Question

y varies directly as x and inversely as the square of z. y=40 when x=64 and z=4. Find y when x=2 and z=6

Answer 1

We are given the following information:

y varies directly as x and inversely as the square of z

y = 40 when x = 64 and z = 4

We are asked to find the value of y when x = 2 and z = 6.

To answer this, we must first express the relationship of x, y, and z:

`y=(kx)/(z^2)`

In the equation above, k is the constant of variation.

We placed x in the numerator as it varies directly as y, meaning, when y increases, x also increases and vice versa.

Meanwhile, z^2 is in the denominator because it varies inversely as y--when y increases, z^2 decreases and vice versa.

Now that we have an equation to work with, we can use the given values of x, y, and z to solve for k.

`\begin{gathered} 40=(k(64))/(4^2) \\ \\ 40=(64k)/(16) \\ \\ 40=4k \\ 10=k \end{gathered}`

Now that we know the value of k, we can solve for y when x =2 and z = 6 using the same equation.

`\begin{gathered} y=(10(2))/(6^2) \\ \\ y=(20)/(36) \\ \\ y=(5)/(9) \end{gathered}`

The answer is 5/9.