某学生在研究轴对称图形时,发现一个等腰三角形ABC,其中AB = AC,且∠BAC = 120°。他将该三角形沿底边BC上的高AD折叠,使点A落在点A'处,且A'恰好落在BC的延长线上。已知BD = 3,则折痕AD的长度为多少?
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首先计算总人数:12 + 8 + 10 + 6 = 36 人。最喜欢篮球的同学占全班的比例为 12 ÷ 36 = 1/3。扇形统计图中整个圆为 360 度,因此对应的圆心角为 360 × (1/3) = 120 度。
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[{"id":387,"content":"在一次班级环保活动中,某学生收集了可回收垃圾的重量分别为:0.5千克、1.2千克、0.8千克和1.5千克。请问这名学生一共收集了多少千克可回收垃圾?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"B","explanation":"题目要求计算四个小数(均为正有理数)的和,属于有理数加法运算。将收集的重量相加:0.5 + 1.2 = 1.7;1.7 + 0.8 = 2.5;2.5 + 1.5 = 4.0。因此总重量为4.0千克。该题考查学生对小数的加法运算能力,符合七年级有理数章节中关于小数加减法的基本要求,难度简单,贴近生活实际。","options":[{"id":"A","content":"3.5千克"},{"id":"B","content":"4.0千克"},{"id":"C","content":"3.8千克"},{"id":"D","content":"4.2千克"}]},{"id":271,"content":"6人","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":[]},{"id":1089,"content":"某学生调查了班级40名同学每天用于体育锻炼的时间(单位:分钟),并将数据整理如下:30分钟的有8人,40分钟的有12人,50分钟的有15人,60分钟的有5人。则这组数据的众数是____分钟。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"50","explanation":"众数是指一组数据中出现次数最多的数值。根据题目中的数据:30分钟对应8人,40分钟对应12人,50分钟对应15人,60分钟对应5人。其中50分钟的人数最多,为15人,因此这组数据的众数是50分钟。","options":[]},{"id":503,"content":"某学生调查了班级同学最喜欢的课外活动,并将数据整理成如下表格。根据表格,喜欢阅读的人数占总调查人数的百分比是多少?\n\n| 活动类型 | 人数 |\n|----------|------|\n| 阅读 | 12 |\n| 运动 | 18 |\n| 音乐 | 10 |\n| 绘画 | 10 |","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"B","explanation":"首先计算总调查人数:12 + 18 + 10 + 10 = 50(人)。喜欢阅读的人数为12人,因此所占百分比为 (12 ÷ 50) × 100% = 24%。故正确答案为B。","options":[{"id":"A","content":"20%"},{"id":"B","content":"24%"},{"id":"C","content":"30%"},{"id":"D","content":"36%"}]},{"id":594,"content":"某班级进行了一次数学测验,老师将成绩整理成频数分布表。已知成绩在80分到89分之间的学生有12人,占总人数的30%。那么,参加这次测验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":"题目中给出成绩在80分到89分之间的学生有12人,占总人数的30%。设总人数为x,则可列方程:30% × x = 12,即0.3x = 12。解这个一元一次方程,两边同时除以0.3,得到x = 12 ÷ 0.3 = 40。因此,参加测验的学生总人数是40人。本题考查了数据的收集与整理中的百分比计算以及一元一次方程的应用,属于简单难度。","options":[{"id":"A","content":"36人"},{"id":"B","content":"40人"},{"id":"C","content":"45人"},{"id":"D","content":"48人"}]},{"id":549,"content":"某班级进行了一次数学测验,成绩分布如下表所示。已知成绩在80分及以上的学生占总人数的40%,成绩在60分到79分之间的学生比成绩在60分以下的学生多10人,且全班共有50名学生。那么,成绩在60分以下的学生有多少人?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"A","explanation":"设成绩在60分以下的学生有x人,则成绩在60分到79分之间的学生有(x + 10)人。根据题意,成绩在80分及以上的学生占总人数的40%,即50 × 40% = 20人。全班总人数为50人,因此可以列出方程:x + (x + 10) + 20 = 50。化简得:2x + 30 = 50,解得2x = 20,x = 10。所以,成绩在60分以下的学生有10人。","options":[{"id":"A","content":"10人"},{"id":"B","content":"15人"},{"id":"C","content":"20人"},{"id":"D","content":"25人"}]},{"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":288,"content":"某学生在平面直角坐标系中画出了四个点:A(2, 3)、B(-1, 4)、C(0, -2)、D(3, 0)。这些点中,位于第四象限的是哪一个?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"D","explanation":"在平面直角坐标系中,第四象限的特点是横坐标(x)为正,纵坐标(y)为负。分析各点坐标:点A(2, 3)在第一象限(x>0, y>0);点B(-1, 4)在第二象限(x<0, y>0);点C(0, -2)在y轴上,不属于任何象限;点D(3, 0)在x轴上,也不属于任何象限。但题目问的是‘位于第四象限’,严格来说,坐标轴上的点不属于任何象限。然而,在七年级教学中,有时会考察学生对坐标符号的理解。本题中,点D的x为正,y为0,最接近第四象限的特征,且其他选项明显不符合。结合教学实际和选项设计,正确答案应为D,强调第四象限x正、y非正的特征。","options":[{"id":"A","content":"点A(2, 3)"},{"id":"B","content":"点B(-1, 4)"},{"id":"C","content":"点C(0, -2)"},{"id":"D","content":"点D(3, 0)"}]},{"id":810,"content":"在一次班级图书捐赠活动中,某学生第一天捐了若干本书,第二天比第一天多捐了5本,两天一共捐了23本。设第一天捐了___本书。","type":"填空题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"9","explanation":"设第一天捐了x本书,则第二天捐了(x + 5)本。根据题意,两天共捐书数量为:x + (x + 5) = 23。解这个一元一次方程:2x + 5 = 23,移项得2x = 18,解得x = 9。因此,第一天捐了9本书。","options":[]},{"id":2362,"content":"如图,在平面直角坐标系中,点A(0, 4),点B(6, 0),点C是线段AB上的一点,且满足AC : CB = 1 : 2。点D是点C关于直线y = x的对称点。若一次函数y = kx + b的图像经过点D和原点O(0, 0),则k的值为多少?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"中等","answer":"B","explanation":"首先根据定比分点公式求出点C的坐标。由于AC:CB = 1:2,即C将AB分为1:2,因此C的坐标为:x = (2×0 + 1×6)\/(1+2) = 6\/3 = 2,y = (2×4 + 1×0)\/3 = 8\/3,故C(2, 8\/3)。点D是C关于直线y = x的对称点,根据轴对称性质,对称点坐标互换,即D(8\/3, 2)。一次函数y = kx + b经过原点O(0,0)和点D(8\/3, 2),代入原点得b = 0,故函数为y = kx。将D点坐标代入得:2 = k × (8\/3),解得k = 2 × 3 \/ 8 = 6\/8 = 3\/4。因此正确答案为B。","options":[{"id":"A","content":"2\/3"},{"id":"B","content":"3\/4"},{"id":"C","content":"4\/5"},{"id":"D","content":"5\/6"}]}]