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6 December, 08:43

Annette drove to shawnee in 4 hours and drove back in 3 hours. what were her speeds if her speed coming back was 11 miles per hour greater than her speed going

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  1. 6 December, 09:12
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    The distance covered in both trips is the same.

    If x mph is the going speed, then coming back speed = (x+11) mph

    Distance = speed * time

    Therefore,

    4*x = 3 * (x+11) = > 4x = 3x+33 = > 4x-3x = 33 = > x = 33 mph, and x+11 = 44 mph

    The going speed is 33 mph while coming back speed is 44 mph
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