Math problem, please show work?

Question

Suppose that x varies directly as the square of y and inversely as z and that x=6 when y=-3 and z=12. Find x when y=9 and z=36

Answer 1

x=ky^2/z

k=xz/y^2
=6*12/9
=8

x=8*y^2/Z

x=8*81/36 =18

Answer 2

x varies directly as the square of y and inversely as z is written algebraically as:
x = k y^2 / z .................(1)
where k is a constant
With x = 6, y = 3 and z = 12, this formula gives:
6 = k 9 / 12
k = 8.
Substituting this in (1) with the values y = 9 and z = 36 gives:
x = 8 x 81 / 36 = 18.

Answer 3

Start with the equation x = a(y^2)/z, a being a constant. Substitute the known values: 6 = a(-3^2)/12. Simplify: 6 = 9a/12. Multiply both sides by 12: 72 = 9a. Divide both sides by 9: 8 = a. Now we have the value of the constant in the first equation, so x = 8(Y^2)/z. Substitute the given values for y and z: x = 8(9^2)/36. Simplify: x = 8(81)/36, then x = 648/36, then x = 18.

Answer 4

x= k (y^2)/z.To calculate k
6=k9/12
so k= 8 and x= 8*(y^2)/z
At the given values
x= 8*81/36=18

Answer 5

x=(ky^2)/z

so... 6=(k(-3)^2)/12
combine like terms... and... k=8

x=((8)(9)^2)/36 and you get x=18

Answer 6

x=ky^2 and xz=72
solve for k solve for k (w/substitution)
-9=k 72=k
x=-9y^2 x36=72
x=-729 x=2

Answer 7

x=(ky^2)/z

6=9k/12
72=9k
k=8

So: x=8(81)/36

x=648/36

x= 18