## Functions Modeling Change: A Preparation for Calculus, 5th Edition

Yes, $R(t)=t^{-\frac{1}{2}}$
Since $\sqrt{ab}=\sqrt{a} \cdot \sqrt{b}$, $$R(t)=\frac{4}{\sqrt{16} \cdot \sqrt{t}}=\frac{4}{4\sqrt{t}}=\frac{1}{\sqrt{t}}$$ Using the fact that $\sqrt{t}=t^\frac{1}{2}$ and that $\frac{1}{a^n}=a^{-n}$, the function becomes $R(t)=t^{-\frac{1}{2}}$, a power function with $k=1$ and $p=-\frac{1}{2}$.