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anonymous
 one year ago
Please help me simplify
anonymous
 one year ago
Please help me simplify

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 3i }{ 2+5i }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0god help me they call everything "simplify" what they mean is "write in standard from as \(a+bi\)" tell your math teacher !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now everything makes sense

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 3i }{ 2+5i}\times \frac{25i}{25i}\] is a start

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the conjugate right?... I don't know if thats how you spell it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to eliminate the i in the bottom

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i.e. multiply by the conjugate of the denominator the conjugate of \(a+bi\) is \(abi\) and this works because \[(a+bi)(abi)=a^2+b^2\] a real number

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0on your case you will have \[\frac{(3i)(25i)}{2^2+5^2}\] all the work is now in the numerator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah that is how you spell it too

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah that would be \[\frac{ 410i }{ 425i^2 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My bad, I mean \[\frac{ 117i }{ 425i^2 }\]

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1\[(a+bi)(abi)=a^2+b^2\] so \[(2+5i)(25i)=2^2+5^2=4+25=29\]

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1dont mess around with any \(i\) stuff when you multiply by the conjugate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0all right so it would be \[\frac{ 1 }{ 29 }\frac{ 17i }{ 29 }\]
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