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This is the formula to calculate the perimeter of a rectangle, where l = length and w = width.

Perimeter = 2l + 2w : This can also be written as 2(l + w).

Q: 2 l plus 2 w is used for what?

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P2 + l 2 + w

P=2L+2W or P= L+L+W+W In conclusion, yes it does.

Suppose the length and width are L and W Perimeter = 2(L+W) = 200 so L+W = 100 so L= 100-W Area = L*W = 2400 so (100-W)*W = 2400 W2 - 100W + 2400 = 0 W = (100 +/- sqrt(10000 - 4*2400))/2 where the minus sign will give W and the plus sign will give L So W = (100 - sqrt(400))/2 = (100 - 20)/2 = 80/2 = 40 feet So L = 100-40 = 60 feet

Perimeter: (2*L)+(2*W) [length times two plus width times two] Area: L*W [length times width]

Stating our known facts Let W = Width, and L = Length L = 2W, and W = W+2 ----What you given is a never ending loop, saying that W always equals itself + 2. I will assume that you meant to put W = L+2 instead. Letting L = 2W and W = L + 2, we can state that x (the total area) is defined asL W(substituted so we're only working for one term)x = (2W)(2W + 2) Simplifyingx = 4W2 + 4W Since we are left with 2 variables, it can be stated that there is not enough information to solve the problem.

Related questions

P2 + l 2 + w

That is the formula for a rectangle. In the case of a square, since length = width, this can be simplified as 4 x base, where "base" is the length of any of the 4 identical sides. In the case of a rectangle, "l+l+w+w" can be simplified to 2l + 2w, or 2(l+w).

A = 2lw + 2lh + 2wh so A - 2lw = 2lh + 2wh = 2h(l+w) = h*2(l+w) so h = 1/2*(A-2lw)/(l+w)

Length (L) x Width (W) = Area 2*L+2*W = Perimeter 48/W=L (solved for L) 2*48/W+2*W=32 (inserted L into perimeter equation) 48+W^2=16*W (quadratic equation or factor) W=12 or 4 Therefore L=4 when W= 12 or L=12 when W=4 Length (L) x Width (W) = Area 2*L+2*W = Perimeter 48/W=L (solved for L) 2*48/W+2*W=32 (inserted L into perimeter equation) 48+W^2=16*W (quadratic equation or factor) W=12 or 4 Therefore L=4 when W= 12 or L=12 when W=4

P=2L+2W or P= L+L+W+W In conclusion, yes it does.

Suppose the length and width are L and W Perimeter = 2(L+W) = 200 so L+W = 100 so L= 100-W Area = L*W = 2400 so (100-W)*W = 2400 W2 - 100W + 2400 = 0 W = (100 +/- sqrt(10000 - 4*2400))/2 where the minus sign will give W and the plus sign will give L So W = (100 - sqrt(400))/2 = (100 - 20)/2 = 80/2 = 40 feet So L = 100-40 = 60 feet

Perimeter: (2*L)+(2*W) [length times two plus width times two] Area: L*W [length times width]

A = Area l = length w = width A = (l)(w) Ex: L = 5 in. W = 2 in. A = ? A = l(w) A = 5(2) A = 10 in2

Stating our known facts Let W = Width, and L = Length L = 2W, and W = W+2 ----What you given is a never ending loop, saying that W always equals itself + 2. I will assume that you meant to put W = L+2 instead. Letting L = 2W and W = L + 2, we can state that x (the total area) is defined asL W(substituted so we're only working for one term)x = (2W)(2W + 2) Simplifyingx = 4W2 + 4W Since we are left with 2 variables, it can be stated that there is not enough information to solve the problem.

Q=L & w=R

Perimeter equals 2 Length plus 2 width so find what l a w is then multiply them

A = 3x2 + 9xy + 6y2l = 3x + 6yA = l×w∴ w = A/l∴ w = (3x2 + 9xy + 6y2) / (3x + 6y)∴ w = x + 1