The table and the graph below each show a different relationship between the same two variables, x and y: A table with two columns and 5 rows is shown. The column head for the left column is x, and the column head for the right column is y. The row entries in the table are 3,240 and 4,320 and 5,400 and 6,480. On the right of this table is a graph. The x axis values are from 0 to 10 in increments of 2 for each grid line. The y axis values on the graph are from 0 to 450 in increments of 90 for each grid line. A line passing through the ordered pairs 2, 90 and 4, 180 and 6, 270 and 8, 360 is drawn. How much more would the value of y be in the table than its value on the graph when x = 11? 110
150
385
450
if anyone gives me a fals answer i will personally make sure your pets suffer :)

b. 153 students do not participate in sports and seventy-five 12th graders were surveyed

3

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.