the length of a rectangle L is 1 more than twice the width, and perimeter is 62.find the width and the width???

polynomials

(0.7x^2 + 0.2x-0.8)-(0.9x^2-1.1x-1.4)=

(3x-6)(2x+4)

4xy^2(7x^3 +3x^2y^2-9y^3)

(4x^3-3x^2+3x-4)(6x-4)

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- Oct 21st 2009, 01:32 PMBeBeMalahelp please
the length of a rectangle L is 1 more than twice the width, and perimeter is 62.find the width and the width???

polynomials

(0.7x^2 + 0.2x-0.8)-(0.9x^2-1.1x-1.4)=

(3x-6)(2x+4)

4xy^2(7x^3 +3x^2y^2-9y^3)

(4x^3-3x^2+3x-4)(6x-4) - Oct 21st 2009, 02:10 PMpickslides

Please post each question in a separate thread with an explanantion on what is required and your attempts at solution.

$\displaystyle P = 2W+2L$

Where $\displaystyle L = 2W+1$ and $\displaystyle P = 62$

therefore $\displaystyle 62 = 2W+2(2W+1)$

$\displaystyle 62 = 2W+4W+2$

$\displaystyle 62 = 6W+2$

Can you solve it from here? - Oct 21st 2009, 02:13 PMpickslides
- Oct 21st 2009, 02:46 PMBeBeMala
polynomial:=0.2x^2 +1.3x +0.6????

the width is 10?? how i find the length?? - Oct 21st 2009, 02:51 PMstapel
You find the length by solving the "width" equation you were given, and plugging the solution value into the "length" formula you were given. (Wink)