Melody

Was it a test Chris ?

Thanks Heureka  

This has been included in our reference material thread :)

Oh yea I forgot to do that bit, thanks Chris.    

Thanks Chris,

I was playing with it too but I didn't really get anywhere.

Here is the Wolfram|Alpha solution 

A beam of light from a sodium street lamp is found to have a frequency of 5.09 x 1014 Hz. What is the wavelength?

Hz is wavelengths per sec

so I use the units in a rather unusual way. I am sure others would use a formua.

we have

 $$\\frequency =\frac{5.09*10^{14}\;\;\lambda}{second}\qquad speed\;of\;light=\frac{299792458\;m}{sec}\qquad we\;want\;\; \frac{m}{\lambda}\\\\
\frac{299792458\;m}{sec}\times \frac{sec}{5.09*10^{14}\;\;\lambda}\qquad $The seconds cancel out$\\\\
=\frac{299792458\;m}{5.09*10^{14}\;\;\lambda}$$

$${\frac{{\mathtt{299\,792\,458}}}{\left({\mathtt{5.09}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{14}}}\right)}} = {\mathtt{0.000\: \!000\: \!588\: \!983\: \!218\: \!1}}$$

So the wavelength is    $$\approx 589*10^{-9}\;\;metres = 589\;nanometres$$    

$$4$$  is rational and  $$\sqrt{11}$$   is irrational.

I assume that you really mean this     

Sin(x)/Cos(x) + Cos(x)/(1 + Sin(x) )

$$\\\frac{Sin(x)}{Cos(x)} + \frac{Cos(x)}{1 + Sin(x) }\\\\
=\frac{Sin(x)}{Cos(x)} + \frac{Cos(x)(1-sin(x)) }{(1 + Sin(x))(1-sin(x)) }\\\\
=\frac{Sin(x)}{Cos(x)} + \frac{Cos(x)(1-sin(x)) }{(1-sin^2(x)) }\\\\
=\frac{Sin(x)}{Cos(x)} + \frac{Cos(x)(1-sin(x)) }{cos^2(x) }\\\\
=\frac{1}{Cos(x)}\\\\
=sec(x)$$

Well I am guessing but I think it would be the sale price minus the cost price of the item.

This is the 'obvious' gross profit.

I suppose that for the net profit you would need to consider all the other cost that are incurred.

e.g.   Storage, transport, refrigeration, electicity, employees, equipment, etc etc.

Not quite anon.

Say you were doing a survey to find out how tall people are in your school.

You randomly choose 50 players off the year 12 boys and record their heights.

Would this be a fair,  representative, selection of the students.

Of course it would not, it was a bisased survey!

#

For an unbias survey you might choose a stratified selection.  

You might randomly choose 4 girls and 4 boys from each year and record their heights.

That would be an unbiased survey

Anon is right BUT you can get h in term of r   

$$\\A=\pi r\sqrt{h^2+r^2}\qquad V=\frac{\pi r^2h}{3}\qquad and \qquad V=3A\\
so \\
3\pi r\sqrt{h^2+r^2}=\frac{\pi r^2h}{3}\\\\
9\sqrt{h^2+r^2}= rh\\\\
81*(h^2+r^2)= r^2h^2\\\\
81h^2+81r^2= r^2h^2\\\\
81h^2-r^2h^2= -81r^2\\\\
h^2(81-r^2)= -81r^2\\\\
h^2= \frac{-81r^2}{81-r^2}\\\\
h^2= \frac{81r^2}{r^2-81}\\\\
h= \frac{9r}{\sqrt{r^2-81}}\;\;cm\\\\$$