Melody

doesn't make sense!

pi/3=60degrees

draw an equilateral triangle of side length 2

cut it in half.

put angles and side lengths on it

find the height using pythagoras

"read" tan 60 off the triangle.

I just realised that I can remove one off the watchlist and then reinstate it and it will  go to the top of the list.

 I did not think that it worked like that before!  I must make use of this feature.    

It does not work that way with the sticky topics.     

It is the question watchlist that you have on the top right hand side of your page.

YOu can save your favourite threads in there so that you can find them easily later.

When you open a thread there is a button at the top of it that says 'save to watchlist' or 'remove from watchlist so that you can keep it all nicely organised.  

Unfortunately once it is in the list it slowly moves down in importance.  There does not seem to be a way of keeping the ones that you want the most at the top of the list.

This is why I often go through them and delete less important ones!

Yep got it.  Thanks Chris.   

$$\\(2* R1* R2) / (R1 + R2 ) \\\\
=\dfrac{\frac{2xy\; km^2}{h^2}}{\frac{x\;km}{h}+\frac{y\;km}{h}}\\\\\\
=\dfrac{\frac{2xy\; km^2}{h^2}}{\frac{x+y\;km}{h}}\\\\\\
=\frac{2xy\; km^2}{h^2}\div \frac{x+y\;km}{h}\\\\
=\frac{2xy\; km^2}{h^2}\times \frac{h}{x+y\;km}\\\\
=\frac{2xy\; km}{h}\times \frac{1}{x+y}\\\\
=\frac{2xy\; km}{(x+y)h}\\\\
=\frac{2xy}{(x+y)}\frac{km}{h}\\\\$$

Umm.  Not a great deal of help although at least the units are right.  

I will have to think about it some more.  

I didn't know that Chris.  Thank you.   I'll have to think about it more when I have more time!  :)

$$\\2^m = 0.04\\
\mbox{when ever you are solving for a power you use logs (unless it is a very easy one)}\\\\
log\;2^m = log 0.04\\\\
mlog\;2 = log 0.04\\\\
m=\frac{log 0.04}{log\;2}$$

And you can finish it 

$$g(x)= \frac{3x+4}{x-1}\\

We are looking for the domain.\\

Well we can't divide by 0 so x cannot be 1, I can see no other restrictions.\\


Domain: x\;\in R \quad where \quad x\ne1$$

Do you want me to use calculus or physics formulas?

Calculus:

Initially  

 $$\\\ddot y=-9.8\;m/s^2, \qquad \dot y=0\;m/s, \qquad y=78.4\;m \\\\
\mbox{Find t when y=0}\\\\$$

Ongoing:

$$\\\ddot y=-9.8\;m/s^2\\\\
\ddot y=-9.8t+C_1 \\\\
c_1 =0 \qquad \mbox{because of initial conditions therefore} \\\\
\dot y=-9.8t\\\\
y=\frac{-9.8t^2}{2}+C_2\\\\
C_2 =78.4 \qquad \mbox{because of initial conditions therefore} \\\\
y=-4.9t^2+78.4$$

find t when y=0

$$\\0=-4.9t^2+78.4\\\\
-78.4=-4.9t^2\\\\
16=t^2\\\\
t=4\;seconds$$

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If you want to do it by physics formula you will use         $$s=ut+\frac{1}{2}at^2$$

and you will get the same result.