Question 88658
Suppose z varies directly as the square of x and inversely as y.  If z = 8 when x = 4 and y = 6, find z when x =6 and y = 12.
You must have meant z instead of x, so I changed the first x to a z.
{{{z=kx^2/y}}}  means z varies directly with the square of x and inversely as y.
{{{(8)=k(4)^2/(12)}}}  Substitute and solve for k.
{{{8=16k/12}}}
{{{8(12)=12(16k/12)}}}
{{{96=16k)}}}
{{{96/16=16k/16}}}
{{{6=k}}}
Your new formula is:
{{{z=6x^2/y}}}  Substitute and solve for z.
{{{z=6(6)^2/12}}}
{{{z=6(36)/12}}}
{{{z=216/12}}}
{{{highlight(z=18)}}}
Happy Calculating!!!