6 7

x

7B y

d

2

(3, 4) 1 2 3 4 5 6

x

−7 −6 −5 −4 −3 −2 −1−2O (−3,−4) −4 −6 −8 −10 −12 −14 −16 −18 −20 −22

1 2

x

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y

Answers

e

6 4 2

20 18 16 14 12 10 8 8.5 6 4 (3, 4) 2 −3 −2−2O y

f

4 2

2 4 6 8

5

O −4 −3 −2 −1 −2 −4 −6 −8 −10 −12 −14

x

x

y 4 2

x

x

y

k

(−2, 1) 2

(1, 3)

O −4 −3 −2 −1 −2 −4 −6 −8 −10 −12

1 2 3 4

18 16 14 12 10 8 6 4 2.5 2 (1, 2) O −5 −4 −3 −2 −1 1 2 3 4 5 6 7 −2

(3, 4)

g

(1, 3)

y

j

O −4 −2−2 2 4 6 8 10 −4 −0.5 −6 −8 −10 −12 −14

h

y

i

1 2 3 4

O −5 −4 −3 −2 −1 1 2 3 −2 −4 −6 −7 −8 −10 −12 −14

x

y

y

l

O −4 −3 −2 −1 −2 −3 1 2 3 4 (1, −4) −4 −6 −8 −10 −12 −14 −16 −18

x

x

4 2

(2, 3)

O −1 1 2 3 4 5 −2 −4 −6 −8 −10 −12 −13 −14

x

840 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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1 a x = -1, x = 2

b x = 3, x = 4

c x = -1, x = -5 e x = 0, x = 5 √ √ g x =± 5 h x =± 7 2 a x = -2, x = -1

d x = 0, x = 3 f x = 0, x = -2 √ i x =± 2 2 b x = -4, x = 2

c x=4 e x = 0, x = 6

d x = 0, x = 4 f x = 0, x = -5

g x =± 3 √ i x = ± 10

h x =± 5

O

−2

x

8

−16 (3, −25)

b -8 g -9

3 a 2 f 0

y

d

Answers

Exercise 7C

c 16 h -25

4 a (x + 6)(x - 8) c x=1

d 0 i -10

e 0

y

e

b x = -6, x = 8 d (1, -49)

y

5 a

−2 O

8

x

4

−8 O

(1, −9)

x

2 4 (3, −1)

y

f

y

b

12 −3

O

x

7

7C

−21 O

c

2

x

6

(2, −25) y

g

(4, −4) y

7

15 −7

−1 O

x

(−4, −9) −5 −3 (−4, −1)

O

x

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y

Answers

h

y

d

20

30

O

2

x

10

(6, −16) y

6 a

y

e

4

20

O

x

4 5 (4.5, −0.25)

−4

x

−1 O

(−2.5, −2.25)

y

b

x

O −6 −5 (−5.5, −0.25)

y

f

6

12

−1 O

−12 O

2 3 (2.5, −0.25)

x

x

y

c

g

(−6.5, −30.25) y

12

O1

12

x

−2 O

6

x

−12 (6.5, −30.25)

(2, −16)

842 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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y

l

O

−11

2

Answers

y

h

x

−22 x

−1 O 2 −2 (0.5, −2.25) y

i

−2 O

(−4.5, −42.25) y

7 a

x

7

−14

j

x

O −2 (−1, −1)

(2.5, −20.25) y

y

b

7C O −4 1 (−1.5, −6.25)

x

−4

k

x

O

−6

y (−3, −9)

−10

O

−30

(−3.5, −42.25)

3

x

c

y

O

4

x

(2, −4)

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y

Answers

d

y

c

25 O

x

5

O

(2.5, −6.25)

x

5

y

e

y

d

100

x

O

−3

x

y

f

y

9 a

x

O

−7

−3

8 a

O

−10

(−1.5, −2.25)

O

x

3

−9

(−3.5, −12.25) y

y

b

4

−4

O

4

x

x

−2 O y

b

16 −16

−4

O

x

844 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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y

d

Answers

y

c

(1.5, 2.25) O

−2

x

2

O

x

3

−4

10 a (3.5, -4.5) b (3.5, -6.75)c (-3, -3) e (0, -196) f (0, 196) i (1, 8)

g (1, 0)

d (-3, -4) h (1, 0)

y

e

11 a a = -1, b = -3, TP (2, -1) b a = 5, b = -1, TP (-2, -9)

8

c a = 2, b = -6, TP (2, -16) √ √ 12 a x-intercepts: 2, - 2; TP (0, -2) √ √ b x-intercepts: 11, - 11; TP (0, -11) √ √ c x-intercepts: 5 2, -5 2; TP (0, -50)

(1, 9)

−2 O

x

4

y

13 a

9 y

f

−3

O

3

(4, 25)

x 9 −1

y

b

1 −1 O

O

9

x

7C

x

1

14 a = -2, TP (1, 18) 15 Coefficient does not change the x-intercept. 16 a y = x2 - 2x + 1 = (x - 1)2 Only one x-intercept, which is the turning point. b Graph has a minimum (0, 2), therefore its lowest point is 2

y

c

O

units above the x-axis. 17 a x = 4, x = -2

(2, 4)

4

x

b (1, -9), (1, 9)

c Same x-coordinate, y-coordinate is reflected in the x-axis. ( ) b b2 18 a 0 b 0, -b c - ,4 2 19 a y = x(x - 4) b y = x(x - 2) c y = x(x + 6) d y = (x + 3)(x - 3) √ √ e y = (x + 2)(x - 2) f y = (x + 5)(x - 5) g y = (x + 4)(x - 2) h y = (x - 1)(x - 5) i y = (x + 1)(x - 3) k y = -(x + 2)(x - 6)

j y = -x(x - 4) √ √ l y = -(x - 10)(x + 10)

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Answers

Exercise 7D

y

b

2

1 a y = x + 2x - 5 ( )2 ( )2 2 2 -5 = x2 + 2x + 2 2

7

= (x + 1)2 - 6 TP (-1, -6)

−7

b y = x2 + 4x - 1 ( )2 ( )2 4 4 -1 = x2 + 4x + 2 2 = (x + 2)2 - 5 TP (-2, -5)

x

O

(−4, −9)

c y = x2 - 6x + 10 ( )2 ( )2 6 6 + 10 = x2 - 6x + 2 2

15

c x = 3, x = -1 √ f x=6± 5

3 a min (3, 5) d min (-2, -5)

c max (-1, -2) f max (7, 2)

b max (1, 3) e min (-5, 10)

g max (3, 8) h min (3, -7) 4 a 6 b -2 c 7 f -55 j -8

5 a x = 5, x = 1 c x = 9, x = -3 √ e x = 1 ± 10 g x=4

−5 −3 (−4, −1)

h 1 l -5

5

O1

j no x-intercept √ l x = 3 ± 10 √ b x =± 7 - 3 √ d x = -1 ± 7 √ f x = 6 ± 41

6 a x = -1, x = -5 √ c x = -4 ± 21 e no x-intercept 7 a y

y

e

80

(−8, 16) O

4

x

y

f

51

(−7, 2) O O

x

5 (3, −4)

b x = -7, x = -1 √ d x = -2 ± 5 √ f x=5± 3 h x = -6

i no x-intercept √ k x=2± 5

x

O

y

d

d 9

g 3 k 13

y

c

= (x - 3)2 + 1 TP (3, 1) d y = x2 - 3x - 7 ( )2 ( )2 3 3 = x2 - 3x + -7 2 2 ( )2 3 37 = x2 4 3 37 TP ( , - ) 2 4 √ 2 a x =± 3 b x =± 3 √ d x = -5, x = 3 e x =± 2 - 4

e -16 i -5

−1

x

x

(2, −4)

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y

8 a

Answers

y

g

5 3

(2, 1)

O −1 (−2, −1) y

O

x

−3

x b

y

h

15

−1 O (3, 6)

x

O y

i

O

3

−3 (1, −4) y

c

(5, −4)

9

x

−29 −3 O

y

d

y

j

7D

x

(−4, −9)

O

−25

y

e

x

4

(−9, 25) y −14

−4 O

x

16

O

k

x

−2 O

x

4

x

−8 −56 l

y

(1, −9) y

f

(2, 4) −3 O

O

4

x

−15

5

x

(1, −16)

847 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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y

Answers

g

y

d

20

7

−7

(4, 4)

x

O

−1

x

O y

e

(−4, −9) y

h

x −2 + 2√2

O

−2 − 2√2

−4 5

−5

−1

y

f

x

O

(−2, −8)

(−3, −4) y

i

O

−12

1

x

O

3,−5 2 4

g

(−6, −36) y

9 a

−5 + √17 2 −5 − √17 2

1 O −2 + √3

−2 − √3

−

x

3 + √5 2

3 − √5 2

x

y 2 x

O

5 , − 17 2 4

y

h

(−2, −3) y

b

O −3 − √14

−5

−1 O −2 1

x −3 + √14

2

,−9

y

i

x

2 4

(−3, −14) y

c

−

3, 3 2 4

3 O

6

x

(1, 5) 10 a 2

O

x

b 1

f 2 √ 11 a x = -1 ± 6 √ d x = -2 ± 10

c 1

d 0

e 0

b x = 3, 1

c x = 7, x = -1 √ e x = -2 ± 11

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√ 3± 5 2 12 a y = -(x - 12)2 + 7

Answers

( )2 5 33 f y=- x+ + 2 4 y 5 33

f x=

− , 2 4

y (−2, 7)

−5 − √33 2

3 −2 + √7 x

−2 − √7 O

13 a k > 0

b y = -(x - 1)2 + 3

y (1, 3) 2 x 1 + √3

O

1 − √3

b k=0

c k<0 ( )2 ( )2 b b +c 14 x2 + bx + c = x2 + bx + 2 2 ( )2 b2 4c b = x+ + 4 2 4 ( )2 2 - 4c b b = x+ 4 2

15 a y = 4(x + 1)2 - 1

c y = -(x - 3)2 + 5

y

2 −5 + √33 2 x O

y

(3, 5)

3 O 3 − √5

x 3 + √5

−1 2

−3 2

x

O

−4

(−1, −1)

d y = -(x - 4)2 + 8

y

7D

b y = 3(x - 2)2 - 2

y

(4, 8) 10

O

4 − 2√2

4 + 2√2 x

2

2 + √3 x

O 2 −√ 23

−8

(2, −2)

( )2 3 11 e y=- x+ 2 4 y

x

(− 32 , − 114) O −5

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( )2 5 191 g y=6 x+ + 12 24 y

Answers

c y = 2(x + 3)2 - 17

y 17

−3, −√ 2

17

−3 + √ 2

1

9

x

O

− 5 , 191 12 24

x

O ( )2 3 131 h y=5 x+ 10 20 y

(−3, −17) ( )2 1 25 d y=2 x+ 4 8 y

7

−3 2

O

x

1

( )2 7 25 e y=2 x4 8 y

3

O

− 12 5

1 2

3

(74 , − 258)

x

O ( )2 6 36 i 5 x+ 5 5 y

− 1 , − 25 4 8

O

(103 , 131 20 )

x

x − 6 , − 36 5 5

( )2 5 25 j y=7 x+ 7 7 y

f y = 4(x - 1)2 + 16

y − 10 7

20

O

x

(1, 16) − 5 , − 25 7 7

O

x

850 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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2

−

2

−

3 √105 + 2 6

(−2, −13) y

b

x

O

O

−2.29

( )2 5 21 l y = -4 x + 4 4 y 5 21 4

c

O

y

x O

−1.35

x

3.35

−9

Exercise 7E 1 a 2 intercepts √ 2 a -1 ± 2 √ -3 ± 15 d 3 3 a zero 4 a 1 intercept d 2 intercepts g 2 intercepts j 0 intercepts 5 a 3 f -10

x 1.15

(0.25, −3.25)

d

O −1

x

(−1, −5) y

4

5 + √21 4

0.29

−2

−0.65 −3

5 − √21 4

x

O 0.55 −5

− 4.55

4

3 √105 − 2 6

y

7 a

Answers

)2 ( 35 3 + k y = -3 x + 2 4 y − 3 , 35

b 0 intercepts b 2.5, -1

d (1, -5) ( ) 3 1 g - , -5 8 ( 4 ) 1 3 j , -2 4 4

y

e

7E

(1, −11)

−1.08 O b positive

c negative

b no intercepts e no intercepts

c 2 intercepts f 1 intercept

h 2 intercepts k 2 intercepts

i 2 intercepts l 2 intercepts

c -2 h 0

b 5 g 0

6 a (-1, 3)

c 1 intercept √ -1 ± 17 c 4

b (-2, ( -5) ) 3 1 e - ,6 ( 2 4) 3 9 h ,8 16 ( ) 1 1 k - , 3 3

d -4 i -7

5.08

x

−11 (2, −19) e 8

y

f

c (2, ( -1) ) 7 1 f ,5 2 4

O

−3.86

x

0.86

−10

i (0, -9) l (0, 2) g

(−1.5, −16.75) y (1, 11) 8 −0.91

O

2.91

x

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Answers

h

y

(−1, 9)

y

b

7

O

−3.12 y

i

x

1.12

(1, 7)

1 O

x

1 3

y

c

3

−

O

−0.32

2.32

x

y

j

(−0.25, 12.13)

−25 y

d

12

O

O

−2.71

2.21

x

5 3

−25

x

y

e

y

k −

x

O

−11 y

f

x

(2, −4)

O

y

l

x

(2, −3)

O

1, 1 3 3

−0.67

x

O

5 2

(−1, 1) O

x −0.55

−1.45

−16 y

g

−4 −

y

8 a

9

−

3 2

O

1, 2

2

3

y

h

x

x

O

−

2, 2 3 3

2 O

x

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h

2 a

Answers

√ 2 3 3 √ 10 b x = -1 ± 2 √ 10 c x=1± 2 √ -3 ± 15 d x= 2 √ 6 e x=2± 2 √ 30 f x=1± 5 10 y = (x + 1)2 - 6 = x2 + 2x - 5

9 a x=1±

(10, 9)

2.75 O

22

d

b 9m c 22 m

d

3 a

11 a anything with b2 - 4ac > 0 b anything with b2 - 4ac = 0 c anything with b2 - 4ac < 0 12 Number under square root = 0, therefore x = (one solution) √ -b ± b2 - 4c 13 x = 2 b2 14 y = +c 4a ( ) b c 15 x2 + x+ =0 a a ( ) b2 b2 b c x2 + + 2 - 2 + =0 4a 4a a a ) ( b2 b 2 c = 2 x+ 2a 4a a ) ( b 2 b2 - 4ac = x+ 2a √ 4a2

O

−9

-b 2a

9

x

−27 b 18 cm c 27 cm 4 a (100, 20) c

b 0 and 200

h

7F

(100, 20)

b2

b - 4ac =± 2a 4a2 √ b b2 - 4ac x+ =± 2a √ 2a -b ± b2 - 4ac as required x= 2a

x+

h

200

x

d 200 m e 20 m 5 a 2 × length = 20 - 2x

Exercise 7F 1 a

O

length = 10 - x b A = x(10 - x) c 0 < x < 10

(2, 20)

d

y (5, 25)

O O

4

t

10

x

e 25 cm2 f 5 cm by 5 cm

b 20 m c 4 seconds

853 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Answers

6 a 100 - 2x d A

b A = x(100 - 2x)

c 0 < x < 50

11 a 1 m b No, 1 metre is the minimum height the kite falls to. 12 P = x(64 - x) so maximum occurs at x = 32. Maximum product = 32(64 - 32) = 1024

(25, 1250)

13 a A = (20 - 2x)(10 - 2x) c A

x 50 f width = 25 m, length = 50 m

O e 1250 m2 7 a 20 - x

b min x = 0, max x = 5

200

x O 5 d Turning point occurs for an x value greater than 5.

b P = x(20 - x) c P

e 1 cm

(10, 100)

14 a 6 m 15 a y

x 0 20 d i x = 0 or 20 ii x = 10 e 100 8 a (20, 0) b h O

20

40

O

x

b i 2

100

x

(50, −250) ii none

c i (27.6, -200) and (72.4, -200) ii (1.0, -10) and (99.0, -10)

−10

d The highway meets the edge of the river (50 metres along). 1 16 5 m 24

c 40 m d 10 m √ 9 a 6 seconds

Exercise 7G

h

b

b No, the maximum height reached is 4.5 m.

30

O √6 √ c 2 seconds 10 a h

1 a one

b zero, one or two

2 a (2, 12) 3 a

b (-1, -3) y 5 y=x+2 4

t y = x2 − x − 1

3 2 1

7

−5 −4 −3 −2 −1 O 1 −1 −2

O1

7

2

3 4

5

x

−3

t b (-1, 1) and (3, 5) c x2 - x - 1 = x - 2 x2 - 2x - 1 = 2 x2 - 2x - 3 = 0

(4, −9) b i 1 second ii 7 seconds c 9 m below sea level d at 3 and 5 seconds

iii 4 seconds

(x - 3)(x + 1) = 0 x - 3 = 0 or x + 1 = 0 x = 3 or x = -1 When x = 3, y = 3 + 2 = 5 When x = -1, y = -1 + 2 = 1

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b x2 - 5x + 3 = 0

5 a b2 - 4ac < 0 c b2 - 4ac = 0

b b2 - 4ac > 0

6 a (-3, 6) and (2, 6) b (-2, 12) and (6, 12) c no solutions ( ) 1 d (-3, -2) and - , -2 2 ( ) 3 e - ,0 2 f no solutions 7 a x = 0, y = 0 and x = 3, y = 9

1 7 12 a x = -1, y = -2 and x = - , y = 2 4 15 5 b x = , y = - and x = 2, y = -4 2 4 c x = 1, y = 8 and x = 2, y = 7 d x = -6, y =(-14 and ) x = 2, y = 2 1 1 13 a (-1, 4) and ,5 b 212 m 2 2 14 a (3, -4)

b x = 0, y = 0 and x = -2, y = 4

b i c > -4 ii c = -4 iii c < -4 15 a 1 + 4k 1 1 1 ii k = iii k < b i k>4 4 4 16 a Discriminant from resulting equation is less than 0. b kÅ2

c x = -3, y = 9 and x = 6, y = 36 d x = 0, y = 5 and x = 3, y = 8

17 a m = 2 or m = -6 b The tangents are on different sides of the parabola, where

Answers

4 a x2 + 3x + 6 = 0 c x2 + 3x - 12 = 0

one has a positive gradient and the other has a negative gradient.

e x = -6, y = 34 and x = -2, y = 22 f x = -2, y = -3 and x = 3, y = 17

c m > 2 or m < -6

g no solutions h no solutions 9 65 i x = - , y = and x = -1, y = 8 2 2 5 25 j x = - , y = - and x = 3, y = 1 3 3 k x = -3, y = 6

Progress quiz y

1 a

5 4 3 2 1

l x = -1, y = 2 8 a x = -4, y = 16 and x = 2, y = 4 b x = -1, y = 1 and x = 2, y = 4 1 1 c x = -1, y = 1 and x = , y = 3 9 13 1 d x = -2, y = 7 and x = - , y = 2 4 2 16 e x = -2, y = 0 and x = , y = 3 9 f x = -8, y = -55 and x = 2, y = 5

(1, 2)

(0, 0) 1 2 −2 −1−1 y

b

5 4 (0, 3) 3 2 1

9 a i no solutions ii x = -0.7, y = 1.5 and x = 2.7, y = 8.5 iii x = -1.4, y = -2.1 and x = 0.4, y = 3.1 iv x = -2.6,√ y = 8.2 and x =√-0.4, y = 3.8 -1 ± 21 -1 ± 21 b i x= ,y = 2 2 √ √ 3± 5 ii x = ,y = 3 ± 5 2√ √ -1 ± 13 iii x = , y = 1 ± 13 2√ √ -1 ± 17 iv x = , y = ± 17 2 10 a 2 b 0 c 2 d 0 e 1 f 2 11 Yes, the ball will hit the roof. This can be explained in a number of ways. Using the discriminant, we can see that the path of the ball intersects the equation of roof y = 10.6.

x

−3 −2 −1−1O −2

1 2 3

x

y

c

10 9 8 7 6 5 4 3 2 1 −1−1O

(0, 9)

(3, 0) 1 2 3 4 5 6 7

x

855 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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b y = (x - 1)2 - 7

y

Answers

d

y

1 −5 −4 −3 −2 −1−1O 1 −2 (−2, −1) −3 −4 (0, −5) −5 −6 2 y= 3 a

3x2

x

3 − − 1 − √7 1 + √7 x O −3 −2 −1 1 2 3 4 5 −3 (0, −6) −6 −9 (1, −7)

+2

7 i x-intercepts ii y-intercept iii Turning point a △ < 0; no x-intercepts (0, 5) (2, 1)

y 3 2 1

(−1, 0)

b △ > 0; two x-intercepts c △ = 0; one x-intercept

(3, 0)

−2 −1−1O

1 2 3 4

x

y

b

6 5 4 3 2 1

(0, 4)

A

(2, 0) 1 2 3 4 5

(11, 242)

x

4 a = 2, b = -4; y = (x + 2)(x - 4); Turning point is at (1, -9). 5 a Turning point is a maximum at (3, 8). b y-intercept is at (0, -10). c x-intercepts at 5 and 1. 6 a y = (x - 2)2 - 1

y

O 22 c 242 m2 ; 11 m by 22 m

x

10 a x = 5, y = 25 and x = -3, y = 9 b x = 2, y = 3 and x = - 1, y = - 8 3 9 11 0

Exercise 7H

5 4 3 2 1 −1−1O −2

x

9 a A = x(44 - 2x) or A = 44x - 2x2 b

−1−1O

(-3, -16) (-4, 0)

y

8

6 5 (0, 5) 4 3 2 1 0.78 3.22 O −1 −1 1 2 3 4 5 −2 −3 (2, −3) −4

−2 (0, −3) −3 −4 (1, −4) −5

(0, -7) (0, -16)

b f(x) = 9 - x2

1 a f(x) = 8x d f(x) = x(2x - 3)

c f(x) =

e f(x) = 2x + 1

2 x

4)2

1 2 3 4 5 (2, −1)

x

f f(x) = (x +1 2 a true b true c false 3 a yÅ0 e yÅ0

b y>0

4 a function d function

c y>9

e true

d 0ÄyÄ1

b function c function e not a function f function

g not a function h function 5 a 4

d false

b 10

c 28

i not a function 1 e -2 2

d 5

f 3a + 4

856 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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d 230

e 0.176

c The y-axis is the axis of symmetry for the function. 17 a i y

7 a f(0) = 0, f(2) = 8, f(-4) = -16, f(a) = 4a, f(a + 1) = 4a + 4

(−1, 2)

b f(0) = 1, f(2) = -3, f(-4) = -15, f(a) = 1 - a2 , f(a + 1) = -a2 - 2a 1 c f(0) = 1, f(2) = 4, f(-4) = , f(a) = 2a , f(a + 1) = 2a+1 16 1 2 d f(0) = undefined, f(2) = 1, f(-4) = - f(a) = , 2 a 2 f(a + 1) = (a + 1) e f(0) = -12, f(2) = 0, f(-4) = -12, f(a) = a2 + 4a - 12,

(1, 2)

d all real x

e all real x 9 a all real y

f all real x b yÅ0

h x¢0 d all real y

e y>0 10 a i 5

f y>0 g yÄ2 h y¢0 ii -2 iii 3 iv -15 v 5

g all real x c yÅ0

y

ii

4 2

−2

O

x

2

−2 y

iii

vi -4 5 b a = represents the x value of the point where the line 3 graphs intersect. 11 a i false ii false b i false c i false

x

O

f(a + 1) = a2 + 6a - 7 f f(0) = 9, f(2) = 25, f(-4) = 73, f(a) = 4a2 + 9, f(a + 1) = 4a2 + 8a + 13 8 a all real x b all real x c all real x

Answers

c -4

6 a 0 b 2 f 2k3 - k2 + k

6

(1, 6)

4

ii true ii false

2

12 4x + 2h - 3 13 a They all pass the vertical line test, as each x value has only

−2

−4

O

one y value. b vertical lines in the form x = a

−2

c The y value of the vertex is the maximum or minimum value of the parabola and therefore is essential when finding the

−4

2

4

x

7I

range. d i y Å -4 ii y Å -12 1 iii y Ä 1.125 4 iv y Å 1 1 14 a x ¢ 1 b x ¢ - c x ¢ 1 2 15 a x Å 0 b x Å 2 c x Å -2 d x Ä 2 1 16 a f(a) = f(-a) = a2 + 2 a b

b iii c i yÅ0

ii 0 Ä y Ä 4

iii y Å -4

d i 8, 8, 2 ii 34, 18, -2

Exercise 7I y

1

2

y 6

−2

O

2

x

4 −2

2 −4

O

−2

2

−2

4

x

√ 2 a x =± 5 √ d y = ± 11

b x =± 4 √ e y = ± 57

√ c x =± 3 f y =± 2

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Answers

3 a (0, 0) 4 a (0, 0) √

b r √b r = 3 27 ±3 3 = d x =± 2 2 e y

y

12

√ c y =± 5

√10

3

O

−√10 O

−3

x

3

(1, 3)

(−1, −3) −√10

−3

y

13 √ b r=5

5 a (0, 0) d x =± 3

c y =±

y

e

19 2

√6 −

5

2√6 √30 √5 5

5

d x2 + y2 = 2601 f x2 + y2 = 10

g x2 + y2 = 1.21 h x2 + y2 = 0.25 √ √ √ √ 8 a (1, 3), (1,)- (3) √ b ) (-1, 3), (-1, - 3) ( √ 1 15 15 1 c , , ,2 2 2 2 ) ( √ ) (√ 15 1 15 1 ,, ,e (0, -2) d 2 2 2 2 f (2, 0), (-2, 0) 9 a x-intercepts: ± 1, y-intercepts: ± 1 b x-intercepts: ± 4, y-intercepts: ± 4 √ √ c x-intercepts: ± 3, y-intercepts: ± 3 √ √ d x-intercepts: ± 11, y-intercepts: ± 11 √ 10 a r = 2 2 b r=2 c r=3 √ √ √ d r = 10 e r=2 3 f r=2 5 11 y

3

( O

−3

3 √2

3

,

x

2√6 − √30 √5 5

y

14

√5

b r=9 √ d r= 5 √ √ f r = 20 = 2 5 b x2 + y2 = 49

c x2 + y2 = 10 000 e x2 + y2 = 6

√6

−√6

x

−5 6 a r=6 c r = 12 √ e r = 14 7 a x2 + y2 = 4

O

−√6

O

−5

x

√10

3 √2

x

(2, 1) −√5 (−1, −2)

O

√5

x

−√5

√ Chord length = 3 2 units √ 15 a m = ± 3 √ √ b m > 3 or m < - 3 √ √ c - 3<m< 3 16 a D d C

b A e F

c E f B

√ √ 17 a y = ± 16 - x2 = ±√ 42 - x2 √ √ b x = ± 3 - y2 = ± ( 3)2 - y2 18 a Radius of graph is 2, so points are 2 units from (0, 0); i.e. < 2. b Radius of graph is 1, so points are 1 unit from (0, 0); i.e. -1

is the leftmost point, which is not as far as -2. 19 a i y

2

) −2

O

2

x

− 3 ,− 3 √2 √2

−3

858 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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y

ix

5

2√3

O

−5

x

5

O

2√3

y

iii

O

−1

√ b i y = 25 - x2 √ iii x = 4 - y2 √ v y = 3 - x2 √ vii x = 10 - y2 √ ix y = - 18 - x2

y O

−√10

√10

x

−√10 y

v

1 a -2 1 4 y

x y

O

4

x

b

-1 1 2

4 3 2 1

y

vi

6

0

1

2

1

2

4

y = 2x

−2 −1 O

x

O

−6

√ ii y = - 16 - x2 √ iv x = - 1 - y2 √ vi y = - 5 - x2 √ viii x = - 8 - y2

Exercise 7J

4

−4

x

−2√3

x

1

−1 iv

Answers

y

ii

7J

x

1 2

2 a -2 1 9 y

x

−6

y

y

vii

b

√7

−√7

10 8 6 4 2

x

O

y

0

1

2

1

3

9

y = 3x

−2 −1 O

−√7 viii

-1 1 3

x

1 2

3 a x

√5

y1 = 2

√5

O

x

x

y2 = –2x y3 = –2

x

-2 1 4 -1 4 4

-1 1 2 -1 2 2

0

1

2

1

2

4

-1

-2

-4

1

1 2

1 4

−√5

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y

b

Answers

y = 2−x

4 3 2 1

2

1

y=

1 ¢ -a2 a2

4 a a-2 =

x

e F

8 6 4 2

−2x

d -9, -125, -

1 . 32

1 4

a

b

d E

y

17 a

y c (1, 5) (1, 4)

c D

15 y = 1 16 It is the asymptote.

b False since 3-2 =

c 5-3 , 3-2 , 2-1 5

b A

f B 14 Substitute (2, 5) into the equation y = 22 = 4 ¢ 5.

−1 −1O −2 −3 −4

−2

12 x = 2.322 13 a C

y = 2x

(1, 2)

x 1 2 3 −3 −2 −1 O They are the same graph. ( )x ( )x ( )x 1 1 1 b i y= ii iii y = 3 5 10 c i y = 4-x ii y)= 7-x iii y = 11-x ( x 1 1 d = a-1 , thus = (a-1 )x = a-x as required a a (or similar)

1 x

O

Exercise 7K 1 a

y

6

O

-1

-1 2

1 2

1

2

y

-1 2

-1

-2

2

1

1 2

-1 3 -9

1 3 9

1

3

3

1

y

b

(1, −2)

(1, −3)

b

c

a

c y

8 a i (0, 1) iv (2, 9) b i (4, -16) iv (2, ( -4)) 1 c i 1, 4 9 a (2, 4) 10 a 1000 b i 2000 c i 2 years

−2 −1−1O −2

(1, 12 )

2 a

(1, ) 1 3

1 O

2 1

a

(1, −5) b

11 a N = 2t

-2

x

−1

7

x

x

( ) 1 ii -1, 3

-3

-1

y

-1

-3

iii (0, -1) (

iii (0, 1) c (1, -4)

iv

x

y 9 6 3

iii (0, 1)

( ) 1 ii -1, 2

b (2, 9)

x

b

(1, 16 )

ii (-3, 64)

1 2

1,

) 1 4

−3 −2 −1−3O −6 −9

1 2 3

x

d (-3, 8)

ii 8000 ii 4 years b N = 210 = 1024

c 14 seconds

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x -4 -2 -1 - 1 - 1 1 1 1 2 4 2 4 4 2 1 y 1 2 4 8 -8 -4 -2 -1 - 1 2 2

y

d

(1, 4)

y

b

x

O

8 6 4 2

Answers

3 a

(−1, −4)

−4 −3 −2 −1−2O −4 −6 −8

1 2 3 4

x

y

e

(−1, 1) x

O (1, −1)

4 a 1 ÷ 0.1, 1 ÷ 0.01, 1 ÷ 0.001, 1 ÷ 0.00001 1 b x= 100 c 0.099 d 998 5 a

y

f

(−1, 2)

y

O

(1, 1)

(−1, −1)

x (1, −2)

x

O

y

g

y

b

7K

(−1, 3) (1, 2) x

O

x

O

(1, −3)

(−1, −2) y

h

y

c

(−1, 4) O

(1, 3) O

x (1, −4)

x

(−1, −3) 6 a (2, 1) c (-1, -2) ( ) 1 7 a 10, 2 ( ) 5 c -7, 7

( ) 1 4, ( 2 ) 1 d -6, 3 ( ) 5 b -4, 4 ( ) 5 d 9, 9 b

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Answers

8 a (1, 3) b (3, 1) ( ) 1 d - , -6 2 9 a yes b yes ( ) 1 10 a ,2 2 c e g h 11 a c e g 12 a

c

) ( 3 - , -2 2

c no d no ( ) 1 b ,6 (6 ) 1 (-1, -1) d - , -10 ( 10 ) ( ) 1 1 (1, 1), (-1, -1) f - , -2 , ,2 2 )( ) 2 ( √ √ 1 1 √ , 2 -√ , - 2 2 2 ( ) ( ) √ √ √1 , 5 , -√1 , - 5 5 5 ( ) ( ) 2 1 , -3 b - ,4 (3 ) ( 2 ) 1 1 4, d -6, 2 ( 3) ( ) 1 1 (1, -2), (-1, 2) f , -4 , - , 4 √2 √ √2 √ (2, -1), (-2, 1) h ( 2, - 2), (- 2, 2) E b C c D

d B

e A

13 Yes, x = 0 or y = 0. 14 a zero b zero

c infinity

c Increasing distance from home, positive constant gradient, higher constant speed. d Increasing distance from home, positive varying gradient, increasing speed, accelerating. e Increasing distance from home, positive varying gradient, decreasing speed, decelerating. f Decreasing distance from home, negative varying gradient, decreasing speed, decelerating. g Decreasing distance from home, negative varying gradient, increasing speed, accelerating. 4 a i p = 4q b i p = 50q

ii p = 60 iii q = 25 ii p = 750 iii q = 4 72 ii y = 2 iii x = 24 5 a i k = 72, y = x 50 ii y = 0.5 iii x = 0.5 b i k = 50, y = x 6 a Positive variable rate of change, increasing speed, accelerating. b Positive constant rate of change, constant speed. c Positive varying rate of change, decreasing speed, decelerating. d Zero rate of change, stationary.

f F

e Negative varying rate of change, increasing speed, accelerating.

d infinity

f Negative constant rate of change, constant speed. g Negative varying rate of change, decreasing speed,

15 Greater the coefficient, the closer the graph is to the asymptote. √ √ 1± 5 -1 ± 5 16 a i x = ,y = 2√ 2√ ii x = 1 ± 2, y = -1 ± 2 √ √ iii x = -1 ± 2, y = 1 ± 2 b No intersection, D < 0. c y = -x + 2, y = -x - 2

decelerating. 7 a y is increasing at an increasing rate. b y is increasing at a decreasing rate. c y is decreasing at an increasing rate. d y is decreasing at a decreasing rate. 8 a k = $244/tonne b P = 244n d 1175 tonnes 74 9 a C= b $4.93 s 10 a y (km/h)

Exercise 7L 1 a direct proportion

c $2.47

y=x

100

b indirect proportion c indirect proportion

c $33184

(100, 100)

80

d direct proportion e neither f indirect proportion 2 a Straight line with y-intercept; neither direct nor inverse (indirect) proportion. b Straight line starting at (0, 0); direct proportion. c Upward sloping curve so as x increases, y increases; neither direct nor inverse (indirect) proportion. d Hyperbola shape so as x increases, y decreases; inverse

60 40 20 0

20

40

60

80

100

x (km)

(indirect) proportion. 3 a Fixed distance from home, zero gradient, stationary. b Decreasing distance from home, negative constant gradient, lower constant speed.

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y=

Cost per person

200

b P

320 x

(2, 160)

150 100

t

c P

50

(8, 40) 0

c

Answers

b

1

3 4 5 6 No. of people

2

y (lbs)

7

8

200

t

d P

(100, 220) y = 2.2x

150 t

100

13 a

Depth

50

0

20

40

d

60

y=

800 x

100

x (kg) Time

(10, 80)

80

b

60

7L

Depth

Time (min)

80

40 20

(80, 10)

Time c

20

60 40 Words/min

80 Depth

0

11 a decreasing at a decreasing rate b increasing at an increasing rate c increasing at a decreasing rate d decreasing at an increasing rate e decreasing at a constant rate

Time d

f increasing at a constant rate

Depth

12 a P

Time t

863 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Continuous motion means that no breaks in the curve are possible.

Depth

Final deceleration segment needs a curve becoming flatter, showing a decreasing gradient. 15 A & d: School bus; distance increases at an increasing rate (acceleration), then a constant rate (steady speed) and then a decreasing rate (deceleration) becoming a zero rate (stopped).

Time f

B & a: Soccer player; distance increases at a constant rate (steady speed), then a zero rate (stopped) and then at an increasing rate (acceleration). C & c: Motor bike; distance increases at a constant rate

Depth

Answers

e

(steady speed), then at an increasing rate (acceleration), and then at a decreasing rate (deceleration) becoming a zero rate

Time 14 Corrected graphs are shown with a dashed line. a

(stopped). D & b: Rocket booster; distance increases at an increasing rate (upward acceleration), then a decreasing rate (deceleration when detached) becoming zero (fleetingly

Distance

stopped). Distance then decreases at an increasing rate (acceleration towards Earth) and finally distance decreases at a constant rate (steady fall to Earth with parachute). 16 Various solutions; check with your teacher.

Time Vertical line incorrect. Can’t change distance instantaneously. b

Exercise 7M 1 a up e right

Distance

i right 2 a

b down f left

c right g up

d left h down

j left

k down

l up

y 3

O −1

x (3, −1)

Time Graph correct.

y

b

c

Distance

(−2, 3) 3 x

−2 O Time Can’t be in two places simultaneously. Curve must increase in gradient, turn, decrease in gradient.

y

c

2√6 − 3 −3 O

Distance

d

5 (1, −3)

x

−2√6 − 3

Time

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y

i

Answers

y

d

6 (−3, 1) (−3, 2)

x

O −2

x

−1 O

−5

√21 − 3

−√21 − 3 y

e

y

3 a

1 + √5 (−2, 1)

2√2 − 2 O

−2

1 − √5

−2√2 − 2

y = −2 y

b

y

f

x

O −1

x

10 2 4

−2√5

y=1

1 O

O

2√5

−2

x y

c

y

g

x

7M

2√2

−1 O

−4

2

x

x

O

−4

−2√2

y = −5

y

h

2 − √39 5 − 2√15

y

d

5 + 2√15

(2, 5) O

2 + √39

x

1 2

O

y=0

x

865 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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y

Answers

e

y

4 a

8 O

y=0

y=2

2

x

x

O

−1 2

y

f

x=0 2

y

b

O

y=0

x

1

O −1

y

g 3 2

y = −1

x=0

y=1

1 O

x

c

x = −3

y

1 3

y

h

x

O

1

y=0

x

x

O

y

d

y = −3

−3 y

i

O −1 2

O −378 4

2

y=0

x

x y = −4

x=2

866 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

y

i

6

y=6

29 5

2 1 O −2−1

y=1

Answers

y

e

x

O

29 6

x

5

x = −1 x=5

y

f

y=1

4 3

O

x

x

O

y = −3

−4

x=3

y

x

O −3

2

y

c

y=0 O h

x = −4

7M

x

3

5 2

y = −3

−4

y=2

5 3

O

y

b

x=1 g

y

5 a

x

y −8 3 −4

− 4 −3

y

d

x

O y = −1

O −1 4

y=0

x

867 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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y

Answers

e

b

x

O

c

y = −1 −3

8 a b

y

f

1 -1 x-2 1 +3 y= x+1 1 3 y= + x 1 √ 2 √ ) ( √ ) √ ( -3 - 5 1 - 5 -3 + 5 1 + 5 , , , 2 2 2 2 √ √ √ √ ( 5, 3 + 5), (- 5, 3 - 5) √ √ ( ) ( ) -1 - 11 √ -1 + 11 √ , 11 , , - 11 2 2 ( ) 3 6 (1, 2), - , 5 5

7 a y=

c

5

y=5 d

e (-6, 3), (-2, -1)

1 x

O

f (3, 0), (-3, -2) 9 a max x = 5, min x = 1

b max y = 0, min y = -4

10 a (x - 2)2 + (y - 1)2 = 8

b (x + 2)2 + y2 = 25 1 c (x + 5)2 + (y + 3)2 = 18 d y = +1 x-1 1 -1 e y= -1 f y= x+2 x+3 1 = -x would require x2 = -1, which is not 11 a Solving x possible.

x = −1 y

6 a

y=2

2 1

O

x = −2

b

−4

y

O −1

−2

x

−1 2

−1

b Circle has centre (1, -2) and radius 2, so maximum y value on the circle is 0, which is less than 1.

y = −1

x

−2

y

c

−4 3

O

2

3

x y = −2

2

x=3

c Exponential graph rises more quickly than the straight line and this line sits below the curve. 2 1 d Solving -1 = gives a quadaratic with △ < 0, thus x+3 3x no points of intersection. 12 a i (x + 2)2 + (y - 1)2 = 4, C(-2, 1), r = 2 ii (x + 4)2 + (y + 5)2 = 36, C(-4, -5), r = 6 iii (x - 3)2 + (y - 2)2 = 16, C(3, 2), r = 4 √ iv (x - 1)2 + (y + 3)2 = 15, C(1, -3), r = 15 √ v (x + 5)2 + (y + 4)2 = 24, C(-5, -4), r = 2 6 √ vi (x + 3)2 + (y + 3)2 = 18, C(-3, -3), r = 3 2 √ ( )2 ( ) 29 3 29 3 vii x + + (y - 3)2 = , C - , 3 , r = 2 2 4 2 ( )2 ( ) 5 49 5 7 viii x + + (y - 2)2 = , C - , 2 , r = 2 4 2 2√ ( )2 ( )2 ( ) 1 3 3 1 3 3 ix x + y+ = ,C ,- ,r = 2 2 2 2 2 2 ( )2 ( )2 ( ) 5 3 5 25 3 5 x x+ y= ,C , ,r = √ 2 2 2 2 2 2 2 2 b (x + 2) + (y - 3) = -2; radius can’t be negative.

Problems and challenges 2 1 3 1 1 a - ÄxÄ b x > - or x < 3√ 2 4 3 √ 7 - 41 7 + 41 c <x< 2 2

868 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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2 a minimum at (2, 0) c maximum at (-1, -2)

y

b maximum at (0, 5) d minimum at (3, 4)

y

3 a

3 + 2√3 (2, 3) 2 − √7

x 2 + √7

O 3 − 2√3

−2 O −4

3 a b2 - 4ac < 0 b b2 - 4ac = 0 c b2 - 4ac > 0 4 (x - 2)2 + (y + 3)2 = -15 + 9 + 4 = -2, which is 5 6

7

8 9 10 11

y

b

impossible. 1 1 1 b k< c k> a k= 3 √ 3 3 √ a k = ± 20 = ± 2 5 √ √ b k > 2 5 or k < -2 5 √ √ c -2 5 < k < 2 5 a y = -(x + 1)(x - 3) 3 b y = (x + 2)2 - 3 4 c y = x2 - 2x - 3 ( ) 3 73 a = 2, b = -3, c = -8; TP , 4 8 20 √ 4 3

16

−4

−2 O

−8

(0, 8) y5 = 8x − 8

x

O

y

c

y

4

x

(1, −9)

4 a i maximum at (1, -3) iii no x-intercepts

ii -4

y

iv

x

(1, 0)

O

x

2

Answers

2 (x - 2)2 + (y - 3)2 Ä 16

O

x (1, −3)

−4

y5 = 8 − 8x (0, −8)

b i minimum at (-3, -8)

Multiple-choice questions 1 B

2D

3E

4D

5A

6A 11 E

7 C 12 C

8D 13 D

9A 14 B

10 D 15 E

16 A

17 C

Short-answer questions 1 a minimum at (1, -4) c -1 and 3

ii 10

iii -1 and -5 iv

y 10

−1 O

−5

x

b x=1 d -3

(−3, −8)

869 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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y

Answers

5 a

c Only x = 1, All real y. d All real x except x = 0, All real y except y = 0. e All real x, y > -3. f All real x, y > 0.

1 O 2 − √3

5

2 + √3

(2, −3) y

b

O 2

−3 − √17 2

y

13 a

x

O

−5

x

5

−5

x

−3 + √17 2

y

b

− 3 , − 17 2 4

6 a 1

b 0

7 a i 5 iv

ii (2, -3)

c 2

d 0

√7

iii 0.8 and 3.2

y

−√7

O

x

√7

−√7 5 y

14

0.8

O

3.2

x

b i 4

y

iv

− 3 ,− 6 √5 √5

4

O

−3

3 , 25 2 4

O

3 6 , √5 √5

iii -1 and 4

4 −1

(

3

(2, −3) ( ) 3 25 ii , 2 4

x

3

) x

−3 y

15 a

(1, 4) 8 a 100 - x d A

b A = x(100 - x)

c 0 < x < 100

1 (50, 2500)

b

9

10 11 12

x

O

O e 2500 m2 f 50 m by 50 m a x = 2, y = 10 and x = -6, y = 10 b no solutions 1 10 c x = , y = and x = -1, y = 2 3 9 Show b2 - 4ac = 0. a 9 b 24 c 3 a All real x, All real y. b All real x, only y = 4.

100

x

y x

O −1 (1, −3)

d 2k2 - k + 3

870 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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(1, 15 )

1

y

c

O

−

x

x −2.5

y

16 a

5 3

O

−2

y = −3

−3 (1, 2)

x = −2 x

O

Extended-response questions b 0 Ä x Ä 400

1 a (200, 30) c 30 Ä h Ä 80

(−1, −2)

h

d

y

b

(400, 80)

80 (−1, 3)

(200, 30) O x

O

Answers

y

c

x

400

e 400 m

f 30 m

g 80 m 2 a 450 g b i 150 g

(1, −3) ) ( √ √ √ √ 4 ,3 b ( 2, 2 2) and (- 2, -2 2) 17 a 3 720 18 a y = 5x b i k = 72 ii 18 iii 7 19 a y

ii 5.6 g

450

t

O

2 + √3 (−1, 2)

8A

c after 2 years d A

e 9.8 years

−1 O 2 − √3

x

Chapter 8 Exercise 8A

y

b

3.5

O

y=3

x

1 a 2 1 d 2 1 2 a 4 3 d 8 1 3 a 7

b {H, T} 1 e 2 1 b 6 2 e 3 2 b 7

c yes f 1 c

1 4

f 0 c

5 7

d

3 7

871 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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2 5 1 b 2 1 f 5 b 0.43

3 10 1 5 a 10 2 e 5 6 a 0.09

Answers

3 5 1 c 2 3 g 10 c 0.47

b

4 a

7 a 0.62 1 8 a 50 1 9 a 2 5 d 24 6 10 a 25 2 d 5 7 11 a i 10 1 b 10 12 a 59

b 0.03 3 b 10 b

c

d

c 0.97

d 0.38

c

A

B

d

A

B

e

A

B

f

A

B

g

A

B

h

A

B

i

A

B

49 50

3 8

c

1 4

e 1

f 0

1 50 2 e 25 1 iii 20

21 25 4 f 25 1 v 20

b

1 ii 5

1 2 1 d 2 1 h 10 d 0.91

c

c

iv 0

b 4, as

2 a ∅ d ∅

c

g AÉBÉC h AÈBÈC 3 a no b yes c no

13 a e 14 a b

41 of 10 is closest to 4. 100 41 8, as of 20 is closest to 8. 100 1 1 1 1 b c d 4 13 52 2 2 4 12 9 f g h 13 13 13 13 7 15 15; any multiple of 15 is a possibility as 3 and 5 must be

factors. 15 a 625π b i 25π ii 200π 1 8 c i ii 25 25 24 17 v vi 25 25 d No it doesn’t.

iii 400π 16 iii 25 vii 1

9 25 17 viii 25 iv

4 a

b AÈB e EÈF

A 4

c AÉB f WÉZ

B 3

3 0

b i A È B = {2, 5, 8} ii A É B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 7 3 c i ii iii 1 10 10 d No, since A È B ¢ φ . 5 a

A 3

B 2

5 0

Exercise 8B

b i A È B = {2, 13}

1 a

A

B

b

A

B

ii A É B = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} 1 7 1 c i ii iii iv 1 2 10 5 6 a F N

25 10 10 5 b i 25 2 c i 5

ii 5 1 ii 5

iii

1 5

872 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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F

15

B

25 10 5

A ∩ B′ A ∩ B A′ ∩ B

5 b i 25 7 2 c i ii 9 9 8 a B BÌ

ii 5 8 iii 9

v 7 1 c i 8 9 a

1 9

v

A′ ∩ B ′ 16 a

AÌ 6

8

5 7

3 9

8 16

A 1

2 9

iv

A 2

b i 2

(A ∪ B)′

M

E

w

v

c n s

m

ii 6

iii 5

iv 3

vi 8 9 ii 16

vii 13 5 iii 16

viii 16

b i

1 3

2 3

ii

17 a

iii

iv

2 3

v

1 3

L S

1 2

2

I

1

2 2

1 1

3 b 1

b B BÌ

A

AÌ

4 1

2 3

6 4

5

5

10

c i 2

b 10, 12

11 a

C O

c i

7 iv 10 d nothing

ii

6 13

N

Y R

iii

10 13

iv

12 a B BÌ

A 3

AÌ 3

6

4 7

1 4

5 11

A

AÌ

2 2

7 1

9 3

4

8

12

b B BÌ 13 3 14 a 1 - a

b a+b

1 3

iii

13 15

L S

2x 7

2 2 x y

iv

1 15

L = Own State S = Interstate I = Overseas

18 − 2x

b i 4 5 c i 19

I

c 0

8C

ii 10 1 ii 19

iii

7 38

iv

35 38

v

25 38

Exercise 8C

A 9 13

ii

3

D

C P M T E E L M

3 5

18 a

2 iii 5 c a, c, e

ii 3

10 a 4

b i

1 6

5

B 4

B

A

Answers

7 a

4 13

v

3 13

1 a i {4, 5, 6} ii {2, 4, 6} iii {2, 4, 5, 6} iv {4, 6} b No, A È B ¢ φ. 2 c 3 2 a 0.8 b 0.8 c 0.7 d 1 3 0.05 4 a i 13 ii 4 1 3 b i ii 4 4 4 c 13 3 d 52 5 a i {3, 6, 9, 12, 15, 18 } 1 13 b i ii 20 20 7 c 20

iii 1 1 iii 52

ii {2, 3, 5, 7, 11, 13, 17, 19 }

873 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Answers

5 24 b 0.2

1 8 7 a 0.1

6 a

b

8 a 0.3 3 9 a 8 4 10 a 13 49 e 52 11 a 0.4

b 0.1 5 b 32 4 b 13 10 f 13 b 0.45

7 a

V 3

P 2

4 7 13 10 g 13

7 13 25 h 26

c

b 4

d

1 d 4

2 c 5

8 a

A

C

14

6

14 Pr(A É B É C) = Pr(A) + Pr(B) + Pr(C)Pr(A È B) - Pr(A È C) - Pr(B È C) + Pr(A È B È C) 1 3 3 b c 15 a 10 4 20 9 3 13 e f d 20 20 5 1 71 33 16 a b c 4 500 500 7 1 7 d e f 100 25 500

B BÌ

b i

A 6

AÌ 9

15

14 20

1 10

15 30

3 10

ii

7 15

2 5 3 d 10 9 a c

Exercise 8D

1 b 5

9 1

12 Because Pr(A È B) = 0 for mutually exclusive events. 13 a Pr(A) < Pr(A È B) b Pr(A) + Pr(B) < Pr(A É B)

1 a i 2 2 b 9 2 a i 7 7 b 10 1 3 a 3 9 4 a i 13 14 b i 17 3 c i 4 7 d i 16 7 5 a i 18 4 b i 9 8 c i 17 3 d i 4 6 a

6

B BÌ

ii 9

A

AÌ

2 3

2 1

4 4

3

8

5 ii 10 7 c 12 1 b 2 3 ii 13 4 ii 17 5 ii 8 1 ii 8 1 ii 9 1 ii 9 7 ii 17 1 ii 4

i 1 b B BÌ 3 7 4 iii 7 5 iii 7 1 iii 4 1 iii 5 1 iii 5 7 iii 10 2 iii 3 iii

B

AÌ 6

15

BÌ

4

1

5

13

7

20

9 d 13

1 3 2 iv 7 5 iv 6 2 iv 7 2 iv 7 1 iv 4 7 iv 8 1 iv 3 iv

10 11 12 13 14 15

iii

A

AÌ

3 5

13 6

16 11

8

19

27

1 2

3 3 i 6 ii iii 8 16 1 1 1 a b c 13 13 4 1 1 a b 3 2 Pr(A|B) = Pr(B|A) = 0 as Pr(A È B) = 0 1 a 1 b 5 a Pr(A È B) = Pr(A) × Pr(B|A) 174 a 329 b 329 24 31 d e 155 231

d

1 2

b 0.18 81 c 329 18 f 31

Progress quiz

A 9

3 c 5

2 ii 5

1 10 2 a 0.17 1 3 a 13 1 a

1 1 c 10 20 b 0.29 c 0.33 1 1 1 b c d 26 52 13 b

2 5 d 0.67 1 e 13

d

e

f

3 5

12 13

874 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Like tennis Dislike tennis Total 3 17 20

Like squash Dislike squash Total

29 32

tennis

1 18

30 50

1 2 3 4

2nd roll

squash

29

3

b 16 1 c 16 1 d i 4 4 a

17

1 b 29 1 c 50

(1, 1) (1, 2) (1, 3) (1, 4)

ii

b 20

13 e 33 6 a 0.83 1 7 a 2 8 a

28 f 33 b 0.17 1 b 6

c 5

d

7 33

1 4

d

5 8

iii

1 2 e 250 5 a

1 3

Like water B Dislike water BÌ

15 20

5 0

20 20

Total

35

5

40

1 4

ii

2nd 1 6 v 1

ii

S

1st E

T

X (S, E) (S, T)

(E, S) X (E, T)

(T, S) (T, E) X

2 3

iv

D O G

2nd

D

1st O

G

(D, D) (D, O) (D, G)

(O, D) (O, O) (O, G)

(G, D) (G, O) (G, G)

D

1st O

G

X (D, O) (D, G)

(O, D) X (O, G)

(G, D) (G, O) X

2nd

ii

D O G

2nd

2 a 9

2 3

iii

1 3

8E

6 a

1 a i

d i 0

13 16

3 4

S E T

b i

Exercise 8E

b i 9 1 c i 3

4 (4, 1) (4, 2) (4, 3) (4, 4)

1st toss H T (H, H) (T, H) (H, T) (T, T)

Like soft drink A Dislike soft drink AÌ Total

b

(3, 1) (3, 2) (3, 3) (3, 4)

b 4 1 c 4 d i

c

(2, 1) (2, 2) (2, 3) (2, 4)

H T

2nd toss

5 a 7

1st roll 2 3

1

Answers

3 a

4 a

ii 6 5 9 2 ii 3 b 6 ii

4 9 1 iii 3 iii

iv

8 9

iv 1

L E V E L

L X (L, E) (L, V) (L, E) (L, L)

E (E, L) X (E, V) (E, E) (E, L)

1st V (V, L) (V, E) X (V, E) (V, L)

E (E, L) (E, E) (E, V) X (E, L)

L (L, L) (L, E) (L, V) (L, E) X

b 20 c i 8 2 d i 5 1 e 5

ii 12 3 ii 5

iii 12 3 iii 5

2 9 1 v 3 v

875 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Exercise 8F

Answers

7 a

Die 2

1 2 3 4 5 6

1

2

Die 1 3 4

2 3 4 5 6 7

3 4 5 6 7 8

4 5 6 7 8 9

b 36 c i 2 ii 6 1 1 ii d i 6 6 1 e . Her guess is wrong. 6 8 a

2nd

b 21 9 a i 100 1 b i 10 19 c 100 1 10 a i 4 2 b i 5 11 a without 12 a 30 1 b i 15 1 c 18 13 a

1 7 ii 90 1 ii 10

6

6 7 8 9 10 11

7 8 9 10 11 12

iii 15 35 iii 36

iv

3 8

1 12

5 8

D (D, C) (D,O) (D, L) (D, L) (D, E) (D, G) (D, E)

3 8

5 8

3 a

1 4

4 iii 5

1 2

ii

ii

1 15

iii

d without

2 15

iv

d

3 8

3 8

M M, M

5 8

F

M

3 ×3= 9 8 8 64

M, F

3 5 15 × = 8 8 64

3 8

M F, M

5 × 3 = 15 8 8 64

5 8

F

F, F

2 7

M

M, M

5 7

F

M, F

3 5 × 8 7

= 15 56

3 7

M

F, M

5 3 × 8 7

= 15 56

4 7

F

F, F

5 4 × 8 7

5 = 14

F

5 × 5 = 25 8 8 64

M

F

Box

1 2

2 iii 3 c with

ii

b

c

5 8 1 ii 10 b with

3 5 3 ii 5 3 ii 4

2 5 2 b i 5 1 c i 4 2 a 1 a i

b

1st L (L,C) (L, O) (L, L) (L, L) (L, E) (L, G) (L, E)

O (O, C) (O, O) (O, L) (O, L) (O, E) (O, G) (O, E)

C O L L E G E c

5 6 7 8 9 10

5

3 4

3 2 × 8 7

3 = 28

Counter

Outcome

Probability

1 4

yellow

(A, yellow)

1 1 1 × = 2 4 8

3 4

orange

(A, orange)

1 3 3 × = 2 4 8

3 4

yellow

(B, yellow)

1 3 3 × = 2 4 8

1 4

orange

(B, orange)

1 1 1 × = 2 4 8

A

B

e

1 2

4 15

1st

2nd

b 16 c i 1 1 d i 16 7 e 16

2.5 5 10 20

2.5

5

10

20

5 7.5 12.5 22.5

7.5 10 15 25

12.5 15 20 30

22.5 25 30 40

ii 8 1 ii 8

iii 8 1 iii 4

iv

3 16

876 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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2nd toss

1

1 4 1 4 1 4

1 4 1 4

2

1 4 1 4 1 4

1 4

3

4

1 4

1 4 1 4 1 4

1 4

1 4

1 4

1 4 1 4 1 4

1 4

Outcome 1

(1, 1)

2 3

(1, 2) (1, 3)

1 2

4 1

(1, 4) (2, 1)

1 2

2 3

(2, 2) (2, 3)

4 1

(2, 4) (3, 1)

2 3

(3, 2) (3, 3)

4 1

(3, 4) (4, 1)

2 3

(4, 2) (4, 3)

4

(4, 4)

1 16 1 d i 16 5 a

1 4 1 ii 4

2 3

1 3

4 15 2 c i 9 6 a

3 7

4 7

1 i 7 9 b i 49

iii

R

silver

(Falcon, silver)

1 1 1 × = 2 4 8

2 3

white (Commodore,white)

1 2 1 × = 2 3 3

1 3

red

1 1 1 × = 2 3 6

3 8 7 iv 24

1 2

1 2

i

5 8

1 6 5 v 6 Outcome

(R, R)

2 3 2 × = 3 5 5

2 5

W

(R, W)

4 5

R

(W, R)

1 4 4 × = 3 5 15

1 5

W

(W, W)

1 1 1 × = 3 5 15

8 15 4 iii 9 Outcome Probabilities

ii

iii

1 3

M

(M, M)

3 1 1 × = 7 3 7

2 3

F

(M, F)

3 2 2 × = 7 3 7

1 2

M

(F, M)

4 1 2 × = 7 2 7

1 2

F

(F, F)

4 1 2 × = 7 2 7

M

F

2 ii 7 16 ii 49

4 iii 7 24 iii 49

3 iv 7 25 iv 49

1 2

W

1 2

R

(R, R)

1 2

W

(R, W)

1 2

R

(W, R)

1 2

W

(W, W)

ii

1 2

W

1 6 1 9 a i 5 b

4 5

1 45 c 62.2% 10 a i 0.17 i

3 4 Outcome

3 4 Probability

iii

R

(R, R)

1 1 1 × = 2 3 6

2 3

W

(R, W)

1 2 1 × = 2 3 3

2 3

R

(W, R)

1 2 1 × = 2 3 3

1 3

W

(W, W)

1 1 1 × = 2 3 6

2 3 4 ii 5 ii

iii

5 6

iv

Outcome

Probability

U

(U, U)

1 1 1 × = 5 9 45

8 9

N

(U, N)

1 8 8 × = 5 9 45

2 9

U

(N, U)

4 2 8 × = 5 9 45

7 9

N

(N, N)

4 7 28 × = 5 9 45

ii

16 45

ii 0.11

b i 0.1445 ii 0.0965 3 4 11 a b 7 7

iii

8F

5 6

1 9

U

N

iv

1 3

R

i

1 5

17 24 1 vi 3 iii

R

1 4

1 2

(Commodore, red)

ii

8 a

2 2 4 × = 3 5 15

2 5 4 ii 9

b i

1 4

b

R

W

1 3 3 × = 2 4 8

Commodore

Outcome Probabilities 3 5

white (Falcon, white)

b i

ii

Probability

3 4

Falcon

b 16 c i

Outcome

7 a

Answers

1st toss

4 a

44 45

iii 0.83 iii 0.8555

877 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Outcome

Answers

12 a 1 4

R

(R, R, R)

1

B

(R, R, B)

R

(R, B, R)

B

(R, B, B)

R

(B, R, R)

2 3 2 3

B

(B, R, B)

R

(B, B, R)

1 3

B

(B, B, B)

R

1 3

R

2 5

B

3 4 1 2

3 5

R

B B

1 2

i

0

1 10

ii

b i 1

3 10 ii

1 3

2 3

iii 0

iv

9 10

v

Probability

9 10

2 5

Outcome Probability

13 a

A 1 3

1 3

B

1 3

C

0 1 10 1 10 1 5 1 10 1 5 1 5 1 10

3 1 ii 8 2 c not independent b i

5 a

2

b c 6 a b c

P

(A, P)

1 1 1 × = 3 4 12

3 4

G

(A, G)

1 3 1 × = 3 4 4

7 a

1 2

P

(B, P)

1 1 1 × = 3 2 6

b

1 2

G

(B, G)

1 1 1 × = 3 2 6

c

3 4

P

(C, P)

1 3 1 × = 3 4 4

(C, G)

1 1 1 × = 3 4 12

G

d

1 12

ii

1 6

iii

d

d 8 a

7

1 4

G

1 7 ii 8 8 b $87.50 to player A, $12.50 to player B

14 a i

iii A $81.25, B $18.75

1 2 yes 1 2 3 i 10 no

b c 2 a b

c no 3 a with

ii

1 2

ii

1 3

A 1

2

T 8

TÌ 2

10

7 15

0 2

7 17

ii

7 17

31 32 1 b 216

iii

4 5

b

c

d

1 36

d 2

b 0.76

Problems and challenges

B 2

15 17 b no 1 9 a 32 1 10 a 216

8

i

13 a 0.24 5 14 6

b without

4 a

iv A $34.38, B $65.62

GÌ

G

31 32 1 c 72 2 11 False; Pr(A | B) = 0 but Pr(A) = . 9 12 a 6 b 22 c 49

Exercise 8G 1 a i

ii A $50, B $50

1

T

1 2

c i A $68.75, B $31.25

2

2 2 ii i 3 3 independent 3 1 i , ii not independent 4 2 1 1 i , ii independent 4 4 1 1 ii independent i , 3 3 2 i ,0 ii not independent 7 1 1 Pr(A) = , Pr(A | B) = , independent 2 2 3 1 Pr(A) = , Pr(A | B) = , not independent 10 4 5 3 Pr(A) = , Pr(A | B) = , not independent 12 20 1 1 Pr(A) = , Pr(A | B) = , independent 9 9

b 6 c i

B

1

1 4

1 4

A

1 a 0.16

2 3

2 0.593 75 7 3 a 8 1 4 a 12

b 0.192 b 1 b

1 2

c 0.144 4 7 3 c 4 c

d

2 3

878 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

6 7 8 9 10

63 64

10 a

1 13 983 816 3 5 1 12 true 8 9

H E Y

2nd

H

A

1st P

P

Y

(H, H) (H, E) (H, Y)

(A, H) (A, E) (A, Y)

(P, H) (P, E) (P, Y)

(P, H) (P, E) (P, Y)

(Y, H) (Y, E) (Y, Y)

b 15 c i

1 15

ii

2 15

iii

1st

11 a

2nd 1 4 1 4

Multiple-choice questions 1 A

2B

3C

4E

5A

6B

7 D

8C

9A

10E

1

1 8 5 2 a 8

1 4 1 b 2 1 ii 4

1 a

3 8 5 c 8

b

2 3 a i 5 3 b i 5 4 a

ii

C 12

c

d

5 8

e

1 2

1 4

1 4 1 4

2

1 iii 5

1 iv 10

1 v 20

1 4

17 20

1 4 1 4

3

1 4 1 4

5 4

C

CÌ

6 12

5 13

11 25

18

18

36

5 a 6

ii b

5 36

6 13 ii 4 1 ii 52

6 a i 13 3 b i 4 4 c 13 10 d 13 7 a 0.1 b 0.5 2 1 8 a b 5 5 5 4 9 a i ii 11 11 b No, Pr(A | B) ¢ Pr(A). 1 1 c i ii 2 4 d Yes, Pr(A|B) = Pr(A).

1 4

1 4

b

c 13 1 d i 6

1 4

1 4

1 4

13

H HÌ

1 4

1 4

H 6

1 4

1 4

1 4

Short-answer questions

iii

1 ii 4

1 b i 16 12

Total 1

2

2 3

3 4

4 1

5 3

2 3

4 5

4 1

6 4

2 3

5 6

4 1

7 5

2 3

6 7

4

8

iii 0

1 4

F

3 4

M

1 2

F

1 2

M

iv 1

F

2 5

1 2

13 15

Answers

5

3 5

M

iii 1 2 5 13 a 0.12 a

b

3 3 c 4 10 b 0.58

d

3 5

e

7 10

Extended-response questions 1 a 3 iii

iii

1 5 1 2

b i

c S SÌ

7 15

1 15

R

RÌ

3 3

1 8

4 11

9

15

6 1 d i 2

ii

3 ii 4

879 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

8 Check with your teacher. 9 a small survey, misinterpreted their data

R S W

2nd

b i

1 9

ii

R

1st S

W

(R, R) (R, S) (R, W)

(S, R) (S, S) (S, W)

(W, R) (W, S) (W, W)

1 3

iii

5 9

c 4 5 d 9 e

iv

b Survey more companies and make it Australia-wide. c No, data suggest that profits had reduced, not necessarily that they were not making a profit. Also, sample size is too small.

4 9

10 a graph A b graph B c The scale on graph A starts at 23, whereas on graph B it starts at 5. d Graph A because the scale expands the difference in

Outcome 1 6

1 3

2 7

column widths, including erroneous data values. 12–14 Research required.

(R, S)

W

(R, W)

1 3

R

(S, R)

Exercise 9B

S

(S, S)

W R

(S, W) (W, R)

1 a 10 2 a

S

(W, S)

1 6

1 2 1 3

1 3

W 1 3

1 i 21

column heights. 11 For example, showing only part of the scale, using different

S

S

3 7

(R, R)

1 2

R 2 7

R

6 iii 7

c 1

Class interval 0–

1 5

10 50

2 10

20 100

Frequency 8

Percentage frequency 16

85– 90–

23 13

46 26

95–100 Total

6 50

12 100

b

Chapter 9

Class interval 80–

Exercise 9A 1 a C

b D

c A

d B

e E

2 a B f A

b E

c C

d D

e F

3 a numerical c categorical

3 a Class interval

b categorical d numerical

4 D 5 a numerical and discrete

Frequency

Percentage frequency

0–

5

25

5– 10–15

9 6

45 30

20

100

Total

b numerical and discrete

Histogram of wins

b

c categorical and nominal d numerical and continuous

Frequency

e categorical and ordinal 6 a D b D is the most representative sample. A may pick out the keen students; B probably are good maths students who like maths; and C will have different-sized classes. 7 a For example, likely to be train passengers. b For example, email will pick up computer users only. c For example, electoral roll will list only people aged 18 years and over.

e 90% Percentage frequency 20

30–40 Total

16 iv 21

d 1

Frequency 2

10– 20–

W (W, W)

10 ii 21

b 1.4

10

50

8

40

6

30

4

20

2

10 0

c

Stem 0 1 d 7.5

5

10 Wins

Percentage frequency

Answers

2 a

15

Leaf 01344556778999 012235

880 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Percentage frequency

16 6

40 15

Car Train Tram Walking

8 5

20 12.5

Bicycle Bus

2 3

5 7.5

Total b i 6

40 ii car

100

iii 40%

12 10 8 6 4 2 0

iv 17.5%

v 42.5% 5 a symmetrical

b negatively skewed

c positively skewed 6 a i 34.3 ii 38

d symmetrical iii 39

b i 19.4 7 a

ii 20

iii no mode

Mass 10–

2 3 1 4 5 Jack’s goal scoring c Well spread performance. d Irregular performance, positively skewed. 8 a Sensor A Sensor B Sensor C frequency frequency frequency 0– 21 12 6 3– 0 1 11 6– 0 1 3 9– 0 1 1 12– 0 0 0 15– 0 2 0 18– 0 2 0 21– 0 1 0 24–26 0 1 0 Total 21 21 21 0

Frequency

Sensor C

0 3 6 9 12 15 18 21 24 27

9 a

b

25 20 15 10 5 0

0 3 6 9 12 15 18 21 24 27

c i insensitive ii very sensitive iii moderately sensitive

0 1 2 3 Nick’s goal scoring

b

Sensor B

Frequency

Frequency

Frequency

Type of transport

Answers

14 12 10 8 6 4 2 0

4 a

Sensor A

Frequency Percentage frequency 3 6

15– 20–

6 16

12 32

25– 30–35

21 4

42 8

Total

50

100

b 50

c 32%

d At least 25 g but less than 30 g. f 94% 10 a Section Strings

e 42%

Frequency 21

Percentage frequency 52.5

Woodwind Brass

8 7

20 17.5

Percussion Total

4 40

b 40

c 52.5%

9B

10 100 d 47.5%

e 9.3%

f 65.6% 11 8 students scored between 20 and 30 and there are 32 students all together, so this class interval makes up 25% of the class. 12 No discrete information, only intervals are given and not individual values. 13 3 Ä a Ä 7, 0 Ä b Ä 4, c = 9

0 3 6 9 12 15 18 21 24 27

881 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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. 5 a i 5.3 b i 2.5

Answers

14 a Percentage cumulative frequency

Frequency

Cumulative frequency

0– 40–

2 1

2 3

5.4 8.1

80– 120–

12 18

15 33

40.5 89.2

3 1

36 37

97.3 100

Bill ($)

160– 200–240 Percentage cummulative frequency

b 15 c

100

40

80

30

60

. c i 2.93 ii 0.5 6 a min = 0, max = 17

7 a i min = 4, max = 14 iii Q1 = 5, Q3 = 9 b i min = 16, max = 31 iii Q1 = 21, Q3 = 27 8 a i min = 25, max = 128 iii Q1 = 38, Q3 = 52.5

0

d i $130 e $180

40

ii $100

80

120 Bill ($)

160

200

240

iii $150

v no outliers

ii 25 iv IQR = 6

v no outliers

ii 47 iv IQR = 14.5

d Outliers are 2, 8. 10 a IQR = 12 b No outliers.

1 a Min, lower quartile (Q1 ), median (Q2 ), upper quartile (Q3 ), max

14 Answers may vary. Examples: a 3, 4, 5, 6, 7

all the data, IQR is the spread of the middle 50% of data. c An outlier is a data point (element) outside the vicinity of the

b 2, 4, 6, 6, 6 c 7, 7, 7, 10, 10

Q1 - 1.5 × IQR. 2 a 0, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 8 ii 3

d 2 e -2, 6

b It is doubled.

13 a It stays the same. b It doubles. c It is reduced by a scale factor of 10.

b Range is max – min; IQR is Q3 - Q1 . Range is the spread of

rest of the data. d If the data point is greater than Q3 + 1.5 × IQR or less than

c 24

d 22 11 1, 2, 3 12 a Increases by 5. c It is divided by 10.

Exercise 9C

f yes; 8 3 a i 10.5

ii 7.5 iv IQR = 4

c A more advanced calculator was used. 9 a no outliers b Outlier is 2. c Outliers are 103, 182.

f approx. 20%

b 2 c i 1

e 0

v yes; 128 vi 51.25 b Median as it is not affected dramatically by the outlier.

10

20

b median = 13

c Q1 = 10, Q3 = 15 d IQR = 5 f Road may have been closed that day.

20

40

ii 2.4 ii 2

15 It is not greatly affected by outliers. 16 Answers will vary

Exercise 9D 1 a 15 e 10

b 5 f 20

c 25 g 10

d 20

2 a 4

b 2

c 18

d 20

e It is. ii 7.5

iii 12

3 a

b 4.5 c 0.75, 18.75

1 2 3 4 5 6 7 8

d no 4 a min = 3, Q1 = 4, median = 8, Q3 = 10, max = 13; range = 10, IQR = 6 b min = 10, Q1 = 10.5, median = 14, Q3 = 15.5, max = 18; range = 8, IQR = 5 c min = 1.2, Q1 = 1.85, median = 2.4, Q3 = 3.05, max = 3.4; range = 2.2, IQR = 1.2 d min = 41, Q1 = 53, median = 60.5, Q3 = 65, max = 68; range = 27, IQR = 12

b

5 6 7 8 9 10 11 12 13 14 15 16 17 4 a i Q1 = 4, Q3 = 7; outlier is 13 ii

2 4 6 8 b i Q1 = 1.6, Q3 = 1.9; outlier is 1.1 ii

10

12

14

1.0 1.2 1.4 1.6 1.8 2.0 c i Q1 = 19, Q3 = 23; outliers are 11 and 31

2.2

882 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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10 12 14 16 18 20 22 24 26 28 30 32 d i Q1 = 0.03, Q3 = 0.05; no outliers ii 0.02

0.04

0.03

0.05

0.06

0 2 4 6 8 10 12 14 16 18 20

44

48

52

60

b A

4 a Gum Heights b Gum Heights 5 a mean = 6, s = 2.2 c mean = 8, s = 3.8

b

40

b smaller

2 a B 3 a A

b The data values in A are spread farther from the mean than the data values in B.

0.07

5 a

36

1 a larger

Answers

Exercise 9E

ii

64

b mean = 3.6, s = 2.6 d mean = 32.5, s = 3.6 b mean = 14.5, s = 6.6

6 a mean = 2.7, s = 0.9 7 a The outer-suburb school has more data values in the higher range. b There is less spread. Data values are closer to the mean.

c

0

1

2

3

4

5

70

80

90

100

110

120

c Students at outer-suburb schools may live some distance from the school. Answers will vary.

d

130

6 a Same minimum of 1. b B c i 5

c true 9 a mean = 2, s = 1.0

ii 10

d Data points for B are more evenly spread than those for A.

Box plot of lemur weights

11 a No, standard deviation reflects the spread of the data

13 14 15 16 17 18 19 20 Lemur weights

12 The IQRs would be the same, making the data more comparable.

values from the mean not the size of the data values. b No. As for part a.

8 a They have the same median and upper quartile. b B

b B

Set 2 Set 1 0 2 4 6 8 10 12 14 16 Spelling errors b Yes, examiner 2 found more errors. 11 Answers may vary. Examples: a i, ii Class results had a smaller spread in the top 25% and bottom 25% performed better. iii State results have a larger IQR. b The class did not have other results close to 0 but the school did.

ii 53.16 vi 21.16

iii 101.16

9E

iv 37.16

iii 100%

ii One SD from the mean = 68% Two SDs from the mean = 95%

c B

Box plot of Set 1, Set 2

10 a

13 a i 85.16 v 117.16

b i 66% ii 96% c i Research required

c i 4 ii 5 d Set B is more spread out. 9 a A

b mean = 5.25, s = 0.7 10 a no b no c yes d Yes, one of the deviations would be calculated using the outlier.

7 a Q1 = 14.6, Q2 = 15.3, Q3 = 15.8 b 19.7 kg c

8 a false b true

Three SDs from the mean = 99.7% Close to answers found.

Progress quiz 1 a numerical and discrete b categorical and nominal 2 a Class interval Frequency 03

Percentage frequency 15

1020-

8 5

40 25

3040-50

3 1

15 5

12 Answers will vary

883 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Travel time from home to school

3 a Population

100

8

75

6 5

50

4 3

25

2

Relative frequency

7

1

c

c i 500 0

Stem

10 20 30 40 Time (in minutes) Leaf

50

0 689 1 24555689 2 03578 3 238 4 4 2 | 3 means 23 18 + 19 d Median = = 18.5 2 3 a Range = 32 - 4 = 28; Q2 = 17, Q1 = 12, Q3 = 23, IQR = 11 b Range = 6.6 - 4.2 = 2.4; Q2 = 5.2, Q1 = 4.5,

4 a Price ($)

0

1000 900 800 700 600 500 400 0 1998 2000 2002 2004 2006 2008 2010 2012 Year

b Generally linear in a positive direction.

5 a Q1 = 16, Q3 = 22, IQR = 6 Q3 + 1.5 × IQR = 31

5 a

Year

b The pass rate for the examination has increased marginally over the 10 years, with a peak in 2001. c 2001 6 a linear 7 a i $6000 b 1

b

12 14 16 18 20 22 24 26 28 30 32

c

Exercise 9F

b 36°C c i 12 p.m. to 1 p.m.

c non-linear d linear iii 33°C iv 35°C ii 3 p.m to 4 p.m.

d Temperature is increasing from 8 a.m. to 3 p.m. in a generally linear way. At 3 p.m. the temperature starts to drop.

Sales ($’000)

6 2.7

b no trend ii 33°C

100 90 80 70 0 1994 1996 1998 2000 2002 2004

d 11%

b i $650 000 ii $750 000

32 is an outlier.

1 a linear 2 a i 28°C

F M A M J J A S O N D Month

November before trending upwards again in the final month.

ii median = 37.5 iii lower quartile = 34, upper quartile = 44

The outlier is 69. b For example, school open day or grandparents day.

J

c $0.21

Q3 = 6.1, IQR = 1.6 4 a i min = 30, max = 69

iv IQR = 10 v Q1 - 1.5 × IQR = 19; Q3 + 1.5 × IQR = 59

1.45 1.4 1.35 1.3 1.25 1.2 0

ii 950

b The share price generally increased until it peaked in June and then continually decreased to a yearly low in

Pass rate (%)

Frequency

Answers

b

20 18 16 14 12 10 8 6 4 2 0

ii $4000

Southbank

City Central

Jul Aug Sept Oct Nov Dec

Month

884 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

b i

10

ii Sales for Southbank peaked in August before taking a downturn.

8 6

e about $5000 8 a i 5.8 km ii 1.7 km

4

b i Blue Crest slowly gets closer to the machine. ii Green Tail starts near the machine and gets further from it.

2 0

c 8:30 p.m. 9 a The yearly temperature is cyclical and January is the next month after December and both are in the same season. b no

0.0 0.5 1.0 1.5 2.0 ii y generally decreases as x increases.

b Compound interest—exponential growth. 11 a Graphs may vary, but it should decrease from room

internal environment of the fridge. 12 a 1

2

3

4

5

6

7

8

9 10

0 b positive 4 a y

Number of runs

120 100 80 60 40 20 0

6

7

8

x

Scatter plot

3 2

Score

0

2

4

6 8 Innings number

10

12

d The moving average graph follows the trend of the score graph but the fluctuations are much less significant.

Exercise 9G 1 a unlikely e likely

b likely f likely

y

c unlikely

d likely

Scatter plot

12

0 b c

ii The graph is fairly constant with small increases and decreases.

10

9G

1

Moving average

c Innings number. i The score fluctuates wildly.

2 a i

1 2 3 4 5 c strong d (8, 1.0)

4

Score 26 38 5 10 52 103 75 21 33 0 Moving average 26 32 23 20 26 39 44 41 40 36 b

x

2.5

Scatter plot

1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 0.9

10 a Increases continually, rising more rapidly as the years progress.

temperature to the temperature of the fridge. b No. Drink cannot cool to a temperature lower than that of the

y

3 a

c Northern hemisphere, as the seasons are opposite, June is summer.

Innings

Scatter plot

y

Answers

d i The sales trend for City Central for the 6 months is fairly constant.

d 5 a

5.0 negative strong (14, 4) negative y

24 22 20 18 16 14 12 10 8

7.5

10.0

12.5

15.0

x

Scatter plot

1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9

x

8 6 4 2 0

2

4

6

8

10

x

ii y generally increases as x increases.

885 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

10 a

Answers

b positive

y

Scatter plot

Scatter plot

4.5

150 Volume (dB)

140 130 120

100

0

2

4

6

8

3.0 2.5

x

10

Scatter plot

y

0

200 400 600 800 1000120014001600 Distance (m)

b negative c As d increases, v decreases.

c none

11 a i Weak, negative correlation.

26 24 22 20 18 16 14 12 10 8

Incidence of crime

Scatter plot

x

15.0 17.5 20.0 22.5 25.0 27.5 30.0 32.5 6 a none b weak negative c positive

Incidence of crime

ii car H

8.5

E

8.0 7.5

B

C

A D 20

25 30 35 Fertiliser (grams per week)

40

b D c Seems likely but small sample size does lead to doubt. 9 a

Scatter plot 2250 2000 1750 1500 1250 1000 750 500 250

20

10

15 20 25 No. of police

30

ii no correlation

Scatter plot

6.0

25

5

8 a

6.5

30

Scatter plot

b decrease c i yes

7.0

35

15

d strong positive 7 a yes

Average diameter (cm)

3.5

2.0 1.5

110

No. of words

4.0

0

1

2

3 4 5 No. of photos b Negative, weak correlation.

6

26 25 24 23 22 21 20 19

7.5 10.0 12.5 15.0 17.5 20.0 22.5 25.0 27.5 No. of police b Survey 1, as this shows an increase in the number of police has seen a decrease in the incidence of crime. 12 The positive correlation shows that as height increases ability to play tennis increases. 13 Each axis needs a better scale. All data are between 6 and 8 hours sleep and show only a minimal change in exam marks. Also, very small sample size. 14 a i students I, T ii students G, S b i students H, C ii students B, N c students C, G, H, S, d students B, I, N, T e no

7

886 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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Exercise 9H

y

b

y

Answers

1 a

8 a y = 5x - 5 b 85 cm c 21 kg 9 a Data do not appear to have any correlation. b Too few data points. 10 a Too few data points to determine a correlation. b The data points suggest that the trend is not linear. 11 a i 50 ii 110

y

d

x

y

x x 1 7 2 17 2 a y= x+ b y=- x+ 2 2 3 3 ii

23 4

28 5

ii

14 5

b i 4 a

180 160 140 120 100 10

20

Scatter plot

6 5 4 3 2

200 180 160 140 120 100

1 0 0 1 2 b positive correlation

3

4

5

6

7

x

c As above. d All answers are approximate. ii 0.9 b ¥6

iii 1.8 c ¥ 0.5

30

b i ¥ 140 c i ¥ 25

40

50 60 Age (years)

70

80

90

9H

ii ¥ 125 ii ¥ 22

d experiment 2 e Research required. iv 7.4 d ¥ 50

Exercise 9I 1 a i 12

3 6 a y = x + 18 5

ii 3.26

b i 7 ii 2 2 a There is no linear correlation.

ii 72 ii 100

7 a, b

b The correlation shown is not a linear shape. 3 A a y

Scatter plot

Growth (cm)

70

220

20

b i 42 c i 30

60

Scatter plot

y

i 3.2 5 a ¥ 4.5

30 40 50 Age (years)

Experiment 2

Max. heart rate (b.p.m.)

3 a i 17

Scatter plot 200

Max. heart rate (b.p.m.)

x c

b It is possible to obtain scores of greater than 100%. 12 a Experiment 1

90 80 70 60 50 40 30 20

8 7 6 5 4 3 400

c i ¥ 25 cm d i ¥ 520 mm

500

600 700 Rainfall (mm) ii ¥ 85 cm

ii ¥ 720 mm

800

2 1 0

0

1

2

3

4

5

6

7

8

x

887 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

5 a, c

d Least squares is the black line, median–median is the grey line. e i 7.3 f i 7.1 B a

Scatter plot Number of jackets

Answers

b y = 0.554762x + 3.45357 c y = 0.42x + 4.17667

ii 10.1 ii 9.2

y 4.0

0 5 10 15 20 25 30 35 40 Temperature (°C)

3.5

b y = -1.72461x + 190.569

3.0

d i 139 6 a, c

2.5

Number of breakdowns

1.5 1.0 0.5 0

5

10

15

20

25

30

x

e i 3.7 f i 3.7

30 25 20 15 10 5

−5

d Least squares is the blue line, median–median is the red line.

0

20

40

60

80 100 120 140 160

Number of copies (× 1000)

b y = 0.25x - 8 d 232 000 copies

ii 3.3 ii 3.3

e The regression line suggests that the photocopier will be considered for scrap because of the number of copies

y

made, as it’s likely to reach 200 000 copies before breaking down 50 times.

8

7 a

Scatter plot 120

6

110

Record (m)

7

5 4

100 90 80

3

70 1970 1975 1980 1985 1990 1995 2000 Year

2

b y = 2.01053x - 3881.6 c i 139 m ii 180 m

1 0

iii 113 Scatter plot

0

b y = -0.077703x + 4.21014 c y = -0.086667x + 4.34333

C a

ii 130 35

2.0

0.0

160 140 120 100 80 60 40 20 0

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6

b y = -3.45129x + 7.41988 c y = -2.85714x + 6.67381

x

d No, records are not likely to continue to increase at this rate. 8 a All deviations are used in the calculation of the least squares regression. b Outliers have little or no effect on the median values used

d Least squares is the blue line, median–median is the red line.

to calculate the median–median regression. 9 A, as it has been affected by the outlier.

e i -16.7 f i -13.3

10 Research required.

ii -34.0 ii -27.6

Problems and challenges

4 a i y = -3.54774x + 43.0398 ii y = -3.28571x + 40.8333 b $32397

c $1405

d 8 years

e 10 years

1 66 kg 2 88%

888 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

5 a larger by 3 b larger by 3 c no change d no change e no change

25

6 y = x2 - 3x + 5 7 5.8 Ä a < 6.2

20

1 D

2B

3D

4C

5A

6C

7 D

8E

9C

10B

Short-answer questions Class interval

Frequency

Percentage frequency

0– 5–

2 7

12.5 43.75

10–

5

31.25

15– 20–25

1 1

6.25 6.25

Total

16

4

25

2

3 d (3, 5)

4

5

3 7 5 a y= x+ 5 5 b i 3.8

ii 7.4

2 c i 2 3

ii 17

2 3

c The Cats

d The Cats

Extended-response questions 1 a i 14

ii 41

b i no outliers ii no outliers c

5 10 15 20 25 Number of hours of TV c It is positively skewed. d Stem Leaf e 8.5 hours 0 135667889 1 012346 2 4 2 a i 10 ii Q1 = 2.5, Q3 = 5.5 iii 3 iv 12 v

Tree 2

0

b i 17

6 8 4 ii Q1 = 15, Q3 = 24

10

12 iii 9

iv none v

c i 2.4 iv 0.7

Tree 1 10

20

30 40 50 60 70 Number of flying foxes

80

90

d More flying foxes regularly take refuge in tree 1 than in tree 2, for which the spread is much greater. 2 a

Scatter plot 1800 Number of shoppers

2

x

9 a y = -3.75x + 25.65 b y = -3.8333x + 25.1667

50

6

2 c weak

8 a non-linear b linear

% Frequency

Frequency

5

6 a mean = 7, s = 2.5 b mean = 4, s = 3.0 7 a The Cats b The Cats

100

8

b

Scatter plot

1 b negative

Multiple-choice questions

d true

10

spread of all results from each class. This number gives a meaningful comparison of results. Physics 1.67, Maths 1.11, Biology 0.

c true

15

8 Physics, Biology. The number of standard deviations from the mean shows the relative position of Emily’s mark within the

1 a

b true

Answers

3 a false 4 a y

3 19 4 1.1

1600 1400 1200 1000 800 600

10

20 25 30 15 ii Q1 = 2.1, Q3 = 2.6 iii 0.50

25.0

27.5 30.0 32.5 35.0 Maximum daily temperature (°C)

37.5

positive correlation b y = 92.8571x - 1491.67 c i 737 ii 32.2°C d y = 74.5585x - 1010.06

v

0.5

1.0

1.5

2.0

2.5

3.0

e i 779

ii 33.7°C

889 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

Chapter 10

log10P

b

5

Exercise 10A

4

1 x 2x

0 1

1 2

2 4

3 8

4 16

5 32

3

3x 4x

1 1

3 4

9 16

27 64

81 256

243 1024

2

5x 10x

1 1

5 10

25 100

125 1000

2 a 4

b 4 1 b 2 1 f 25

1 10 000 1 e 27

3 a

625 3125 10 000 100 000

c 3 1 c 4 1 g 64

d 4 1 d 32 1 h 36

4 a 24 = 16 1 d 2-2 = 4

b 102 = 100

5 a log2 8 = 3

b log3 81 = 4 c log2 32 = 5 1 1 e log10 = -1 f log5 = -3 10 125

e

d log4 16 = 2

10-1

= 0.1

c 33 = 27 1 f 3-2 = 9

1 O

1 2 3 Graph is a straight line. c log10 P = h 13 a i 10 ii 100 b i 10 ii 6 14 Yes. 0 < b < 1 Û loga b < 0, 1 e.g. log2 = -2. 4 1 4 1 f 3 6 k 5

15 a

1 5 2 g 3 4 l 7 b

4

h

5

iii 10 000 iii 2 when a > 1;

1 2 4 h 3

1 3 1 i 2

c

d

6 a 4

b 2

c 6

d 3 g 3

e 1 h 3

f 2 i 2

j 2

k 5

l 3

m0 p undefined

n 0

o 0

7 a -3 d -3

b -2 e -2

c -2 f -4

g -4 j -3

h -1 k -5

i -1 l -1

m -3 p -2

n -1

o -2

8 a 0.699 d -0.097

b 1.672 e -0.770

c 2.210 f -1.431

2 a 2 e 12

b 1 f 3

c 2 g 4

d 4 h -1

9 a 3 d 4

b 5 e 3

c 6 f 2

i -1 3 a 2

b 5

c 3

d -4

g 16 1 j 9 m3

h 81 1 k 4 n 2

i 1000 1 l 343 o 4

e 12 4 a loga 6

f 0 b loga 15

c loga 28

d logb 18

p 8 s 2

q 3 t -1

r 10

10 a

Time (min)

0

1

2

3

4

5

Population

1

2

4

8

16

32

b P = 2t 11 a 16 12 a

c 256

d 13 min

b 26

c 6

h 0 1

2

3

4

e log2 10 000 5

P 1 10 100 1000 10000 100000 log10 P 0 1 2 3 4 5

890

1 2 1 j 2 e

Exercise 10B 1 a logb xy = logb x + logb y x b logb = logb x - logb y y c loga bm = m × loga b d loga a = 1 e logc 1 = 0 1 f logb = -loga b b

e logb 15 f logb 17 5 a loga 2 b loga 3 c loga 10 ( ) ( ) 3 7 e logb f logb 2 5

d logb 2

6 a loga 9 e loga 32

b loga 25 c loga 27 f loga 1000

d loga 16

7 a 0

b 0

c 0

f 1

g 0

h 3

d 1 1 i 3

k 1

l

e 1 2 3

j

1 2

8 a -2

b -3

c -3

d -1

e -2 9 a 1

f -5 b 1

c 3

d 2

e 2

f 2 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

e log3 8 3 2 4 f 5

11 a

b log10 48 f 0 b

5 2

8 a 14.21 years 9 a 10.48 years

c log10 2 d log7 2 ( ) 3 h log5 6 g log2 4 c

4 3

d

3 2

e

1 3

1 = loga 1 - loga x = 0 - loga x = -loga x as x required (using 2nd log law) 1 b loga = loga x-1 = -loga x as required (using x 3rd log law) 1 √ loga x 1 13 loga n x = loga x n = loga x = as required n n (using 3rd log law)

12 a loga

10 a A = 2000 × 1.1n 11 a F = 300 000 × 0.92n 12 a 69 years

c 47.19 years c 91.17 years

b 7 years b 8.3 years

b 1386 years

13 a i

log10 7 log10 2

ii

log10 16 log10 3

iii

b i

1 log10 5

ii

3 log10 2

iii - 1 log10 3

c i 1.631

log10 1.3 log10 5

ii 1.167

iii -0.196

ii -2

iii 1

Exercise 10D 1 a 5 b i 3

14 a Recall index law 1: am × an = am + n Now let x = am and y = an (1) so m = loga x and n = loga y From (1), xy = am × an = am + n

b 23.84 years b 22.20 years

c 2 2 a linear e constant i cubic

(2)

So m + n = loga xy From (2), m + n = loga x + loga y

3 a 4 e -9

iv -1

b quadratic c quartic

d quadratic

f linear

g constant

h quartic

b 3 f 2

c -2

d -2

So loga xy = loga x + loga y, as required. b Recall index law 2: am ÷ an = am-n

4 a, b, f are polynomials. 5 a 14 b 92

c 8

d 4

Now let x = am and y = an (1) so m = loga x and n = loga y (2) x From (1), x ÷ y = = am ÷ an = am-n y x So m - n = loga y From (2) m - n = loga x - loga y x So loga = loga x - loga y, as required. y c Recall index law 3: (am)n = amn Let x = am

6 a -5 7 a 0

b 11 b 92

c 1 c -4

d -45 d 42

8 a -2 e 17

b 25 f -351

c -22

d -17

b -1

c

b 4 ii 24 m

c -108 iii 0 m

9 a -

So mn = loga From (1): nloga x = loga xn , as required.

Exercise 10C

d 8

3 a 0.845 e 0.780

b -0.222 f 0.897

c -0.125

d 1.277

4 a 1.465 e 1.177

b 3.459 f 2

c 1.594

d 6.871

5 a 1 e 6.579

b 1 f 1.528

c 3.969

d 1.727

6 a 2.585 e 1.559

b 1.893 f 6.579

c 1.209 g 3.322

i 0.356 7 a 2 days

j 3.969 k 3.106 b 2.548 days

Essential Mathematics for the Victorian Curriculum Year 10 & 10A

b n+1

c 1

d 1

9 13 a 8 16 e 27

20 b 27 216 f 125

5 c 8

d

g -

1 2

27 64 9 h 8

b 2b3 - b2 - 5b - 1 d -2a3 - a2 + 5a - 1

e -16a3 - 4a2 + 10a - 1 f -54k3 - 9k2 + 15k - 1 g 2a3 b3 - a2 b2 - 5ab - 1 h -2a3 b3 - a2 b2 + 5ab - 1

e log7 2 = x f log1.1 7 = x c 2

12 a 8

14 a 2k3 - k2 - 5k - 1 c 16a3 - 4a2 - 10a - 1

1 d log3 10 = x 2

b 4

10C

1 2

b Yes, when 5 < x < 7.

xn

2 a 3 f 0.1

1 2

10 a 0 11 a i 30 m

So m = loga x (1) xn = amn using index law 3

1 a log2 8 = 3 b log5 25 = 2 c log4 2 =

Answers

10 a log3 20

e 2

15 a i 10 v -9

ii 2 vi -18

b i 3 ii -11 c a = 2 and b = -1

iii 1

iv -13

iii -22

Exercise 10E

d 1.129 h 1.262 l 1.137 c 3.322 days

1 a x2 + 2x

b x - 3x2

e 6x2 - 13x - 5 2 a x4 - 5x3 + 4x2 - 3 c -3x8 - x6 - 6x + 3

c x2 - 1

d x2 + 6x - 55

f 8x2 - 26x + 15 b -x6 - 3x4 + x3 - x2 + 13

3 a, b, c are true.

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891

Answers

4 a x3 - 3x2 d x3 - x4

b x4 - x2 e x5 + 3x4

c 2x2 + 6x3 f -3x6 + 3x3

g -2x5 - 2x4 h -x7 + x4 5 a x5 + x3 + 2x2 + 2

i -4x7 + 8x10

e - 2x3 - 2x2 - 5x + 7 = (x + 4)(-2x2 + 6x - 29) + 123 f

- 5x3 + 11x2 - 2x - 20 = (x - 3)(-5x2 - 4x - 14) - 62

b x5 - x c x5 - x4 - 3x3 + 3x2

6 a 6x4 - x3 + 2x2 - x + 2 = (x - 3)(6x3 + 17x2 + 53x + 158) + 476

d x5 - x3 - 2x2 - 2x + 4 e x5 + 2x4 + 2x3 - 2x2 - 3x

b 8x5 - 2x4 + 3x3 - x2 - 4x - 6

f x5 - 2x4 + 5x3 - 4x2 g x6 - x5 + x4 - 4x3 + 2x2 - x + 2

= (x + 1)(8x4 - 10x3 + 13x2 - 14x + 10) - 16

h x6 - 5x5 - x4 + 8x3 - 5x2 - 2x + 2 i x8 - x6 + x5 - 2x4 - x3 + 3x2 + x - 3

7 a x2 - 2x + 3 -

c x4 - 4x3 + 6x2 - 4x + 1 7 a x5 + 3x4 - x3 - 9x2 - 2x + 8 b x4 + 2x3 - 3x2 - 4x + 4 c x6 + 4x5 + 2x4 - 12x3 - 15x2 + 8x + 16

6x3 - 30x2 -7x2 + 32x

5x2

c + - 23x + 5 d -x5 + 5x4 - 2x3 + 5x2 - x + 1

-7x2 + 35x -3x + 15

e -x6 - 2x4 - x2 + 4 f -x6 - x4 - 10x3 + 26x2 - 10x + 1 9

=

-3x + 15

x8 + 4x7 + 2x6 - 8x5 - 5x4 + 8x3 + 2x2 - 4x + 1

10 (x2 - x - 1)2 - (x2 - x + 1)2 = x4 - 2x3 - x2 + 2x + 1 - (x4 2x3 + 3x2 - 2x + 1) = 4x - 4x2 as required (or could use DOPS) 11 Yes. Multiplicative axiom ab = ba. 12 a 3 b 5 c 7 13 a m e 2m

b m f 3n

14 a x4 - x3 + x2 - x c x3 + 4x2 + x - 6

c m+n

d 12 d 2m

b x5 + 2x4 - 3x3 d 6x3 + 23x2 - 5x - 4

e 15x3 - 11x2 - 48x + 20 f x5 + 3x4 - x3 - 3x2 - 2x - 6

b 3

c 0

2 a If 182 ÷ 3 = 60 remainder 2 then 182 = 3 × 60 + 2. b If 2184 ÷ 5 = 436 remainder 4 then 2184 = 5 × 436 + 4. c If 617 ÷ 7 = 88 remainder 1 then 617 = 7 × 88 + 1. 3 P(x) = (x - 1)(x2 + 2x) + 3 4 P(x) = (x + 1)(3x2 - 4x + 5) - 3 5 a 2x3 - x2 + 3x - 2 = (x - 2)(2x2 + 3x + 9) + 16 b 2x3 + 2x2 - x - 3 = (x + 2)(2x2 - 2x + 3) - 9 c 5x3 - 2x2 + 7x - 1 = (x + 3)(5x2 - 17x + 58) - 175 d - x3 + x2 - 10x + 4 = 2

(x - 4)(-x - 3x - 22) - 84

0 Remainder of 0, as required. 13 10 a 4 b 8 -253 41 11 a -8 b c 16 27 12 a x3 - x2 + 3x + 2 = (x2 - 1)(x - 1) + 4x + 1 b 2x3 + x2 - 5x - 1 = (x2 + 3)(2x + 1) - 11x - 4 c 5x4 - x2 + 2 = 5x(x3 - 2) - x2 + 10x + 2

Progress quiz

Exercise 10F 1 a 1

1 x-1

79 x+3 240 d x3 + 4x2 + 15x + 60 + x-4 8 -1, 1, 2 6x2 - 7x - 3 ) 9 x-5 6x3 - 37x2 + 32x + 15

8 a x3 + x2 - 4x + 1 b x3 - x2 + 6x - 1

(x2 + x - 1)4

b x2 + 2x + 2 -

c x3 - 3x2 + 9x - 27 +

6 a x5 - 2x4 + 2x3 - 3x2 + 3x - 1 b x6 + 2x4 - 2x3 + x2 - 2x + 1

2x3

5 x+2

1 a log2 32 = 5 2 a 102 = 100

b log10 1000 = 3 b 23 = 8

c loga a = 1 c 70 = 1

3 a x=4 d x = 625

b x=4 e x=5

c x=0 f x = -6

4 a 3

b 2 c 2 1 e f 1 2 5 a x = 1.771 b x = 29.060 c x = 8.043 √ 6 The term involving x has a fractional index when written in index notation. d -2

7 a 8 8 a 3

b 7 b 1

9 a x7 - 2x5 + x4 10 a x4 + x3 + 3x + 2

c 3k4 - 2k3 + k2 + 7k + 8 c 3 d 4x3 b x4 + 2x3 + 5x2 - 2x - 6 b -x4 + x3 - x + 2

c x8 + 4x5 + 4x2 d x7 + x5 + 4x4 + 2x2 + 4x 2 11 P(x) = (x + 2)(2x - 5x + 13) - 27

892 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

1 a -1

b 41

2 a 3 3 0

b -2

4 a 3 e -127

b 11 f -33

c -19

d -141

c 27 g -13

d 57 h -69

c x+2

d x-2 9 a x - 2, x - 1 and x + 1

b x - 3, x - 1 and x + 2

c x - 3, x - 2 and x + 1 10 a -4 b -2

d x - 5, x - 1 and x + 4 c -14 d 96

11 -38 12 a 5

c 5

b 1

13 a -2 b 23 14 a a = -1 and b = 2

−1 O

−3

d 5

b x - 1, x + 1 or x + 2

d -3

b a = 3 and b = -4

Exercise 10H

x

2

y

b

3 O

−5

2

x

5

2 a y-intercept is 12. x-intercepts are -1, 3, 4. b y-intercept is -21. x-intercepts are -3, -1, 7. c y-intercept is 0. x-intercepts are -2, 0, 4.

1 P(-1) = 0 2 a x = -3 or 1 1 c x = - or 4 2 e x = -3

y

1 a

3

5 a 3 b 20 c 36 6 b, c and e are factors of P(x). 7 b, d, f, g 8 a x+1

Answers

Exercise 10I

Exercise 10G

d y-intercept is 0. x-intercepts are -7, 0, 5. 3 a, b, c:

b x = -2 or 2 d x = -3 or 4 f x = -4 or -3

3 a -3, 1, 2 b -7, -2, 1 2 1 1 1 e - ,- ,3 d - ,- ,3 2 3 3 2 12 1 11 3 2 1 g - ,- ,h - ,- , 11 2 3 5 19 2 4 a (x - 3)(x - 2)(x + 1); -1, 2, 3

c -4, 3, 4 2 1 2 f - , , 7 4 5

b (x + 1)(x + 2)(x + 3); -3, -2, -1 c (x - 3)(x - 2)(x - 1); 1, 2, 3

x

–2

–1

y = x2

4

1

y = x3

–8

–1

y = x4

16

1

b x = -2

b x = -3, -2 or 1 b -2, -1 or 3

8 a 3 b 4 9 a x2 (x - 1); 0, 1

c n b x2 (x + 1); -1, 0

c x(x - 4)(x + 3); -3, 0, 4 10 0 = x4 + x2 = x2 (x2 + 1)

d 2x3 (x + 1)2 ; -1, 0

0 0

2

1

4

1

8

1

16

10G

−2

O −1 −5 −10

1

x

2

4 a y-intercept: 6 x-intercepts: -2, 1, 3

y

6

No solution to x2 + 1 = 0. Thus, x = 0 is the only solution. 11 The discriminant of the quadratic is negative, implying solutions from the quadratic factor are not real. x = 2 is the

c x = -3, -2, 1 or 3

0

1

10 5

6 a x = -1, 3 or 5 7 a x = -4, 1 or 3

only solution. 12 a x = -4, -3, -2 or 1

1 2 1 4 1 8 1 16

0

y

d (x - 4)(x - 3)(x - 1); 1, 3, 4 e (x - 6)(x + 1)(x + 2); -2, -1, 6 f (x - 2)(x + 3)(x + 5); -5, -3, 2 √ √ 5 a x = 1 or 1 + 5 or 1 - 5

–1 2 1 4 –1 8 1 16

−2

O

1

3

x

b x = -2 or 3 1 d x = -2, , 1 or 2 2

893 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Answers

h y-intercept: 0 x-intercepts: -1, 0, 3

b y-intercept: 12 x-intercepts: -1, 3, 4 y

y

−1 O

12

−1

O

3

x

4

i y-intercept: 8

y

x-intercepts: -4, -2, 1

8

c y-intercept: 10 x-intercepts: -2, 1, 5

y

−4

10 −2 O

1

j y-intercept:

x

5

d y-intercept: 3 x-intercepts: -3, 1, 2

O

−2

x

1

3 2

x-intercepts: -3, -1,

1 2

y

3 2

−3

x

O 0.5

−1

y

O

−3

y

5 a

3 1

2

1 −1

y

x-intercepts: -3, 0, 2

O

−3

x

2

y

5

O

−1 1

O

1

x

−1

x

−1

−2

−2

−3

6 a y = (x - 2)(x + 1)(x + 4) b y = (x + 3)(x - 1)(x - 3) 1 c y = x(x - 2)(x + 3) 2 1 d y = - (x + 3)(x + 1)(x - 2) 2 7 a y

f y-intercept: 0

−1 O

y

b

x

e y-intercept: 0

x-intercepts: -1, 0, 5

x

3

x

60 40 30 20

g y-intercept: 0 x-intercepts: -3, 0, 1

y

−2 −1O

1 2

3 4 5

x

−20 −3

O

1

x

894 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

y

9 a

y

b

Answers

y

b

80 60 40

O

3

x

−1 O

d

y

x

20 y

c

x

O1 2 3 4

−4 −3 −2 −1

4

−2 y

8 a

b

2 −1−2O −4 −6 −8 −9 −10

8 6 4 2

x

2

12

y

−2O

x

3

y

e

y

d

O

−2

x

x −2 −2 −4

−4

y

c

−1 O

−4

2

y

f

x 2

2

−2 −2 −1

−4

O1 2

x

−1

O

x

2

−2

O

2

x

−2 y

e

10 a i y-intercept = (0, -6) ii y = (x - 1)(x + 2)(x + 3) iii x-intercepts: (-3, 0), (-2, 0), (1, 0)

4

y

iv

2 −4 −3 −2 −1

10I

x

O

O − 3 −2

1

x

y

f

15 13 10

−6

5 O −5

5 (2, −3)

x

b i y-intercept = (0, 15) ii y = (x - 5)(x - 3)(x + 1) iii x-intercepts: (-1, 0), (3, 0), (5, 0)

895 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

y

Answers

iv

11 a 2 Ä x Ä 5 or x Ä -1 b -4 < x < 1 or x > 4 1 12 y = (x - 3)2 (x + 2) 9 13 16 1 14 y = - x2 (x - 3)(x + 3) 10

15

Multiple-choice questions

x

−1 O 3 5

c i y-intercept = (0, 8)

y 8 −1 O

−4

2B

3E

4A

5D

6D

7 A

8B

9E

10E

Short-answer questions

ii y = (x - 2)(x - 1)(x + 1)(x + 4) iii x-intercepts: (-4, 0), (-1, 0), (1, 0), (2, 0) iv

1 C

x

1 2

1 a log2 16 = 4 1 c log3 = -2 9

b log10 1000 = 3

2 a 34 = 81

b 4-2 =

c 10-1 = 0.1 3 a 3

b 4

c 4

e -3 h -4

f -3 i -2

d 0 g -1 4 a loga 8 e loga 4 3 i 2

d i y-intercept = (0, 225) ii y = (x - 5)(x - 3)(x + 3)(x + 5) iii x-intercepts: (-5, 0), (-3, 0), (3, 0), (5, 0) iv y

b logb 21 f loga 1000

6 a

O

3

log10 13 log10 2

b

b 8

2 a 1.43

b -1.43

1 2 c -2.71 c

3 x=6 4 2p - q + 2 5 a x > 2.10 6 a=2× 7

1 34 ,

5

x

b x Å -2.81 1 b = log2 3 4

9 a x3 + x2 + 2x + 3 = (x - 1)(x2 + 2x + 4) + 7 b x3 - 3x2 - x + 1 = (x + 1)(x2 - 4x + 3) - 2

d 3

10 a -3 b -39 11 b, c and d are factors.

10 Proof using long division required. a (x3 - a3 ) ÷ (x - a) = x2 + ax + a2 b

+

÷ (x + a) =

- ax +

d 41

1 3 b x = - , or 5 3 2 13 a (x - 1)(x + 2)(x + 3) = 0 x = -3, -2 or 1 b (x + 2)(x - 5)(x - 6) = 0 x = -2, 5 or 6 12 a x = -2, 1 or 3

y

14 a

−1

x2

c -91

O

y

b

1

8 -2 9 a = 5, b = -2

a3 )

c -2 d -34 b x5 - x4 - 3x3

c x5 + x4 - 3x3 - x2 - x + 3 d x6 + 2x4 - 4x3 + x2 - 4x + 3

log10 2 = log1.1 2 log10 1.1

(x3

log10 2 log10 0.8

c 2x3 - x2 + 4x - 7 = (x + 2)(2x2 - 5x + 14) - 35 d -2x3 - x2 - 3x - 4 = -(x - 3)(2x2 + 7x + 24) - 76

Problems and challenges 1 a 2

d loga 10 h 1

j 4 loga 2 = loga 16

7 a -1 b 1 8 a x4 + 3x2 + 2

−3

c logb 144 g 2

5 a x = log3 6 b x = log1.2 2

225

−5

1 16

5 4 1

x

−1O

−1

−5

−2

−10

4 5

x

a2

896 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

Parabolas and other graphs

y

d

10

Answers

y

c

Multiple-choice questions

5

1 D

10

−5 −2 O

3

5

x

3D

4C

5B

Short-answer questions

5

−5

2A

x

O

1 a Dilated by a factor of 3.

y

−10 3

Extended-response questions 1 a B b i $121 000 ii $15 369 c i 7.27 2 a i 32

ii 6.17 ii 0

iii $272 034

(1, 3)

x

O

iii 16.89

b Reflected in x-axis and translated 2 units left.

y

b There is no remainder; i.e. P(1) = 0. c x2 - 4x - 21 d (x - 7)(x - 1)(x + 3) e x = 7, 1 or -3

x

O

−2

f P(0) = 21 g y

−4

21 −3 O 1

7

c Translated 5 units up.

x

y

SR2 Semester review 2 Measurement 2D

3B

4A

5D

2 a -5 d

b (1, -5)

−5

1 a 36 cm, 52 cm2 b 1.3 m, 0.1 m2 c 220 mm, 2100 mm2 2 a 188.5 m2 , 197.9 m3 c 6.8 3 √ a 1.8 cm 27 cm 4 π

c (-2, -9)

y

Short-answer questions

m2 , 1.3

x

O

Multiple-choice questions 1 B

(1, 6)

5

O 1

x

−5 b 50.3 cm2 , 23.7 cm3

m3 b 58.8 cm2

(−2, −9) −9 3 a Maximum at (3, 8).

b -10

c (1, 5)

Extended-response question a 753.98 cm3

b 206.02 cm3

c 17 cm

d 1.79 cm

897 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

y (3, 8)

Answers

d

y

c

(−1, 5)

(1, 3) O 1

y

d

x

5

1 x

O

1

−10

(1, 15 )

O y

e 4 a y = (x + 3)2 - 7

(1, 2)

y

x

O

2

O

−3 − √7

x

x

(−1, −2)

−3 + √7

y

f

(−1, 6) ( b y=

(−3, −7) )2 5 7 x+ 2 4 y

x

O (1, −6)

8

( 52 , 74 )

O

√ √ √ √ 8 a (( 3, 2) 3), ( (- 3, -2 ) 3) 1 1 c , 4 , - , -4 2 2 9 a y

x

b (4, 16)

−1 + 2√3 5 a Discriminant = 72, thus two x-intercepts.

x 2 + √15

O

2 − √15

y

b

(2, −1)

−1 − 2√3 −1.1 O

x

3.1

y

b

(1, 17)

(1, −9) c (1, -9) and (-2, 9) 6 a Each x value produces a unique y value (any vertical line will cut the graph at most once) b i 3 ii 12 iii (a - 2)2 + 3 c All real x, y Å 3 7 a y

−2

O

1 O

y

b

2

9 y=1

x

√10

2

−2

x

−√10

O

√10

x

−√10

898 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

4 a

x = −2

Die 2

x

5 −2 − 3 O − 52

1 2 3 4 5

2 3 4 5 6

3 4 5 6 7

4 5 6 7 8

b 16 3 16

c i

y = −3

−3

5 8

ii

5 a

A 4

Extended-response question

1 5

iii

B 4

2 2

h

a

Die 1 1 2 3 4

Answers

y

c

2 ii 3

2 b i 3

(200, 427.5)

c Yes they are since Pr(A | B) = Pr(A).

27.5

Extended-response question

x

O 10

110 (60, −62.5)

Outcome

c 10 m and 110 m from start

d 427.5 m

e 62.5 m

5 8

Multiple-choice questions 1 C

2E

250 g

3B

4D

5 7

250 g

3 7

200 g 450 g

4 7

250 g

15 ii 28 3 c 5

3 a i 28 9 b 14

Probability

200 g 400 g

200 g

3 8

b 27.5 m

2 7

450 g

500 g

5 iii 14

SR2

5C

Statistics Short-answer questions 1 a

x

A g

u

e

r

d

f

i

a

o

p

b h

y

1 E

z s

2 B 3 C

q

4 B 5 A

B

w v

Multiple-choice questions

c

n

t

j

k

5 3 b i ii 26 26 c No, A È B ¢ ∅

l

m 9 iii 26

iv

19 26

2 a A AÌ b i 3 1 c i 4 3 a 0.18

B 3

BÌ 1

4

4 7

4 5

8 12

ii 4 1 ii 12 b 0.37

iii 5 7 iii 12

iv 8 3 iv 4

Short-answer questions 1 a Class interval

Frequency

Percentage frequency

05-

2 4

10% 20%

1015-

4 3

20% 15%

2025 - 30

6 1

30% 5%

Total

20

100%

899 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

30

6

Frequency

5 4

20

3 10

2 1

0

0 5

0

c i 14

10 15 20 Number of days

25

Extended-response questions

Percentage frequency (%)

Answers

b

a i y = 1.50487x + 17.2287 ii y = 1.55x + 16.0833 b 41 cm c 22 mL

Logarithms and polynomials

30

Multiple-choice questions 1 E

ii 50%

iii 20–24 days, those that maybe catch public transport to work or school each week day.

5

10

15

20

4

2 3 2500 2000 1500 1000

Balance

5

6

7

25

8

c 3 g 1

d 0 h 3

iv -5

b i 2x6 + 6x5 - 11x4 - 25x3 + 34x + 24 ii 4x6 - 12x4 - 16x3 + 9x2 + 24x + 16 5 P(x) = (x - 3)(x2 - x - 1) + 4 6 a -24, not a factor b 0, a factor c -40, not a factor 5 2 7 a x = -1, 3, -6 b x = 0, , 2 3 1 c x = -4, -2, 1 d x = -1, , 2 2

c May and June d increase of $500

Extended-response question a P(-3) = 0 b P(x) = -(x + 3)(2x - 1)(x - 4) c P(x)

y

6 5 4 3 2

b positive d i ¥ 3.2

5D

b i x = 2.460 ii x = 9.955 4 a i 6 ii 0 iii -49

9

b Balance fluctuated throughout the year but ended up with more money after 12 months.

0

4C

2 a x=3 b x=3 c x = 81 3 a i x = log3 30 ii x = log2.4 4

J F M A M J J A S O N D Months

4 a, c

b -2 f 2

1 a 3 e 2

b

3 a

3D

Short-answer questions

2 a

0

2A

(−5, 198)

x

1 2 3 4 5 6 7 8 9 10 11 12 13 ii ¥ 11.5

O −5

−3

−12

1 2

4

6

x

5 a i under 40 ii over 40 b Over 40: mean = 11, standard deviation = 7.3; Under 40: mean = 24.1, standard deviation = 12.6

(6, −198)

900 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.

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901 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

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902 Essential Mathematics for the Victorian Curriculum Year 10 & 10A

Cambridge University Press ISBN 978-1-316-62367-1 © Greenwood et al. 2016 Photocopying is restricted under law and this material must not be transferred to another party.