The soccer team collected $800 at a car wash fundraiser. they charged $5.00 for small vehicles and $10.00 for larger vehicles. the amount collected can be modeled by the equation 5x + 10y = 800, where x represents the number of small vehicles and y represents the number of larger vehicles. if the number of larger vehicles washed was 50, how many small vehicles were washed in total?

Number of small vehicles washed were 60.

Step-by-step explanation:

Soccer team collected $800 by charging $5 for the small vehicles and $10 for larger vehicles.

Amount collected was modeled by the equation

5x + 10y = 800

Where x = Number of small vehicles

y = Number of larger vehicles

If y = 5 then we have to calculate the number of small vehicles washed.

By putting y = 50 in the equation,

5x + 10(50) = 800

5x + 500 = 800

5x = 800 - 500

5x = 300

x =

x = 60

Therefore, number of small vehicles washed were 60.

there would only be 50 odd numbers between 1 and 100 so therefor there would be 50 even so that would only be 50 odd numbers.

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.