a math course is open for first year and second year students with an enrollment capacity of 30. assume that 6 first year students. how many first year and second year students must enroll of gender and class are to be independent

m a t h w a y. c o m

step-by-step explanation:

x coordinate can be written by using the formula, x= x1 + 2/5(x2 - x1)

step-by-step explanation:

formula

x = x1 + p/w(x2 - x1)

let the line segment ab with a(x1,y1) and b(x2,y2)

a point(x,y) divide ab in the ratio, 2: 3 the x coordinate can be written as

x = x1 + p/w(x2 - x1)

here p=2 and w = 2+3

therefore,

x coordinate can be written by using the formula, x= x1 + 2/5(x2 - x1)

similarly y coordinate y = y1 + 2/5(y2 - y1)

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.