Please help! The first one to answer with a detailed, thorough, likely Expert Verified answer gets Brainliest! Please don't copy someone else's answer; give a non-detailed answer or spam. Thank you! Figure ABCDE has vertices A(−2, 3), B(2, 3), C(5, −2), D(0, −3), and E(−2, −2). Plot the points on your own coordinate grid and connect the alphabetical order points. Decompose Figure ABCDE into rectangles and triangles.

Part A: How many triangles and rectangles did you make? (1 point)
Part B: Use Figure ABCDE created on your coordinate grid to find the lengths, in units, of Sides AB and AE. (4 points)
Part C: What is the area of Figure ABCDE? Show your work. (5 points)

A: I drew 3 triangles and one rectangle.  

B: The distance between A and B = 4. The distance between A and E = 5.

C: For the rectangle, it's 5 x 4 equals 20. For the first triangle, 1/2 3 x 5 equals 7.5. For the second triangle, 1/2 1 x 5 equals 2.5. For the last triangle, 1/2 1 x 2 equals 1. This all together equals 31.

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.

answer: 7. if the functions had a negative leading coefficient and is of even degree, which statement about the graph is true?

step-by-step explanation: