Algebra 2
posted by Anonymous .
An employee drove to work on Monday at 45 mi/h and arrived one minute early. The employee drove to work on Tuesday, leaving home at the same time driving 40 mi/h and arriving one minute late.
a. How far does the employee live from work?
b. At what speed should the employee drive to arrive five minutes early?

Using V = D/T
60D/45 = T  1
60D/40 = T + 1
Solving for T and equating yields
60(40D  45D)/1800 = 2 or 5D = 60 making D = 12 miles
In order to arrive in 17 minutes,
60(12)/V = 17 yielding = 42.35mph.
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