A bottling company makes decorative bottles and ships them to their customers. Each bottle has a height of 13 inches and a radius of 2.5 inches. The manufacturer places each bottle in cylinder packaging and adds cushioning to the insides of the cylinder. The cushioning is an inch thick on each side of the bottle. If the height of the cylinder is the same as the height of the bottle, what is the volume of the cylinder used to package the bottles? (Round to the nearest whole number. Use π = 3.14.)

The volume of the cylinder used to package the bottles is 500.045 in³

Step-by-step explanation:

Hi, to answer this question we have to analyze the information given:

So, we have to apply the formula: Volume = πr²h

Where the radius (r) is 2.5 + 1=3.5 (the radius of the bottle plus the thickness of the cushion), and the height is 13.

Mathematically speaking:

Volume =πr²h = 3.14 x 3.5² x13 =500.045 in³

The volume of the cylinder used to package the bottles is 500.045 in³

Feel free to ask for more if needed or if you did not understand something.

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step-by-step explanatio89

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.