Melody

Hi Anonymous,

If you are a conscientious objector we can respect that.

If you post a lot why don't you give yourself a unofficial username - you can put it at the end of all your posts and we will slowly learn who you are.

We would really like you to be a proper 'member' of the forum even if it is unofficial.   

No it is not

$$5=11+2(x-1)\\
5-11=11+2(x-1)-11\\
-6=2x-2\\
-6+2=2x-2+2\\
-4=2x\\
-4/2=x\\
x=-2$$

COOL!!! 

That's brilliant!  Thanks Bertie.  I'll give you another 3 points.  For sure!!!

@@ End of Day Wrap - Wednesday 21/5/14   Sydney, Australia   Time 19:35

Hi Everybody,

Great answers were posted today by CPhill, problem, admin, zegroes, Bertie, Reinout-g, GoldenLeaf, Heureka and Kitty

Zegroes also reported a problem with the message centre by posting it on the forum.  This was a great idea.  Admin saw the post and responded quite quickly.  Thanks to both of you.

There were plenty of questions but I don't think any really stood our for their mathematical content.

These ones may provide some entertainment.

The fun continued with this first one:

That's scarecrows for you!  They just cannot be trusted 

Good thought Bertie.  

Very impressive CPhill.

My turn

A pythagorean formula is one that satisfies Pythagoras' Theorem.

Pythagoras' Theorem states that in any right angled triangle the square on the hypotenuse is equal to the sum of the squares on the other 2 sides.

If the hypotenuse is h and the other 2 sides are a and b then         $$h^2=a^2+b^2$$

In all fairness GoldenLeaf you have answered this question very well.

HOWEVER

The question that the asker intended is not the one that you answered. 

The intended question was 

6872=14592(1+r/1)^11

I know this because the word exponential was used in the question.

$$6872=14592(1+r)^{11}\\\\
\frac{6872}{14592}=(1+r)^{11}\\\\
log\left(\frac{6872}{14592}\right)=log(1+r)^{11}\\\\
log\left(\frac{6872}{14592}\right)=11log(1+r)\\\\
\dfrac{log\left(\frac{6872}{14592}\right)}{11}=log(1+r)\\\\
10^{\frac{log\left(\frac{6872}{14592}\right)}{11}}=10^{log(1+r)}\\\\
10^{\frac{log\left(\frac{6872}{14592}\right)}{11}}=1+r\\\\
10^{\frac{log\left(\frac{6872}{14592}\right)}{11}}-1=r\\\\$$

$${{\mathtt{10}}}^{\left({\frac{{log}_{10}\left({\frac{{\mathtt{6\,872}}}{{\mathtt{14\,592}}}}\right)}{{\mathtt{11}}}}\right)}{\mathtt{\,-\,}}{\mathtt{1}} = -{\mathtt{0.066\: \!165\: \!642\: \!977\: \!294\: \!8}}$$

$$r\approx -0.066$$