1. The problem states: The variable $x$ and $y$ are inversely proportional, which means $$x \propto \frac{1}{y}$$ This implies there exists a constant $k$ such that $$x = \frac{k}{y}$$. 2. The third variable $z$ is proportional to the square root of $y$, so $$z \propto \sqrt{y}$$ which means there exists a constant $m$ such that $$z = m \sqrt{y}$$. 3. To express $x$ in terms of $z$, solve for $y$ from the second equation: $$z = m \sqrt{y} \implies \sqrt{y} = \frac{z}{m} \implies y = \left(\frac{z}{m}\right)^2 = \frac{z^2}{m^2}$$. 4. Substitute this into the expression for $x$: $$x = \frac{k}{y} = \frac{k}{\frac{z^2}{m^2}} = k \cdot \frac{m^2}{z^2} = \frac{k m^2}{z^2}$$. 5. Therefore, $x$ expressed in terms of $z$ is $$\boxed{x = \frac{K}{z^2}}$$ where $K = k m^2$ is a constant composed of the original proportionality constants. This shows that $x$ is inversely proportional to the square of $z$.