1. **State the problem:** We know that $y$ is inversely proportional to the square of $x$, which means $y \propto \frac{1}{x^2}$. Given $y=4$ when $x=10$, we want to find: a) $y$ when $x=5$. b) $x$ when $y=25$ and $x>0$. 2. **Write the formula:** Since $y$ is inversely proportional to $x^2$, we write: $$y = \frac{k}{x^2}$$ where $k$ is the constant of proportionality. 3. **Find $k$ using the given values:** $$4 = \frac{k}{10^2} = \frac{k}{100}$$ Multiply both sides by 100: $$4 \times 100 = \cancel{100} \times \frac{k}{\cancel{100}} \Rightarrow 400 = k$$ 4. **Rewrite the formula with $k=400$:** $$y = \frac{400}{x^2}$$ 5. **Part a: Find $y$ when $x=5$:** $$y = \frac{400}{5^2} = \frac{400}{25}$$ Simplify: $$y = 16$$ 6. **Part b: Find $x$ when $y=25$ and $x>0$:** $$25 = \frac{400}{x^2}$$ Multiply both sides by $x^2$: $$25x^2 = 400$$ Divide both sides by 25: $$\cancel{25}x^2 = \frac{400}{\cancel{25}} \Rightarrow x^2 = 16$$ Take the positive square root (since $x>0$): $$x = 4$$ **Final answers:** - a) $y=16$ when $x=5$. - b) $x=4$ when $y=25$ and $x>0$.