ashontae19
  • ashontae19
square root of -200 in standard form
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ashontae19
  • ashontae19
@Daddysgirl14
ashontae19
  • ashontae19
@ybarrap
xapproachesinfinity
  • xapproachesinfinity
iroot(200)=10iroot(2)

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More answers

xapproachesinfinity
  • xapproachesinfinity
i is imaginary unit
xapproachesinfinity
  • xapproachesinfinity
\(\sqrt{-200}=i\sqrt{200}=i10\sqrt{2}\)
ashontae19
  • ashontae19
oh ok i was correct i have no more if u dnt mind
xapproachesinfinity
  • xapproachesinfinity
squared by two what do you mean lol
xapproachesinfinity
  • xapproachesinfinity
post your other questions in different posts pls!!
Michele_Laino
  • Michele_Laino
hint: we can write this: \[\Large - 200 = 200\left\{ {\cos \left( {\pi + 2k\pi } \right) + i\sin \left( {\pi + 2k\pi } \right)} \right\}\]
xapproachesinfinity
  • xapproachesinfinity
polar form..
xapproachesinfinity
  • xapproachesinfinity
well do the same process you did with the first
xapproachesinfinity
  • xapproachesinfinity
-7 you don't have to worry about
xapproachesinfinity
  • xapproachesinfinity
just root(-96)
xapproachesinfinity
  • xapproachesinfinity
how did root (96) simplify to 4?
Michele_Laino
  • Michele_Laino
we have two square roots: \[\Large \begin{gathered} \sqrt { - 200} = \sqrt {200} \left\{ {\cos \left( {\frac{\pi }{2} + k\pi } \right) + i\sin \left( {\frac{\pi }{2} + k\pi } \right)} \right\}, \hfill \\ \hfill \\ k = 0,1 \hfill \\ \end{gathered} \]
xapproachesinfinity
  • xapproachesinfinity
root(96)=root(16x6)=4root(6)
xapproachesinfinity
  • xapproachesinfinity
oh you have root(6) my bad i miss read your reply
Michele_Laino
  • Michele_Laino
\[\Large \begin{gathered} \sqrt { - 96} = \sqrt {96} \left\{ {\cos \left( {\frac{\pi }{2} + k\pi } \right) + i\sin \left( {\frac{\pi }{2} + k\pi } \right)} \right\}, \hfill \\ \hfill \\ k = 0,1 \hfill \\ \end{gathered} \]
xapproachesinfinity
  • xapproachesinfinity
your answer is good!
ashontae19
  • ashontae19
yes so i was correct?
xapproachesinfinity
  • xapproachesinfinity
yes!
ashontae19
  • ashontae19
thank youu i do not know which one to give medel to since yall both helped me
xapproachesinfinity
  • xapproachesinfinity
does not matter :)

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