answer:Answer: a1 b3 c2 d5 e6 f4.Solution. Note that the number 6 can be uniquely represented as the sum of three numbers from the set of numbers from 1 to 6, which is 6=1+2+3 (or the same numbers in a different order).Now let's look at the numbers B, D, and E. The maximum value of the sum D+E is the sum 5+6=11, and the maximum value of B is 3. According to the condition, the sum of these numbers is exactly 14. Thus, B=3, D and E are 5 and 6 (or vice versa), and A and C are 1 and 2 (or vice versa). By the method of elimination, we get that F=4.The maximum value of the sum C+E is the sum 6+2=8, and according to the condition, 12=C+E+4. From this, we can uniquely restore the entire arrangement of numbers.

answer:3. Let the areas of the triangles be n, n+1, n+2, n+3. Then the area of the quadrilateral A B C D is 4 n+6. It is easy to see that the area of triangle B C D is four times the area of triangle E C F, so this area is at least 4 n. Therefore,S_{A B D}=S_{A B C D}-S_{B C D} leqslant(4 n+6)-4 n=6 .Equality is achieved if triangle E C F has the smallest areaFig. 151 of the four triangles mentioned.It remains to prove that the value 6 is possible. An example is an isosceles trapezoid with bases A D=6, B C=4 and height 2 (the numbers in Fig. 151 denote the areas of the triangles).Comment. It is easy to see that the area of triangle A B D is always either 2 or 6.

answer:Analysis:This problem primarily examines the negation of a universal statement resulting in an existential statement, which is a fundamental concept.Step-by-step solution:1. Identify the given statement P: forall x in mathbb{R}, x geqslant 2.2. To negate this statement, we need to find a counterexample that disproves the original statement.3. In other words, we need to find an x_0 in mathbb{R} such that x_0 does not satisfy the condition x geqslant 2.4. The correct negation of the statement P is exists {{x}_{0}} in mathbb{R}, x_{0} < 2.Therefore, the correct answer is boxed{B}.

answer:Solution.begin{aligned}& frac{2^{-2}+5^{0}}{(0.5)^{-2}-5(-2)^{-2}+left(frac{2}{3}right)^{-2}}+4.75=frac{frac{1}{2^{2}}+1}{-frac{1}{(0.5)^{2}}-frac{5}{(-2)^{-2}}+left(frac{3}{2}right)^{2}}+4.75= & =frac{frac{1}{4}+1}{frac{1}{0.25}-frac{5}{4}+frac{9}{4}}+4.75=frac{-frac{4}{4}}{4+1}+4.75=frac{1}{4}+4 frac{3}{4}=5 .end{aligned}Answer: 5.

answer: - We need to find the minimum ( N ) such that every possible good group is covered. - Consider the worst-case scenario where each passenger covers a unique pair ((i, j)). The total number of such pairs is given by the binomial coefficient ( binom{2018}{2} ).4. Calculating the Minimum ( N ): - The number of ways to choose 2 stations out of 2018 is: [ binom{2018}{2} = frac{2018 times 2017}{2} = 2037173 ] - However, we need to ensure that for any four stations ( i_1, i_2, j_1, j_2 ), there is a passenger covering either ( (i_1, j_1) ) or ( (i_2, j_2) ). - This implies that we need to cover all possible pairs ((i, j)) with the minimum number of passengers.5. Finding the Minimum ( N ): - To cover all possible pairs, we need to ensure that each pair is covered by at least one passenger. - The minimum number of passengers required to cover all pairs can be found by considering the maximum number of pairs a single passenger can cover. - Each passenger can cover pairs from their boarding station to any de-boarding station after it.6. Conclusion: - The minimum number of passengers ( N ) required to ensure that every good group is covered is: [ N = 1009 ] - This is because each passenger can cover pairs from their boarding station to any de-boarding station after it, and we need to ensure that all pairs are covered.The final answer is ( boxed{1009} ).

answer:(1) + (4) gives 8(w+z)+10(x+y)=0.(2) + (3) gives 10(w+z)+10(x+y)=0.Thus, w+z=0 and x+y=0.Substituting y=-x and z=-w into equations (1) and (2) yields 5 x-4 w=20, w+5 x=-20.Eliminating x gives -5 w=40, i.e., w=-8.From this, we get x=-frac{12}{5}, y=frac{12}{5}, z=8.