Question about Texas Instruments TI-84 Plus Calculator

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The width is 96. Since this sounds like a homework problem, you'll have to show your work.

Posted on May 19, 2011

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Posted on Jan 02, 2017

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Be sure to show your work.

Be sure to show your work.

Jan 04, 2017 | Homework

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

Rectangle has two pairs of equal sides.

L=length=3 units

w=width =1 unit

**Perimeter = sum of the measures of all sides.**

2 sides with measure L: sum of lengths=**2*L**

2 sides with measure w: sum of widths=**2*w**

Perimeter = 2L+2w=2(L+w)=2(1+3)=8 units.

L=length=3 units

w=width =1 unit

2 sides with measure L: sum of lengths=

2 sides with measure w: sum of widths=

Perimeter = 2L+2w=2(L+w)=2(1+3)=8 units.

Oct 29, 2013 | Mathsoft StudyWorks! Mathematics Deluxe...

Rectangle has two pairs of equal sides.

L=length=3 units

w=width =1 unit

**Perimeter = sum of the measures of all sides.**

2 sides with measure L: sum of lengths=**2*L**

2 sides with measure w: sum of widths=**2*w**

Perimeter = 2L+2w=2(L+w)=2(1+3)=8 units.

L=length=3 units

w=width =1 unit

2 sides with measure L: sum of lengths=

2 sides with measure w: sum of widths=

Perimeter = 2L+2w=2(L+w)=2(1+3)=8 units.

Oct 29, 2013 | 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

Translate the English into Mathematics.

W=L-5 (in inches)

P=2(L+W)=2(L+L-5)=2(2L-5)

Use distirbutive property of multiplication with respect to addition to open up the parentheses (brackets)

P=4L-10.

Set P= 50 (inches), to get 4L-10=50

**Solve for L: Do it'! **

Find W= L (the one you just found) -5 =

Now, with W the value you just calculated

the new length is** L'=-4+3W**

and the new perimeter is** P'=2(L'+W)**

Now your turn to do some work.

W=L-5 (in inches)

P=2(L+W)=2(L+L-5)=2(2L-5)

Use distirbutive property of multiplication with respect to addition to open up the parentheses (brackets)

P=4L-10.

Set P= 50 (inches), to get 4L-10=50

Find W= L (the one you just found) -5 =

Now, with W the value you just calculated

the new length is

and the new perimeter is

Now your turn to do some work.

Jan 02, 2012 | 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 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...

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