If the domain of the function f(x) = frac{sqr | vedclass

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(C) The function is $f(x) = \frac{\sqrt{x^2-25}}{4-x^2} + \log_{10}(x^2+2x-15)$.For the square root term,$x^2-25 \geq 0 \Rightarrow x \in (-\infty, -5

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$150$

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(C) The function is $f(x) = \frac{\sqrt{x^2-25}}{4-x^2} + \log_{10}(x^2+2x-15)$.For the square root term,$x^2-25 \geq 0 \Rightarrow x \in (-\infty, -5] \cup [5, \infty)$.For the denominator,$4-x^2 
eq 0 \Rightarrow x 
eq \pm 2$.For the logarithm,$x^2+2x-15 > 0$ $\Rightarrow (x+5)(x-3) > 0$ $\Rightarrow x \in (-\infty, -5) \cup (3, \infty)$.Taking the intersection of all conditions:$x \in ((-\infty, -5] \cup [5, \infty)) \cap (-\infty, -5) \cup (3, \infty) \cap \mathbb{R} \setminus \{-2, 2\}$.This simplifies to $x \in (-\infty, -5) \cup [5, \infty)$.Comparing with $(-\infty, \alpha) \cup [\beta, \infty)$,we get $\alpha = -5$ and $\beta = 5$.Therefore,$\alpha^2 + \beta^3 = (-5)^2 + 5^3 = 25 + 125 = 150$.

If the domain of the function $f(x) = \frac{\sqrt{x^2-25}}{4-x^2} + \log_{10}(x^2+2x-15)$ is $(-\infty, \alpha) \cup [\beta, \infty)$,then $\alpha^2 + \beta^3$ is equal to:

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