Question

Suppose that y varies directly as the square root of x, and that y = 48 when x = 64. What is y when x = 712 Round your answer to two decimal places if necessary.

Answer 1

we have the following:

$\begin{gathered} y=k\cdot\sqrt{x} \\ 48=k\cdot\sqrt{64} \\ k=\frac{48}{8} \\ k=6 \\ y=6\cdot\sqrt{x} \end{gathered}$

therefore, x = 712

$\begin{gathered} y=6\cdot\sqrt{712}=6\cdot26.68 \\ y=160.1 \end{gathered}$

therefore, the answer is 160.1