Question 1162790
If z varies directly with x and inversely with the square of y,
{{{Z=k*x*(1/y^2)}}} for all (x,y,z) points in the relationship,
for some constant {{{k>0}}} .
 
We just have to find {{{k}}} .
 
If z=9 when x=3 and y=3, we know that {{{9=k*3*(1/3^2)}}}
{{{9=k*3*(1/3^2)}}}
{{{9=k*3/3^2}}}
{{{9*3=k}}}
{{{k=27}}}
 
What is z when x=11 and y=6?
{{{z=k*x*(1/y^2)}}}
{{{z=27*11*(1/6^2)}}}
{{{z=27*11*(1/36)}}}
{{{z=27*11/36}}}
{{{highlight(z=33/4=8.25)}}}