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If the perimeter of a rectangle is 58m and length is 2m more than twice the width what is length and width?
Length = 20 m and width = 9 m
How do you get the answer to the math problem The perimeter of a rectangle is 58m and the length is 2m more than twice the width find the dimension for the length and the width?
Perimeter of a rectangle is 2 times (length + width) P = 2 (L+W) In this case the length is 2m more than twice the width. That is written L = 2 + 2W P = 2 (L + W) 58 = 2 (2+ 2W + W) 58 = 2 (2 + 3W) 58/2 = (2(2+3W))/2 29 = 2 + 3W 29 - 2 = 2 + 3W - 2 27 = 3W 9 = W L = 2 + 2W L = 2 +2(9) L = 2 + 18 L = 20 For a rectangle with a perimeter of 58 m and a length that is 2 m more than twice the width, the length is 20 m and the width is 9 m.
What is the side length of a square with an area of 40 m squared?
it is 2
If a scale is given as 1cm to 20m find the real length of a line if its scale drawing size is 6cm?
1cm:15m If 1 cm = 20 m then 6 cm = 120 m real length
What is Length and width of a rectangular lot with a perimeter of 88 meters if the length is 4 meters more than the width?
P = 2(l + w) & l = w + 4 Substitute P = 2((w + 4) + w) P = 2(2w + 4) P = 4w + 8 4w = P - 8 w = (P - 8) / 4 Substitute w = (88 - 8) / 4 w = 80/4 w = 20 m Hence l = 24 m
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