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Jim has designed a rectangle with an area of 100 square feet and a perimeter of 401 feet.
What are the dimensions of Jim's rectangle?please show me how u got the answer.

A = LW

Let's see which factors of 100 would add up to 401.

10 * 10
2 * 50
4 * 25
5 * 20

P = 2L + 2W

None of those come close to a perimeter of 401.

Are you sure you copied the problem correctly?

Yes and there is only one way

lw = 100
and 2l + 2w = 401
from the 1st .... l = 100/w

into the 2nd:
2(100/w) + 2w = 401
times w
200 + 2w^2 = 401w

2w^2 - 401w + 200 = 0
by the formula:

w = (401 ± √159201)/4
= (401 ± 399)/4
= 200 or .5

or , it factors to
(w-200)(2w - 1) = 0
w = 200 or w = 1/2

if w=200, the l = 100/200 = 1/2
if w = 1/2, the l = 100/(1/2) = 200

The rectangle is 200 feet by 1/2 ft

check:
area = (1/2)(200) = 100 , checks!
perimeter = 200 + 200 + 1/2 + 1/2 = 401, checks!

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