Which explanation can be used to derive the formula for the circumference of a circle? Find the length of the radius. Then find the circumference. Use the square of the radius to set up an equation showing the relationship that the square of the radius equals the ​π​ times circumference. Then rearrange the equation to equal the circumference.

Find the length of the diameter and double this amount. Subtract it from the length of the circumference. Set an equation to show the difference equaling the ratio between a circumference and radius. Then rearrange the equation to solve for the circumference.

Find the length of the diameter and the length of the circumference of the circle. Divide the length of the circumference by the length of the diameter. Set up an equation showing the ratio of the circumference to the diameter equal to ​π​. Then rearrange the equation by solving it for the circumference. Substitute 2 times the radius for the diameter.

Find the length of the radius and the area of the circle. Divide the area by the radius to get a quotient of ​π​. Write an equation showing the area of a circle equals this quotient times the radius. Then substitute the area for 3 times the circumference and rearrange the equation to equal the circumference.

I uploaded the answer to a file hosting. Here's link:

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i dont know sorry

step-by-step explanation:

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.