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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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)
Quote:
Originally Posted by BeBeMala
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?
group like terms here, expand the negative through the second expression first.Quote:
Originally Posted by BeBeMala
$\displaystyle (0.7x^2 + 0.2x-0.8)-(0.9x^2-1.1x-1.4)=0.7x^2 + 0.2x-0.8-0.9x^2+1.1x+1.4 $ $\displaystyle =(0.7x^2 -0.9x^2)+(1.1x+ 0.2x)+(1.4-0.8)=\dots$
Can you finish it?
polynomial:=0.2x^2 +1.3x +0.6????
the width is 10?? how i find the length??
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)
