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Jamierox4ev3r
 one year ago
Simplify completely:
Jamierox4ev3r
 one year ago
Simplify completely:

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Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0\(\Large 9(\frac{t2}{2t+1})^{8}\times \frac{(2t1)2(t2)}{(2t+1)^{2}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think u know me as GTA_Hunter35 that account kinda crashed

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0^^There's the thing I have to simplify. Not quite sure where to begin.

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.0Parentheses Exponents Multiplication Division Addition Subtraction FROM LEFT TO RIGHT

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2all 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
 one year ago
Best ResponseYou've already chosen the best response.0Is that all we can do?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2therefore, 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
 one year ago
Best ResponseYou've already chosen the best response.2yes! I think so!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what about squaring now?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2better is if we don't develop the powers of binomials, because we have to do a very long computation @please_help_me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well.... ur right ithink thats it

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0wait. For the top part, (2t+1)2(t2) = 2t2t+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
 one year ago
Best ResponseYou've already chosen the best response.0basically, a 5 instead of a 3?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0distribute, then ull get 3

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.01? oo 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
 one year ago
Best ResponseYou've already chosen the best response.0in the original problem, it is 2t+1, not 2t1.

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0oh wow. I wrote it out wrong. So sorry XD

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2ok! then you are right, it is 5

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2here 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
 one year ago
Best ResponseYou've already chosen the best response.0to clarify, let me type this out correctly. *cracks knuckles* \(\LaTeX\) is hard \(\Large 9(\frac{t2}{2t+1})^{8} \times \frac{(2t+1)2(t2)}{(2t+1)^{2}}\)

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0So I don't have to factor out the denominator or do anything fancy like that?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2no, since numerator and denominator don't contain common divisors

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0oh, 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
 one year ago
Best ResponseYou've already chosen the best response.0and yeah you're right, I don't see any common divisors. That makes sense

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I have applied the rule of multiplication of powers with same basis at denominator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02t1 was to the 8th power and the other 2t1 was to the2nd power

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0multiplying numbers with exponents adds up the exponents

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2here 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
 one year ago
Best ResponseYou've already chosen the best response.0which created 2t1^10

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much, it makes sense now :D
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