## College Algebra 7th Edition

(a) $z=k \frac{x^{3}}{y^{2}}$ (b) $\frac{27}{4}$
(a) Since $z$ is directly proportional to the cube of $x$ and inversely proportional to the square of $y$, we have: $z=k \frac{x^{3}}{y^{2}}$ (b) We are given that $x$ is tripled and $y$ is doubled, so we have: $z=k \frac{(3x)^{3}}{(2y)^{2}}=k\frac{27}{4}\frac{x^{3}}{y^{2}}$ Therefore, $z$ changes by a factor of $\frac{27}{4}$.