On a coordinate plane, a line goes through (0, 4) and (4, 16). The line representing the first table is shown in the graph. What is the solution to the system of linear equations represented by both tables? 2 tables. A 2-column table with 4 rows. Column 1 is labeled x with entries negative 2, 0, 2, 4. Column 2 is labeled y with entries negative 2, 4, 10, 16. A 2-column table with 4 rows. Column 1 is labeled x with entries negative 2, 0, 2, 4. Column 2 is labeled y with entries 18, 16, 14, 12.

seems like d

step-by-step explanation:

dilation scale factor 2: starting with δabc, draw the dilation image of the triangle with a center at the origin and a scale factor of two. notice that every coordinate of the original triangle has been multiplied by the scale factor (x2). dilation with scale factor 2, multiply by 2.

i think the answer would be c

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.