Melody

$$\begin{array}{rlll}
100*0.3636363636... =&36.363636...\\
1*0.363636..... =&\;\;0.36363636 \qquad \mbox{now subtract on both sides}\\
99*0.36363636=&36 \qquad \mbox{so}\\\\
0.36363636=&\frac{36}{99} \qquad \mbox{which can be simpified to}\\\\
0.36363636=&\frac{12}{33} \qquad \mbox{}\\
\end{array}$$

You can also do this as the sum of a GP  

Thank you David.

You have expressed is a wonderful show of appreciation.

I really appreciate it :)

WOW Honga,

This is really impressive.

To be truthful I spend so much time on administrating this site that i do not always look at answers that I am sure I would find greatly illuminating.  

I am very glad that I looked at this answer.  Your explanation is brilliant.

I hope that you stick around to WOW us a lot more.

Thank you.

Tues 2/9/14

*1) Finding the last 2 digits of a large power or a multiple.   Some very impressive number work here (Thanks Honga)

@@ End of Day Wrap:    Mon 1/9/14       Sydney, Australia      Time 1:30am (Really Tuesday morning)     ♬

Hi all,

Today our great answers were supplied by CPhill, AzizHusain, Bertie, Honga, Heureka and TakahiroMaeda.  A very big thank you to each of you.  

Now straight to our interest posts:

1) This one looks interesting.

Thanks Chris 

I have been asked to finish this one - I am your humble servant (only this time lol) so I shall be pleased to do so.   

$$\begin{array}{rlll}
\frac{12}{sinB}&=&\frac{11}{sin30}\\\\
\frac{sinB}{12}&=&\frac{sin30}{11}\\\\
SinB&=&\frac{12sin30}{11}\\\\
B&=&asin\left( {\frac{12sin30}{11}}\right)\\\\
B&=&33.056^0\qquad &\mbox{First quadrant solution correct to 3 dec places} \\\\
B&=&180-33.056=146.944\qquad &\mbox{Second quadrant solution correct to 3 dec places}\\\\
\end{array}$$

$$If B is $33^0$ then \\
$C=180-30-33=117^0$\\\\
If B is $147^0$ then \\
$C=180-30-147=3^0$\\\\
So the 2 possible solutions are $

I think that means "how much more will i get paid"

40% of 12000=40/100*12000 

You can finish it on the site calc on the home page 

I am not sure if you want me to do this with calculus or with physics formulas.

I'll do it with physics formula.

$$\\u=80km/hour\;=\frac{80000}{60*60}m/sec = \frac{200}{3*3}m/sec= \frac{200}{9}m/sec\\\\
v=20km/hour\;=\frac{20000}{60*60}m/sec = \frac{50}{9}m/sec\\\\
a=-2m/s^2\\\\
find\;t\\\\
v=u+at\\\\
\frac{50}{9}=\frac{200}{9}+-2t\\\\
\frac{-250}{9}=-2t\\\\
\frac{250}{18}=t\\\\
t=13.9\;sec \qquad \mbox{to nearest one tenth of a second}\\\\$$

I've have constructed the triangle to show you that there are exactly 2 solutions solutions.

This triangle is drawn to scale.

We need to find angle B first.

You can do this using the sine rule but remember, since sin is positive in the first and second quadrant there will be an accute angle B and an obtuse angle B - Two possible solutions for B

Once you have angle B you can find Angle C by using angle sum of a triangle.  Once agian there will be 2 solutions.

You try finding this these angles.

If you get stuck again put another post after this one to ask for more help.