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August 29, 2014

August 29, 2014

Posted by **Stella** on Sunday, April 22, 2012 at 3:36pm.

The speed on the first part is?

The speed on the second part is? Someone was kind enough to help, but it did not work into the equation. Help? Please..

- Algebra -
**Henry**, Tuesday, April 24, 2012 at 7:46pmV1 = X mi/h on 1st part of trip.

t1 = d1/V1 = 57 / X hrs.

t2 = d2/V2 = 13 / (x-5) hrs.

t1 + t2 = 5 hrs.

57/x + 13/(x-5) = 5.

Multiply both sides by x(x-5).

57(x-5) + 13x = 5x(x-5).

57x - 285 + 13x = 5x^2 - 25x.

-5x^2 + 57x +13x +25x = 285.

-5x^2 + 95x = 285.

-5x^2 + 95x - 285 = 0.

Divide both sides by -5:

x^2 - 19x + 57 = 0.

Use Quadratic Formula.

X = 15.3 mi/h = Speed on 1st part of trip.

x-5 = 15,3-5 = 10.3 mi/h = Speed on 2nd

part of trip.

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