oh god oh god oh god no not sigfigs what did we do to deserve this. . . NO
*melts*
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oh god oh god oh god no not sigfigs what did we do to deserve this. . . NO
*melts*
Oops. Did I do that?
^Tbh, I actually prefer decimals. Im a physicist so I dont mind rounding. Its only those neeky mathematicians who say pi/3 instead of 1.05
Because decimals are, most of the time, approximations, and not exact values of a given situation. Decimals also introduce more room for potential errors; when exact values are mistaken, usually the answer goes from correct to really wrong really fast, and is a form of consistency for our theoretical, perfect little worlds we have in our heads. Also, we hate significant figures. To the death.
ugh...when I enter a science classroom it's just 2+2=5 for me all of the sudden ;_;
You should do your own homework. But I will do it for you this once. The easiest method is to use DeMoivre's Theorem, which presumably you know,
exp(iy) = (cos(y + 2(PI)n) + i sin(y + 2(PI)n)). n is an integer.
put x^3 = a*exp(iy). So y = 0 and a = 192/81.
Then
x = (a^(1/3))*exp(i(y/3))
x = (a^(1/3))*(cos(y/3 + 2(PI)n/3) + i sin(y/3 + 2(PI)n/3))
Here we have
y = 0
so
x = (a^(1/3))*(cos(2(PI)n/3) + i sin(2(PI)n/3))
Then stick in integers for n,
n = 0
x = (a^(1/3))*(cos(0) + i sin(0))
x = (a^(1/3)) = 4/3
n = 1
x = (a^(1/3))*(cos(2(PI)/3) + i sin(2(PI)/3))
x = (2/3)*(1 + i Sqrt(3))
n = 2
x = (a^(1/3))*(cos(4(PI)/3) + i sin(4(PI)/3))
x = (2/3)*(1 - i Sqrt(3))
n >= 3 just cycles through these solutions.
These solutions define an equilateral triangle in the complex plane. PuPu did it a really nice way, and yeah you have to get 3 answers.