Apacking company is doing an inventory of boxes. their most popular box is display below: 2ft,(3x-5)ft,(2x-1)ft you can use the formula v= lwh to find the volume of the box. the volume of the box is 40ft3.what is the value of x ? find the length and width of the box.

The value of x  = 3.

Length = 4 feet and Width = 5 feet

Step-by-step explanation:

Height of the box = 2 ft

Length of the box = (3x-5) ft

Width of the box  = (2x-1) ft

Volume of the box = 40 cubic feet

Now, VOLUME OF CUBE  = LENGTH x WIDTH x HEIGHT

Now, solving for the value of x, we get:

⇒ Either x-3 = 0, or ( 6x+5) = 0

⇒ Either x  = 3, or x = -5/6

As x = SIDE OF A CUBE ,so it CANNOT be negative.

⇒x  = 3

So, length = (3x-5)  = 3(3) -5 = 9-5 = 4 units

Width = (2x-1) = 2(3) -1  = 5 units

mx+89b(-98)nb

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.