danielbarriosr1
  • danielbarriosr1
Please help me simplify
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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.
chestercat
  • chestercat
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danielbarriosr1
  • danielbarriosr1
\[\frac{ 3-i }{ 2+5i }\]
anonymous
  • anonymous
god help me they call everything "simplify" what they mean is "write in standard from as \(a+bi\)" tell your math teacher !
danielbarriosr1
  • danielbarriosr1
Now everything makes sense

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More answers

anonymous
  • anonymous
\[\frac{ 3-i }{ 2+5i}\times \frac{2-5i}{2-5i}\] is a start
danielbarriosr1
  • danielbarriosr1
the conjugate right?... I don't know if thats how you spell it
danielbarriosr1
  • danielbarriosr1
to eliminate the i in the bottom
anonymous
  • anonymous
i.e. multiply by the conjugate of the denominator the conjugate of \(a+bi\) is \(a-bi\) and this works because \[(a+bi)(a-bi)=a^2+b^2\] a real number
anonymous
  • anonymous
on your case you will have \[\frac{(3-i)(2-5i)}{2^2+5^2}\] all the work is now in the numerator
anonymous
  • anonymous
yeah that is how you spell it too
anonymous
  • anonymous
you got the top?
danielbarriosr1
  • danielbarriosr1
yeah that would be \[\frac{ 4-10i }{ 4-25i^2 }\]
danielbarriosr1
  • danielbarriosr1
right? @satellite73
danielbarriosr1
  • danielbarriosr1
My bad, I mean \[\frac{ 1-17i }{ 4-25i^2 }\]
danielbarriosr1
  • danielbarriosr1
right? @satellite73
misty1212
  • misty1212
\[(a+bi)(a-bi)=a^2+b^2\] so \[(2+5i)(2-5i)=2^2+5^2=4+25=29\]
misty1212
  • misty1212
dont mess around with any \(i\) stuff when you multiply by the conjugate
danielbarriosr1
  • danielbarriosr1
all right so it would be \[\frac{ 1 }{ 29 }-\frac{ 17i }{ 29 }\]
danielbarriosr1
  • danielbarriosr1
right? @misty1212

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