Game boy cartridges are on sale for 15 percent off. How much will a cartridge cost that normally sells for $19.99? ( Round to the nearest cent.)

$16.99

Step-by-step explanation:

The first step into solving this requires you to find out what 15% of $19.99 is. In order to do that, you must multiply the number (19.99) by the decimal form of the percentile (15%). To find a decimal form of a percent sign, you must simply move the decimal two spots to the left, to change the denominator to 100. A number always has a denominator, even if it isn't shown. in this case, it is 15/1, and we have to convert it to 15/100 so all you have to do is divide 15 by 100, or move the decimal to the left 2 spots. The answer you should get is 0.15

I normally don't round until I have the final answer, just so the numbers aren't off. So then, what you need to do, is multiply 19.99 by your newly found decimal. You should get 2.9985. Now, the question asks for how much the cartridge would cost if it were 15% off, so therefore you must subtract your answer from the original number (19.99), and should get 16.9915. Now is the time you should simplify to the nearest hundredth, and should get $16.99 as your final answer.

56

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.