Question about Prentice Hall College Algebra Enhanced with Graphing Utilities Value Package (includes Ma

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Posted on Jan 02, 2017

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It has 4 sides, two long sides are the same, two short ends are the same. Take 160 from the total, and divide by 4 which gives you a square. Add 80 to the long sides.

Easier than this

The perimeter of rectangle is 800 yards What are the dimensions of the...

SOLUTION the perimeter of rectangle is 800 yards what are the dimenision...

Easier than this

The perimeter of rectangle is 800 yards What are the dimensions of the...

SOLUTION the perimeter of rectangle is 800 yards what are the dimenision...

Jun 18, 2018 | The Computers & Internet

the perimeter of any square sided object is 2 times the height added to 2 times the width: (2h)+(2w). The actual answer for your sizes is 46 yards given by 2 times 11 plus 2 times 12. Hope this helps.

Aug 03, 2017 | The Computers & Internet

Let's break down the question to its components.

It is a rectangular pool. It has four sides, with a length and width and the angles in corners being 90 degrees.

Let l be the length of the pool.

Let w be the width of the pool.

The perimeter of the pool is l + l + w + w or 2l + 2w

So 2l + 2w = 55 metres

You could make up a table for all the possible values, starting with a width of 1. 2(1) + 2w = 55

2 + 2w = 55

2w = 53

w = 53 /2

Good luck,

Paul

It is a rectangular pool. It has four sides, with a length and width and the angles in corners being 90 degrees.

Let l be the length of the pool.

Let w be the width of the pool.

The perimeter of the pool is l + l + w + w or 2l + 2w

So 2l + 2w = 55 metres

You could make up a table for all the possible values, starting with a width of 1. 2(1) + 2w = 55

2 + 2w = 55

2w = 53

w = 53 /2

Good luck,

Paul

Feb 09, 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

160 by 140

The perimeter is 600, so 2w+2l=600 where w is the width and l is the length.

Divide both sides by 2: w+l=300

The length is 20 more than the width: l=w+20

Substituting in the previous equation: w+(w+20)=300

Collecting terms: 2w+20=300

Subtract 20 from both sides: 2w=280

Divide by 2: w=140

Thus the width is 140. Substituting into the equation for length: l=140+20

Simplifying: l=160

The width is 140 and the length is 160

The perimeter is 600, so 2w+2l=600 where w is the width and l is the length.

Divide both sides by 2: w+l=300

The length is 20 more than the width: l=w+20

Substituting in the previous equation: w+(w+20)=300

Collecting terms: 2w+20=300

Subtract 20 from both sides: 2w=280

Divide by 2: w=140

Thus the width is 140. Substituting into the equation for length: l=140+20

Simplifying: l=160

The width is 140 and the length is 160

Sep 15, 2014 | MathAid Algebra II

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

Assuming it's rectangular, the perimeter is 100 feet.

Aug 21, 2013 | Computers & Internet

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

May 03, 2011 | Texas Instruments TI-30 XIIS Calculator

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...

Let x = the short side. Then the long side is the short side plus 6 meters

now we have 2 sides that are x meters long, add those together gives us 2x for the length

of both short sides

The length of the long side is x + 6 for one side and x + 6 for the other side. Added together we have x + 6 + x + 6 = 2x + 12

Adding the 4 sides together we get 2x ( the 2 short sides) + 2x + 12 (the long sides) we get 4x + 12 = the perimeter or 628. 4x + 12 = 628. Subtract 12 from both sides of the equation leaves 4x = 616. divide both sides by 4 leaves x = 154 So the width is 154 meters and the length is 160 meters.

Hope this helps

Good luck Loringh

now we have 2 sides that are x meters long, add those together gives us 2x for the length

of both short sides

The length of the long side is x + 6 for one side and x + 6 for the other side. Added together we have x + 6 + x + 6 = 2x + 12

Adding the 4 sides together we get 2x ( the 2 short sides) + 2x + 12 (the long sides) we get 4x + 12 = the perimeter or 628. 4x + 12 = 628. Subtract 12 from both sides of the equation leaves 4x = 616. divide both sides by 4 leaves x = 154 So the width is 154 meters and the length is 160 meters.

Hope this helps

Good luck Loringh

Oct 06, 2008 | SoftMath Algebrator - Algebra Homework...

Aug 19, 2018 | Computers & Internet

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