Aline is represented by the equation y/x-2=3/11. what are the coordinates of some points that line on the line?

Some points that lie on the line are required.

Some of the points are , and .

The equation of the line is

This can be written as

Finding points

The point is

The point is

The point is

The points are , and .

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y/x-2=3/11 would be easier to work with if we were to put it into a more standard form, e. g., y = mx + b.

First, add 2 to both sides, to isolate y/x:

y/x = 3/11 + 22/11 = 25/11.

Next, mult. both sides by x, to get y by itself:  y = (25/11)x.  

This is your function.

Now make a table.  You can choose any x values you want, and then calculate y for each one.

x   y

0  0

1   25/11

3  (25/11)*3 = 75/11

Then we have three points on this line: (0,0), (1, 25/11), 3, 75/11).  You could obtain more by choosing additional x values.

nope lol

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.