Certainly! Let's work through the problem step by step.
1. Identify the sets:
- [tex]\( S = \{4, 9, 14, 19, 24, 29\} \)[/tex]
- [tex]\( E = \{9, 19, 29\} \)[/tex]
- [tex]\( F = \{4, 14, 24\} \)[/tex]
- [tex]\( G^ = \{24, 29\} \)[/tex]
2. Find the intersection of sets [tex]\( E \)[/tex], [tex]\( F \)[/tex], and [tex]\( G^ \)[/tex]:
The intersection of sets [tex]\( E \)[/tex], [tex]\( F \)[/tex], and [tex]\( G^* \)[/tex] is the set of elements that are in all three sets simultaneously.
Let's evaluate:
[tex]\[
E \cap F = \{9, 19, 29\} \cap \{4, 14, 24\} = \emptyset \quad (\text{since there are no common elements})
\][/tex]
Since [tex]\( E \cap F \)[/tex] is empty, any intersection with another set will also be empty:
[tex]\[
E \cap F \cap G^* = \emptyset \cap \{24, 29\} = \emptyset
\][/tex]
Therefore, the intersection [tex]\( E \cap F \cap G^ \)[/tex] is [tex]\(\emptyset\)[/tex].
3. Find the complement of [tex]\( E \cap F \cap G^ \)[/tex] in set [tex]\( S \)[/tex]:
The complement of a set [tex]\( A \)[/tex] in a universal set [tex]\( U \)[/tex] is the set of all elements in [tex]\( U \)[/tex] that are not in [tex]\( A \)[/tex]. Here, our universal set is [tex]\( S \)[/tex], and [tex]\( A \)[/tex] is [tex]\( E \cap F \cap G^* \)[/tex].
So we need to find:
[tex]\[
(E \cap F \cap G^)^C = S \setminus (E \cap F \cap G^)
\][/tex]
Since [tex]\( E \cap F \cap G^* = \emptyset \)[/tex]:
[tex]\[
S \setminus (E \cap F \cap G^*) = S \setminus \emptyset = S
\][/tex]
Therefore, the complement of the intersection is the set [tex]\( S \)[/tex]:
[tex]\[
(E \cap F \cap G^*)^C = \{4, 9, 14, 19, 24, 29\}
\][/tex]
So, to summarize the solutions:
- The intersection [tex]\( E \cap F \cap G^* \)[/tex] is [tex]\(\emptyset\)[/tex].
- The complement of the intersection in set [tex]\( S \)[/tex] is \{4, 9, 14, 19, 24, 29\}.
