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

Answer: F (False)

Step-by-step explanation:

 Jointly variation has the following form:

y=kxz

Where k is a constant of propotionality.

Substitute values:

If y=16, x=4 and z=2, then k is:

$16=k(4)(2)^{3}\\k=1/2$

If x=-8 and z=-3 the the value of y is:

$y=(1/2)(-8)(-3)^{3}\\y=-108$

Then the answer is FALSE.

Answer 2

Answer:

False (F)

Step-by-step explanation:

Given: Y varies jointly as x and the cube of z.

y = kxz^3

Now we have to find k, when y = 16, when x = 4 and z = 2

16 = k . 4. 2^3

16 = k.4. 8

16 = 32k

k = 16/32

k = 1/2

Now we have the equation y = 1/2 . x. z^3

Now we have to check for y =0.5 when x = -8 and z = -3

Now we have to plug in x = -8 and z = -3 and see if we get y = 0.5

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

y = -4*-27

y = -108

We got y = -108 which is not equal to y = 0.5

Therefore, answer if false (F)

Thank you.