"x varies directly as y ^2 and inversely as z"
----> x = k(y^2 * 1/z)
repeat the method I used for your other two problems.
Suppose that x varies directly as y ^2 and inversely as z, and x is 5 when y is 3 and z is 3. What is z when x is 10 and y is 3?
2 answers
You can do it without finding k explicitly. You do know that
x = ky^2z, so x/(y^2z) = k, a constant. Thus, you just need to find z such that
10/(9z) = 5/(9*3)
or, even more simply in this case, since y has not changed, if x is twice as big, so is z.
x = ky^2z, so x/(y^2z) = k, a constant. Thus, you just need to find z such that
10/(9z) = 5/(9*3)
or, even more simply in this case, since y has not changed, if x is twice as big, so is z.
6 answers