anonymous
  • anonymous
2a^-3/(2a)^-3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
first you need to Arrange no. with alphabet and then go thru on latter and you can solve it
anonymous
  • anonymous
i mean first you need to Arrange no. with alphabet and then go thru on Number and you can solve it
anonymous
  • anonymous
1

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Jhannybean
  • Jhannybean
\[\large \frac{2a^{-3}}{(2a)^{-3}}\] first expand \(\large (2a)^{-3}\)\[\large (2a)^{-3} = 2^{-3}\cdot a^{-3}\] so now evaluate what you can. \[\large \frac{2 \cdot \cancel{a^{-3}}}{2^{-3}\cdot \cancel {a^{-3}}}\]You're left with \[\large \frac{2}{2^{-3}}= 2^{1-(-3)} = 2^{4} =16\]
whpalmer4
  • whpalmer4
Another approach: \[\frac{2a^{-3}}{(2a)^{-3}} = \frac{2(2a)^3}{a^3} = \frac{2*2^3*\cancel{a^3}}{\cancel{{a^3}}}=2*8=16\]because \[a^{-n}=\frac{1}{a^n}\]

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