I wanna share awarness about this app.. many kids come on here to get help with work and many kids to come here to online date... and some grown men come here to be per-verts And im sick and tired of coming on this app asking for help and i see kids who are 18 and younger online dating on this app... and im tired of joing z0oms and grown men trying to take avenge of me... well some times i join a z0om and some people is actually sweet but anyways im tired of grown men saying "hey show ur body" and let get upset if i say " UNDERAGE i will report you to the feds" (feds are 12 and FBI and 12 are officers) and i don't like to snitch but... yeah be careful on this app and yeah have a nice day my LUVS <3 AND DON'T EVER LET A BUMMY ()ITCH EVER TRY TO TAKE ADVANGE OF YOU PERIODT !

Thankk yo soo muc because there are bad things happening and I saw things on a zo om that was horrible

have a good day :)

Step-by-step explanation:

finally someone talking about this ! i saw a 13 year old post a photo saying "who wants to be my friend?" and some gross dudes were commenting on it. like this app is not a dating app.

the answer is d. 75%

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.