《2013年广西玉林市中考数学真题及答案.docx》由会员分享,可在线阅读,更多相关《2013年广西玉林市中考数学真题及答案.docx(18页珍藏版)》请在淘文阁 - 分享文档赚钱的网站上搜索。
1、well, with the effectiveness of services to defend the interests of the masses. Third, we should strive to do well. To achieve good practical results, the key is to know the law, to grasp the laws and using laws. Office of economic development, social progress, there are rules to follow. Only act ac
2、cording to the law, to overcome blindness and strengthening initiative, creative. Working in the Office, we should be good at analyzing the essence of things, to find regular thing, change from passive to active, to seek one. Investee 2. establishment of evaluation system. Evaluation system is essen
3、tially an incentive mechanism, the evaluation was objective and fair, reasonable, and can stimulate a persons energy, mobilizing peoples enthusiasm or be misleading, dampened the enthusiasm of people. At present, the concept of people-oriented people, but to establish and perfect evaluation system i
4、s still lagging behind. Work in this area should be seriously caught up. Three emphasis on to take advantage of. Is a focus on standards and scientific. A fundamental point of the evaluation criteria, is to keep contact, development, comprehensive eye evaluation of cadres. Office work, both record a
5、nd potential performance; both subjective efforts, take another look at the objective condition; both pragmatic enough, take another look at the retreat level. Second, focus on participating the breadth of the subject. Evaluation of cadres must give full play to democracy, cadres and the masses hand
6、ing over evaluation. Office work services leadership and service base, serve the people, then nominated the main object will contain a variety of services, so that assessment was accurate and to avoid one-sidedness. Third, pay attention to the seriousness of the conclusion. The use of evaluation fin
7、dings is mainly embodied in his direction. Permitted to choose one, it would be tantamount to establishing a banner. Evaluation cant take care of the balance, otherwise it would undermine the offset the positive significance of evaluation. Office of the party committees at all levels should take the
8、 findings as on the cadres bonus-penalty, an important basis for future movements, incentive and restraint effect of making evaluations. 3. Prevention of understanding infidelity. People-centered development, concrete is dialectical,. In practice in the process, to prevent one-sided and still look,
9、from actual people. One is to use dialectical point of view to understand. Humanist is the core of the scientific Outlook on development, remains the development of the scientific Outlook on development emphasizes, is still taking economic construction as the Center. Any departure from the developme
10、nt of thought and action, is a misinterpretation and misunderstanding of people-oriented. Adhere to people-oriented both to overcome one-sidedness of the old, but also to prevent one-sidedness, not accelerating the development of people and the opposition. Must be very clear, people oriented and spe
11、eding up development is not opposed, but consistent, we should never bestrengthened password centralized management, enhanced password communication network and the construction of electronic Government Affairs network, strengthen the translation of code cables work, strengthen the management of imp
12、ortant files, make sure that the password is absolutely safe, ensure that Cryptography is absolutely clear. History of work to write a history of the original, compiled a book called changde Yearbook, out of a journal of the history of the publication, production of a song of the past after reading
13、forums, changde, a story of peoples hard work and exhibition. Archives to go the path of managing archives according to law, strengthen the construction of all types of archives, with the focus on key project archives, private enterprises, credit and Community Archives Archives, making full use of e
14、lectronic means of technology and network, and promote the standardization and legalization of archives and information. Investee 4. insist on the logistics, will be around strong strong reforms, cadres of service management. Office workers is to be for this, deepen reform, strengthen management, an
15、d continuously meet the cadres of the organs function properly and the need for material and cultural life. One is to enhance the logistical support capability. Adapt to the new situation of the market economy, making full use of social power, flexibility in the use of market mechanisms and accelera
16、te the process of marketization and socialization of logistic services. Adhere to the institutionalization of the logistics management, establish and implement a financial management system, trolley management systems, health management systems, security systems, such as the logistics management sys
17、tem, strengthen the management of human, financial, material, and logistical services standardized and institutionalized track. Secondly, we should strengthen the management of construction. Increasing investment, improving the working conditions of organs; overall civilization create an activity, a
18、nd constantly improve the level of civilization; to promote greening, lighting, purification, flowers, landscaping, Office workers create a good environment for work and life. Third, to improve staff welfare. To in policy allows of range within, trying to for Office cadres workers solution housing,
19、and couples separation, and children employment, aspects of actual difficult provides help, has conditions of place and units, also should appropriate to increased cadres workers of welfare, real do career left people, feelings left people, appropriate treatment left people, let General Office staff
20、 in pay hard efforts of while, feel organization of warm, get due of affordable. Investee 5. insist on the team, will focus on stimulating vitality who care and respect for people. To Office cadres for the cut . Learning style, strengthen the theoretical knowledge, optimize the structure of knowledg
21、e, and continuously improve the quality of cadres and workers. Attention to their physical health, and actively carry out regular wholesome recreational2013年广西玉林市中考数学真题及答案一、选择题(共12小题,每小题3分,满分36分)1(3分)2的相反数是()A2B2CD考点:相反数分析:根据相反数的定义求解即可解答:解:2的相反数为:2故选B点评:本题考查了相反数的知识,属于基础题,掌握相反数的定义是解题的关键2(3分)若=30,则的补角
22、是()A30B60C120D150考点:余角和补角专题:计算题分析:相加等于180的两角称作互为补角,也作两角互补,即一个角是另一个角的补角因而,求这个角的补角,就可以用180减去这个角的度数解答:解:18030=150故选D点评:本题主要是对补角概念的考查,是需要在学习中识记的内容3(3分)我国第一艘航母“辽宁舰”最大排水量为67500吨,用科学记数法表示这个数字是()A6.75103吨B67.5103吨C6.75104吨D6.75105吨考点:科学记数法表示较大的数分析:科学记数法的表示形式为a10n的形式,其中1|a|10,n为整数确定n的值是易错点,由于67500有5位,所以可以确定n
23、=51=4解答:解:67 500=6.75104故选C点评:此题考查科学记数法表示较大的数的方法,准确确定a与n值是关键4(3分)直线c与a,b均相交,当ab时(如图),则()A12B12C1=2D1+2=90考点:平行线的性质分析:根据平行线的性质:两直线平行,内错角相等可得答案解答:解:ab,1=2,故选:C点评:此题主要考查了平行线的性质,关键是掌握两直线平行,内错角相等5(3分)在数轴上表示不等式x+51的解集,正确的是()ABCD考点:在数轴上表示不等式的解集;解一元一次不等式专题:计算题分析:求出不等式的解集,表示在数轴上即可解答:解:不等式x+51,解得:x4,表示在数轴上,如图
24、所示:故选B点评:此题考查了在数轴上表示不等式的解集,把每个不等式的解集在数轴上表示出来(,向右画;,向左画),数轴上的点把数轴分成若干段,如果数轴的某一段上面表示解集的线的条数与不等式的个数一样,那么这段就是不等式组的解集有几个就要几个在表示解集时“”,“”要用实心圆点表示;“”,“”要用空心圆点表示6(3分)已知一组从小到大的数据:0,4,x,10的中位数是5,则x=()A5B6C7D8考点:中位数分析:根据中位数是5,得出(4+x)2=5,求出x的值即可解答:解:一组从小到大的数据:0,4,x,10的中位数是5,则(4+x)2=5,x=6;故选B点评:此题考查了中位数,中位数是将一组数据
25、从小到大(或从大到小)重新排列后,最中间的那个数(最中间两个数的平均数),叫做这组数据的中位数,是一道基础题7(3分)某几何体的三视图如图所示,则组成该几何体共用了()小方块A12块B9块C7块D6块考点:由三视图判断几何体分析:观察该几何体的三视图发现该几何体共有三层,第一层有三个,第二层有两个,第三层也有两个,由此可以得到答案解答:解:观察该几何体的三视图发现该几何体共有三层,第一层有三个,第二层有两个,第三层也有两个,该几何体共有3+2+2=7个,故选C点评:本题考查了由三视图判断几何体的知识,解题的关键是会利用物体的三视图判断出该几何体的形状8(3分)如图是某手机店今年15月份音乐手机
26、销售额统计图根据图中信息,可以判断相邻两个月音乐手机销售额变化最大的是()A1月至2月B2月至3月C3月至4月D4月至5月考点:折线统计图分析:根据折线图的数据,分别求出相邻两个月的音乐手机销售额的变化值,比较即可得解解答:解:1月至2月,3023=7万元,2月至3月,3025=5万元,3月至4月,2515=10万元,4月至5月,1914=5万元,所以,相邻两个月中,用电量变化最大的是3月至4月故选C点评:本题考查折线统计图的运用,折线统计图表示的是事物的变化情况,根据图中信息求出相邻两个月的音乐手机销售额变化量是解题的关键9(3分)方程的解是()Ax=2Bx=1Cx=Dx=2考点:解分式方程
27、专题:计算题分析:分式方程去分母转化为整式方程,求出整式方程的解得到x的值,经检验即可得到分式方程的解解答:解:去分母得:x+13(x1)=0,去括号得:x+13x+3=0,解得:x=2,经检验x=2是分式方程的解故选A点评:此题考查了解分式方程,解分式方程的基本思想是“转化思想”,把分式方程转化为整式方程求解解分式方程一定注意要验根10(3分)如图,在给定的一张平行四边形纸片上作一个菱形甲、乙两人的作法如下:甲:连接AC,作AC的垂直平分线MN分别交AD,AC,BC于M,O,N,连接AN,CM,则四边形ANCM是菱形乙:分别作A,B的平分线AE,BF,分别交BC,AD于E,F,连接EF,则四
28、边形ABEF是菱形根据两人的作法可判断()A甲正确,乙错误B乙正确,甲错误C甲、乙均正确D甲、乙均错误考点:菱形的判定分析:首先证明AOMCON(ASA),可得MO=NO,再根据对角线互相平分的四边形是平行四边形可判定判定四边形ANCM是平行四边形,再由ACMN,可根据对角线互相垂直的四边形是菱形判定出ANCM是菱形;四边形ABCD是平行四边形,可根据角平分线的定义和平行线的定义,求得AB=AF,所以四边形ABEF是菱形解答:解:甲的作法正确;四边形ABCD是平行四边形,ADBC,DAC=ACN,MN是AC的垂直平分线,AO=CO,在AOM和CON中,AOMCON(ASA),MO=NO,四边形
29、ANCM是平行四边形,ACMN,四边形ANCM是菱形;乙的作法正确;ADBC,1=2,6=7,BF平分ABC,AE平分BAD,2=3,5=6,1=3,5=7,AB=AF,AB=BE,AF=BEAFBE,且AF=BE,四边形ABEF是平行四边形,AB=AF,平行四边形ABEF是菱形;故选:C点评:此题主要考查了菱形形的判定,关键是掌握菱形的判定方法:菱形定义:一组邻边相等的平行四边形是菱形(平行四边形+一组邻边相等=菱形);四条边都相等的四边形是菱形对角线互相垂直的平行四边形是菱形(或“对角线互相垂直平分的四边形是菱形”)11(3分)一列数a1,a2,a3,其中a1=,an=(n为不小于2的整数
30、),则a100=()AB2C1D2考点:规律型:数字的变化类专题:规律型分析:根据表达式求出前几个数不难发现,每三个数为一个循环组依次循环,用100除以3,根据商和余数的情况确定a100的值即可解答:解:根据题意得,a2=2,a3=1,a4=,a5=2,依此类推,每三个数为一个循环组依次循环,1003=331,a100是第34个循环组的第一个数,与a1相同,即a100=故选A点评:本题是对数字变化规律的考查,计算并观察出每三个数为一个循环组依次循环是解题的关键12(3分)均匀地向一个瓶子注水,最后把瓶子注满在注水过程中,水面高度h随时间t的变化规律如图所示,则这个瓶子的形状是下列的()ABCD
31、考点:函数的图象分析:根据图象可得水面高度开始增加的快,后来增加的慢,从而可判断容器下面粗,上面细,结合选项即可得出答案解答:解:因为水面高度开始增加的快,后来增加的慢,所以容器下面粗,上面细故选B点评:本题考查了函数的图象,要能根据函数图象的性质和图象上的数据分析得出函数的类型和所需要的条件,结合实际意义得到正确的结论二、填空题(共6小题,每小题3分,满分18分)13(3分)|1|=1考点:绝对值分析:计算绝对值要根据绝对值定义去掉这个绝对值的符号解答:解:|1|=1故答案为:1点评:此题考查了绝对值的性质,要求掌握绝对值的性质及其定义,并能熟练运用到实际运算当中绝对值规律总结:一个正数的绝
32、对值是它本身;一个负数的绝对值是它的相反数;0的绝对值是014(3分)化简:=考点:分母有理化分析:根据的有理化因式是,进而求出即可解答:解:=故答案为:点评:此题主要考查了分母有理化,正确根据定理得出有理化因式是解题关键15(3分)(2013平凉)分解因式:x29=(x+3)(x3)考点:因式分解-运用公式法分析:本题中两个平方项的符号相反,直接运用平方差公式分解因式解答:解:x29=(x+3)(x3)点评:主要考查平方差公式分解因式,熟记能用平方差公式分解因式的多项式的特征,即“两项、异号、平方形式”是避免错用平方差公式的有效方法16(3分)如图,实线部分是半径为15m的两条等弧组成的游泳
33、池,若每条弧所在的圆都经过另一个圆的圆心,则游泳池的周长是40m考点:弧长的计算分析:如图,连接O1O2,CD,可求得C02O1=60,C02D=120,再由弧长公式l=求得答案解答:解:如图,连接O1O2,CD,CO2,O1O2=C02=CO1=15cm,C02O1=60,C02D=120,则圆O1,O2的圆心角为360120=240,则游泳池的周长为=2=2=40(m)故答案为:40点评:本题考查了弧长的计算,解答本题的关键是根据弧长公式计算,在计算的过程中首先要利用圆的半径的关系求出圆心角17(3分)如图,在直角坐标系中,O是原点,已知A(4,3),P是坐标轴上的一点,若以O,A,P三点
34、组成的三角形为等腰三角形,则满足条件的点P共有6个,写出其中一个点P的坐标是(5,0)考点:等腰三角形的判定;坐标与图形性质专题:数形结合分析:作出图形,然后利用数形结合的思想求解,再根据平面直角坐标系写出点P的坐标即可解答:解:如图所示,满足条件的点P有6个,分别为(5,0)(8,0)(0,5)(0,6)(5,0)(0,5)故答案为:6;(5,0)(答案不唯一,写出6个中的一个即可)点评:本题考查了等腰三角形的判定,坐标与图形的性质,利用数形结合的思想求解更简便18(3分)如图,ABC是O内接正三角形,将ABC绕点O顺时针旋转30得到DEF,DE分别交AB,AC于点M,N,DF交AC于点Q,
35、则有以下结论:DQN=30;DNQANM;DNQ的周长等于AC的长;NQ=QC其中正确的结论是(把所有正确的结论的序号都填上)考点:圆的综合题分析:连结OA、OD、OF、OC、DC、AD、CF,根据旋转的性质得AOD=COF=30,再根据圆周角定理得ACD=FDC=15,然后根据三角形外角性质得DQN=QCD+QDC=30;同理可得AMN=30,由DEF为等边三角形得DE=DF,则弧DE=弧DF,得到弧AE=弧DC,所以ADE=DAC,根据等腰三角形的性质有ND=NA,于是可根据“AAS”判断DNQANM;利用QD=QC,ND=NA可判断DNQ的周长等于AC的长;由于NDQ=60,DQN=30
36、,则DNQ=90,所以QDNQ,而QD=QC,所以QCNQ解答:解:连结OA、OD、OF、OC、DC、AD、CF,如图,ABC绕点O顺时针旋转30得到DEF,AOD=COF=30,ACD=AOD=15,FDC=COF=15,DQN=QCD+QDC=15+15=30,所以正确;同理可得AMN=30,DEF为等边三角形,DE=DF,弧DE=弧DF,弧AE+弧AD=弧DC+弧CF,而弧AD=弧CF,弧AE=弧DC,ADE=DAC,ND=NA,在DNQ和ANM中,DNQANM(AAS),所以正确;ACD=15,FDC=15,QD=QC,而ND=NA,ND+QD+NQ=NA+QC+NQ=AC,即DNQ的
37、周长等于AC的长,所以正确;DEF为等边三角形,NDQ=60,而DQN=30,DNQ=90,QDNQ,QD=QC,QCNQ,所以错误故答案为点评:本题考查了圆的综合题:弧、弦和圆心角之间的关系以及圆周角定理在有关圆的几何证明中经常用到,同时熟练掌握三角形全等的判定、等边三角形的性质以及旋转的性质三、解答题(共8小题,满分66分)19(6分)计算:+2cos60(21)0考点:实数的运算;零指数幂;负整数指数幂;特殊角的三角函数值分析:分别进行三次根式的化简、零指数幂的运算,然后特殊角的三角函数值后合并即可得出答案解答:解:原式=2+21=2点评:本题考查了实数的运算,涉及了零指数幂及特殊角的三
38、角函数值,特殊角的三角函数值是需要我们熟练记忆的内容20(6分)如图,AB=AE,1=2,C=D求证:ABCAED考点:全等三角形的判定专题:证明题分析:首先根据1=2可得BAC=EAD,再加上条件AB=AE,C=D可证明ABCAED解答:证明:1=2,1+EAC=2+EAC,即BAC=EAD,在ABC和AED中,ABCAED(AAS)点评:此题主要考查了三角形全等的判定方法,判定两个三角形全等的一般方法有:SSS、SAS、ASA、AAS、HL注意:AAA、SSA不能判定两个三角形全等,判定两个三角形全等时,必须有边的参与,若有两边一角对应相等时,角必须是两边的夹角21(6分)已知关于x的方程
39、x2+x+n=0有两个实数根2,m求m,n的值考点:根与系数的关系分析:利用根与系数的关系知2+m=1,2m=n,据此易求m、n的值解答:解:关于x的方程x2+x+n=0有两个实数根2,m,解得,即m,n的值分别是1、2点评:本题考查了根与系数的关系,属于基础题解题过程中,需要熟记公式x1+x2=,x1x2=22(8分)某小区为了促进生活垃圾的分类处理,将生活垃圾分为:可回垃圾、厨余垃圾、其他垃圾三类,分别记为A,B,C:并且设置了相应的垃圾箱,依次记为a,b,c(1)若将三类垃圾随机投入三个垃圾箱,请你用树形图的方法求垃圾投放正确的概率:(2)为了调查小区垃圾分类投放情况,现随机抽取了该小区
40、三类垃圾箱中总重500kg生活垃圾,数据如下(单位:)abcA401510B6025040C151555试估计“厨余垃圾”投放正确的概率考点:列表法与树状图法;利用频率估计概率分析:(1)根据题意画出树状图,由树状图可知总数为9,投放正确有3种,进而求出垃圾投放正确的概率;(2)由题意和概率的定义易得所求概率解答:解:(1)如图所示:共有9种情况,其中投放正确的有3种情况,故垃圾投放正确的概率:=;(2)“厨余垃圾”投放正确的概率为:=点评:本题考查的是用列表法或画树状图法求概率列表法或画树状图法可以不重复不遗漏的列出所有可能的结果用到的知识点为:概率=所求情况数:总情况数23(9分)如图,以
41、ABC的BC边上一点O为圆心的圆,经过A,B两点,且与BC边交于点E,D为BE的下半圆弧的中点,连接AD交BC于F,若AC=FC(1)求证:AC是O的切线:(2)若BF=8,DF=,求O的半径r考点:切线的判定分析:(1)连接OA、OD,求出D+OFD=90,推出CAF=CFA,OAD=D,求出OAD+CAF=90,根据切线的判定推出即可;(2)OD=r,OF=8r,在RtDOF中根据勾股定理得出方程r2+(8r)2=()2,求出即可解答:(1)证明:连接OA、OD,D为弧BE的中点,ODBC,DOF=90,D+OFD=90,AC=AF,OA=OD,CAF=CFA,OAD=D,CFA=OFD,
42、OAD+CAF=90,OAAC,OA为半径,AC是O切线;(2)解:O半径是r,当F在半径OE上时,OD=r,OF=8r,在RtDOF中,r2+(8r)2=()2,r=,r=(舍去);当F在半径OB上时,OD=r,OF=r8,在RtDOF中,r2+(r8)2=()2,r=,r=(舍去);即O的半径r为点评:本题考查了切线的判定,等腰三角形的性质和判定,勾股定理等知识点的应用,主要考查学生的推理和计算的能力24(9分)工匠制作某种金属工具要进行材料煅烧和锻造两个工序,即需要将材料烧到800,然后停止煅烧进行锻造操作,经过8min时,材料温度降为600煅烧时温度y()与时间x(min)成一次函数关
43、系;锻造时,温度y()与时间x(min)成反比例函数关系(如图)已知该材料初始温度是32(1)分别求出材料煅烧和锻造时y与x的函数关系式,并且写出自变量x的取值范围;(2)根据工艺要求,当材料温度低于480时,须停止操作那么锻造的操作时间有多长?考点:反比例函数的应用;一次函数的应用分析:(1)首先根据题意,材料加热时,温度y与时间x成一次函数关系;停止加热进行操作时,温度y与时间x成反比例关系;将题中数据代入用待定系数法可得两个函数的关系式;(2)把y=480代入y=中,进一步求解可得答案解答:解:(1)停止加热时,设y=(k0),由题意得600=,解得k=4800,当y=800时,解得x=6,点B的坐标为(6,800)材料加热时,设y=ax+32(a0),由题意得800=6a+32,解得a=128,材料加热时,y与x的函数关系式为y=128x+32(0x5)停止加热进行操作时y与x的函数关系式为y=(5x20);(2)把y=480代入y=,得x=10,故从开始加热到停止操作,共经历了10分钟答:从开始加热到停止操作,共经历了10分钟点评:考查了反比例函数和一次函数的应用,现实生活中存在大量成反比例函数
限制150内