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the venn diagram shows sets a, b, c, and the universal set u. shade (c …

Question

the venn diagram shows sets a, b, c, and the universal set u. shade (c ∩ a) ∪ b on the venn diagram.

Explanation:

Step1: Find \( C \cap A \)

The intersection \( C \cap A \) is the region common to both set \( C \) and set \( A \). In the Venn diagram, this is the area where circles \( A \) and \( C \) overlap (excluding the part that also overlaps with \( B \) for now, but just the core overlap of \( A \) and \( C \)).

Step2: Find \( (C \cap A)' \)

The complement \( (C \cap A)' \) includes all elements in the universal set \( U \) that are not in \( C \cap A \). So we shade everything except the region we found in Step 1.

Step3: Find \( (C \cap A)' \cup B \)

The union \( (C \cap A)' \cup B \) includes all elements that are in \( (C \cap A)' \) or in set \( B \) (or both). So we take the shaded region from Step 2 and add (or keep, since union includes both) the entire region of set \( B \). Visually, this means:

  • The entire circle \( B \) (including its overlaps with \( A \) and \( C \)) is shaded.
  • Additionally, any part of \( U \) that is not in \( C \cap A \) (i.e., not in the overlap of \( A \) and \( C \) only) and not already in \( B \) is also shaded. So outside of \( B \), we shade everywhere except the small region that is only the overlap of \( A \) and \( C \) (not including \( B \)'s overlap with them).

Answer:

To shade \( (C \cap A)' \cup B \):

  1. Shade the entire circle \( B \) (including its overlaps with \( A \) and \( C \)).
  2. Shade all regions in \( U \) except the region that is the intersection of \( A \) and \( C \) but not in \( B \) (the small "lens" - shaped area where only \( A \) and \( C \) overlap, excluding \( B \)'s overlap with them).

(Note: Since this is a visual task, the key is to include all of \( B \) and all parts of \( U \) not in \( C \cap A \) (non - \( B \) parts outside \( A \cap C \) and non - \( A \cap C \) parts within \( B \) are already included as \( B \) is fully shaded).)