answer:Given vectors overrightarrow{m}=(1,a) and overrightarrow{n}=(2b-1,3) where a>0 and b>0, and the dot product overrightarrow{m} cdot overrightarrow{n}=1, we start by calculating the dot product: [ overrightarrow{m} cdot overrightarrow{n} = (1 cdot (2b-1)) + (a cdot 3) = 2b - 1 + 3a = 1 ] This simplifies to: [ 3a + 2b = 2 ] Next, we aim to find the minimum value of frac{1}{a}+frac{2}{b}. We manipulate the equation 3a + 2b = 2 to assist in this: [ frac{1}{a}+frac{2}{b} = frac{1}{2} left( frac{1}{a}+frac{2}{b} right) (3a+2b) ] Expanding the right-hand side, we get: [ = frac{1}{2} left( 3 + frac{2b}{a} + 4 + frac{6a}{b} right) ] This simplifies to: [ = frac{7}{2} + frac{1}{2} left( frac{2b}{a} + frac{6a}{b} right) ] By the AM-GM inequality, we know that: [ frac{frac{2b}{a} + frac{6a}{b}}{2} geqslant sqrt{frac{2b}{a} cdot frac{6a}{b}} ] Simplifying this gives: [ frac{1}{2} left( frac{2b}{a} + frac{6a}{b} right) geqslant 2sqrt{3} ] Thus, we have: [ frac{1}{a}+frac{2}{b} geqslant frac{7}{2} + 2sqrt{3} ] Equality holds when frac{2b}{a} = frac{6a}{b}, which implies b = sqrt{3}a. Therefore, the minimum value of frac{1}{a}+frac{2}{b} is boxed{frac{7}{2}+2sqrt{3}}. Hence, the answer is: boxed{B}.

answer:To determine the number of polynomials in the given expression, we need to understand what a polynomial is. A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. Let's examine each term in the given expression: 1. -7 is a constant polynomial because it can be considered a polynomial of degree 0. 2. x is a polynomial of degree 1. 3. m^{2}+frac{1}{m} is not a polynomial because it contains m in the denominator. 4. x^{2}y+5 is a polynomial. 5. frac{{x+y}}{2} is not a polynomial because it involves division by a variable expression. 6. -5ab^{3}c^{2} is a polynomial. 7. frac{1}{y} is not a polynomial because it contains y in the denominator. Counting the terms that are polynomials, we have: - -7 - x - x^{2}y+5 - -5ab^{3}c^{2} This gives us a total of 4 polynomials. However, this contradicts the provided solution, which states there are 5 polynomials. Upon reevaluation, it's clear that the term frac{{x+y}}{2}, despite involving division, is actually a polynomial because the division is by a constant, not a variable, making it a polynomial as well. Therefore, the correct count is: - -7 - x - x^{2}y+5 - frac{{x+y}}{2} - -5ab^{3}c^{2} This gives us a total of 5 polynomials, which matches the provided solution. Therefore, the correct answer is boxed{C}.

answer:1. **Define the problem in terms of graph theory:** Let ( G(V, E) ) be a graph with ( m+n ) vertices, where each vertex represents a card. We do not know which vertices correspond to cards with 0 or 1 on their back. Placing two cards into the machine is equivalent to drawing an edge between two vertices and labeling that edge 0 or 1 depending on the machine output. 2. **Determine the worst-case scenario:** Suppose we add edges in ( G ) arbitrarily until we find an edge that gets labeled 1. At this point, we know that both vertices incident with that edge correspond to cards with 1 on their back. However, if ( G ) still contains an independent set of size ( n ), we could shuffle the cards around so that the cards with label 1 are on the vertices of the independent set. This implies that all edges in ( G ) would have label 0, meaning those moves won't suffice on this graph. 3. **Find the smallest number of edges in a graph ( G ) that contains no independent set of size ( n ):** This is equivalent to the complement graph ( overline{G} ) having the maximum number of edges and no ( n )-clique. According to Turán's theorem, the Turán graph ( T(n+m, n-1) ) is an ( (n-1) )-partite graph with the size of the parts differing by at most 1. 4. **Calculate the number of edges in the complement of ( T(n+m, n-1) ):** The complement of ( T(n+m, n-1) ) is the disjoint union of ( n-1 ) cliques whose sizes differ by at most 1. If ( n+m = q(n-1) + s ), then this graph has: [ binom{q+1}{2}s + binom{q}{2}(n-s-1) ] edges. The strategy is to split the cards up into ( n-1 ) piles whose sizes differ by at most 1 and test all pairs of cards in each group. 5. **Determine the minimum number of tests required:** To ensure that we get at least one edge labeled 1, we need to test all pairs of cards in each group. The number of tests required is given by the number of edges in the complement of ( T(n+m, n-1) ). The final answer is ( boxed{ binom{q+1}{2}s + binom{q}{2}(n-s-1) } ).

answer:Part (a): 1. **Problem Recap and Given Facts**: - We need to construct the triangle (ABC) knowing the three points (A', B', C'), where the bisectors of its angles intersect the circumcircle of (ABC). Both triangles are given to be acute. 2. **Reference to Relevant Theorem**: - As per Problem 2.19 (a), the points (A, B), and (C) are the points of intersection of the extensions of the altitudes of the triangle (A'B'C') with its circumcircle. 3. **Steps to Construct Triangle (ABC)**: - **Step 1**: Draw the triangle (A'B'C') with the given points (A', B', C'). - **Step 2**: Draw the circumcircle of (A'B'C') using the perpendicular bisectors of the sides (A'B'), (B'C'), and (C'A'). The point where these perpendicular bisectors intersect is the center of the circumcircle. - **Step 3**: Extend the altitudes from each vertex across the opposite side. Each of these extended lines will intersect the circumcircle at two points. - **Step 4**: Identify the appropriate intersection points as (A, B), and (C) such that they form a triangle (ABC) with the given conditions. Conclusion for Part (a): Given the correct construction steps, we will achieve the desired triangle (ABC). ( blacksquare ) Part (b): 1. **Problem Recap and Given Facts**: - Construct the triangle (ABC) given points (A', B', C'), where the altitudes intersect the circumcircle of (ABC). Both triangles are given to be acute. 2. **Reference to Relevant Theorem**: - As per Problem 2.19 (b), the points (A, B), and (C) are the points of intersection of the extensions of the bisectors of the angles of the triangle (A'B'C') with its circumcircle. 3. **Steps to Construct Triangle (ABC)**: - **Step 1**: Draw the triangle (A'B'C') with the given points (A', B', C'). - **Step 2**: Draw the circumcircle of (A'B'C') using the perpendicular bisectors of the sides (A'B'), (B'C'), and (C'A'). The point where these perpendicular bisectors intersect is the center of the circumcircle. - **Step 3**: Extend the angle bisectors from each vertex across the opposite side. Each of these extended lines will intersect the circumcircle at two points. - **Step 4**: Identify the appropriate intersection points as (A, B), and (C) such that they form a triangle (ABC) with the given conditions. Conclusion for Part (b): Given the correct construction steps, we will achieve the desired triangle (ABC). ( blacksquare )