25 POINTS AND BRAINLIEST 17. A group of college students is volunteering for Help the Homeless during their spring break. They are putting the finishing touches on a house they built. Working alone, Irina can paint a certain room in 9 hours. Paulo can paint the same room in 8 hours. Write an equation that can be used to find how long it will take them working together to paint the room. How many hours will it take them to paint the room? If necessary, round your answer to the nearest hundredth.

4.24 hours

Step-by-step explanation:

Irina can paint 1/9 of a room in 1 hour since she can paint a room in 9 hours. 1/9x, where x is the number of hours she works and 1/9 of a room per hour is her speed, would be her part of the calculation.

Paulo can paint 1/8 of a room in 1 hour since he can paint an entire room in 8 hours. 1/8x, where x is the number of hours he works and 1/8 of a room per hour is his speed, would be his part of the equation.

1/9x + 1/8x = 1 (Irina's portion of the room plus Paulo's portion of the room equals one complete room) would be the equation.

Look for a denominator that has the same value as the numerator. Both 9 and 8 split evenly into 72 as the initial number. We multiply the top of 1/9 by 8 to convert the fraction and get 8/72x because 9*8 = 72. We multiply the top of 1/8 by 9 to convert the fraction and get 9/72x because 8*9 = 72. We have 8/72x+9/72x=1 currently.

17/72x=1

÷ both sides by 17/72:

17/72x ÷ 17/72 = 1÷17/72

∴ x=1/1 * 72/17

∴ x=72/17= 4.24

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.