Ajogger ran 3 miles due east of his house. then he ran 5 miles at a heading of 30 east of north. how far is he from his house after running 8 miles?

Let's say her speed was x miles/hour during the first 3 miles runThen, time = distance/speedt1 = 3/x       eq1 In the next 4 miles run, her speed = x-1 miles/hourTime taken:t2 = 4/(x-1)      eq2 Now, total time:t1 + t2 = 1 3/5 hourssubstitute t1 and t2 from eqs. 1 and 2 3/x + 4/(x-1) = 1 3/5=> 3/x + 4/(x-1) = 8/5
=> 3(x-1) + 4x = 8x(x-1)/5=> 35x - 15 = 8x2 - 8x=> 8x2 - 43x + 15 = 0=> (8x-3)*(x-5) = 0=> x = 3/8 or 5 miles/hourx can not be 3/8 miles/hour because in that case, the speed during 4 miles run would be 3/8-1 = negative numberi.e. speed during 3 miles segment = 5 miles/hourand speed during 4 miles segment = 5-1 = 4 miles/hour

The third question is correct.