1. **State the problem:** We are given distances traveled by the hare and the tortoise and need to find the ratio of their distances, complete a table, graph the data, find an equation relating their distances, and determine the tortoise's distance when the hare finishes a 9000-foot race. 2. **Part (a): Find the ratio of the hare's distance to the tortoise's distance as a unit rate.** The hare has run 315 feet, and the tortoise has run 70 feet. The ratio is $$\frac{\text{hare's distance}}{\text{tortoise's distance}} = \frac{315}{70}$$ Simplify the fraction: $$\frac{315}{70} = \frac{315 \div 35}{70 \div 35} = \frac{9}{2} = 4.5$$ So, the hare runs 4.5 feet for every 1 foot the tortoise runs. **Answer:** The unit rate is 4.5 feet per 1 foot. 3. **Part (b): Complete the table.** We use the ratio from part (a) to find the hare's distance $H$ for each tortoise distance $T$: $$H = 4.5 \times T$$ Calculate for each $T$: - For $T=2$: $H = 4.5 \times 2 = 9$ - For $T=4$: $H = 4.5 \times 4 = 18$ - For $T=6$: $H = 4.5 \times 6 = 27$ - For $T=8$: $H = 4.5 \times 8 = 36$ - For $T=10$: $H = 4.5 \times 10 = 45$ | Tortoise, T (feet) | Hare, H (feet) | |--------------------|----------------| | 2 | 9 | | 4 | 18 | | 6 | 27 | | 8 | 36 | | 10 | 45 | 4. **Part (c): Graph the information.** Plot the points $(T, H)$ from the table on the graph with $T$ on the horizontal axis and $H$ on the vertical axis. The points are (2,9), (4,18), (6,27), (8,36), and (10,45). The points lie on a straight line because the relationship is linear. 5. **Part (d): Create an equation relating $T$ and $H$.** From the ratio and table, the equation is: $$H = 4.5T$$ This means the hare's distance is 4.5 times the tortoise's distance. 6. **Part (e): Find the tortoise's distance when the hare finishes a 9000-foot race.** Given $H = 9000$, solve for $T$: $$9000 = 4.5T$$ Divide both sides by 4.5: $$T = \frac{9000}{4.5} = 2000$$ **Answer:** The tortoise has traveled 2000 feet when the hare finishes the 9000-foot race. This completes the problem.