Mr. cooper has 3pails with golf balls in them. the first pail has 87ball. the second has 105golf balls and the third has 113golf balls. what is the total number of golf balls that mr. cooper has in the 3pails

The total number of golf balls that Mr.Cooper has in the 3 pails is 305

It is given that Mr. Cooper has 3 pails with golf balls in them.

The first pail has 87 ball

The second has 105 golf balls and the third has 113 golf balls.  

We need to find the total number of golf balls that Mr. Cooper has in the 3 pails.

To find the total number of golf balls we need to add the number of balls in each pail.

Total number of golf balls in 3 pails = number of balls in first pail + number of balls in second pail + number of balls in third pail

Total number of golf balls = 87+105+113

= 305 balls

Therefore, the total number of golf balls in all 3 pails combined it 305

i think the answer is true

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.