look at part c of the question. i accidently put that in. cause on the way there the speed is determined by the time so u just add both speeds and divide by two. but on the way back its determined by distance so u have to do a bunch of other math. but for part c.. how can the avg of the whole speed be72.5? i accidently put that in cause i thought the way there was 72.5 and back was 72.5 so that avg was obviously 72.5.. but since the way back was based on distance i had to rethink it... anyways how is the avg of the whole trip 72.5??