上海市奉贤区中考数学二模(高清打印)2

精品文档，欢迎下载！2019学年奉贤区质量调研九 年级 六年级体育公开课教案九年级家长会课件PPT下载六年级家长会PPT课件一年级上册汉语拼音练习题六年级上册道德与法治课件 数学202005（满分150分，考试时间100分钟）考生注意：1.本试卷含三个大题，共25题．答题时，考生务必按答题要求在答题纸规定的位置上作答，在草稿纸、本试卷上答题一律无效．2.除第一、二大题外，其余各题如无特别说明，都必须在答题纸的相应位置上写出证明或计算的主要步骤．一、选择题（本大题共6题，每题4分，满分24分）1.下列计算中，结果等于a2m的是（▲）（A）amam；（B）ama2；（C）(am)m；（D）(am)2．2.下列等式成立的是（▲）（A）（3）23；（B）（3）23；（C）333；（D）（-3）23．3.如果关于x的一元二次方程x2﹣2x+m＝0有两个不相等的实数根，那么实数m的值可以是（▲）（A）0；（B）1；（C）2；（D）3．4.甲、乙、丙、丁四位同学本学期5次50米短跑成绩的平均数x(秒)及方差S2(秒2)如表1所示.如果从这四位同学中选出一位成绩较好且状态稳定的同学参加学校比赛，那么应该选的同学是（▲）表1：甲乙丙丁x777.57.5S22.11.921.8（A）甲；（B）乙；（C）丙；（D）丁．5.四边形ABCD的两条对角线AC、BD互相平分．添加下列条件，一定能判定四边形ABCD为菱形的是（▲）（A）ABDBDC；（B）ABDBAC；（C）ABDCBD；（D）ABDBCA．6.如果线段AM和线段AN分别是△ABC边BC上的中线和高，那么下列判断正确的是（▲）（A）AMAN；（B）AMAN；（C）AMAN；（D）AMAN．二、填空题（本大题共12题，每题4分，满分48分）7．计算：9a3b3a2=▲．28.如果代数式在实数范围内有意义，那么实数x的取值范围是▲．3x9.方程x14的解是▲．第-1-页精品文档，欢迎下载！10.二元一次方程x+2y＝3的正整数解是▲．11.从分别写有数字1，2，4的三张相同卡片中任取两张，如果把所抽取卡片上的两个数字分别作为点M4的横坐标和纵坐标，那么点M在双曲线y上的概率是▲.x12.如果函数ykx(k0)的图像经过第二、四象限，那么y的值随x的值增大而▲．（填“增大”或“减小”）13.据国家统计局数据，2019年全年国内生产总值接近100万亿，比2018年增长6.1%．假设2020年全年国内生产总值的年增长率保持不变，那么2020年的全年国内生产总值将达到▲万亿．14.已知平行四边形ABCD，E是边AB的中点．设ABa，BCb，那么DE＝▲．（结果用a、b表示）．15.某校计划为全体1200名学生提供以下五种在线学习的方式：在线听课、在线答题、在线讨论、在线答疑和在线阅读．为了解学生需求，该校随机对部分学生进行了“你对哪类在线学习方式最感兴趣”的调查，并根据调查结果绘制成扇形统计图（如图1）．由这个统计图可知，全校学生中最喜欢“在线答疑”的学生人数约为▲人．16.如图2，一艘轮船由西向东航行，在A处测得灯塔P在北偏东60°的方向，继续向东航行40海里后到B处，测得灯塔P在北偏东30°的方向，此时轮船与灯塔之间的距离是▲海里．抽取的学生最感兴趣的学习方式的扇形图EA在线听课A10%20%B在线答题AC在线讨论DBPDD在线答疑C25%E在线阅读15%ABCB图1图2图317.在矩形ABCD中，AB＝5，BC＝12．如果分别以A、C为圆心的两圆外切，且圆A与直线BC相交，点D在圆A外，那么圆C的半径长r的取值范围是▲．18.如图3，在Rt△ABC中，∠ACB＝90°，∠B＝35°，CD是斜边AB上的中线，如果将△BCD沿CD所在直线翻折，点B落在点E处，联结AE，那么∠CAE的度数是▲度．三、解答题（本大题共7题，满分78分）19．（本题满分10分）1计算：82222220200．第-2-页20．（本题满分10分）x36先化简，再求值：(1)，其中x3．x26x9x321．（本题满分10分，每小题满分5分）已知：如图4，在平面直角坐标系xOy中，直线AB与x轴交于点A（-2，0），与y轴的正半轴交于m点B，与反比例函数y＝（x0）的图像交于点C，且AB=BC，点C的纵坐标为4．xy(1)求直线AB的表达式；mC(2)过点B作BD∥x轴，交反比例函数y＝的图像于点D，x求线段CD的长度．BAox图422．（本题满分10分，每小题满分5分）如图5，由于四边形具有不稳定性，因此在同一平面推矩形的边可以改变它的形状（推移过程中边的长度保持不变）．已知矩形ABCD，AB=4cm，AD=3cm，固定边AB，推边AD，使得点D落在点E处，点C落在点F处．（1）如图5-1，如果∠DAE=30°，求点E到边AB的距离；（2）如图5-2，如果点A、E、C三点在同一直线上，求四边形ABFE的面积．DCDCEFEFABA图5图5-1图5-2B23．（本题满分12分，每小题满分6分）已知：如图6，在梯形ABCD中，CD∥AB，∠DAB=90°，对角线AC、BD相交于点E，AC⊥BC，垂足为点C，且BC2CECA．DC（1）求证：AD=DE；（2）过点D作AC的垂线，交AC于点F，E求证：CE2AEAF．AB图6第-3-页24．（本题满分12分，每小题满分4分）1如图7，在平面直角坐标系xOy中，抛物线yx2bx经过点A(2，0)．直线yx2与x轴交2于点B，与y轴交于点C．求这条抛物线的表达式和顶点的坐标；（1）y（2）将抛物线yx2bx向右平移，使平移后的抛物线经过点B，求平移后抛物线的表达式；（3）将抛物线yx2bx向下平移，使平移后的抛物线交y轴于点D，交线段BC于点P、Q，（点P在点Q右侧），平移后抛物线的顶点为M，如果DP∥x轴，求∠MCP的oABx正弦值．C图725．（本题满分14分，第(1)小题满分4分，第(2)小题满分5分，第(3)小题满分5分）如图8，已知半圆⊙O的直径AB=10，弦CD∥AB，且CD=8，E为弧CD的中点，点P在弦CD上，联结PE，过点E作PE的垂线交弦CD于点G，交射线OB于点F．（1）当点F与点B重合时，求CP的长；（2）设CP=x，OF=y，求y与x的函数关系式及定义域；（3）如果GP=GF，求△EPF的面积．EEGCDCDPAAOFBOB图8备用图第-4-页奉贤区2019学年度九年级数学质量调研 参考答案 有机化学期末考试题统计学b答案数学分析3答案计算机必考试卷02新大家的日语参考答案 一、选择题：（本大题共6题，每题4分，满分24分）1．D；2．A；3．A；4．B；5．C；6．B.二、填空题：（本大题共12题，每题4分，满分48分）7.3ab；8.x3；9.x15；1x111．；12.减小；10.；3y1113．106.1；14．ab；15．360；216．40；17．1r8；18．125．三、解答题（本大题共7题，其中19-22题每题10分，23、24题每题12分，25题14分，满分78分）119．解原式=22(2-2)1··································（每个2分，共8分）423=22121．······························································（2分）22x3x3620.解原式=····························································（4分）(x3)2x3x3x31=．······················（3分）(x3)2x3x3当x3时，原式=133．··················································（3分）33621.（1）解：过点C作CHy轴，垂足为H，得CH//x轴．∴BCCH．···················································································（1分）ABAO∵A（-2，0），∴AO=2，∴CH=2．∵点C的纵坐标为4，∴点C的坐标为（2，4）．······································（2分）设直线AB的表达式ykxb(k0)，2kb0k1由它经过点A、C，得，解得．···········（2分）2kb4b2∴直线AB的表达式yx2．m（2）∵反比例函数y＝的图像交于点C（2，4），∴m8．·······················（1分）x第-5-页∵直线AB与与y轴的正半轴交于点B，∴点B的坐标为（0，2）．··············（1分）∵BD∥x轴，∴点D纵坐标为2．··························································（1分）∵点D在反比例函数y＝8的图像上，∴点D坐标为（4，2）．····················（1分）x∴CD(2-4)2(4-2)222．·················································（1分）············22.（1）过点E作EHAB轴，垂足为H．·············································（1分）∵四边形ABCD是矩形，∴∠DAB=90°，∴AD//EH．∴∠DAE=∠AEH．··············································································（1分）∵∠DAE=30°，∴∠AEH=30°．在直角△AEH中，∠AHE=90°，∴EHAEcosAEH．·········（2分）··················23323∵AD=AE=3cm，∴EH3cm．································33即点E到边AB的距离是cm．2（2）过点E作EHAB，垂足为H．∵四边形ABCD是矩形，∴AD=BC．∵AD=3cm，∴BC=3cm．在直角△ABC中，∠ABC=90°，AB=4cm，，∴ACAB2BC25cm．································································（1分）AEEH∵EH//BC，∴．ACBC39∵AE=AD=3cm，∴EH．∴EHcm．···········································（2分）545∵推移过程中边的长度保持不变，∴ADAEBF,ABDCEF．∴四边形ABCD是平行四边形．·····························································（1分）936cm2．··············································（1分）∴SABEH4平行四边形ABFE55BCCA23．证明：（1）∵BC2CECA，∴．···········································（1分）CEBC∵ECBBCA，∴△BCE∽△ACB．············································（1分）∴CBECAB．·······················（1分）∵AC⊥BC，∠DAB=90°，∴BECCBE90，DAECAB90．∴BECDAE．·········································································（1分）第-6-页∵BECDEA，∴DAEDEA．··············································（1分）∴ADDE．··················································································（1分）（2）∵DF⊥AC，AC⊥BC，∴∠DFE=∠BCA=90°．∴DF//BC．CEBE∴．·················································································（2分）EFDEBEAE∵∴．·····························································（1分）DC//AB，DECECEAE∴．····················································································（1分）EFCE∵ADDE，DF⊥AC，∴AFEF．····················································（1分）∴CE2AEEF．··············································································（1分）24.解：（1）由题意，抛物线yx2bx经过点A(2，0)，得042b，解得b2·····················································（2分）∴抛物线的表达式是yx22x.··························································（1分）它的顶点C的坐标是（1，-1）.······························································（1分）1（2）∵直线yx2与x轴交于点B，∴点B的坐标是(4，0).··················（1分）2①将抛物线yx22x向右平移2个单位，使得点A与点B重合，此时平移后的抛物线表达式是y（x3）21·······································（2分）②将抛物线yx22x向右平移4个单位，使得点O与点B重合，此时平移后的抛物线表达式是y（x5）21········································（1分）（3）设向下平移后的抛物线表达式是：yx22xn，得点D（0，n）.∵DP∥x轴，∴点D、P关于抛物线的对称轴直线x1对称，∴P（2，n）.1∵点P在直线BC上，∴n221.2∴平移后的抛物线表达式是：yx22x2.··········································（2分）∴新抛物线的顶点M的坐标是（1，-2）.················································（1分）∴MC//OB，∴∠MCP=∠OBC．OC在Rt△OBC中，sinOBC，BC第-7-页由题意得：OC=2，BC25，25∴sinMCPsinOBC.·····················································（1分）2555即∠MCP的正弦值是.525.解：（1）联结EO，交弦CD于点H.∵E为弧CD的中点，∴EO⊥AB．······························································（1分）1∵CD∥AB，∴OH⊥CD．∴CH=CD．2联结CO，∵AB=10，CD=8，∴CO=5，CH4．∴OHCO2CH23．··········································································（1分）∴EHEOOH2．∵点F与点B重合，∴OBEHGE45．∵PE⊥BE，∴HPEHGE45，∴PEGE．········································（1分）∴PHHG2．∴CPCHPH2．··············································································（1分）（2）∵∠PEH+∠OEF=90°，∠OFE+∠OEF=90°，∴∠PEH=∠OFE．∵∠PHE=∠EOF=90°，∴PEH∽EFO．···············································（2分）EHPH∴．FOEO24x∵EH2，FOy，PH4x，EO5，∴．···········（1分）y510∴y（0x3）．···········································································（2分）4x（3）过点P作PQAB，垂足为Q．∵GP=GF，∴∠GPF=∠GFP．·································································（1分）∵CD∥AB，∴∠GPF=∠PFQ．∵PE⊥EF，∴·····································（1分）PQ=PE．······································第-8-页微信公众号：上海教学案中心PEPH由（2）可知，PEH∽EFO，∴．EFEO∵PQ=OH=3，∴PE=3．∵，∴PH22．EH2PEEH535∴．EF5∴EF35．························································································（2分）1195∴SPEEF33．··················································（1分）5EPF222第-9-页