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Jamierox4ev3r

  • one year ago

Simplify completely:

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  1. anonymous
    • one year ago
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    hi again

  2. Jamierox4ev3r
    • one year ago
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    \(\Large 9(\frac{t-2}{2t+1})^{8}\times \frac{(2t-1)-2(t-2)}{(2t+1)^{2}}\)

  3. anonymous
    • one year ago
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    i think u know me as GTA_Hunter35 that account kinda crashed

  4. Jamierox4ev3r
    • one year ago
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    ^^There's the thing I have to simplify. Not quite sure where to begin.

  5. Jamierox4ev3r
    • one year ago
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    @please_help_me if you'd like to address something to me, please contact me via pm. Do not spam my post, thank you :)

  6. anonymous
    • one year ago
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    Parentheses Exponents Multiplication Division Addition Subtraction FROM LEFT TO RIGHT

  7. Michele_Laino
    • one year ago
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    all we can do is: \[\Large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t - 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{3}{{{{(2t + 1)}^2}}} \hfill \\ \end{gathered} \]

  8. Jamierox4ev3r
    • one year ago
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    Is that all we can do?

  9. Michele_Laino
    • one year ago
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    therefore, we can write this: \[\Large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t - 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{3}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 27\frac{{{{\left( {t - 2} \right)}^8}}}{{{{\left( {2t + 1} \right)}^{10}}}} \hfill \\ \end{gathered} \]

  10. Michele_Laino
    • one year ago
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    yes! I think so!

  11. anonymous
    • one year ago
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    what about squaring now?

  12. Michele_Laino
    • one year ago
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    better is if we don't develop the powers of binomials, because we have to do a very long computation @please_help_me

  13. anonymous
    • one year ago
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    well.... ur right ithink thats it

  14. Jamierox4ev3r
    • one year ago
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    wait. For the top part, (2t+1)-2(t-2) = 2t-2t+1+4? so then we would have \(\Large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t - 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{\color{red}{5}}{{{{(2t + 1)}^2}}} \hfill \\ \end{gathered}\)

  15. Jamierox4ev3r
    • one year ago
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    basically, a 5 instead of a 3?

  16. Michele_Laino
    • one year ago
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    I got 2t-1-2t+4

  17. anonymous
    • one year ago
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    which equals 3

  18. anonymous
    • one year ago
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    distribute, then ull get 3

  19. Jamierox4ev3r
    • one year ago
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    -1? o-o oh. wait, i think i see something. \(\Large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t \color{red}+ 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{\color{red}{5}}{{{{(2t + 1)}^2}}} \hfill \\ \end{gathered}\)

  20. Jamierox4ev3r
    • one year ago
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    in the original problem, it is 2t+1, not 2t-1.

  21. Jamierox4ev3r
    • one year ago
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    oh wow. I wrote it out wrong. So sorry XD

  22. anonymous
    • one year ago
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    np was a typo

  23. Michele_Laino
    • one year ago
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    ok! then you are right, it is 5

  24. Michele_Laino
    • one year ago
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    here are the updated steps: \[\Large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t + 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{5}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 45\frac{{{{\left( {t - 2} \right)}^8}}}{{{{\left( {2t + 1} \right)}^{10}}}} \hfill \\ \end{gathered} \]

  25. Jamierox4ev3r
    • one year ago
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    to clarify, let me type this out correctly. *cracks knuckles* \(\LaTeX\) is hard \(\Large 9(\frac{t-2}{2t+1})^{8} \times \frac{(2t+1)-2(t-2)}{(2t+1)^{2}}\)

  26. anonymous
    • one year ago
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    can u guys help me

  27. Jamierox4ev3r
    • one year ago
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    So I don't have to factor out the denominator or do anything fancy like that?

  28. anonymous
    • one year ago
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    i dont think so

  29. Michele_Laino
    • one year ago
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    no, since numerator and denominator don't contain common divisors

  30. Jamierox4ev3r
    • one year ago
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    oh, and how did you get from the penultimate step to the last step? Like what are the exact operations that are involved in combining this multiplication problem into one fraction?

  31. Jamierox4ev3r
    • one year ago
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    and yeah you're right, I don't see any common divisors. That makes sense

  32. anonymous
    • one year ago
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    see the 2t-1?

  33. Michele_Laino
    • one year ago
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    I have applied the rule of multiplication of powers with same basis at denominator

  34. anonymous
    • one year ago
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    2t-1 was to the 8th power and the other 2t-1 was to the2nd power

  35. anonymous
    • one year ago
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    multiplying numbers with exponents adds up the exponents

  36. Michele_Laino
    • one year ago
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    here are more steps: \[\large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t + 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{5}{{{{(2t + 1)}^2}}} = 45\frac{{{{\left( {t - 2} \right)}^8}}}{{{{\left( {2t + 1} \right)}^8}}} \cdot \frac{1}{{{{\left( {2t + 1} \right)}^2}}} \hfill \\ \hfill \\ = 45\frac{{{{\left( {t - 2} \right)}^8}}}{{{{\left( {2t + 1} \right)}^{10}}}} \hfill \\ \end{gathered} \]

  37. anonymous
    • one year ago
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    which created 2t-1^10

  38. Jamierox4ev3r
    • one year ago
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    Oh I see! I was just kind of wondering how the numerator of five contributed to the solid number of 45 in the mixed fraction. I see it now

  39. Michele_Laino
    • one year ago
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    ok! :)

  40. Jamierox4ev3r
    • one year ago
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    Thank you so much, it makes sense now :D

  41. Michele_Laino
    • one year ago
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    :)

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