温州外国语学校第三次中考模拟试卷
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2020年浙江省温州外国语学校九年级第三次模拟考试英语试卷(WORD含答案)2020年温州市初中毕业升学考试模拟检测英语试题卷2020. 06 考生须知:欢迎参加考试!请你认真审题,积极思考,细心答题,发挥最佳水平。
答题时,请注意以下几点:1. 全卷共8页,有五大题,56小题。
全卷满分120分,考试时间90分钟。
2. 请用2B铅笔将选择题的答案填涂在答题卡相应位置上,用0. 5 毫米及以上黑色字迹的钢笔或签字笔将主观题的答案写在答题卡相应的答题区内。
写在试题卷、草稿纸上均无效。
3. 打*的单词可以在小词典里查找意思。
祝你成功!说明:全卷书写分4 分。
一、单项填空(本题有10小题,每小题2分,共20分)请从A、B、C、D四个选项中选出可以填入空白处的最佳答案。
1. -How was your summer vacation?-Great! We went to Vancouver _________ big city in Canada.A. aB. anC. theD. /2. --What smells ___________,Ted?-I'm sorry. I’ll take off my socks and put them in the washer.A. niceB. wellC. terribleD. terribly3. -Vicky, why don't you join us in Hawaii?- Thank you, John. It's a very kind_________, but I have to stay and take care of my mother.A. decisionB. promiseC. orderD. offer4. –If you always ___________ yourself with others, you may have tons of pressure.--I agree, we should believe in ourselves.A. considerB. complainC. compareD. connect5. --Mike, let's prepare for our baseball game.--OK. We’ll lose the game _______________ we try our best.A. unlessB. ifC. afterD. since6. --I can't stop playing computer games.--For your health, my boy, I'm afraid you___________.A. CanB. mayC. mustD. will7. --This blue dress looks nice on you!- My aunt _________it to me as a present on my 15th birthday.A. givesB. gaveC. has givenD. will give8. --The more I get to know about Nancy, the more I can realize that we have a lot _______.--No wonder she is your best friend.A. in needB. in commonC. in troubleD. in peace9. -Do you know_________ this evening?-Let me see. Oh, at 7 o'clock!A. how long the bite of China Ⅱwill lastB. how long will the bite of China ⅡlastC. w hen the bite of China Ⅱwill beginD. when will the bite of China Ⅱbegin10. --Carl had an accident. He hurt one of his legs yesterday. ______!I hope he gets better soon.A. No wayB. Hurry upC. Look outD. Bad luck二、完形填空(本题有15小题,每小题1分,共15分)阅读下面短文,掌握大意,然后从每题所给的A、B、C、D 四个选项中选出最佳选项。
2023-2024学年浙江温州外国语学校中考语文全真模拟试卷注意事项:1.答题前,考生先将自己的姓名、准考证号码填写清楚,将条形码准确粘贴在条形码区域内。
2.答题时请按要求用笔。
3.请按照题号顺序在答题卡各题目的答题区域内作答,超出答题区域书写的答案无效;在草稿纸、试卷上答题无效。
4.作图可先使用铅笔画出,确定后必须用黑色字迹的签字笔描黑。
5.保持卡面清洁,不要折暴、不要弄破、弄皱,不准使用涂改液、修正带、刮纸刀。
一、积累与运用1.下列词语中加点的字,每对读音都不同的一项是()A.沉淀./破绽.懈怠./百战不殆.循规蹈矩./目光如炬.B.凭据./拮据.卷.曲/手不释卷.排除万难./患难.与共C.创.新/悲怆.弹劾./言简意赅.含辛.茹苦/莘.莘学子D.晕.车/黄晕.倔强./强.词夺理宁.缺毋滥/宁.死不屈2.下列词语中加点字的注音完全正确的一项是()A.热忱.(chén)骊.歌(lí)肆.虐(sì)针砭.时弊(biān)B.渲.染(xuān)藐.视(miǎo)旖.旎(yǐ)参差.不齐(cī)C.重荷.(hè)翘.首(qiào)莅.临(lì)咄.咄逼人(duō)D.矜.持(jīn)偌.大(ruò)峥.嵘(zhēnɡ)束.手无策(sù)3.下列说法不正确的一项是()A.古人对“死”有许多讳称,如对天子、太后、公卿王侯之死称为薨、百岁等;对父母之死称见背、崩等;对佛道徒之死称坐化、仙游等。
B.黄金台,亦称招贤台,又称金台、燕台。
据史料考证,战国时期燕昭王有感于千金买骨的故事,高筑“黄金台”以招贤纳士,以致名将乐毅、剧辛先后投奔燕国。
C.传说中的“三山”因是神仙居住的地方,格外受到古人的神往。
《史记》载:“齐人徐福等上书,言海中有三神山,名曰蓬莱、方丈、瀛洲”。
D.古代人作揖的方式有很多,男子右手握拳左手包于其上是“吉拜”,表示尊重,用于见面、告别等场合;相反的手势则是“凶拜”,一般用于吊丧。
中考数学三模试卷题号一二三总分得分一、选择题(本大题共10小题,共40.0分)1.-2019的相反数是( )A. 2019B. -2019C.D. -2.如图所示的几何体的左视图是( )A.B.C.D.3.鞋店要进一批新鞋,你是店长,应关注下列哪个统计量( )A. 平均数B. 方差C. 众数D. 中位数4.下列四幅图案中,既是轴对称图形又是中心对称图形的是( )A. B. C. D.5.下列运算正确的是( )A. x3+x2=x5B. (x-3)2=x2-9C. (x2)3=x5D. 5x2•x3=5x56.一个圆锥的高是4cm,底面半径是3cm,那么这个圆锥的侧面积为( )A. 15cm2B. 12cm2C. 15πcm2D. 12πcm27.某公司承担了制作300个道路交通指引标志的任务,原计划x天完成,实际平均每天多制作了5个,因此提前10天完成任务.根据题意,下列方程正确的是( )A. B.C. D.8.已知m是方程x2-2019x+1=0的一个根,则代数式m2-2018m++2的值是( )A. 2018B. 2019C. 2020D. 20219.如图,将矩形ABCD的四边BA,CB,DC,AD分别延长至点EF,G,H,使得AE=BF=CG=DH.已知AB=1,BC=2,∠BEF=30°,则tan∠AEH的值为( )A. 2B.C. -1D. +110.如图,一次函数分别与x轴,y轴交于AB两点,与反比例函数交于C、D两点,若CD=5AB,则k的值是( )A. B.6 C. 8 D. -4二、填空题(本大题共6小题,共30.0分)11.因式分解:a2+2ab= ______ .12.不等式的解集是______.13.如图,AB∥CD,EF平分∠AEC,EG⊥EF.若∠C=110°,则∠BEG的度数为______度.14.如图,已知直线y=+b交y轴正半轴于点B,在x轴负半轴上取点A,使2BO=3AO,AC⊥x轴交直线y=+b于点C,若△OAC的面积为,则b的值为______.15.如图,在直角坐标系中,⊙A的圆心坐标为(,a)半径为,函数y=2x-2的图象被⊙A截得的弦长为2,则a的值为______.16.如图,在正方形ABCD中,AB=3,点E是对角线BD上的一点,连结AE,过点E作EF垂直AE交BC于点F,连结AF,交对角线BD于G.若三角形AED与四边形DEFC的面积之比为3:8,则cos∠GEF=______.三、解答题(本大题共8小题,共80.0分)17.(1)计算:2-1++(2019+π)0-7sin30°(2)先化简,再求值:(x+4)2-x(x-3),其中x=18.两块完全相同的直角三角形纸板ABC和DEF,按如图所示的方式叠放,其中∠ABC=∠DEF=90°,点O为边BC和EF的交点.(1)求证:△BOF≌△COE.(2)若∠F=30°,AE=1,求OC的长.19.在一个不透明的布袋里装有4个球,其中3个白球,1个红球,它们除颜色外其余都相同.(1)若从中任意摸出一个球,求摸出白球的概率;(2)若摸出1个球,记下颜色后不放回,再摸出1个球,求两次摸出的球恰好颜色相同的概率(要求画树状图或列表)20.已知网格的小正方形的边长均为1,格点三角形ABC如图所示,请仅使用无刻度的直尺,且不能用直尺中的直角,画出满足条件的图形(保留作图痕迹)(1)在图甲AB边上取点D,使得△BCD的面积是△ABC的;(2)在图乙中,画出△ABC所在外接圆的圆心位置.21.如图,在△ABC中,AB=BC,以AB为直径的⊙O交BC于点D,交AC于点F,过点C作CE∥AB,与过点A的切线相交于点E,连接AD.(1)求证:AD=AE.(2)若AB=10,sin∠DAC=,求AD的长.22.如图,过抛物线y=ax2+bx上一点A(4,-2)作x轴的平行线,交抛物线于另一点B,点C在直线AB上,抛物线交x轴正半轴于点D(2,0),点B与点E关于直线CD对称.(1)求抛物线的表达式;(2)①若点E落在抛物线的对称轴上,且在x轴下方时,求点C的坐标.②AE最小值为______.23.某水产经销商从批发市场以30元每千克的价格收购了1000千克的虾,了解到市场价在一个月内会以每天0.5元每千克的价格上涨,经销商打算先在塘里放养几天后再出售(但不超过一个月).假设放养期间虾的个体质量保持不变,但每天有10千克的虾死去.死去的虾会在当天以20元每千克的价格售出.(1)若放养10天后出售,则活虾的市场价为每千克______元.(2)若放养x天后将活虾一次性售出,这1000千克的虾总共获得的销售额为36000元,求x的值.(3)若放养期间,每天会有各种其他的各种费用支出为a元,经销商在放养x天后全部售出,当20≤x≤30时,经销商日获利的最大值为1800元,则a的值为______(日获利=日销售总额-收购成本-其他费用)24.如图,在ABC中,已知AB=BC=10,AC=4,AD为边BC上的高线,P为边AD上一点,连结BP,E为线段BP上一点,过D、P、E三点的圆交边BC于F,连结EF.(1)求AD的长;(2)求证:△BEF∽△BDP;(3)连结DE,若DP=3,当△DEP为等腰三角形时,求BF的长;(4)把△DEP沿着直线DP翻折得到△DGP,若G落在边AC上,且DG∥BP,记△APG、△PDG、△GDC的面积分别为S1、S2、S3,则S1:S2:S3的值为______.答案和解析1.【答案】A【解析】解:因为a的相反数是-a,所以-2019的相反数是2019.故选:A.根据相反数的意义,直接可得结论.本题考查了相反数的意义.理解a的相反数是-a,是解决本题的关键.2.【答案】B【解析】解:从左边看第一层是两个小正方形,第二层是一个小正方形,故选:B.根据左边看得到的图形是左视图,可得答案.本题考查了简单组合体的三视图,从左边看得到的图形是左视图.3.【答案】C【解析】解:由于众数是数据中出现次数最多的数,故应最关心这组数据中的众数.故选:C.平均数、中位数、众数是描述一组数据集中程度的统计量;方差是描述一组数据离散程度的统计量.既然是对该鞋子销量情况作调查,那么应该关注那种尺码销的最多,故值得关注的是众数.此题主要考查统计量的旋转,数据的平均数、众数、中位数是描述一组数据集中趋势的特征量,极差、方差是衡量一组数据偏离其平均数的大小(即波动大小)的特征数,描述了数据的离散程度.4.【答案】D【解析】解:A、不是轴对称图形,也不是中心对称图形,不符合题意;B、是轴对称图形,不是中心对称图形,不符合题意;C、不是中心对称图形,是轴对称图形,不符合题意;D、是轴对称图形,也是中心对称图形,符合题意.故选:D.根据轴对称图形与中心对称图形的概念求解.此题考查了转轴对称及中心对称的知识,轴对称图形的关键是寻找对称轴,图形两部分折叠后可重合,中心对称图形是要寻找对称中心,旋转180度后两部分重合.5.【答案】D【解析】解:A、x3和x2不能合并同类项,故本选项不符合题意;B、结果是x2-6x+9,故本选项不符合题意;C、结果是x6,故本选项不符合题意;D、结果是5x5,故本选项,符合题意;故选:D.根据合并同类项法则、完全平方公式、单项式乘以单项式、幂的乘方和积的乘方分别求出每个式子的值,再判断即可.本题考查了合并同类项法则、完全平方公式、单项式乘以单项式、幂的乘方和积的乘方等知识点,能正确求出每个式子的值是解此题的关键.6.【答案】C【解析】【分析】先根据勾股定理计算出圆锥的母线长,然后根据圆锥的侧面展开图为一扇形,这个扇形的弧长等于圆锥底面的周长,扇形的半径等于圆锥的母线长和扇形的面积公式求解.本题考查了圆锥的计算:圆锥的侧面展开图为一扇形,这个扇形的弧长等于圆锥底面的周长,扇形的半径等于圆锥的母线长.【解答】解:圆锥的母线长==5(cm),所以这个圆锥的侧面积=×5×2π×3=15π(cm2).故选C.7.【答案】B【解析】解:设原计划x天完成,根据题意得:-=5.故选:B.根据制作的个数为等量关系得出等式即可.此题主要考查了由实际问题抽象出分式方程,正确找出等量关系是解题关键.8.【答案】C【解析】解:∵m是方程x2-2019x+1=0的一个根,∴m2-2019m+1=0,∴m2=2019m-1,∴m2-2018m++2=2019m-2018m-1++2=m++1=+1=+1=2019+1=2020.故选:C.利用一元二次方程的解的定义得到m2=2019m-1,利用整体代入的方法变形得到m2-2018m++2=m++1,然后通分后再利用整体代入的方法计算.本题考查了一元二次方程的解:能使一元二次方程左右两边相等的未知数的值是一元二次方程的解.9.【答案】C【解析】解:设AE=BF=CG=DH=x,∵四边形ABCD是矩形,∴∠ABC=∠BAD=90°,∴∠EAD=∠EBF=90°,∵AB=1,∠BEF=30°,∴BE=BF,∴x+1=x,解得:x=,∴AE=BF=CG=DH=,∴AH=AD+DH=2+=,∴tan∠AEH===2-1,故选:C.设AE=BF=CG=DH=x,根据矩形的性质得出AD=BC=2,∠ABC=∠BAD=90°,求出∠EAD=∠EBF=90°,解直角三角形求出x,求出AH,解直角三角形求出即可.本题考查了矩形的性质和解直角三角形,能解直角三角形求出x是解此题的关键.10.【答案】B【解析】解:作CE⊥y轴于E,DF⊥x轴于F,连接EF,DE、CF,设D(x,),则F(x,0),由图象可知x>0,k>0,∴△DEF的面积是וx=k,同理可知:△CEF的面积是k,∴△CEF的面积等于△DEF的面积,∴边EF上的高相等,∴CD∥EF,∵BD∥EF,DF∥BE,∴四边形BDFE是平行四边形,∴BD=EF,同理EF=AC,∴AC=BD,∵CD=5AB,∴AD=3AB,由一次函数分别与x轴,y轴交于AB两点,∴A(-1,0),B(0,),∴OA=1,OB=,∵OB∥DF,∴===,∴DF=3,AF=3,∴OF=3-1=2,∴D(2,3),∵点D在反比例函数图象上,∴k=2×=6,故选:B.作CE⊥y轴于E,DF⊥x轴于F,连接EF,DE、CF,设D(x,),得出F(x,0),根据三角形的面积求出△DEF的面积,同法求出△CEF的面积,即可得到△CEF的面积等于△DEF的面积,证出平行四边形BDFE和平行四边形ACEF,得到BD=AC,则AD=3AB,根据平行线分线段成比例定理即可求得D点的坐标,代入反比例函数,即可求得k的值.本题考查了一次函数和反比例函数的交点问题,平行四边形的性质和判定,三角形的面积,相似三角形的判定等知识点的运用,关键是检查学生综合运用定理进行推理的能力,题目具有一定的代表性,有一定的难度,是一道比较容易出错的题目.11.【答案】a(a+2b)【解析】解:原式=a(a+2b),故答案为:a(a+2b)原式提取公因式即可得到结果.此题考查了因式分解-提公因式法,熟练掌握提取公因式的方法是解本题的关键.12.【答案】0<x≤【解析】解:,由①得:x≤,由②得:x>0,∴不等式组的解集为:0<x≤.故答案为:0<x≤.首先分别解出两个不等式的解集,再根据解集的规律:同大取大;同小取小;大小小大中间找;大大小小找不到确定其公共解集即可.此题主要考查了不等式组的解法,关键是正确解出两个不等式的解集.13.【答案】55【解析】解:∵AB∥CD,∴∠C+∠AEC=180°,∵∠C=110°,∴∠AEC=70°,∵EF平分∠AEC,∴∠AEF=35°,∵EF⊥EG,∴∠FEG=90°,∴∠BEG=90°-35°=55°,故答案为:55想办法求出∠AEF,再根据∠AEF+∠BEG=90°,即可求出∠BEG.本题考查平行线的性质,角平分线的定义等知识,解题的关键是熟练掌握基本知识,属于中考常考题型.14.【答案】【解析】解:∵y=+b交y轴正半轴于点B,∴B(0,b),∵在x轴负半轴上取点A,使2BO=3AO,∴B(0,b),当x=-时,y=2b,∴C(-,2b),∴△OAC的面积=×2b=,∴b=,故答案为.根据条件求出B(0,b),B(0,b),C(-,2b),再由△OAC的面积=×2b=,即可求b的值.本题考查一次函数的图象及性质;熟练掌握函数图象及性质,会求三角形的面积是解题的关键.15.【答案】4-2【解析】解:作AC⊥x轴于C,交CB于D,作AE⊥CB于E,连结AB,如图,∵⊙A的圆心坐标为(,a),∴OC=,AC=a,把x=代入y=2x-2得y=2-2,∴D点坐标为(,2-2),∴CD=2-2,∵AE⊥CB,∴CE=BE=BC=1,在Rt△ACE中,AC=,∴AE===2,∵y=2x-2,当x=0时,y=-2;当y=0时,x=1,∴G(0,-2),F(1,0),∴OG=2,OF=1,∵AC∥y轴,∴∠ADE=∠CDF=∠OGF,∴tan∠ADE==tan∠OGF==,∴DE=2AE=4,∴AD===2,∴a=AC=AD+CD=2+2-2=4-2,故答案为:4-2.作AC⊥x轴于C,交CB于D,作AE⊥CB于E,连结AB,由题意得出OC=,AC=a,把x=代入y=2x-2得y=2-2,得出D点坐标为(,2-2),得出CD=2-2,由垂径定理得出CE=BE=BC=1,由勾股定理得出AE==2,求出直线y=2x-2与坐标轴的交点坐标,得出OG=2,OF=1,由平行线的性质得出∠ADE=∠CDF=∠OGF,求出DE=2AE=4,由勾股定理得出AD==2,即可得出结果.本题考查了垂径定理、坐标与图形性质、一次函数的应用、勾股定理、平行线的性质、解直角三角形等知识.本题综合性强,有一定难度.16.【答案】【解析】解:连接CE,作EH⊥CD于H,EM⊥BC于M,如图所示:则四边形EMCH是矩形,∴EM=CH,CM=EH,∵四边形ABCD是正方形,∴BC=CD=3,∠ABC=90°,AB=CB,∠ABE=∠CBE=∠BDC=45°,在△ABE和△CBE中,,∴△ABE≌△CBE(SAS),∴EA=EF,∠BAE=∠BCE,同理:△ADE≌△CDE,∴△ADE的面积=△CDE的面积,∵△AED与四边形DEFC的面积之比为3:8,∴△CDE:△CEF的面积=3:5,∵EF⊥AE,∴∠AEF=90°,∴∠ABC+∠AEF=180°,∴A、B、F、E四点共圆,∴∠GEF=∠BAF,∠EFC=∠BAE=∠BCE,∴EF=EC,∵EM⊥BC,∴FM=CM=EH=DH,设FM=CM=EH=DH=x,则FC=2x,EM=HC=3-x,∵△CDE:△CEF的面积=3:5,∴,解得:x=,∴FC=1,BF=BC-FC=2,∴AF==,∴cos∠GEF=cos∠BAF===;故答案为:.连接CE,作EH⊥CD于H,EM⊥BC于M,则四边形EMCH是矩形,得出EM=CH,CM=EH,由正方形的性质得出BC=CD=3,∠ABC=90°,AB=CB,∠ABE=∠CBE=∠BDC=45°,证明△ABE≌△CBE得出EA=EF,∠BAE=∠BCE,同理:△ADE≌△CDE,得出△ADE的面积=△CDE的面积,由已知得出△CDE:△CEF的面积=3:5,证明A、B、F、E四点共圆,由圆周角定理得出∠GEF=∠BAF,∠EFC=∠BAE=∠BCE,得出EF=EC,由等腰三角形的性质得出FM=CM=EH=DH,设FM=CM=EH=DH=x,则FC=2x,EM=HC=3-x,由△CDE:△CEF的面积=3:5得出方程,解得:x=,得出FC=1,BF=BC-FC=2,由勾股定理求出AF==,即可得出结果.本题考查了正方形的性质、全等三角形的判定与性质、矩形的性质、四点共圆、圆周角定理、勾股定理、等腰三角形的判定与性质、解直角三角形等知识;本题综合性强,有一定难度,证明三角形全等是解题的关键.17.【答案】解:(1)原式=+2+1--=2-2;(2)原式=x2+8x+16-x2+3x=11x+16,当x=时,原式=11×+16=25.【解析】(1)先根据负整数指数幂、二次根式的性质、零指数幂、特殊角的三角函数值进行计算,再求出即可;(2)先算乘法,再合并同类项,最后代入求出即可.本题考查了负整数指数幂、二次根式的性质、零指数幂、特殊角的三角函数值、整式的混合运算和求值,能求出每一部分的值是解(1)的关键,能正确根据整式的运算法则进行化简是解(2)的关键.18.【答案】(1)证明:∵△ABC≌△DEF,∴AB=DE,AC=DF,∠F=∠C,∴BF=CE,在△BOF与△EOC中,,∴△BOF≌△COE(AAS);(2)解:∵∠ABC=∠DEF=90°,∠F=30°,AE=1,∴∠C=∠F=30°,∴AC=2AE=2,∴CE=1,∵∠CEO=∠DEO=90°,∴OC==.【解析】(1)根据三角形全等的性质得到AB=DE,AC=DF,∠F=∠C,根据全等三角形的判定定理得到△BOF≌△COE(AAS);(2)解直角三角形得到AC=2AE=2,求得CE=1,根据三角函数的定义即可得到结论.此题主要考查了全等三角形判定与性质,解答此题的关键是根据题意得出AF=DC.19.【答案】解:(1)若从中任意摸出一个球,则摸出白球的概率为;(2)树状图如下所示:∴两次摸出的球恰好颜色相同的概率为=.【解析】(1)直接利用概率公式计算可得;(2)列举出所有情况,看两个球都颜色相同的情况数占总情况数的多少即可.此题考查的是用列表法或树状图法求概率.注意画树状图法与列表法可以不重复不遗漏的列出所有可能的结果,列表法适合于两步完成的事件;树状图法适合两步或两步以上完成的事件.20.【答案】解:(1)如图点D即为所求.(2)如图点O即为所求.【解析】(1)利用平行线等分线段定理,把线段AB三等分即可.(2)作出线段AB,AC的垂直平分线,两条垂直平分线的交点O即为所求.本题考查作图-应用与设计,平行线等分线段定理,垂径定理,三角形的外接圆的圆心等知识,解题的关键是灵活运用所学知识解决问题.21.【答案】(1)证明:∵AE与⊙O相切,AB是⊙O的直径∴∠BAE=90°,∠ADB=90°,∴∠ADC=90°,∵CE∥AB,∴∠BAE+∠E=180°,∴∠E=90°,∴∠E=∠ADB,∵在△ABC中,AB=BC,∴∠BAC=∠BCA,∵∠BAC+∠EAC=90°,∠ACE+∠EAC=90°,∴∠BAC=∠ACE,∴∠BCA=∠ACE,在△ADC和△AEC中,,∴△ADC≌△AEC(AAS),∴AD=AE;(2)解:连接BF,如图所示:∵∠CBF=∠DAC,∠AFB=90°,∴∠CFB=90°,sin∠CBF==sin∠DAC=,∵AB=BC=10,∴CF=2,∵BF⊥AC,∴AC=2CF=4,在Rt△ACD中,sin∠DAC==,∴CD=×4=4,∴AD===8.【解析】(1)由切线的性质和圆周角定理得出∠BAE=90°,∠ADB=∠ADC=90°,由平行线的性质得出∠E=∠ADB,证出∠BCA=∠ACE,证明△ADC≌△AEC,即可得出结论;(2)连接BF,由圆周角定理得出∠CBF=∠DAC,∠AFB=90°,得出∠CFB=90°,由三角函数求出CF=2,由等腰三角形的性质得出AC=2CF=4,在Rt△ACD中,由三角函数求出CD=×4=4,再由勾股定理即可得出结果.本题考查了切线的性质,平行线的性质,等腰三角形的性质,圆周角定理,全等三角形的性质及判定,勾股定理,解直角三角形等知识点,综合程度较高.22.【答案】2-2【解析】解:(1)将点A(4,-2)、D(2,0)代入,得:,解得:,∴抛物线的表达式为y=-x2+x;(2)①如图1,连接BD、DE,作EP⊥AB,并延长交OD于Q,∵抛物线的对称轴为直线x=-=1,∴点A(4,-2)关于对称轴对称的点B坐标为(-2,-2),∴BD==2,设C(m,-2),则BC=CE=m+2,DE=BD=2,∵QD=1,PQ=2,∴PE=QE-PQ=-1=-1,∵PC=1-m,∴由PC2+PE2=CE2可得(1-m)2+(-1)2=(m+2)2,解得m=,∴点C的坐标为(,-2);②如图2,∵DB=DE=2,∴点E在以D为圆心、2长为半径的⊙D上,连接DA,并延长交⊙D于点E′,此时AE′取得最小值,∵DA==2,则AE的最小值为DE-DA=2-2,故答案为:2-2.(1)将点A(4,-2)、D(2,0)代入求出a、b的值即可得;(2)①连接BD、DE,作EP⊥AB,并延长交OD于Q,先求出B(-2,-2)、BD=2,设C(m,-2),知BC=CE=m+2,DE=BD=2,由QD=1,PQ=2知PE=QE-PQ=-1,由PC=1-m及PC2+PE2=CE2可得m的值,从而得出答案;②由DB=DE=2,知点E在以D为圆心、2长为半径的⊙D上,连接DA,并延长交⊙D于点E′,此时AE′取得最小值,根据AE的最小值为DE-DA可得答案.本题是二次函数的综合问题,解题的关键是掌握待定系数法求函数解析式、轴对称的性质、勾股定理等知识点.23.【答案】35 210【解析】解:(1)30+0.5×10=35元,答:放养10天后出售,则活虾的市场价为每千克35元,故答案为:35;(2)由题意得,(30+0.5x)(1000-10x)+200x=36000,解得:x1=20,x2=60(不合题意舍去),答:x的值为20;(3)设经销商销售总额为y元,根据题意得,y=(30+0.5x)(1000-10x)+200x-30000-ax,且20≤x≤30,整理得y=-5x2+(400-a)x,对称轴x=,当0≤a≤100时,当x=30时,y有最大值,则-4500+30(400-a)=1800,解得a=190(舍去);当a≥200时,当x=20时,y有最大值,则-2000+20(400-a)=1800,解得a=210;当100<a<200时,当x=时,y取得最大值,y最大值=(a2-800a+16000),由题意得(a2-800a+16000)=1800,解得a=400(均不符合题意,舍去);综上,a的值为210.故答案为:210.(1)原价格加上这10天增加的价格即可得;(2)根据活虾的销售额+死吓的销售额=36000列方程求解可得;(3)设经销商销售总额为y元,根据题意得出y=(30+0.5x)(1000-10x)+200x-30000-ax且20≤x≤30,整理成一般式后得出对称轴x=,再根据20≤x≤30及二次函数的性质分类讨论即可得.本题主要考查了二次函数的综合应用,解题时要利用图表中的信息,学会用待定系数法求解函数解析式,并将实际问题转化为求函数最值问题,从而来解决实际问题.24.【答案】3:3:2【解析】解:(1)设CD=x,则BD=10-x,在Rt△ABD和Rt△ACD中,AD2=AB2-BD2=AC2-CD2,依题意得:,解得x=6,∴AD==8.(2)∵四边形BFEP是圆内接四边形,∴∠EFB=∠DPB,又∵∠FBE=∠PDB,∴△BEF∽△BDP.(3)由(1)得BD=6,∵PD=3,∴BP==,∴cos∠PBD=,当△DEP为等腰三角形时,有三种情况:Ⅰ.当PE=DP=3时,BE=BP-EP=,∴BF===.Ⅱ.当DE=PE时,E是BP中点,BE=,∴BF===,Ⅲ.当DP=DE=3时,PE=2×PD cos∠BPD==,∴BE=3,∴BF===,若DP=3,当△DEP为等腰三角形时,BF的长为、、.(4)连接EG交PD于M点,∵DG∥BP∴∠EPD=∠EDF=∠PDG,∴PG=DG,∵EP=PG,ED=DG,∴四边形PEDG是菱形,∴EM=MG,PM=DM,EG⊥AD,又∵BD⊥AD,∴EG∥BC,∴EM=,∴,∴AM=6,∴DM=PM=2,∴PD=4,AP=4,∴S△APG==×4×3=6,S△PDG==×4×3=6,S△GDC===4.∴S1:S2:S3=6:6:2=3:3:2.(1)设CD=x,则BD=10-x,在Rt△ABD和Rt△ACD中利用勾股定理列方程即可求出x ,进而求出AD,(2)由圆内接四边形性质可知∠BFE=∠BPD,即可证明△BEF∽△BDP(3)因为DP=3,由②BP=3,可得分三种情况PE=DP、DE=PE、DP=DE利用直角三角形和等腰三角形性质先求出EB,再根据BF=即可求解;(4)连接EG交PD于M点,DG∥BP和折叠的性质可得∠EPD=∠EDF=∠PDG,EP=PG=ED=DG,即可得出E是BP中点,进而求出EM=GM==3,由,DM=,即可求出PM=2,PD=4,AP=4,再利用三角形面积求法即可解答.此题主要考查了圆的综合应用以及相似三角形的性质和勾股定理等知识,圆与相似三角形,及三角函数相融合的解答题、根据图形分类讨论是近几年中考的热点,故要求学生把所学知识融汇贯穿,灵活运用.。
2024年浙江省温州外国语学校九年级中考三模数学试题一、单选题12-,0,17中,最大的数是( ) AB .2-C .0D .172.截至2024年4月,“协同体”成员学校在校人数约为22400人,则数据22400用科学记数法表示( )A .522410⨯B .42.2410⨯C .52.2410⨯D .50.22410⨯ 3.如图所示的几何体是由一个圆柱和一个长方体组成的,则该几何体的俯视图是( )A .B .C .D . 4.4月23日是世界读书日,学校举行“快乐阅读,健康成长”读书活动.小明随机调查了本校七年级30名同学近1个月内每人阅读课外书的数量,统计结果如下:则这组数据的中位数和众数分别是( )A .3,4B .3,2C .2,3D .2,25.如图,矩形ABCD 为一个正在倒水的水杯的截面图,杯中水面与CD 的交点为E ,当水杯底面BC 与水平面的夹角为27°时,∠AED 的大小为( )A .27°B .53°C .57°D .63°6.如图,点A 在O e 上,OD ⊥弦BC 于点D .若45BAC ∠=︒,1OD =,则BC =( )AB .C .2D 7.我国古代数学经典著作《九章算术》中记载:“今有黄金九枚,白银一十一枚,称之重适等,交易其一,金轻十三两,问金、银各重几何?”意思是:甲袋中装有黄金9枚(每枚黄金重量相同),乙袋中装有白银11枚(每枚白银重量相同),称重两袋相等,两袋互相交换1枚后,甲袋比乙袋轻了13两(袋子的重量忽略不计),问黄金、白银每枚各重多少两?设每枚黄金重x 两,每枚白银重y 两,则可列方程组( )A .91181310x y x y y x =⎧⎨++=+⎩B .11981310x y x y y x=⎧⎨++=+⎩ C .91181013x y x y y =⎧⎨+=+⎩D .11981310x y x y x =⎧⎨+=+⎩ 8.已知直线y cx c =+与抛物线2y ax bx c =++交于A ,B 两点,则抛物线()2y ax c b x =-+-的图象可能..是( )A .B .C .D .9.如图,在ABC V 中,已知5cos 7B =,点P ,Q 在边AC ,BC 上,PQ AB ∥.设AP x =,PQ y =,若14y x =-+,则BC =( )A .7B .14C .D .2010.如图,在矩形ABCD 中,2AB =,E ,F 分别是AD ,OC 的中点.若EF BD ⊥,则BF =( )A B C .D .3二、填空题11.因式分解:24x -=.12.一个不透明的布袋里装有3个红球,2个白球,1个绿球,它们除颜色外其余都相同.从布袋里任意摸出1个球,是红球的概率为.13.已知关于x 的二次函数()26950y ax ax a a =-++<,该函数的最大值为.14.如图,在菱形ABCD 中,80BAD ∠=︒,以A 为圆心,AB 长为半径画弧,交对角线AC 所在直线于点P ,连结PB ,则PBC ∠=.15.如图,在平面直角坐标系中,已知ABC V 是等腰直角三角形,90C ∠=︒,点A ,B 分别在x 轴、y 轴上,点C 在函数()0k y k x=>的图象上.若tan BAO ∠1OA =,则k =.16.如图,在Rt ABC △中,90ACB ∠=︒,以其三边为边在AB 的同侧作三个正方形,点I 在DE 上,以EF 为直径的圆交直线AB 于点M ,N .若I 为DE 的中点,5AB =,则MN =.三、解答题17.计算:(1)()1013π 3.142-⎛⎫-+-- ⎪⎝⎭ (2)解不等式组:2312136x x x x -<⎧⎪+⎨-≤⎪⎩ 18.健康的体魄对中学生的身心成长有重要意义.某校为了解今年九年级学生的身体素质,随机抽取部分九年级学生的长跑测试成绩作为样本,按A ,B ,C ,D 四个等级进行统计,制成了如下不完整的统计图.(1)在扇形统计图中,C 对应的扇形的圆心角是______︒;请补全条形统计图.(2)该校九年级有300名学生,请估计长跑测试成绩达到A 级的学生有多少人.19.图1是某小型汽车的侧面示意图,其中矩形ABCD 表示该车的后备箱,在打开后备箱的过程中,箱盖ADE 可以绕点A 逆时针旋转,落在AD E ''的位置(如图2所示).已知90cm AD =,30cm DE =,40cm EC =.(1)当旋转角为60︒时,求点E 到点E '的距离.(2)已知箱盖可旋转的最大角度是75︒,求此时点D ¢到BC 的距离.(参考数据:sin 750.97︒≈,cos750.26︒≈,tan 75 3.73︒≈)20.如图,在ABC V 和ADE V 中,BAC DAE ∠=∠,AB AC =,AD AE =,点D 在线段BC 上,DE ,AC 交于点O ,连结CE .(1)求证:AC 平分BCE ∠.(2)若8AO AC ⋅=,求AD 的值.21.为了解新建道路的通行能力,查阅资料获知:在某种情况下,车流速度V (单位:千米/时)是车流密度x (单位:辆/千米)的函数,其函数图象如图所示.(1)当28188x ≤≤时,求V 关于x 的函数表达式.(2)车流量是单位时间内通过观测点的车辆数,计算公式为:车流量P =车流速度V ⨯车流密度x .若车流速度V 不超过80千米/时,求当车流密度x 为多少时,车流量P (单位:辆/时)达到最大,并求出这一最大值.22.【推理】(1)如图1,在平行四边形ABCD 中,点E ,F 在对角线BD 上,且::1:2:1BE EF FD =.求证:四边形AECF 为平行四边形.【应用】(2)如图2,在平行四边形ABCD 中,点E ,F 在对角线BD 上,且::1:2:3BE EF FD =.在AB ,BC 上分别找一点P ,Q ,使四边形PEQF 为平行四边形.若4AB =,求BP 的长.23.在平面直角坐标系中,已知抛物线()2233y mx m x m =--+-(m 是常数,且0m ≠)经过点()2,4,且与y 轴交于点A ,其对称轴与x 轴交于点B .(1)求出二次函数的表达式.(2)垂直于y 轴的直线l 与抛物线交于点(),P a p 和(),Q b q ,与直线AB 交于点(),c n ,若a c b <<,直接写出a b c ++的取值范围.(3)当13x t =-,2x t =,33x t =+时,对应的函数值分别为1y ,2y ,3y .求证:123454y y y ++≥. 24.如图,以AB 为直径作圆,弦CD AB ⊥于点E ,点P 在弧AC 上,点C 关于直线DP 的对称点为F .连接AP ,AC ,AD ,CF ,DF .(1)如图1,当点F 与点A 重合时,求证:60P ∠=︒.(2)如图2,当点F 落在直径AB 上时,若8=CF ,tan 2P ∠=,求直径AB 的长.(3)如图3,当点F 落在AD 的中点时,求出AE BE的值.。
2024年温州市初中英语学业水平考试适应性卷(三)亲爱的同学:1. 全卷共10页,有七大题,76小题。
全卷满分120分。
考试时间100分钟。
2. 欢迎参加考试!请你认真审题,积极思考,细心答题,发挥最佳水平。
祝你成功!第一部分听力部分(共20分)一、听力(本题有15小题。
第1-10小题,每题1分;第11-15小题,每题2分。
共20分)第一节:听小对话,请从A、B、C三个选项中选出最佳选项。
1. How does Tim improve English?A. B. C.2. What sign did Mum see?A. B. C.3. How much should the woman pay if she buys two hats?A. B. C.4. Which map does Helen suggest putting on the wall?A. B. C.5. What will the speakers probably watch at last?A. B. C.第二节:听较长对话,请从A、B、C三个选项中选出问题的答案。
听第一段较长对话,回答6-7小题。
6. Where will people live in the future according to the book?A. Under the sea.B. In the sky.C. On the moon.7. What will the robots do?A. Make machines.B. Buy things.C. Write books.’听第二段较长对话,回答8-10小题。
8. Why is Mark going to Beijing?A. To attend a meeting.B. To go to university.C. To take a vacation.9. What’s the weather like in Beijing?A. Sunny.B. Rainy.C. Windy.10. What time is Mark’s flight?A. At two o’clock.B. At three o’clock.C. At four o’clock.第三节:听一段独白,请根据内容从A、B、C三个选项中选择正确的选项,完成信息记录表。
浙江温州外国语学校2018-2019学年九年级下学期英语第三次模拟考试考生须知1.本卷满分90分,考试时间90分钟:2.本卷共8页,请在答题卷答题区域作答,不得超出谷腰区域边框线:亲爱的同学,如果把这份试卷比作一片湛蓝的海,那么,我们现在启航,展开你自信和智慧的双翼,乘风踏浪,你定能收获无限风光!一单项选择(本题有10小题,每小题1分,共10分)请从A,B,C、D四个选项中选出可以填入空白处的最佳答案。
1.This is_______movie I've told you about.Isn't it fantastic?A.a B an C.the D.\2.I've just bought two pairs of shoes. ______of them are made in China.A.AllB.BothC.EitherD.None3.-Did you pass the test?-Yes,I'm very_______for your advice.A.thankfulB.carefulefulD.helpful4.Mrs.Smith showed great ______as her son was presented medals at the sports meeting.A.fearB.prideC.courageD.energy5 _______cross the road when the light is red.Or you may have accidents.A.AlwaysB.OftenC.HardlyD.Never6.The girl weighs only 35 kg.She is so thin that she needs to _______more weight.A.put onB.put upC.take offD.take up7.--Do you have any plans for the coming holiday?I'm not sure.I______go to the countryside to see my grandparents.A.canB.mustC.mayD.need8.--Shall we go shopping now,dear?-Sorry,I can't.I______my clothes.A.was washingB.have washedC.washedD.am washing9.--I wonder what you usually do in the course of life management.We learn how to plan our life in the future and______to make it.A.who we should ask for helpB.what we should doC.whether we should study hardD.where we should go10.-Lots of luck for the Year of the Pig._______A.I hope so.B.You're welcome.C.The same to you.D.That's OK.二完形填空(本题有15小题,每小题1分,共15分)阅读读下面的短文,从每小题所给的A,B.C.D四个选项中选出最佳选项It seems that some people never try hard,but always get perfect results.However,no one is born a winner.People make themselves into winners through their own __11____ .I learned this lesson from an experience many years ago.I took the head coaching job at a school in Baxley, Georgia.It was a small school with a(n)__12_____football program.It was a tradition for the school's old team to play ___13____the new team at the end of spring practice.The old team had no coach,and they didn't even practice to__14_____the game. Being the coach of the new team,I was excited ___15____I knew we were going to win,but to my disappointment we were beaten.I couldn't __16_____I had got into such a situation.Thinking hard about it,I realized that my team might not be the number one team in Georgia,,but they were depending on____17____.I started doing anything I could to help them build a little confidence,after I changed my___18____about their ability and potential*(潜力).Most importantly,I began to treat them like winners.That summer,when the other teams enjoyed their.vacations,we met every day and__19_____passing and kicking the football.Six months after suffering our failure on the spring practice field,we won our first game and our second,and continued to improve.___20_____,we faced the number one team in the state.I felt that it would be a(n)____21____ for us even if we lost the game.But that wasn't _____22___ happened.Giving me one of the greatest excitements of my life,my boys beat the ___23____ team in Georgia!From the experience,I learned a lot about how the attitude of the leader can affect the members of a team.___24____ seeing my boys as losers,I pushed,supported and cheered for them.I helped them to see themselves ___25_____ ,and they built themselves into winners.I witnessed how hard they had tried,how much progress they had made,and how fearlessly they had headed for what they wanted.Winners are made,not born.11.A.luck B.test C.effort D.nature12.A.interesting B.boring C.strong D.weak13.A.for B.against C.with D.on14.A.cheer for B.prepare for C.help with D.finish with15.A.because B.so C.while D.though16.A.believe B.agree C.describe D.regret17.A.it B.you C.me D.them18.A.plan B.mind C.attitude D.thought19.A.risked B.missed C.considered D.practiced20.A.Secondly B.Finally C.Luckily D.Sadly21.A.shame B.shock C.victory D.occasion22.A.how B.when C.what D.where23.A.worst B.best C.biggest D.smallest24.A.Not only B.Instead of C.Even though D.As if25.A.honestly B.individually C.calmly D.differently三,阅读理解(本题有15小题,第26-28题每小题1分,第40题5分,其余每小题2分,共30分)阅读下面短文:客观题请从每小题所给的A,B.C,D四个选顶中透出最佳选项:主观题请在答题纸规定区域作答。
温州外国语2022九年级第三次模拟试卷语文答案一、积累运用(38分)选择题1―6题。
(1 4分)1、下列加点字的注音,全部正确的一组是(2分)A、氛围(fēn)脊梁(jǐ)污秽(suì)扣人心弦(xián)B、迸溅(bèng) 嶙峋(lín) 潜行(qiǎn) 恹恹欲睡(yān)C、香醇(chún)跻身(jī)飞窜(cuàn) 争妍斗艳(yán)D、相称(chèn) 模样(mú) 聆听(lín溃卷帙浩繁(yì)2、下列文句的空缺处,依次填入的词语恰当的一组是(2分)①春天来了,天气变暖,冰雪开始融化,树枝上也绽出了嫩芽。
②如果对中国人民的严正声明和强烈抗议,一意孤行,必将自食其果。
③大自然用“死”的物质了这样丰富多彩的生命,而人类却不能出一个哪怕是最简单的生物。
A、逐步置之不理创造制造B、逐渐置之度外制造创造C、逐步置之度外制造创造D、逐渐置之不理创造制造3、下列各句中,加点的成语使用错误的一项是(2分)A、从新城区到老城区,沿街庄重典雅的建筑,精致中透着秀气的园林,温文尔雅、朴实守礼的市民使外地游客的喜悦之情油然而生。
B、新来的王老师为人不苟言笑,只有校长有时还能跟他开点无伤大雅的玩笑。
C、网络交友已是许多人玩腻了的游戏,可有些年轻人依然乐此不疲,一个个前赴后继地扎进去。
D、有些干部不注重调查研究,到了基层,下车伊始就发议论,提意见。
4、下面句子中,没有语病的一句是(2分)A、为了防止地球环境不恶化,我们应站在可持续发展的高度,合理开发和利用自然资源。
B、居里夫人为科学而献身的精神和品质是值得我们学习的榜样。
C、“中国梦”记录着中华民族从饱受屈辱到赢得独立解放到改革开放。
D、生活有多广阔,语文就有多广阔。
因此,我们不仅要在课堂上学语文,还要在生活中学语文。
5、给下列句子排序,最合理的一项是(3分)①其中包括他们的肤色、相貌、身材、线条、姿态、气质、风度等许多方面。
全身被毛,体温恒定,胎生,为国家一级保护动物。
在动物分类中,年上海国际车展中,新能源汽车成为绝对主角。
某电动汽车采用磷酸亚铁锂电池其充电反应原理为: 状曲面玻璃幕墙采用了碲化镉发电玻璃,为场馆提供电力支持。
镉元素在元素周期溶液aA.有机物B.碱C.氧化物D.盐9.、下列科学教材中涉及到的实验不能用气压与流速关系解释的是(▲)A B C D10.今年5月,温外赛艇队亮相第五届南京国际水上运动节。
参赛过程中,运动员身体有许多器官协同完成各项生命活动。
下列参与各项生理活动的器官及其功能表述错误的是(▲)A.肺------与外界进行气体交换B.胃------消化和吸收的主要场所C.肾------排出血液中的代谢废物D.心脏------人体血液流动的动力11.如图是新石器时代遗址发掘时发现的陶罐-----“火种器”。
把燃烧的火炭置于陶罐内并覆盖黑炭,适度封闭上端口部。
采火种时,打开口部并轻轻吹,即可获得火苗。
其原因是(▲)A.降低了可燃物的着火点B.提高了可燃物的温度C.提供了充足的氧气D.降低了可燃物周围的二氧化碳浓度12.通常智能手机是通过指纹开关S1或密码开关S2来解锁的,若其中任一方式解锁失败后,锁定开关S3则会断开而暂停手机解锁功能,S3将在一段时间后自动闭合而恢复解锁功能。
若用灯泡L发光模拟手机解锁成功,则符合要求的模拟电路是(▲)A B C D13.我校项目化小组同学利用纸盒和凸透镜等材料制作成像仪。
焦距为15cm的凸透镜固定在纸盒侧面,F光源可以在纸盒中的轨道上移动,轨道上标注了离纸盒侧面O点的距离,如图所示。
若光屏上呈如图所示的像,则F 光源可能处在轨道的哪一位置(▲)A.a点位置B.b点位置C.c点位置D.d点位置第13题图第14题图”或“等臂”)。
乙神舟十五号乘组三名宇航员在轨飞行六个月时间。
为了对抗失重生理效应,航天员每天都需要进行锻炼。
小脑”或“脑干”)。
,也是中国乃至亚洲第一个世界农业文化遗产。
2024年浙江省温州外国语学校中考三模英语试题一、阅读理解Need money to support your project?! Want to offer help? This is the right website for you both!This Week’s ProjectsSuper kitchen!A year ago, we made homemade food from our own kitchens for school restaurants of our city. Our food is not only low-cost, but also healthy for the students. We have been so successful that we need a larger kitchen now. Support us and you will receive VIP cards for our top ten dishes!★Help us put on our play!The play we wrote last year is a new look at the family life. It has won first prize in a national play writing competition. Now we want to take it to the International Youth Theatre Festival, but we don’t have enough money for our trip to the festival. Support our project, and you will receive free tickets to the opening night!★Give a positive message!I am an art student who has designed various posters with positive messages. Now I need to print and sell them to art shops around the country. Invest in me and you will receive a free warm poster—and you will help make my dream come true.★A tree house for everyone to share!We have got a plan to build an imaginative tree house in our local park. It will be a place for children to have adventures or for families to have a picnic. When completed, the tree house will show our talents and enable us to realize our dreams. Support us and you will receive a 20%discount on your first tree house order!1.Who is the project “super kitchen” for?A.Excellent actors.B.School students.C.Poor housewives.D.Shop assistants.2.Sam got a theatre ticket for free. Which project did he offer money to?A.Super kitchen!B.Give a positive message!C.Help us put on our play!D.A tree house for everyone to share! 3.According to the text, what can we know about the website?A.It is mainly for teenagers.B.It shows how to do projects.C.It’s a notice about helping others.D.It helps raise money for projects.On the last Sunday of every month, my mother would sit my sister and me down and force us to write letters to our family in India. Although my mom was an orphan (孤儿), she came from a large Indian family with lots of caring and friendly relatives. At the time, we were the only ones who had left India to go abroad.It was the 1970s, and my mother raised two little girls on her own, living a hard life in a new country. We lived in a small apartment and couldn’t afford the phone calls. Letter-writing was the cheapest and the only way for us to keep in touch.The last thing I wanted to do was to sit down for an hour after dinner and write letters to people I hardly knew. But over time, my sister and I came to enjoy it. My mother would tell us stories about each family member, and it was interesting to get to know some people who looked and sounded just like us.When my mother died, my sister and I took her ashes (骨灰) back to India. We spent a month visiting relatives. Each family would take out photo albums full of photographs of us as well as the letters they’d received over the years. They’d saved them all and wanted to show us that they’d never forgotten usAs we managed to five abroad, my mother made sure that we always knew who we were, where we came from and where we were going. She kept us connected to a family that was thousands of miles away through a lifetime of letter-writing. It’s a gift that we are always proud of. I’ll be thankful for it forever.4.What can we learn about the writer's mother from the text?A.She left India to study abroad.B.She raised her children alone.C.She was born into a happy family.D.She once lost touch with her relatives. 5.How might the writer first get to know about her relatives?A.By listening to Mom’s stories.B.By looking at family photos.C.By talking to them on the phone.D.By visiting them on a holiday.6.Which shows the writer’s change of feelings towards writing letters?A.Bored→excited→regretful.B.Interested→happy→sorry.C.Bored→interested→thankful.D.Interested→bored→proud.7.What is the best title of the passage?A.My Relatives in India B.The Power of Letter-writingC.My Hard-working Mother D.The Hard Life of Living AbroadFor thousands of years, humans have used names to communicate with one another. We also give names to our pets. Until now there has been few animals naming one another, but a new study suggests that elephants use certain noises to call other elephants.A few animals, including dolphins (海豚), use sounds that are similar to names. Each dolphin invents a special sound, and other members communicate with it by copying this special call. The new study, led by Michael Pardo, shows that wild African elephants use names in a way that is not just copying sounds and is much closer to the way humans use names.For the study, the researchers recorded 625 sounds made by wild African elephants in Kenya that they called “rumbles”. This is the most common kind of call produced by elephants, and it can travel as far as 3.7 miles. It takes place at a very low frequency (频率), which means humans can’t hear it.The scientists studied the sounds and found that some rumbles were directed at certain elephants to get their attention. They found that all the elephants in the group used the same call to get a certain elephant’s attention. Also, unlike the way dolphins communicate, the rumbles were not just imitations (模仿) of the elephants they were trying to communicate with.The researchers then played back some of the recorded rumbles to the elephants. They found that elephants responded more to their own name than to other calls. Caitlin, an elephantexpert, said the study shows that elephants can still keep in touch with one another even across a large area. She told Live Science, “It allows them to spread out much further and still communicate with each other.”8.Why is the example of dolphins used in Paragraph 2?A.To explain why dolphins are so smart.B.To prove dolphins can invent a special sound.C.To compare with the way elephants use names.D.To show how dolphins communicate with others.9.What can we learn about “rumbles”?A.Elephants usually make “rumbles”.B.Humans can hear “rumbles” easily.C.There are 625 kinds of “rumbles” in total.D.Dolphins make “rumbles” to communicate.10.What does the underlined word “It” refer to in Paragraph 5?A.The fact that elephants copy the sounds of other ones.B.The fact that elephants can communicate far and wide.C.The fact that elephants answer their names more often.D.The fact that elephants use the same call to get attention.11.What’s the writer’s purpose in writing this text?A.To explain how “rumbles” work properly.B.To describe the sounds of different animals.C.To introduce how elephants name each other.D.To call on more people to protect elephants.As your alarm clock rings, you slowly open one eye. It seems like it was only a few hours ago that you finished your homework. You close your eyes for just five more minutes, but then your mom knocks on the door. “Time to get up, or you’ll miss the bus!” she says.If this sounds familiar, you’re not alone. According to a survey, nearly 60 percent of middle school students in the U. S. aren’t getting the recommended 8.5 to 9.5 hours of sleep a night.How does it happen? In fact, starting the school day too early can take away much-neededsleep from teenagers. Studies show that well-rested teenagers are more likely to get good grades and less likely to suffer from depression (抑郁). Besides, teenagers fall asleep later at night by nature. As a result, some students may only get as few as five hours of sleep.Many people have realised the importance of enough sleep. Schools in several states have changed to later start times in recent years. In New Jersey, a group of students have formed an organization called Later School Start Times. It wants local middle schools to start 30 minutes later.▲ Many schools say later start times would create a number of problems. There’s a worry that some kids will walk back home in the dark if the school starts later. It will also leave teens with fewer hours for after-school activities and homework. That’s what Erin Isherwood, a parent from California, is worried about. “My son has an exercise class that he really loves at a gym right after school,” she says. “Now he can’t go.”Considering the advantages and disadvantages of later school start times, it seems a hard choice to make. The early bird catches the worm (虫子). But what if the bird is too sleepy and tired to catch one?12.Why does the writer describe the situation in Paragraph 1?A.To show a teenager’s busy school life.B.To show a common problem for teenagers.C.To tell how hard it is to get up in the morning.D.To tell why teenagers get on badly with parents.13.Why can’t most American students get enough sleep?★The school starts too early.★They suffer from depression.★They fall asleep later by nature.★They stay up late to get good grades.A.★★B.★★C.★★D.★★14.Which can be the best to fill in the “ ▲ ” in Paragraph 5?A.And all the efforts have made a difference.B.And lots of parents shared the same opinion.C.But not everyone thinks school should start late.D.But few people take action to change the situation.15.Which of the following might the writer agree on?A.Early school days can help kids develop hobbies.B.Early school days can cause more harm than good.C.Later school start times can cause safety problems.D.Later school start times can bring few advantages.二、任务型阅读阅读下面材料,从方框中所给的A—E 五个选项中选择正确的选项(其中一项是多余选项),将其序号填入第1—4题,并回答第5题。
温州外国语学校第三次中考模拟试卷1.全卷共5页,有三大题,24小题,满分为150分。
考试时间为120分钟。
本次考试采用闭卷形式。
2.全卷分试卷Ⅰ(选择题)和试卷Ⅱ(非选择题)两部分。
3.试卷Ⅰ(选择题)请用2B 铅笔在答题卡上将答案对应的方框涂黑、涂满; 试卷Ⅱ(非选择题)请用钢笔或圆珠笔在密封线外每题的相应位置上答题。
温馨提示:请仔细审题,细心答题,相信你一定会有出色的表现! 参考公式: 二次函数2y ax bx c =++图象的顶点坐标是)44 ,2(2ab ac a b --试 卷 Ⅰ说明:本卷共有一大题,10小题,共40分。
一、选择题(本题共10小题,每小题4分,共40分.请选出各题中一个符合题意的正确选项,不选、多选、错选均不给分) 1、-2的相反数为( )A .2B .-2C .12 D . 12- 2、直角坐标系中,点P(2,-4)在( )A. 第一象限B.第二象限C.第三象限D.第四象限 3、下图能说明∠1>∠2的是( )A B C D 4、如图,已知圆心角∠AOB 的度数为100°,则圆周角∠ACB 的度数是( )A.80°B.100°C.50°D.40° 5、因式分解a a -3的结果是( )A. 2a B. )1(2-a a C. )1)(1(+-a a a D. 2)1(-a a6、抛物线4)3(22+-=x y 的顶点坐标是 ( )A.(3,4)B.(4,3)C.(—3,4)D.(—3,—4)7、已知圆锥的底面半径为9㎝,母线长为30㎝,则圆锥的侧面积为( )2cm 。
A.270π B.360π C.450π D.540π8、如图是一些相同的小正方体构成的几何体,则它的俯视图为( )9、已知⊙O 1与⊙O 2的圆心距是3,两圆的半径分别是2和5则这两个圆的位置关系是( ) A .外离 B .外切 C .相交 D .内切 10、如图是一张简易活动餐桌,现测得OA=OB=30cm , OC=OD=50cm ,现要求桌面离地面的高度为40cm ,那么 两条桌腿的张角∠COD 的大小应为…………………( )OBC AABC D EF GA .100°B .120°C .135°D .150°试 卷 Ⅱ11、不等式组⎩⎨⎧≤-073x >x 的解是 。
12、甲、乙两人进行射击比赛,在相同条件下各射击10次.他们的平均成绩均为7环,10次射击成绩的方差分别是:S 甲2=3,S 乙2=1.2。
成绩较为稳定的是 .(填“甲”或“乙”)。
13、已知在Rt △ABC 中,∠C=90°,AC=1,BC=3,那么=A tan 。
14、右边是三种化合物的结构式及分子式,请按其规律,写出后一种化合物的分子式... . 15、《某省工伤保险条例》规定:职工有依法享受工伤保险待遇的权利,某单位一名职工因公受伤住院治疗了一个月(按30天计),用去医疗费7000元,伙食费500元,工伤保险基金按规定给他补贴医疗费6300元,其单位按因公出差标准(每天50元)的百分之七十补助给他做伙食费,则在这次工伤治疗中他自己只需支付 元。
16、如图,点E 、F 分别是矩形ABCD 的边AB 、BC 的中点,连AF 、CE 交于点G ,则=ABCDAGCD S S 矩形四边形 。
三、解答题(共80分)17、(1)(5分)计算:++--02)32()21(tan60° (2)(5分)解方程:3231+=-x x 18、(本题8分)如图,在□ABCD 中,BE ⊥AC 于点E ,DF ⊥AC 于点F ,求证:AE=CF 证:19、(8分)在右图所示的5×5出一个格点△ABC,使10,13==BC AB 。
(画出一个三角形即可,不必写画图步骤,并在图上标出相应的字母。
)20、(10分)下图是某班学生外出乘车、步行、骑车的人数分布直方图和扇形分布图。
(1)求该班有多少名学生?(2分) (2)补上步行分布直方图的空缺部分;(2分)C 3H 8C 2H 6CH 4HH H H H H H H HH H H H H CCC CCH H HH C图①(3)在扇形统计图中,求骑车人数所占的圆心角度数。
(3分) (4)若全年级有500人,估计该年级步行人数。
(3分)解:21、(8分)如图,直线b x y +=21与x 轴、y 轴交于 A 、B ,与双曲线xky =2(x <0)点交于点C 、D ,已知 点C 的坐标为(一1,4).(1)求直线和双曲线的解析式; (2)利用图象,说出x 在什么范围内取值时,有1y >2y 。
22、(10分)有四张背面相同的纸牌A ,B ,C ,D ,其正面分别画有四个不同的几何图形(如图).小华将这4张纸牌背面朝上洗匀后摸出一张,放回洗匀后再摸出一张.(1)用树状图(或列表法)表示两次模牌所有可能出现的结果(纸牌可用A 、 B 、C 、D 表示);(2)求摸出两张牌面图形都是中心对称图形的纸牌的概率.23、(本题12分)某水果店有200个菠萝,原计划以2.6元/千克的价格出售,现在为了满足市场需要,水果店决定将所有的菠萝去皮后出售。
以下是随机抽取的5个菠萝去皮前后相应的质量统计表:(单位:千克) (1)计算所抽取的5个菠萝去皮前的平均质量和去皮后的平均质量,并估计这200个菠萝去皮前的总质量和去皮后的总质量。
(2)根据(1)的结果,要使去皮后这200个菠萝的销售总额与原计划的销售总额相同,那么去皮后的菠萝的售价应是每千克多少元? 24、(本题14分)如图①,矩形ABCD 被对角线AC 分为两个直角三角形,AB=3,BC=6.现将Rt △ADC 绕点C 顺时针旋转90º,点A 旋转后的位置为点E,点D 旋转后的位置为点F.以C 为原点,以BC 所在直线为x 轴,以过点C 垂直于BC 的直线为y 轴,建立如图②的平面直角坐标系. (1) 求直线AE 的解析式;(2) 将Rt △EFC 沿x 轴的负半轴平行移动,如图③.设OC=x (09x <≤),Rt △EFC 与Rt △ABO 的重叠部分面积为s ; ① 当x =1与x =8时,分别求出s 的值;② S 是否存在最大值?若存在,求出这个最大值及此时x 的值;若不存在,请说明理去皮前各菠萝的质量 1.0 1.1 1.4 1.2 1.3去皮后各菠萝的质量 0.6 0.7 0.9 0.8 0.9乘车50%步行 20% 骑车30%人数骑车步行乘车48161220图②图③由.温州外国语学校第三次模拟考数学答案一、本题共10小题,每小题4分,共40分A D C C C C A C DB 11、0<37≤x12、乙 13、314、84H C 15、15016、32 三、17、(1)++--02)32()2(tan60°解:原式=4-1+3 …………………(3分)=3+3 …………………(2分)(2)3231+=-x x 解:方程两边同乘以)3)(3(+-x x 得,623-=+x x …………………(2分)9=x …………………(2分)经检验:原方程的解是9=x 。
…………………(1分) 18、如图,在□ABCD 中,BE ⊥AC 于点E ,DF ⊥AC 于点F ,求证:AE=CF证明:∵四边形ABCD 是平行四边形,∴AB=CD ,AB ∥CD …………………(2分) ∴∠BAC=∠DCA …………………(1分) 又∵BE ⊥AC ,DF ⊥AC∴∠AEB=Rt ∠=∠DFC …………………(2分) ∴△ABE ≌△CDF …………………(2分) ∴AE=CF …………………(1分)AEFA B C A B CACB19、(8分)△ABC 就是所求的三角形。
(画对一图即可) 20、解:(1)20÷50%=40(人)……………(2分)(2)如图,40×20%=8(人)…………(2分) (3)360°×30%=108° ……………(3分)(4)500×20%=100(人)……………(3分)21、解:(1)将C (一1,4)分别代入b x y +=21、xky =2得k= 一4,b=6,……………(2分)∴621+=x y ,xy 42-=。
……………(2分)(2)解⎪⎩⎪⎨⎧-=+=x y x y 462得⎩⎨⎧=-=4111y x ,⎩⎨⎧=-=2222y x ……………(2分) ∴由图象可知当12-- x 时,21y y 。
……………(2分) 22、解:(1)画树状图或列表正确的……………(5分)(2)P=81162=……………(5分)23、解:(1)5个菠萝去皮前的平均质量为2.153.12.14.11.10.1=++++(kg )……(2分)5个菠萝去皮后的平均质量为78.059.08.09.07.06.0=++++(kg )……………(2分)200个菠萝去皮前的总质量约为1.2200=(kg )……………(2分) 200个菠萝去皮后的总质量约为0.78×200=156(kg )。
……………(2分) (2)去皮后的菠萝的售价应是2.6×240÷156=4(元)……………(4分) 24、 解:(1)∵A 点坐标为(-6,3),E 点坐标为(3,6)……………(2分)∴直线AE 的解析式为531+=x y ……………(2分)(2)①当x =1时,如图,重叠部分为△POC 可得: Rt △POC ∽Rt △BOA , ∴2()AOBs OC SAO= 即:2(935s =……(直接写出此关系式不扣分)(1分) 解得:S =15.…………………………………………………………………………(1分) ②当x =8时,如图,重叠部分为梯形FQAB可得:OF =5,BF =1,FQ =2.5 ………………(1分) ∴S =1111()(2.53)1224FQ AB BF +•=+⨯= ……………………………………………………(1分) (3)解法一:①显然,画图分析,从图中可以看出:当03x <≤与7.59x <≤时,不会出现s 的最大 值.……………………………………………………………………………(2分) ②当36x <≤时,由图可知:当6x =时,s 最大.此时,365OBN S=,94OFM S = ∴S =369995420OBN OFM S S -=-=.………………(1分) ③当67.5x <≤时,如图25OCNx S=,2(3)4OFMx S -=,2(6)BCGSx =-∴S =OCN OFMBCG SSS --=222(3)(6)54x x x ----∴S = 222127153214536()20242077x x x -+-=--+ ∴当457x =时,S 有最大值,367S =最大……………………………………………(1分)综合得:当457x =时,存在S 的最大值,367S =最大.………………………………(2分)解法二:同解法一③可得:若03x <≤,则当3x =时,S 最大,最大值为95;………………………………(1分) 若36x <≤,则当6x =时,S 最大,最大值为9920;………………………………(1分)若67.5x <<,则当457x =时,S 最大,最大值为367;…………………………(1分)若7.59x ≤≤,则当7.5x =时,S 最大,最大值为6316;…………………………(1分) 综合得:当457x =时,存在S 的最大值,367S =最大………………………………(2分)。