Melody

Thanks for the thumbs up 

$$\[ \left( \begin{array}{cc}
1 & 2 \\
3 & 4
\end{array} \right)
%
\left( \begin{array}{cc}
2 & 6 \\
-1 & 0
\end{array} \right)=
%
\left( \begin{array}{cc}
0 & 6 \\
2 & 18
\end{array} \right)
\]
\[ \left( \begin{array}{cc}
0 & 6 \\
2 & 18
\end{array} \right)+
%
\left( \begin{array}{cc}
2 & 6 \\
-1 & 0
\end{array} \right)=
%
\left( \begin{array}{cc}
0 & 12 \\
1 & 18
\end{array} \right)
\]$$

\[ \left( \begin{array}{cc}
1 & 2 \\
3 & 4
\end{array} \right)
%
\left( \begin{array}{cc}
2 & 6 \\
-1 & 0
\end{array} \right)=
%
\left( \begin{array}{cc}
0 & 6 \\
2 & 18
\end{array} \right)
\]
\[ \left( \begin{array}{cc}
0 & 6 \\
2 & 18
\end{array} \right)+
%
\left( \begin{array}{cc}
2 & 6 \\
-1 & 0
\end{array} \right)=
%
\left( \begin{array}{cc}
0 & 12 \\
1 & 18
\end{array} \right)
\]

$$\[ \left( \begin{array}{cc}
1 & 2 \\
3 & 4
\end{array} \right)
%
\left( \begin{array}{cc}
2 & 6 \\
-1 & 0
\end{array} \right)=
%
\left( \begin{array}{cc}
0 & 6 \\
2 & 18
\end{array} \right)
\]
\[ \left( \begin{array}{cc}
0 & 6 \\
2 & 18
\end{array} \right)+
%
\left( \begin{array}{cc}
2 & 6 \\
-1 & 0
\end{array} \right)=
%
\left( \begin{array}{cc}
0 & 12 \\
1 & 18
\end{array} \right)
\]$$

I've never done that before - I'm quite pleased with myself!

Hang on I'm trying to work out how to do matrix input.    Ummm

You just add the ones that are in the same position.

sin45=1/(sqrt2)

sine 90=1

When x is on the bottom it is easiest just to take the reciprocal of both sides

$$\frac{45}{sin45}=\frac{x}{sin90}\\\\
45\sqrt2=x\\\\
x=45\sqrt2$$

I believe that the answer is a half because the other side has to be either white or red.

the fact that it is red on one side is a given so the all white card has got nothing to do with it.

I guess that we do too.  thanks Rosala lol.

$$97000=71127e^{0.073t}\\\\
\frac{97000}{71127}=e^{0.073t}\\\\
ln\left(\frac{97000}{71127}\right)=ln e^{0.073t}\\\\
ln\left(\frac{97000}{71127}\right)=0.073t(lne)\\\\
ln\left(\frac{97000}{71127}\right)=0.073t\\\\$$

You should be able to finish it.  If you want me to check your answer I can to that. 

Thanks NinjaDevo,

This looks great.

I've decided that this needs to go into the Sticky Notes (to replace the other one)

If it is not there no-one will use it and that defeats the whole purpose.

I will comment on it as I use it.

Please, NinjaDevo and I want your feed back.

Please use this information thread and add any relevant comments.

People are welcome to add their own reference / comment page BUT

Please just use one post as your own and edit it when needed.