## Algebra 2 (1st Edition)

$\approx 26.28$Hz
General equation for inverse variation is given by $y=\dfrac{k}{x}$ Here, we have $f=k \dfrac{\sqrt T}{Ld}$ ...(1) Plugging in the data, we have $262=k \dfrac{\sqrt {670}}{(62)(0.1025)}$ and $K \approx 64.3249482$ Thus, equation (1) becomes: $f=64.3249482 \dfrac{\sqrt {1629}}{(201.6)(0.49)} \approx 26.28$Hz