SOLUTION: The length of a rectangle is 2 cm more than 5 times its width. If the area of the rectangle is 65 cm2, find the dimensions of the rectangle to the nearest thousandth Hint: Cal

Question 89473: The length of a rectangle is 2 cm more than 5 times its width. If the area of the rectangle is 65 cm2, find the dimensions of the rectangle to the nearest thousandth
Hint: Call the width x. Then the length is 5x + 2. Now write your equation and solve.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The length of a rectangle is 2 cm more than 5 times its width. If the area of the rectangle is 65 cm2, find the dimensions of the rectangle to the nearest thousandth
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Area = lw
Let width = x
Then length = 5x+2
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EQUATION:
x(5x+2) = 65
5x^2+2x = 65
5x^2+2x-65=0
x=[-2+-sqrt(4-4*5*-65)]/10
x=[-2+-sqrt1304]/10
x=[-2+-36.11..]/10
Positive solution:
x=[-2+36.11...]/10
x=4.411 (width)
5x+2 - 19.06 (length)
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Cheers,
stan H.

Positive solution:
x=-1+sqrt14 = 2.74... (width)
5x+2 = 15.71..

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