Question about The Computers & Internet

35 x 35 x .305 = 373.625

May 22, 2017 | In Office Equipment & Supplies

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

3600

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

See if this works

=IF(J28<17,0,IF(J28<=25, J27*0.12, IF(J28<=30, J27*0.15, IF(J28<=35, J27*0.2, IF(J28<=40, J27*0.25, J27*0.3)))))

You may have to adjust for your boundary conditions. For example, if the value is 25, do you want to multiply by 12% or 15%?

Good luck,

Paul

=IF(J28<17,0,IF(J28<=25, J27*0.12, IF(J28<=30, J27*0.15, IF(J28<=35, J27*0.2, IF(J28<=40, J27*0.25, J27*0.3)))))

You may have to adjust for your boundary conditions. For example, if the value is 25, do you want to multiply by 12% or 15%?

Good luck,

Paul

Nov 28, 2016 | Office Equipment & Supplies

This is a dealer specific part and should by no means be repaired by a home handyman, even with decent welding experience. there are many complicated steps involved with multiple specialized tools. please call your local HVAC tecnician.

Dec 31, 2014 | Convair Portable Air Conditioner

Infinite.

A one meter by 48 meter rectangle has an area of 48 square meters and a perimeter of 98 meters. A one centimeter by 4800 meter rectangle has the same area and a perimeter of 9600.02 meters. A one millimeter by 48,000 meter rectangle has the same area and a perimeter of 96,000.002 meters. A rectangle one micron by 48,000,000 meters has an area of 48 square meters and a perimeter of 96,000,000.000002 meters. Keep making the rectangle skinnier and skinnier without changing the area, and the perimeter keeps getting longer and longer.

A one meter by 48 meter rectangle has an area of 48 square meters and a perimeter of 98 meters. A one centimeter by 4800 meter rectangle has the same area and a perimeter of 9600.02 meters. A one millimeter by 48,000 meter rectangle has the same area and a perimeter of 96,000.002 meters. A rectangle one micron by 48,000,000 meters has an area of 48 square meters and a perimeter of 96,000,000.000002 meters. Keep making the rectangle skinnier and skinnier without changing the area, and the perimeter keeps getting longer and longer.

May 01, 2014 | Computers & Internet

AT&T 2Wire routers function as both routers and modems. You
can disable the router functionality by placing the router in bridge
mode. In bridge mode, the 2Wire device functions only as a modem and
forwards all incoming traffic to the device behind the 2Wire router. For
example, you can use a different router instead of the 2Wire's built-in
router by enabling bridging mode -- the router behind the 2Wire device
will get the external IP address, and the 2Wire router won't interfere.

Step 1Type "Homeportal/management" (without quotes) into your Web browser's address bar and press "Enter" to access the Management page in the 2Wire router's Web interface.

Step 2Provide your 2Wire router's custom password, if you set one. The router comes with no password set.

Step 3Click "Configure" under Broadband on the Management page.

Step 4Type a period (".") into the "Disable PVC Search" box.

Step 5Set the value of the "VPI" box to "0" and the value of the "VCI" box to "35".

Step 6Set the "Connection Type" to "Direct IP" and click "Submit."

Step 7Click the "Configure Services" link on the Management page.

Step 8Uncheck "Enable Routing" and click "Submit."

Step 1Type "Homeportal/management" (without quotes) into your Web browser's address bar and press "Enter" to access the Management page in the 2Wire router's Web interface.

Step 2Provide your 2Wire router's custom password, if you set one. The router comes with no password set.

Step 3Click "Configure" under Broadband on the Management page.

Step 4Type a period (".") into the "Disable PVC Search" box.

Step 5Set the value of the "VPI" box to "0" and the value of the "VCI" box to "35".

Step 6Set the "Connection Type" to "Direct IP" and click "Submit."

Step 7Click the "Configure Services" link on the Management page.

Step 8Uncheck "Enable Routing" and click "Submit."

Jan 31, 2013 | Computers & Internet

the area of your roof is equal to the base (B) times the hight (H) divided by 2.

A = ( B x H )/ 2 = 120

To get the Perimeter of your roof you need to sum the base plus the sides (S)

So P = B + 2 x S

if you have a slope degree of 30 then

B/2 = cos (30) x S

so P = B + 2 x B / 2 / cos(30)

P = B ( 1 + cos (30) )

we also know that

H = tg (30) x B/2

from this last one we can go back to the Area and get a value for B

120 = ( tg (30) x B / 2 x B) / 2

480 x SQRT(3) = B^2

B = SQRT(480/tg(30))

So:

P = SQRT(480/tg(30)) x ( 1 + cos (30))

tg (30) = 1 / SQRT(3) cos (30) = SQRT(3)/2

P = 2 SQRT(30 x (12 + 7 x SQRT(3) ) ) or aproximately 53.8m

A = ( B x H )/ 2 = 120

To get the Perimeter of your roof you need to sum the base plus the sides (S)

So P = B + 2 x S

if you have a slope degree of 30 then

B/2 = cos (30) x S

so P = B + 2 x B / 2 / cos(30)

P = B ( 1 + cos (30) )

we also know that

H = tg (30) x B/2

from this last one we can go back to the Area and get a value for B

120 = ( tg (30) x B / 2 x B) / 2

480 x SQRT(3) = B^2

B = SQRT(480/tg(30))

So:

P = SQRT(480/tg(30)) x ( 1 + cos (30))

tg (30) = 1 / SQRT(3) cos (30) = SQRT(3)/2

P = 2 SQRT(30 x (12 + 7 x SQRT(3) ) ) or aproximately 53.8m

Mar 03, 2011 | Texas Instruments TI-83 Plus Calculator

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...

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