Y varies directly as x and inversely as the square of z, and when x = 32, y = 6 and z = 4. Find x when y = 3 and z = 10.
y = k x / z^2
6 = k (32) / (4)^2
6 = (32/16)k
6 = 2k
k = 3 = variation constant
So......
3 = 3 x / (10)^2
1 = x / 10^2
x = 10^2 = 100
