If the length of fred's piece of paper is represented by 2x-6 ad the width is represented by 3x-5, then what is the perimeter and area of fred's paper?

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Answer 1 · 2018-03-26T02:03:35

Area $=6x^2-28x+30$

Perimeter $=10x-22$

So to start off, the perimeter is

$P = 2l+2w$

Then you input the width for $w$ and the length for $l$ . You get

$P = 2(2x-6) + 2(3x - 5)$

$P = 4x - 12 + 6x - 10$

$P = 10x - 22$

for the perimeter. For the area, you multiply.

$A=L*W$

So

$A = (2x-6)(3x-5)$

$= 6x^2-10x-18x+30$

$= 6x^2-28x+30$