Find the equation of the line with exact numbers

In this question, we're trying to find the equation of the given line.

In this case, we would represent the line as:

y = mx + b

m = slope

b = beginning point

Lets find the equation.

We know that the beginning point is on the y-axis on the graph.

We can see that 5 is on the y-axis. Therefore, our beginning point would be 5.

Your equation should look like this:

y = mx + 5

Now, lets find the slope.

We know that the line is decreasing, so it would be a negative slope.

To find the slope, we would use the rise over run strategy. In this case, we would go down and run since it's a negative.

We go down 1 and go to the right 3 times. This means that our slope of the line would be represented as a fraction.

The slope of the line would be 1/3

When you put it into the equation, it should look like:

y = -1/3x + 5

The equation of the line is y = -1/3x + 5

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.