Question

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

Answer 1

Step-by-step explanation:

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

$p = x \times \sqrt[3]{q}$

, 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:

$x = \frac{4}{ \sqrt[3]{8} } = \frac{4}{2} = 2$

As a result we get that for Q=64:

$p = 2 \times \sqrt[3]{64} = 2 \times 4 = 8 \\ since \: {4}^{3} = 4 \times 4 \times 4 = \\ 16 \times 4 = 64$

So for q=64,p=8.

Answer 2

Answer:

8

Step-by-step 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?

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

$P=2 \cdot 4$

$P=8$