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April 24, 2014

April 24, 2014

Posted by **Anonymous** on Monday, April 18, 2011 at 3:02pm.

- Algebra -
**Henry**, Wednesday, April 20, 2011 at 5:26pmRate of 2nd cyclist = X mi/h.

Rate of 1st cyclist = (x+10) mi/h.

Eq1: d2 = Xt = 210mi.

X = 210 / t.

Eq2: d1 = (X+10)(t-2.4) = 210mi.

In Eq2, substitute 210/t for X:

(210/t+10)(t-2.4) = 210,

210-504/t+10t-24 = 210,

-504/t+10t-24 = 210-210,

-504/t+10t-24 = 0,

Multiply both sides by t:

10t^2-24t-504 = 0,

Divide both sides by 2:

5t^2-12t-252 = 0.

Solve using Quadratic Formula and get:

t = 8.4, and -6. Select positive t.

t = 8.4h.

X = 210/t = 210 / 8.4 = 25mi/h.

X+10 = 25+10 = 35mi/h = Rate of 1st cyclist.

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