Afloor refinishing company charges $1.83 per square foot to strip and refinish a tile floor for up to 1000 square feet. there is an additional charge of $350 for toxic waste disposal for any job which includes more than 150 square feet of tile.
a) express the cost, y, of refinishing a floor as a function of the number of square feet, x, to be refinished.
b) graph the function, give the domain and range.

Here x represents the number of square feet to be refinished and y represents the cost of refinishing the floor,

Given,

The cost of a tile floor for up to 1000 square feet is $1.83 per square,

So, the cost of x square feet of tile = 1.83x for x ≤ 1000

⇒ y = 1.83x for x ≤ 1000

Since, there is an additional charge of $350 for toxic waste disposal for any job which includes more than 150 square feet of tile.

That is, y = 1.83x + 350, for x > 150

So, y must be 1.83x for x ≤ 150.

A) Hence, the function that express the cost, y, of refinishing a floor as a function of the number of square feet, x, to be refinished, is,

B) The domain of the function =  all possible value of x

⇒ Domain = 0 ≤ x ≤ 1000

Range = All possible value of y,

Since, the range of function y=1.83x, 0≤ x ≤ 150 is [0, 274.5]

While the range of function y = 1.83x + 350, for x > 150 is (624.5, 2180]

Hence, the range of the function (1) = [0, 274.5]∪(624.5, 2180]

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

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.