Ayouth organization collected nickels and dimes for a charity drive. by the end of the 1-day drive, the youth had collected $63.80. if there were five times as many dimes as nickels how many of each type of coin was collected

Let, the numbers of nickels be x.

Given, there were 5 times as many dimes as nickels.

So, the number of dimes = 5x

We know that 1 nickel = $(0.05)

So x nickels = $(0.05x)

Also we know that 1 dime = $(0.1)

So 5x dimes = $(0.1 ×5x) = $(0.5x)

Given total value of dimes and nickels at the end of day 1 is $63.80

So we can write the equation

Now we to find x, we have to move 0.55 to the other side. As 0.55 is multiplied there so to move it we need to divide it to both sides.

So we have got the number of nickels = 116

Number of dimes = 5x = 5(116) = 580

So we have got the required answer.

Number of nickels = 116, number of dimes = 580.

the thermometer is precise but not accurate

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.