Here is a question just for LaTex beginners (Like DragonSlayer)

Had to look up a few things, but here it is!

$$\Big(\overline{x} \pm z\frac{s}{\sqrt{n}}\Big)$$

Or, alternatively:

$$\left(\overline{x} \pm z\frac{s}{\sqrt{n}}\right)$$

$$\frac{3}{5}$$

Who says an old dog can't learn new tricks????

Where's my points ??.......Where's my points ??.....Where's my points ??..........Where's my points ??..........Where's my points ??.........

$$\frac{3}{5}$$ 

Ok you two - fair enough.

Dragon was complaining that if he only answered validly then there would be nothing there for him at all.

BUT I want you two to learn as well.  

Something harder.

okay see if you can write the quadratic formula in LaTex. All set out beautifully.

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THIS NEXT ONE IS JUST FOR DRAGON!    Write 6 and three quarters in LaTex (as a whole number and an upright fraction).

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Aw, heck....!!! You mean Aziz and I can't play, too???

I'm taking my marbles and going home........

YES I WANT YOU TO PLAY TOO!  CAN'T YOU DO THE HARD ONE!

ARN'T YOU UP FOR A CHALLENGE?

YESTERDAY I TOLD YOU YOUR BLOOD WAS WORTH BOTTLING AN NOW YOU ARE PIKING!!!

COME ON - GIVE IT A GO!

$$\begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}$$

What a genius I am....!!!!   You know....this copying and pasting code stuff is pretty brain-draining ....!!!!

EXCELLENT CHRIS - I KNEW YOU COULD DO IT!!!!

It gets much easier with practice.

Here is a very special gift just of you!  I hope that you don't already have one like this - It is rare!

Do you want to play again?

WOW...that's a pretty cool image!!!...I might even consider that as a new avatar - it's time to change mine, anyway......thanx......

That would make a great avatar for you!  Although I really like the one that you have now.  It is so you!

CHRIS CHEATED (NOT REALLY)  - I DIDN'T SPELL OUT THE RULES PROPERLY DID I CHRIS

You can keep your points and your marble.  So please stay and play  

Chris can you please copy that code and paste it straight onto the post underneath the output.

Then I will talk you through it - It is more complicated then it needed to be. 

Let me try.

$$\frac{3}{5}$$

$$x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

:D

That is great Ninja.  That is all you needed to do Chris.

Chris got his from the site entitled. "How to get from Minesota to Louisiana via Sydney."

Although the other bit that Chris wrote was also relevant.  That is what the quadratic equation is used to solve!

But...I'm still in Sydney !!  ....I just took the "scenic route"

Yes that is true - look at Ninja's code.  It is nice and simple.

Actually I'd still like you to display your code properly.  I'd like to take a look at it.

Thanks.    

Here's the code......

\begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}

Oh I wasn't aware that you needed that other part that Chris had.

This is fun though. Got any other challenges?

\begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}

This is all array stuff that you do not need. It has been used as a way to separate the three parts, that is foumula, when and initial equation

the 2 backslashes before it ate only a line break so they are definitely not needed.

\begin{array}{*{20}c}    all the ampersands &  and   \end{array}

So I'm going to see what is left

 x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when} ax^2 + bx + c = 0

I don't know what the \rm{ } does.  I looks like it is meant to be short for remark but i use \mbox{  } for that

\mbox is short for message box.

If we get rid of all that then this is what I have - the output is underneath

x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{2a}}} when ax^2 + bx + c = 0

$$x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{2a}}} when ax^2 + bx + c = 0$$

Now I want a space between the three sections so I am going to use the function \qquad  (quad means 4 so I think this means leave 8 spaces. I could use \quad for a smaller gap

x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{2a}}} \qquad when\qquad  ax^2 + bx + c = 0

$$x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{2a}}} \qquad when\qquad ax^2 + bx + c = 0$$

There you go - that is looking better.

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Now, for my own sake I want to look at this bit

\begin{array}{*{20}c}    all the ampersands &  and   \end{array}

in particular this bit  {*{20}c}  it goes with the ampersands

It means something like  

Put the formula at the begining&use the next 20 spaces for when&centre the last bit

I am not really sure what it does and it would be really good if HEUREKA tells us.

 He know everything about LaTex.    

Alright Ninja,

Here is another one - try reproducing this - not this big!

If you get it out then try enclosing it in big brackets

This is how I would have done it - it is slightly different from yours

This proves that there is more than one way to skin a cat - oh sorry kitty.

I new about using Big but it has never worked fo me when I wanted to use it!

The \left and \right is probably better most of the time because it makes it the correct size, as opposed to a fixed large size.

\left(\bar{x} \pm z\frac{s}{\sqrt{n}}\right)

$$\left(\bar{x} \pm z\frac{s}{\sqrt{n}}\right)$$

Here is a stack of them that you can practice on

Hint:  the flatterned s at the front of each one is an integral sign.

 {*{20}c}    ?

Hi Melody,

I will answer your question:  the meening of {*{20}c} 

Generally :

*{count}{form}

Example:

{*{20}c}   means  {cccccccccccccccccccc}    "c" twenty times

{*{3}{|r}}  means  {|r|r|r}                        "|r" three times

OK Thanks Heureka   

Melody, can I try?

Look here

http://web2.0calc.com/questions/here-is-a-question-just-for-latex-beginners-like-draagonslayer#r3

Never mind this was YOUR question Dragon - And I quote.

THIS NEXT ONE IS JUST FOR DRAGON!    

Write 6 and three quarters in LaTex (as a whole number and an upright fraction).

$$6\frac{3}{4}$$

I did it!!!!!

Melody, can I do another one please?

I solved the one you told me to do.

Melody, the first one is REALLY EASY.

$${\frac{{\mathtt{3}}}{{\mathtt{5}}}}$$