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

Perimeter $= 10 x - 22$ $=10x-22$

So to start off, the perimeter is

$P = 2 l + 2 w$ $P = 2l+2w$

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

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

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

$P = 10 x - 22$ $P = 10x - 22$

for the perimeter. For the area, you multiply.

$A = L \cdot W$ $A=L*W$

So

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

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

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