Melody

This answer is quite similar to mine Nauseated.

You may have made it a smidge easier by getting a common denominator in the first place but otherwise our answers are the same.    

SEE great minds do think alike!  

Thanks Rosala - It is Australia Day too !!

Our Great nations have much in common :))

@@ End of Day Wrap    Mon 26 /1/15     Sydney, Australia      Time 9:25 pm

How did you guess  -   It was a Public Holiday in Australia today.     

Australia Day marks the anniversary of the 1788 arrival of the First Fleet in Sydney Cove.

(Well the first boat anyway)  

Now down to business.

There were a lot of questions and some very busy  answerers today.  Thank you Heureka, Sasini, CPhill, MsTeya, Saseflower, Newbie, Alan, SevenUP, and I think I just saw a pop-up for Nauseated.  

And the cheer goes out.   

Now for the interest Posts

Tues 27/1/15

1) Chris's retort was great - I had not heard this saying before.  Thanks Chris

Another of Heureka's masterpeices.   (26/1/15)

$$\left[\ 1+\frac{1}{3}+\frac{2}{3}+\frac{2}{9}+\frac{4}{9}+\frac{4}{27}+\frac{8}{27} +\frac{8}{81} +\frac{16}{81}+\dots \ \right] \\\\ \\= 1+ \frac{1}{3} + \frac{2}{3} + \underbrace{\frac{1}{3}*\frac{2}{3} }_{=\frac{2}{9}}+ \underbrace{\frac{2}{3}*\frac{2}{3} }_{=\frac{4}{9}}+ \underbrace{\frac{1}{3}*\frac{4}{3^2} }_{=\frac{4}{27}}+ \underbrace{\frac{2}{3}*\frac{4}{3^2} }_{=\frac{8}{27}}+ \underbrace{ \frac{1}{3}* \frac{8}{3^3}}_{=\frac{8}{81}}+ \underbrace{\frac{2}{3}*\frac{8}{3^3}}_{=\frac{16}{81}}+\dots \ \\\\\\= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{4}{3^2}}+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{8}{3^3}}+\dots \ \\\\\\= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^2}{3^2}}+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^3}{3^3}}+\dots \ \\\\\\= 1+ 1+ \frac{2}{3}}+ \frac{2^2}{3^2}}+ \frac{2^3}{3^3}}+\dots \ \\\\\\s=1+1*( \frac{2}{3} ) ^0+1*(\frac{2}{3})^1+1*(\frac{2}{3})^2+1*(\frac{2}{3})^3+1*(\frac{2}{3})^4+\dots \$$

\left[\ 1+\frac{1}{3}+\frac{2}{3}+\frac{2}{9}+\frac{4}{9}+\frac{4}{27}+\frac{8}{27} +\frac{8}{81} +\frac{16}{81}+\dots \ \right] \\\\\\

= 1+ \frac{1}{3} + \frac{2}{3} + \underbrace{\frac{1}{3}*\frac{2}{3} }_{=\frac{2}{9}}+ \underbrace{\frac{2}{3}*\frac{2}{3} }_{=\frac{4}{9}}+ \underbrace{\frac{1}{3}*\frac{4}{3^2} }_{=\frac{4}{27}}+ \underbrace{\frac{2}{3}*\frac{4}{3^2} }_{=\frac{8}{27}}+ \underbrace{ \frac{1}{3}* \frac{8}{3^3}}_{=\frac{8}{81}}+ \underbrace{\frac{2}{3}*\frac{8}{3^3}}_{=\frac{16}{81}}+\dots \ \\\\\\

= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{4}{3^2}}+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{8}{3^3}}+\dots \ \\\\\\

= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^2}{3^2}}+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^3}{3^3}}+\dots \ \\\\\\

= 1+ 1+ \frac{2}{3}}+ \frac{2^2}{3^2}}+ \frac{2^3}{3^3}}+\dots \ \\\\\\s=1+1*( \frac{2}{3} ) ^0+1*(\frac{2}{3})^1+1*(\frac{2}{3})^2+1*(\frac{2}{3})^3+1*(\frac{2}{3})^4+\dots \

Very nice Heureka     

The two questions are not the same for the bottom question

5x-5=0  or  x+3=0

so x=1  or    x=-3

how does the discount rate fit into the question?

Use this site - I cannot make it easier than this

That would depend on the diameter of the coin