1. **Problem Statement:** We are given a rational function for braking distance $D(x) = \frac{2500}{30(0.3 + x)}$ where $x$ is the grade of the hill expressed as a decimal. We need to: (a) Evaluate $D(0.05)$ and interpret it. (b) Describe how braking distance changes as the hill gets steeper. (c) Find the grade $x$ when braking distance is 220 ft. 2. **Formula and Explanation:** The braking distance function is: $$D(x) = \frac{2500}{30(0.3 + x)}$$ Here, $x$ is the hill grade (rise over run), and $D(x)$ is the braking distance in feet. 3. **Part (a) Evaluate $D(0.05)$:** Substitute $x=0.05$: $$D(0.05) = \frac{2500}{30(0.3 + 0.05)} = \frac{2500}{30 \times 0.35} = \frac{2500}{10.5} \approx 238.10$$ Rounded to the nearest foot, $D(0.05) \approx 238$ ft. **Interpretation:** At a 5% uphill grade, the braking distance is about 238 feet on a wet road at 50 mph. 4. **Part (b) Behavior as hill gets steeper:** As $x$ increases, the denominator $30(0.3 + x)$ increases, so $D(x)$ decreases because the function is inversely proportional to $0.3 + x$. This means the braking distance gets shorter as the hill becomes steeper uphill. **Driving experience:** This makes sense because going uphill helps slow the car, reducing braking distance. 5. **Part (c) Find $x$ for $D(x) = 220$ ft:** Set $D(x) = 220$: $$220 = \frac{2500}{30(0.3 + x)}$$ Multiply both sides by $30(0.3 + x)$: $$220 \times 30(0.3 + x) = 2500$$ $$6600(0.3 + x) = 2500$$ Divide both sides by 6600: $$0.3 + x = \frac{2500}{6600} \approx 0.3788$$ Solve for $x$: $$x = 0.3788 - 0.3 = 0.0788$$ Convert to percent: $$x = 7.9\%$$ **Answer:** The grade associated with a braking distance of 220 ft is approximately 7.9%.