Start with a circle. If you make 1 cut, you have 2 pieces. If you make 2 cuts, you have a pieces. If you make 3 cuts, you can have a maximum of 7 pieces.
1. Draw a picture to show how 3 cuts can give 7 pieces.
2. Find the maximum number of pieces you can get from 4 cuts.
3. From 5 cuts.
4. Can you find a function that gives the maximum number of pieces from n cuts?

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ummm seems that u didnt post the attachement !

answer: (yeees, you need a poet.)

//blooming serenity//

beautiful in the morning

mysterious at night

wonderous nature souring

with all it's mystical might

the flower that blooms

to the fox that looms

nocturnal creatures

with various features

a blooming flower soon in sight

as a new little hatchling fears to take flight  

afraid to leave it's calming nest,

but dear little bird, mother knows best

as sure as there are stars in the sky,

life is not about who, where what, when or why-

it's the journey that follows

adventures we swallow

of   nature's deep wonders,

natural disasters and storm's thunders.

blooming serenity

life and nature, truly is heavenly.

-kerstien lehman

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.