z is directly proportional to the square of x and inversely proportional to the square root of y. If z=12 when x=2 and y=4, find z when x=6

We have  \(z=k\frac{x^2}{\sqrt{y}}\)   where k is a constant

 

Find k as  \(k=12\frac{\sqrt{4}}{2^2}\rightarrow6\)

 

So when x = 6 and y = 32 we have \(z=6\frac{6^2}{\sqrt{32}}\rightarrow27\sqrt2\)

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