Which is equivalent to 5−∣x−2∣≥15 - |x - 2| \ge 15−∣x−2∣≥1?A−2≤x≤6-2 \le x \le 6−2≤x≤6B∣x−2∣≥4|x - 2| \ge 4∣x−2∣≥4CBoth B and Ccheck_circleD∣x−2∣≤4|x - 2| \le 4∣x−2∣≤4Explanation−∣x−2∣≥−4⇒∣x−2∣≤4-|x - 2| \ge -4 \Rightarrow |x - 2| \le 4−∣x−2∣≥−4⇒∣x−2∣≤4, i.e., −4≤x−2≤4⇒−2≤x≤6-4 \le x - 2 \le 4 \Rightarrow -2 \le x \le 6−4≤x−2≤4⇒−2≤x≤6.