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

k = 1/2

y = - 108

Step-by-step 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

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

Step-by-step 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.