answer:First, let's find out how much work Y does in one day. Since Y can finish the work in 16 days, Y's work rate is 1/16 of the work per day. Now, let's calculate how much work Y has done in 12 days. Y's work for 12 days is: 12 * (1/16) = 12/16 = 3/4 of the work. This means that Y has completed 3/4 of the work, and there is 1/4 of the work remaining. Now, let's find out how much work X does in one day. Since X can finish the work in 20 days, X's work rate is 1/20 of the work per day. To find out how many days X needs to finish the remaining 1/4 of the work, we divide the remaining work by X's work rate: (1/4) / (1/20) = (1/4) * (20/1) = 20/4 = 5 days. So, X alone needs boxed{5} days to finish the remaining work.

answer:Since f(x) is an odd function and is increasing on (-infty, 0), it follows that f(x) is also increasing on (0, +infty). Since f(-1) = 0, it implies f(1) = 0. Therefore, the inequality f(x) > 0 is equivalent to: 1. When x > 0, f(x) > f(1), which means x > 1; 2. When x < 0, f(x) > f(-1), which means -1 < x < 0; Combining the above, the range of x is (-1, 0) cup (1, +infty). Hence, the answer is boxed{(-1, 0) cup (1, +infty)}. Based on the odd function f(x) being increasing on (-infty, 0) and f(-1) = 0, we can determine the monotonicity and zero point of the function f(x) on (0, +infty), thereby converting the inequality f(x) > 0 into a variable inequality using the monotonicity of the function. This problem examines the odd-even properties and monotonicity of functions, and the conversion of inequalities based on the monotonicity of functions, reflecting the idea of transformation and the concept of case discussion, making it a medium-level question.

answer:Since A={-1, -2, 0}, and B={-3, -4, 0}, Therefore, C_UA={-3, -4}, Thus, (C_UA) cap B={-3, -4}. Hence, the answer is boxed{text{B}}.

answer:``` Problem: 9, 甲, 乙, 丙, 丁与小强五位同学一起比赛象棋， 每两个人都比赛一盘。到现在为止， 甲已经赛了 4 盘, 乙赛了 3 盘, 丙赛了 2 盘, 丁赛了 1 盘。 问小强赛了几盘? Solution: ``` : 1. Since "甲已经赛了 4 盘", it means 甲 has played 4 games with 乙, 丙, 丁, and 小强 respectively (i.e., 甲 has played one game with each of them). 2. Considering "丁赛了 1 盘", it implies that 丁 has only played with 甲. 3. Given that "乙赛了 3 盘", we deduce that 乙 has played with 甲, 丙, and 小强 each once (i.e., 小强 has played 1 game with 乙). 4. Since "丙赛了 2 盘", we deduce that 丙 has played with 甲 and 乙 each once. Given this, we have accounted for the number of games played by all except for 小强: - 甲 has played with 乙, 丙, 丁, and 小强. - 乙 has played with 甲, 丙, and 小强. - 丙 has played with 甲 and 乙. - 丁 has only played with 甲. 5. Now, for 小强, to find the total number of games: - 小强 has played with 甲 (one game), - 小强 has played with 乙 (one game), - 小强 should also play with 丙 to sum up to 2 games (since these are the remaining reasonable and unplayed matches left). Conclusion: Thus, 小强 has played 2 games in total. boxed{2} ```