Question

If y varies jointly as x and the cube of z and y=16 when x=4 and z=2 then y=0.5 when x=-8 and z=-3

Answer 1

y = 1/2xz³ and y = 108 when = -8 and z = -3

Explanation:

From question statement, we observe that

y ∝ xz³

y = kxz³ eq(1)

Where k is proportionality constant.

Given that

y = 16 when x = 4 and z = 2

k = ?

Putting the given values in above eq(1), we have

16 = k(4)(8)

16 = 32k

k = 16/32

k = 1/2

putting the value of k in eq(1), we have

y = 1/2xz³ eq(2)

Putting x = -8 and z = -3 in eq(2), we have

y = 1/2(-8)(-3)³

y = 1/2(-8)(-27)

y = 1/2(216)

y = 108 when x = -8 and z = -3.

Answer 2

k = 1/2

y = -108

Explanation:

y varies jointly as x and the cube of z

y =kxz^3

We know y=16 when x=4 and z=2 so we can determine k

16 = k (4) 2^3

16 = k*4*8

16 = 32k

Divide each side by 32

16/32 = 32k/32

1/2 = k

y = 1/2 xz^3

Now y=? when x=-8 and z=-3

y = 1/2 * 8 (-3)^3

y = (1/2) *8 (-27)

y =4 (-27)

y = -108