2023年秋长郡双语九年级数学开学摸底考试

这是一份2023年秋长郡双语九年级数学开学摸底考试，文件包含2023年秋九年级开学摸底考试数学参考答案doc、2023年秋九年级开学摸底考试数学doc、2023年秋九年级开学摸底考试数学答题卡pdf等3份试卷配套教学资源，其中试卷共14页， 欢迎下载使用。
2023年秋九年级开学摸底考试数学 参考答案一、选择题1.答案：D解析：观察知D是轴对称图形，故选D.2.答案：D解析：，，，故选D.3.答案：D解析：由表知，这组数据5.2出现次数最多，所以众数为5.2，中位数为，极差为，平均数为，故选D.4.答案：C解析：由题意，对于A选项，，A选项错误，对于B选项，，B选项错误，对于C选项，，C选项正确，对于D选项，与不是同类项不能合并，D选项错误，故选C.5.答案：D解析：一次函数过点，，解得，一次函数为，y随x增大而减小，直线经过一、二、四象限，故A，B错误；当时，，因此直线不过点，故C错误；该一次函数与x轴交于点，与y轴交于点，与坐标轴围成的三角形面积为，故D正确；故选D.6.答案：D解析：如图，过点B作，垂足为C，延长ND交AC于M.观察图形可知，.在中，，所以，即门口A到藏宝点B的直线距离是.故选D.7.答案：B解析：由题意知，MN是线段BC的垂直平分线，，四边形ABCD是平行四边形，，由勾股定理得，，故选B.8.答案：D解析：，抛物线开口向下，对称轴为y轴，顶点坐标为，在对称轴的右侧，y随x的增大而减小，选项A，B，C中的说法都不正确.，抛物线与x轴有两个交点.故选D.9.答案：D解析：连接AM，如图所示.，，，.，，四边形AFPE是矩形，A，M，P三点共线，，与AP互相平分，M为AP的中点，.当时，AP最小，此时PM也最小，，的最小值为4.8，当PM最小时，.故选D.10.答案：B解析：如图，正方形ABCD和正方形CEFG中，点D在CG上，，，，，.延长AD交EF于M，连接AC，CF，则，，.四边形ABCD和四边形GCEF是正方形，，.H为AF的中点，.在中，由勾股定理得，，故选B.二、填空题11.答案：，解析：由题意得，，或，解得，.12.答案：1解析：.13.答案：解析：将抛物线向下平移2个单位，再向左平移1个单位，所得的抛物线的解析式为.14.答案：解析：如图，连接CE交AB于点O.中，，，，.若平行四边形CDEB为菱形，则，，.，，，.故答案为.15.答案：解析：如图，作点A关于y轴的对称点N，连接BN交y轴于一点，即为点P，此时值最大，，，设直线BN的解析式为，将，代入，得，解得，直线BN的解析式为，当时，，，故答案为：.三、解答题16.解析：，将方程变为一般形式为：，，················································2分故方程有两个实数根为：，，，故方程的解为：，．······················································4分17.解析：原式.········································································3分当，时，原式.················································6分18.解析：四边形ABCD是平行四边形，，，，········································································2分点G，H分别是AB，CD的中点，，在和中，，，·······························································5分，，，，又，四边形EGFH是平行四边形.·························································7分19.解析：（1）设每辆B货车的运费为x元，则每辆A货车的运费为元，则由题意得，，解得，，每辆A货车，B货车的运费分别为600，1000元.·····························3分（2）设A货车有a辆，则B货车有辆，总费用为y，则由题意得，，解得，总费用，整理得，，················6分，y随a的增大而减小，当时，y最小，，安排A货车5辆，B货车5辆时费用最少.·········································8分20.解析：（1）10天里香槟玫瑰的销售额中360出现的次数最多，故众数；把10天里铃兰的销售额从小到大排列，排在中间的两个数是370，370，故中位数.故答案为：360；370.······························································4分（2）(天)，估计“时光花店”本月的铃兰销售额达到“A”等级的天数约10天.··········6分（3）铃兰的销售情况更好，理由如下：因为铃兰销售额的平均数，中位数，众数均大于香槟玫瑰，铃兰销售额的方差小于香槟玫瑰，所以铃兰的销售情况更好.································8分21.解析：(1)，，四边形BDCE为平行四边形.，为AC边上的中线，，四边形BDCE为菱形.········································3分(2)连接DE交BC于O点.四边形BDCE为菱形，，，，.，，······································5分，，.······································································8分22.解析：(1)因为在中，，所以，所以.··········································2分在中，，所以，所以供水点M到喷泉A，B需要铺设的管道总长为.··································································································4分(2)因为，，，，所以是直角三角形，即，··········································6分则喷泉B到小路AC的最短距离为BM的长，所以喷泉B到小路AC的最短距离是.··········································8分23.解析：(1).·················································2分(2)，·····················································4分.·······························································································6分(3)，①，②，······································10分得：，.············································································13分24.解析：(1)，，················································3分理由：在正方形ABCD中，，，又，，，，又，，，.(2)在正方形ABCD中，，，，，，，··································································5分，，，，，················6分，，，，.············································································8分(3)在正方形ABCD中，，，，，点H是BP的中点，，，，········································································9分，，，，，··········································································10分在四边形QABC中，设，，则，，，，，，，即，.············································································13分