Q38. - Algebra 1

Question

When air is pumped into a ball, the pressure required can be computed by using the formula $\mathit{P}\mathbf{=}\frac{\mathbf{3412}\mathbf{.94}}{\frac{\mathbf{4}\mathbf{\pi }{\mathbf{r}}^{\mathbf{3}}}{\mathbf{3}}}$, where P represents the pressure in pound per square inch (psi), and r is the radius of the ball in inches.

a. Simplify the complex fraction.

b. Suppose the air pressure inside the ball is 8 psi. Approximate the radius of the ball to the nearest hundredth.

Step-by-Step Solution

Answer

a. The simplified fraction is $P=\frac{\mathbf{814}\mathbf{.45}}{r^3}$

b. Radius of the ball is 4.67 inches.

Step 1. State the concept of Simplification.

To reduce a fraction to its lowest terms by canceling to the lowest common factor for both numerator and denominator or to condense an algebraic expression by grouping and combining similar terms.

Step 2. Simplify the expression.

The given fraction can be simplified as follows:

$\begin{array}{c}P=\frac{3412.94}{\frac{4\pi r^3}{3}}\\ =3412.94\times \frac{3}{4\pi r^3}\\ =3412.94\times \frac{3}{4\left(\frac{22}{7}\right)r^3}\\ =3412.94\times \frac{3}{\frac{88}{7}{r}^3}\\ =3412.94\times \frac{3\times 7}{88r^3}\\ =\frac{71671.74}{88r^3}\\ =\frac{814.45}{r^3}\end{array}$

Step 3. Find the value of r .

The radius of the ball is calculated as follows:

$\begin{array}{c}P=\frac{814.45}{r^3}\\ 8=\frac{814.45}{r^3}\\ 8\times r^3=814.45\\ \frac{8r^3}{8}=\frac{814.45}{8}\\ r^3=101.81\\ r=4.6694\\ r\approx 4.67\end{array}$

Step 4 . Write the conclusion.

Therefore, the simplified fraction is $\frac{814.45}{r^3}$ and the radius of the ball is 4.67 inches.