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

  • A
    $0.2+0.1^{\frac{1}{3}}$
  • B
    $0.1+0.1^{\frac{1}{3}}+0.1^{\frac{1}{2}}$
  • C
    $0.1+0.1^{\frac{4}{3}}+0.1^{\frac{1}{2}}$
  • D
    $0.1+0.1^{\frac{1}{4}}+0.1^{\frac{1}{2}}$

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