Asked by Anonymous

the area of a certain rectangle is 288 yd squared. the perimeter is 68 yards. f you double the length and width what will be the area and perimeter of the new rectangle
Answered by helper
Area (A) = Lw = 288
Perimeter (P) = 2L + 2w = 68
L = length, w = width
Solve 2L + 2w = 68 for either L or w
I'll choose L
2L + 2w = 68
2L = 68 - 2w
2L = 2(34 - w)
L = 34 - w

Substitute L = 34 - w, for L in
A = Lw = 288
(34 - w)(w) = 288

34w - w^2 = 288
w^2 - 34w + 288 = 0
w = 18, 16
So, this certain rectangle has dimensions of 18 by 16.

you should be able to calculate the area and perimeter of the new rectangle now. (double the length and width and solve the equations)





Answered by Ashley
r = 2/3t + v, for t
Answered by Mckenlie Allen
A=1120yd squared P= 134yd squared
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