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July 24, 2014

July 24, 2014

Posted by **Greg** on Sunday, July 24, 2011 at 8:26am.

(type an integer or a decimal. Round to the nearest hundredth)

- Algebra -
**Henry**, Sunday, July 24, 2011 at 4:47pm98mi @ X mi/hr.

10mi @ (x-5) mi/hr.

T = 98/x + 10/(x-5) = 3hrs.

Multiply both sides by x(x-5):

98(x-5) + 10x = 3x(x-5),

98x - 490 + 10x = 3x^2 - 15x,

-3x^2 + 108x + 15x - 490 = 0,

-3x^2 + 123x -490 = 0,

Solve using Quadratic Formula and get:

X = 36.53, and 4.47.

The required value of x is > 5.

X = 36.53 mi/hr = Speed during 1st part of trip.

X-5 = 31.53 = Speed during 2nd part of trip.

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