answer:To solve the problem, we follow these steps closely related to the given solution:1. Identify the geometric solid formed by rotating the right triangle ABC around line AC. Since angle C=90^{circ}, rotating the triangle around AC forms a cone.2. Determine the dimensions of the cone. The height of the cone is equal to the length of side AC, which is 2. The radius of the base of the cone is equal to the length of side BC, which is 1.3. Calculate the volume of the cone using the formula for the volume of a cone, V = frac{1}{3} pi r^2 h, where r is the radius of the base, and h is the height of the cone. Substituting r = 1 and h = 2 into the formula gives: [ V = frac{1}{3} pi cdot 1^2 cdot 2 = frac{1}{3} pi cdot 2 = frac{2pi}{3} ]Therefore, the volume of the resulting geometric solid, which is a cone, is boxed{frac{2pi}{3}}.Hence, the correct answer is boxed{C}.

answer:AnalysisThis question examines the normal vectors of planes, involving the positional relationship between planes, which is a basic question. By the operation of the dot product, we can find that the dot product is (0), which means the normal vectors are perpendicular. Therefore, the planes are perpendicular.SolutionGiven the problem, we have ((1,2,0) cdot (2,-1,0) = 1 times 2 - 2 times 1 + 0 times 0 = 0),Therefore, the normal vectors of the two planes are perpendicular, which means the positional relationship between plane (alpha) and plane (beta) is perpendicular.Thus, the correct choice is boxed{C}.

answer:[Solution] First, consider the following problem: If there must be 3 students among the group such that for each question, at least 2 of the 3 have the same answer, what is the minimum number of students who should participate in the exam?Obviously, as long as there are 4 people participating in the exam, it can be ensured that two people have the same answer to the first question. Secondly, as long as there are 5 people participating, it can be guaranteed that 4 people have only two different answers to the second question, thus 3 people can be selected so that their answers to the first and second questions have at most two different answers. Furthermore, by the pigeonhole principle, as long as there are 7 people participating, it can be guaranteed that 5 people have only two different answers to the third question. Finally, by the pigeonhole principle, as long as there are 10 people participating in the exam, it can be ensured that 7 people have only two different answers to the fourth question. Thus, 3 people can definitely be selected whose answers to each question have at most two different answers. This indicates that the number of people participating in the exam is at most 9.For the case where 9 people participate in the exam, the following table gives a satisfactory answer sheet scheme: In summary, the maximum number of people participating in the exam is 9.

answer:1. Answer: 8 pencils.

answer:## Solutiony^{prime}=left(x^{2} arccos x-frac{x^{2}+2}{3} sqrt{1-x^{2}}right)^{prime}==2 x cdot arccos x+x^{2} cdot frac{-1}{sqrt{1-x^{2}}}-frac{2 x}{3} sqrt{1-x^{2}}-frac{x^{2}+2}{3} frac{1}{2 sqrt{1-x^{2}}} cdot(-2 x)==2 x cdot arccos x-frac{x^{2}}{sqrt{1-x^{2}}}-frac{2 xleft(1-x^{2}right)}{3 sqrt{1-x^{2}}}+frac{left(x^{2}+2right) x}{3 sqrt{1-x^{2}}}==2 x cdot arccos x-frac{3 x^{2}+2 x-2 x^{3}-x^{3}-2 x}{3 sqrt{1-x^{2}}}==2 x cdot arccos x-frac{3 x^{2}-3 x^{3}}{3 sqrt{1-x^{2}}}=2 x cdot arccos x-frac{x^{2}(1-x)}{sqrt{1-x^{2}}}==2 x cdot arccos x-x^{2} cdot sqrt{frac{1-x}{1+x}}## Problem Kuznetsov Differentiation 14-26

answer:7. D.Since the average of the seven numbers is x, thenbegin{array}{l}frac{60+100+x+40+50+200+90}{7}=x Rightarrow 7 x=540+x Rightarrow x=90 .end{array}Substituting x=90 we get40, 50, 60, 90, 90, 100, 200 text {. }At this point, 90 is also the median, so x=90.