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).
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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.