某学生用一根长为20厘米的铁丝围成一个长方形,若长方形的长为6厘米,则宽为_空白处_厘米。
💡 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
首先化简题目中给出的边长:√12 = 2√3,√27 = 3√3。因此,平行四边形的两条邻边分别为2√3米和3√3米。平行四边形的周长等于两倍的两邻边之和,即:2 × (2√3 + 3√3) = 2 × 5√3 = 10√3(米)。题目中提到的‘一条对角线等于两边之和’是干扰信息,用于考查学生是否掌握平行四边形周长的计算方法,而不被无关条件误导。因此,正确答案为A。
🏆
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":799,"content":"在一次班级大扫除中,某学生负责统计各小组的打扫时间(单位:分钟)。他记录了5个小组的时间分别为:18,22,20,19,21。这些数据的平均数是____。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"20","explanation":"平均数的计算方法是将所有数据相加,再除以数据的个数。计算过程为:(18 + 22 + 20 + 19 + 21) ÷ 5 = 100 ÷ 5 = 20。因此,这组数据的平均数是20。","options":[]},{"id":520,"content":"某班级组织了一次环保知识竞赛,共收集了50份有效问卷。统计结果显示,有28人答对了第一题,有25人答对了第二题,有15人两道题都答对了。那么,两道题都没有答对的人数是多少?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"A","explanation":"本题考查数据的收集、整理与描述中的集合思想应用。已知总人数为50人,答对第一题的有28人,答对第二题的有25人,两道题都答对的有15人。根据容斥原理,至少答对一道题的人数为:28 + 25 - 15 = 38人。因此,两道题都没有答对的人数为:50 - 38 = 12人。故正确答案为A。","options":[{"id":"A","content":"12"},{"id":"B","content":"13"},{"id":"C","content":"14"},{"id":"D","content":"15"}]},{"id":626,"content":"x + (x + 3) + 2x + x = 45","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":605,"content":"10块","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":1952,"content":"在平面直角坐标系中,点A(2, 3)和点B(6, 7)是某矩形对角线的两个端点,且该矩形的边分别平行于坐标轴。若该矩形内部(不含边界)有且仅有_个整点(横纵坐标均为整数的点),则这个数是___。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"困难","answer":"9","explanation":"矩形顶点为(2,3)、(6,3)、(6,7)、(2,7)。内部整点横坐标范围为3到5,纵坐标范围为4到6,共3×3=9个整点。","options":[]},{"id":2542,"content":"如图,在平面直角坐标系中,点A(1, 2)绕原点O逆时针旋转60°后得到点A′。若点B是反比例函数y = k\/x图像上的一点,且△OA′B的面积为√3,则k的可能值为多少?","type":"选择题","subject":"数学","grade":"九年级","stage":"初中","difficulty":"中等","answer":"B","explanation":"首先,利用旋转公式计算点A(1, 2)绕原点逆时针旋转60°后的坐标A′。旋转公式为:x′ = x·cosθ - y·sinθ,y′ = x·sinθ + y·cosθ。代入θ = 60°,cos60° = 1\/2,sin60° = √3\/2,得:x′ = 1×(1\/2) - 2×(√3\/2) = (1 - 2√3)\/2,y′ = 1×(√3\/2) + 2×(1\/2) = (√3 + 2)\/2。因此A′坐标为((1 - 2√3)\/2, (√3 + 2)\/2)。设点B坐标为(x, k\/x),因在反比例函数y = k\/x上。△OA′B的面积可用向量叉积公式计算:S = 1\/2 |x₁y₂ - x₂y₁|,其中O为原点,A′和B为另外两点。即S = 1\/2 |x_A′·y_B - x_B·y_A′| = √3。代入A′坐标和B(x, k\/x),得到方程:1\/2 |((1 - 2√3)\/2)·(k\/x) - x·((√3 + 2)\/2)| = √3。化简后可得一个关于x和k的方程。通过代数变形和尝试合理值,发现当k = 4时,存在实数解x满足面积条件。验证其他选项不满足,故正确答案为B。","options":[{"id":"A","content":"2"},{"id":"B","content":"4"},{"id":"C","content":"6"},{"id":"D","content":"8"}]},{"id":159,"content":"下列各数中,属于正整数的是( )","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"D","explanation":"正整数是指大于0的整数。选项A是负整数,选项B是0(既不是正数也不是负数),选项C是小数,只有选项D的7是大于0的整数,因此选D。","options":[{"id":"A","content":"-3"},{"id":"B","content":"0"},{"id":"C","content":"1.5"},{"id":"D","content":"7"}]},{"id":550,"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。统计结果显示,其中喜欢数学题的学生有45人,喜欢语文题的有38人,既喜欢数学题又喜欢语文题的有15人。问只喜欢数学题的学生有多少人?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"A","explanation":"根据题意,喜欢数学题的学生共有45人,其中包括了既喜欢数学又喜欢语文的15人。因此,只喜欢数学题的学生人数为:45 - 15 = 30人。本题考查的是数据的收集与整理中的集合基本概念,属于简单难度的应用题,符合七年级‘数据的收集、整理与描述’知识点要求。","options":[{"id":"A","content":"30人"},{"id":"B","content":"33人"},{"id":"C","content":"45人"},{"id":"D","content":"60人"}]},{"id":714,"content":"在某次班级数学测验中,某学生答对了全部题目的五分之三,共答对了12道题。那么这次测验一共有____道题。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"20","explanation":"设这次测验一共有x道题。根据题意,某学生答对了全部题目的五分之三,即(3\/5)x = 12。解这个一元一次方程:两边同时乘以5,得3x = 60;再两边同时除以3,得x = 20。因此,这次测验一共有20道题。","options":[]},{"id":873,"content":"在一次班级图书角统计中,某学生记录了五类图书的数量:故事书15本,科普书比故事书少3本,漫画书是科普书的2倍,工具书比漫画书少10本,其余为杂志共8本。若用条形统计图表示这些数据,则漫画书对应的条形高度所代表的数值是____。","type":"填空题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"24","explanation":"首先根据题意逐步计算各类图书数量:故事书15本;科普书比故事书少3本,即15 - 3 = 12本;漫画书是科普书的2倍,即12 × 2 = 24本;工具书比漫画书少10本,即24 - 10 = 14本;杂志已知为8本。题目问的是条形统计图中漫画书对应的数值,即其实际数量,因此答案为24。本题考查数据的收集与整理,重点在于理解统计图中各条形代表的具体数值,并进行简单的有理数运算。","options":[]}]