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

This Question is Closed

dellzasaurBest ResponseYou've already chosen the best response.1
\[\sqrt{7}\] = 7i or \[i \sqrt{7}\]
 one year ago

dellzasaurBest ResponseYou've already chosen the best response.1
\[\sqrt{81}\] = 3i
 one year ago

dellzasaurBest ResponseYou've already chosen the best response.1
\[\sqrt{16}\] = \[2i \sqrt{4}\]
 one year ago

dellzasaurBest ResponseYou've already chosen the best response.1
\[3\sqrt{9}\] = \[3i\]
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
The last one is wrong.
 one year ago

dellzasaurBest ResponseYou've already chosen the best response.1
@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: \[2i \sqrt{3}\]
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
The last one should be 9i.
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Well think about it WITHOUT the negative.
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Now what would you do?
 one year ago

dellzasaurBest ResponseYou've already chosen the best response.1
is that a 4 and 18? I would think to factor them to 2 x 2 and 9 x 2 maybe
 one year ago

dellzasaurBest ResponseYou've already chosen the best response.1
but inside of the radicals
 one year ago

dellzasaurBest ResponseYou've already chosen the best response.1
dw:1352593862710:dw
 one year ago

dellzasaurBest ResponseYou've already chosen the best response.1
@Dido525 Oh ok! That made since. I was using 8, 9, and i to make 72
 one year ago

dellzasaurBest ResponseYou've already chosen the best response.1
@Dido525 Thank you so much for all your help :) Makes me feel more confident about simplying them!
 one year ago

dellzasaurBest ResponseYou've already chosen the best response.1
*simplifying whoops
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
\[\sqrt{72}=\sqrt{(1)\cdot2\cdot 36}\] \[=6i\sqrt{2}\]
 one year ago

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

dellzasaurBest ResponseYou've already chosen the best response.1
@chaguanas Would I do \[3i \sqrt{3}\] for 81? and 4i or \[4i \sqrt{4}\] for 16?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
do you know what \(\sqrt{81}\)is?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
well...\[\sqrt{81}=\sqrt{(1)81}=i9=9i\]
 one year ago

dellzasaurBest ResponseYou've already chosen the best response.1
@Zarkon Ohh so that's how you do it. You wouldn't factor the 9 out to a 3?
 one year ago

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

ZarkonBest ResponseYou've already chosen the best response.1
\[\sqrt{16}=\sqrt{4^2}=4i\]
 one year ago

dellzasaurBest ResponseYou've already chosen the best response.1
@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.
 one year ago
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