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In a rectangle: P = 2(L + W) and A = LW P = 2(L + W) (replace P with 20) 20 = 2(L + W) (solve for L; divide by 2 to both sides) 10 = L + W (subtract W to both sides ) 10 - W = L A = LW (substitute 10 - W for L) A = (10 - W)W A = 10W - W2
8m Let the width of the table be w; then its length is w + 2. Area = length x width 80 = (w + 2)w w2 + 2w - 80 = 0 (w + 10)(w - 8) = 0 w = -10 or 8 As the width must be positive, it is 8m.
W = j + 10
Perimeter is 70, length is width + 5 so 2L + 2W = 70 and L = W + 5. So 2 (W + 5) + 2W = 70 ie 2W + 10 + 2W = 70 ie 4W = 70 - 10 ie 4W = 60 ie W = 15 and L = 20 Alternative method: L + W = 35, L = W + 5 W + 5 + W = 35 2W = 30, W =15
if width is equal to W and length is equal to L, then W x L = 80 L = W +2 so W x (W+2) = 80 solve to find W=8 and L= 10