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A和点B之间的一个点,且AC:CB = 2:5,则点C表示的数是___。
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温度下降应使用负数表示。题目中明确指出气温下降了8℃,因此应记作-8℃。选项B正确。其他选项要么符号错误,要么数值错误,不符合正负数表示实际意义的要求。
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[{"id":2004,"content":"某学生测量了一块直角三角形铁片的边长,其中两条直角边分别为5 cm和12 cm,他需要计算斜边的长度以确定是否适合放入一个边长为13 cm的正方形槽中。请问这块铁片的斜边长度是多少?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"简单","answer":"B","explanation":"根据勾股定理,在直角三角形中,斜边的平方等于两条直角边的平方和。设斜边为c,则有:c² = 5² + 12² = 25 + 144 = 169。因此,c = √169 = 13(cm)。所以斜边长为13 cm,正好可以放入边长为13 cm的正方形槽中。","options":[{"id":"A","content":"10 cm"},{"id":"B","content":"13 cm"},{"id":"C","content":"15 cm"},{"id":"D","content":"17 cm"}]},{"id":2181,"content":"在一次数学测验中,一名学生记录了连续五天的气温变化情况(单位:摄氏度),以0℃为标准,高于0℃记为正,低于0℃记为负。这五天的气温分别为:+3,-2,+1,-4,+2。若将这五个有理数按从小到大的顺序排列,则排在第三位的数是( )。","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"中等","answer":"B","explanation":"首先将五个有理数按从小到大的顺序排列:-4,-2,+1,+2,+3。其中-4最小,其次是-2,第三位是+1。因此,排在第三位的数是+1。本题考查有理数的大小比较及排序能力,符合七年级学生对有理数顺序的理解要求。","options":[{"id":"A","content":"-2"},{"id":"B","content":"+1"},{"id":"C","content":"-4"},{"id":"D","content":"+2"}]},{"id":326,"content":"某学生调查了班级同学最喜欢的运动项目,并将数据整理成如下表格。已知喜欢篮球的人数比喜欢足球的多6人,喜欢乒乓球的人数是喜欢羽毛球的2倍,且总人数为40人。如果喜欢足球的有8人,那么喜欢羽毛球的有多少人?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"B","explanation":"根据题意,喜欢足球的有8人,喜欢篮球的比足球多6人,所以喜欢篮球的有 8 + 6 = 14 人。设喜欢羽毛球的有 x 人,则喜欢乒乓球的有 2x 人。总人数为40人,因此可以列出方程:足球人数 + 篮球人数 + 羽毛球人数 + 乒乓球人数 = 总人数,即 8 + 14 + x + 2x = 40。化简得 22 + 3x = 40,解得 3x = 18,x = 6。所以喜欢羽毛球的有6人,正确答案是B。","options":[{"id":"A","content":"5人"},{"id":"B","content":"6人"},{"id":"C","content":"7人"},{"id":"D","content":"8人"}]},{"id":710,"content":"在一次班级环保活动中,某学生收集了若干个塑料瓶,若每5个装一袋,则最后剩下3个;若每7个装一袋,则刚好装完。该学生至少收集了___个塑料瓶。","type":"填空题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"28","explanation":"设该学生收集的塑料瓶总数为x。根据题意,x除以5余3,即x ≡ 3 (mod 5);同时x能被7整除,即x ≡ 0 (mod 7)。我们寻找满足这两个条件的最小正整数。从7的倍数开始尝试:7、14、21、28……检查这些数除以5的余数。7÷5余2,14÷5余4,21÷5余1,28÷5余3,符合条件。因此,最小的x是28。本题考查一元一次方程与同余思想的初步应用,结合生活情境,适合七年级学生理解。","options":[]},{"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":2533,"content":"如图,一个圆锥的底面半径为3 cm,高为4 cm。若将该圆锥沿一条母线展开,得到的扇形圆心角为θ度。已知圆锥的侧面积公式为πrl(其中r为底面半径,l为母线长),则θ的值最接近以下哪个选项?","type":"选择题","subject":"数学","grade":"九年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"首先,根据勾股定理计算圆锥的母线长l:l = √(r² + h²) = √(3² + 4²) = √(9 + 16) = √25 = 5 cm。圆锥的底面周长为2πr = 2π×3 = 6π cm。展开后的扇形弧长等于底面周长,即6π cm。扇形的半径为母线长5 cm,因此扇形所在圆的周长为2π×5 = 10π cm。圆心角θ占整个圆的比例为弧长与圆周长之比:θ\/360 = 6π \/ 10π = 3\/5。解得θ = 360 × 3\/5 = 216°。因此,正确答案为A。","options":[{"id":"A","content":"216°"},{"id":"B","content":"180°"},{"id":"C","content":"144°"},{"id":"D","content":"120°"}]},{"id":200,"content":"一个长方形的长是8厘米,宽是5厘米,它的周长是______厘米。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"26","explanation":"长方形的周长计算公式是:周长 = 2 × (长 + 宽)。将已知的长8厘米和宽5厘米代入公式,得到:2 × (8 + 5) = 2 × 13 = 26(厘米)。因此,这个长方形的周长是26厘米。","options":[]},{"id":1937,"content":"某学生在平面直角坐标系中绘制了一个三角形,其三个顶点分别为 A(2, 3)、B(5, -1)、C(-1, -1)。若将该三角形沿 x 轴方向平移 _ 个单位长度后,点 A 的对应点 A' 恰好落在 y 轴上,则平移的单位长度为 ___。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"困难","answer":"2","explanation":"点 A 的横坐标为 2,要使其平移到 y 轴上(横坐标为 0),需向左平移 2 个单位。平移不改变纵坐标,仅改变横坐标。","options":[]},{"id":2467,"content":"如图,在平面直角坐标系中,点A(0, 4),点B(6, 0),点C在x轴正半轴上,且△ABC是以∠ACB为直角的直角三角形。点D是线段AB上一点,过点D作DE⊥AC于点E,DF⊥BC于点F,使得四边形DECF为矩形。已知矩形DECF的面积S与点D的横坐标x满足关系式:S = -x² + 6x。若点P是该矩形对角线交点,求当点P到原点的距离最小时,点P的坐标。","type":"解答题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"中等","answer":"待完善","explanation":"解析待完善","options":[]},{"id":145,"content":"已知一个三角形的两边长分别为3cm和7cm,第三边的长度可能是以下哪一个?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"B","explanation":"根据三角形三边关系定理:任意两边之和大于第三边,任意两边之差小于第三边。设第三边为x,则需满足:7 - 3 < x < 7 + 3,即4 < x < 10。选项中只有5cm在这个范围内,因此正确答案是B。","options":[{"id":"A","content":"3cm"},{"id":"B","content":"5cm"},{"id":"C","content":"10cm"},{"id":"D","content":"11cm"}]}]