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danielbarriosr1

  • one year ago

Please help me simplify

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  1. danielbarriosr1
    • one year ago
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    \[\frac{ 3-i }{ 2+5i }\]

  2. anonymous
    • one year ago
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    god help me they call everything "simplify" what they mean is "write in standard from as \(a+bi\)" tell your math teacher !

  3. danielbarriosr1
    • one year ago
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    Now everything makes sense

  4. anonymous
    • one year ago
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    \[\frac{ 3-i }{ 2+5i}\times \frac{2-5i}{2-5i}\] is a start

  5. danielbarriosr1
    • one year ago
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    the conjugate right?... I don't know if thats how you spell it

  6. danielbarriosr1
    • one year ago
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    to eliminate the i in the bottom

  7. anonymous
    • one year ago
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    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

  8. anonymous
    • one year ago
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    on your case you will have \[\frac{(3-i)(2-5i)}{2^2+5^2}\] all the work is now in the numerator

  9. anonymous
    • one year ago
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    yeah that is how you spell it too

  10. anonymous
    • one year ago
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    you got the top?

  11. danielbarriosr1
    • one year ago
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    yeah that would be \[\frac{ 4-10i }{ 4-25i^2 }\]

  12. danielbarriosr1
    • one year ago
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    right? @satellite73

  13. danielbarriosr1
    • one year ago
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    My bad, I mean \[\frac{ 1-17i }{ 4-25i^2 }\]

  14. danielbarriosr1
    • one year ago
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    right? @satellite73

  15. misty1212
    • one year ago
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    \[(a+bi)(a-bi)=a^2+b^2\] so \[(2+5i)(2-5i)=2^2+5^2=4+25=29\]

  16. misty1212
    • one year ago
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    dont mess around with any \(i\) stuff when you multiply by the conjugate

  17. danielbarriosr1
    • one year ago
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    all right so it would be \[\frac{ 1 }{ 29 }-\frac{ 17i }{ 29 }\]

  18. danielbarriosr1
    • one year ago
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    right? @misty1212

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