horsegal244
  • horsegal244
Pic Below Its An Essay Will Medal And Fan for Best Answer!
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
horsegal244
  • horsegal244
horsegal244
  • horsegal244
@Michele_Laino
horsegal244
  • horsegal244
@dan815 @Luigi0210

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Michele_Laino
  • Michele_Laino
first step: since \(9^0=1\) the I can write this: \[\huge \begin{gathered} {\left( {{3^8} \cdot {2^{ - 5}} \cdot {9^0}} \right)^{ - 2}} = {\left( {{3^8} \cdot {2^{ - 5}} \cdot 1} \right)^{ - 2}} = \hfill \\ \hfill \\ = {\left( {{3^8} \cdot {2^{ - 5}}} \right)^{ - 2}} = {3^{8 \cdot \left( { - 2} \right)}} \cdot {2^{ - 5 \cdot \left( { - 2} \right)}} = ...? \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
oops.. I rewrite such expression: \[\Large \begin{gathered} {\left( {{3^8} \cdot {2^{ - 5}} \cdot {9^0}} \right)^{ - 2}} = {\left( {{3^8} \cdot {2^{ - 5}} \cdot 1} \right)^{ - 2}} = \hfill \\ \hfill \\ = {\left( {{3^8} \cdot {2^{ - 5}}} \right)^{ - 2}} = {3^{8 \cdot \left( { - 2} \right)}} \cdot {2^{ - 5 \cdot \left( { - 2} \right)}} = ...? \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
please apply these identities: \[\Large {\left( {{3^m}} \right)^n} = {3^{m \cdot n}},\quad {\left( {{2^k}} \right)^q} = {2^{k \cdot q}}\] where \(m,\;n,\;k,\;q\) are integers
horsegal244
  • horsegal244
@Michele_Laino Sorry i had to go somewhere
Michele_Laino
  • Michele_Laino
I'm very sorry I have to go now, since I have to give a tutoring lesson here at my home. I will return between 2 hours
horsegal244
  • horsegal244
ok
Michele_Laino
  • Michele_Laino
thanks!! :)
horsegal244
  • horsegal244
@dan815 @nincompoop @Hero

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