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Google translation (from Mongolian)

Three externally indistinguishable from a genuine coin, the coin weight differences were counterfeit coins. Using the pan without the scale, weighing only a few counterfeit coin can be found?

Sorry but I do not understand the question. 

Let x = 1 + δ, where δ is very small and tends to zero as x tends to 1.

Then x^1/m = (1 + δ)^1/m → 1 + δ/m as δ → 0

Simnilarly x^1/n → 1 + δ/n

This means (x^1/m -1)/(x^1/n - 1) → (1 + δ/m - 1)/(1 + δ/n - 1) → (δ/m)/(δ/n) → n/m

So the limit of (x^1/m -1)/(x^1/n - 1) as x → 1 is n/m

.

Гадаад байдлаараа ялгагдахгүй гурван зоосны нэг нь жинхэнэ зоосноос жингээрээ ялгаатай  хуурамч зоос байв. Туухайгүй тавган жинлүүрийн тусламжтайгаар хамгийн цөөн хэдэн жинлэлтээр хуурамч зоосыг олж болох вэ?

I actually can't take much credit for this one.......the answer comes straight from a "formula" supplied by  Nauseated....

For any R x C grid arrangement where R represents the rows and C represents the columns.....the number of distinct paths from one corner of the grid to a diagonally opposite corner is given by :

(R + C ) !   / [R! * C! ]

CPhill and I got the same answer, wonders never cease!

Mmm  Well there are 4 ways to get from E to F

Now going from F to G

If you horizontal first then there are 4 ways

if you go down1 then across there are 3 ways

if you go down 2 then across there ae 2 ways

if you go all the way down first then there is 1 way.

4+3+2+1=10 ways

If that is correct then there would be   4*10=40 ways to go from  E to F to G

Yeah.....and a lot more, too.....it's a non-repeating, non-terminating decimal......

2\5\8 * 5\1\3  =

(2/5) / (8/1) *  (5/1) /(3/1)  =

(2/5) * (1/8)  *   (5/1) * (1/3)  =

10/120 =

1/12

4-5[8-2((-4)^2+5)]    simplify

4 -5[8 -2( 16 + 5) ] =

4 -5[8 - 2(21)]  =

4 -5 [8 - 42] =

4 - 5[ -34]  =

4 + 170  =

174

The square root of 1 is one

They have no common factors.....{they are relatively prime }

If you mean "least common multiple,"  it's 36

Not sure about the  "13/& pi"  part, but the slope is the thing in front of x, i.e.,  (5/4)

We can use some polynomial "long division" to help us with this one.....

                          -2x^2          + [c- 4]x          + [2c - 13]

----------------------------------------------------------------

x - 2     -2x^3 + cx^2            - 5x                       + 2

            -2x^3  + 4x^2

----------------------------------------------------------------

                          (c -4)x^2      -5x

                          (c - 4)x^2    -2(c - 4)x

----------------------------------------------------------------

                                              [2c - 13]x               + 2                                        

                                              [2c - 13]x              -2[2c - 13]

-----------------------------------------------------------------------

                                                                             4c  - 24

                             -2x^2          + [c + 2]x          - [ 7 + c]

             -------------------------------------------------------------

x + 1    -2x^3  + cx^2             - 5x                      + 2

             -2x^3  -  2x^2

-----------------------------------------------------------------------

                           [c + 2)x^2     -5x                                                             

                           [c + 2]x^2    + [c + 2)x

------------------------------------------------------------------------

                                                 - [7 + c] x            + 2

                                                 - [7 + c] x          - [7 + c]

--------------------------------------------------------------------------

                                                                              9 + c

And it's obvious that the remainders will be equal when.....

4c - 24 = 9 + c      subtract c from both sides and add 24 to both sides

3c  =  33   .....  so .......

c = 11

Thank You for the answer that there is infinite solutions

"Anchor' the Democrats in any 5 consecutive seats. And there are 5! = 120 ways to arrange them.

Next, the Republicans can either be on the left or right of the Dems......so that's 2 ways of choosing where they sit. And they can be individually arranged in 5! ways, too = 120 ways.

The independent gets the remaining chair by default.....so we have

5! x 2 x 5!  =  120^2 x 2  = 28,800 possible arrangements

Another way to do this is to "anchor" the independent first.......the same logic would still apply.....

It can't be "solved"......at least not in terms of a single solution.....there are more variables (2) than equations (1).....when that happens, we have infinite solutions.....

cos 0°  = 1

Melody is correct....let me expand on the second part of her answer, since it might not be clear how she arrived at that.....

Notice that if we "anchor" the 3 people in any three consecutive chairs in the circle.....there are 5! ways to arrange the other people in the remaining chairs.....but....there are also 3! ways to arrange the people occupying the consecutive seats.....!!!

So  5! x 3!   = 120 x 6   = 720

LOL!!!.....I'm not even sure it gave me the correct interpretation....!!!!

The ways to get from E to F =  [1 + 3] !  / [1! * 3!]  = 4!/3!  = 4 ways

Ways to get from F to G  = [3 + 2]! / [3! * 2! ] =   5! / 12 =  120 / 12 = 10 ways

So

4 ways  x  10 ways  = 40 ways to go from E to G  through F 

OK....I'll take a leap at this one....again!!!

Possible

Splits

  ↓

5 - 0            put the 5 b***s in 1 box or the other  = 2 ways

4 - 1           select 4 of 5 b***s  and 1 of 2 boxes  = C(5,4) x 2  = 10 ways

3 - 2           select 3 of 5 b***s  and 1 of two boxes  = C(5,3) x  2   = 20 ways

2 + 10 + 20  = 32 ways

im almost positive the answer is 0 but try a calculator

hope this helps

A= pi*r^2

Example:

Find the area of a circle with a radius of 3. 

Formula: A= pi*r^2

A= pi*3^2

A= pi*9

A= 28.27

Give or Take. Make sense? 

Hope it helps!

It is just as well that you have that decoder ring Chris!