answer:To solve the problem, we need to find the maximum number of intersection points between the circle and the lines, and between the lines themselves: 1. **Intersections between a line and a circle**: Each line can intersect the circle at most at two points. With three lines, the intersection points with the circle are at most (3 times 2 = 6). 2. **Intersections between the lines**: Since no two lines are parallel and no three lines intersect at a single point, each pair of lines intersects exactly once. With three lines, we can form three pairs of lines (1-2, 1-3, 2-3), so there are (3) intersection points among the lines. 3. **Total intersections**: The sum of intersections between the lines and the circle and among the lines themselves is (6 + 3 = 9). Thus, the largest possible number of points of intersection of the circle and the three lines is 9. Conclusion: The problem setup and solution are coherent and valid. The non-parallel condition and the specification that not all three lines meet at one point allow us to uniquely determine the number of intersections. The final answer is boxed{D}.

answer:1. **Renaming and Initial Considerations**: - Rename ( N ) to ( n ). - We need to show that the maximum possible weight of bamboo Panda can get is ( (n-1)^2 ) kilograms. - For ( n = 1 ), ( M = m ), so ( M - m = 0 ). Hence, the statement holds trivially. - Assume ( n geq 2 ). 2. **Constructing a Valid Tree**: - Consider a path graph on ( n ) vertices labeled ( 0 ) to ( n-1 ). - Write the number ( i^2 ) on the ( i )-th vertex. - Here, ( m = 0 ) and ( M = (n-1)^2 ), so ( M - m = (n-1)^2 ). 3. **Verifying the Average Condition**: - For terminal vertices: - ( 0^2 + 1 geq 1^2 ) and ( (n-1)^2 + 1 geq (n-2)^2 ). - For a vertex ( 1 leq i leq n-2 ): [ i^2 + 1 geq frac{(i-1)^2 + (i+1)^2}{2} ] - This simplifies to: [ i^2 + 1 geq frac{i^2 - 2i + 1 + i^2 + 2i + 1}{2} = i^2 + 1 ] - The equality holds, so the construction is valid. 4. **Proving the Upper Bound**: - Root the tree at a vertex labeled ( M ). - Let ( n_u ) denote the size of the subtree of vertex ( u ). - Let ( p_u ) denote the parent of vertex ( u ). - Define ( x_u ) as the real number written on vertex ( u ). 5. **Claim**: - For any vertex ( u ) other than the root, ( x_{p_u} - x_u leq 2n_u - 1 ). 6. **Proof of the Claim**: - **Base Case**: If ( v ) is a leaf: [ x_v + 1 geq x_{p_v} implies x_{p_v} - x_v leq 1 ] - Since ( n_v = 1 ), the bound holds. - **Inductive Step**: Suppose ( v ) has ( k ) children ( v_1, v_2, ldots, v_k ): [ x_v + 1 geq frac{x_{p_v} + sum_{i} x_{v_i}}{k+1} ] - From the inductive hypothesis: [ x_{v_i} geq x_v - 2n_{v_i} + 1 ] - Substituting: [ x_v + 1 geq frac{x_{p_v} + sum_{i} (x_v - 2n_{v_i} + 1)}{k+1} = frac{x_{p_v} + kx_v - sum_{i} (2n_{v_i} - 1)}{k+1} ] - Since ( n_{v_1} + ldots + n_{v_k} = n_v - 1 ): [ x_v + 1 geq frac{x_{p_v} + kx_v + k - 2n_v + 2}{k+1} ] - Simplifying: [ (k+1)x_v geq x_{p_v} + kx_v - 2n_v + 1 implies x_v geq x_{p_v} - 2n_v + 1 ] - Thus, ( x_{p_v} - x_v leq 2n_v - 1 ). 7. **Bounding ( M - m )**: - Let the path from ( M ) to ( m ) be ( M_0 = M, M_1, ldots, M_k = m ). - From the claim: [ x_{M_{i-1}} - x_{M_i} leq 2n_{M_i} - 1 ] - Summing up: [ M - m = M_0 - M_k leq sum_{i=1}^k (2n_{M_i} - 1) ] - Since ( n_{M_i} leq n - i ): [ M - m leq sum_{i=1}^k (2(n-i) + 1) leq sum_{i=1}^n (2(n-i) + 1) = (n-1)^2 ] 8. **Conclusion**: - The maximum weight of bamboo Panda can get is ( (n-1)^2 ) kilograms. The final answer is ( boxed{ (N-1)^2 } ) kilograms.

answer:Since overrightarrow{a}=(4,2) and overrightarrow{b}=(x,3), From overrightarrow{a} parallel overrightarrow{b}, we get 4 times 3 - 2x = 0. Solving for x, we obtain: boxed{x = 6}. We directly use the determinant of the coordinate representation of collinear vectors to solve for the value of x. This question examines the concepts of parallel and collinear vectors, focusing on the coordinate representation of collinear vectors. It is a basic computational problem.

answer:Applying modular arithmetic to each number individually: - 2021 equiv 6 pmod{7} (since 2021 = 289 times 7 + 6) - 2022 equiv 0 pmod{7} (since 2022 = 289 times 7 + 0) - 2023 equiv 1 pmod{7} (since 2023 = 289 times 7 + 1) - 2024 equiv 2 pmod{7} (since 2024 = 289 times 7 + 2) Next, calculate the product modulo 7: [ (2021 cdot 2022 cdot 2023 cdot 2024) equiv (6 cdot 0 cdot 1 cdot 2) pmod{7} ] [ equiv 0 cdot 1 cdot 2 pmod{7} ] [ equiv 0 pmod{7} ] So the result of 2021 cdot 2022 cdot 2023 cdot 2024 mod 7 is boxed{0}.