anonymous
  • anonymous
2-5i/3i
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
\[\frac{ 2-5i }{ 3i }\]
anonymous
  • anonymous
I understand you are supposed to multiply by the conjugate, but what is the conjugate of 3i
anonymous
  • anonymous

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

zepdrix
  • zepdrix
The conjugate of \(\large\rm 0+3i\) is \(\large\rm 0-3i\)
zepdrix
  • zepdrix
So therefore the conjugate of \(\large\rm 3i\) is simply \(\large\rm -3i\) k? :)
anonymous
  • anonymous
don't be a slave to method you can multiply by \(i\) top and bottom if you like or \(-i\)
zepdrix
  • zepdrix
Ya that'll save you a couple steps :D Seems like a good idea
anonymous
  • anonymous
either way the denominator will become a real number, either 3 or -3 depending on which you pick
anonymous
  • anonymous
I see thank you so much
anonymous
  • anonymous
Wait, I get a different answer from when i multiply -3i to when i multiply from just i
anonymous
  • anonymous
not after you cancel the common factor of 3
anonymous
  • anonymous
which is why multiplying by \(-3i\) is silly in this case
anonymous
  • anonymous
ohhh so should the final answer be -2i -5
anonymous
  • anonymous
idk i didn't do it want me to check?
anonymous
  • anonymous
yes please
anonymous
  • anonymous
no actually that can't be it
anonymous
  • anonymous
\[\frac{ 2-5i }{ 3i }\times \frac{i}{i}\] \[=\frac{2i+5}{-3}\]
anonymous
  • anonymous
i got -6i -15 but then factored out 3
anonymous
  • anonymous
yes but my answer has to be in complex number form i'm sorry i forgot to mention that
anonymous
  • anonymous
now i am confused @satellite73 @zepdrix
anonymous
  • anonymous
you mean in the form \(a+bi\)?
anonymous
  • anonymous
yes
anonymous
  • anonymous
break it in to two pieces is all
anonymous
  • anonymous
for example \[\frac{7+8i}{5}=\frac{7}{5}+\frac{8}{5}i\]
anonymous
  • anonymous
oh i understand now can i get back to you in a minute with my answer?
anonymous
  • anonymous
is this correct? \[\frac{ 2i }{ -3 }+\frac{ 5 }{ -3 }\]
beginnersmind
  • beginnersmind
Yes, although you might want to write it as \[-\frac{ 5 }{ 3 } -\frac{ 2 }{ 3 }i\] or even \[\frac{ -5-2i }{ 3 }\] but it's just cosmetical at this point.
anonymous
  • anonymous
thank you so much
beginnersmind
  • beginnersmind
No problem, satellite did all the work though. ;)

Looking for something else?

Not the answer you are looking for? Search for more explanations.