Question

P varies directly as the cube root of Q. If P = 4 when Q = 8, find P when Q = 64
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Answer 1

Explanation:

P varies directly as the cubic root of Q means that:

`p = x * \sqrt[3]{q}`

, where x is a real constant number.

Which means that:

`x = \frac{p}{ \sqrt[3]{q} }`

for q ≠0.

So we get that for q=8,p=4 which means:

`x = \frac{4}{ \sqrt[3]{8} } = (4)/(2) = 2`

As a result we get that for Q=64:

`p = 2 * \sqrt[3]{64} = 2 * 4 = 8 \\ since \: {4}^(3) = 4 * 4 * 4 = \\ 16 * 4 = 64`

So for q=64,p=8.

Answer 2

8

Explanation:

If P varies directly with the cube root of Q, then there is a constant k such that:

`P=k\sqrt[3]{x}`

So we are given P=4 when Q=8. Plug this into find our constant, k.

`4=k \cdot \sqrt[3]{8}`

`4=k \cdot 2`

Divide both sides by 2:

`2=k`

So the equation no matter the P and the Q is:

`P=2 \cdot \sqrt[3]{x}`

What is P when Q=64?

`P=2 \cdot \sqrt[3]{64}`

`P=2 \cdot 4`

`P=8`