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Answer 1

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Answer 2

$\Large{\underbrace{\sf{\red{Required\:Solution:}}}}$

$\displaystyle{\bf{ \dfrac{ 3^{-5} \times 10^{-5} \times 125}{ 5^{-7} \times 6^{-5}}}}$

Sol'n. $\displaystyle{\sf{\dfrac{ 3^{-5} \times 10^{-5} \times 125}{ 5^{-7} \times 6^{-5}}}}$

$\longrightarrow{\sf{ \dfrac{ 3^{-5} \times (2\times 5)^{-5} \times 5^{3}}{ 5^{-7} \times (2 \times 3)^{-5}}}}$

$\longrightarrow{\sf{ \dfrac{3^{-5} \times 2^{-5} \times 5^{-5} \times 5^{3}}{ 5^{-7} \times 2^{-5} \times 3^{-5}}}}$

• Exponential law used - $\displaystyle{\bf{ a^{m} \times a^{n} = a^{m+n}}}$

$\longrightarrow{\sf{ \dfrac{3^{-5} \times 2^{-5} \times 5^{-5+3}}{ 5^{-7} \times 2^{-5} \times 3^{-5}}}}$

$\longrightarrow{\sf{ \dfrac{3^{-5} \times 2^{-5} \times 5^{-2}}{ 5^{-7} \times 2^{-5} \times 3^{-5}}}}$

• Law of exponent used - $\displaystyle{\bf{ \dfrac{a^{m}}{a^{n}} = a^{m-n}}}$

$\longrightarrow{\sf{ 3^{-5-(-5)} \times 2^{-5-(-5)} \times 5^{-2-(-7)}}}$

$\longrightarrow{\sf{ 3^{0} \times 2^{0} \times 5^{-2+7}}}$

• Law of exponent used - $\displaystyle{\bf{a^{0} =1}}$

$\longrightarrow{\sf{ 1\times 1 \times 5^{5}}}$

$\large{:}\implies{\boxed{\sf{\red{3,125}}}}$

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