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Destinyyyy

  • one year ago

perform the addition.. How do I solve this? The answer I got was incorrect.

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  1. Destinyyyy
    • one year ago
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    (-5+5i) + (9+8i)

  2. anonymous
    • one year ago
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    4+13i

  3. Destinyyyy
    • one year ago
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    Yes I know that thats the correct answer. Im not asking for that. Im asking how to solve it

  4. DecentNabeel
    • one year ago
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    \[\left(5i-5\right)+\left(8i+9\right)\] \[\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\] \[=\left(5i-5\right)+8i+9\] \[\mathrm{Add/Subtract\:the\:numbers:}\:-5+9=4\] \[=8i+5i+4\] \[\mathrm{Add\:similar\:elements:}\:8i+5i=13i\] \[=13i+4\] that is the answer

  5. DecentNabeel
    • one year ago
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    are you understand @Destinyyyy

  6. Destinyyyy
    • one year ago
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    Sorry one second while I read it.. There was a spider in my room :(

  7. DecentNabeel
    • one year ago
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    hahaha no problem @Destinyyyy

  8. Destinyyyy
    • one year ago
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    My example said to remove the i and put it outside of the equation.. (-5 +5) +(9+8)i

  9. Destinyyyy
    • one year ago
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    It was fluttering huge... Death by broom

  10. Destinyyyy
    • one year ago
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    it change f-ing to fluttering -.-

  11. DecentNabeel
    • one year ago
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    \[\mathrm{Simplify}\:\left(-5+5\right):\quad 0\] \[=i\left(8+9\right)+0\] \[\mathrm{Simplify}\:i\left(8+9\right):\quad 17i\] 17i+0=17i

  12. anonymous
    • one year ago
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    The way your book does it is still true but i think its in this format (a+bi) + (d+ ci) = (a+d) (b+c) i

  13. anonymous
    • one year ago
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    To add or subtract we use Add (a+bi) + (c +di) = (a+b) + (c +d)i Subtract (a+bi) (c +di) = (a+b) - (c +d)i So in your case (-5+5i) + (9+8i) = (-5+5) + (9+8)i -5+9 + 5+8 4 + 13i

  14. DecentNabeel
    • one year ago
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    any confusion @Destinyyyy

  15. DecentNabeel
    • one year ago
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    @Deeezzzz -5+9=4 not -4

  16. Destinyyyy
    • one year ago
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    Okay.. I get it now.. I multiplied the parenthesis... Thank you everyone!!

  17. anonymous
    • one year ago
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    Makes sense?

  18. Destinyyyy
    • one year ago
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    Yes.. Can I get help with another one?

  19. DecentNabeel
    • one year ago
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    yes tell me @Destinyyyy

  20. anonymous
    • one year ago
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    Thanks! @DecentNabeel simple arithmetic mistake

  21. DecentNabeel
    • one year ago
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    no problem @Deeezzzz

  22. Destinyyyy
    • one year ago
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    Perform the addition (-9/2 +1/2i) + (3/2 - 7/2i)

  23. DecentNabeel
    • one year ago
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    \[\left(-\frac{9}{2}+\frac{i}{2}\right)+\left(\frac{3}{2}-\frac{7i}{2}\right)\] \[\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\] \[=\left(-\frac{9}{2}+\frac{i}{2}\right)+\frac{3}{2}-\frac{7i}{2}\] \[\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}\] \[=\frac{-7i+i+3-9}{2}\] refine \[=\frac{-6i-6}{2}\] factor out -6 \[=-\frac{6\left(i+1\right)}{2}\] \[\mathrm{Divide\:the\:numbers:}\:\frac{6}{2}=3\] \[=-\left(3\left(i+1\right)\right)\] \[\mathrm{Negate}\:-\left(3\left(i+1\right)\right)=-3\left(i+1\right)\] \[=-3\left(i+1\right)\] that is the answer

  24. DecentNabeel
    • one year ago
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    ok @Destinyyyy

  25. Destinyyyy
    • one year ago
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    ??? Why is the 1 now i and the 7 now 7i?

  26. Destinyyyy
    • one year ago
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    My examples show to put the i on the outside of the equation like @Nixy did

  27. DecentNabeel
    • one year ago
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    -3i+3

  28. Destinyyyy
    • one year ago
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    Still confused

  29. DecentNabeel
    • one year ago
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    why confused tell me

  30. DecentNabeel
    • one year ago
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    you said i on the out sides .. so that.. 3(i+1) =3i+3

  31. Destinyyyy
    • one year ago
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    No I mean like this (-9/2 +1/2) + (3/2 - 7/2)i for the first step

  32. DecentNabeel
    • one year ago
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    \[\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\] \[=\left(-\frac{9}{2}+\frac{i}{2}\right)+\frac{3}{2}-\frac{7i}{2}\]

  33. DecentNabeel
    • one year ago
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    alright @Destinyyyy

  34. Destinyyyy
    • one year ago
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    I dont understand why 1 is now i and why its 7i

  35. DecentNabeel
    • one year ago
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    ok i expla it

  36. anonymous
    • one year ago
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    \( \huge (-\frac{9}{2} +\frac{1}{2}i) + (\frac{3}{2} - \frac{7}{2}i) \) Is the above how it looks in your book?

  37. DecentNabeel
    • one year ago
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    (-9/2 +1/2i) + (3/2 - 7/2i) that is your question

  38. anonymous
    • one year ago
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    \( \huge (-\frac{9}{2} +\frac{1}{2}i) + (\frac{3}{2} - \frac{7}{2}i) \) \( \huge (-\frac{9}{2} +\frac{1}{2}) + (\frac{3}{2} - \frac{7}{2}) i\) \( \huge (-\frac{9}{2} +\frac{3}{2}) + (\frac{1}{2} - \frac{7}{2}) i\) \( \huge -\frac{6}{2} + (-\frac{6}{2}) i\) \( \huge -3 -3i\)

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