Quadrilateral ABCD with coordinates A (-6, 3). B (-4, 6), C (-2,5), and D (-3, 2) is mapped
onto quadrilateral A'B'C'D' with coordinates
A' (8, 1), B' (6, 4), C' (4, 3), and D' (5, 0). Are the
quadrilaterals congruent?

37

step-by-step explanation:

answer: ay  

fonsi  

dy  

oh

oh no, oh no

oh yeah

diridiri, dirididi daddy  

go

sí, sabes que ya llevo un rato mirándote  

tengo que bailar contigo hoy (dy)  

vi que tu mirada ya estaba llamándome  

muéstrame el camino que yo voy (oh)

tú, tú eres el imán y yo soy el metal  

me voy acercando y voy armando el plan  

solo con pensarlo se acelera el pulso (oh yeah)

ya, ya me está gustando más de lo normal  

todos mis sentidos van pidiendo más  

esto hay que tomarlo sin ningún apuro

despacito  

quiero respirar tu cuello despacito  

deja que te diga cosas al oído  

para que te acuerdes si no estás conmigo

despacito  

quiero desnudarte a besos despacito  

firmo en las paredes de tu laberinto  

y hacer de tu cuerpo todo un manuscrito (sube, sube, sube)

(sube, sube)

answer: 30

step-by-step explanation: its 30 because im a basketball player and stephen curry hits most of his shots so if you put 30 you might make the shot goodluck

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.