Roxanne is going to build a rectangular prism out of 36 one- inch cubes. She wants the height of the rectangular prism to be 3 inches, and she wants the width to be shorter than the length. How many rectangular prisms with different dimensions can she build?

3

Step-by-step explanation:

Since Roxanne has 36 one inch cubes, the maximum volume of rectangular prism she can have is 36 in³ = 1 in × 1 in × 36 in.

Now, if Roxanne wants a rectangular prism with height 3 inches, this implies that the area of its base is 36 in³/3 in = 12 in².

We now find all the factors of 12 = 1 × 12, 2 × 6, 3 × 4.

So, we have 3 different areas of base of rectangular prism in which the width is shorter than its length the width being the smaller dimension.

So, the different rectangular prisms in which the volume = 36 in³ in which the width is shorter than its length and has a height of 3 in is

1 × 12 × 3, 2 × 6 × 3, 3 × 4 × 3.

So we have 3 different rectangular prisms

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.