Question

Write an equation describing the relationship of the given variables. Use it to determine y. y varies jointly as z and of the square of x and inversely as t and of the square root of w. When x=2, z=3, w=16, and t=3, then y=1. Find y when x=3, z=2, w=36 and t=1.The value for y is:Answer

Answer 1

From the question, we have that y varies jointly as z and of the square of x and inversely as t and of the square root of w.

This is mathematically written as

`\begin{gathered} y\propto zx^2\text{ and } \\ y\propto\frac{1}{t\sqrt[]{w}} \end{gathered}`

Combining we have

`\begin{gathered} y\propto\frac{zx^2}{t\sqrt[]{w}} \\ \propto\text{ is the proportionality sign } \end{gathered}`

Removing the proportionality sign and introducing a constant k, we have

`\begin{gathered} y=\frac{kzx^2}{t\sqrt[]{w}} \\ \text{cross multiplying we have } \\ kzx^2=yt\sqrt[]{w} \\ k=\frac{yt\sqrt[]{w}}{zx^2} \end{gathered}`

Now, plugging in the initial values of y, t, w, z and x, we have k as

`\begin{gathered} k=\frac{yt\sqrt[]{w}}{zx^2} \\ k=\frac{1*3\sqrt[]{16}}{3*2^2} \\ k=(3*4)/(3*4) \\ k=(12)/(12) \\ k=1 \end{gathered}`

So, the relationship is

`\begin{gathered} y=\frac{kzx^2}{t\sqrt[]{w}} \\ y=\frac{1zx^2}{t\sqrt[]{w}} \\ y=\frac{zx^2}{t\sqrt[]{w}} \end{gathered}`

Using the relationship

`y=\frac{zx^2}{t\sqrt[]{w}}`

to find y, we plug in the second values of x, z, t and w, we have

`\begin{gathered} y=\frac{zx^2}{t\sqrt[]{w}} \\ y=\frac{2*3^2}{1\sqrt[]{36}} \\ y=(2*9)/(1*6) \\ y=(18)/(6) \\ y=3 \end{gathered}`

Hence the answer is y = 3