Melody

Can we use other numbers as well?  I'm not sure I understand.

I think that this is really your intended question

f(x)=x^2+2x-3      find f(0)

well what you do is replace all the xes with 0

f(0)=0^2+2*0-3=0+0-3=-3

f(0)=-3

this video looks pretty good.

I just did a huge answer here and I have lost the lot!   

6! reads as 6 factorial

6!=1*2*3*4*5*6

34! / (34-6)! 6! = can be entered into any calculator but you will need to put the bottom in brackets

like this  34!/((34-6)!*6!)=

$${\frac{{\mathtt{34}}{!}}{\left((\left({\mathtt{34}}{\mathtt{\,-\,}}{\mathtt{6}}\right)){!}{\mathtt{\,\times\,}}{\mathtt{6}}{!}\right)}} = {\mathtt{1\,344\,904}}$$

I had writen a great deal more than this.  If you would like more explanation or you need to know how to do it by hand, please ask and i will show you.   

Excellent answer Pokemonfan58  

I did notice that too Chris but it may not have been a gender change maybe just a quick change of name - Maybe a border was crossed and Jenny I mean Jimmy was a wanted man I mean woman er I don't know what I mean,  Neither does Jinny.  

I am just speaking for myself but I do not think that this makes sense.

Can you word it better?

For instance,

Is the first digit the one in the 10thousand column or the one in the units column?

But I think even more explanation than this is necessary.

Sorry.

why dont you do it yourself on the site calculator

click over to here to see how it is done.

@@ End of Day Wrap - Sunday 29/6/14   Sydney, Australia   Time 10:10pm   ♬ 

Hi everybody,  ♬

Now that was one quiet day.  Actually there were a couple of interesting questions. 

Anyway answerers included Alan, CPhill, NinjaDevo and Rosala.  Thank you.

Here are some interesting posts.     ♬

Rosala wants to talk to us all.  (Precurser)

f(54) = (1+sqrt(5))^54-(1-sqrt(5))^54/2^54*sqrt(5)  -------?

$$\begin{array}{rll}
f(54)&=&(1+\sqrt5)^{54}-\frac{(1-\sqrt5)^{54}}{2^{54}}\times\sqrt5
\end{array}$$

Before I even think about going further, is this your intended question?