The perimeter of a rectangular lot is 88 m, and its area is 480 m^2. Find the width and the length.
2006-12-31 04:15:32 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
the sum of two uneven sides is 44m. So (44-x)(44-y) = 480 sq m.
find y to create a polynomial:
y = 44-x
so x(44-x) = 480
44x - x^2 = 480
Create our standard zero polynomial:
-x^2 + 44x -480 = 0
(-x + 20)(x - 24) = 0
So x can equal 20 or 24.
(-20 + 20)(20 - 24) = 0 * -4 = 0 or
(-24 + 20)(24 - 24) = -4 * 0 = 0
So the width would be 20 m and the length would be 24 m.
2006-12-31 05:07:58 · answer #1 · answered by Rockstar 6 · 0⤊ 0⤋
The answer is length=24 width=20
The way to go about is
X^2-44x+480=0 solve for x
then X*y=480 solve for y
2006-12-31 04:28:51 · answer #2 · answered by Suhas 2 · 0⤊ 0⤋
20 x 24 m
2006-12-31 04:25:44 · answer #3 · answered by Jay M 4 · 0⤊ 0⤋
L for the length
W for the width
L*W=480
2*(L+W)=88
L+W=44
L=44-W
(44-W)*W=480
44*W-W^2=480
W^2-44*W+480=0
W=22-2=20
L=44-20=24
2006-12-31 04:32:57 · answer #4 · answered by T.M.M. 4 · 0⤊ 1⤋
let width be x m
and length be y cm
2(x+y)=88
x+y=44
xy = 480
x=24m
y=20m
2006-12-31 04:27:04 · answer #5 · answered by pratish 1 · 0⤊ 0⤋