Questions: Sets (A, B), and (C) are subsets of the universal set (U). These sets are defined as follows. (U=f, g, h, p, q, r, x, y, z) (A=g, h, q, r, z) (B=q, r, x, y, z) (C=h, p, q, y) Find ((B cap A)^prime cup C). Write your answer in roster form or as (varnothing). ((B cap A) cup C=)

Sets (A, B), and (C) are subsets of the universal set (U). These sets are defined as follows.
(U=f, g, h, p, q, r, x, y, z)
(A=g, h, q, r, z)
(B=q, r, x, y, z)
(C=h, p, q, y)

Find ((B cap A)^prime cup C).
Write your answer in roster form or as (varnothing).
((B cap A) cup C=)

Transcript text: Sets $A, B$, and $C$ are subsets of the universal set $U$. These sets are defined as follows. \[ \begin{array}{l} U=\{f, g, h, p, q, r, x, y, z\} \\ A=\{g, h, q, r, z\} \\ B=\{q, r, x, y, z\} \\ C=\{h, p, q, y\} \end{array} \] Find $(B \cap A)^{\prime} \cup C$. Write your answer in roster form or as $\varnothing$. \[ (B \cap A) \cup C= \]

Solution

Solution Steps

Step 1: Find $ B \cap A $

The intersection of sets $ B $ and $ A $ is the set of elements that are common to both $ B $ and $ A $. From the given sets: \[ B = \{q, r, x, y, z\}, \quad A = \{g, h, q, r, z\} \] The common elements are $ q, r, z $. Thus: \[ B \cap A = \{q, r, z\} \]

Step 2: Find $ (B \cap A)^{\prime} $

The complement of $ B \cap A $, denoted as $ (B \cap A)^{\prime} $, is the set of all elements in the universal set $ U $ that are not in $ B \cap A $. The universal set $ U $ is: \[ U = \{f, g, h, p, q, r, x, y, z\} \] Since $ B \cap A = \{q, r, z\} $, the complement is: \[ (B \cap A)^{\prime} = \{f, g, h, p, x, y\} \]

Step 3: Find $ (B \cap A)^{\prime} \cup C $

The union of $ (B \cap A)^{\prime} $ and $ C $ is the set of all elements that are in $ (B \cap A)^{\prime} $ or $ C $ or both. From Step 2 and the given set $ C $: \[ (B \cap A)^{\prime} = \{f, g, h, p, x, y\}, \quad C = \{h, p, q, y\} \] The union is: \[ (B \cap A)^{\prime} \cup C = \{f, g, h, p, x, y, q\} \]