某学生在平面直角坐标系中绘制了一个三角形,其三个顶点分别为 A(2, 3)、B(5, -1)、C(-1, -1)。若将该三角形沿 x 轴方向平移 _ 个单位长度后,点 A 的对应点 A' 恰好落在 y 轴上,则平移的单位长度为 ___。
💡 提示:点击下方 "查看答案" 查看解析,或 "提交答案" 后自动显示结果
众数是一组数据中出现次数最多的数。观察数据:2.5 出现了两次,3.0、2.8、3.2、3.1、2.9 各出现一次。因此,2.5 是出现次数最多的数,即这组数据的众数是 2.5。
🏆
练习完成!
恭喜您完成了本次练习,继续加油提升!
💡 学习建议:您在一元一次方程的应用方面掌握良好,但仍有提升空间。建议重点复习方程求解步骤和实际应用问题。
[{"id":2151,"content":"在一次数学测验中,某学生解答一道关于一元一次方程的题目时,列出了方程:3x + 5 = 20。该方程的解表示的意义是:某数的三倍加上5等于20,那么这个数是多少?解这个方程得到的正确结果是:","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"B","explanation":"解方程 3x + 5 = 20,首先两边同时减去5,得到 3x = 15,然后两边同时除以3,得到 x = 5。因此,这个数是5,对应选项B。该题考查一元一次方程的基本解法,符合七年级数学课程内容。","options":[{"id":"A","content":"4"},{"id":"B","content":"5"},{"id":"C","content":"6"},{"id":"D","content":"7"}]},{"id":584,"content":"某学生在整理班级同学的课外阅读时间时,随机抽取了30名学生进行调查,发现每天阅读时间在0.5小时到1.5小时之间。他将这些数据分为5组,并制作了频数分布表。若每组组距相同,则每组的组距是多少小时?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"B","explanation":"题目中给出的数据范围是从0.5小时到1.5小时,因此全距为1.5 - 0.5 = 1.0小时。将数据分为5组,且每组组距相同,则组距 = 全距 ÷ 组数 = 1.0 ÷ 5 = 0.2小时。因此正确答案是B选项。","options":[{"id":"A","content":"0.1"},{"id":"B","content":"0.2"},{"id":"C","content":"0.3"},{"id":"D","content":"0.4"}]},{"id":156,"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一个?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"B","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。已知两边为5cm和8cm,则第三边x应满足:8 - 5 < x < 8 + 5,即3 < x < 13。选项中只有5cm在这个范围内,因此正确答案是B。","options":[{"id":"A","content":"3cm"},{"id":"B","content":"5cm"},{"id":"C","content":"13cm"},{"id":"D","content":"15cm"}]},{"id":517,"content":"在一次环保活动中,某班级收集了废旧纸张共120千克。第一周收集了总量的1\/3,第二周收集了剩余部分的1\/2。请问第二周收集了多少千克废旧纸张?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"C","explanation":"首先,第一周收集的废旧纸张为总量的1\/3,即120 × 1\/3 = 40千克。剩余部分为120 - 40 = 80千克。第二周收集了剩余部分的1\/2,即80 × 1\/2 = 40千克。因此,第二周收集了40千克,正确答案是C。","options":[{"id":"A","content":"20千克"},{"id":"B","content":"30千克"},{"id":"C","content":"40千克"},{"id":"D","content":"60千克"}]},{"id":400,"content":"某学生在整理班级同学的课外阅读情况时,统计了每人每周阅读课外书的平均时间(单位:小时),并将数据分为5组,绘制成频数分布直方图。已知前四组的频数分别为3、7、10、5,第五组的频率为0.2,则该班级参与调查的学生总人数是多少?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"B","explanation":"题目考查的是数据的收集、整理与描述中的频数与频率概念。已知第五组的频率为0.2,即第五组人数占总人数的20%。设总人数为x,则第五组人数为0.2x。前四组频数之和为3 + 7 + 10 + 5 = 25,因此总人数为前四组人数加上第五组人数:25 + 0.2x = x。解这个方程:25 = x - 0.2x → 25 = 0.8x → x = 25 ÷ 0.8 = 30。所以参与调查的学生总人数是30人。","options":[{"id":"A","content":"25"},{"id":"B","content":"30"},{"id":"C","content":"35"},{"id":"D","content":"40"}]},{"id":2216,"content":"某学生在记录一周气温变化时,发现某天的气温比前一天下降了5℃,记作-5℃。如果第二天的气温又比当天上升了8℃,那么第二天的气温变化应记作____℃。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"3","explanation":"题目中气温先下降5℃,记作-5℃,第二天又上升8℃,即进行加法运算:-5 + 8 = 3。因此第二天的气温变化应记作+3℃,通常简写为3℃。这体现了正负数在表示相反意义的量时的实际应用,符合七年级学生对正负数加减运算的理解水平。","options":[]},{"id":178,"content":"小明去文具店买笔记本,每本笔记本的价格是8元。他买了3本,付给收银员50元,应找回多少钱?","type":"选择题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"B","explanation":"首先计算3本笔记本的总价:8元\/本 × 3本 = 24元。小明付了50元,所以应找回的钱为:50元 - 24元 = 26元。因此正确答案是B选项。本题考查的是基本的整数乘法与减法运算,符合七年级数学中关于有理数运算的实际应用要求。","options":[{"id":"A","content":"24元"},{"id":"B","content":"26元"},{"id":"C","content":"34元"},{"id":"D","content":"42元"}]},{"id":1060,"content":"在一次班级环保活动中,某学生收集了废旧纸张和塑料瓶共12件,其中废旧纸张比塑料瓶多4件。设塑料瓶的数量为x件,则根据题意可列出一元一次方程:_x + (x + 4) = 12_,解得x = _4_,因此塑料瓶有_4_件,废旧纸张有_8_件。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"x + (x + 4) = 12;4;4;8","explanation":"设塑料瓶数量为x件,则废旧纸张数量为x + 4件。根据总数量为12件,可列方程x + (x + 4) = 12。解这个方程:2x + 4 = 12 → 2x = 8 → x = 4。因此塑料瓶有4件,废旧纸张有4 + 4 = 8件。本题考查一元一次方程的建立与求解,符合七年级数学课程内容。","options":[]},{"id":2146,"content":"某学生在解方程时,将方程 2x + 3 = 9 的解题步骤写为:第一步,两边同时减去3,得到 2x = 6;第二步,两边同时除以2,得到 x = 3。这名学生使用的解方程依据是___。","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"B","explanation":"该学生在解方程过程中,第一步使用了等式的基本性质:两边同时减去3,保持等式成立;第二步两边同时除以2(不为0),也符合等式的基本性质。因此正确依据是选项B所描述的内容。选项C和D虽然也是方程变形中的方法,但不是本题中直接体现的依据。","options":[{"id":"A","content":"等式两边同时加上同一个数,等式仍然成立"},{"id":"B","content":"等式两边同时减去同一个数,等式仍然成立,且等式两边同时除以同一个不为0的数,等式仍然成立"},{"id":"C","content":"移项时符号要改变"},{"id":"D","content":"合并同类项法则"}]},{"id":339,"content":"20","type":"选择题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"答案待完善","explanation":"解析待完善","options":[]}]