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Diana.xL
 one year ago
help?
Diana.xL
 one year ago
help?

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zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Post the question where you put "help?"

Diana.xL
 one year ago
Best ResponseYou've already chosen the best response.0my answer is b. but im confused.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, B is the answer. What confuses you?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2almost, you do want to square root both sides, but when we induce the square, we get a positive and negative answer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If 16 is X squared, then X is the square root of 16. Ok, now I'm confused.

Diana.xL
 one year ago
Best ResponseYou've already chosen the best response.0so it would be c yea?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2i.e. \[(4)^2=(4)(4)=16=4*4=4^2=\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Confusion is occuring.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't know what I'm doing it appears. I'm just gonig to go because this seems to be much more complicated than I thought :I

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2the answer to \(x=\sqrt{16}\) is \(x=4\) The answer to \(x^2=16\) is \(x=\pm 4\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2@FitzIncorporated it is a common mistake. when we write \(\sqrt{something}\) we mean the positive square root(just like if we write 1). But when we induce a square (when we take square root of both sides) we get two answers. In other words, the square root of 4 is 2 but there are two answers if I ask you what number squared is 4, they are 2 and 2

Diana.xL
 one year ago
Best ResponseYou've already chosen the best response.0so its c? its lagging a bit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@zzr0ck3r Thank you :)
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