在一次班级数学测验中,某学生记录了五名同学的数学成绩(单位:分)分别为:85,92,78,90,85。这组数据的众数是____。
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设笔记本的价格为x元,则钢笔的价格为2x元。根据题意,钢笔和笔记本共花费18元,可列出方程:x + 2x = 18,即3x = 18。解得x = 6。因此,笔记本的价格是6元。选项A正确。
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[{"id":2131,"content":"某学生在解方程 2(x - 3) = 4 时,第一步将方程两边同时除以2,得到 x - 3 = 2。接下来他应该进行的正确步骤是:","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"B","explanation":"方程 x - 3 = 2 中,为了求出 x,需要将 -3 消去。根据等式性质,应在等式两边同时加上3,得到 x = 5。这是七年级一元一次方程求解中的基本步骤,符合课程标准要求。","options":[{"id":"A","content":"两边同时减去3,得到 x = -1"},{"id":"B","content":"两边同时加上3,得到 x = 5"},{"id":"C","content":"两边同时乘以3,得到 x = 6"},{"id":"D","content":"两边同时除以3,得到 x = 2\/3"}]},{"id":2327,"content":"某学生在研究轴对称图形时,发现一个四边形ABCD关于直线MN对称,其中点A与点C对称,点B与点D对称。若∠ABC = 70°,则∠ADC的度数为多少?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"由于四边形ABCD关于直线MN轴对称,且点A与点C对称,点B与点D对称,说明图形在对称轴两侧完全重合。因此,对应角相等。∠ABC与∠ADC是关于对称轴对应的角,故∠ADC = ∠ABC = 70°。本题考查轴对称图形的性质:对称点所连线段被对称轴垂直平分,且对称图形中对应角、对应边相等。","options":[{"id":"A","content":"70°"},{"id":"B","content":"110°"},{"id":"C","content":"90°"},{"id":"D","content":"140°"}]},{"id":567,"content":"平均数是5.2,中位数是5,众数是5","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":1023,"content":"某学生在整理班级同学的身高数据时,将150厘米到160厘米之间的身高记录为一个区间。如果一名学生的身高是155.3厘米,那么这个数据应被归入该区间的第___个十分位段(将150到160平均分成10段,每段为1厘米)。","type":"填空题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"6","explanation":"将150厘米到160厘米的区间平均分成10段,每段为1厘米,分别对应第1段(150≤身高<151)、第2段(151≤身高<152)……第6段(155≤身高<156)。因为155.3厘米满足155 ≤ 155.3 < 156,所以它属于第6个十分位段。本题考查数据的收集与整理中对数据区间的划分与归类,属于‘数据的收集、整理与描述’知识点,符合七年级数学课程内容。","options":[]},{"id":2360,"content":"在一次校园绿化设计中,某学生需要计算一个由两个全等直角三角形拼接而成的菱形花坛的对角线长度。已知每个直角三角形的两条直角边分别为√12米和√27米,且这两个直角边分别作为菱形的两条对角线的一半。求该菱形花坛的面积。","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"中等","answer":"C","explanation":"首先化简已知的直角边:√12 = 2√3,√27 = 3√3。根据题意,这两个直角边分别是一条对角线的一半,因此菱形的两条对角线长度分别为2 × 2√3 = 4√3(米)和2 × 3√3 = 6√3(米)。菱形的面积公式为:面积 = (对角线1 × 对角线2) ÷ 2。代入得:面积 = (4√3 × 6√3) ÷ 2 = (24 × 3) ÷ 2 = 72 ÷ 2 = 36(平方米)。因此正确答案为C。本题综合考查了二次根式的化简、勾股定理背景下的几何理解以及菱形面积公式的应用,难度适中。","options":[{"id":"A","content":"18平方米"},{"id":"B","content":"27平方米"},{"id":"C","content":"36平方米"},{"id":"D","content":"54平方米"}]},{"id":2259,"content":"在数轴上,点A表示的数是-3,点B与点A的距离为5个单位长度,且点B在原点右侧,则点B表示的数是___。","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"B","explanation":"点A表示的数是-3,点B与点A的距离为5个单位长度,说明点B可能在-3的左边或右边5个单位。若在左边,则为-3 - 5 = -8;若在右边,则为-3 + 5 = 2。题目说明点B在原点右侧,即表示的数大于0,因此点B表示的数是2。","options":[{"id":"A","content":"-8"},{"id":"B","content":"2"},{"id":"C","content":"8"},{"id":"D","content":"-2"}]},{"id":2226,"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上升了5℃,记作+5℃。如果第二天的气温又比这一天下降了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":"-8","explanation":"根据正负数表示相反意义的量的知识点,气温上升用正数表示,下降则用负数表示。题目中气温下降了8℃,因此应记作-8℃。本题考查学生对正负数在实际情境中应用的理解,符合七年级课程标准要求。","options":[]},{"id":376,"content":"某学生在平面直角坐标系中描出三个点 A(2, 3)、B(-1, 4)、C(0, -2),然后画出由这三个点组成的三角形。请问这个三角形的周长最接近下列哪个数值?(单位:长度单位)","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"B","explanation":"首先计算三角形三条边的长度。使用两点间距离公式:若两点坐标为 (x₁, y₁) 和 (x₂, y₂),则距离为 √[(x₂−x₁)² + (y₂−y₁)²]。\n\n1. 计算 AB 的长度:A(2,3) 到 B(-1,4)\n AB = √[(-1−2)² + (4−3)²] = √[(-3)² + (1)²] = √(9 + 1) = √10 ≈ 3.16\n\n2. 计算 BC 的长度:B(-1,4) 到 C(0,-2)\n BC = √[(0−(-1))² + (-2−4)²] = √[(1)² + (-6)²] = √(1 + 36) = √37 ≈ 6.08\n\n3. 计算 AC 的长度:A(2,3) 到 C(0,-2)\n AC = √[(0−2)² + (-2−3)²] = √[(-2)² + (-5)²] = √(4 + 25) = √29 ≈ 5.39\n\n将三边相加得周长:3.16 + 6.08 + 5.39 ≈ 14.63\n\n最接近的整数是 14,因此正确答案是 B。","options":[{"id":"A","content":"12"},{"id":"B","content":"14"},{"id":"C","content":"16"},{"id":"D","content":"18"}]},{"id":2256,"content":"在数轴上,点A表示的数是-3,点B与点A之间的距离为5个单位长度,且点B在原点的右侧。那么点B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A表示的数是-3,点B与点A相距5个单位长度。由于在数轴上向右移动表示数值增大,且点B在原点右侧,说明点B的数值大于0。从-3向右移动5个单位:-3 + 5 = 2,因此点B表示的数是2。选项B正确。","options":[{"id":"A","content":"-8"},{"id":"B","content":"2"},{"id":"C","content":"5"},{"id":"D","content":"8"}]},{"id":2359,"content":"某学生在一张方格纸上画了一个等腰三角形ABC,其中AB = AC,且顶点A位于坐标原点(0, 0),底边BC关于y轴对称。已知点B的坐标为(-3, 4),点C的坐标为(3, 4)。该学生想验证△ABC是否为直角三角形,并计算其面积。以下结论正确的是:","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"中等","answer":"C","explanation":"首先,根据题意,点A(0,0),点B(-3,4),点C(3,4)。由于B和C关于y轴对称,且AB = AC,符合等腰三角形特征。计算各边长度:AB = √[(-3-0)² + (4-0)²] = √(9+16) = √25 = 5;同理AC = 5;BC = √[(3+3)² + (4-4)²] = √36 = 6。三边为5、5、6。验证是否满足勾股定理:若为直角三角形,则应有某两边平方和等于第三边平方。检查:5² + 5² = 50 ≠ 36;5² + 6² = 25 + 36 = 61 ≠ 25。因此不满足勾股定理,不是直角三角形。面积可用底×高÷2计算:以BC为底,长度为6,高为A到BC的垂直距离。由于BC在y=4上,A在(0,0),高为4,故面积为(6×4)\/2 = 12。综上,△ABC不是直角三角形,面积为12,正确答案为C。","options":[{"id":"A","content":"△ABC是直角三角形,且直角位于顶点A,面积为12"},{"id":"B","content":"△ABC是直角三角形,且直角位于底边BC的中点,面积为24"},{"id":"C","content":"△ABC不是直角三角形,但面积为12"},{"id":"D","content":"△ABC是直角三角形,且直角位于点B,面积为6"}]}]