Two familes are taking a vacation together. They each leave at 9 a.m. on Saturday morning, driving their cars along the same route to a cabin on - The Jacksons Ive 40 miles closer to the cabin than the Miligans. - The Jacksons drive at an average speed of 50 miles per hour, and the Miligans drive at an average speed of 60 miles per hour. An equation can be written to represent this stuation for each family relating the time, \( x \), they drive and the distance, \( y \), they are from the cabin. Decide if each statement is true or false. The equation for the Jacksons is \( y=40 x+50 \) and the equation for the Chases is \( y=x+60 \). True False The cars will reach the same point in 4 hours. It the cabin is 400 mies away, it will take the Jacksons 7 hours and 12 minutes to get there. The two cars will reach the same point at noon.

Let's analyze each statement based on the given information: ### Given: - **Jacksons** are **40 miles closer** to the cabin than the **Miligans**. - **Jacksons' speed**: **50 mph** - **Miligans' speed**: **60 mph** - Both families leave at **9 a.m.** on Saturday. ### 1. **The equation for the Jacksons is \( y = 40x + 50 \) and the equation for the Miligans is \( y = x + 60 \). True or False?** **False.** **Explanation:** - Let \( D \) be the initial distance of the Jacksons from the cabin. Then, the Miligans are \( D + 40 \) miles away. - For the Jacksons: Distance from cabin after \( x \) hours is \( y = D - 50x \). - For the Miligans: Distance from cabin after \( x \) hours is \( y = (D + 40) - 60x \). The proposed equations \( y = 40x + 50 \) and \( y = x + 60 \) incorrectly model the relationship because they suggest the distance from the cabin **increases** with time, whereas it should **decrease**. Additionally, the coefficients and constants do not align with the given speeds and distance differences. ### 2. **The cars will reach the same point in 4 hours. True or False?** **True.** **Explanation:** To find when both cars are at the same distance from the cabin: - **Jacksons**: \( y = D - 50x \) - **Miligans**: \( y = (D + 40) - 60x \) Setting them equal: \[ D - 50x = D + 40 - 60x \\ 10x = 40 \\ x = 4 \text{ hours} \] Thus, both cars will be at the same point **4 hours** after departure, which is at **1 p.m.**. ### 3. **If the cabin is 400 miles away, it will take the Jacksons 7 hours and 12 minutes to get there. True or False?** **True.** **Explanation:** Assuming the cabin is **400 miles** away from the **Miligans' starting point**, the Jacksons are **40 miles closer**, so: \[ \text{Jacksons' distance to cabin} = 400 - 40 = 360 \text{ miles} \] Time taken by Jacksons: \[ \frac{360 \text{ miles}}{50 \text{ mph}} = 7.2 \text{ hours} = 7 \text{ hours and } 12 \text{ minutes} \] ### 4. **The two cars will reach the same point at noon. True or False?** **False.** **Explanation:** The families depart at **9 a.m.**, so **noon** is **3 hours** later. According to statement 2, they reach the same point in **4 hours** (1 p.m.), not at noon. Furthermore, checking their positions at **3 hours**: - **Jacksons**: \( y = 360 - 50 \times 3 = 210 \text{ miles} \) from cabin. - **Miligans**: \( y = 400 - 60 \times 3 = 220 \text{ miles} \) from cabin. They are **not** at the same point at noon. --- ### **Summary of Statements:** 1. **False** 2. **True** 3. **True** 4. **False** **Answer:** 1. False 2. True 3. True 4. False