If A = { x : x text{ is an integer and } x^2 | vedclass

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(A) First,we determine the elements of each set: $A = \{ x : x 	ext{ is an integer and } x^2 - 9 = 0 \} = \{ -3, 3 \}$. $B = \{ x : x 	ext{ is a nat

Vedclass · Accepted Answer

$\{ -3, 3, 4 \}$

Vedclass · Answer

(A) First,we determine the elements of each set: $A = \{ x : x 	ext{ is an integer and } x^2 - 9 = 0 \} = \{ -3, 3 \}$. $B = \{ x : x 	ext{ is a natural number and } 2 \leq x $C = \{ x : x 	ext{ is a prime number } \leq 4 \} = \{ 2, 3 \}$. Next,calculate $(B - C)$: $B - C = \{ 2, 3, 4 \} - \{ 2, 3 \} = \{ 4 \}$. Finally,calculate $(B - C) \cup A$: $(B - C) \cup A = \{ 4 \} \cup \{ -3, 3 \} = \{ -3, 3, 4 \}$.

If $A = \{ x : x \text{ is an integer and } x^2 - 9 = 0 \}$,$B = \{ x : x \text{ is a natural number and } 2 \leq x < 5 \}$,and $C = \{ x : x \text{ is a prime number } \leq 4 \}$,then $(B - C) \cup A$ is:

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