Noah bought 4 tacos and paid $6. At this rate, how many tacos could he buy for $15? Jada's family bought 50 tacos for a party and paid $72. Were Jada's tacos the same price as Noah's tacos?

-Noah could buy 10 tacos for $15.

-Jada's tacos were not the same price as Noah's tacos.

Step-by-step explanation:

You can use a rule of three to find the amount of tacos that Noah could buy with $15:

$6  →  4

$15 →  x

x=(15*4)/6=10

At this rate, Noah could buy 10 tacos for $15.

Now, to find out if Jada's tacos were the same price as Noah's tacos, you have to find the price per taco in each case by dividing the amount paid by the number of tacos bought:

Noah: 15/10=1.5

Jada: 72/50=1.44

According to this, Jada's tacos were not the same price as Noah's tacos.

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.