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Solve Using the Addition Principle $x+\frac{1}{11} \geq \frac{4}{11}$ A) $\left\{x \mid x<\frac{2}{11}\right\}$

Question 123

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Solve using the addition principle. Graph and write set-builder notation for the answer.
- $x+\frac{1}{11} \geq \frac{4}{11}$
$Solve using the addition principle. Graph and write set-builder notation for the answer. - x+\frac{1}{11} \geq \frac{4}{11} A) \left\{x \mid x<\frac{2}{11}\right\} B) \left\{\mathrm{x} \mid \mathrm{x} \leq \frac{3}{11}\right\} C) \left\{x \mid x \geq \frac{3}{11}\right\} D) \left\{\mathrm{x} \mid \mathrm{x}>\frac{2}{11}\right\}$

A) $\left\{x \mid x<\frac{2}{11}\right\}$
$Solve using the addition principle. Graph and write set-builder notation for the answer. - x+\frac{1}{11} \geq \frac{4}{11} A) \left\{x \mid x<\frac{2}{11}\right\} B) \left\{\mathrm{x} \mid \mathrm{x} \leq \frac{3}{11}\right\} C) \left\{x \mid x \geq \frac{3}{11}\right\} D) \left\{\mathrm{x} \mid \mathrm{x}>\frac{2}{11}\right\}$
B) $\left\{\mathrm{x} \mid \mathrm{x} \leq \frac{3}{11}\right\}$
$Solve using the addition principle. Graph and write set-builder notation for the answer. - x+\frac{1}{11} \geq \frac{4}{11} A) \left\{x \mid x<\frac{2}{11}\right\} B) \left\{\mathrm{x} \mid \mathrm{x} \leq \frac{3}{11}\right\} C) \left\{x \mid x \geq \frac{3}{11}\right\} D) \left\{\mathrm{x} \mid \mathrm{x}>\frac{2}{11}\right\}$
C) $\left\{x \mid x \geq \frac{3}{11}\right\}$
$Solve using the addition principle. Graph and write set-builder notation for the answer. - x+\frac{1}{11} \geq \frac{4}{11} A) \left\{x \mid x<\frac{2}{11}\right\} B) \left\{\mathrm{x} \mid \mathrm{x} \leq \frac{3}{11}\right\} C) \left\{x \mid x \geq \frac{3}{11}\right\} D) \left\{\mathrm{x} \mid \mathrm{x}>\frac{2}{11}\right\}$
D) $\left\{\mathrm{x} \mid \mathrm{x}>\frac{2}{11}\right\}$
$Solve using the addition principle. Graph and write set-builder notation for the answer. - x+\frac{1}{11} \geq \frac{4}{11} A) \left\{x \mid x<\frac{2}{11}\right\} B) \left\{\mathrm{x} \mid \mathrm{x} \leq \frac{3}{11}\right\} C) \left\{x \mid x \geq \frac{3}{11}\right\} D) \left\{\mathrm{x} \mid \mathrm{x}>\frac{2}{11}\right\}$

Solve Using the Addition Principle x+111≥411x+\frac{1}{11} \geq \frac{4}{11}x+111​≥114​ A) {x∣x<211}\left\{x \mid x<\frac{2}{11}\right\}{x∣x<112​}

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Solve Using the Addition Principle $x+\frac{1}{11} \geq \frac{4}{11}$ A) $\left\{x \mid x<\frac{2}{11}\right\}$

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