answer:AnalysisThis question tests students' understanding of the properties of arithmetic sequences and their ability to flexibly use the formula for the sum of the first n terms of an arithmetic sequence to simplify and solve for values. It is considered a medium-difficulty question.We can obtain: a_{m-1}+a_{m+1}=2a_m, by substituting into a_{m-1}+a_{m+1}-a_m^2=0, we can solve for the value of the m-th term, and then by substituting into the sum formula with the given information, we can derive the equation for m and solve it.SolutionBased on the properties of an arithmetic sequence, we have: a_{m-1}+a_{m+1}=2a_m, then a_{m-1}+a_{m+1}-a_m^2=a_m(2-a_m)=0, solving this gives: a_m=0 or a_m=2, if a_m=0, it is clear that S_{2m-1}= frac{(2m-1)(a_1+a_{2m-1})}{2} =(2m-1)a_m=18 does not hold,thus, we have a_m=2, therefore S_{2m-1}=(2m-1)a_m=4m-2=18, solving this gives m=5. Therefore, the correct choice is boxed{text{D}}.

answer:Five, the selection method is as follows:1) Select 1500 integers from 1501 to 3000 to form A_{1}.2) From 1 to 1500, remove numbers 751 to 1500 (these numbers multiplied by 2 are in A_{1}), and select 376 to 750, a total of 375 integers to form A_{2}.3) From 1 to 375, remove 188 to 375 (these numbers multiplied by 2 are in A_{2}), and select 94 to 187 from 1 to 187, a total of 94 integers to form A_{3}.4) From 1 to 93, remove 47 to 93 (these numbers multiplied by 2 are in A_{3}), and select 24 to 46 from 1 to 46, a total of 23 integers to form A_{4}.5) From 1 to 23, remove 12 to 23 and select 6 to 11 and 1 (or 2), a total of 7 integers to form A_{5}.Let A=A_{1} cup A_{2} cup A_{3} cup A_{4} cup A_{5}, then the number of elements in A is: 1500+375+94+23+7 =1999 (elements).

answer:Since the distance AB is given by the formula sqrt{b^2 - 4ac}, where a, b, and c are the coefficients of the quadratic equation, for our parabola, this distance is sqrt{k^2 + 2k + 5}. The y-coordinate of the vertex C is -frac{1}{4}(k^2 + 2k + 5).Therefore, the area of triangle ABC is frac{1}{2} times sqrt{k^2 + 2k + 5} times -frac{1}{4}(k^2 + 2k + 5). Since the minimum value of k^2 + 2k + 5 is 4, the minimum area of triangle ABC is boxed{1}.Hence, the correct answer is boxed{text{A}}.

answer:1. Find all pairs (r, s) of real numbers such that the zeros of the polynomialsf(x)=x^{2}-2 r x+randg(x)=27 x^{3}-27 r x^{2}+s x-r^{6}are all real and nonnegative.Answers: (0,0),(1,9)Solution:Let x_{1}, x_{2} be the zeros of f(x), and let y_{1}, y_{2}, y_{3} be the zeros of g(x).By Viete's relation,begin{aligned}x_{1}+x_{2} & =2 r x_{1} x_{2} & =rend{aligned}andbegin{aligned}y_{1}+y_{2}+y_{3} & =r y_{1} y_{2}+y_{2} y_{3}+y_{3} y_{1} & =frac{s}{27} y_{1} y_{2} y_{3} & =frac{r^{6}}{27}end{aligned}Note thatbegin{aligned}left(frac{x_{1}+x_{2}}{2}right)^{2} geq x_{1} x_{2} & Rightarrow r^{2} geq r frac{y_{1}+y_{2}+y_{3}}{3} & geq sqrt[3]{y_{1} y_{2} y_{3}} frac{r}{3} & geq sqrt[3]{frac{r^{6}}{27}} r & geq r^{2}end{aligned}Hence r=r^{2}, and consequently x_{1}=x_{2} and y_{1}=y_{2}=y_{3}. Moreover, r=0,1.- If r=0, then f(x)=x^{2} with x_{1}=x_{2}=0. And since y_{1}=y_{2}=y_{3} with y_{1}+y_{2}+y_{3}=0, then ultimately s=0.- If r=1, then f(x)=x^{2}-2 x+1=(x-1)^{2} with x_{1}=x_{2}=1. And since y_{1}=y_{2}=y_{3} with y_{1}+y_{2}+y_{3}=1 then y_{1}=y_{2}=y_{3}=frac{1}{3}. Therefore s=9.Thus, the possible ordered pairs (r, s) are (0,0) and (1,9).

answer:Answer: 360Let R be the region in question. Then we have[-1,1]^{2} times[-sqrt{2014}, sqrt{2014}] subset R subset[-sqrt[2016]{2016}, sqrt[2016]{2016}]^{2} times[sqrt{2016}, sqrt{2016}]We find some bounds: we havesqrt{2016}355and upper-bounded by2^{2}left(1+frac{0.8}{100}right)^{2} cdot 2(45)=360left(1+frac{6}{1000}right)^{2}<365Thus, R rounded to the nearest ten is 360 .

answer:360