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(\(\frac{5^{55 }}{10^{ 30}}\)) (\(\frac{2^{30 }}{5^{25 }}\)) =

A. 1

B. \(\frac{5^{30}}{2^{60}}\)

C. \(\frac{5^{30}}{2^{30}}\)

D. \(\frac{(5^ {55}) (2^ {30})}{10^{30}}\)

E. \(\frac{10^{85}}{50^{55}}\)
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AbdurRakib
(\(\frac{5^{55 }}{10^{ 30}}\)) (\(\frac{2^{30 }}{5^{25 }}\)) =

A. 1

B. \(\frac{5^{30}}{2^{60}}\)

C. \(\frac{5^{30}}{2^{30}}\)

D. \(\frac{(5^ {55}) (2^ {30})}{10^{30}}\)

E. \(\frac{10^{85}}{50^{55}}\)

Merging topics. Please refer to the discussion above.
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Bunuel
\((\frac{5^{55}}{10^{30}})(\frac{2^{30}}{5^{25}})=\)

A. 1
B. 5^30/2^60
C. 5^30/2^30
D. 5^55*2^30/10^30
E. 10^85/50^55

This question uses following formulae of indices:

\(a^m\)*\(a^n\) = \(a^{m+n}\)

\(\frac{1}{{a^m}}\) = \(a^{-m}\)

\(a^m\) / \(a^n\) = \(a^{m-n}\)


\((\frac{5^{55}}{10^{30}})(\frac{2^{30}}{5^{25}})=\)

\(10^{30}\) can be written as \(5^{30}\) * \(2^{30}\)

Powers of 5 can now be brought together: +55-30-25 = 0

\(5^0\) = 1

Similarly let us arrange power of 2:

-30+30 = 0


\(2^0\) = 1


Answer: 1*1 = 1. A is the answer.
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AbdurRakib
(\(\frac{5^{55 }}{10^{ 30}}\)) (\(\frac{2^{30 }}{5^{25 }}\)) =

A. 1

B. \(\frac{5^{30}}{2^{60}}\)

C. \(\frac{5^{30}}{2^{30}}\)

D. \(\frac{(5^ {55}) (2^ {30})}{10^{30}}\)

E. \(\frac{10^{85}}{50^{55}}\)

(\(\frac{5^{55 }}{10^{ 30}}\)) (\(\frac{2^{30 }}{5^{25 }}\)) = (5^25/ 2^30) * (2^30/5^25) = 1

Correct Option: A

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