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math reading......

Thank you supermanaccz and guest,

Superman, I have never seen it done your way, there are usually several ways to solve the same problem and it is interesting to see alternate routes.    

There are six different sixth roots of 64. That is, there are six complex numbers that solve x^6=0, find them.

The 6 points will be equally distributed on the complex plane at an absolute distance of 2 units.

A revolution is 2pi radians so the points will be   2pi/6 = pi/3 radians apart.

Also you know that one of them is   \(2 = 2e^0 = 2e^{(0\pi/3)i}\)

so here are your roots.

\(\sqrt[6]{2}=2e^{(0\pi/3)i}, \quad2 e^{1(\pi/3)i}, \quad2 e^{2(\pi/3)i}, \quad2 e^{3(\pi/3)i},\ \quad2 e^{4(\pi/3)i}, \quad2 e^{5(\pi/3)i}\\ \sqrt[6]{2}=2e^{(\pi/3)i}, \quad2 e^{(\pi/3)i}, \quad2 e^{(2\pi/3)i}, \quad2 e^{\pi i},\ \quad2 e^{(4\pi/3)i}, \quad2 e^{(5\pi/3)i}\\\)

I've done it exactly the same way as our guest only he/she has gone from -pi to +pi  and I have gone from 0 to 2pi.   I am not sure if there is any convention on this, but the answers are the same.

3x^2 -24x + 55       complete the square

3 (x^2 - 8x   +  55/3)

3(x^2  - 8x  +  16  +  55/3  - 16)

3 [  (x -4)^2   + 55/3  - 48/3 ]

3[ (x - 4 )^2  +  7/3)

3(x - 4)^2  +  7

So

a =  3

d  = -4

e  = 7

And their sum  =  6

\(\frac{\pm\sqrt{1-\sin^2\theta}}{\sin\theta}\)

*edit*

I don't think that this is correct anymore, but I don't know what the correct answer is.

can u write this equation in the latex thingy so i can understand more clearly

yea i got the same answer but if u want the answer as a whole then either the value of a is wrong or u wrote the wrong equation

Well, thank you for the catered meal, Mr. BB! Crow is a delicacy for us chimps, and plantains are a fine complement for such a meal, especially when boiled and reduced in blarney sauce.   I usually dress for fine dining, so I will don my gay apparel and Troll the ancient Mr. BB’s carol: Troll-la-Troll-la-Troll, la-Troll-Troll, Troll!

Lancelot Link had Stu, and Nauseated had Sorasyn, but they left, taking their dumbness elsewhere.  I have you, Mr. BB, the stubborn, relentless, intractable Blarney Banker of lore: a pseudo intellectual with a multiplicity of advanced dimwit degrees in arrogant stupìdity, a professor of misinformation, who teaches with authority and irritation.

One usually needs a scanning electron microscope to find an error by Sir CPhill, That’s why the errors are “itsy-bitsy, teeny-weeny, microscopic, nearly infinitesimally small,” but still calculable. For you, I need no such instrument. A casual stroll through the forum shows a path, clearly marked by stupìdity, mòronic analysis, and toxic misinformation, leading right to your hovel.

You may not believe this, but I’m glad you’re here.  The most successful Trolls have a reliable parade of stubborn dùmb dùmbs, or a relentless, indefatigable dùmb dùmb who always gives the wrong answer to “What is the airspeed velocity of an unladen swallow?”  No matter how many times my troll character tosses you over the bridge railing into the river, you always come back for more.  With this, you have personified a modern attachment to the ancient legends of Camelot and its Trolls.   

We all owe you a debt of pseudo gratitude for the unwitting humor you cause.

GA

As many times as Sisyphus has pushed that bolder up the hill, you’d think both he and the rock would be reduced to near ZERO by now.

Yes, that makes sense! 

Thanks, Julius....look at the logic...if  t  =  1620, we have

A(1/2)^1620/1620  = 

A (1/2)¹  =

(1/2)A.........which is exactly what we would expect....1/2 of the amount, A, is what remains after 1620 years

CPhill : Just a slight error in obtaining b:

b = 2 a + 7/2,  where a = 3

b =2*3 + 7/2

b =6 + 7/2

b=[12 + 7] / 2

b = 19/2

Hmm, this seems to make much more sense to me. I thought it should be x/1620, but this other website said x+1620 which wasn't making any sense to me... 

Thanks so much for your time! 

OK....we can solve this :

8  =  12* (1/2)^t/1620        where t is the time (in years) we are looking for

Divide both sides by  12

8/12  = (1/2)^t/1620

2/3  =  (1/2)^t/1620        take the log of both sides

log (2/3)  =  log(1/2)^t/1620          and we can write

log(2/3)  =  (t / 1620) * log (1/2)     divide both sides by log(1/2)

log (2/3) / log (1/2)  =  t  / 1620      multiply both sides by 1620

1620 * log (2/3) / log (1/2)  =  t   ≈  947.6  ≈   948  years

I see :D Thank you so much! 

And sorry if I caused any confusion. 

I realize this is the THIRD time you have posted the SAME question....PLEASE reply to the original thread with follow-up questions to avoid multiple people answering in multiple places....avoids confusion     THANX ! 

   ~EP

90 = 10 log (I/(10^-12)

9=log (I/(10^-12)

I = .001  =1 x 10^-3  W/m^2

Excellent, Julius and superman !!!!

= +, - [sqrt(1 - sin^2(theta))] / sin(theta)

64^(1/6) =

+ OR - 2

2 e^((i π)/3)≈1.0000 + 1.7321 i

2 e2 e^(-(2 i π)/3)≈-1.0000 + 1.7321 i

2 e^(-(2 i π)/3)≈-1.0000 - 1.7321 i

2 e^(-(i π)/3)≈1.0000 - 1.7321 i

Both 15^6 and  7^6 are squares, so we can factor 15^6 - 7^6  as the difference of squares:

15^6-7^6 = (15^3-7^3)(15^3+7^3)

We can pound out the computation from here (yuck!), or we can factor 15^3-7^3  as a difference of cubes and 15^3+7^3 as a sum of cubes:

[(15^3-7^3)(15^3+7^3) = [(15-7)(15^2+15*7 + 7^2)][(15+7)(15^2-15*7 + 7^2)]

The largest of these factors is 379. 

I did a better solution on paper, but I can't attach a file, sorry!