初中
数学
中等
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(-3) = 5 + 3 = 8。因此,两天气温相差8℃。本题考查正负数在实际情境中的意义及简单运算,符合七年级学生对正负数应用的理解水平。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 14:27:19","updated_at":"2026-01-09 14:27:19","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":1786,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在平面直角坐标系中绘制了一个四边形ABCD,已知点A的坐标为(0, 0),点B的坐标为(4, 0),点C的坐标为(5, 3),点D的坐标为(1, 3)。该学生想判断这个四边形是否为平行四边形,并计算其面积。以下说法正确的是:","answer":"A","explanation":"首先判断四边形是否为平行四边形。根据坐标,可计算各边向量:向量AB = (4, 0),向量DC = (5-1, 3-3) = (4, 0),故AB与DC平行且相等;向量AD = (1, 3),向量BC = (5-4, 3-0) = (1, 3),故AD与BC也平行且相等。因此两组对边分别平行且相等,四边形ABCD是平行四边形。接着计算面积:可利用底乘高。以AB为底,长度为4,点D到AB(x轴)的垂直距离为3,故面积为4 × 3 = 12。或者用向量叉积法:|AB × AD| = |4×3 - 0×1| = 12。因此正确答案为A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 15:56:29","updated_at":"2026-01-06 15:56:29","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":"四边形ABCD是平行四边形,面积为12平方单位","is_correct":1},{"id":"B","content":"四边形ABCD是平行四边形,面积为10平方单位","is_correct":0},{"id":"C","content":"四边形ABCD不是平行四边形,但面积为12平方单位","is_correct":0},{"id":"D","content":"四边形ABCD不是平行四边形,面积为10平方单位","is_correct":0}]},{"id":2765,"subject":"历史","grade":"七年级","stage":"初中","type":"选择题","content":"唐朝时期,一位外国使节来到长安,看到城内市场繁荣、街道整齐,还有来自不同国家的人穿着各异、使用不同语言交流。他惊叹于唐朝的开放与包容。这种局面最能体现唐朝哪一方面的特点?","answer":"C","explanation":"题目描述的是唐朝都城长安中外人士云集、市场繁荣、文化多元的场景,这直接反映了唐朝对外开放、积极与外国进行经济和文化交流的特点。唐朝实行开明的对外政策,长安作为国际大都市,吸引了大量外国商人、使节和留学生,体现了其文化包容性和中外交流的频繁。选项A、B、D虽然也是唐朝的特点,但与题干中‘外国使节’‘不同国家的人’等关键词不符,因此正确答案为C。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-12 10:40:18","updated_at":"2026-01-12 10:40:18","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":"选项A","is_correct":0},{"id":"B","content":"选项B","is_correct":0},{"id":"C","content":"选项C","is_correct":1},{"id":"D","content":"选项D","is_correct":0}]},{"id":413,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生调查了班级同学每天使用手机的时间(单位:分钟),并将数据整理成如下频数分布表:\n\n| 使用时间区间 | 频数(人数) |\n|---------------|--------------|\n| 0–30 | 8 |\n| 31–60 | 12 |\n| 61–90 | 15 |\n| 91–120 | 10 |\n| 121以上 | 5 |\n\n请问这组数据的中位数最可能落在哪个区间?","answer":"C","explanation":"首先计算总人数:8 + 12 + 15 + 10 + 5 = 50人。中位数是第25和第26个数据的平均值。累计频数:0–30分钟有8人,31–60分钟累计为8+12=20人,61–90分钟累计为20+15=35人。由于第25和第26个数据都落在累计频数超过25的区间,即61–90分钟区间内,因此中位数最可能落在61–90分钟。故正确答案为C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:30:07","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":"A","content":"0–30分钟","is_correct":0},{"id":"B","content":"31–60分钟","is_correct":0},{"id":"C","content":"61–90分钟","is_correct":1},{"id":"D","content":"91–120分钟","is_correct":0}]},{"id":2535,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生在研究二次函数 y = x² - 4x + 3 的图像时,发现该抛物线与x轴有两个交点。若将该抛物线绕其顶点旋转180°,则旋转后的抛物线解析式为( )","answer":"A","explanation":"原函数 y = x² - 4x + 3 可配方为 y = (x - 2)² - 1,其顶点为 (2, -1)。绕顶点旋转180°后,开口方向改变,二次项系数变为相反数,但顶点不变。因此新函数为 y = -(x - 2)² - 1,展开得 y = -x² + 4x - 4 - 1 = -x² + 4x - 5。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 16:28:33","updated_at":"2026-01-10 16:28:33","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 + 4x - 5","is_correct":1},{"id":"B","content":"y = -x² + 4x - 3","is_correct":0},{"id":"C","content":"y = -x² - 4x - 3","is_correct":0},{"id":"D","content":"y = -x² + 4x + 3","is_correct":0}]},{"id":17,"subject":"历史","grade":"初二","stage":"初中","type":"选择题","content":"工业革命首先发生在哪个国家?","answer":"A","explanation":"工业革命首先发生在18世纪的英国。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-08-29 16:33:04","updated_at":"2025-08-29 16:33:04","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":"英国","is_correct":1},{"id":"B","content":"法国","is_correct":0},{"id":"C","content":"德国","is_correct":0},{"id":"D","content":"美国","is_correct":0}]},{"id":1915,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"在一次环保活动中,某班级收集了可回收垃圾和不可回收垃圾共30千克。已知可回收垃圾比不可回收垃圾多6千克,设不可回收垃圾为x千克,则可列出的方程是:","answer":"A","explanation":"题目中设不可回收垃圾为x千克,根据‘可回收垃圾比不可回收垃圾多6千克’,可知可回收垃圾为(x + 6)千克。两者总重量为30千克,因此方程为:x + (x + 6) = 30。选项A正确。选项B错误地将可回收垃圾表示为比不可回收少6千克;选项C忽略了不可回收垃圾的重量;选项D的表达式不符合题意且结果为负数,不合理。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 13:12:36","updated_at":"2026-01-07 13:12:36","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":"x + (x + 6) = 30","is_correct":1},{"id":"B","content":"x + (x - 6) = 30","is_correct":0},{"id":"C","content":"x + 6 = 30","is_correct":0},{"id":"D","content":"x - (x + 6) = 30","is_correct":0}]},{"id":999,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次环保主题活动中,某学生记录了连续5天每天收集的废旧电池数量(单位:节),分别为:-3,5,0,-2,7。这里规定:收集到电池记为正数,丢失或损坏电池记为负数。这5天该学生实际收集的电池总数为___节。","answer":"7","explanation":"题目中给出的数据是有理数,包含正数、负数和零。根据题意,正数表示收集到的电池数量,负数表示丢失或损坏的数量,因此需要将所有数值相加得到净收集量。计算过程为:(-3) + 5 + 0 + (-2) + 7 = (5 + 7) + (-3 - 2) + 0 = 12 - 5 = 7。所以这5天实际收集的电池总数为7节。本题考查有理数的加法运算,结合生活情境,帮助学生理解有理数在实际问题中的意义。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 04:51:06","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":146,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"下列各数中,属于正整数的是( )。","answer":"D","explanation":"正整数是大于0的整数,如1, 2, 3, …。选项A是负整数,选项B是零,既不是正数也不是负数,选项C虽然是正数,但5也是正整数,但题目要求选择‘属于正整数’的一项,D选项2符合定义。注意:虽然C和D都是正整数,但题目为单选题,D为正确答案。此处设计意图是考察学生对正整数概念的理解,2是最典型且无争议的正整数代表。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-24 11:30:06","updated_at":"2025-12-24 11:30:06","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":"-3","is_correct":0},{"id":"B","content":"0","is_correct":0},{"id":"C","content":"5","is_correct":0},{"id":"D","content":"2","is_correct":1}]}]