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Alona [7]
1 year ago
11

Which expression is equivalent to StartFraction (3 m Superscript negative 1 Baseline n squared) Superscript negative 4 Baseline

Over (2 m Superscript negative 2 Baseline n) cubed EndFraction?
Mathematics
2 answers:
koban [17]1 year ago
8 0

Answer:

3m^{10}n^{-11}

Step-by-step explanation:

Given the expression \frac{(3m^{-1}n^{2})^{-4}   }{(2m^{-2}n)^{3}  }, we will use laws of indices to get the equivalent expression as shown below;

According to one of the law of indices,

\frac{a^{m} }{a^{n} } = a^{m-n}  \ and\ (a^{m})^{n} = a^{mn}

\frac{(3m^{-1}n^{2})^{-4}   }{(2m^{-2}n)^{3}  }\\= \frac{3m^{4}n^{-8}   }{2m^{-6}n^{3}  }\\= 3m^{(4-(-6))} * n^{-8-3}\\ = 3m^{10}n^{-11}

This gives the required expression

Guest1 year ago
0 0

Which expression is equivalent to StartFraction (3 m Superscript negative 1 Baseline n squared) Superscript negative 4 Baseline Over (2 m Superscript negative 2 Baseline n) cubed EndFraction? Assume mc024-2.jpg.
2 m squared n Superscript 5
StartFraction 81 m squared n Superscript 5 Baseline Over 8 EndFraction
2 m squared n squared
StartFraction 81 m squared n squared Over 8 EndFraction

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-------------------------------

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