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1. \(A \cup B = \{1, 2, 3, 4, 5, 6\} = A\) since everything in \(B\) is already in \(A\text{.}\)

2. \(A \cap B = \{2, 4, 6\} = B\) since everything in \(B\) is in \(A\text{.}\)

3. \(B \cap C = \{2\}\) as the only element of both \(B\) and \(C\) is 2.

4. \(A \cap D = \emptyset\) since \(A\) and \(D\) have no common elements.

5. \(\bar{B \cup C} = \{5, 7, 8, 9, 10\}\text{.}\) First we find that \(B \cup C = \{1, 2, 3, 4, 6\}\text{,}\) then we take everything not in that set.

6. \(A \setminus B = \{1, 3, 5\}\) since the elements 1, 3, and 5 are in \(A\) but not in \(B\text{.}\) This is the same as \(A \cap \bar B\text{.}\)

7. \((D \cap \bar C) \cup \bar{A \cap B} = \{1, 3, 5, 7, 8, 9, 10\}\text{.}\) The set contains all elements that are either in \(D\) but not in \(C\) (i.e., \(\{7,8,9\}\)), or not in both \(A\) and \(B\) (i.e., \(\{1,3,5,7,8,9,10\}\)).

8. \(\emptyset \cup C = C\) since nothing is added by the empty set.

9. \(\emptyset \cap C = \emptyset\) since nothing can be both in a set and in the empty set.