Question

Please solve this :- (0.0010×0.004^-3)÷(4^2+3^2)^0

Answer 1

$(0.0010 \times 0.004^{- 3}) \div (4^2 + 3^2)^0 \\ \\ = (0.0010 \times (\frac{4}{1000})^{- 3}) \div (16 + 9)^0 \\ \\ = (0.001 \times (\frac{1000}{4})^3) \div 25^0 \\ \\ = (\frac{1}{1000} \times \frac{1000}{4} \times \frac{1000}{4} \times \frac{1000}{4}) \div 25^0 \\ \\ = (\frac{1}{4} \times \frac{1000}{4} \times \frac{1000}{4}) \div 25^0 \\ \\ = ((\frac{1}{2})^2 \times (\frac{1000}{4})^2) \div 25^0 \\ \\ = ((\frac{1}{2} \times \frac{1000}{4})^2) \div 25^0 \\ \\ = ((\frac{1000}{8})^2) \div 25^0$

$\\ \\ = 125^2 \div 25^0 \\ \\ = 15625 \div 1 \\ \\ = 15625$

15625 is the answer.

Done on the basis of BODMAS.

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