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Solve the resulting quadratic equation

2X^2+3X-90=0

Discriminant: (3)^2-4(2)(-90)=729=(27)^2

Two roots

X_1=(1/4)*(-3+27)=6

X-2=(1/4)*(-3-27) =-(15/2), negative

Since the width must be positive, reject the negative root and keep X_1=6

**Width =6
Length=2(6)+3=15**

Check 6*(15)=90. Checks OK

Posted on Apr 23, 2014

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Posted on Jan 02, 2017

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SOURCE: area of rectangle having length 1 fit and width 3

Hi amita_mediai

Do this:

1*3*20

Explanation

Multiply the length and the width. This gives the area in sq ft. Multiply this with 20. This will be the cost in rs.

Good luck

luciana44

Posted on Sep 19, 2009

SOURCE: A one meter tall cedar fence surrounds a yard. The

The dimensions of the yard are 13.5m x 10.5m.

Posted on May 03, 2011

area = length x width If you assign the height (width) as X then the base (Length) would then be 2X. So if you go

50 = X x 2X

50 = 2X squared

25 = X squared

5 = X

So your height is 5 and the base is 10

50 = X x 2X

50 = 2X squared

25 = X squared

5 = X

So your height is 5 and the base is 10

Oct 22, 2017 | The Office Equipment & Supplies

We are given the following data:

length = 5/2 * width

length = 10 inWe know through the transitive

property of equality that the following must be true:

5/2 * width = 10 in

We then solve this equation algebraically for width:

width = (10 in) * 2/5

width = 4 in

The perimeter of the rectangle is given by substituting our known values of length and width into the following general equation:

Perimeter = 2 * length + 2 * width

Perimeter = 2(10 in) + 2(4 in) = 28 in

In a similar fashion, we substitute these known values into the general equation for the area of the rectangle to solve for the same:

Area = length * width

Area = (10 in)(4 in)=40 in^2

length = 5/2 * width

length = 10 inWe know through the transitive

property of equality that the following must be true:

5/2 * width = 10 in

We then solve this equation algebraically for width:

width = (10 in) * 2/5

width = 4 in

The perimeter of the rectangle is given by substituting our known values of length and width into the following general equation:

Perimeter = 2 * length + 2 * width

Perimeter = 2(10 in) + 2(4 in) = 28 in

In a similar fashion, we substitute these known values into the general equation for the area of the rectangle to solve for the same:

Area = length * width

Area = (10 in)(4 in)=40 in^2

Nov 29, 2016 | The Computers & Internet

The rectangle is 11cm by 19cm.

The perimeter is 60, so the width and length must add to 30. The length is 8 more than the width, or width+width+8=30. Solve that for width, then you can calculate the length based on the width.

If this is homework, be sure to show your work.

The perimeter is 60, so the width and length must add to 30. The length is 8 more than the width, or width+width+8=30. Solve that for width, then you can calculate the length based on the width.

If this is homework, be sure to show your work.

Oct 11, 2013 | Lands Phones

Area = length X width, and you know the area and the width. So,

4122 sq cm = length X 3 cm

length = 4122 sq cm / 3 cm

length = 1374 cm

(That is one long, skinny rectangle, isn't it?)

4122 sq cm = length X 3 cm

length = 4122 sq cm / 3 cm

length = 1374 cm

(That is one long, skinny rectangle, isn't it?)

Mar 02, 2011 | Office Equipment & Supplies

1. Well this one is pretty simple. Since you are not supposed to trim the length trim the width instead.

to make the perimeter 0.4 m shorter cut the cloth by 0.2 m width the picture will make it clear.

2. let the length of rectangle be x cm

and breadth = 20 cm

now original perimeter = 2(x+20) cm

new length = x-30 cm

breadth remains same = 20 cm

new perimeter = 2(x-30+20) cm = 2(x-10) cm

According to question

new perimeter = old perimeter /2

2(x-10) = 2(x+20) /2

2x -20 = x + 20

x = 40 cm

hence length of longer side 40 cm.

If you want to directly comm ankitsharma.220@gmail.com

to make the perimeter 0.4 m shorter cut the cloth by 0.2 m width the picture will make it clear.

2. let the length of rectangle be x cm

and breadth = 20 cm

now original perimeter = 2(x+20) cm

new length = x-30 cm

breadth remains same = 20 cm

new perimeter = 2(x-30+20) cm = 2(x-10) cm

According to question

new perimeter = old perimeter /2

2(x-10) = 2(x+20) /2

2x -20 = x + 20

x = 40 cm

hence length of longer side 40 cm.

If you want to directly comm ankitsharma.220@gmail.com

Feb 13, 2011 | Vivendi Excel@ Mathematics Study Skills...

The diameter of the circle is the diagonal of the rectangle use Pythagoras this rectangle is a made from two 3,4,5 triangles, so the diameter is 5cm, the radius will be 2.5cm don't forget the BODMAS ordering rule (Brackets of Division Multiplication Addition and Subtraction) Also rounding to 2 decimal places. This solution is based on all four corners of the rectangle touching the circumference of the circle.

Area of a circle is P (22/7) i x Radius (squared ^2) ====> 22/7 x 2.5cm ^ 2 ====> 22/7 x 6.25cm

TOTAL AREA of CIRCLE = 19.64 cm^2

Area of the rectangle is length x width ====> 4 cm x 3 cm

TOTAL AREA of RECTANGLE = 12 cm ^2

The remaining portion is the circle area minus the rectangle 19.64-12 = 7.64 cm^2

SIMPLES brought to you by SpideRaY @ http://www.windowstipsclub.com

Area of a circle is P (22/7) i x Radius (squared ^2) ====> 22/7 x 2.5cm ^ 2 ====> 22/7 x 6.25cm

TOTAL AREA of CIRCLE = 19.64 cm^2

Area of the rectangle is length x width ====> 4 cm x 3 cm

TOTAL AREA of RECTANGLE = 12 cm ^2

The remaining portion is the circle area minus the rectangle 19.64-12 = 7.64 cm^2

SIMPLES brought to you by SpideRaY @ http://www.windowstipsclub.com

Jan 31, 2011 | Computers & Internet

Hi amita_mediai

Do this:

1*3*20

Explanation

Multiply the length and the width. This gives the area in sq ft. Multiply this with 20. This will be the cost in rs.

Good luck

luciana44

Do this:

1*3*20

Explanation

Multiply the length and the width. This gives the area in sq ft. Multiply this with 20. This will be the cost in rs.

Good luck

luciana44

Sep 19, 2009 | Casio Office Equipment & Supplies

The length of a rectangle is twice its width w. a second rectangle, which is 8 cm longer and 3 cm narrower than the first rectangle, has perimeter 154 cm. Make a sketch of the rectangles expressing all dimensions of w. Then find the dimensions of each rectangle.

Oct 03, 2008 | Texas Instruments BA Real Estate...

Let x = the width of the women's court. Then 2x is the length of the women's court.

the area of the womens court is x times 2x = 2x^2

the length of the mens court is 4 ft longer than the women's or 2x + 4

The width of the men's court is 5 ft wider than the womwn's or x + 5

The are of the mens court is 650 square feet larger that the womens so we have to add 650 to the area of the womens court so the men's and women's will be equal

Now we have area of the women's court = area of the men's court

2x^2 + 650 = (x + 5) (2x + 4) = 2x^2 + 14x + 20

The x^2 cancel out and we are left with 14X + 20 + 650 solving for x we get x = 45

Plugging x back into the length and width equations we get the length is 94 and the width is 50

Hope this helped

Good luck Loringh

the area of the womens court is x times 2x = 2x^2

the length of the mens court is 4 ft longer than the women's or 2x + 4

The width of the men's court is 5 ft wider than the womwn's or x + 5

The are of the mens court is 650 square feet larger that the womens so we have to add 650 to the area of the womens court so the men's and women's will be equal

Now we have area of the women's court = area of the men's court

2x^2 + 650 = (x + 5) (2x + 4) = 2x^2 + 14x + 20

The x^2 cancel out and we are left with 14X + 20 + 650 solving for x we get x = 45

Plugging x back into the length and width equations we get the length is 94 and the width is 50

Hope this helped

Good luck Loringh

Oct 03, 2008 | Computers & Internet

Nov 19, 2017 | Epson Office Equipment & Supplies

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