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The area of a rectangle is 45 yd2 , and the length of the rectangle is 1 yd more than twice the width. Find the dimensions of the rectangle.
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Asked 9/27/2014 8:55:40 PM
Updated 9/28/2014 1:20:31 AM
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Flagged by yeswey [9/28/2014 1:17:08 AM]
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User: The area of a rectangle is 45 yd2 , and the length of the rectangle is 1 yd more than twice the width. Find the dimensions of the rectangle.

Weegy: rectangle...area=length*wide=63 length=2*wide+11 area=wide*(2*wide+11)=63 2*wide^2 +11wide -63=0 wide=3.5 or -9...use 3.5 length=2*3.5+11=11+7=18 18*3.5=63
natacia|Points 0|

Question
Asked 9/27/2014 8:55:40 PM
Updated 9/28/2014 1:20:31 AM
1 Answer/Comment
This conversation has been flagged as incorrect.
Flagged by yeswey [9/28/2014 1:17:08 AM]
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The area of a rectangle is 45 yd2 , and the length of the rectangle is 1 yd more than twice the width. Let x be the width, then the length is 2x + 1.
The area = (2x + 1)x = 45
2x^2 + x - 45 = 0
(x + 5)(2x - 9) = 0
x = -5 (impossible) or x = 9/2.
The width of the rectangle is 9/2 yards and the length is 10 yards.
Added 9/28/2014 1:20:31 AM
This answer has been confirmed as correct and helpful.
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Questions asked by the same visitor
The length of a rectangle is 11 yd more than twice the width, and the area of the rectangle is 63 yd2 . Find the dimensions of the rectangle.
Weegy: rectangle...area=length*wide=63 length=2*wide+11 area=wide*(2*wide+11)=63 2*wide^2 +11wide -63=0 wide=3.5 or -9...use 3.5 length=2*3.5+11=11+7=18 18*3.5=63 User: The area of a rectangle is 27 yd2 , and the length of the rectangle is 3 yd less than double the width. Find the dimensions of the rectangle. (More)
Question
Expert Answered
Updated 9/28/2014 1:36:24 AM
1 Answer/Comment
The area of a rectangle is 27 yd2 , and the length of the rectangle is 3 yd less than double the width. Let x be the width, the length is 2x - 3.
the area = (2x - 3)x = 27
2x^2 - 3x - 27 = 0
(x + 3)(2x - 9) = 0
x = -3 or x = 9/2
So the width of the rectangle is 9/2 yards and the length is 6 yards.
Added 9/28/2014 1:36:24 AM
This answer has been confirmed as correct and helpful.
The length of a rectangle is 3 ft more than double the width, and the area of the rectangle is 77 ft2 . Find the dimensions of the rectangle.
Weegy: Suppose the width of the rectangle is w, then its length is 2w - 1. [ its area = w(2w - 1) = 21, solve the equation we get w = -3 or 7/2, since its a rectange the side can not be negative so its width is 7/2 feets, and the length is 2(7/2) - 1 = 6 feets. ] (More)
Question
Not Answered
Updated 9/28/2014 1:33:02 AM
1 Answer/Comment
The length of a rectangle is 3 ft more than double the width, and the area of the rectangle is 77 ft2. Let x be the width of the rectangle, then the length is 2x + 3.
The area = (2x + 3)x = 77
2x^2 + 3x - 77 = 0
(x + 7)(2x - 11) = 0
x = -7 (impossible) or x = 11/2
So the width of the rectangle is 11/2 feet and the length is 14 feet.
Added 9/28/2014 1:33:02 AM
This answer has been confirmed as correct and helpful.
Confirmed by andrewpallarca [9/28/2014 10:02:50 AM]
The area of a rectangle is 66 yd2 , and the length of the rectangle is 7 yd less than three times the width. Find the dimensions of the rectangle.
Weegy: rectangle...area=length*wide=63 length=2*wide+11 area=wide*(2*wide+11)=63 2*wide^2 +11wide -63=0 wide=3.5 or -9...use 3.5 length=2*3.5+11=11+7=18 18*3.5=63 (More)
Question
Not Answered
Updated 9/28/2014 1:24:57 AM
1 Answer/Comment
The area of a rectangle is 66 yd2 , and the length of the rectangle is 7 yd less than three times the width. Let x be the width of the rectangle, then the length is 3x - 7.
The area = (3x - 7)x = 66
3x^2 - 7x - 66 = 0
(x - 6)(3x + 11) = 0
x = 6 or x = -11/3 (impossible)
The width of the rectangle is 6 yards and the length is 11 yards.
Added 9/28/2014 1:24:44 AM
This answer has been confirmed as correct and helpful.
The length of a rectangle is 5 yd more than twice the width, and the area of the rectangle is 42 yd2 . Find the dimensions of the rectangle.
Weegy: rectangle...area=length*wide=63 length=2*wide+11 area=wide*(2*wide+11)=63 2*wide^2 +11wide -63=0 wide=3.5 or -9...use 3.5 length=2*3.5+11=11+7=18 18*3.5=63 (More)
Question
Updated 9/27/2014 9:55:35 PM
1 Answer/Comment
The length of a rectangle is 5 yd more than twice the width, and the area of the rectangle is 42 yd2 . Let x be the width, the length is 2x + 5.
The rectangle area = x (2x + 5) = 42
2x^2 + 5x - 42 = 0
(x + 6)(2x - 7) = 0
x + 6 = 0 or 2x - 7 = 0
x = -6 or x = 7/2 yard.
The width of the rectangle is 7/2 yards and the length is 12 yards.
Added 9/27/2014 9:55:35 PM
This answer has been confirmed as correct and helpful.
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