Melody

@@ What is Happening?  [Wrap4]   Sat 14/11/15   Sydney, Australia Time 9:00pm   ♪ ♫

Hi all,

It was a very quiet Fri/Sat today.  Tha only answerers, other than myself, were TMga2tyb, CPhill, Alan and guests.  Thank you.   

Repeat request:

In the post below where I have invited people to discuss their views of the forum.  All sensible thoughts will be greatly appreciated.

Hi Guest,  I looked through your answer today and was most impressed.

It made perfect sense - thanks for that.

It would have been good if you had identified yourself - you could give yourself a name even if you don't want to become a member or if you cannot sign in.

Our only member who you may be is Bertie.  Did you answer this for me Bertie?

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

A general comment and question.

 I only started learning modular arithmetic from this forum so I have never formally learned the rules.

This seems to be one of the main ones.

\(\boxed{ mod\;b\;(t*q)=mod\;b\;[mod\;b\;(t)*mod\;b\;(q) ] }\)

This seems to be reasonable but I would  be interested in seeing a proof.

What are the other 'rules'

I suppose there is lots of information on this on other sites.  I am being lazy.

I want the short answer, not the long one.  LOL

x(2ln(x)+1), x=1/(e)^(1/2)

\(x(2ln(x)+1),\qquad x=1/(e)^{1/2}=e^{-0.5}\\ =e^{-0.5}(2lne^{-0.5}+1)\\ =e^{-0.5}((2*-0.5)lne+1)\\ =e^{-0.5}((-1)*1+1)\\ =e^{-0.5}(0)\\ =0 \)

No inequality is shown.     

(sqrt(3)-sqrt(2))^10

\((\sqrt3-\sqrt2)^{10}\\ =(-\sqrt2)^{10}+10C1(-\sqrt2)^9(\sqrt3)+10C2(-\sqrt2)^8(\sqrt3)^2+10C3(-\sqrt2)^7(\sqrt3)^3+10C4(-\sqrt2)^6(\sqrt3)^4+10C5(-\sqrt2)^5(\sqrt3)^5+10C6(-\sqrt2)^4(\sqrt3)^6+10C7(-\sqrt2)^3(\sqrt3)^7+10C8(-\sqrt2)^2(\sqrt3)^8+10C9(-\sqrt2)^1(\sqrt3)^9 +(\sqrt3)^{10} \\ =32+10(-16\sqrt2)(\sqrt3)+10C2*16*(\sqrt3)^2+10C3(-8\sqrt2)(\sqrt3)^3+10C4*8(\sqrt3)^4+10C5(-4\sqrt2)(\sqrt3)^5+10C6*4(\sqrt3)^6+10C7(-2\sqrt2)(\sqrt3)^7+10C8*2(\sqrt3)^8+10C9(-\sqrt2)(\sqrt3)^9 +(\sqrt3)^{10} \\ =32+10(-16\sqrt2)(\sqrt3)+10C2*16*3+10C3(-8\sqrt2)(3\sqrt3)+10C4*8*9+10C5(-4\sqrt2)(9\sqrt3)+10C6*4*27+10C7(-2\sqrt2)(27\sqrt3)+10C8*2*81+10C9(-\sqrt2*81\sqrt3) +243 \\ =32+10(-16\sqrt2)(\sqrt3)+10C2*16*3+10C3(-8\sqrt2)(3\sqrt3)+10C4*8*9+10C5(-4\sqrt2)(9\sqrt3)+10C6*4*27+10C7(-2\sqrt2)(27\sqrt3)+10C8*2*81+10C9(-\sqrt2*81\sqrt3) +243 \\ =32+10(-16\sqrt6)+10C2*48+10C3(-24\sqrt6)+10C4*72+10C5(-36\sqrt6)+10C6*108+10C7(-54\sqrt6)+10C8*162+10C9(-81\sqrt6) +243 \\ =32+48*10C2 +72*10C4 +108*10C6 +162*10C8 +243 - [160 +10C3*24+10C5*36+10C7*54+10C9*81]\sqrt6 \\ \)

You can simplify it the rest of the way.

If there are no careless errors (which is relatively unlikely) then it should simplify to

\(47523-19402\sqrt6\qquad \qquad \mbox{Ref: Wolfram Alpha}\)

I assume that this is you again NeedSupply.

Your questions are unmistakable.  Posting anonymously does not make any difference.

Post one question at a time like 'guest' told you to and work WITH us to show us that you are participating in your own learning.

We are not here to do all the questions in your textbook for you!  You will learn nothing that way anyhow!

NeedSupply,

You have posted a lot of questions now and many have been answered.

Please show us what you can do yourself on the rest of your questions and we can help with the bits that you get stuck on.

There must be some bits that you can do from every questions.  Even if you get what you do wrong it will give us a better idea on how we can help you.  

Thanks Guest, you are right, people need to thoughtful about there questioning requests.

I do want to point out though tat there are no STAFF on here.

Staff implies that people get paid.  NO ANSWERER here receives any renumeration whatsoever. !!

All answerers are generous people who give their time and share their expertise freely.      

The equal payments are made at the END of the time periods SO it is an ORDINARY annuity.

The debt is at the BEGINNING at the end the value is 0.  So it is a PRESENT VALUE question.

  i = 0.1.     n=4

That should be ample help.  :))