Question

How long is the edge of a cube whose volume is twice that of the cube below? Round to the nearest tenth.

Answer 1

Answer:

$\large\boxed{10.1\ cm}$

Step-by-step explanation:

The formula of a volume of a cube:

$V=a^2$

a - length of the edge

We have a = 8cm. Substitute:

$V=8^3=512\ cm^3$

The volume of second cube:

$V'=2V\to V'=2(512\ cm^3)=1024\ cm^3$

The length of the edge of the second cube (b):

$b^3=1024\to b=\sqrt[3]{1024}\ cm^3$

$1024=2^{10}=2^{9+1}=2^9\cdot2$

Used $a^n\cdot a^m=a^{n+m}$

$b=\sqrt[3]{1024}=\sqrt[3]{2^{10}}=\sqrt[3]{2^9\cdot2}=\sqrt[3]{2^9}\cdot\sqrt[3]{2}$

Use $(a^n)^m=a^{nm}$

$=\sqrt[3]{2^{3\cdot3}}\cdot\sqrt[3]{2}=\sqrt[3]{(2^3)^3}\cdot\sqrt[3]{2}$

Use $\sqrt[3]{a^3}=a$

$=2^3\sqrt[3]{2}=8\sqrt[3]{2}\ cm$

$\sqrt[3]{2}\approx1.26\to b\approx8(1.26)\approx10.1\ cm$

Other method:

$V'=2V\\\\V=a^3,\ V'=b^3\to b^3=2a^3\to b=\sqrt[3]{2a^3}\\\\b=a\sqrt[3]2$

$a=8\ cm\to b=8\sqrt[3]2\ cm\approx10.1\ cm$