Pls need hurry need pls
a paper manufacturer is evaluating their packing procedures. currently, they package twelve reams of paper in a box—in three stacks of four reams. however, their company has been getting complaints from the retail stores that the box is too large and heavy for the average consumer to carry by their self. so, they have decided to reduce the weight of each box by ten pounds. you have been asked to explore the different packing options based on the new weight condition and present your recommendation to the ceo of the paper company.

note: in your evaluation, make sure to consider the ease with which one person can carry the box, as well as the amount of cardboard required to make the box. in addition, label all measurements.

make note!
a ream of paper weighs 5 pounds and has dimensions of 8.5 inches by 11 inches by 2 inches.

what is the volume of the current-sized box? a0 cubic inches
make note!
a ream of paper weighs 5 pounds and has dimensions of 8.5 inches by 11 inches by 2 inches.

40 minutes

step-by-step explanation:

add the time and multiply by 2

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.

but there’s no any figure there!