answer:If a neq 0, to ensure that the graph of the function f(x) = ax^2 + bx + c is always above the x-axis, it is necessary to ensure that the parabola opens upwards and does not intersect with the x-axis; Then a > 0 and Delta = b^2 - 4ac < 0. However, if a = 0, when b = 0 and c > 0, the graph of the function f(x) = ax^2 + bx + c = c is always above the x-axis, which does not lead to Delta = b^2 - 4ac < 0; Conversely, "b^2 - 4ac < 0" does not necessarily lead to "the graph of the function f(x) = ax^2 + bx + c is always above the x-axis", such as when a < 0. Therefore, "b^2 - 4ac < 0" is neither a sufficient nor a necessary condition for "the graph of the function f(x) = ax^2 + bx + c is always above the x-axis". Hence, the correct choice is boxed{text{D}}.

answer:Sean started with 27 houses and traded in 8 houses, so he had 27 - 8 = 19 houses left after the trade. After collecting rent, Sean had 31 houses. To find out how many houses he bought after collecting rent, we subtract the number of houses he had left after the trade from the number of houses he had after buying more: 31 (houses after buying more) - 19 (houses left after the trade) = 12 houses. So, Sean bought boxed{12} houses after collecting rent.

answer:The problem requires us to remove 4 matchsticks from the given figure such that a large square and 3 small squares remain. We will provide two distinct solutions to achieve this goal. 1. **First Solution**: The goal is to identify which 4 matchsticks can be removed so that the remaining matchsticks form the desired pattern. - Remove the matchstick at the lower-left corner. - Remove the matchstick directly to the right of the previously removed matchstick. - Remove the matchstick at the lower-right corner. - Remove the matchstick directly above the previously removed matchstick. The resulting figure should leave us with a large square occupying the entire outer boundary and three smaller squares within it. 2. **Second Solution**: We will again identify another possible set of 4 matchsticks that can be removed to reach the same goal. - Remove the matchstick at the lower-left corner. - Remove the matchstick directly above the previously removed matchstick. - Remove the matchstick at the lower-right corner. - Remove the matchstick directly to the left of the previously removed matchstick. With these sets of removals, we end up having one large square and three smaller squares inside. The visuals (drawings or illustrations) of the final figures after removing the specified 4 matchsticks for both solutions can be fronted or shown based on the descriptions provided above. **Conclusion**: ( boxed{text{Two solutions provided as described}} )

answer:Solution: (1) From f(x) leqslant 2, we get |x-a| leqslant 2, Solving this gives a-2 leqslant x leqslant a+2, And since the solution set of the inequality f(x) leqslant 2 is {x|1 leqslant x leqslant 5}, We have begin{cases} a-2=1 a+2=5 end{cases}, Solving this gives a=3; (2) When a=3, f(x)=|x-3|, Let g(x)=f(2x)+f(x+2), Then g(x)=f(2x)+f(x+2)=|2x-3|+|x-1|=begin{cases} 3x-4, & xgeqslant dfrac {3}{2} 2-x, & 1 < x < dfrac {3}{2} -3x+4, & xleqslant 1 end{cases}, Therefore, the minimum value of g(x) is gleft( dfrac {3}{2}right)= dfrac {1}{2}, Hence, when the inequality f(2x)+f(x+2) geqslant m holds true for all real numbers x, The range of values for the real number m is m leqslant dfrac {1}{2}. Thus, the answers are: (1) a=boxed{3}; (2) The range of m is m leqslant boxed{dfrac {1}{2}}.