yea, sometimes Illuminati can be evil

but, it is also inefficient to trust Illuminati according to the Holy Bible of Life

Chris, even though you didn't write pi in roman numerals, you can finally get ice cream

anyways, that doesn't mean you're off the hook. still have to write it in Roman Numerals

I wish you all the luck in the world Zegroes :)

That is how I apprached it and Heureka too I think.

I tried to use the table fuction on the forum for layout.  But it does not work anywhere good enough and it looks like a mess.

I expect Heureka's and mine are very similar but mine is probably riddled with errors. :)

x^2+ x +32 = 0    this will  not factor and there are no real solutions, because b² - 4ac < 0

Here are the solutions :

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{32}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{127}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{127}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5.634\: \!713\: \!834\: \!793\: \!303}}{i}\right)\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5.634\: \!713\: \!834\: \!793\: \!303}}{i}\\
\end{array} \right\}$$

Good luck, Z-Man  !!!!

I need to derivative a difficult function where there's several unknown units. And to go forward I need to simplify that. It's called times sinus of a. 

I found it to be 6-x^2*sin^2a but been told it's wrong 

Subtract both sides by 2g: -9g>=63.

Divide both sides by -9, but flip the sign because you are dividing a negative number.

Correct answer is A.

Here you go, DB.......not too difficult to understand, I don't think.......

Yes, Cphill is right, but maybe there are guys/ girls that just give you dislikes... 

We're even......once.....my homework ate my dog.......!!!!

Yeah...the "down points" can definitely affect your score.......you shouldn't have many, DB...... most of your answers usually look good.....!!!

{Avoid getting into controversies on the "social" posts...this is where people will "zap" you }

(x+5)=x^2+2(5)(x)+5^2=x^2+10x+25.

Multiplying by 36: 36x^2+360x+900

You must have given yourself the " -5 "   ?????

3((x-1)/y)  =

3(x -1)  / y   =

[3x - 3 ] / y

Every thing that you can give it a name, is a thing... 'cause if you give it a name, it means that you believe in this thing, so...

Sorry....I don't understand what the term xsinx(a)  means.......???

I have "whited" this one until MG comes on....it will be a good one to practice his Latex (and Algebra) on......!!!

Why not? 

Very nice, Melody and heureka......!!!

This WAS a tough one   !!!!

One observation......  if we laid out ABCDES in a "line"......we only need to choose 4 of any of the 6 letters. Then, putting each resulting set in rotational order will  give us all the possible cyclic quads.

I've tackled it from the other end:

.

7y-3x=-27 and 4x-y-11=0

Using the second equation, we can write it as  y = 4x - 11    .....and substituting this in for "y" in  the frist equation, we have

7(4x - 11) - 3x = -27

28x - 77 - 3x  = -27

25x  - 77  = -27      add 77 to both sides

25x  = 50         divide both sides by 2

x = 2       and y = 4(2) - 11  = 8 - 11  = -3

So    (x, y)  =  (2, -3)

"We don't know".....

It depends upon what "a" is....if positive,  ax > ay

But....if negative......ax < ay

Let's call x, the total amount he started with before his unfortunate fate  {I'm assuming that a coin = a pound}

The first thief gets  (1/2)x + 20.......and he is left with x - [(1/2)x + 20)]  = (1/2)x - 20

The next thief gets (1/2)[ (1/2)x - 20] + 20   = (1/4)x + 10 .....and he is left with :

x - [ (1/2x + 20] - [(1/4)x + 10 ] = (1/4)x -30

The third thief gets  (1/2)[(1/4)x - 30)] + 20  = (1/8)x  + 5   and he is left with

x - [(1/2)x + 20] - [(1/4)x + 10] - [(1/8)x  + 5] =  (1/8)x - 35

The fourth thief gets (1/2)[(1/8)x - 35] + 20  = (1/16)x + 2.5    and he is left with

x - [(1/2)x + 20] - [(1/4)x + 10] - [(1/8)x  + 5] - [(1/16)x + 2.5]  = (1/16)x  - 37.5

The last thief gets (1/2)[(1/16)x - 37.5] + 20  = (1/32)x + 1.25

And he is left with....uh......nothing !!!!

So we need to solve the following equation :

x - [(1/2)x + 20] - [(1/4)x + 10] - [(1/8)x  + 5] - [(1/16)x + 2.5]  - [(1/32)x + 1.25] = 0

And....{with a little help from WolframAlpha}......x = 1240 coins  {pounds  ???}

Proof.....

The first thief gets  640  coins

The second thief gets  320 coins

The third theif gets  160 coins

The fourth thief gets 80 coins

And the last thief gets 40

So .... 640 + 320 + 160 + 80 + 40  = 1240 ......yep...that seems correct....

Thanks to Alan, I spotted an earlier mistake.....we both got the same answer approaching from "different poles"......