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karinewoods17 Group TitleBest ResponseYou've already chosen the best response.1
The answer in my book is \[6i  2i \sqrt{10}\] I dont know how they got that answer.
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
You can't take the square root of a negative number
 one year ago

ChmE Group TitleBest ResponseYou've already chosen the best response.0
@ksaimouli is incorrect
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.1
thats why you multiply it by i which is equal to 1
 one year ago

Ahaanomegas Group TitleBest ResponseYou've already chosen the best response.0
@ksaimouli That's totally wrong. sqrt(a)+sqrt(b) is NOT sqrt(a+b). @ilikephysics2 Yes, You can. That's what imaginary numbers are there for. @karinewoods17 Notice that \[ \Large {\sqrt {8} = 2\sqrt{2}} \]and \[ \Large {\sqrt {32} = 4\sqrt{2}}, \]so \[ \Large {\sqrt {8} = 2i\sqrt{2}} \]and \[ \Large {\sqrt {32} = 4i\sqrt{2}} \]Do you see how that works? Now, just add to get 6isqrt(2).
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.1
wouldnt it be \[6i \sqrt{2}\] ??
 one year ago

ChmE Group TitleBest ResponseYou've already chosen the best response.0
\[\sqrt{8}+\sqrt{32}=\sqrt{8}+\sqrt{8\times4}= \sqrt{8} + 2\sqrt{8}\]
 one year ago

ChmE Group TitleBest ResponseYou've already chosen the best response.0
\[3\sqrt{8}=3i \sqrt{8}=3i \sqrt{2\times4}=6i \sqrt{2}\]
 one year ago

ChmE Group TitleBest ResponseYou've already chosen the best response.0
Yes you are correct
 one year ago

Ahaanomegas Group TitleBest ResponseYou've already chosen the best response.0
Yep. That's exactly what I said, except I explained everything. =P
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.1
but... the answer in my book says the answer is \[6i  2i \sqrt{10}\] How did they get that?
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.1
Im sorry:/ Im making coffee and waiting for you:)
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.1
im not! sorry!
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.1
I was making breakfast for my boyfriend.
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.1
umm. \[\sqrt{32}\] breaks into \[\sqrt{32} = \sqrt{4} * \sqrt{8}\] which simplifies to \[4i \sqrt{2}\] ?
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.1
And yes I try:)
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.1
now what. lol
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.1
@nincompoop please do not ignore me;)
 one year ago
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