Answer :
Let's solve each part step by step.
### Step 1: Finding [tex]\( A' \)[/tex] and [tex]\( B' \)[/tex]
The universal set [tex]\( U \)[/tex] is given by:
[tex]\( U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \)[/tex]
Given sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\( A = \{1, 4, 7, 10\} \)[/tex]
[tex]\( B = \{1, 3, 4, 7, 10\} \)[/tex]
The complement of [tex]\( A \)[/tex], denoted [tex]\( A' \)[/tex], is the set of elements in [tex]\( U \)[/tex] that are not in [tex]\( A \)[/tex]:
[tex]\[ A' = U - A = \{2, 3, 5, 6, 8, 9\} \][/tex]
So,
[tex]\[ A' = \{2, 3, 5, 6, 8, 9\} \][/tex]
The complement of [tex]\( B \)[/tex], denoted [tex]\( B' \)[/tex], is the set of elements in [tex]\( U \)[/tex] that are not in [tex]\( B \)[/tex]:
[tex]\[ B' = U - B = \{2, 5, 6, 8, 9\} \][/tex]
So,
[tex]\[ B' = \{2, 5, 6, 8, 9\} \][/tex]
### Step 2: Finding [tex]\( A \cup B \)[/tex] and [tex]\( A \cap B \)[/tex]
The union of [tex]\( A \)[/tex] and [tex]\( B \)[/tex], denoted [tex]\( A \cup B \)[/tex], is the set containing all elements that are in [tex]\( A \)[/tex], [tex]\( B \)[/tex], or both. Since:
[tex]\[ A = \{1, 4, 7, 10\} \][/tex]
[tex]\[ B = \{1, 3, 4, 7, 10\} \][/tex]
[tex]\[ A \cup B = \{1, 3, 4, 7, 10\} \][/tex]
So,
[tex]\[ A \cup B = \{1, 3, 4, 7, 10\} \][/tex]
The intersection of [tex]\( A \)[/tex] and [tex]\( B \)[/tex], denoted [tex]\( A \cap B \)[/tex], is the set containing all elements that are both in [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ A = \{1, 4, 7, 10\} \][/tex]
[tex]\[ B = \{1, 3, 4, 7, 10\} \][/tex]
[tex]\[ A \cap B = \{1, 4, 7, 10\} \][/tex]
So,
[tex]\[ A \cap B = \{1, 4, 7, 10\} \][/tex]
Thus, the detailed solutions are:
1. [tex]\( A' = \{2, 3, 5, 6, 8, 9\} \)[/tex]
2. [tex]\( B' = \{2, 5, 6, 8, 9\} \)[/tex]
3. [tex]\( A \cup B = \{1, 3, 4, 7, 10\} \)[/tex]
4. [tex]\( A \cap B = \{1, 4, 7, 10\} \)[/tex]
Hence,
[tex]\[ \begin{align*} A^{\prime} &= \{2,3,5,6,8,9\} \quad \checkmark \\ B^{\prime} &= \{2,5,6,8,9\} \quad \checkmark \\ A \cup B &= \{1, 3, 4, 7, 10\} \quad \text{(fill this in \( \square \))} \\ A \cap B &= \{1, 4, 7, 10\} \quad \text{(fill this in \( \square \))} \end{align*} \][/tex]
### Step 1: Finding [tex]\( A' \)[/tex] and [tex]\( B' \)[/tex]
The universal set [tex]\( U \)[/tex] is given by:
[tex]\( U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \)[/tex]
Given sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\( A = \{1, 4, 7, 10\} \)[/tex]
[tex]\( B = \{1, 3, 4, 7, 10\} \)[/tex]
The complement of [tex]\( A \)[/tex], denoted [tex]\( A' \)[/tex], is the set of elements in [tex]\( U \)[/tex] that are not in [tex]\( A \)[/tex]:
[tex]\[ A' = U - A = \{2, 3, 5, 6, 8, 9\} \][/tex]
So,
[tex]\[ A' = \{2, 3, 5, 6, 8, 9\} \][/tex]
The complement of [tex]\( B \)[/tex], denoted [tex]\( B' \)[/tex], is the set of elements in [tex]\( U \)[/tex] that are not in [tex]\( B \)[/tex]:
[tex]\[ B' = U - B = \{2, 5, 6, 8, 9\} \][/tex]
So,
[tex]\[ B' = \{2, 5, 6, 8, 9\} \][/tex]
### Step 2: Finding [tex]\( A \cup B \)[/tex] and [tex]\( A \cap B \)[/tex]
The union of [tex]\( A \)[/tex] and [tex]\( B \)[/tex], denoted [tex]\( A \cup B \)[/tex], is the set containing all elements that are in [tex]\( A \)[/tex], [tex]\( B \)[/tex], or both. Since:
[tex]\[ A = \{1, 4, 7, 10\} \][/tex]
[tex]\[ B = \{1, 3, 4, 7, 10\} \][/tex]
[tex]\[ A \cup B = \{1, 3, 4, 7, 10\} \][/tex]
So,
[tex]\[ A \cup B = \{1, 3, 4, 7, 10\} \][/tex]
The intersection of [tex]\( A \)[/tex] and [tex]\( B \)[/tex], denoted [tex]\( A \cap B \)[/tex], is the set containing all elements that are both in [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ A = \{1, 4, 7, 10\} \][/tex]
[tex]\[ B = \{1, 3, 4, 7, 10\} \][/tex]
[tex]\[ A \cap B = \{1, 4, 7, 10\} \][/tex]
So,
[tex]\[ A \cap B = \{1, 4, 7, 10\} \][/tex]
Thus, the detailed solutions are:
1. [tex]\( A' = \{2, 3, 5, 6, 8, 9\} \)[/tex]
2. [tex]\( B' = \{2, 5, 6, 8, 9\} \)[/tex]
3. [tex]\( A \cup B = \{1, 3, 4, 7, 10\} \)[/tex]
4. [tex]\( A \cap B = \{1, 4, 7, 10\} \)[/tex]
Hence,
[tex]\[ \begin{align*} A^{\prime} &= \{2,3,5,6,8,9\} \quad \checkmark \\ B^{\prime} &= \{2,5,6,8,9\} \quad \checkmark \\ A \cup B &= \{1, 3, 4, 7, 10\} \quad \text{(fill this in \( \square \))} \\ A \cap B &= \{1, 4, 7, 10\} \quad \text{(fill this in \( \square \))} \end{align*} \][/tex]
