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Solve Using the Addition Principle $x-\frac{2}{11} \geq-\frac{8}{11}$ A) $\left\{x \mid x \leq-\frac{6}{11}\right\}$

Question 28

Multiple Choice

Solve using the addition principle. Graph and write set-builder notation for the answer.
- $x-\frac{2}{11} \geq-\frac{8}{11}$
$Solve using the addition principle. Graph and write set-builder notation for the answer. - x-\frac{2}{11} \geq-\frac{8}{11} A) \left\{x \mid x \leq-\frac{6}{11}\right\} B) \left\{|x| x<-\frac{1}{11}\right\} C) \left\{x \mid x \geq-\frac{6}{11}\right\} D) \left\{x \mid x>-\frac{1}{11}\right\}$

A) $\left\{x \mid x \leq-\frac{6}{11}\right\}$
$Solve using the addition principle. Graph and write set-builder notation for the answer. - x-\frac{2}{11} \geq-\frac{8}{11} A) \left\{x \mid x \leq-\frac{6}{11}\right\} B) \left\{|x| x<-\frac{1}{11}\right\} C) \left\{x \mid x \geq-\frac{6}{11}\right\} D) \left\{x \mid x>-\frac{1}{11}\right\}$
B) $\left\{|x| x<-\frac{1}{11}\right\}$
$Solve using the addition principle. Graph and write set-builder notation for the answer. - x-\frac{2}{11} \geq-\frac{8}{11} A) \left\{x \mid x \leq-\frac{6}{11}\right\} B) \left\{|x| x<-\frac{1}{11}\right\} C) \left\{x \mid x \geq-\frac{6}{11}\right\} D) \left\{x \mid x>-\frac{1}{11}\right\}$
C) $\left\{x \mid x \geq-\frac{6}{11}\right\}$
$Solve using the addition principle. Graph and write set-builder notation for the answer. - x-\frac{2}{11} \geq-\frac{8}{11} A) \left\{x \mid x \leq-\frac{6}{11}\right\} B) \left\{|x| x<-\frac{1}{11}\right\} C) \left\{x \mid x \geq-\frac{6}{11}\right\} D) \left\{x \mid x>-\frac{1}{11}\right\}$ D) $\left\{x \mid x>-\frac{1}{11}\right\}$
$Solve using the addition principle. Graph and write set-builder notation for the answer. - x-\frac{2}{11} \geq-\frac{8}{11} A) \left\{x \mid x \leq-\frac{6}{11}\right\} B) \left\{|x| x<-\frac{1}{11}\right\} C) \left\{x \mid x \geq-\frac{6}{11}\right\} D) \left\{x \mid x>-\frac{1}{11}\right\}$

Solve Using the Addition Principle x−211≥−811x-\frac{2}{11} \geq-\frac{8}{11}x−112​≥−118​ A) {x∣x≤−611}\left\{x \mid x \leq-\frac{6}{11}\right\}{x∣x≤−116​}

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Solve Using the Addition Principle $x-\frac{2}{11} \geq-\frac{8}{11}$ A) $\left\{x \mid x \leq-\frac{6}{11}\right\}$

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