1. **State the problem:** We need to find the weight of 5 washers. Each washer is a cylindrical ring with an inner diameter of $\frac{1}{4}$ inch, outer diameter of $\frac{3}{4}$ inch, thickness of $\frac{1}{4}$ inch, and density $0.285$ pounds per cubic inch. 2. **Formula for volume of a washer:** The volume $V$ of a washer (a cylindrical ring) is the volume of the outer cylinder minus the volume of the inner cylinder: $$V = \pi h (R^2 - r^2)$$ where $h$ is thickness, $R$ is outer radius, and $r$ is inner radius. 3. **Calculate radii:** $$R = \frac{3}{4} \div 2 = \frac{3}{8}$$ $$r = \frac{1}{4} \div 2 = \frac{1}{8}$$ 4. **Calculate volume of one washer:** $$V = \pi \times \frac{1}{4} \times \left(\left(\frac{3}{8}\right)^2 - \left(\frac{1}{8}\right)^2\right)$$ Calculate squares: $$\left(\frac{3}{8}\right)^2 = \frac{9}{64}, \quad \left(\frac{1}{8}\right)^2 = \frac{1}{64}$$ Subtract inside parentheses: $$\frac{9}{64} - \frac{1}{64} = \frac{8}{64} = \frac{1}{8}$$ So volume: $$V = \pi \times \frac{1}{4} \times \frac{1}{8} = \pi \times \frac{1}{32}$$ 5. **Calculate volume numerically:** $$V = \pi \times \frac{1}{32} \approx 3.1416 \times 0.03125 = 0.09817 \text{ cubic inches}$$ 6. **Calculate weight of one washer:** Weight $= \text{density} \times \text{volume} = 0.285 \times 0.09817 = 0.02797$ pounds 7. **Calculate weight of 5 washers:** $$5 \times 0.02797 = 0.13985$$ 8. **Round to nearest hundredth:** $$0.13985 \approx 0.14$$ **Final answer:** The weight of 5 washers is approximately **0.14 pounds**.