Question

y varies jointly as x and z, and y = 32 m, x = 6 m, and z = 8 m. What is the value of y when x = 5 m and z = 12 m

Answer 1

We are given y = kxz
We also know 32 = k×6×8, so k = 32/48 = 2/3.
We are then asked to solve:


`y = (2)/(3) * 5 * 12 = 40`

Answer 2

y = 40

Explanation:

If y varies jointly as x and z, the equation is

y = k* xz

We know y x and z so we can substitute them in

32 = k * 6*8

32 = k*48

Divide by 48

32/48 = k*48/48

32/48 =k

2/3 =

So the equation is y =2/3 x*z

Since we know x=5 and z =12, we can find y

y = 2/3 * 5 * 12

y = 40