Question 1067293
let k = the constant
 y varies directly as x and inversely as the square of z.
y = {{{(kx)/z^2}}}
:
 "y=6 when x=8 and z=2." Find k
{{{(8k)/2^2}}} = 6
{{{(8k)/4}}} = 6
Cancel 4
2k = 6
k = 6/2
k = 3
The equation then
y = {{{(3x)/z^2}}}
:
" find y when x=75 and z=5"
y = {{{(3*75)/5^2}}} 
y = 225/25
y = 9