Pedro tiene una tienda donde vende juguetes de colección, entre ellos las esferas del dragón. Se sabe que, cada esfera viene en una caja de forma
cúbica cuya arista mide (a + 2) cm.
Carlos hace un pedido de cierta cantidad de esferas, y estas son colocadas
en una caja cuya base tiene un área de (9a

2 + 36a + 36) cm2 y una altura de

(3a + 6) cm.
A. ¿Cuál es el volumen de una caja que contiene una esfera del dragón?

B. ¿Cuál es el volumen de la caja donde se ubicará el pedido? Representa
algebraicamente de dos formas.

C. ¿Cuántas cajas con las esferas del dragón compró Carlos, si se sabe qué la caja
donde se colocará el pedido se llena totalmente?
● Representa esta situación mediante una fracción algebraica y redúcela de ser
posible.

● Si el valor de “a” fuera 6 ¿cambiaría la cantidad de cajas con esferas del
dragón pedidas por Carlos? justifica tu respuesta

c

step-by-step explanation:

it would be see because your formula is f(x)=6x, so 6 times 1 is 6 and you go on and the best answer is c.

a

step-by-step explanation:

strong negative correlation

let's fu(!

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.