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14 a i (−2, 17)(−1, 14)(0, 11)(1, 8)(2, 5)(3, 2)(4, −1)(5, −4)

ii (−2, −3)(−1, −1)(0, 1)(1, 3)(2, 5)(3, 7)(4, 9)(5, 11)

b (2, 5) y = 11 − 3x y = 2x + 1

5 = 11 − 3 × 2 5 = 2 × 2 + 1

5 = 11 − 6 5 = 4 + 1

5 = 5 True 5 = 5 True

c It is the only shared point.

15 a i Any answer with y as a whole number is correct, e.g.

2x − 1 = −3, 2x − 1 = −1, 2x − 1 = 3

ii No. Many possible correct examples,

e.g. 2x − 1 = 2.5, 2x − 1 = −1.75, 2x − 1 = 2.8

b i Any answer with the y-value to one decimal place

is correct, e.g. 2x − 1 = −2.6, 2x − 1 = −1.8,

2x − 1 = 0.7

ii No. Many possible correct answers,

e.g. 2x − 1 = 0.42, 2x − 1 = −1.68, 2x − 1 = 2.88

c i 2x − 1 = 2.04, 2x − 1 = 2.06

ii Many possible correct answers, e.g. (1.521, 2.042)

(1.529, 2.058); 2x − 1 = 2.042, 2x − 1 = 2.058

iii Yes, for every two points on a line another point can

be found in between them, so there are an infinite

number of points on a line. Also an infinite number

of equations can be solved from the points on a

straight line if the graph has a suitable scale (digitally

possible).

16 a i x = 2, x = −2 ii x = 3, x = −3

iii x = 4, x = −4

iv x = 5, x = −5

b For each y-coordinate there are two different points, so

two different solutions.

c i x = 2.24, x = −2.24 ii x = 2.61, x = −2.61

iii x = 0.7, x = −0.7 iv x = 3.57, x = −3.57

v x = 4.34, x = −4.34

d The graph of y = x 2 does not include a point where y = −9.

e Many correct answers all with x 2 equal to a negative

number, e.g. x 2 = −5, x 2 = −10, x 2 = −20.

f Positive numbers or zero.

g x = −1, x = 2

Exercise 7H

1 a 15 b 45 c 5 d 125

2 a m = 3, b = 2 b m = −2, b = −1

c m = 4, b = 2

d m = −10, b = −3

e m = −20, b = 100 f m = 5, b = −2

3 a 60 cm b 150 cm c 330 cm

4 a 28 L b 24 L c 10 L

5 a

t 0 1 2 3

d 0 6 12 18

b

d

18

12

6

O

t

1 2 3

c d = 6t d 9 km e 2 hours

6 a

t 0 1 2 3 4 5 6 7 8

d 0 5 10 15 20 25 30 35 40

b d

40

30

20

10

O

t

2 4 6 8

c d = 5t d 22.5 km e 4 hours

7 a

t 0 1 2 3 4 5

v 20 16 12 8 4 0

b

V

20

16

12

8

4

O

t

1 2 3 4 5

c V = −4t + 20 d 11.2 L e 3 seconds

8 a

t 0 1 2 3 4

h 500 375 250 125 0

b

h

h i x = −2.77, x = 5.77

ii x = −8.27, x = 3.27

iii no solution

iv x = 3

500

375

250

125

O

t

1 2 3 4

c h = −125t + 500 d 275 m e 3 minutes

9 a M = −0.5t + 3.5 b 7 hours

c 4.5 hours

Cambridge Maths NSW

Stage 4 Year 8 Second edition

ISBN 978-1-108-46627-1 © Palmer et al. 2018

Cambridge University Press

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