Step-by-step explanation:
P varies directly as the cubic root of Q means that:
[tex]p = x \times \sqrt[3]{q} [/tex]
, where x is a real constant number.
Which means that:
[tex]x = \frac{p}{ \sqrt[3]{q} } [/tex]
for q ≠0.
So we get that for q=8,p=4 which means:
[tex]x = \frac{4}{ \sqrt[3]{8} } = \frac{4}{2} = 2[/tex]
As a result we get that for Q=64:
[tex]p = 2 \times \sqrt[3]{64} = 2 \times 4 = 8 \\ since \: {4}^{3} = 4 \times 4 \times 4 = \\ 16 \times 4 = 64[/tex]
So for q=64,p=8.
Answer:
8
Step-by-step explanation:
If P varies directly with the cube root of Q, then there is a constant k such that:
[tex]P=k\sqrt[3]{x}[/tex]
So we are given P=4 when Q=8. Plug this into find our constant, k.
[tex]4=k \cdot \sqrt[3]{8}[/tex]
[tex]4=k \cdot 2[/tex]
Divide both sides by 2:
[tex]2=k[/tex]
So the equation no matter the P and the Q is:
[tex]P=2 \cdot \sqrt[3]{x}[/tex]
What is P when Q=64?
[tex]P=2 \cdot \sqrt[3]{64}[/tex]
[tex]P=2 \cdot 4[/tex]
[tex]P=8[/tex]
