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Mathematics 26 Online
OpenStudy (anonymous):

Alg 2 help? Will fan, medal and testify.

OpenStudy (anonymous):

Simplify the expression \[\frac{ -5+i }{ 2i }\] Show your work

OpenStudy (xapproachesinfinity):

i suggest that you multiply top and bottom by i first

OpenStudy (nicholausblackmon):

what do you mean testify?

OpenStudy (xapproachesinfinity):

\(\large \frac{-5+i}{2i}=\large \frac{(-5+i)i}{-2}\)

OpenStudy (anonymous):

\[\frac{ (-5i+i^2 }{ 2i}\] Like this? And I meant a testimonial

OpenStudy (xapproachesinfinity):

Take it from there....

OpenStudy (xapproachesinfinity):

look at my reply i said top and bottom by i not just top

OpenStudy (anonymous):

Why would multiplying the bottom by i remove the i? Wouldn't it just stay the same because 1xanything= the other number?

OpenStudy (xapproachesinfinity):

because \(\large i^2=-1\) you need to rationalize the bottom

OpenStudy (anonymous):

\[\frac{ -5i+i^2 }{ -2}\] Ok so thats what it would end up as then?

OpenStudy (xapproachesinfinity):

i^2=-1 you have i^2 on top and simplify more

OpenStudy (anonymous):

\[\frac{ -5i-1 }{ -2 }\] which would then simplify to \[-5i+\frac{ 1 }{ 2 }\]?

OpenStudy (xapproachesinfinity):

you have separated the fraction but not correctly 5/2 i +1/2

OpenStudy (anonymous):

Oh ok. Sorry, simple mistake I should't have missed. So is that the final answer?

OpenStudy (xapproachesinfinity):

yes! it preferably to right a complex number in this form z=a+bi so 1/2+5/2 i but either ways is correct

OpenStudy (anonymous):

Thanks a million! You helped a lot!

OpenStudy (xapproachesinfinity):

no problem

OpenStudy (anonymous):

I will post ur testimony later tonight, my family is yelling for me to get upstairs for family night! Thanks again!

OpenStudy (xapproachesinfinity):

No worries about that^_^

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