Write the equation of the line containing the points (2,6)(2,6)and(5,0)(5,0)
. Do not include spaces in your answers. Use the forward slash (i. e. "/") for all fractions (e. g. -1/2 is the same as −12−12)
=

Step-by-step explanation:

Given the coordinate points (2, 6) and (5, 0)

We are to find the equation of the line passing through the points

The standard form of the equation is expressed as y = mx+c

m is the slope

c is the y intercept

Get the slope

m = y2-y1/x2-x1

m = 0-6/5-2

m = -6/3

m = -2

Get the intercept

Substitute m = -2 and the point (5, 0) into the equation y = mx+c

0 = -2(5) + c

0 = 10 + c

c = 10

Get the equation

Recall that y = mx+c

y = -2x + 10

Hence the required equation is y = -2x + 10

3

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.