The set A = { x: x in R, x^2 = 16 text{ and } | vedclass

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(A) The set $A$ is defined by the intersection of two conditions: $x^2 = 16$ and $2x = 6$.First,solving $x^2 = 16$ gives $x = 4$ or $x = -4$.Second,so

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$\phi$

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(A) The set $A$ is defined by the intersection of two conditions: $x^2 = 16$ and $2x = 6$.First,solving $x^2 = 16$ gives $x = 4$ or $x = -4$.Second,solving $2x = 6$ gives $x = 3$.Since there is no value of $x$ that satisfies both conditions simultaneously,the set $A$ contains no elements.Therefore,$A = \phi$ (the empty set).

The set $A = \{ x: x \in R, x^2 = 16 \text{ and } 2x = 6 \}$ equals

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