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(7a) Let y = ax + b where a and b are constants.

When x = 1/2, y = 11 and so a/2 + b = 11 ... (1)

When x = 1/4, y = 21 and so a/4 + b = 21 ... (2)

Solving, we have a = -40 and b = 31

(b) When x = 1/8, y = -40/8 + 31 = 26

(15a) Let a = kb/√c where k is a constant.

When b = 8 and c = 4, 6 = 8k/√4

k = 3/2

(b i) a = (3/2) x 5/√9 = 5/2

(ii) a/b = 2/3

k/√c = 2/3

(3/2) (1/√c) = 2/3

c = 2/3

(27a) Let C = aV + bA where a and b are constants.

When V = 27 and A = 6 x 9 = 54 (a cube of side length = 3 m), C = 324 and then 324 = 27a + 54b ... (1)

When V = 64 and A = 6 x 16 = 96 (a cube of side length = 4 m), C = 608 and then 608 = 64a + 96b ... (2)

Solving, we have a = 3 and b = 5.

(b) V = 125 and A = 6 x 25 = 150, so

C = 3 x 125 + 5 x 150 = 1125

(29a) From the given, we have E = kN + C where k is a constant.

Comparing to the graph of y = mx + c, wher m is the slope and c is the y-intercept of the line, we have, from the given graph, C = 600

(b) From the graph, slope = 10 and therefore k = 10.

So E = 10N + 600.

(c) When N = 30, E = 900 and hence each player will pay \$30.

Source(s): My Maths knowledge