Answer:
[tex]\sf (A' \cap B) \cup (A' \cap C') =\{1,6,7\}[/tex]
Step-by-step explanation:
[tex]\begin{array}{|c|c|l|} \cline{1-3} \sf Symbol & \sf N\:\!ame & \sf Meaning \\\cline{1-3} \{ \: \} & \sf Set & \sf A\:collection\:of\:elements\\\cline{1-3} \cup & \sf Union & \sf A \cup B=elements\:in\:A\:or\:B\:(or\:both)}\\\cline{1-3} \cap & \sf Intersection & \sf A \cap B=elements\: in \:both\: A \:and \:B} \\\cline{1-3} \sf ' \:or\: ^c & \sf Complement & \sf A'=elements\: not\: in\: A \\\cline{1-3} \sf - & \sf Difference & \sf A-B=elements \:in \:A \:but\: not\: in \:B}\\\cline{1-3} \end{array}[/tex]
Given sets:
- Universal = {1, 2, 3, 4, 5, 6, 7, 8}
- A = {2, 4, 5, 8}
- B = {1, 4, 6}
- C = {1, 2, 3, 4, 5}
Therefore, the complement sets are:
[tex]\begin{aligned}\sf A' & = \text{U}-\sf A\\& = \{1,2,3,4,5,6,7,8 \}- \{2,4,5,8 \}\\& = \{1,3,6,7 \}\end{aligned}[/tex]
[tex]\begin{aligned}\sf B' & = \text{U}-\sf B\\& = \{1,2,3,4,5,6,7,8 \}- \{1,4,6 \}\\& = \{2,3,5,7,8 \}\end{aligned}[/tex]
[tex]\begin{aligned}\sf C' & = \text{U}-\sf C\\& = \{1,2,3,4,5,6,7,8 \}- \{1,2,3,4,5 \}\\& = \{6,7,8 \}\end{aligned}[/tex]
Solution
[tex]\begin{aligned}\sf (A' \cap B) \cup (A' \cap C') & = \sf \left(\{1,3,6,7 \} \cap \{1,4,6\} \right) \cup \left(\{1,3,6,7 \} \cap \{6, 7, 8 \} \right)\\\\& = \sf \{1,6\} \cup \{6,7 \} \\\\& = \sf \{1,6,7\} \end{aligned}[/tex]
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