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What are the dimensions of a rectangle with an area of fifty square feet and a perimeter of 201 feet?
Area of a rectangle = length(L) x width(W) Perimeter of a rectangle = 2L + 2W. 50 = LW : therefore W = 50/L 201 = 2L + 2 x (50/L) = 2L + 100/L : Multiply by L 201L = 2L2 + 100 : Therefore 2L2 - 201L + 100 = 0 This quadratic equation factorises (2L - 1)(L -100) = 0 When 2L - 1 = 0 then L = 1/2 When L - 100 = 0 then L = 100 The dimensions of the rectangle are therefore length = 100 ft, width = 1/2 foot
What are the dimensions of a rectangle with a perimeter of 401 feet and an area of 100 feet?
The rectangle cannot have an area of 100 feet, since that is a measure of distance and not area. Assume, therefore, that the area is 100 SQUARE feet. Suppose the rectagle has length L and width W then 2(L + W) = 401 and L*W = 100 So L = 100/W Then 2(100/W + W) = 401 Multiply by W: 2(100 + W2) = 401W or 2W2 - 401W + 200 = 0 or (2W - 1)(W-200) = 0 So W = 0.5 ft or W = 200 ft which imply that L = 200 ft or 0.5 ft respectively. So, the dimensions are 200 ft by 0.5 ft (or 6 inches).
Give the dimensions of the rectangle with an area of 100 square units and whole-number side lengths that has the largest perimeter?
w 20; l 30
What is the equivilant to 100 m into pounds?
There is no sensible answer to this question. A metre is a measure of length, with dimensions [L]. A pound is a measure of mass (or currency), with dimensions [M or £]. Basic dimensional analysis teaches that you cannot convert between measures with different dimensions without additional information.
What are the dimensions of mass density?
Density has dimensions of (mass) divided by (volume) = M L-3
