0.02 min–1
2.7 min–1
0.063 min–1
6.3 min–1
\(A_0=4250\)
\(A=2250\)
\(A=A_0e^{−λt}\)
\(2250=4250e−λt\)
\(\frac {2250}{4250}=e^{−λt}\)
\(λ(10)=ln (\frac {4250}{2250})\)
\(λ(10)=0.636\)
\(λ=0.063\)
So, the correct option is (C): \(0.063\ min^{-1}\)
If $10 \sin^4 \theta + 15 \cos^4 \theta = 6$, then the value of $\frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta}$ is:
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Read More: Nuclei