💡 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
首先计算图书总数:12 + 8 + 10 + 6 = 36(本)。数学类图书占总数的比例为 8 ÷ 36 = 2/9。扇形统计图中整个圆为360度,因此数学类对应的圆心角为 360 × (2/9) = 80(度)。
🏆
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":2047,"content":"某公园计划修建一个菱形花坛,设计图纸上标注了两条对角线的长度分别为6米和8米。施工过程中,工人需要在外围铺设一圈装饰灯带,灯带必须沿着菱形的四条边铺设。已知每米灯带的成本为15元,则铺设完整圈灯带的总成本是多少元?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"中等","answer":"D","explanation":"本题考查菱形的性质与勾股定理的应用。菱形的两条对角线互相垂直且平分,因此可以将菱形分成四个全等的直角三角形。每条对角线的一半分别为3米和4米,根据勾股定理,菱形边长为√(3² + 4²) = √(9 + 16) = √25 = 5米。菱形周长为4 × 5 = 20米。每米灯带15元,总成本为20 × 15 = 300元。因此正确答案为D。","options":[{"id":"A","content":"120元"},{"id":"B","content":"150元"},{"id":"C","content":"180元"},{"id":"D","content":"300元"}]},{"id":2223,"content":"某学生在记录一周气温变化时,发现某地周一的气温比标准气温低3℃,记作-3℃;周三的气温比标准气温高5℃,记作+5℃。如果标准气温为0℃,那么周一和周三的气温相差___℃。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"8","explanation":"周一气温为-3℃,周三气温为+5℃。求两天气温的差值,即计算5 - (-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。","options":[]},{"id":189,"content":"小明去文具店买笔记本,每本笔记本的价格是5元。他买了3本,付给收银员20元,应找回多少钱?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"A","explanation":"首先计算小明购买3本笔记本的总花费:每本5元,3本就是 5 × 3 = 15 元。他付了20元,所以应找回的钱是 20 - 15 = 5 元。因此正确答案是A。","options":[{"id":"A","content":"5元"},{"id":"B","content":"10元"},{"id":"C","content":"15元"},{"id":"D","content":"20元"}]},{"id":2293,"content":"如图,在△ABC中,AB = AC,∠BAC = 120°,D为BC边上一点,且AD ⊥ BC。若BD = 2,则△ABC的面积为多少?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"因为AB = AC,所以△ABC是等腰三角形,顶角∠BAC = 120°。由于AD ⊥ BC,且D在BC上,根据等腰三角形三线合一的性质,AD既是高也是底边BC的中线,因此BD = DC = 2,故BC = 4。在直角三角形ABD中,∠BAD = 60°(等腰三角形顶角平分线将120°分为两个60°),BD = 2。利用tan(60°) = √3 = AD \/ BD,可得AD = 2√3。因此,△ABC的面积为(1\/2) × 底 × 高 = (1\/2) × BC × AD = (1\/2) × 4 × 2√3 = 4√3。","options":[{"id":"A","content":"4√3"},{"id":"B","content":"6√3"},{"id":"C","content":"8√3"},{"id":"D","content":"12√3"}]},{"id":2143,"content":"某学生在解一个关于一元一次方程的问题时,列出了方程 3(x - 2) = 2x + 1。该方程的解是以下哪一个?","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"B","explanation":"解方程 3(x - 2) = 2x + 1:首先去括号得 3x - 6 = 2x + 1,然后将含x的项移到左边,常数项移到右边,得 3x - 2x = 1 + 6,即 x = 7。因此正确答案是B。","options":[{"id":"A","content":"x = 5"},{"id":"B","content":"x = 7"},{"id":"C","content":"x = -5"},{"id":"D","content":"x = -7"}]},{"id":831,"content":"某学生测量了一个长方体的长、宽、高分别为 3 厘米、4 厘米和 5 厘米,则该长方体的体积是 _ 立方厘米。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"60","explanation":"长方体的体积计算公式为:体积 = 长 × 宽 × 高。将已知数据代入公式:3 × 4 × 5 = 60。因此,该长方体的体积是 60 立方厘米。本题考查几何图形初步中的立体图形体积计算,属于七年级数学基础知识点。","options":[]},{"id":2263,"content":"在数轴上,点P表示的数是-3,点Q表示的数是5。一名学生从点P出发,先向右移动8个单位长度,再向左移动4个单位长度,最终到达的位置所表示的数是多少?","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"B","explanation":"点P表示-3,向右移动8个单位长度到达-3 + 8 = 5;再向左移动4个单位长度,即5 - 4 = 1。因此最终位置表示的数是1,正确答案是B。","options":[{"id":"A","content":"-7"},{"id":"B","content":"1"},{"id":"C","content":"4"},{"id":"D","content":"9"}]},{"id":2265,"content":"在数轴上,点A表示的数是-3,点B与点A的距离为7个单位长度,且点B在原点右侧。若点C是点B关于原点的对称点,则点C表示的数是___。","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"中等","answer":"B","explanation":"点A表示-3,点B与点A的距离为7个单位长度,且点B在原点右侧。因此点B可能在-3的右侧7个单位,即-3 + 7 = 4,所以点B表示4。点C是点B关于原点的对称点,即与4到原点距离相等但方向相反,因此点C表示-4。故正确答案为B。","options":[{"id":"A","content":"4"},{"id":"B","content":"-4"},{"id":"C","content":"10"},{"id":"D","content":"-10"}]},{"id":963,"content":"在一次班级环保活动中,某学生收集的可回收物品数量比班级平均数量多3件。如果班级平均每人收集5件,那么这名学生实际收集了___件可回收物品。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"8","explanation":"题目中给出班级平均每人收集5件可回收物品,而该学生比平均数量多3件。因此,只需将平均数量加上多出的部分:5 + 3 = 8。所以这名学生实际收集了8件可回收物品。本题考查有理数中的加法运算,结合生活情境,帮助学生理解正数在实际问题中的应用。","options":[]},{"id":667,"content":"在一次环保活动中,某学生收集了若干个废旧电池,其中可回收电池比不可回收电池多8个。如果可回收电池的数量是15个,那么不可回收电池有___个。","type":"填空题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"7","explanation":"题目中已知可回收电池比不可回收电池多8个,且可回收电池为15个。设不可回收电池的数量为x,根据题意可得方程:15 = x + 8。解这个一元一次方程,两边同时减去8,得到x = 7。因此,不可回收电池有7个。本题考查了一元一次方程的实际应用,属于七年级数学课程中的重点内容。","options":[]}]