w x 2w = 63. 2w^2 = 63 w^2 = 31.5 w = sqrt 31.5 = 5.6125 so L = 11.225 which gives area of 63.0003, which is near enough for Wiki!
From the statement of the problem, if w is the width, the area is 2w2 , the product of the width and the length, which is stated to be twice the width. Since 2w2 must be less than 50, w2 < 25, and the width must be less than 5 meters.
4 cm
9" x 4"
Let length be denoted by l, and width by w. Length is 4 ft more than twice the width => l = 2w + 4 five times the width is the same as twice the length increased by 10 feet. =>5w = 2l + 10 Solving the two equations we get l = 40 w = 18 So, the rectangle is 40 ft long and 18 ft wide.
you can't, unless the area was an odd number you can't, unless the area was an odd number
From the statement of the problem, if w is the width, the area is 2w2 , the product of the width and the length, which is stated to be twice the width. Since 2w2 must be less than 50, w2 < 25, and the width must be less than 5 meters.
3 and 7
Badly expressed! Do you mean "what is the area of a rectangle with a length that is 3 more than twice its width?" ? If you have a width, then presumably you have a length, in which case you presumably have a rectangle. But what do we want to know about this rectangle? You haven't said. Its area? Its perimeter? Its color? Its shoe size?
Factors of 28 include 7 x 4. 7 is 1 less than twice 4...
Length = 20 m and width = 9 m
lenght=15, width= 10
15
12 meters
154 Divided by8
Area of a rectangle is length times width, therefore if the length is two feet longer than the width then the length is 9 and the width is 7, when they are multiplied you get 63
width=15 length=35
4 cm