Which statements must be true about the image of
AMNP atter a reflection across El? Select thres options,
The image will be congruent to AMNP,
The orientation of the image will be the same as the
orientation of AMNP.
Et will be perpendicular to the line segments connecting
the corresponding vertices.
The line segments connecting the corresponding vertices
will all be congruent to each other,
The line segments connecting corresponding vertices will
all be parallel to each other.

The image will be congruent to AMNP.

The orientation of the image will be the

same as the orientation of AMNP.

The line segments connecting corresponding vertices will all be parallel to each other.

Step-by-step explanation:

Reflection is a rigid transformation which maps a shape across a line such that the image will look like the mirror image of the pre-image.

It preserves the size and orientation of the shape such that the pre- image will be congruent to the image formed.

Also, if we connect corresponding vertices, then they will all be parallel to each other.

ratio is 2 : 9

the answer is three

13.5 : 3

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.