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I have answered this one - it may not be correct of course.     

I don't know Mellie, your questions are hard.  :))

Fristly any quadrilateral that is formed will be a convec quadrilateral because opposite angels in a cyclic quads are supplementary - so none can be greater than 180 degrees.

First lets see how many diagonals that there are 

From S there are 5, then from A there are 4, then 3 then 2 then 1 = 5+4+3+2+1=15 diagonal

How many ways can we choose 4 of these 15 diagonals.    15C4

 $${\left({\frac{{\mathtt{15}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)} = {\mathtt{1\,365}}$$         ways

I'm going to attempt to list all the cyclic quad possibilities

SABCS,SABDS,SABES, SACDS, SACES, SADES, SBCDS, SBCES, SBDE, SCDES      that is 10

Now  I think I have exhasted all the ones with the point S in them  

ABCDA, ABCEA, ABDEA, ACDEA     that is 4

Now  I think I have exhasted all the ones with the point S or A in them.

BCDEB  that is the only one left    

Total = 10+4+1 = 15

So I get     $${\frac{{\mathtt{15}}}{{\mathtt{1\,365}}}} = {\frac{{\mathtt{1}}}{{\mathtt{91}}}} = {\mathtt{0.010\: \!989\: \!010\: \!989\: \!011}}$$

That is my best guess at the moment.   It might even be correct     

hmmmmmmmm i don't know

titanium rome is good at these problems

he's just too busy watching Battle Royale (currently, he is in the nanahara Escapes from Kiriyama scene, before lighthouse)

anyways, he will come here eventually and decide who is the victor of this match by math skills and illuminati triangle

The ratio between the circumference of a circle and its diameter.

Yes that is good.

Your coding is also good.

Untill I am totally convinced that you are not fluking it I would like you to put ALL the working in.

So lets have a look at you last answer

$$\begin{array}{rrlll}\\\frac{x}{3}&=&-4\qquad&(Times\;by\;3,\;because\;the\;inverse\;of\;divide\;is\;times)\\x&=&-12\\\end{array}$$

I would like to see you put an extra line of working in...like this

$$\begin{array}{rlll}\\\frac{x}{3}&=&-4\qquad&(Times\;by\;3,\;because\;the\;inverse\;of\;divide\;is\;times)\\\\
\frac{x}{3}\times 3&=&-4\times 3\\\\
x&=&-12\\
\end{array}$$

-----------------------------------------

If you want to get fancy with your LaTex you could even show the cancelling

$$\frac{x}{\not{3}}\times \not{3}&=&-4\times 3\\\\$$

\frac{x}{\not{3}}\times \not{3}&=&-4\times 3\\\\

Here is another one.  There is no nedd for you to rqce back with an answer, answer when (and if) you want :)

$$-6x+4=-2$$

remember, you get rid of tack ons first, then you get x on its own second.  :)

It might just be me, but I do not understand the question. Squared or 2?

140 divided by 20 will equal 7. It's basically 14/2=7 but in larger numbers.

$$\left({\frac{{\mathtt{3}}}{{\mathtt{8}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{5}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right) = {\frac{{\mathtt{1}}}{{\mathtt{20}}}} = {\mathtt{0.05}}$$

$${\mathtt{111\,111\,111}}{\mathtt{\,\times\,}}{\mathtt{111\,111\,111}} = {\mathtt{12\,345\,678\,987\,654\,321}}$$

I need a really good explanation here. Because it is a bit hard for me to understand. And thanks if you have helped me all this time (CPhill, Melody, Geno, Alan, Rosala, Heureka, and Happy7) Thanks!!!

Thank you both, but Melody's answer was correct. Thanks guys so so much once again!!

odd integer +odd integer= even integer

even interger + odd interger = odd integer

1 3 5 7 9 11 13 15 all are odd integer

It doesnt matter what you choose from first two list ,when you add it up, it always give you a even number.

and in the third list, all are odd interger,so when calculate the sum of your integer that you get and the integer

that you choose in third list,you always get a odd integer.

So I think it is impossible to do this. 

s = the size of the side of the Pentagon

d = the size of the diagonal

the central angle of the pentagon is $$\small{\text{$\dfrac{360\ensurement{^{\circ}}}{5}=72 \ensurement{^{\circ}}
$}}$$

the incribed angle is half of the central angle is $$\small{\text{$\dfrac{72\ensurement{^{\circ}}}{2}=36 \ensurement{^{\circ}}
$}}$$

so

$$\cos{(36\ensurement{^{\circ}} ) = \dfrac{\frac{d}{2} } {s }
\qquad \rm{or} \qquad \dfrac{ d } {s } = 2\cdot \cos{(36\ensurement{^{\circ}} ) \qquad | \qquad \cos{(36\ensurement{^{\circ}} ) =\dfrac{1}{4}\cdot(1+\sqrt{5})$$

$$\\\dfrac{ d } {s } = 2\cdot \dfrac{1}{4}\cdot(1+\sqrt{5})\\\\\\
\dfrac{ d } {s } = \dfrac{1}{2}\cdot(1+\sqrt{5})=\varphi$$

Sorry for taking so long i had to clean.

Well, now that you put it that way it sounds a lot easier!

$$\begin{array}{rrlll}\\
\frac{x}{3}&=&-4\qquad&(Times\;by\;3,\;because\;the\;inverse\;of\;divide\;is\;times)\\
x&=&-12\\
\end{array}$$

Hmm..Also i'm not sure about the coding.

I have a question for Nauseated or Heureka

I just made a LaTex box for MathsGod1 but there was a lot of coding involved that I just copied from our LaTex thread.  I seemed to recall that Nauseated made a box like this quite recently and that the coding for his was much simpler.  Do either of you know the coding that I am talking about?

Nope not right MG - lets think about this

$${\frac{{\mathtt{x}}}{{\mathtt{3}}}} = -{\mathtt{4}}$$

is  the same as    x divided by 3

you need to get rid of the divided by 3 to get x by itself.

what is the opposite of dividing by 3?

If I am confusing you, think about this.

$$6\div3 \;\;\fbox{\begin{minipage}{0.6cm}\hfill\vspace{0.6cm}\end{minipage}}\;\;3=6$$      What symbol goes into the box to make this statement true?

A task is always great to do when you're making progress!

(In my head)

$$\begin{array}{rrlll}\\
\frac{x}{3}&=&-4\qquad&(divide\;both\;side\;by\;3)\\
x&=&-12\\\\
\end{array}$$

I'm kind of curious on this question, isn't it already done, like the last question when i was left in that state i just simplified not divide, so the answer is (maybe) -4.

Oops EDIT:

$$\begin{array}{rrlll}\\
\frac{x}{3}&=&-4\qquad&(Divide\;by\;3)\\
x&=&-1.3\\
\end{array}$$

It seems like the answer is -4...

EDIT: I'm sure the answer is not -4.

I'll be back I need to have breakfast

(I wake up to maths)

(PS Its a good thing :D)

Thank you MathGod1.

I have never had such a rave report in my life before.

I am speechless - I am over the top thrilled           

Now, I want contiune my work.

In my last post ,we got cos 72 degrees =(-1+$${\sqrt{}}$$5)/4

Any interior angle of regular pentagon=(5-2)*180/5 degrees=108 degrees

cos 108 degrees =-cos（$${\frac{{\mathtt{\pi}}}{{\mathtt{2}}}}$$-72 degrees) =-cos72 degrees=(1-$${\sqrt{}}$$5)/4

In any pentagon ,every sides are equal and every diagonals are equal.

Lets say we are fing the ratio between AB and AC

According to law of cosine 

In triangle ABC,we have AB^2=AC^2+BC^2-2*AC*BC*cosC, rearrange we have

                     2*AC*BC*cosC=AC^2+BC^2-AB^2

                    and because AC and BC are the sides of regular pentagon,so

                    2*AC^2*cosC=2*AC^2-AB^2

Both side divide by 2*AC^2,then we get

                               cosC=1-AB^2/2*AC^2

                AB^2/2*AC^2=1-cosC

                AB^2/2*AC^2=1-(-cos72 degrees)

                AB^2/2*AC^2=1+cos72 degrees

               AB^2/2*AC^2=1+(-1+5)/4

$${\frac{\left({\mathtt{sqrt5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}} = {\mathtt{1.618\: \!033\: \!988\: \!749\: \!894\: \!8}}$$ 

Now,I am going to rearrange the Alan's equation and check

sin18 degrees=s/2d

2*sin 18 degrees=s/d

1/(2*sin18 degrees)=d/s

d/s=1/(2*sin18 degrees)=1.6180339887496196

Hi Melody, 

I just want to say, Thank You.

These past months on this forum on my learning is all  because of you! :)

No teacher could teach me the way you do, I've learnt alot and it has been easy, your teaching 

is the reason why.

I could probably learn A-levels without any question, if I could go to Australia every day to be in your class i would lol.

But seriously, thank you, i have never ever in my life enjoyed learning so much!

Thank you .

Yes - that is excellent MG   (the mathematics AND the Latex)

Here is another if you are interested

$$\frac{x}{3}=-4$$

Remember I want the logic behind what you do :)

Before i start, i want clarify something, when you got to the fraction, you simplied it to a improper fraction then to a mixed number, right?

:)

$$\begin{array}{rrlll}\\\\
5x-8&=&20\qquad&(To\;get\;X\;alone\;+\:8\;so\;-8\;goes\;away)\\\\
5x&=&28\qquad&(5x\:=\;x\times5\;so\;\div\;by\;5)\\\\
\frac{5x}{5}&=&\frac{28}{5}\qquad&(Now\;cancel\;out)\\\\
x&=&\frac{28}{5}\qquad&(Improper\;fraction\;to\;Mixed\;number)\\\\
x&=&5\frac{3}{5}\\\\
\end{array}$$

I don't know if there is anything i have missed out or added, but i think this is right.

CODE:

\begin{array}{rrlll}\\\\5x-8&=&20\qquad&(To\;get\;X\;alone\;+\:8\;so\;-8\;goes\;away)\\\\5x&=&28\qquad&(5x\:=\;x\times5\;so\;\div\;by\;5)\\\\\frac{5x}{5}&=&\frac{28}{5}\qquad&(Now\;cancel\;out)\\\\x&=&\frac{28}{5}\qquad&(Improper\;fraction\;to\;Mixed\;number)\\\\x&=&5\frac{3}{5}\\\\\end{array}

@@ End of Day Wrap   Sun 17/5/15   Sydney, Australia Time   9:05pm  ♪ ♫

Hi all,

This weekend we have had fabulous answers from Alan, Gibsonj338, Headingnorth, Heureka, MathsGod1, Geno3141, CPhill, EinsteinJr, BrittanyJ, Fiora, Zacismyname and HeatherF.  Thank you.

Yes，φ is the simple ratio between the size of the diagonal and the the size of the sides of any regular pentagon.

So,what?Do you know the background information of φ?Do you know understand why the φ is ratio betweeenthe diagonal and the sides?Can you prove it?DO you know how to use φ in a math problem or ideal life? Might be not.

And I want to say there is no always a shortcut to solve a problem.Every formula and shortcut is come from 

ancient peoples through thinking and hard working to figure it out,and reserve it for us.They might figured out the ratio between the diagonal and of any regular pentagon just like I did,or may be in a easier way.

That is it.

Wow, thanks herueka and Melody, I shall visit the links!