Question

1. ( 1 < - 1 + r leq 8 )
2. ( - 1 < \frac { b } { 8 } leq 0 )
3. ( \frac { b } { 4 } > - 1 ) and ( b + 2 leq 11 )
4. ( 4 + b geq 8 ) or ( b + 7 < 6 )
5. ( - 12 < x - 4 leq 3 )
6. ( \frac { n } { 4 } > 1 ) or ( \frac { n } { 8 } leq - 1 )
7. ( 7 + x > - 3 ) and ( 5 + x < 1 )
8. ( 9 b > 36 ) or ( b - 2 leq - 3 )
9. ( - 1 < b - 1 < 4 )
10. ( 2 k leq 4 ) or ( k - 6 > 0 )
11. ( \frac { v } { 5 } < 0 ) and ( - 10 + v geq - 14 )

Explanation:

Let's solve problem 15: \(1 < -1 + r \leq 8\)

Step 1: Add 1 to all parts

To isolate \(r\), we add 1 to each part of the compound inequality.
\(1 + 1 < -1 + r + 1 \leq 8 + 1\)
\(2 < r \leq 9\)

Answer:

The solution to the inequality \(1 < -1 + r \leq 8\) is \(2 < r \leq 9\). On the number line, we would mark an open circle at 2 (since \(r > 2\)) and a closed circle at 9 (since \(r \leq 9\)) and shade the region between them.