Melody

Thanks very much Alan :)

I looked at Matlab by Mathworks. It is $219 in Australia. 

But then they say  that to buy Simulink and add ons they are $65 each.

So this means I do not really know how much it will cost as I don't know if I need these extras.

The videos you sent have been written in Python programming language which appears to be a free download. In the very distance past I was relatively proficient at FORTRAN so I have some of the basic concepts and may be able to learn this language. I will admit though that getting my head around the logic could be a major problem for me these days.

I also looked at a Matlab video. It was showing many fabulous practical applications but I know I will never master any of those. At this instant in time I am only interested in running MonteCarlo simulations so that i can check my own probability answers.  

I do not think I am interested in learning to do this in EXCEL. I think it would be super clumsy.

So this has led me to 3 options:

1) Get Matlab and learn to use it.

2) Download Python and learn to use it.

3) Keep asking you to check my answer :))   

Mmmm

Do you have any input for me?

Don't appologise, your repost request was done exactly as I always asked for it to be done.

I had not looked at your question before then. :)

Thanks Chris,

The bit that follows is exactly the same as Chris said and then i have continued the proof   

I am going to call the angles in the original triangle 

All angles in the 3 other triangles are all 60 degrees because they are equilateral triangles.

Consider

  \(\triangle ACX\;\; and \;\; \triangle YCB \\ AC=b, \;\;CX=a\;\; and \;\; \angle ACX=\angle C+60^\circ\\ YC=b, \;\;CB=a\;\; and \;\; \angle YCB=\angle C+60^\circ\\ \therefore \triangle ACX\cong \triangle YCB\qquad \text{(2 sides and included angle test)} \\ \therefore AX=BY \qquad \text{Corresponding sides in congruent triangles.}\)

\(\text{Now consider} \triangle ABY\;\; and \;\; \triangle AZC \\ AB=AZ=c\\ AY=AC=b\\ \angle YAB=\angle CAZ=\angle A+60^\circ\\ \therefore \triangle ABY\cong \triangle AZC\qquad \text{(2 sides and included angle test)} \\ \therefore BY=ZC \qquad \text{Corresponding sides in congruent triangles.}\\ \)

\(\therefore AX=BY=ZC\\~\\ QED\)

I've drawn the pic but not answered the question yet  

Chris you are better at these than I am, why don't you give it a go?

Here is the pic but I don't have an answer yet.

Hectictar's Web2 anniversary - posted by GingerAle  7/2/18

Nice answer Heureka :)

find the base of numeration in which 135/21 has a zero remainder

Let the base be x

\(\frac{x^2+3x+5}{2x+1}=n \qquad n\in Z\\ \)

I am going to try trial and error.

x has to be 6 or bigger because of the 5.

If x=6 this becomes

\(\frac{36+16+5}{13}=\frac{57}{13}\ne n\)

If x=7

\(\frac{49+21+5}{14+1}=\frac{75}{15}=5=n\)

So base 7 works