Melody

just type into the calculator 

atan(0.5)

Hi MG,

Yes basically you do factor then cancel that is.

I want to go back a step and see how much you understand of numbers.  If I was tutoring you face to face this would be so much easier and quicker for both of us but I really need to work out your level.  It is easy to rote learn, you obvioulsy have a good memory, but i need to be certain that you really understand each step and you are just not mimicking what i have told you.  (you may not even realise that you really did not understand)

So 

Please be patient with me and show me how to do these.  If you can't do them say so but it would be good if you have a go as well.

Please do all the cancelling that you can and do Q2 and Q3 in a few steps so I can see how you have done them.    

$$\\1)\;\;\frac{2}{3}\times \frac{5}{7}\\\\
2)\;\; \frac{16}{12}\times \frac{15}{10} \\\\
3)\;\;6\frac{2}{5}\times 1\frac{2}{6}\\\\$$

ok we might as well all answer this question!

This is how I was trained to do it

$$\\T\alpha \frac{L}{N} \\\\
T=k* \frac{L}{N}\qquad but\\\\
5=k* \frac{200}{3}\qquad so\\\\
k=\frac{3}{40}\qquad so\\\\
T=\frac{3L}{40N}\\\\
$When N=5 and L=500$\\\\
T=\frac{3*500}{40*5}\\\\
T=7.5 \;\;hours$$

$$\\\left(x-\frac{2}{(x+1)}\right)\div \left(1-\frac{4x+7}{x^2+4x+3} \right) \\\\
=\left(\frac{x(x+1)-2}{(x+1)}\right)\div \left(\frac{x^2+4x+3-(4x+7)}{x^2+4x+3} \right) \\\\
=\left(\frac{x^2+x-2}{(x+1)}\right)\div \left(\frac{x^2-4}{x^2+4x+3} \right) \\\\
=\frac{x^2+x-2}{(x+1)}\times \frac{x^2+4x+3} {x^2-4}\\\\
=\frac{(x+2)(x-1)}{(x+1)}\times \frac{(x+3)(x+1)} {(x-2)(x+2)} \\\\
=\frac{(x-1)}{1}\times \frac{(x+3)} {(x-2)} \\\\
=\frac{(x+1)(x+3)} {(x-2)} \\\\
=\frac{x^x+4x+3} {x-2} \\\\$$

If I haven't made any stupid mistakes then that will be ok :)

You need to check it!

What music Dbg ?   You did not post any music clips    

Did you look at my clip?

It is 1- the probability that they all have different birthdays

$$\\=1 - (\frac{365}{365}*\frac{364}{365}*\frac{363}{365}*\frac{362}{365}*\frac{361}{365}*......*\frac{366-25}{365})\\\\
=1- (\frac{365}{365}*\frac{364}{365}*\frac{363}{365}*\frac{362}{365}*\frac{361}{365}*......*\frac{365-24}{365})\\\\
=1- (\frac{365P24}{365^{25}})$$

$${\mathtt{1}}{\mathtt{\,-\,}}{\frac{{\left({\frac{{\mathtt{365}}{!}}{({\mathtt{365}}{\mathtt{\,-\,}}{\mathtt{25}}){!}}}\right)}}{{{\mathtt{365}}}^{{\mathtt{25}}}}} = {\mathtt{0.568\: \!699\: \!703\: \!969\: \!463\: \!9}}$$

So P(2 or more have the same birthday) is  approx 0.57      or    57% 

$$\\\frac{-1}{8}\times\frac{-4}{5}=\frac{-1*-4}{8*5}=\frac{+4}{+40}=\frac{2}{20}=\frac{1}{10}$$

There you go - any questions ?

Here was your question from your last thread that you have not answered yet :)

Can you simplify this one for me ?  Do it with cancelling.  

Can you remember how to do that or do you want me to show you?

$$\frac{21x}{y}\times \frac{8y^2x}{14t}$$