Question

solve the absolute value inequality. other than \\( \varnothing \\), use interval notation to express the solution set and graph the solution set on a number line.\\( |x + 5| > 0 \\)\\(\

$$\begin{array}{c}\ \\leftarrow\\hspace{-0.5em}\\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}\\hline\\multicolumn{1}{|c|}{-12}&\\multicolumn{1}{c|}{-10}&\\multicolumn{1}{c|}{-8}&\\multicolumn{1}{c|}{-6}&\\multicolumn{1}{c|}{-4}&\\multicolumn{1}{c|}{-2}&\\multicolumn{1}{c|}{0}&\\multicolumn{1}{c|}{2}&\\multicolumn{1}{c|}{4}&\\multicolumn{1}{c|}{6}&\\multicolumn{1}{c|}{8}&\\multicolumn{1}{c|}{10}&\\multicolumn{1}{c|}{12}\\\\ \\hline\\end{array}$$

\hspace{-0.5em}\
ightarrow\\\\\
\end{array}\\)\\(\bigcirc\\) \\( (-\infty, -5) \cup (-5, \infty) \\)\\(\

$$\begin{array}{c}\ \\leftarrow\\hspace{-0.5em}\\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}\\hline\\multicolumn{1}{|c|}{-12}&\\multicolumn{1}{c|}{-10}&\\multicolumn{1}{c|}{-8}&\\multicolumn{1}{c|}{-6}&\\multicolumn{1}{c|}{-4}&\\multicolumn{1}{c|}{-2}&\\multicolumn{1}{c|}{0}&\\multicolumn{1}{c|}{2}&\\multicolumn{1}{c|}{4}&\\multicolumn{1}{c|}{6}&\\multicolumn{1}{c|}{8}&\\multicolumn{1}{c|}{10}&\\multicolumn{1}{c|}{12}\\\\ \\hline\\end{array}$$

\hspace{-0.5em}\
ightarrow\\\\\
\end{array}\\)\\(\bigcirc\\) \\( (-5, \infty) \\)\\(\

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\hspace{-0.5em}\
ightarrow\\\\\
\end{array}\\)\\(\bigcirc\\) \\( \varnothing \\)\\(\

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\hspace{-0.5em}\
ightarrow\\\\\
\end{array}\\)\\(\bigcirc\\) \\( (-5, 5) \\)\\(\

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\hspace{-0.5em}\
ightarrow\\\\\
\end{array}\\)\\(\bigcirc\\) bad question. \\( c>0 \\)

Explanation:

Step1: Recall absolute value property

For $|A|>0$, $A
eq0$.

Step2: Set expression inside not equal to 0

$x+5
eq0$

Step3: Solve for x

$x
eq-5$

Step4: Write in interval notation

All real numbers except $-5$ is $(-\infty, -5) \cup (-5, \infty)$

Answer:

A. $(-\infty, -5) \cup (-5, \infty)$
(With the corresponding number line that has an open circle at $-5$ and shaded lines extending to both the left and right)