If mn=3 and frac{1}{m}+frac{1}{n}=frac{4}{3}, | vedclass

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(A) Given,$mn=3$ and $\frac{1}{m}+\frac{1}{n}=\frac{4}{3}$.From the second equation,$\frac{m+n}{mn}=\frac{4}{3}$.Substituting $mn=3$,we get $\frac{m+n

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$0.2+0.1^{\frac{1}{3}}$

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(A) Given,$mn=3$ and $\frac{1}{m}+\frac{1}{n}=\frac{4}{3}$.From the second equation,$\frac{m+n}{mn}=\frac{4}{3}$.Substituting $mn=3$,we get $\frac{m+n}{3}=\frac{4}{3}$,which implies $m+n=4$.We have the system $m+n=4$ and $mn=3$. The roots of the quadratic equation $x^2-4x+3=0$ are $m$ and $n$.$(x-1)(x-3)=0$,so $m=1, n=3$ or $m=3, n=1$.Now,evaluate the expression $0.1+0.1^{\frac{1}{m}}+0.1^{\frac{1}{n}}$.Substituting $m=1$ and $n=3$,we get $0.1+0.1^1+0.1^{\frac{1}{3}} = 0.1+0.1+0.1^{\frac{1}{3}} = 0.2+0.1^{\frac{1}{3}}$.

If $mn=3$ and $\frac{1}{m}+\frac{1}{n}=\frac{4}{3}$,then the value of $0.1+0.1^{\frac{1}{m}}+0.1^{\frac{1}{n}}$ is

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