某学生在平面直角坐标系中画了一个点 P,其横坐标是 -3,纵坐标比横坐标大 5,则点 P 的坐标是( )
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灯到地面的距离是2.8米,课桌高度为0.75米,课桌桌面距离地面0.75米。因此灯到桌面的垂直距离为2.8减去0.75,即2.8 - 0.75 = 2.05(米)。本题考查有理数的减法在实际生活中的应用,属于简单难度的计算题。
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[{"id":258,"content":"某学生在解方程 3(x - 2) + 5 = 2x + 7 时,第一步将等式左边展开得到 3x - 6 + 5,合并后变为 3x - 1。接着他将等式右边的 2x 移到左边,常数项 7 移到右边,得到 3x - 2x = 7 + 1。按照这个步骤继续解下去,最终 x 的值是___","type":"填空题","subject":"数学","grade":"初一","stage":"初中","difficulty":"中等","answer":"8","explanation":"根据题目描述的解题步骤:原方程为 3(x - 2) + 5 = 2x + 7。第一步展开括号得 3x - 6 + 5,合并常数项后为 3x - 1。此时方程变为 3x - 1 = 2x + 7。将 2x 移到左边变为 -2x,将 -1 移到右边变为 +1,得到 3x - 2x = 7 + 1,即 x = 8。因此,最终解为 x = 8。该题考查一元一次方程的解法,包括去括号、移项和合并同类项,符合七年级数学课程内容。","options":[]},{"id":237,"content":"某学生在计算一个数减去 35 时,误将减法当作加法计算,结果得到 82。那么正确的计算结果应该是____。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"12","explanation":"该学生误将减法当作加法,即把原数加上 35 得到 82。设原数为 x,则有 x + 35 = 82,解得 x = 82 - 35 = 47。正确的计算应是 47 减去 35,即 47 - 35 = 12。因此正确答案是 12。","options":[]},{"id":2525,"content":"如图,在水平地面上竖立着一个圆形转盘,其中心为O,半径为2米。转盘绕点O顺时针旋转90°后,点P落在点P'的位置。若点P初始位置在转盘的最右端,则点P到点P'的直线距离为多少?","type":"选择题","subject":"数学","grade":"九年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"点P初始位于圆盘最右端,即坐标为(2, 0)。圆盘绕中心O顺时针旋转90°后,点P移动到P',相当于将点(2, 0)绕原点顺时针旋转90°。根据旋转公式,顺时针旋转90°后的新坐标为(0, -2)。因此,点P(2, 0)与点P'(0, -2)之间的距离为√[(2-0)² + (0+2)²] = √(4 + 4) = √8 = 2√2(米)。本题考查旋转与坐标结合的距离计算,属于简单综合应用。","options":[{"id":"A","content":"2√2米"},{"id":"B","content":"4米"},{"id":"C","content":"2米"},{"id":"D","content":"√2米"}]},{"id":175,"content":"小明买了3支铅笔和2本笔记本,共花费18元。已知每支铅笔的价格是2元,那么每本笔记本的价格是多少元?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"A","explanation":"首先计算3支铅笔的总价格:3 × 2 = 6(元)。小明总共花费18元,因此2本笔记本的价格为:18 - 6 = 12(元)。那么每本笔记本的价格是:12 ÷ 2 = 6(元)。所以正确答案是A。","options":[{"id":"A","content":"6元"},{"id":"B","content":"5元"},{"id":"C","content":"4元"},{"id":"D","content":"3元"}]},{"id":435,"content":"90","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":976,"content":"某学生测量教室地面的长方形区域,测得长为 (2x + 3) 米,宽为 (x - 1) 米,若该区域的周长为 26 米,则 x 的值为 ___。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"11\/3","explanation":"长方形的周长公式为:周长 = 2 × (长 + 宽)。根据题意,长为 (2x + 3) 米,宽为 (x - 1) 米,周长为 26 米。代入公式得:2 × [(2x + 3) + (x - 1)] = 26。先化简括号内:2x + 3 + x - 1 = 3x + 2。然后计算:2 × (3x + 2) = 6x + 4。列方程:6x + 4 = 26。解方程:6x = 22,x = 22 ÷ 6 = 11\/3。因此,x 的值为 11\/3。","options":[]},{"id":2333,"content":"某公园内有一块三角形花坛ABC,工作人员在边AB外侧作等边三角形ABD,在边AC外侧作等边三角形ACE。连接BE和CD,交于点F。若∠BFC = 120°,则△ABC的形状最可能是以下哪种?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"中等","answer":"A","explanation":"本题综合考查全等三角形与轴对称思想的应用。由于△ABD和△ACE均为等边三角形,可得AB = AD,AC = AE,且∠BAD = ∠CAE = 60°。因此∠DAC = ∠BAE(同加∠BAC),从而可证△DAC ≌ △BAE(SAS),进而推出∠ABE = ∠ADC。进一步分析可知,BE与CD的交角∠BFC与∠BAC互补。题目给出∠BFC = 120°,故∠BAC = 60°。同理可推∠ABC = ∠ACB = 60°,因此△ABC为等边三角形。此结论也符合几何构造中的旋转对称性——将△ABE绕点A逆时针旋转60°可与△ADC重合,进一步验证了结论。","options":[{"id":"A","content":"等边三角形"},{"id":"B","content":"等腰直角三角形"},{"id":"C","content":"含30°角的直角三角形"},{"id":"D","content":"一般锐角三角形"}]},{"id":1949,"content":"某学生在平面直角坐标系中绘制了一个四边形ABCD,其顶点坐标分别为A(1, 2)、B(5, 2)、C(7, -1)、D(3, -1)。若将该四边形沿x轴正方向平移3个单位,再沿y轴负方向平移4个单位,则平移后点C的坐标为____。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"困难","answer":"(10, -5)","explanation":"平移规则:横坐标加3,纵坐标减4。原C(7, -1) → 7+3=10,-1-4=-5。","options":[]},{"id":861,"content":"某学生在调查班级同学最喜欢的课外活动时,收集了以下数据:阅读、运动、绘画、音乐、编程。他将每种活动的人数整理成频数分布表后发现,喜欢运动的人数是喜欢绘画人数的2倍,喜欢音乐的人数比喜欢绘画的多3人,喜欢编程的人数最少,为4人,而喜欢阅读的人数与喜欢音乐的人数相同。如果总共有35人参与调查,那么喜欢绘画的人数是____人。","type":"填空题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"6","explanation":"设喜欢绘画的人数为x人,则喜欢运动的人数为2x人,喜欢音乐的人数为x+3人,喜欢编程的人数为4人,喜欢阅读的人数与音乐相同,也为x+3人。根据总人数为35,列出方程:x + 2x + (x+3) + 4 + (x+3) = 35。化简得:5x + 10 = 35,解得5x = 25,x = 6。因此,喜欢绘画的人数是6人。","options":[]},{"id":174,"content":"小明去文具店买笔记本,每本笔记本的价格是8元。他带了50元,买完笔记本后还剩下10元。请问小明买了多少本笔记本?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"A","explanation":"小明一共带了50元,买完笔记本后剩下10元,说明他花了 50 - 10 = 40 元买笔记本。每本笔记本8元,所以买的本数为 40 ÷ 8 = 5(本)。因此正确答案是A。本题考查的是简单的整数除法在实际生活中的应用,符合七年级数学中‘有理数的运算’和‘列方程解应用题’的基础知识。","options":[{"id":"A","content":"5本"},{"id":"B","content":"6本"},{"id":"C","content":"4本"},{"id":"D","content":"7本"}]}]