## Intermediate Algebra for College Students (7th Edition)

$50$
The equation here is $y=\frac{kab}{\sqrt c}$. Thus $12=\frac{2\cdot3k}{\sqrt{25}}\\12=\frac{6}{5}k\\k=10$. Hence $y=\frac{10ab}{\sqrt c}=\frac{10\cdot5\cdot3}{\sqrt 9}=50.$