Question

square root of -200 in standard form

Answer 1

@Daddysgirl14

Answer 2

@ybarrap

Answer 3

iroot(200)=10iroot(2)

Answer 4

i is imaginary unit

Answer 5

\(\sqrt{-200}=i\sqrt{200}=i10\sqrt{2}\)

Answer 6

oh ok i was correct i have no more if u dnt mind

Answer 7

squared by two what do you mean lol

Answer 8

post your other questions in different posts pls!!

Answer 9

hint:
we can write this:
\[\Large - 200 = 200\left\{ {\cos \left( {\pi + 2k\pi } \right) + i\sin \left( {\pi + 2k\pi } \right)} \right\}\]

Answer 10

polar form..

Answer 11

well do the same process you did with the first

Answer 12

-7 you don't have to worry about

Answer 13

just root(-96)

Answer 14

how did root (96) simplify to 4?

Answer 15

we have two square roots:
\[\Large \begin{gathered}
\sqrt { - 200} = \sqrt {200} \left\{ {\cos \left( {\frac{\pi }{2} + k\pi } \right) + i\sin \left( {\frac{\pi }{2} + k\pi } \right)} \right\}, \hfill \\
\hfill \\
k = 0,1 \hfill \\
\end{gathered} \]

Answer 16

root(96)=root(16x6)=4root(6)

Answer 17

oh you have root(6)
my bad i miss read your reply

Answer 18

\[\Large \begin{gathered}
\sqrt { - 96} = \sqrt {96} \left\{ {\cos \left( {\frac{\pi }{2} + k\pi } \right) + i\sin \left( {\frac{\pi }{2} + k\pi } \right)} \right\}, \hfill \\
\hfill \\
k = 0,1 \hfill \\
\end{gathered} \]

Answer 19

your answer is good!

Answer 20

yes so i was correct?

Answer 21

yes!

Answer 22

thank youu i do not know which one to give medel to since yall both helped me

Answer 23

does not matter :)