part 1. Use side ratios to determine if the triangles in each pair of similar. If so complete the similarity statement. ​NO LINKS!!!

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  4) ΔEFG ~ ΔECD

  6) Not Similar

Step-by-step explanation:

4) Shortest to longest sides are ...

  EF : FG : GE = 14 : 16 : 20 = 7 : 8 : 10 (reduced)

  EC : CD : DE = 7 : 8 : 10

The triangles have the same side length ratios, so are similar.

  ΔEFT ~ ΔECD

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6) Shortest to longest, the given sides have ratios ...

  UT : TV = 71 : 84 (reduced)

  GT : TH = 60 : 72 = 5 : 6 (reduced)

The side ratios are different, so the triangles are NOT SIMILAR.

part 1. Use side ratios to determine if the triangles in each pair of similar. If so complete the similarity statement. NO LINKS!!!

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.