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

1 Answer/Comment

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Flagged by yeswey [9/28/2014 1:17:08 AM]

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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.

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.

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.

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)

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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.

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.

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.

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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