• October 21st 2009, 01:32 PM
BeBeMala
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)
• October 21st 2009, 02:10 PM
pickslides
Quote:

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

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

$P = 2W+2L$

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

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

$62 = 2W+4W+2$

$62 = 6W+2$

Can you solve it from here?
• October 21st 2009, 02:13 PM
pickslides
Quote:

Originally Posted by BeBeMala

polynomials

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

group like terms here, expand the negative through the second expression first.

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

Can you finish it?
• October 21st 2009, 02:46 PM
BeBeMala
polynomial:=0.2x^2 +1.3x +0.6????

the width is 10?? how i find the length??
• October 21st 2009, 02:51 PM
stapel
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)