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Das ist gleichbedeutend mit:

$${f}{\left({\mathtt{x}}\right)} = {\mathtt{k}}{\mathtt{\,\times\,}}{\mathtt{x}}$$

Dies wiederum ist eine ganz normale Funktion. Das mit dem Pfeil ist nur eine etwas andere Schreibweise.

Yes you can do an enormous amount of things with the array function.

It may be time for you to really practice the LaTex that you have already acquired for a while before you try too hard to add to it.  You have learned an enormous amount in a very short space of time.   

Have a look through the LaTex thread or just look at mine and Heureka's answers if you want to add more to your knowledge base.

Also there are the automated functions that admin added in the first place - like colours, italics, bold and others.  Have a look at what is easily available to you in the LaTex ribbon.

You are very welcome :)

Sorry didn't know it was called factorize, english is not my native language, thanks melody! ^^

Your formula is not correct

$$\boxed{(a+b)^2=a^2+2ab+b^2}$$

I can factorize it for you

$$(7x+1)^2$$

with decompose I mean like a²+ab+b²=(a+b)²

Are G&C s like G&T s Chris ?

Mmm both sound ok but I think I'd rather a G&T       

Thanks Headingnorth.   Well done!

This one I definitely want to come back to.   

That is pure logic.

It's because $$(n-1)!=\frac{n!}{n}$$

That means :

$$\\6!=1\times 2\times 3\times 4\times 5\times 6=720
\\(6-1)!=5!=\frac{6!}{6}=120
\\(5-1)!=4!=\frac{5!}{5}=24
\\(4-1)!=3!=\frac{4!}{4}=6
\\(3-1)!=2!=\frac{3!}{3}=2
\\(2-1)!=1!=\frac{2!}{2}=1
\\\textcolor[rgb]{0,0,1}{(1-1)!=0!=\frac{1!}{1}=1}$$

Well the chimp should be back by now because the God has already returned :/

Wow the array displays algebra nicely!

I actually wanted to post the question on the forum but I didn't know how to display the question so I asked you for help .

You may be right; wait, I'll remove things in the wording...

it is 1 1/8

You are a master of these graphical solutions Alan  

pH is defined as -log₁₀(hydrogen ion concentration).

True, but what I meant was that you had supplied all the necessary formulas, so it's just a matter of substituting the numbers into them (though I suppose you have to know how to get the molar mass of H₂O from the individual masses supplied)

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Civonamzuk, thanks for these interesting questions.

If you write serial questions like this could you please link them with the web address of the others.

So a person can see the whole thing?

This was the original post.

I have included this question in the Puzzle Thread

This question is continued here

Yes I think this is quite hard M.G.  

It looks like a good puzzle.  I will add it to our puzzles thread   

$${{\mathtt{27}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)} = {\left({{\mathtt{27}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}^{{\mathtt{2}}} = {{\mathtt{3}}}^{{\mathtt{2}}} = {\mathtt{9}}$$

the answer is 9

The 1st sum is 1-1+1-1+1-1+... and not 1+0+1+0+1+0+...