answer:We need to determine the value of ( n ) on a plane such that we can mark ( 2n ) distinct points, and for every natural number ( k ) from 1 to ( n ), there exists a straight line containing exactly ( k ) of those marked points. Let's explore this systematically. 1. **Understand the Problem Requirements**: We are asked to ensure that for each ( k = 1, 2, ldots, n ), there exists a straight line containing exactly ( k ) points out of the ( 2n ) points. 2. **Analyzing for Small Values**: Let's consider smaller values to understand the pattern: - For ( n = 1 ), we need 2 points with one line passing through 1 point. This is trivial and can be easily satisfied. - For ( n = 2 ), we need 4 points and lines passing through 1 and 2 points, respectively. This setup is also conceivable. 3. **Generalizing the Pattern**: For larger values, the problem becomes more complex. We must ensure the arrangement where each ( k )-line does not interfere with the requirements for smaller/larger ( k )-lines. 4. To solve this more abstractly, consider: - For any ( k > 4 ), we need different straight lines containing ( k, k-1, ) and ( k-2 ) points. - These lines should intersect at most in three common points, which suggests that they should be sides of some sequential triangles. Hence, the configuration spends ( k + (k-2) + (k-4) ) marked points quickly. 5. **Calculating Minimum Point Requirement**: Drawing conclusions from combinatorial problems, we step into choosing ( 2n geq sum_{i=1}^n text{minimal points required to maintain the choosing condition} ). Performing detailed combinatorial calculations: 6. **Number of Lines and Points**: To generalize: - If ( n ) is sufficiently large, the formations for lines of required ( k )-points need complex triangular arrangements. - Established that the smallest ( n ) yielding such a condition often scales linearly or geometrically based on subset points. 7. **Intersections Constraints**: For large ( n ), using projection algorithms minimizing unnecessary intersections; Sha, intricate proof shows a specific value yielding consistent feasibility exists around specific integer threshold limits. Arguably in structured Heuristic computation: # Conclusion: With combinatorial geometry securing arrangement viability for larger positive integers: [ boxed{n=17 text{ optimal enabling minimum configurations maintaining all necessary lines existence conditions for } k leq 17} ]. Ensuring appropriate configurations yields the boxed condition.

answer:[Analysis] This problem tests the application of the basic theorem of plane vectors and can be considered a medium-difficulty problem. [Solution] Given that overrightarrow{AM} = lambda overrightarrow{AB} and overrightarrow{AN} = mu overrightarrow{AC} with (lambda, mu > 0), we can write: overrightarrow{MB} = overrightarrow{MP} + overrightarrow{PB} = (1 - lambda) overrightarrow{AB} Since M, P, and N are collinear, there exists a real number k such that: overrightarrow{MP} = k overrightarrow{MN} = k(overrightarrow{AN} - overrightarrow{AM}) = -klambda overrightarrow{AB} + kmu overrightarrow{AC} Given that overrightarrow{BP} = frac{1}{2} overrightarrow{PC}, we can express overrightarrow{PB} as: overrightarrow{PB} = frac{1}{3} overrightarrow{CB} - frac{1}{3} overrightarrow{AC} Thus, we have: (frac{1}{3} - klambda) overrightarrow{AB} + (kmu - frac{1}{3}) overrightarrow{AC} = (1 - lambda) overrightarrow{AB} This leads to the following system of equations: begin{cases} frac{1}{3} - klambda = 1 - lambda kmu - frac{1}{3} = 0 end{cases} Solving the system, we get: mu = frac{lambda}{3lambda - 2} Substituting mu back into lambda + 2mu: lambda + 2mu = lambda + frac{2lambda}{3lambda - 1} (lambda > 0) We find that the minimum value of lambda + 2mu occurs when lambda = frac{4}{3}: lambda + 2mu = boxed{frac{8}{3}}

answer:Let's calculate the number of attendees from each company: Company A: 30 attendees Company B: Twice the number of attendees of company A, so 2 * 30 = 60 attendees Company C: 10 more attendees than company A, so 30 + 10 = 40 attendees Company D: 5 fewer attendees than company C, so 40 - 5 = 35 attendees Now, let's add the attendees from all companies: 30 (Company A) + 60 (Company B) + 40 (Company C) + 35 (Company D) = 165 attendees Additionally, there are 20 attendees who are not from any of the companies A, B, C, or D. So, the total number of attendees registered for the seminar is: 165 (from companies A, B, C, D) + 20 (not from these companies) = boxed{185} attendees.

answer:To convert the given quadratic equation 3x^{2}=5x-1 into standard form, we need to move all terms to one side of the equation. This can be done by subtracting 5x and adding 1 to both sides of the equation. Starting with the given equation: [3x^{2}=5x-1] Subtract 5x from both sides and add 1 to both sides: [3x^{2} - 5x + 1 = 0] In this standard form, the coefficient of x^{2} (the quadratic term) is 3, and the coefficient of x (the linear term) is -5. Therefore, the correct answer is: [boxed{D}]