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The respective numbers of significant figures in 23.023 , 0.0003 , 2.1 × 1 0 3 23.023 , 0.0003 , 2.1\times10^{-3} are A , B , C A, B ,C ,then find the value of A 3 + B 2 + C A^3+B^2+C


The answer is 128.

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1 solution

Akshat Sharda
Oct 8, 2015

23.023 A = 5 23.023\Rightarrow A=5

0.0003 B = 1 0.0003\Rightarrow B=1

2.1 × 1 0 3 C = 2 2.1×10^-3\Rightarrow C=2

A 3 + B 2 + C = 125 + 1 + 2 = 128 \Rightarrow A^{3}+B^{2}+C=125+1+2=\boxed{128}

A damn easy question

Mohit Gupta - 5 years, 8 months ago

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