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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) \( (A \cup B)^{2} \).

Ask by Greene Bowman. in Zambia
Mar 13,2025

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Tutor-Verified Answer

Answer

\( A' = (-\infty, -7] \cup [8, 11] \cup [11, \infty) \) \( A \cap B = [0, 8) \cup (11, 20] \) \( (A \cup B)^{2} = (49, 400) \cup (121, \infty) \)

Solution

To solve the problem, we will first define the sets \( A \) and \( B \) based on the given information: - \( A = (-7, 8) \cup (11, \infty) \) - \( B = [0, 20] \) Now, let's find the required sets step by step. ### Step 1: Find \( A' \) (the complement of \( A \)) The complement of \( A \) in the universal set \( R \) consists of all real numbers that are not in \( A \). 1. The interval \( (-7, 8) \) includes all numbers between -7 and 8, not including -7 and 8. 2. The interval \( (11, \infty) \) includes all numbers greater than 11. Thus, the complement \( A' \) can be expressed as: - All numbers less than or equal to -7 - All numbers between 8 and 11 (not including 8 and 11) - All numbers greater than or equal to 11 So, we can write: \[ A' = (-\infty, -7] \cup [8, 11] \cup [11, \infty) \] ### Step 2: Find \( A \cap B \) (the intersection of \( A \) and \( B \)) The intersection \( A \cap B \) consists of all elements that are in both \( A \) and \( B \). 1. From \( A = (-7, 8) \cup (11, \infty) \): - The part \( (-7, 8) \) intersects with \( B = [0, 20] \) in the interval \( [0, 8) \). - The part \( (11, \infty) \) intersects with \( B = [0, 20] \) in the interval \( (11, 20] \). Thus, the intersection can be expressed as: \[ A \cap B = [0, 8) \cup (11, 20] \] ### Step 3: Find \( (A \cup B)^{2} \) (the square of the union of \( A \) and \( B \)) First, we need to find \( A \cup B \): 1. The union \( A \cup B \) combines all elements from both sets: - From \( A = (-7, 8) \cup (11, \infty) \) - From \( B = [0, 20] \) The union can be expressed as: \[ A \cup B = (-7, 8) \cup [0, 20] \cup (11, \infty) \] This simplifies to: \[ A \cup B = (-7, 20) \cup (11, \infty) \] Now, squaring the union means we take the union of the intervals and square the endpoints: \[ (A \cup B)^{2} = (-7^2, 20^2) \cup (11^2, \infty^2) = (49, 400) \cup (121, \infty) \] ### Summary of Results 1. \( A' = (-\infty, -7] \cup [8, 11] \cup [11, \infty) \) 2. \( A \cap B = [0, 8) \cup (11, 20] \) 3. \( (A \cup B)^{2} = (49, 400) \cup (121, \infty) \) Now, let's display these sets on the number line. ### Number Line Representation - **For \( A' \)**: - Shade the regions \( (-\infty, -7] \), \( [8, 11] \), and \( [11, \infty) \). - **For \( A \cap B \)**: - Shade the regions \( [0, 8) \) and \( (11, 20] \). - **For \( (A \cup B)^{2} \)**: - Shade the regions \( (49, 400) \) and \( (121, \infty) \). This completes the solution. If you need a visual representation, please let me know!

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Bonus Knowledge

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!

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