Question 1032107
I think you mean
{{{z = kx^3y^2}}}
If so, we plug in the original data to find k, then go from there...thus we have...
{{{172 = k(6^2)(6^3)}}}
which gives us
k = .02212
Now we plug this in to the top equation and plug in the new x and y...
{{{z = (.02212)(4^3)(3^2)}}} =
12.74