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Solve for rr.\newliner+113|r| + 1 \leq 13\newlineWrite a compound inequality like 1 < x < 3 or like x < 1 or x > 3. Use integers, proper fractions, or improper fractions in simplest form.\newline______

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Q. Solve for rr.\newliner+113|r| + 1 \leq 13\newlineWrite a compound inequality like 1<x<31 < x < 3 or like x<1x < 1 or x>3x > 3. Use integers, proper fractions, or improper fractions in simplest form.\newline______
  1. Isolate absolute value expression: First, we need to isolate the absolute value expression on one side of the inequality.\newliner+113|r| + 1 \leq 13\newlineSubtract 11 from both sides to isolate r|r|.\newliner+11131|r| + 1 - 1 \leq 13 - 1\newliner12|r| \leq 12
  2. Consider absolute value definition: Now, we need to consider the definition of absolute value, which states that r)isthedistanceof$r|r| ) is the distance of \$r from 00 on the number line. This means that rr can be either positive or negative, but not greater than 1212 in magnitude.\newlineSo we have two cases:\newline11. r12r \leq 12 (when rr is positive or zero)\newline22. r12-r \leq 12 (when rr is negative)
  3. Solve for negative values: For the second case, we need to solve for rr. To do this, we multiply both sides by 1-1, remembering to reverse the inequality sign because we are multiplying by a negative number.\newliner12-r \leq 12\newline(1)(r)(1)(12)(-1)(-r) \geq (-1)(12)\newliner12r \geq -12
  4. Combine both cases: Combining both cases, we get the compound inequality that represents all possible values of rr.12r12-12 \leq r \leq 12This inequality shows that rr can be any number between 12-12 and 1212, inclusive.

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