Question

A rectangle has a length of the cube root of 81 inches and a width of 3 to the 2 over 3 power inches. Find the area of the rectangle.

3 to the 2 over 3 power inches squared
3 to the 8 over 3 power inches squared
9 inches squared
9 to the 2 over 3 power inches squared

Answer 1

$A= \sqrt[3]{81}*3^{ \frac{2}{3} }= \sqrt[3]{3^4}*3^{ \frac{2}{3} } = 3^{ \frac{4}{3} }*3^{ \frac{2}{3} }=3^{ \frac{4}{3} + \frac{2}{3} }=3^2=9$

9 inches squared

Answer 2

Answer:

9 square inches.

Step-by-step explanation:

We have been given that a rectangle has a length of the $\sqrt[3]{81}$ inches and a width of $3^{\frac{2}{3}}$ power inches. We are asked to find the area of given rectangle.

We know that area of rectangle in length times width of rectangle.

$\text{Area of rectangle}=\sqrt[3]{81}\times 3^{\frac{2}{3}}$

We can write 81 as $3^4$ as:

$\text{Area of rectangle}=\sqrt[3]{3^4}\times 3^{\frac{2}{3}}$

Using exponent rule $\sqrt[n]{a^m}=a^{\frac{m}{n}}$, we can write $\sqrt[3]{3^4}=3^{\frac{4}{3}}$.

$\text{Area of rectangle}=3^{\frac{4}{3}}\times 3^{\frac{2}{3}}$

Using exponent rule $a^b\cdot a^c=a^{b+c}$, we will get:

$\text{Area of rectangle}=3^{\frac{4}{3}+\frac{2}{3}}$

$\text{Area of rectangle}=3^{\frac{4+2}{3}}$

$\text{Area of rectangle}=3^{\frac{6}{3}}$

$\text{Area of rectangle}=3^{2}$

$\text{Area of rectangle}=9$

Therefore, the area of given rectangle is 9 square inches.