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Quadratic Equations

TS EAMCET 2020 (Online) 14th September Evening Shift

MCQ+1 / -02020

When $\mathbf{R}$ is the set of all real numbers,


\(\left\{x \in \mathbf{R}: \frac{\sqrt{12-x-x^2}}{x+10} \leq \frac{\sqrt{12-x-x^2}}{2 x+9}\right\}=\)


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