Melody

$$\\log(8x)-log(1+x^{1/2})=2\\\\
log\frac{8x}{1+x^{0.5}}=2\\\\
10^{log\frac{8x}{1+x^{0.5}}}=10^2\\\\
\frac{8x}{1+x^{0.5}}=100\\\\
8x=100(1+x^{0.5})\\\\
0.08x=1+\sqrt{x}\\\\
0.08x-1=\sqrt{x}\\\\
0.0064x^2-0.16x+1=x\\\\
0.0064x^2-1.16x+1=0\\\\$$

x is approx  0.866  or  180.384  but these must be checked. :)

$${\mathtt{0.006\: \!4}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{1.16}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{125}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{33}}}}{\mathtt{\,-\,}}{\mathtt{725}}\right)}{{\mathtt{8}}}}\\
{\mathtt{x}} = {\frac{\left({\mathtt{125}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{33}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{725}}\right)}{{\mathtt{8}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{0.866\: \!208\: \!647\: \!843\: \!302\: \!2}}\\
{\mathtt{x}} = {\mathtt{180.383\: \!791\: \!352\: \!156\: \!697\: \!8}}\\
\end{array} \right\}$$

log(8x)-log(1+x^1/2)=2

check

$${log}_{10}\left({\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{0.866\: \!208\: \!643}}\right){\mathtt{\,-\,}}{log}_{10}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{0.866\: \!208\: \!643}}}^{{\mathtt{0.5}}}\right) = {\mathtt{0.554\: \!996\: \!959\: \!640\: \!612\: \!2}}$$

$${log}_{10}\left({\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{180.383\: \!791\: \!352}}\right){\mathtt{\,-\,}}{log}_{10}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{180.383\: \!791\: \!352}}}^{{\mathtt{0.5}}}\right) = {\mathtt{1.999\: \!999\: \!999\: \!999\: \!798\: \!4}}$$

$$So\;\; x\approx 180.184$$

summation notation of -1+2+7+14+23...

What I noticed is that if I add 2 to each of these terms I get

1+4+9+16+     familiar?

sum of  n^2-2

$$\displaystyle\sum_{n=1}^\infty\;(n^2-2)$$

Google translation (from Mongolian)

Three externally indistinguishable from a genuine coin, the coin weight differences were counterfeit coins. Using the pan without the scale, weighing only a few counterfeit coin can be found?

Sorry but I do not understand the question. 

CPhill and I got the same answer, wonders never cease!

Mmm  Well there are 4 ways to get from E to F

Now going from F to G

If you horizontal first then there are 4 ways

if you go down1 then across there are 3 ways

if you go down 2 then across there ae 2 ways

if you go all the way down first then there is 1 way.

4+3+2+1=10 ways

If that is correct then there would be   4*10=40 ways to go from  E to F to G

It is just as well that you have that decoder ring Chris!  

8 people are sitting at a round table. 3 of them are chosen at random to give a presentation. What is the probability that the three chosen people were sitting in consecutive seats?

the number of ways that 8 people can sit around a table is   $${\mathtt{7}}{!} = {\mathtt{5\,040}}$$

the number of ways 3 distinct people can sit together around the table is  $$(\left({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{1}}\right)){!}{\mathtt{\,\times\,}}{\mathtt{3}}{!} = {\mathtt{720}}$$

so I think the prob will be    $${\frac{{\mathtt{1}}}{{\mathtt{7}}}}$$

$${\frac{{\mathtt{720}}}{{\mathtt{5\,040}}}} = {\frac{{\mathtt{1}}}{{\mathtt{7}}}} = {\mathtt{0.142\: \!857\: \!142\: \!857\: \!142\: \!9}}$$

the meaning is not clear - brackets are needed BUT

inverse tan is the same as arc tan

enter this into the web2 calc.    put the brackets in the correct place of course.  

atan(4.7/3.5)

6 people are sitting around a table. Let x be the number of people sitting next to at least one woman and y be the number of people sitting next to at least one man. How many possible values of the ordered pair (x,y) are there? (For example, (6,0) is the pair if all 6 people are women, since all 6 people are sitting next to a woman, and 0 people are sitting next to a man.)

6 people

x be the number of people sitting next to at least one woman

y be the number of people sitting next to at least one man

I tried to do some of this with combinations but it didn't work.

I ended up drawing lots of hexagons, and just looking at what happens.

I think it is correct. 

(6,0) if all are women

(5,2)   if 1 man

(4,4)   if 2 men sitting together

(6,3)   if the 2 men and they are separated by 1 seat

(6,4)   if the 2 men are opposite each other

This pattern must be symmetrical

(0,6) if all are men

(2,5) if 1 woman

(4,4) if 2 woman and they sit together

(3,6) if the 2 mwomen and they are separated by 1 seat

(4,6)   if the 2 women are opposite each other

This one was harder.  But after drawing pics it seems to me that there are only 3 posibilities for 3 men and 3 women.

(5,5)  The men all sit together

(5,5)  A pair of men, a pair of women, a man, a woman

(6,6)  Alternating

so what do we have

(6,0), (0,6)

(5,2)(2,5)

(4,4)

(6,3)(3,6)

(6,4)(4,6)

(5,5)

(6,6)

11 pairs

Sorry Chris, I misread your first answer - but I am still really dubious.

I like the working of my answer but I don't understand why  the number would be so much bigger (or bigger at all) than if everyone could just sit where they wanted.  

Both your answer and mine have a bigger value than if everyone was individuals and to me this seems a little crazy.

There has to be doubling up happening.  !!