Question

Suppose that y varies directly with x and inversely with z^2, & x=48 when y=8 and z=3. How do you find x when y=12 & z=2?

Answer 1

`x=32`

Explanation:

Equation can be built
`y=k*x/z^2`

we will find `k`

`8=k*48/3^2 => k=(9*8)/48 = 9/6=3/2`

now solve for 2nd part

`12 = 3/2 * x/2^2 => 12 = (3x)/8`
`4=x/8`

`x=32`