Part A Start by creating two different circles:

Create a point, and label it A. To make the remainder of the activity easier, choose integers for the x and y coordinates of point A.
Create a circle with its center at point A and with a radius of your choice. To make the remainder of the activity easier, choose an integer value for the radius.
Create another point, and label it B. To make the remainder of the activity easier, choose integers for the x and y coordinates of the point.
Create a circle with its center at point B and with a radius of your choice that is different from the radius chosen for circle A. To make the remainder of the activity easier, choose an integer value for the radius.
Proving That All Circles Are Similar
In this activity, you will explore how similarity transformations establish similarity for all circles. Open the GeoGebra geometry tool, and complete each step below. If you need help, follow these instructions for using GeoGebra.

54

step-by-step explanation:

6x9

6 and 24

step-by-step explanation:

6 x 4 = 24

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.

i will with my problems and all the other things that need to   be on by the way

step-by-step explanation: