Melody

\(f(x) = \begin{cases} k(x) &\text{if }x>3, \\ x^2-6x+12&\text{if }x\leq3. \end{cases}\)

The LaTex for this question is really cool.  I've not seen this code before.

f(x) = \begin{cases} k(x) &\text{if }x>3, \\ x^2-6x+12&\text{if }x\leq3. \end{cases}

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

\(y=x^2-6x+12\\ y=x^2-6x+9+3\\ y=(x-3)^2+3\\ y-3=(x-3)^2\\ \pm \sqrt{y-3}=x-3 \qquad where\;\;y\ge3\\ x=\pm \sqrt{y-3}+3\\ \text{The desired inverse will be one of these.}\\ y=\pm \sqrt{x-3}+3\\ \text{BUT is it the plus or the minus?}\)

Consider f(0)=12  

The reflection of (0,12) in the line y=x (which is the inverse) is  (12,0)

This works when the sign is negative.

so this underneath will be its own inverse.

\(f(x) = \begin{cases}3 -\sqrt{x-3} &\text{if }x>3, \\ x^2-6x+12&\text{if }x\leq3. \end{cases} \)

And here is the graph:

The denominator of a fraction is 3 more than its numerator.

When  12   is added to this fraction, the resulting fraction's denominator is twice the denominator of the original fraction and its numerator is 1 more than its denominator.

What is the numerator of the original fraction?

Let the fraction be    \(\frac{a}{a+3}\)

\(\frac{a}{a+3}+12 =\frac{a+12(a+3)}{a+3}=\frac{13a+36}{a+3}=\frac{2(13a+36)}{2(a+3)}\)

The meaning of the question is very unclear.

I've tried a couple of different versions an none make sense.

This question,as far as I can tell, is a garbage question.

A collection of nickels, dimes, and quarters is worth 5.30. There are two more dimes than nickels and four more quarters than dimes. How many quarters were in the collection?

this is an international forum. You should spell out the value of your currency.

I think...

You mean 

$5.30

a nickel is 5c 

a dime is 10c

a quarter is 25c

Let N be the number of nickels, D be the number of dimes and Q be the number of quarters.

There are two more dimes than nickels

D=N+2

and four more quarters than dimes.

Q=D+4

so

Q=N+2+4=N+6

-----

5N+10D+25Q=530

N+2D+5Q      = 106

N+2(N+2)+5(N+6)=106

N+2N+4+5N+30=106

8N+34               =  106

             8N        = 72

                     N   =9

so

N=9      D=11   and   Q=15

check  9*5+11*10+15*25 = 530         Das ist ausgezichnet!

So wir haben

9 nickels,  11 dimes, and  15 quarters.

Mary has six cards whose front sides show the numbers  1,2,3,4,5and 6 . She turns the cards face-down, shuffles the cards until their order is random, then pulls the top two cards off the deck. What is the probability that at least one of those two cards shows a square number?

The only square numbers are 1 and 4

P(one of them is 4) = \(\frac{1}{6}\times \frac{5}{5}+\frac{5}{6}\times \frac{1}{5}=\frac{2}{6}\)

P(one of them is 1) = \(\frac{2}{6}\)

P(one is 1 and the other is 4)=\(2*\frac{1}{6}\times\frac{1}{5}=\frac{2}{30}\)

so 

P(at least one is either a 1 or  a 4 is) =  \(\frac{4}{6}-\frac{2}{30}=\frac{18}{30}=\frac{3}{5}\)

Happy New Year to you too   

It is nice to see you back again Heureka. 

And with another masterpeice of presentation to start the forum year  

I just made this for you

Thanks Rom, much appreciated.

I thought about it but I am not good with mod questions.

I assume the fact that 143=11*13   can somehow be used.  ??

Just replace the x with 3

\(f(3)=\frac{20}{4+3e^{-0.2*3}}\\ f(3)=\frac{20}{4+3e^{-0.6}}\\\)

20/(4+3e^-0.6) approx = 3.5420580109166572