某学生计算一个多边形的内角和时,误将其中一个内角重复加了一次,得到的结果是1440度。这个多边形正确的内角和应该是______度。
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首先,全班共有40名学生。成绩在80分及以上的学生占总人数的3/8,因此人数为:40 × 3/8 = 15人。题目说明成绩在60分到79分之间的学生比80分及以上的多10人,所以该分数段人数为:15 + 10 = 25人。那么,成绩低于60分的学生人数为总人数减去前两个分数段的人数:40 - 15 - 25 = 5人。因此,正确答案是A。
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[{"id":2242,"content":"某学生在数轴上从原点出发,先向右移动5个单位,再向左移动8个单位,然后向右移动3个单位,最后向左移动6个单位。此时该学生所在位置表示的数是___。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"困难","answer":"-6","explanation":"根据正负数在数轴上的移动规则,向右为正,向左为负。起始位置为0,第一次移动+5,第二次移动-8,第三次移动+3,第四次移动-6。计算过程为:0 + 5 - 8 + 3 - 6 = (5 + 3) - (8 + 6) = 8 - 14 = -6。因此最终位置表示的数是-6。","options":[]},{"id":2412,"content":"某学生在研究两个三角形时发现,△ABC 和 △DEF 中,∠A = ∠D,AB = DE,且 ∠B = ∠E。若他想证明这两个三角形全等,应使用以下哪个判定定理?此外,若 AC = 5 cm,BC = 7 cm,∠C = 60°,则根据全等性质,DF 的长度应为多少?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"中等","answer":"A","explanation":"题目中给出 ∠A = ∠D,AB = DE,∠B = ∠E,即两个角和它们的夹边分别相等,符合 ASA(角-边-角)全等判定定理。由于 AB 是 ∠A 与 ∠B 的夹边,对应边 DE 是 ∠D 与 ∠E 的夹边,因此 △ABC ≌ △DEF(ASA)。根据全等三角形的性质,对应边相等,AC 对应 DF,已知 AC = 5 cm,故 DF = 5 cm。因此正确答案为 A。","options":[{"id":"A","content":"ASA,DF = 5 cm"},{"id":"B","content":"AAS,DF = 7 cm"},{"id":"C","content":"SAS,DF = 5 cm"},{"id":"D","content":"ASA,DF = 7 cm"}]},{"id":935,"content":"在一次班级视力情况调查中,共收集了40名学生的视力数据。其中,视力在4.8及以上的学生有25人,视力低于4.8的有___人。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"15","explanation":"题目考查的是数据的收集与整理。总人数为40人,已知视力在4.8及以上的有25人,要求视力低于4.8的人数,只需用总人数减去已知部分:40 - 25 = 15。因此,视力低于4.8的学生有15人。","options":[]},{"id":628,"content":"某次环保活动中,某班学生收集废旧纸张和塑料瓶进行回收。已知每3千克废旧纸张和每2千克塑料瓶可兑换15元环保基金。如果该班共收集了9千克废旧纸张和6千克塑料瓶,那么他们可以兑换多少元环保基金?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"B","explanation":"根据题意,每3千克废旧纸张和2千克塑料瓶可兑换15元。观察所收集的数量:9千克废旧纸张是3千克的3倍,6千克塑料瓶是2千克的3倍,说明收集的总量正好是基本兑换单位的3倍。因此,兑换金额为15元 × 3 = 45元。本题考查学生对比例关系的理解与简单整数倍的应用,属于有理数在实际问题中的简单运用。","options":[{"id":"A","content":"30元"},{"id":"B","content":"45元"},{"id":"C","content":"60元"},{"id":"D","content":"75元"}]},{"id":631,"content":"某班级组织了一次环保知识竞赛,共收集了120份有效问卷。在整理数据时,发现有15%的学生选择了‘垃圾分类’作为最关注的环保问题,有40人选择了‘节约用水’,其余学生选择了‘减少塑料使用’。请问选择‘减少塑料使用’的学生人数是多少?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"C","explanation":"首先计算选择‘垃圾分类’的学生人数:120 × 15% = 120 × 0.15 = 18人。已知选择‘节约用水’的有40人。那么选择‘减少塑料使用’的人数为总人数减去前两项:120 - 18 - 40 = 62人。因此正确答案是C。本题考查数据的收集与整理,涉及百分数的基本计算和简单减法运算,符合七年级数学中‘数据的收集、整理与描述’知识点,难度为简单。","options":[{"id":"A","content":"52"},{"id":"B","content":"58"},{"id":"C","content":"62"},{"id":"D","content":"68"}]},{"id":428,"content":"86.2分","type":"选择题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"待完善","explanation":"解析待完善","options":[]},{"id":2292,"content":"某学生测量了一块直角三角形纸片的两条直角边,分别为5 cm和12 cm。若要用一根细线沿着纸片的边缘完整绕一圈,所需细线的最短长度是多少?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"题目要求计算直角三角形纸片的周长,即三条边之和。已知两条直角边分别为5 cm和12 cm,首先利用勾股定理求斜边:斜边 = √(5² + 12²) = √(25 + 144) = √169 = 13 cm。因此,周长为 5 + 12 + 13 = 30 cm。所需细线的最短长度即为周长,故正确答案为A。","options":[{"id":"A","content":"30 cm"},{"id":"B","content":"25 cm"},{"id":"C","content":"17 cm"},{"id":"D","content":"13 cm"}]},{"id":306,"content":"某学生在平面直角坐标系中描出三个点 A(2, 3)、B(5, 3) 和 C(4, 6),然后连接这三个点形成一个三角形。若将该三角形向下平移 4 个单位长度,则点 C 的新坐标是?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"A","explanation":"在平面直角坐标系中,将一个点向下平移 4 个单位长度,意味着其纵坐标减少 4,横坐标保持不变。点 C 的原坐标是 (4, 6),向下平移 4 个单位后,纵坐标变为 6 - 4 = 2,因此新坐标为 (4, 2)。选项 A 正确。其他选项中,B 是向上平移,C 和 D 改变了横坐标或方向错误,均不符合平移规则。","options":[{"id":"A","content":"(4, 2)"},{"id":"B","content":"(4, 10)"},{"id":"C","content":"(8, 6)"},{"id":"D","content":"(0, 6)"}]},{"id":143,"content":"已知一个三角形的两边长分别为5cm和8cm,第三边的长度可能是以下哪个值?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"D","explanation":"根据三角形三边关系定理:任意两边之和大于第三边,任意两边之差小于第三边。设第三边为x,则需满足:8 - 5 < x < 8 + 5,即3 < x < 13。选项中只有10cm在这个范围内,因此正确答案是D。","options":[{"id":"A","content":"3cm"},{"id":"B","content":"4cm"},{"id":"C","content":"13cm"},{"id":"D","content":"10cm"}]},{"id":2221,"content":"某学生在记录一周内每天气温变化时,发现某天的气温比前一天上升了5℃,记作+5℃;第二天又下降了3℃,记作-3℃。如果这两天的温度变化总和用正负数表示,那么这两天的总变化是___℃。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"2","explanation":"根据正负数表示相反意义的量,温度上升记为正,下降记为负。两天的变化分别为+5℃和-3℃,总变化为+5 + (-3) = 2℃,因此答案是2。","options":[]}]