Simplify completely:

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

hi again
\(\Large 9(\frac{t-2}{2t+1})^{8}\times \frac{(2t-1)-2(t-2)}{(2t+1)^{2}}\)
i think u know me as GTA_Hunter35 that account kinda crashed

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question