How do you divide #5/(-5i)#?

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Answer 1 · 2016-11-12T13:02:06

Cancel out the 5's and end up with #i#

We have:

#5/(-5i)#

We can rewrite the denominator to make it easier to work with:

#5/(-1xx5xxi)#

and to see that we can cancel out the 5 in the numerator and denominator:

#cancel5/(-1xxcancel5xxi)=1/(-i)#

And now let's clean this fraction up:

#1/(-i)=1/(-i)(1)=1/(-i)(i/i)=i/(-i^2)=i/(-(-1))=i/1=i#