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Needamedal Group Title

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

  • one year ago
  • one year ago

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  1. Needamedal Group Title
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    @amistre64 :) can you help with another couple ?

    • one year ago
  2. AravindG Group Title
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    nice name :)

    • one year ago
  3. Needamedal Group Title
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    @AravindG thanks ... can you help ?

    • one year ago
  4. josie_2015 Group Title
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    450^2

    • one year ago
  5. Needamedal Group Title
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    @josie_2015 can you give the work with it ?

    • one year ago
  6. amistre64 Group Title
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    the 5th roots are essentially adding 2pi/5 to what you have

    • one year ago
  7. amistre64 Group Title
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    243(cos ((5/3+2n/5) pi) + i sin((5/3+2n/5)pi ) ) n = 0 to 4

    • one year ago
  8. Needamedal Group Title
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    so thats the final answer ?

    • one year ago
  9. amistre64 Group Title
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    im not sure what a "final answer" should be formated as; but yes ....

    • one year ago
  10. amistre64 Group Title
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    5th roots are just |dw:1367951499404:dw|

    • one year ago
  11. amistre64 Group Title
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    so start where you are at; 5pi/3 and add on each root

    • one year ago
  12. amistre64 Group Title
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    72 degrees might be easier to read tho

    • one year ago
  13. amistre64 Group Title
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    \[\frac{x}{x-5}\] \[\frac{x+0}{x-5}\] \[\frac{x+(5-5)}{x-5}\] \[\frac{x-5}{x-5}+\frac{5}{x-5}\] \[1+\frac{5}{x-5}\]

    • one year ago
  14. Needamedal Group Title
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    would my response be correct ?

    • one year ago
  15. Needamedal Group Title
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    confused lol

    • one year ago
  16. amistre64 Group Title
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    this is approaching -inf, yes

    • one year ago
  17. Needamedal Group Title
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    thank you so much one more?

    • one year ago
  18. amistre64 Group Title
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    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

    • one year ago
  19. amistre64 Group Title
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    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

    • one year ago
  20. amistre64 Group Title
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    and since the center of the moon is 1000 further than its surface, that should make the correct choice rather obvious

    • one year ago
  21. Needamedal Group Title
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    option D?

    • one year ago
  22. Needamedal Group Title
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    @amistre64

    • one year ago
  23. amistre64 Group Title
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    option D is not correct, since the under y is smaller than the under x

    • one year ago
  24. Needamedal Group Title
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    so then my last educated guess is C @amistre64

    • one year ago
  25. amistre64 Group Title
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    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 ...

    • one year ago
  26. amistre64 Group Title
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    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

    • one year ago
  27. amistre64 Group Title
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    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

    • one year ago
  28. Needamedal Group Title
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    WOW THANKS SO MUCH I SEE NOW !

    • one year ago
  29. amistre64 Group Title
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    good luck ... ;)

    • one year ago
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