I'm so confused right now. I don't need the answer, I just don't know how to put it into equation form.

Let x be the number of students in the senior class.

Four-fifths of this number x is equal to 324 based on the first sentence and the beginning of the second sentence.

So the equation to set up is and it's the same as writing

If you want, you can stop reading here and attempt to solve on your own.

------------------------

On the other hand, if you want to know how to solve, then read on.

The '5' in the denominator means we're dividing 4x over 5.

To get that 5 to the other side, we'll multiply both sides by 5 to undo the division operation.

Then to fully isolate x, i.e. get x all by itself, then we'll divide both sides by 4. This is to undo the multiplication of 4 onto x.

So,

This means there are 405 seniors overall.

4/5 of them went to prom, so (4/5)*405 = 324 went to prom. This confirms the answer.

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.

1. 2

2. 11

3. 5

4. 6

5. 7

6. 5

7. -1

8. 1

step-by-step explanation: