Tuesday

March 31, 2015

March 31, 2015

Posted by **Anonymous** on Tuesday, November 1, 2011 at 1:30am.

a. How far does the employee live from work?

b. At what speed should the employee drive to arrive five minutes early?

- Algebra 2 -
**tchrwill**, Tuesday, November 1, 2011 at 10:30amUsing 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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