When $R$ is the set of all real numbers,$\{x \in R: \frac{\sqrt{12-x-x^2}}{x+10} \leq \frac{\sqrt{12-x-x^2}}{2x+9}\} = $

  • A
    $(-4, 1] \cup \{3\}$
  • B
    $[-4, 1]$
  • C
    $[-4, 1] \cup \{3\}$
  • D
    $\phi$,the empty set

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