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The answer is option D) r < 5 or r > – 1.
I'm going to graph each inequality below on a number line.
A) r > 5 or r > – 1.
[tex]\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r>5~~or~~r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}[/tex]
The result is found just by joining those two intervals together. Actually that compound inequality only implies
r > – 1
which does not represent all real numbers.
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B) r > 5 or r < – 1.
[tex]\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r<-1}&\qquad\mathsf{\overset{~~~*****}{\textsf{|||}}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\textsf{|||}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r>5~~or~~r<-1}&\qquad\mathsf{\overset{~~~*****}{\textsf{|||}}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}[/tex]
Numbers between – 1 and 5 (including them) are not included in the union, so you don't have all real numbers represented there either.
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C) r < 5 or r < – 1.
[tex]\large\begin{array}{cl} \mathsf{r<5}&\qquad\mathsf{\overset{~~~********************}{\textsf{|||}\!\!\!\underset{-1}{\bullet}\!\!\!\textsf{|||||}}\!\!\underset{5}{\circ}\!\!\textsf{|||}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r<-1}&\qquad\mathsf{\overset{~~~*****}{\textsf{|||}}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\textsf{|||}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r<5~~or~~r<-1}&\qquad\mathsf{\overset{~~~********************}{\textsf{|||}\!\!\!\underset{-1}{\bullet}\!\!\!\textsf{|||||}}\!\!\underset{5}{\circ}\!\!\textsf{|||}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}[/tex]
Numbers that are greater or equal to 5 are not in the union. So it does not represent all real numbers.
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D) r < 5 or r > – 1.
[tex]\large\begin{array}{cl} \mathsf{r<5}&\qquad\mathsf{\overset{~~~********************}{\textsf{|||}\!\!\!\underset{-1}{\bullet}\!\!\!\textsf{|||||}}\!\!\underset{5}{\circ}\!\!\textsf{|||}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r<5~~or~~r>-1}&\qquad\mathsf{\overset{~~~**************************~~~}{\textsf{|||}\!\!\!\underset{-1}{\bullet}\!\!\!\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}[/tex]
Now all real numbers are included in the union. So this is the right choice.
Answer: option D) r < 5 or r > – 1.
I hope this helps. =)
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The answer is option D) r < 5 or r > – 1.
I'm going to graph each inequality below on a number line.
A) r > 5 or r > – 1.
[tex]\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r>5~~or~~r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}[/tex]
The result is found just by joining those two intervals together. Actually that compound inequality only implies
r > – 1
which does not represent all real numbers.
—————
B) r > 5 or r < – 1.
[tex]\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r<-1}&\qquad\mathsf{\overset{~~~*****}{\textsf{|||}}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\textsf{|||}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r>5~~or~~r<-1}&\qquad\mathsf{\overset{~~~*****}{\textsf{|||}}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}[/tex]
Numbers between – 1 and 5 (including them) are not included in the union, so you don't have all real numbers represented there either.
—————
C) r < 5 or r < – 1.
[tex]\large\begin{array}{cl} \mathsf{r<5}&\qquad\mathsf{\overset{~~~********************}{\textsf{|||}\!\!\!\underset{-1}{\bullet}\!\!\!\textsf{|||||}}\!\!\underset{5}{\circ}\!\!\textsf{|||}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r<-1}&\qquad\mathsf{\overset{~~~*****}{\textsf{|||}}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\textsf{|||}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r<5~~or~~r<-1}&\qquad\mathsf{\overset{~~~********************}{\textsf{|||}\!\!\!\underset{-1}{\bullet}\!\!\!\textsf{|||||}}\!\!\underset{5}{\circ}\!\!\textsf{|||}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}[/tex]
Numbers that are greater or equal to 5 are not in the union. So it does not represent all real numbers.
—————
D) r < 5 or r > – 1.
[tex]\large\begin{array}{cl} \mathsf{r<5}&\qquad\mathsf{\overset{~~~********************}{\textsf{|||}\!\!\!\underset{-1}{\bullet}\!\!\!\textsf{|||||}}\!\!\underset{5}{\circ}\!\!\textsf{|||}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r<5~~or~~r>-1}&\qquad\mathsf{\overset{~~~**************************~~~}{\textsf{|||}\!\!\!\underset{-1}{\bullet}\!\!\!\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}[/tex]
Now all real numbers are included in the union. So this is the right choice.
Answer: option D) r < 5 or r > – 1.
I hope this helps. =)