answer:To determine the range of values for k that allows the quadratic equation kx^{2}-4x+2=0 to have real roots, we use the discriminant condition from the quadratic formula. The discriminant Delta must be non-negative (Delta geqslant 0) for the equation to have real roots. The discriminant of a quadratic equation ax^2 + bx + c = 0 is given by Delta = b^2 - 4ac. For the given equation, a = k, b = -4, and c = 2. Plugging these values into the discriminant formula gives:[Delta = (-4)^2 - 4 cdot k cdot 2 geqslant 0]Simplifying this inequality:[16 - 8k geqslant 0]Adding 8k to both sides and then subtracting 16 from both sides gives:[8k leqslant 16]Dividing both sides by 8 yields:[k leqslant 2]However, we must also consider that for a quadratic equation, a neq 0, which means k neq 0 in this context. Combining these two conditions, we find that k must be less than or equal to 2 but not equal to 0.Therefore, the correct answer is boxed{D}.

answer:## Solutionbegin{aligned}& int_{0}^{1} frac{x}{x^{4}+1} d x=frac{1}{2} cdot int_{0}^{1} frac{dleft(x^{2}right)}{x^{4}+1}=frac{1}{2} cdot int_{0}^{1} frac{dleft(x^{2}right)}{x^{4}+1}=left.frac{1}{2} cdot operatorname{arctg} x^{2}right|_{0} ^{1}= & =frac{1}{2} cdot operatorname{arctg} 1^{2}-frac{1}{2} cdot operatorname{arctg} 0^{2}=frac{1}{2} cdot frac{pi}{4}-frac{1}{2} cdot 0=frac{pi}{8}end{aligned}Source — "http://pluspi.org/wiki/index.php/������������ %D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD % mathrm{D} 1 % 82 % mathrm{D} 0 % mathrm{~B} 5 % mathrm{D} 0 % mathrm{~B} 3 % mathrm{D} 1 % 80 % mathrm{D} 0 % mathrm{~B} 0 % mathrm{D} 0 % mathrm{BB} % mathrm{D} 1 % 8 mathrm{~B} 4-9 "Categories: Kuznetsov Problem Book Integrals Problem 4 | IntegralsUkrainian Banner Network- Last edited on this page: 16:43, 3 March 2009.- Content is available under CC-BY-SA 3.0.Created by GeeTeatoo## Problem Kuznetsov Integrals 4-10## Material from PlusPi

answer:To determine the specific location based on the given statements, let's analyze each option step by step:- Option A: The statement "The third row of Classroom 3 in Class 7" specifies a location within a particular classroom in a school. However, without knowing the specific school or its location, we cannot determine the exact geographical location. Therefore, option A cannot determine the specific location.- Option B: "People's East Road, Kunming City" indicates a road within Kunming City. While it provides a general area within the city, it does not pinpoint a specific geographical point because a road can span several kilometers. Thus, option B cannot determine the specific location.- Option C: The direction "Southward and westward by 45^{circ}" gives a relative direction from an unspecified starting point. Without a starting point or distance, this information cannot identify a specific location on the globe. Hence, option C cannot determine the specific location.- Option D: The coordinates "East longitude 102^{circ}, north latitude 24^{circ}" provide a precise geographical location on the Earth. These coordinates specify a unique point that can be identified on a map, meeting the requirement for determining a specific location. However, there seems to be a discrepancy in the solution provided. The solution mentions "East longitude 118^{circ}, north latitude 51^{circ}" which is different from the coordinates given in the problem statement. Assuming the correct coordinates are those given in the problem statement, "East longitude 102^{circ}, north latitude 24^{circ}" indeed specifies a unique location.Therefore, based on the analysis and correcting the discrepancy in the solution, the correct answer is:boxed{D} East longitude 102^{circ}, north latitude 24^{circ} can determine the specific location, so this option meets the requirements of the question.

answer:1. We start by noting that the mean of the five integers is 100. Therefore, the sum of the five integers is: [ 5 times 100 = 500 ]2. When the median is removed, the mean of the remaining four integers increases by 5. Thus, the new mean is: [ 100 + 5 = 105 ] Therefore, the sum of the remaining four integers is: [ 4 times 105 = 420 ]3. The number that was removed (the median) must be: [ 500 - 420 = 80 ]4. Let the five integers be (a, b, 80, c, d) where (a < b < 80 < c < d). After removing 80, the remaining integers are (a, b, c, d).5. We are given that the median of the remaining four numbers is now 75. Since the median of four numbers is the average of the two middle numbers, we have: [ frac{b + c}{2} = 75 implies b + c = 150 ]6. We need to maximize (d). To do this, we should minimize the other numbers. The smallest positive integer is 1, so let (a = 1).7. Now, we have the equation for the sum of the remaining four numbers: [ a + b + c + d = 1 + 150 + d = 420 ] Solving for (d): [ 151 + d = 420 implies d = 420 - 151 = 269 ]Conclusion:[d = boxed{269}]

answer:34560

answer:Solution: Let 1996 points be colored with n colors, where each point is colored one of the n colors, and let the number of points of each color be m_{1}, m_{2}, m_{3}, m_{4}, cdots, m_{n}. Without loss of generality, assume m_{1} geq 1, and m_{i} - m_{i-1} leq 2 (i=2,3, cdots, n), otherwise, if m_{i} - m_{i-1} geq 3, take m^{prime}_{i-1} = m_{i-1} + 1, m^{prime}_{i} = m_{i} - 1, then m^{prime}_{i-1} cdot m^{prime}_{i} = (m_{i-1} + 1)(m_{i} - 1) = m_{i-1} cdot m_{i} + m_{i} - m_{i-1} - 1 > m_{i-1} cdot m_{i}. This implies that S cannot be maximized.Also, m_{i} - m_{i-1} = 2 can occur at most once, otherwise, if m_{i} - m_{i-1} = 2, m_{j} - m_{j-1} = 2 (1 < i < j leq n), then m_{i-1} + 2 + m_{j-1} = m_{i} + m_{j}, but m_{i-1} cdot 2 cdot m_{j-1} > m_{i} cdot m_{j}, which means such an S cannot be maximized.We have m_{1} leq 3, otherwise, if m_{1} geq 5, then 2 + (m_{1} - 2) = m_{1}, but 2(m_{1} - 2) = m_{1} + (m_{1} - 4) > m_{1}, thus S' = 2(m_{1} - 2) cdot m_{2} cdot m_{3} cdots cdot m_{n} > m_{2} cdot m_{1} cdots cdot m_{n} = S, i.e., S is not maximized. If m_{1} = 4, then m_{2} = 5 or 6. When m_{1} = 4, m_{2} = 5, 2 + 3 + 4 = m_{1} + m_{2}, but 2 times 3 times 4 > m_{1} cdot m_{2}; when m_{1} = 4, m_{2} = 6, 2 + 3 + 5 = m_{1} + m_{2}, but 2 times 3 times 5 > m_{1} cdot m_{2}; therefore, when m_{1} geq 4, S cannot be maximized, i.e., 2 leq m_{1} leq 3.The sum of n consecutive numbers starting with 2 or 3 cannot be 1996. Otherwise, if the first term is 2, then frac{1}{2} cdot n cdot (n + 3) = 1996, i.e., n(n + 3) = 8 times 499, and n and n + 3 have different parities and n < 499. If the first term is 3, then m_{1} + m_{2} + cdots + m_{i-1} + m_{i} + cdots + m_{n} = 2 + m_{1} + m_{2} + cdots + m_{i-1} + [m_{i+1} - 2] + m_{i} + m_{i+2} + m_{i+3} + cdots + m_{n}, since 2(m_{i+1} - 2) = m_{i+1} + (m_{i+1} - 4) > m_{i+1}, then 2 m_{1} cdot m_{2} cdots cdot m_{i-1} [m_{i+1} - 2] cdot m_{i} cdot m_{i+2} cdots cdot m_{n} > m_{1} cdot m_{2} cdots cdot m_{n}, i.e., if m_{1} = 3, S cannot be maximized. Therefore, to maximize S, we must have m_{1} = 2, and m_{i} - m_{i-1} = 2 occurs exactly once, and the rest m_{j} - m_{j-1} = 1, then we can set m_{1}, m_{2}, cdots, m_{n} as 2, 3, 4, cdots, i-1, i+1, i+2, cdots, n+1, n+2, their sum is frac{(1+n)(n+4)}{2} - i, thus (1+n)(n+4) = 2 times 1996 + 2i (i < n), solving this gives n = 61, i = 19. Therefore, 1996 points can be colored with 61 different colors, the number of points of each color being 2, 3, 4, cdots, 18, 20, 21, cdots, 61, 62, 63, at this point the polygon is a 61-sided polygon, and the number is maximized.