7) Let \( R \), the set of real numbers be the yeiversal set. If \( A=(-7,8) \cup(11, \infty) \) and \( B=[0,20] \), find the following sets and display them on the number line: Scanned with CamScanner (i) \( \quad A \) ' (ii) \( A \cap B \). (iii) \( (A \cup B)^{2} \).

To tackle the problem, let’s break it down step by step! 1. **Finding \( A' \)**: The complement of the set \( A = (-7,8) \cup (11, \infty) \). Since \( A \) consists of all real numbers between -7 and 8, and all real numbers greater than 11, the complement will include everything outside these intervals. Thus, \( A' = (-\infty, -7] \cup [8, 11] \). 2. **Finding \( A \cap B \)**: This intersection represents the numbers that are in both \( A \) and \( B \). We have \( A = (-7,8) \cup (11, \infty) \) and \( B = [0,20] \). The overlap is only in the interval \( (0, 8) \) since \( (11, \infty) \) does not intersect with \( [0, 20] \). Therefore: \( A \cap B = (0, 8) \). 3. **Finding \( (A \cup B)^2 \)**: First, we find \( A \cup B \). This union combines all components from both sets, resulting in: \( A \cup B = (-7, 8) \cup (11, \infty) \cup [0, 20] = (-7, 20] \cup (11, \infty) \) as \( [0, 20] \) encompasses parts of \( A \). The notation \( (A \cup B)^2 \) usually means we are squaring the elements in the set. Therefore, we take all the elements in the defined intervals and consider their squares, which will be a bit more complex and affect the numerical boundaries. Now, let’s summarize the findings for display on the number line: - \( A' = (-\infty, -7] \cup [8, 11] \) - \( A \cap B = (0, 8) \) - \( (A \cup B)^2 \) results in: \( [0, 400] \) assuming we consider squaring the boundaries inclusive of both intervals from \( A \) and \( B \). **Number Line Representation**: - For \( A' \): It includes segments \( (-\infty, -7] \) and \( [8, 11] \). - For \( A \cap B \): Shown with the segment between (0, 8). - For \( (A \cup B)^2 \): Illustrate from 0 up to 400, showing all numbers squared. Have fun drawing these sets on your number line!