某学生在研究一个等腰三角形时,测得其底边长为8,两腰的长度均为√41。若该学生想计算这个三角形的高,他应该使用以下哪个结果?
💡 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
根据题意,文学类书籍为x本,科普类比文学类多8本,即为(x + 8)本。两类书总数为32本,因此可列方程:x + (x + 8) = 32。解这个方程:2x + 8 = 32 → 2x = 24 → x = 12。所以文学类有12本,科普类有12 + 8 = 20本。本题考查一元一次方程的建立与求解,属于七年级上册重点内容。
🏆
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":759,"content":"在一次班级大扫除中,某学生负责统计各小组的垃圾重量。已知第一组收集的垃圾比第二组多3.5千克,两组共收集了12.7千克。设第二组收集的垃圾重量为x千克,则可列出一元一次方程:x + (x + 3.5) = 12.7。解这个方程,第二组收集的垃圾重量为___千克。","type":"填空题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"4.6","explanation":"根据题意,设第二组收集的垃圾重量为x千克,则第一组为(x + 3.5)千克。两组共收集12.7千克,因此可列方程:x + (x + 3.5) = 12.7。化简得:2x + 3.5 = 12.7。两边同时减去3.5,得2x = 9.2。再两边同时除以2,得x = 4.6。所以第二组收集的垃圾重量为4.6千克。","options":[]},{"id":803,"content":"在一次环保活动中,某学校七年级学生共收集了120千克废旧纸张。已知男生收集的纸张比女生多20千克,设女生收集的纸张为x千克,则可列出一元一次方程:_x + (x + 20) = 120_,解得女生收集了___千克。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"50","explanation":"根据题意,女生收集x千克,男生比女生多20千克,即男生收集(x + 20)千克。总重量为120千克,因此方程为x + (x + 20) = 120。解这个方程:2x + 20 = 120 → 2x = 100 → x = 50。所以女生收集了50千克。","options":[]},{"id":2529,"content":"如图,一个圆形花坛被三条等距的半径分成三个扇形区域,分别种植不同花卉。若在花坛边缘随机抛掷一粒石子,落在任意一个扇形区域的概率相等。现将整个花坛绕圆心顺时针旋转60°,此时原位于正北方向的标记点A移动到了点B的位置。若点B恰好落在其中一个扇形区域的边界上,则这个旋转后的图形与原图形重合部分所对应的圆心角是多少度?","type":"选择题","subject":"数学","grade":"九年级","stage":"初中","difficulty":"简单","answer":"C","explanation":"花坛被三条等距半径分成三个扇形,说明每个扇形的圆心角为360° ÷ 3 = 120°。旋转60°后,原标记点A移动到点B,而点B落在某个扇形边界上,说明旋转角度60°正好是两个相邻半径夹角(120°)的一半。由于图形具有120°的旋转对称性,旋转60°后,原图形与旋转后图形的重合部分由两个相邻扇形重叠构成。通过几何分析可知,重合部分的圆心角为120°,即一个完整扇形的角度。因此,正确答案为C。","options":[{"id":"A","content":"60°"},{"id":"B","content":"90°"},{"id":"C","content":"120°"},{"id":"D","content":"180°"}]},{"id":999,"content":"在一次环保主题活动中,某学生记录了连续5天每天收集的废旧电池数量(单位:节),分别为:-3,5,0,-2,7。这里规定:收集到电池记为正数,丢失或损坏电池记为负数。这5天该学生实际收集的电池总数为___节。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"7","explanation":"题目中给出的数据是有理数,包含正数、负数和零。根据题意,正数表示收集到的电池数量,负数表示丢失或损坏的数量,因此需要将所有数值相加得到净收集量。计算过程为:(-3) + 5 + 0 + (-2) + 7 = (5 + 7) + (-3 - 2) + 0 = 12 - 5 = 7。所以这5天实际收集的电池总数为7节。本题考查有理数的加法运算,结合生活情境,帮助学生理解有理数在实际问题中的意义。","options":[]},{"id":173,"content":"小明买了3支铅笔和2本笔记本,共花费18元。已知每本笔记本比每支铅笔贵3元。设每支铅笔的价格为x元,则下列方程正确的是?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"A","explanation":"设每支铅笔的价格为x元,则每本笔记本的价格为(x + 3)元。根据题意,3支铅笔的总价是3x元,2本笔记本的总价是2(x + 3)元,两者相加等于总花费18元。因此,正确的方程为:3x + 2(x + 3) = 18。选项A正确。选项B错误地将笔记本总价写成2x + 3,忽略了是每本贵3元;选项C颠倒了铅笔和笔记本的单价关系;选项D没有正确表示笔记本的价格,且等式右边错误地加了3。","options":[{"id":"A","content":"3x + 2(x + 3) = 18"},{"id":"B","content":"3x + 2x + 3 = 18"},{"id":"C","content":"3(x + 3) + 2x = 18"},{"id":"D","content":"3x + 2x = 18 + 3"}]},{"id":1934,"content":"在平面直角坐标系中,点A(2, 3)、B(5, -1)、C(-1, -4)构成三角形ABC。若点D是线段AB的中点,点E在y轴上,且△CDE的面积为15,则点E的纵坐标为______。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"困难","answer":"6或-12","explanation":"先求D点坐标((2+5)\/2, (3+(-1))\/2) = (3.5, 1)。设E(0, y),利用向量法或坐标面积公式S = 1\/2|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)|,代入C、D、E坐标解得|y−1|=18,故y=6或−12。","options":[]},{"id":1978,"content":"某学生在一张纸上画了一个边长为5 cm的正方形,然后以正方形的一个顶点为圆心,以正方形的边长5 cm为半径画了一个扇形。若将该扇形剪下并绕其圆心顺时针旋转60°,则扇形扫过的区域面积是多少?(π取3.14)","type":"选择题","subject":"数学","grade":"九年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"本题考查扇形旋转过程中扫过区域的面积计算,结合圆与旋转的知识点。初始扇形是以正方形顶点为圆心、半径为5 cm、圆心角为90°的扇形(因为正方形内角为90°)。当该扇形绕圆心顺时针旋转60°时,其扫过的区域是两个扇形之间的环形扇面,即圆心角为60°、半径为5 cm的扇形面积。计算公式为:S = (θ\/360) × πr² = (60\/360) × 3.14 × 5² = (1\/6) × 3.14 × 25 ≈ 13.08 cm²。因此正确答案为A。","options":[{"id":"A","content":"13.08 cm²"},{"id":"B","content":"15.70 cm²"},{"id":"C","content":"18.84 cm²"},{"id":"D","content":"21.98 cm²"}]},{"id":3,"content":"二元一次方程组{x + y = 5, 2x - y = 1}的解是?","type":"选择题","subject":"数学","grade":"初二","stage":"初中","difficulty":"中等","answer":"C","explanation":"使用加减消元法,将两个方程相加消去y:(x + y) + (2x - y) = 5 + 1,得到3x = 6,解得x = 2。将x = 2代入第一个方程:2 + y = 5,解得y = 3。","options":[{"id":"A","content":"x = 1, y = 4"},{"id":"B","content":"x = 3, y = 2"},{"id":"C","content":"x = 2, y = 3"},{"id":"D","content":"x = 4, y = 1"}]},{"id":1922,"content":"某学生在整理班级同学的课外阅读时间时,收集了以下数据(单位:小时):2.5,3,1.5,4,2.5,3.5,2。若将这组数据按从小到大的顺序排列,则位于中间位置的数是:","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"首先将数据从小到大排列:1.5,2,2.5,2.5,3,3.5,4。共有7个数据,为奇数个,因此中位数是正中间的那个数,即第4个数。第4个数是2.5,所以答案是A。本题考查数据的整理与描述中的中位数概念,属于简单难度,符合七年级学生认知水平。","options":[{"id":"A","content":"2.5"},{"id":"B","content":"3"},{"id":"C","content":"2.75"},{"id":"D","content":"3.5"}]},{"id":2149,"content":"某学生在解一元一次方程时,将方程 3(x - 2) = 2x + 5 的括号展开后得到 3x - 6 = 2x + 5,接着移项合并同类项。该学生下一步的正确操作是什么?","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"B","explanation":"解一元一次方程时,移项要变号。原方程展开后为 3x - 6 = 2x + 5。将 2x 移到左边变为 -2x,将 -6 移到右边变为 +6,因此得到 3x - 2x = 5 + 6。选项B正确体现了移项变号的规则,符合七年级一元一次方程的解法要求。","options":[{"id":"A","content":"将 2x 移到左边,-6 移到右边,得到 3x - 2x = 5 - 6"},{"id":"B","content":"将 2x 移到左边,-6 移到右边,得到 3x - 2x = 5 + 6"},{"id":"C","content":"将 3x 移到右边,5 移到左边,得到 -6 - 5 = 2x - 3x"},{"id":"D","content":"两边同时除以 x,得到 3 - 6\/x = 2 + 5\/x"}]}]