Melody

Study this and see if you can work out what I am trying to show you :)

they are all the area of rectangles. 

okay lets look at factoring algebra 6x+4

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

do you know this

have you done it at school yet?

3(3x+2) = 3*3x+3*2 = 9x+6 

well you could go one more step

$$\\\frac{(3x+2)x^3}{12}\\\\
=\frac{x^3(3x+2)}{12}\\\\
=\frac{3x^4+2x^3}{12}\\\\$$

I am not sure that that answers your question :/

Does that help?

@@ End of Day Wrap   Sun 29/3/15   Sydney, Australia   Time 10:27 pm  ♪ ♫

Good evening everyone,

Today our wonderful answerers were Alan, Geno3141, CPhill, Gibsonj338, Bertie, Heureka, Nauseated and Civonamzuk.  A very big thank you to each of you. :)

Interest Posts:

 FTJ means "for the juniors"

Mon 30/3/15

FTJ means "For the Juniors"

Thanks Heureka, 

I have never seen this before.  I am going to try it too :))

I am trying to use acos(0.5)  your way and get the different quadrant answers but it is not working for me.

Can you show me how to do this please Heureka ?

Thanks Bertie,

I just want to look at this simlified scenario,

5 computers 3 printers.

Now it seems to me that the number of ways any computer can be net worked to printers is as follows:

any one printer+ any 2 printers+ all three printers = 3C1+3C2+3C3 = 3+3+1 = 7 ways

There are 5 computers so the number of ways is   $${{\mathtt{7}}}^{{\mathtt{5}}} = {\mathtt{16\,807}}$$

Ok so here every computer is connected to at least one printer BUT for some of these outcomes 1 or more computers are not connected to any printer.  So some of these outcomes are invalid.  The number is too big.

IS THAT STATEMENT CORRECT BERTIE?  Perhaps i see a glimmer of light.  :/

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

What was your formula Nauseated? 

$$\displaystyle \sum \limits_{i=0}^{\textcolor[rgb]{1,0,0}{N-1}} *(-1)^i*\binom{N}{i}* (N-i)^k$$

The solution to this is found using this formula. k = number of computers and N = number of peripheral graphics devices.    N=3,    k=5

$$\displaystyle \sum \limits_{i=0}^{\textcolor[rgb]{1,0,0}{3-1}} *(-1)^i*\binom{3}{i}* (3-i)^5 \hspace{15pt}| \hspace{15pt} \Text {N=3 \; k=5} \\\$$                           k is the big one

I have a question for you too Nauseated :)

How do you know which number is N and which one is k.   Should N be the smallest one, it was for your other example.   If that is so then I have the 5 and 3 around the wrong way here.

PLUS Nauseated, I just realised that you used some LaTex that I didn't know. \binom{5}{i}  also \hspace {15}

Thanks for that - I will add it to our very disorganised LaTex thread.  :))

surface area of what?

arc tan is the same as inverse tan.

Just use atan