Question

y varies jointly as x and z z. If y = 6 when x = 3 and z = 9, find y when x = 4 and z = 3.6.
but z is in square root, just the symbol doesn't show.

Answer 1

Answer:

  y = (8/5)√10

Step-by-step explanation:

You seem to have ...

  $y=kx\sqrt{z}$

We can find k from the given values ...

  $\dfrac{y}{x\sqrt{z}}=k=\dfrac{6}{3\sqrt{9}}=\dfrac{6}{3\cdot 3}=\dfrac{2}{3}$

Then for the new values, ...

  $y=\dfrac{2}{3}\cdot 4\sqrt{3.6}=\dfrac{8\sqrt{90}}{3\cdot 5}=\dfrac{8\sqrt{10}}{5}$