Your mother gave you $13.32 with which to buy a present. this covered 3/5 of the cost. how much did the present cost

$22.20

Step-by-step explanation:

We are told that $13.32 only covered 3/5 of the cost. Does this mean it covered the whole cost? No. The 5 in the denominator represents the whole cost, and the 3 represents the fraction of the cost. In this case, 13.32 represents the 3, as it did not cover the whole cost.

Let's set up our equation by making 13.32 into a fraction.

13.32/x

We will represent the present's actual cost with the letter x.

13.32/x = 3/5

We want to make sure the 13.32 is in the numerator, equal to the 3 in the numerator in the 3/5 number, because we know that $13.32 only covered for a fraction of the present's cost, not the whole cost.

We solve this by doing cross multiplication. Multiply the 13.32 by the 5.

13.32 x 5 = 66.6

Next we divide 66.6 by 3 to get 22.2

66.6/3 = 22.2

This gives us our final answer of $22.20. This means that the cost of the original present was $22.20

answer: a

step-by-step explanation:

if xy ~ yz therefore xy would be the same length as yz

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.