If |x-2| geq 8,then x in | vedclass

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(C) The given inequality is $|x-2| \geq 8$. By the property of absolute value inequalities,$|u| \geq a$ implies $u \leq -a$ or $u \geq a$. Applying th

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$(-\infty, -6] \cup [10, \infty)$

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(C) The given inequality is $|x-2| \geq 8$. By the property of absolute value inequalities,$|u| \geq a$ implies $u \leq -a$ or $u \geq a$. Applying this to the given inequality: $x - 2 \leq -8$ or $x - 2 \geq 8$. Solving the first part: $x \leq -8 + 2 \implies x \leq -6$. Solving the second part: $x \geq 8 + 2 \implies x \geq 10$. Thus,$x \in (-\infty, -6] \cup [10, \infty)$.

If $|x-2| \geq 8$,then $x \in$

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