5. The washer in the image has an inner diameter of \( 1 / 4 \) inch. The outer diameter \( 3 / 4 \) measures inch, and the washer is \( 1 / 4 \) inch thick. The density of the metal the washers are made of is 0.285 pounds per cubic inch. How much do 5 washers weigh, in pounds? Round your answer to the nearest hundredth.

First, let's calculate the volume of one washer. The volume of a cylindrical washer can be found using the formula for the volume of a cylinder, \( V = \pi h (R^2 - r^2) \), where \( R \) is the outer radius, \( r \) is the inner radius, and \( h \) is the thickness. Here, the outer radius \( R \) is \( \frac{3}{4} \) inches divided by 2 equals \( \frac{3}{8} \) inches, and the inner radius \( r \) is \( \frac{1}{4} \) inches divided by 2 equals \( \frac{1}{8} \) inches. Therefore, the volume of one washer is: \[ V = \pi \times \frac{1}{4} \times \left( \left( \frac{3}{8} \right)^2 - \left( \frac{1}{8} \right)^2 \right) \] Calculating the areas: \[ \left( \frac{3}{8} \right)^2 = \frac{9}{64}, \quad \left( \frac{1}{8} \right)^2 = \frac{1}{64} \] Thus, \[ \frac{9}{64} - \frac{1}{64} = \frac{8}{64} = \frac{1}{8} \] Now putting that back into the volume formula: \[ V = \pi \times \frac{1}{4} \times \frac{1}{8} = \frac{\pi}{32} \text{ cubic inches} \] Now, let's find the weight of one washer: \[ \text{Weight} = \text{Volume} \times \text{Density} = \frac{\pi}{32} \times 0.285 \] This results in: \[ \text{Weight of one washer} \approx 0.028 \text{ pounds (using } \pi \approx 3.14\text{)} \] Finally, for 5 washers: \[ 5 \times 0.028 \approx 0.140 \text{ pounds} \] Rounded to the nearest hundredth, the weight of the 5 washers is approximately \( \text{0.14 pounds} \).