Answer
(a) $z=k \frac{x^{2}}{y^{4}}$
(b) $\frac{9}{16}$
Work Step by Step
(a) Since $z$ is directly proportional to the square of $x$ and inversely proportional to the fourth power of $y$, we have:
$z=k \frac{x^{2}}{y^{4}}$
(b) We are given that $x$ is tripled and $y$ is doubled, so we have:
$z=k \frac{(3x)^{2}}{(2y)^{4}}=k\frac{9}{16}\frac{x^{2}}{y^{4}}$
Therefore, $z$ changes by a factor of $\frac{9}{16}$.
