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m=2-sqrt(3)       

\(m.(4-m).(2-m)/sqrt(3)\\~\\ =\frac{m.(4-m).(2-m)}{\sqrt3}\\ \mbox{sub in }m=2-\sqrt3\\ =\frac{(2-\sqrt3)(4-(2-\sqrt3)).(2-(2-\sqrt3))}{\sqrt3}\\ =\frac{(2-\sqrt3)(2+\sqrt3).\sqrt3}{\sqrt3}\\ =(2-\sqrt3)(2+\sqrt3)\\ =4-3\\ =1\)

 (5 - 3i)^1/3.

\(r = mod z = |z| = \sqrt{5^2+(-3)^2}=\sqrt{34}\)

\(5-3i=\sqrt{34}(\frac{5}{\sqrt{34}}-\frac{3i}{\sqrt{34}})\\ 4th\;quad\\ argument \;of\; z = arg(z) =\theta=2\pi-acos(\frac{5}{\sqrt{34}})\\ \theta\approx 5.7427658\;radians\\ 5-3i=\sqrt{34}\;e^{5.7427658i} \)
 

\(|z^{1/3}| = \sqrt{34}^{1/3}=\sqrt[6]{34}\)

\(1)  arg(z^{1/3})=5.7427658/3 = 1.914255\\ 2)  arg(z^{1/3}) = 1.914255 +\frac{2 \pi}{3} = 4.008650\\ 3)  arg(z^{1/3}) = 1.914255 +2*\frac{2 \pi}{3}= 6.103045\\\)

So the 3 cubed roots of are 5-3i  are

\(\sqrt[6]{34}*e^{1.914255i}\quad and \quad \sqrt[6]{34}*e^{4.008650i}\quad and \quad \sqrt[6]{34}*e^{6.103045i}\)

cos(1.914255) = -0.336745771286

sin(1.914255) = 0.94159560615

\(1st\;root=\sqrt[6]{34}\;(-0.336746\;+\;0.941596\;i)\)

etc

check the first root

(34^(1/6)*e^(1.914255*i))^3 = 4.9999975792700509-3.0000040345460377i      Near enough    

Well to the nearest 10 

28 becomes 30

Is that what you mean?

.........

Solve for x:
sqrt(1-x^2) = i

Raise both sides to the power of two:
1-x^2 = -1

Subtract 1 from both sides:
-x^2 = -2

Multiply both sides by -1:
x^2 = 2

Take the square root of both sides:
Answer: | x = sqrt(2)                  or x = -sqrt(2)

nvm

Here's one answer :

1^2 = 1

2^2  = 4

3^2  = 9

so

s^2   = n

Here's another using summation notation :

1 = 1

1 + 3  = 4

1 + 3 + 5  = 9

so

∞

Σ  [2s - 1]      =   1 + 3 + 5 + 7 + ..............

s=1

That awkward moment when your waiting for someone to answer your question.

HAN DIES IN STAR WARS

HAHAHAHAHAHAHAHAHAHA

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^FOLLOW ME

I AM ENTROPY  {nl} I AM DEATH {nl} I AM...YOUR FATHER {nl} NO, WAIT

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KNEEL BEFORE ME

Math Facts: Googol= 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.

.999999999999999999999999999999999999999999999... = 1

1/3=.33333333333333333333333333333333333333333...

1/3 X 3= 1

.3333333333333333333333333333333333333333333333=.9999999999999999999999999=1

you said you wanted math facts.

9 MINUTES.

USE YOUR BRAIN AND FINGERS FOR THAT.

Mmmmmm.....the confusion over "formulas"  arises from the initial value of "s"

Actually, Complex"......if s = the stage number, then your formula  of 3s + 1  is incorrect, because at stage 1, s = 1 and we should have 4  blocks instead of just 1

I made a slight mistake in my formula  by using n instead of s  .......

I should have been....

3s - 2  = n,      assuming we can set s = 1   at the start

But....as to Melody's  assertion that it's the "same" formula re-written, we can see that it's not true because if it were, then

3s + 1  =  3s - 2   which implies that

1 = -2      !!!!

Melody's  "formula"  calculates the first three terms correctly......the 4th term would be 28 and the 5th term would be 85

If the intent of the sequence is to have a difference of 3 blocks at every stage, then her formula would not be correct.......if something else was intended, her formula might be correct and mine might be incorrect....

Who knows???

Nevermind I got it!

for example

look at this picture

5-5-5-85-55-8-8-8-8-8-8-8--66-6-5---5-22-36-5

4 apples...

no i just solved the question :)

What do you mean you found it? 

Do you mean you found it by yourself ?

Fri 25/12/15

Merry Christmas

https://web2.0calc.com/questions/merry-christmas-2015

1)  A Christmas Story from Dragonlance's Journal.  Very sweet and very funny, thanks Dragonlance.

     http://web2.0calc.com/questions/a-christmas-story-from-dragonlance-s-journal/new#edit

2)  Ouija Board Story by DragonLance:  Put this on your 'must read' list. It is hilarious.

     http://web2.0calc.com/questions/dragonlance-s-journal-stories#r1

3)  Mathematica Talent Contest       CPhill wins here !

     http://web2.0calc.com/questions/mathematica-talent-contest

4)  An interesting substitution queston.     Melody

     http://web2.0calc.com/questions/m-2-sqrt-3

5)  Easy complex number equation   Thanks Guest

     http://web2.0calc.com/questions/find-x-in-sqrt-1-x-2-i

6)  One more complex root , just for good measure

     http://web2.0calc.com/questions/complex-operation#r4

Thanks Melody! i found the answer :)

Hi Solveit :)

what does that ~ mean

I wasn't sure so I looked it up :)

It is used in geometry and it means "is similar to"

Two shapes are similar if they have the same angles and the sides are in the same proportions.

That is, one is an enlargement of the other one.  (It doesn't matter if the second one is turned or reflected)

@@ What is Happening?  [Wrap4]   Thurs 25/12/15   Sydney,  Australia Time 11:59 am   ♪ ♫

Enjoy your Chrismas Eve everyone, don't forget to leave out milk and cookies for Santa !    

Hi all,

It has been a week since my last wrap.  That is largely because I was away on holidays. This also means that there are a number of threads and answers that I have not seen :(  Still I have collected many answerer names: Credits go to Alan, CPhill, anonymous4338, Complex, Heureka, Olivia 3.141592654, Cookiesandcream, Alleycat2, gibsonj338 and Dragonlance.  Thank you    

Thanks for a appreciating my colour scheme too Hayley  LOL   

Stranger danger reminder

1/2((0)+(0)+(0)+(3*-2.2033)+(4,652*2,5783)+(4,652*-2,5783)+(4,371*-2,1633)+(4,371*-2,21233)+(0)+(2,65*-2,1633)+(-2,1633*-2,6183)+(-3,269*-2,6183)+(-3,65*-1,3913)-(-3,65*-2,6183)-(-3,65*-2,6183)-(-3,265*-2,1633)-(-3,269*-2,1503)-(2,65*-2,1233)-(0)-(4,371*-2,5783)-4,371*-2,5783)-(4,652*-2,2033)-(0)-(0)-(0)-(-3*-1,3913)) =SOMETHING REALLY STUPID!!!!!!!!!!.

Please evaluate the following expression:

ln{[(5280×(236674+30303×√61))³ +744]² − 393768} ÷ sqrt(1708)=

Also, can you please explain why it gives the result that it does? Thanks a lot.

3.14159265358979323846264338327950288419716939937510582097494459230781640628620872...

This gives one of the BEST approximation to π ever known. It is accurate to 78 decimal places, and was discovered by the great Indian mathematician, S. Rammanujan, based on his work on "modular equations".