Given the set, A = {-4, -2, 0, 2, 4, 6}, what are the members of x in the proposition,(∃𝑥 ∈ 𝐴)(𝑥 + 2 ≤ 5)

The proposition (∃𝑥 ∈ 𝐴)(𝑥 + 2 ≤ 5) translates to "There exists an x in set A such that x + 2 is less than or equal to 5".

To find the members of x that satisfy this condition, we need to subtract 2 from both sides of the inequality, which gives us x ≤ 3.

Now, we compare this result with each member of set A to find the members that satisfy the condition.

-4 + 2 = -2, which is less than 3. So, -4 is a member. -2 + 2 = 0, which is less than 3. So, -2 is a member. 0 + 2 = 2, which is less than 3. So, 0 is a member. 2 + 2 = 4, which is less than 3. So, 2 is not a member. 4 + 2 = 6, which is less than 3. So, 4 is not a member. 6 + 2 = 8, which is less than 3. So, 6 is not a member.

Therefore, the members of x in the proposition (∃𝑥 ∈ 𝐴)(𝑥 + 2 ≤ 5) are -4, -2, and 0.