Area= length * width length= width + 5 so.... Area= width * (width + 5) Area= width^2 + 5width (going to use w for width.) 66= w^2 + 5w w^2 + 5w - 66= 0 Now factor! (w - 6)(w + 11) Solve for w, and since area can't be negative, you get.. w=6 length= 6+5= 11
The dimensions are W and W+M where W is the width.
w=width
The width would be: P - (L*2) = W*2 Perimeter = P Length = L Width = W
Perimeter = length + length + width + width Let "w" = width, then "7w" must equal the length (because it's 7 times the width) 7w + 7w + w + w = perimeter 16w = perimeter
It is (W - 6) feet where W is the width measured in feet.
Area= length * width length= width + 5 so.... Area= width * (width + 5) Area= width^2 + 5width (going to use w for width.) 66= w^2 + 5w w^2 + 5w - 66= 0 Now factor! (w - 6)(w + 11) Solve for w, and since area can't be negative, you get.. w=6 length= 6+5= 11
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.
The dimensions are W and W+M where W is the width.
w=width
An adult emporer penguin's width woul be about 2-3 feet.
Generally - width is denoted by 'w' as inthe unit measurements are l:36", h: 24", w:18"Where l = length, h = height, and w - width.
The width would be: P - (L*2) = W*2 Perimeter = P Length = L Width = W
Perimeter = length + length + width + width Let "w" = width, then "7w" must equal the length (because it's 7 times the width) 7w + 7w + w + w = perimeter 16w = perimeter
The abbreviation for "width" is usually W, as in "H x D x W x", where "H" is height, "D" is depth, and "W" is width, when describing the size of a 3 dimensional object.
Suppose the width is W yards.Then twice the width is 2W yards3 yards less that twice the width is (2W - 3) yards.That is, the length is (2W - 3) yardsSo, the area = Length *Width = (2W - 3)*W which is known to be 27Therefore 2W2 - 3W = 27or 2W2 - 3W - 27 = 0so (W + 3)*(2W - 9) = 0so W = -3 or W = 4.5Since W cannot be negative, W = 4.5 yards and then L = 6 yards.Suppose the width is W yards.Then twice the width is 2W yards3 yards less that twice the width is (2W - 3) yards.That is, the length is (2W - 3) yardsSo, the area = Length *Width = (2W - 3)*W which is known to be 27Therefore 2W2 - 3W = 27or 2W2 - 3W - 27 = 0so (W + 3)*(2W - 9) = 0so W = -3 or W = 4.5Since W cannot be negative, W = 4.5 yards and then L = 6 yards.Suppose the width is W yards.Then twice the width is 2W yards3 yards less that twice the width is (2W - 3) yards.That is, the length is (2W - 3) yardsSo, the area = Length *Width = (2W - 3)*W which is known to be 27Therefore 2W2 - 3W = 27or 2W2 - 3W - 27 = 0so (W + 3)*(2W - 9) = 0so W = -3 or W = 4.5Since W cannot be negative, W = 4.5 yards and then L = 6 yards.Suppose the width is W yards.Then twice the width is 2W yards3 yards less that twice the width is (2W - 3) yards.That is, the length is (2W - 3) yardsSo, the area = Length *Width = (2W - 3)*W which is known to be 27Therefore 2W2 - 3W = 27or 2W2 - 3W - 27 = 0so (W + 3)*(2W - 9) = 0so W = -3 or W = 4.5Since W cannot be negative, W = 4.5 yards and then L = 6 yards.
A rectangle with length L and width W has perimeter = 2*(L+W) So, 2(6 + W) = 220 or 6 + W = 110 so that W = 104 Usually, Length > Width but clearly not in this case.