Jamierox4ev3r
  • Jamierox4ev3r
Simplify completely:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
hi again
Jamierox4ev3r
  • Jamierox4ev3r
\(\Large 9(\frac{t-2}{2t+1})^{8}\times \frac{(2t-1)-2(t-2)}{(2t+1)^{2}}\)
anonymous
  • anonymous
i think u know me as GTA_Hunter35 that account kinda crashed

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Jamierox4ev3r
  • Jamierox4ev3r
^^There's the thing I have to simplify. Not quite sure where to begin.
Jamierox4ev3r
  • Jamierox4ev3r
@please_help_me if you'd like to address something to me, please contact me via pm. Do not spam my post, thank you :)
anonymous
  • anonymous
Parentheses Exponents Multiplication Division Addition Subtraction FROM LEFT TO RIGHT
Michele_Laino
  • Michele_Laino
all we can do is: \[\Large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t - 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{3}{{{{(2t + 1)}^2}}} \hfill \\ \end{gathered} \]
Jamierox4ev3r
  • Jamierox4ev3r
Is that all we can do?
Michele_Laino
  • Michele_Laino
therefore, we can write this: \[\Large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t - 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{3}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 27\frac{{{{\left( {t - 2} \right)}^8}}}{{{{\left( {2t + 1} \right)}^{10}}}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
yes! I think so!
anonymous
  • anonymous
what about squaring now?
Michele_Laino
  • Michele_Laino
better is if we don't develop the powers of binomials, because we have to do a very long computation @please_help_me
anonymous
  • anonymous
well.... ur right ithink thats it
Jamierox4ev3r
  • Jamierox4ev3r
wait. For the top part, (2t+1)-2(t-2) = 2t-2t+1+4? so then we would have \(\Large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t - 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{\color{red}{5}}{{{{(2t + 1)}^2}}} \hfill \\ \end{gathered}\)
Jamierox4ev3r
  • Jamierox4ev3r
basically, a 5 instead of a 3?
Michele_Laino
  • Michele_Laino
I got 2t-1-2t+4
anonymous
  • anonymous
which equals 3
anonymous
  • anonymous
distribute, then ull get 3
Jamierox4ev3r
  • Jamierox4ev3r
-1? o-o oh. wait, i think i see something. \(\Large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t \color{red}+ 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{\color{red}{5}}{{{{(2t + 1)}^2}}} \hfill \\ \end{gathered}\)
Jamierox4ev3r
  • Jamierox4ev3r
in the original problem, it is 2t+1, not 2t-1.
Jamierox4ev3r
  • Jamierox4ev3r
oh wow. I wrote it out wrong. So sorry XD
anonymous
  • anonymous
np was a typo
Michele_Laino
  • Michele_Laino
ok! then you are right, it is 5
Michele_Laino
  • Michele_Laino
here are the updated steps: \[\Large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t + 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{5}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 45\frac{{{{\left( {t - 2} \right)}^8}}}{{{{\left( {2t + 1} \right)}^{10}}}} \hfill \\ \end{gathered} \]
Jamierox4ev3r
  • Jamierox4ev3r
to clarify, let me type this out correctly. *cracks knuckles* \(\LaTeX\) is hard \(\Large 9(\frac{t-2}{2t+1})^{8} \times \frac{(2t+1)-2(t-2)}{(2t+1)^{2}}\)
anonymous
  • anonymous
can u guys help me
Jamierox4ev3r
  • Jamierox4ev3r
So I don't have to factor out the denominator or do anything fancy like that?
anonymous
  • anonymous
i dont think so
Michele_Laino
  • Michele_Laino
no, since numerator and denominator don't contain common divisors
Jamierox4ev3r
  • Jamierox4ev3r
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?
Jamierox4ev3r
  • Jamierox4ev3r
and yeah you're right, I don't see any common divisors. That makes sense
anonymous
  • anonymous
see the 2t-1?
Michele_Laino
  • Michele_Laino
I have applied the rule of multiplication of powers with same basis at denominator
anonymous
  • anonymous
2t-1 was to the 8th power and the other 2t-1 was to the2nd power
anonymous
  • anonymous
multiplying numbers with exponents adds up the exponents
Michele_Laino
  • Michele_Laino
here are more steps: \[\large \begin{gathered} 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t + 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\ \hfill \\ = 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{5}{{{{(2t + 1)}^2}}} = 45\frac{{{{\left( {t - 2} \right)}^8}}}{{{{\left( {2t + 1} \right)}^8}}} \cdot \frac{1}{{{{\left( {2t + 1} \right)}^2}}} \hfill \\ \hfill \\ = 45\frac{{{{\left( {t - 2} \right)}^8}}}{{{{\left( {2t + 1} \right)}^{10}}}} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
which created 2t-1^10
Jamierox4ev3r
  • Jamierox4ev3r
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
Michele_Laino
  • Michele_Laino
ok! :)
Jamierox4ev3r
  • Jamierox4ev3r
Thank you so much, it makes sense now :D
Michele_Laino
  • Michele_Laino
:)

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