Respuesta :
Answer:
(a) B ∩ D' = {k}
(b) (B ∪ D)' = {s}
Step-by-step explanation:
Set Notation
[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:
- U (universal) = {f, k, q, s, x, y, z}
- B = {k, q, x, y}
- D = {f, q, x, y, z}
Part (a)
[tex]\begin{aligned}\sf B \cap D' & = \sf B \cap \left( \text{U}-D\right)\\ &= \sf \{k, q, x, y \} \cap \left(\{f, k, q, s, x, y, z \} - \{f, q, x, y, z \}\right)\\& = \sf \{k, q, x, y \} \cap \{ k, s \}\\ & = \sf \{k \}\end{aligned}[/tex]
Part (b)
[tex]\begin{aligned}\sf \left(B \cup D\right)' & = \sf \text{U}-\left(B \cup D\right)\\& = \sf \{f, k, q, s, x, y, z \}-\left(\{k, q, x, y\} \cup \{f, q, x, y, z\}\right)\\& = \sf \{f, k, q, s, x, y, z \}-\{ f, k, q, x, y, z\}\\& = \sf \{s\}\\\end{aligned}[/tex]
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