Melody

You can use the quadratic formula to solve quadratic equations.

This is a quadratic equation. $$a,b$$ and $$c$$ are constants and $$a\ne 0$$  

$$ax^2+bx+c=0$$

Now this will help you remember it!

Thanks Alan,  I don't disagree with your answer but I don't understand why you introduced 'a'.

I'm going to do it my way. Plus I am experimenting with LaTex.  I wrote this using texmaker and then tried to copy it straight over into the forum.  It didn't work.  I used begin{align}and also the begin{equarray} and I don't think either of these are recognised commands in the forum. So I have copied the output to a .jpg file and then uploaded it using Tiny pic.  This worked okay.  

Lots of experimenting happening here. 

Actually, I am very familiar with equarray but I have not used align very much.  I would have preferred it if the output was closer together.  I wonder if Rom can help me with this.  I think he is our most capable user of LaTex.

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Think about this

$$\mbox{}\\
\frac{8^4}{8^2}=8^{4-2}=8^2\\\\
\frac{8^4}{8^3}=8^{4-3}=8^1\\\\
\frac{8^4}{8^4}=8^{4-4}=8^0\\\\
\mbox{But}\\\\
\frac{8^4}{8^4}=\frac{8*8*8*8}{8*8*8*8}=\frac{1}{1}=1\\\\
\mbox{therefore}\\\\
8^0=1$$

Any number or pronumerals (letter) raised to the power of 0 is 1 EXCEPT  $$0^0$$  is indeterminant

There is no root equivalent.  A power of 0 is not a root.  

It is not anything because you have not included any units of measure. 

@@ End of Day Wrap - Thursday 1/4/14   Sydney, Australia Time 20:06

Hi everyone,

Once again there were many great answers given by Bertie, Alan, CPhill, BluePhoenix912, jmanf729, matisonc17035, alanb1291, BrainChap21, MacGamer04 and Edison.  (I am sorry if I forgot anyone) Thank you all. 

I think when Andre (the site owner and developer) finally gets the members section working properly we will be astonished at how many new members we have.  It is great to see that a number of these are also answering questions.  We encourage everyone to answer.  Obviously if you are unsure you should state this, but ultimately it is up to the question asker to decide to accept or reject an answer.  I also encourage people, especially the mathematicians, to check answers, including each others, so most mistakes are detected.

If new people would like to introduce themselves to the forum in general, by way of a public post, I encourage you to do so.  If you would like to introduce yourself to me privately you can do so via the forum message service or by an email to the address that I will include at the end of this wrap.  I love getting mail and I am sure other members would also welcome your contact.

We passed our 10,000th question today!  BluePhoenix912 bought this to my attention with this post.

Thanks BluePhoenix, but you might like to edit your post, you left out a 0!  

I had some really good thread addresses on my 'information pages worth keeping or developing' sticky thread. Now most don't work.  They are full of old downloads.  I wanted to use my 'dealing with negative indices thread' and could not.  NOT GOOD!  I suppose I'll redo them if I ever get time.

I can't recall any other memorable posts.   If you can think of something or some post please let me know with a message. Then i can edit this wrap or include the omitted content tomorrow.

I am having intermittant problems with the \frac command in LaTex.   Sometimes it works and sometimes it doesn't.

That is it for this Wednesday/Thursday.

Take care,

Melody

melodymathforum@gmail.com

Thanks All SmilesAlways  

And thank you Alan. 

Cute 

None of those answers have an = sign

Therefore none of them is an equation.  

Therefore none of them is the answer.

I appreciate your nice manners but I do not understand your question? 

Yes, I didn't notice that - thanks Alan.

 I assume the question asker actually meant 

$$tan(x)=\frac{7}{11}$$

therefore 

$$x=tan^{-1}\left(\frac{7}{11}\right)$$

This is the question that CPhill answered but it is not actually the question that you asked!