Jake drove 300 miles from his house to a soccer tournament. this graph shows the distance jake had traveled (m) and the time that had passed (h) since his trip started at 8: 00 am on friday.

Part A. The trip starts at 8am which corresponds to 0 hrs, point (0hr, 0mi)

2hrs later it's 10am. .point (2hr, 140mi)

The average speed is the slope between 0 and 2 hrs. Remember the slope formula m = Δy/Δx  

m = (140 - 0) / (2 - 0)  

m = 70 mph

Part B. Average speed from 11am - 2pm  

11am is point (3hr, 140mi)  

2pm is point (6hr, 300mi)  

As you can see from the graph, the speed or slope changes at 1pm (5,260). You Can just use the start and end points.

m  = (300-140) / (6-3)  

m = 160/3

53.3 mph

* It comes out the same solution as if you average the two different slopes. 2hrs at 60mph + 1 hr at 40mph = (120 + 40)/3 = 160/3  

Part C.  Total average speed = total distance / total time driving

He went 70 mph for 2 hrs

stopped for an hour (slope is zero, no speed)

60 mph for 2hrs

40mph for 1 hr

300mi /5hr = 60mph

Part D. No Question

the 2 figures are congruent bc if you reflect abcd over the y axis then over the d axis the figures overlap

1. minimum

2. 6400

3. eight

hope this !