2009年广西省梧州市初中毕业升学考试数学试题

第1页共8页2009年梧州市初中毕业升学考试试题卷数学说明：1．本试卷共8页（试题卷4页，答题卷4页），满分120分，考试时间120分钟．2．答卷前，将准考证号、姓名写在答题卷密封线内，答案请写在答题卷相应的区域内，在试题卷上答题无效．．．．．．．．．．一、填空题（本大题共10小题，每小题3分，共30分．）1．6的相反数是★．2．比较大小：－3★－4．（用“＞”“=”或“＜”表示）3．一组数据为1，2，3，4，5，6，则这组数据的中位数是★．4．因式分解：1822x＝★．5．如图（1），△ABC中，∠A＝60°，∠C＝40°，延长CB到D，则∠ABD＝★度．6．将点A（1，－3）向右平移2个单位，再向下平移2个单位后得到点B（a，b），则ab＝★．7．某蔬菜基地的圆弧形蔬菜大棚的剖面如图（2）所示，已知AB＝16m，半径OA＝10m，则中间柱CD的高度为★m．8．在△ABC中，∠C＝90°，BC＝6cm，53sinA，则AB的长是★cm．9．一个扇形所在圆的半径为3cm，扇形的圆心角为120°，则扇形的面积是★cm2．（结果保留π）10．图（3）是用火柴棍摆成的边长分别是1，2，3根火柴棍时的正方形．当边长为n根火柴棍时，设摆出的正方形所用的火柴棍的根数为s，则s＝★．（用n的代数式表示s）二、选择题（本大题共8小题，每小题3分，共24分．在每小题给出的四个选项中，只有一项是正确的，每小题选对得3分，选错、不选或多选均得零分．）11．在函数21xy中，自变量x的取值范围是（）A．2xB．2xC．x≤2D．x≥2DBAOC图（2）图（3）……n=1n=2n=3ABCD图（1）第2页共8页12．下列运算正确的是（）A．632aaaB．422aaaC．632)(aaD．aaa313．一个布袋中有4个除颜色外其余都相同的小球，其中3个白球，1个红球．从袋中任意摸出1个球是白球的概率是（）A．43B．41C．32D．3114．不等式组2201xx≥的解集在数轴上表示为（）A．B．C．D．15．在下列对称图形中，对称轴的条数最少的图形是（）A．圆B．等边三角形C．正方形（D）正六边形16．在一个仓库里堆放有若干个相同的正方体货箱，仓库管理员将这堆货箱的三视图画出来，如图（4），则这堆货箱共有（）A．6个B．5个C．4个D．3个17．已知点A（11xy，）、B（22xy，）是反比例函数xky（0k）图象上的两点，若210xx，则有（）A．210yyB．120yyC．021yyD．012yy18．如图（5），正方形ABCD中，E为AB的中点，AF⊥DE于点O，则DOAO等于（）A．352B．31C．32D．21三、解答题（本大题共8小题，满分66分．）19．（本题满分6分）计算：11122sin60220．（本题满分6分）解方程：0)3(2)3(2xxx21．（本题满分6分）为了解全市太阳能热水器的销售情况，某调查公司对人口为100万人的某县进行调查，对调查所得的数据整理后绘制成如图（6）所示的统计图．请据图解答下列问题：图（4）主视图左视图俯视图123－10－2123－10－2123－10－2123－10－2图（5）ABFCDEO第3页共8页（1）2008年该县销售中档．．太阳能热水器★台．（2）若2007年销售太阳能热水器的台数是2005年的1.5倍，请补全图（6）-2的条形图．（3）若该县所在市的总人口约为500万人，估计2008年全市销售多少台高档太阳能热水器．22．（本题满分8分）某工厂要招聘甲、乙两种工种的工人150人，甲、乙两种工种的工人的月工资分别为600元和1000元．（1）设招聘甲种工种工人x人，工厂付给甲、乙两种工种的工人工资共y元，写出y（元）与x（人）的函数关系式；（2）现要求招聘的乙种工种的人数不少于甲种工种人数的2倍，问甲、乙两种工种各招聘多少人时，可使得每月所付的工资最少？23．（本题满分8分）如图（7），△ABC中，AC的垂直平分线MN交AB于点D，交AC于点O，CE∥AB交MN于E，连结AE、CD．（1）求证：AD＝CE；（2）填空：四边形ADCE的形状是★．24．（本题满分10分）由甲、乙两个工程队承包某校校园绿化工程，甲、乙两队单独完成这项工程所需时间比是3︰2，两队合做6天可以完成．2005-2008年该县销售太阳能热水器的数量统计图图(6)-21000700600台年20052006200720086007001000DBCAENMO图（7）低档占30%高档占10%中档2008年该县销售高、中、低档太阳能热水器的数量统计图图（6）-1图（6）·第4页共8页OEDBAC·（1）求两队单独完成此项工程各需多少天?（2）此项工程由甲、乙两队合做6天完成任务后，学校付给他们20000元报酬，若按各自完成的工程量分配这笔钱，问甲、乙两队各得到多少元?25．（本题满分10分）如图（8）所示，△ABC内接于⊙O，AB是⊙O的直径，点D在⊙O上，过点C的切线交AD的延长线于点E，且AE⊥CE，连接CD．（1）求证：DC=BC；（2）若AB=5，AC=4，求tan∠DCE的值．26．（本题满分12分）如图（9）-1，抛物线23yaxaxb经过A（1，0），C（3，2）两点，与y轴交于点D，与x轴交于另一点B．（1）求此抛物线的解析式；（2）若直线)0(1kkxy将四边形ABCD面积二等分，求k的值；（3）如图（9）-2，过点E（1，1）作EF⊥x轴于点F，将△AEF绕平面内某点旋转180°得△MNQ（点M、N、Q分别与点A、E、F对应），使点M、N在抛物线上，作MG⊥x轴于点G，若线段MG︰AG＝1︰2，求点M，N的坐标．2009年梧州市初中毕业升学考试图（8）DOBAxyCy=kx+1图（9）-1EFMNGOBAxy图（9）-2Q第5页共8页数学参考答案及评分标准一、填空题（本大题共10小题，每小题3分，共30分．）二、选择题（本大题共8小题，每小题3分，共24分．）三、解答题（本大题共8小题，满分66分．）19．解:原式＝232232·····································································3分＝3232········································································4分＝23···················································································6分20．解:0)23)(3(xxx········································································2分0)33)(3(xx············································································3分03x或033x········································································4分即31x或12x··············································································6分21．解：（1）600···················································································2分（2）在右图上补全条形图如图．·····································································4分（3）500÷100×1000×10%＝500······································································6分22．解：（1）)150(1000600xxy·····························································2分150000400xy································································3分（2）依题意得，1502xx≥·······························································5分50x≤············································································6分题号12345答案6＞3.52（x＋3）（x－3）100题号678910答案154103π2(1)nn题号1112131415161718答案BCADBCAD图(6)-21000700600台年20052006200720086007001000900第6页共8页因为－400＜0，由一次函数的性质知，当x=50时，y有最小值···················7分所以150－50=100答:甲工种招聘50人，乙工种招聘100人时可使得每月所付的工资最少．（8分）23．（1）证明：∵MN是AC的垂直平分线························1分∴OA＝OC∠AOD＝∠EOC=90°···············3分∵CE∥AB∴∠DAO＝∠ECO·······························4分∴△ADO≌△CEO·································5分∴AD＝CE········································6分（2）四边形ADCE是菱形．··································8分（填写平行四边形给1分）24．解:（1）设甲队单独完成此项工程需x天，由题意得······································1分13266xx·····················································································3分解之得15x·······················································································4分经检验，15x是原方程的解．·····························································5分所以甲队单独完成此项工程需15天，乙队单独完成此项工程需15×32=10（天）·················································6分（2）甲队所得报酬:8000615120000（元）···········································8分乙队所得报酬:12000610120000（元）····················································10分25．（1）证明：连接OC··················································································1分∵OA＝O