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I GIVE MEDALS Find the fifth roots of 243(cos 300° + i sin 300°).

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@amistre64 :) can you help with another couple ?
nice name :)
@AravindG thanks ... can you help ?

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Other answers:

@josie_2015 can you give the work with it ?
the 5th roots are essentially adding 2pi/5 to what you have
243(cos ((5/3+2n/5) pi) + i sin((5/3+2n/5)pi ) ) n = 0 to 4
so thats the final answer ?
im not sure what a "final answer" should be formated as; but yes ....
5th roots are just |dw:1367951499404:dw|
so start where you are at; 5pi/3 and add on each root
72 degrees might be easier to read tho
\[\frac{x}{x-5}\] \[\frac{x+0}{x-5}\] \[\frac{x+(5-5)}{x-5}\] \[\frac{x-5}{x-5}+\frac{5}{x-5}\] \[1+\frac{5}{x-5}\]
would my response be correct ?
confused lol
this is approaching -inf, yes
thank you so much one more?
we could say that the slope of 1+5/(x-5) = -5/(x-5)^2 is always negative as x approaches 5 from the left ... but sure, one more
well, since the picture shows the distance along the y axis to be bigger than the x axis .. we want the under ys to be bigger than the under xs by default. that narrow it to 2 options
and since the center of the moon is 1000 further than its surface, that should make the correct choice rather obvious
option D?
option D is not correct, since the under y is smaller than the under x
so then my last educated guess is C @amistre64
i would deduce that you need better education then :) C still has the under y smaller than the under x, AND it is missing the 1000 extra stuff for the radius ...
assume for a moment we have this setup:|dw:1367953733151:dw| in order for this to make any sense:\[\frac{x^2}{n^2}+\frac{y^2}{m^2}=1\]when x=0, we have y=2\[\frac{0^2}{n^2}+\frac{2^2}{m^2}=1~:~m=2\] when y=0, we have x=1\[\frac{1^2}{n^2}+\frac{0^2}{2^2}=1~:~n=1\] therefore our setup has to be:\[\frac{x^2}{1^2}+\frac{y^2}{2^2}=1\] the part under the y has to be greater than the part under the x
in your case: 953 is bigger than 466, but thats not accounting for the extra 1000 radius, sooo.... y x 1953 is bigger than 1466
good luck ... ;)

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