When R is the set of all real numbers,{x in R | vedclass

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(C) The given inequality is $\frac{\sqrt{12-x-x^2}}{x+10} \leq \frac{\sqrt{12-x-x^2}}{2x+9}$.First,for the square root to be defined,we must have $12-

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$[-4, 1] \cup \{3\}$

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(C) The given inequality is $\frac{\sqrt{12-x-x^2}}{x+10} \leq \frac{\sqrt{12-x-x^2}}{2x+9}$.First,for the square root to be defined,we must have $12-x-x^2 \geq 0$,which implies $x^2+x-12 \leq 0$,so $(x+4)(x-3) \leq 0$,giving $x \in [-4, 3]$.Also,the denominators must be non-zero: $x+10 
eq 0 \implies x 
eq -10$ and $2x+9 
eq 0 \implies x 
eq -4.5$.Case $1$: If $12-x-x^2 = 0$,then $x = -4$ or $x = 3$. Both satisfy the inequality $0 \leq 0$.Case $2$: If $12-x-x^2 > 0$,we can divide by $\sqrt{12-x-x^2}$:$\frac{1}{x+10} \leq \frac{1}{2x+9} \implies \frac{1}{x+10} - \frac{1}{2x+9} \leq 0 \implies \frac{2x+9-x-10}{(x+10)(2x+9)} \leq 0 \implies \frac{x-1}{(x+10)(2x+9)} \leq 0$.Using the sign scheme for $\frac{x-1}{(x+10)(2x+9)} \leq 0$ with $x \in (-4, 3)$:The critical points are $-10, -4.5, 1$. Within $(-4, 3)$,the relevant interval is $(-4, 1]$.Combining Case $1$ and Case $2$,we get $x \in [-4, 1] \cup \{3\}$.

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}\} = $

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