SOLUTION: Eric leaves ashville at the rate of 65 mph, bound for Madison. At the same time Beth leaves Madison at the rate of 55mph, bound for ashville. If the cities are 540 miles apart how
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Question 505891: Eric leaves ashville at the rate of 65 mph, bound for Madison. At the same time Beth leaves Madison at the rate of 55mph, bound for ashville. If the cities are 540 miles apart how long will it take eric and beth to meet? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let t=amount of time it takes for them to meet
Now we know that Eric and Beth will have met when their combined distances add up to 540 mi.
Distance Eric travels=65t
Distance Beth travels=55t
Sooooo:
65t+55t=540
120t=540
t=4.5 hr
CK
Distance Eric travels=65*4.5=292.5 mi
Distance Beth travels=55*4.5=247.5 mi
292.5+247.5=540
540=540
Hope this helps----ptaylor
