Can someone help solve this?

x = 8

y = 18

Step-by-step explanation:

First, it would be easiest to start by drawing a picture. To do so, draw two simple Isosceles triangles. Then, label their points. One having the points F, U, and N while the other has B, A, and T. Now, since both triangles are congruent, that means that F≅B, U≅A, and N≅T. Because of this knowledge, FN would be congruent to BT. So, you would be able to write that 52 = 14x - 60. By solving that, you would get the answer to x. Moving onto y, you would see that since the triangles are congruent, the interior angles are too. So, you would be able to use angles F and N to find A. By adding them to 180°, the equation would look like: 71° + 36° + (3y + 19)° = 180°. So, you would be able to find the answer to y. Hope this helped!

goog may have it for you

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.