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anonymous
 3 years ago
Can someone please check my work?? Wanna make sure I'm doing this right! Simply each number by using the imaginary number i.
thanks!
anonymous
 3 years ago
Can someone please check my work?? Wanna make sure I'm doing this right! Simply each number by using the imaginary number i. thanks!

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{7}\] = 7i or \[i \sqrt{7}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{16}\] = \[2i \sqrt{4}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[3\sqrt{9}\] = \[3i\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The last one is wrong.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@dido525 Oh wow!! Thank you. I can't believe I'm doing these somewhat right lol. Would it be \[3i \sqrt{3}\] ?? And there was also a problem that was \[\sqrt{72}\] I got: \[2i \sqrt{3}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The last one should be 9i.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well think about it WITHOUT the negative.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352593737003:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352593754329:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now what would you do?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is that a 4 and 18? I would think to factor them to 2 x 2 and 9 x 2 maybe

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but inside of the radicals

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352593862710:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352594047004:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Dido525 Oh ok! That made since. I was using 8, 9, and i to make 72

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Dido525 Thank you so much for all your help :) Makes me feel more confident about simplying them!

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1\[\sqrt{72}=\sqrt{(1)\cdot2\cdot 36}\] \[=6i\sqrt{2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Getting the i right, but need to look at \[\sqrt{81} and \sqrt{16}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@chaguanas Would I do \[3i \sqrt{3}\] for 81? and 4i or \[4i \sqrt{4}\] for 16?

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1do you know what \(\sqrt{81}\)is?

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1well...\[\sqrt{81}=\sqrt{(1)81}=i9=9i\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Zarkon Ohh so that's how you do it. You wouldn't factor the 9 out to a 3?

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1for any \(a>0\) \[\sqrt{a^2}=a\cdot i\] so \[\sqrt{81}=\sqrt{9^2}=9i\]

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1\[\sqrt{16}=\sqrt{4^2}=4i\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Zarkon Oh alright! I'm not sure if my teacher taught me about the a>0 thing but that makes sense. Thank you very much :) Explained a lot.
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