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√(x) \\ 48=k\cdot√(64) \\ k=(48)/(8) \\ k=6 \\ y=6\cdot√(x) \end{gathered}`

therefore, x = 712

`\begin{gathered} y=6\cdot√(712)=6\cdot26.68 \\ y=160.1 \end{gathered}`

therefore, the answer is 160.1