Isabels competition who sells books and no other products charges a shipping fee of $1.99 per book plus a fixed fee of $3 for each order. let x be the number of books purchased and f (x) be the price of shipping one order of books. write a function that expresses f (x) and explain and explain the value of f (x) for x=0

f(x)= 1.99x +3

f(0) =3

For F(0) there is a order of 0 books.  

Step-by-step explanation:

Hi, to answer this question we have to write an equation with the information given:

The price of the shipping ( f(x)) is equal to the cost of the fixed fee per order (3) plus the product of the number of books ordered (x) and the cost of the fee per book (1.99).

f(x)= 1.99x +3

for f(0) = 1.99(0)+3

f(0) =3

For F(0) there is a order of 0 books.  The price is only for the shippping.

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.

v=1788.75

step-by-step explanation:

i belive that is the answer