Shelby’s total closed-end credit for her car (including all interest) cost $9,478.58. The table shows the schedule of her payments for the first 4 years.

A 4-column table with 4 rows. Column 1 is labeled Year number with entries 1, 2, 3, 4. Column 2 is labeled Balance with entries 7,000 dollars, 5,997 dollars and 39 cents, 4,617 dollars and 69 cents, 3,016 dollars and 21 cents. Column 3 is labeled monthly payment with entries 180 dollars, 180 dollars, 180 dollars, 180 dollars. Column 4 is labeled End of Year Balance with entries 5,997 dollars and 39 cents, 4,617 dollars and 69 cents, 3,016 dollars and 21 cents, 1,157 dollars and 28 cents

What is the total amount of interest Shelby will pay for the car loan?

How many more years will it take Shelby to pay off her loan?

i think it would be 42cm

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.