PLSSS HELP A box has a volume of 1008 cubic inches. Your friend told you that he found the polynomial that represents the volume of the box: 4x³-28x² +144x. If x is the height of the box how tall is the box?

  7 inches tall

Step-by-step explanation:

A graphing calculator quickly finds the solution to the cubic equation. It is convenient to define ...

  f(x) = 4x^3 -28x^2 +144x -1008

so f(x) = 0 gives the value of x that is of interest. That value is x=7.

The box is 7 inches tall.

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The equation f(x) = 0 can be divided by 4 to give ...

  x^3 -7x^2 +36x -252 = 0

This can be factored by grouping:

  (x^3 -7x^2) +(36x -252) = 0

  x^2(x -7) +36(x -7) = 0

  (x -7)(x^2 +36) = 0

The only real root is x = 7, which makes a factor zero.

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.