Question

Suppose z varies directly with y and directly with the cube of x. If z = 2560 when = 4 and y = 10, what is z when I 4? 8 and y z = I Question Help: D Video

Answer 1

It is given that z varies directly with y and directly with the cube of x.

`z=kyx^3`

Where k is the constant of proportionality.

First, we need to find the value of k.

It is given that z = 2560 when x = 4 and y = 10

`\begin{gathered} z=kyx^3 \\ (z)/(yx^3)=k \\ k=(z)/(yx^3) \\ k=(2560)/(10\cdot(4)^3) \\ k=4 \end{gathered}`

So, the value of k is 4

The relation becomes

`z=4yx^3`

Now we can easily find the value of z when x = 8 and y = 4

`\begin{gathered} z=4(4)(8)^3 \\ z=4(4)(512) \\ z=8192 \end{gathered}`

Therefore, the value of z is 8192