The variable \(x\) varies directly as the square of \(y\), and \(y\) varies directly as the cube of \(z\). If \(x\) equals \(-16\) when \(z\) equals 2, what is the value of \(x\) when \(z\) equals \(\frac{1}{2}\)?
+36916 x = ky^2 y = k/z^3 combined x = k /z^6
-16 = k/ 2^3
k = -128
x = -128 / z^6
x = -128 / (1/2)^3 = -1024
