The number of people, N, that Denis can invite to his party is dependent on the amount of money, x, in dollars,
that he saved up for the party and can be modeled by the function N(x)=x/25.
The amount of money that
Denis saved for the party, M, in dollars, is dependent on the number of overtime hours, h, he worked this week
and can be modeled by the function M(h) = 50h + 150.
Find an explicit expression that models the number of people that Denis can invite to his party if he worked h
overtime hours this week.

N(h) = 2h + 6

Step-by-step explanation:

A model of the number of people that Denis can invite to his party:

N(x) = x/25.............(1)

Where x = amount of money saved

Amount of money saved can be modeled by the equation:

M(h) = 50h + 150...........(2)

In the above model, h = number of overtime hours

M(h) = amount of money saved = x

Therefore equation (1) can be re - written as:

N(x) = M(h)/25.................................(3)

Substituting equation (2) into equation (3)

Find an explicit expression that models the number of people that Denis can invite to his party if he worked h  overtime hours this week can be expressed as:

N(h) = 2h + 6

nothjing

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.