[{"id":2518,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生设计了一个圆形花坛，其边缘由一段抛物线形状的装饰带和一段圆弧拼接而成。已知抛物线的顶点在原点，且经过点 (2, -4)，而圆弧所在的圆以原点为圆心，半径为 2。若装饰带与圆弧在点 (2, -4) 处平滑连接，则该抛物线的解析式为（ ）。","answer":"A","explanation":"题目中说明抛物线的顶点在原点，因此可设其解析式为 y = ax²。又已知该抛物线经过点 (2, -4)，代入得：-4 = a × 2² → -4 = 4a → a = -1。因此抛物线的解析式为 y = -x²。虽然题目提到与圆弧连接，但问题仅要求求出抛物线解析式，且点 (2, -4) 确实在 y = -x² 上，而半径为 2 的圆上点 (2, -4) 并不在圆上（因为 2² + (-4)² = 20 ≠ 4），这说明‘平滑连接’在此题中仅为情境设定，不影响抛物线解析式的求解。关键信息是顶点在原点且过 (2, -4)，由此唯一确定解析式为 y = -x²。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:49:55","updated_at":"2026-01-10 15:49:55","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"y = -x²","is_correct":1},{"id":"B","content":"y = -2x²","is_correct":0},{"id":"C","content":"y = -x² + 4","is_correct":0},{"id":"D","content":"y = -2x² + 4","is_correct":0}]},{"id":2318,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某校八年级学生进行体质健康测试，随机抽取了10名学生的1分钟跳绳成绩（单位：次）如下：120, 135, 140, 145, 150, 150, 155, 160, 165, 170。这组数据的中位数和众数分别是多少？","answer":"A","explanation":"首先将数据从小到大排列（已排好）：120, 135, 140, 145, 150, 150, 155, 160, 165, 170。共有10个数据，为偶数个，因此中位数是第5个和第6个数据的平均数，即(150 + 150) ÷ 2 = 150。众数是出现次数最多的数，150出现了两次，其余数均只出现一次，因此众数为150。故正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:47:56","updated_at":"2026-01-10 10:47:56","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"中位数150，众数150","is_correct":1},{"id":"B","content":"中位数147.5，众数150","is_correct":0},{"id":"C","content":"中位数150，众数145","is_correct":0},{"id":"D","content":"中位数147.5，众数145","is_correct":0}]},{"id":516,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"72°","answer":"待完善","explanation":"解析待完善","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:20:15","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":327,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在平面直角坐标系中描出点 A(2, 3) 和点 B(5, 7)，然后他计算了这两点之间的距离。请问他计算出的距离最接近下列哪个数值？","answer":"B","explanation":"根据平面直角坐标系中两点间距离公式：若两点坐标为 (x₁, y₁) 和 (x₂, y₂)，则距离为 √[(x₂ - x₁)² + (y₂ - y₁)²]。将点 A(2, 3) 和点 B(5, 7) 代入公式得：√[(5 - 2)² + (7 - 3)²] = √[3² + 4²] = √[9 + 16] = √25 = 5。因此，两点之间的距离为 5，最接近的选项是 B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:38:53","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"4","is_correct":0},{"id":"B","content":"5","is_correct":1},{"id":"C","content":"6","is_correct":0},{"id":"D","content":"7","is_correct":0}]},{"id":931,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生测量了一个三角形的三条边长，分别为 5 cm、12 cm 和 13 cm。他发现这个三角形是一个直角三角形，因为 5² + 12² = ___。","answer":"13²","explanation":"根据勾股定理，在直角三角形中，两条直角边的平方和等于斜边的平方。题目中给出的三边为 5 cm、12 cm 和 13 cm，其中 5² = 25，12² = 144，25 + 144 = 169，而 13² = 169，因此 5² + 12² = 13²，验证了该三角形为直角三角形。空白处应填写 13²。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 03:01:12","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":869,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生在整理班级同学的课外阅读情况时，发现喜欢阅读小说、科普、漫画的人数分别为12人、8人和10人。若用扇形统计图表示这三类阅读喜好，则代表‘科普’类别的扇形圆心角的度数是____度。","answer":"96","explanation":"首先计算总人数：12 + 8 + 10 = 30人。‘科普’类人数占总人数的比例为8 ÷ 30 = 4\/15。扇形统计图中整个圆为360度，因此‘科普’类对应的圆心角为360 × (4\/15) = 96度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 01:22:07","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":1861,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在平面直角坐标系中绘制了一个四边形ABCD，四个顶点的坐标分别为A(2, 3)、B(5, 7)、C(9, 4)、D(6, 0)。该学生想验证这个四边形是否为平行四边形，并进一步判断它是否为矩形。已知：若一个四边形的对角线互相平分，则它是平行四边形；若平行四边形的对角线长度相等，则它是矩形。请通过计算说明该四边形是否为平行四边形，如果是，再判断它是否为矩形。","answer":"解：\n\n第一步：判断四边形ABCD是否为平行四边形。\n\n根据题意，若对角线互相平分，则四边形为平行四边形。\n\n计算对角线AC和BD的中点坐标：\n\n对角线AC的两个端点为A(2, 3)、C(9, 4)，其中点坐标为：\n((2 + 9)\/2, (3 + 4)\/2) = (11\/2, 7\/2) = (5.5, 3.5)\n\n对角线BD的两个端点为B(5, 7)、D(6, 0)，其中点坐标为：\n((5 + 6)\/2, (7 + 0)\/2) = (11\/2, 7\/2) = (5.5, 3.5)\n\n因为两条对角线的中点相同，均为(5.5, 3.5)，所以对角线互相平分。\n\n因此，四边形ABCD是平行四边形。\n\n第二步：判断该平行四边形是否为矩形。\n\n根据题意，若平行四边形的对角线长度相等，则它是矩形。\n\n计算对角线AC和BD的长度：\n\nAC的长度：\n√[(9 - 2)² + (4 - 3)²] = √[7² + 1²] = √(49 + 1) = √50\n\nBD的长度：\n√[(6 - 5)² + (0 - 7)²] = √[1² + (-7)²] = √(1 + 49) = √50\n\n因为AC...","explanation":"本题综合考查平面直角坐标系中点的坐标、中点公式、两点间距离公式以及平行四边形和矩形的判定定理。解题关键在于：首先利用中点公式验证两条对角线是否互相平分，从而判断是否为平行四边形；若是，则进一步计算两条对角线的长度，若相等，则可判定为矩形。整个过程需要准确进行有理数运算和实数开方，体现了坐标几何与几何性质的综合应用。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 09:39:37","updated_at":"2026-01-07 09:39:37","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":469,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次环保知识问卷调查中，某班级共发放了60份问卷，回收有效问卷54份。请问该问卷的有效回收率是多少？","answer":"B","explanation":"有效回收率的计算公式为：有效回收率 = (有效问卷数量 ÷ 发放问卷总数) × 100%。根据题意，有效问卷为54份，发放总数为60份，因此有效回收率为 (54 ÷ 60) × 100% = 0.9 × 100% = 90%。故正确答案为B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:53:49","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"85%","is_correct":0},{"id":"B","content":"90%","is_correct":1},{"id":"C","content":"95%","is_correct":0},{"id":"D","content":"100%","is_correct":0}]},{"id":129,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"小明在解一个一元一次方程时，将方程 3x + 5 = 20 的解写成了 x = 6。他检查后发现，自己在移项时把常数项 5 移到了等号右边，但忘记变号。如果按照他错误的步骤继续计算，他实际上解的是哪一个方程？","answer":"D","explanation":"小明在解方程 3x + 5 = 20 时，错误地将 +5 移到右边却未变号，即写成了 3x = 20 - 5，而不是正确的 3x = 20 - 5（实际应为 3x = 20 - 5，但此处强调的是他的错误操作逻辑）。虽然结果数值上巧合正确（x=5），但题目问的是他‘实际上解的是哪一个方程’，即他错误操作所对应的方程变形。他写的是 3x = 20 - 5，这等价于原方程为 3x + 5 = 20，但移项错误地写成了减5，因此他实际执行的步骤对应的是将 +5 当作 -5 移项，即他潜意识里解的是 3x = 20 - 5 这个式子，所以正确答案是 D。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-24 08:59:12","updated_at":"2025-12-24 08:59:12","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"3x + 5 = 20","is_correct":0},{"id":"B","content":"3x - 5 = 20","is_correct":0},{"id":"C","content":"3x = 20 + 5","is_correct":0},{"id":"D","content":"3x = 20 - 5","is_correct":1}]},{"id":2516,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生设计了一个圆形花坛，其周长为6π米。现计划在花坛外侧修建一条宽度为1米的环形步道，则这条步道的面积是多少平方米？","answer":"A","explanation":"首先根据圆的周长公式C = 2πr，由已知周长6π米可得：2πr = 6π，解得半径r = 3米。这是花坛的内半径。步道宽1米，因此包含步道后的外圆半径为3 + 1 = 4米。步道的面积等于外圆面积减去内圆面积：π×(4²) - π×(3²) = 16π - 9π = 7π（平方米）。因此正确答案是A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:47:00","updated_at":"2026-01-10 15:47:00","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"7π","is_correct":1},{"id":"B","content":"8π","is_correct":0},{"id":"C","content":"9π","is_correct":0},{"id":"D","content":"10π","is_correct":0}]}]