Question 1163254
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If the one going 70 mph leaves at 10 and the other leaves at 11 am, the first one will have travelled 70 miles in that first hour, so that when they are both moving, they will start at 150 miles apart.  Since they are traveling in opposite directions, their speed of approach is the sum of their speeds, namely 135 miles per hour.  So they will meet at *[tex \Large \frac{150}{135}\ =\ \frac{10}{9}] hours after 11 am, or 12:06:40 pm.  A travels a total of 70 miles plus *[tex \Large \frac{10}{9}] hours times 70 miles per hour.  B travels a total of *[tex \Large \frac{10}{9}] hours times 65 miles per hour.  
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
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