Respuesta :
L=W-2
A=LW and using L from above and A=15
(W-2)W=15
W^2-2W-15=0
W^2-5W+3W-15=0
W(W-5)+3(W-5)=0
(W+3)(W-5)=0, and W>0 so
W=5, and since L=W-2
L=3
The perimeter P is
P=2L+2W=2(L+W)
P=2(3+5)
P=16 units
A=LW and using L from above and A=15
(W-2)W=15
W^2-2W-15=0
W^2-5W+3W-15=0
W(W-5)+3(W-5)=0
(W+3)(W-5)=0, and W>0 so
W=5, and since L=W-2
L=3
The perimeter P is
P=2L+2W=2(L+W)
P=2(3+5)
P=16 units
L = W - 2
A = L x W
A = (W - 2) x W
A = W^2 - 2W
15 = W^2 - 2W
W^2 - 2W - 15 = 0
(W -5)(W + 3 ) = 0
W - 5 = 0; W = 5
W - 3 = 0; W = -3
so Width = 5
Length = 5 - 2 = 3
P = 2(5 +3)
P = 2 (8)
P = 16
answer: P = 16 units