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some.random.cool.kid

  • one year ago

New question If anyone has time to spare I have a few questions I need help with

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  1. some.random.cool.kid
    • one year ago
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    When subtracted, what is the simplified form of (54 − 27i) − (39 + 5i)? Write the answer in standard form, a + bi, where a and b are real numbers.

  2. some.random.cool.kid
    • one year ago
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    I know I have to distribute and also I can do like terms

  3. Nnesha
    • one year ago
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    distribute 2nd parentheses by negative one

  4. some.random.cool.kid
    • one year ago
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    yeah igot tha tpart already im confused as to the rest...

  5. Nnesha
    • one year ago
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    alright show ur work

  6. some.random.cool.kid
    • one year ago
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    well this is what I have so far 54 − 27i) − (39 + 5i) 27{-1} - 5{-1} thats the part I was stuck on and Dont I have to change the sign when canceling or something like that?

  7. some.random.cool.kid
    • one year ago
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    then 54 minus the 39 and now im stuck.

  8. some.random.cool.kid
    • one year ago
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    I bleive its 15 addition to something

  9. Nnesha
    • one year ago
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    well you didn't distribute the 2nd parentheses

  10. Nnesha
    • one year ago
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    why is it 27{-1} = ?

  11. some.random.cool.kid
    • one year ago
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    sq to power of 2?

  12. some.random.cool.kid
    • one year ago
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    -1 x -1 imaginary number

  13. Nnesha
    • one year ago
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    oh well no i = sqrt{-1} i^2 = -1 but there isn't any i^2

  14. some.random.cool.kid
    • one year ago
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    so its not power of two?

  15. Nnesha
    • one year ago
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    |dw:1442334862171:dw| well we don't have to apply the foil method distribute 2nd parenthesizes by -1 and then combine like terms

  16. some.random.cool.kid
    • one year ago
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    uh -5?

  17. some.random.cool.kid
    • one year ago
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    i

  18. some.random.cool.kid
    • one year ago
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    so 27i + -5i?

  19. Nnesha
    • one year ago
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    welll leave the first parentheses alone for first step \[\huge\rm -1(39+5i)\] distribute multiply both terms in the parentheses by negative 1

  20. some.random.cool.kid
    • one year ago
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    so -39 + -5i?

  21. Nnesha
    • one year ago
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    yes right

  22. some.random.cool.kid
    • one year ago
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    i thought you needed like terms?

  23. Nnesha
    • one year ago
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    yes right now we have like terms |dw:1442335143368:dw| combine like terms

  24. some.random.cool.kid
    • one year ago
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    32i btw

  25. Nnesha
    • one year ago
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    if we have negative sign outside the parentheses we must hve to distribute first

  26. Nnesha
    • one year ago
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    hmm -27i-5i = ?

  27. some.random.cool.kid
    • one year ago
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    i guess so

  28. some.random.cool.kid
    • one year ago
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    but it has minus

  29. Nnesha
    • one year ago
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    yes it supposed to be -32 -# -# = -#

  30. some.random.cool.kid
    • one year ago
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    so far i se 15 - 32i? am I steearing in the righ tdirection?

  31. Nnesha
    • one year ago
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    yes that's right

  32. some.random.cool.kid
    • one year ago
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    - hashtag lol

  33. some.random.cool.kid
    • one year ago
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    So am I done now ? because I need help with more if you have time to spare explaining a few with me.

  34. Nnesha
    • one year ago
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    yes that's itdone!

  35. some.random.cool.kid
    • one year ago
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    would you mind helping me with this one too? The number root of order seven of fifty three cubed can be written as fifty three to the power of start fraction cap a over cap b end fraction end power. What is the value of A?

  36. some.random.cool.kid
    • one year ago
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    ooo that came out messy maybe I can attach a file

  37. some.random.cool.kid
    • one year ago
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  38. some.random.cool.kid
    • one year ago
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  39. some.random.cool.kid
    • one year ago
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    ok done. :D

  40. Nnesha
    • one year ago
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    ah okay so can convert root to an exponent form exponent rule \[\huge\rm \sqrt[n]{x^m}=x^\frac{ m }{ n }\] index becomes the denominator of the fraction

  41. some.random.cool.kid
    • one year ago
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    sorry my internet went bye bye lol

  42. some.random.cool.kid
    • one year ago
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    ok so now what?

  43. Nnesha
    • one year ago
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    that's it

  44. some.random.cool.kid
    • one year ago
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    im lost?

  45. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Nnesha ah okay so can convert root to an exponent form exponent rule \[\huge\rm \sqrt[n]{x^m}=x^\frac{ m }{ n }\] index becomes the denominator of the fraction \(\color{blue}{\text{End of Quote}}\) read this

  46. some.random.cool.kid
    • one year ago
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    nvm i got it 7/3?

  47. Nnesha
    • one year ago
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    no that's not right

  48. some.random.cool.kid
    • one year ago
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    hmm?

  49. Nnesha
    • one year ago
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    index would be the `denominator`

  50. some.random.cool.kid
    • one year ago
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    ooohh ohhh sorry

  51. some.random.cool.kid
    • one year ago
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    get it now

  52. Nnesha
    • one year ago
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    alright so A would be what ?

  53. some.random.cool.kid
    • one year ago
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    isnt the 3 the den?

  54. Nnesha
    • one year ago
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    yes right

  55. some.random.cool.kid
    • one year ago
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    because the numer became the power

  56. some.random.cool.kid
    • one year ago
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    alright thanks

  57. Nnesha
    • one year ago
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    excellent! yw good work!

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