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Let L be the length and W be the width of the rectangle. Given: 2(W+L) = 18 -- (1) L=2W-3 -------- (2) Solution: From (2), 2W = L+3 --- (3) substitute (3) into (2). We get (L+3)+2L = 18. 3L + 3 = 18 L = 5. Substitute into (2) W = 4.
5.00 L = 305.12 cubic inches.
The figure could be a rectangle with length 6.79 and width 2.21(2dp). Area (A) of a rectangle = Length (L) x Width (W) = 15 : Then W = 15/L Perimeter = 2(L + W) = 18 : substituting for W then :- 2(L + 15/L) = 18 : 2L + 30/L = 18 : 2L2 - 18L + 30 = 0 Solving for a quadratic equation gives L = (18 ± √84) ÷ 4 = 4.5 ± √5.25 = 4.5 ± 2.29(2dp)
If the length of the ramp is L feet, then tan(20) = 18/L so that L = 18/tan(20) = 49.45 feet.
Perimeter = 2*(Length + Width) 54 = 2*(L + 9) = 2L + 18 So 2L = 54 - 18 = 36 and therefore L = 18 inches.