Find the value of v. v 2 m v 17 m Perimeter=35 meters v= meters Submit

I could just give you the answer of 140 cm

but I think you need to know how I got this answer
A square is an object that has equal sides
area is calculated by length multiplied by width
so 1225 cm^2 area by using the square root of 1225 cm^2 gives you 35 cm
so each side is 35 cm and there are 4 sides so 35 cm +35 cm + 35 cm + 35 cm = 140 cm
If you are doing homework always show how you got the answer because in an exam having the working out shown can sometimes result in getting some marks even if you slipped up slightly and got the wrong answer.

Check this out Uxhcd. Area and Perimeter of Rectangles and Squares Home Campus

35 x 35 x .305 = 373.625

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

3600

Be sure to show your work.

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

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

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.

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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."

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