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anonymous

  • one year ago

Algebra 2 help? I'll loose my life if I do't get this done http://prntscr.com/81bhcy

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  1. anonymous
    • one year ago
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    answer briefly please

  2. jim_thompson5910
    • one year ago
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    |dw:1438817092059:dw|

  3. anonymous
    • one year ago
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    I don't mean to be a fanboy, but holy shiz, I can't believe I'm meeting Jim like this

  4. jim_thompson5910
    • one year ago
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    The bases are the same (both x), so we subtract the exponents |dw:1438817143395:dw|

  5. anonymous
    • one year ago
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    Okay

  6. jim_thompson5910
    • one year ago
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    |dw:1438817164689:dw|

  7. jim_thompson5910
    • one year ago
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    now simplify (4/3) - (5/6)

  8. anonymous
    • one year ago
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    LCD?

  9. jim_thompson5910
    • one year ago
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    yes you'll need to find that first, what is the LCD in this case?

  10. anonymous
    • one year ago
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    3

  11. jim_thompson5910
    • one year ago
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    no

  12. jim_thompson5910
    • one year ago
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    what is the lowest common multiple of 3 and 6?

  13. anonymous
    • one year ago
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    1?

  14. anonymous
    • one year ago
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    idk

  15. jim_thompson5910
    • one year ago
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    multiples of 3: 3, 6, 9, 12, ... multiples of 6: 6, 12, 18, ...

  16. anonymous
    • one year ago
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    nvm its 6

  17. jim_thompson5910
    • one year ago
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    yes

  18. jim_thompson5910
    • one year ago
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    the LCD is 6, so you have to transform each fraction to make sure they all have denominators equal to 6

  19. anonymous
    • one year ago
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    I'm no good with these, can you give me a walkthrough?

  20. jim_thompson5910
    • one year ago
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    |dw:1438817391007:dw|

  21. jim_thompson5910
    • one year ago
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    what can we multiply top and bottom by to get 6 in the denominator?

  22. anonymous
    • one year ago
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    2?

  23. jim_thompson5910
    • one year ago
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    yes |dw:1438817451986:dw|

  24. jim_thompson5910
    • one year ago
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    so 4/3 is the same as 8/6

  25. anonymous
    • one year ago
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    mk

  26. jim_thompson5910
    • one year ago
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    \[\Large {\color{red}{\frac{4}{3}}} - \frac{5}{6} = {\color{red}{\frac{8}{6}}} - \frac{5}{6} =??\]

  27. anonymous
    • one year ago
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    3/6

  28. jim_thompson5910
    • one year ago
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    which reduces to what

  29. anonymous
    • one year ago
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    1/2

  30. jim_thompson5910
    • one year ago
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    so we can say this |dw:1438817675532:dw|

  31. anonymous
    • one year ago
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    mk so am I done with that one?

  32. jim_thompson5910
    • one year ago
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    yes, now simplify the other expression

  33. jim_thompson5910
    • one year ago
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    |dw:1438817769472:dw|

  34. jim_thompson5910
    • one year ago
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    |dw:1438817801342:dw|

  35. anonymous
    • one year ago
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    All i remember is that you add the exponents, at least those 2 inside

  36. jim_thompson5910
    • one year ago
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    yes, so what is \(\Large x*x^3*x^4\) equal to?

  37. anonymous
    • one year ago
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    x^7 or 8

  38. jim_thompson5910
    • one year ago
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    recall that \(\Large x = x^1\)

  39. anonymous
    • one year ago
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    I'd go with 8

  40. jim_thompson5910
    • one year ago
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    \[\Large x*x^3*x^4 = x^1*x^3*x^4 = ?\]

  41. anonymous
    • one year ago
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    x^8

  42. jim_thompson5910
    • one year ago
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    yes

  43. anonymous
    • one year ago
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    mk, now for tht 16 on the outside

  44. jim_thompson5910
    • one year ago
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    |dw:1438817927856:dw|

  45. jim_thompson5910
    • one year ago
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    now use this rule \[\LARGE \sqrt[n]{x^m} = x^{m/n}\]

  46. anonymous
    • one year ago
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    so x^8/16

  47. jim_thompson5910
    • one year ago
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    then 8/16 reduces to ?

  48. anonymous
    • one year ago
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    1/2

  49. jim_thompson5910
    • one year ago
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    |dw:1438818127887:dw|

  50. anonymous
    • one year ago
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    Okay, so both were x ^1/2 lol

  51. jim_thompson5910
    • one year ago
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    correct

  52. anonymous
    • one year ago
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    were they equivalent at start or no

  53. jim_thompson5910
    • one year ago
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    they are completely equivalent as long as x is not 0 if x = 0, then you have a division by zero error in the first expression

  54. anonymous
    • one year ago
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    Okay, so basically equivalent

  55. jim_thompson5910
    • one year ago
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    yes

  56. anonymous
    • one year ago
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    Thanks man.

  57. jim_thompson5910
    • one year ago
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    no problem

  58. anonymous
    • one year ago
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    Got time for one more? Be honest, Idm if you don't

  59. jim_thompson5910
    • one year ago
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    go ahead

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