Melody

Mmm lets see.  I don't know if you have been given a technique Rosala.

$$\sqrt[3]{4913}$$

What digit cubed ends in 3 . 

1^3=1, 2^3=8, 3^3=27, 4^3=64, 5^3 no, 6^3 no, 7^3 yes no 8^3no 9^3=no

I know the number ends in 7 because 

7*7=47 9*7=63  No other's will end in 3

10^3=1000 too small

20^3=2^3*10^3=8000 too big

So the number has to be between 10 and 20 and end in 7

So the number has to be 17      

$$\frac{64b*3}{4}=16b*3=3*16b$$

and you can finish it!

Look at this post. 

You are confusing me with the deltas    Is this what you want?

$$\lim\limits_{\Delta x\rightarrow0} \frac{\Delta x^2-2\Delta x}{\Delta x}\\\\
=\lim\limits_{\Delta x\rightarrow0} \frac{\Delta x(\Delta x-2)}{\Delta x}\\\\
=\lim\limits_{\Delta x\rightarrow0} \frac{(\Delta x-2)}{1}\\\\
=\lim\limits_{\Delta x\rightarrow0} \Delta x-2\\\\
=0-2\\\\
=-2$$

That categorizes me pretty convincingly - I see John Lennon    

That emu is my new desktop background - thanks Rosala. 

Thank you anonymous,

Why don't you join up and answer other questions?  

The regular answerers have a fun little community and it is very easy to join up.     

Chris has assumed that this is a co-ordinate geometry question.

I did not assume this,  I would have just graphed y>3 on a number line and labelled the number line y.

Most likely Chris provided the desired answer.

It is important that askers word their questions as clearly as possible.

Usually that means use brackets but other things need to be clear too.   

Probably temporary I want to get back to if possible.     30/8/14

Physics:

Thanks Rosala.  

Dragon, It is -11/3  !