Melody

1/4 = 2/8

so

$${\frac{{\mathtt{5}}}{{\mathtt{8}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2}}}{{\mathtt{8}}}} = {\frac{{\mathtt{7}}}{{\mathtt{8}}}}$$

why is  

 $$2x\times x=2x^2$$

because

$$2x\times x=2\times x\times x$$        there is an invisable times sign between the 2 and the x

and

$$2x^2=2\times x^2 = 2\times x \times x$$                 because  $${{\mathtt{x}}}^{{\mathtt{2}}}$$   means  $${\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}$$

Does that help?

Thanks flflvm97

I have given this the green Puzzle icon. :)

And I have added it to the Sticky Topic "Puzzles"

Maybe when you become comfortable with the forum you might like to organise our puzzles for us.

I have saved most of the good ones but I just 'shove them in'  They are not actually organised at all. :(

If you - or anyone else want a job like this then message me please.  

I have added this to the Sticky Topic - "Great Answers to Learn From"

I don't know Chris,

This might be the answer wanted but

Newton's law of cooling says that 

$$\\\mbox{T is temperature, t is time, and S is the surrounding temperature}\\\\
\frac{dT}{dt}=-k(t-S)\\\\
$I think this means$\\\\
T(t)-S=(T_0-S)e^{-kt}\\\\$$

I don't think that we are given enough information to solve this question.   

$${\mathtt{a}} = {\sqrt{{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}}}$$     

 find      $${{\mathtt{a}}}^{{\mathtt{2}}}$$

$${{\mathtt{a}}}^{{\mathtt{2}}} = {{\sqrt{{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}}}}^{\,{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\sqrt{{\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}}}}^{\,{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}}}$$

$${{\mathtt{a}}}^{{\mathtt{2}}} = \left({\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}\right){\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{\left(\left({\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}\right){\mathtt{\,\times\,}}\left({\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}\right)\right)}}$$

$${{\mathtt{a}}}^{{\mathtt{2}}} = {\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{\left({\mathtt{36}}{\mathtt{\,-\,}}{\mathtt{11}}\right)}}$$

$${{\mathtt{a}}}^{{\mathtt{2}}} = {\mathtt{12}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{25}}}}$$

$${{\mathtt{a}}}^{{\mathtt{2}}} = {\mathtt{12}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{5}}$$

$${{\mathtt{a}}}^{{\mathtt{2}}} = {\mathtt{2}}$$

and that means that   $$a=\pm \sqrt{ 2}$$          COOL !!!

$${\frac{{\sqrt[{{\mathtt{{\mathtt{7}}}}}]{{{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\sqrt[{{\mathtt{{\mathtt{5}}}}}]{{{\mathtt{a}}}^{{\mathtt{4}}}}}}}}{{\sqrt[{{\mathtt{{\mathtt{5}}}}}]{{\mathtt{a}}}}}}$$

$$\\=\frac{\sqrt[7]{a^2\times a^{4*(1/5)}}}{a^{1/5}}\\\\
=\frac{\sqrt[7]{a^2\times a^{0.8}}}{a^{0.2}}\\\\
=\frac{\sqrt[7]{a^{2.8}}}{a^{0.2}}\\\\
=\frac{(a^{2.8})^{1/7}}{a^{0.2}}\\\\
=\frac{a^{2.8/7}}{a^{0.2}}\\\\
=\frac{a^{0.4}}{a^{0.2}}\\\\
=a^{0.4-0.2}\\\\
=a^{0.2}\\\\
=a^{1/5}\\\\
=\sqrt[5]{a}$$

That is ok.   We are here to teach so if you still have problems we encourage you to tell us.

I have now included the final answer.  If you still have concerns let us know :)

I have now included the final answer as you requested.

If you still have a problem let me know and I will try to talk you through it :)

That is really cool.  Thanks Chris :)

I am going to add this thread into  the      Sticky Topic :  Reference Material     :)