武汉二中广雅中学九年级第一次月考试题
湖北省武汉二中广雅中学2023-2024学年九年级上学期月考数学试题

湖北省武汉二中广雅中学2023-2024学年九年级上学期月考数学试题学校:___________姓名:___________班级:___________考号:___________一、单选题.....如图,O 的半径3=,OAB ∠AB =(A .1.2344.一元二次方程10=配方后可化为()A .2(2)3x +=.2(2)5x +=2(2)x -2(2)x -5.若函数2y x =,则c b -=(A .3-1-16.抛物线1y x =A .4B .28.在平面直角坐标系中,二次函数y 上有三点()3,A y -,()21,B y -,(33,y A .123y y y <<B .213y y y <<9.飞机着陆后滑行的距离s (m )与滑行的时间飞机滑行中最后2s 的滑行距离为(A .600mB .20m10.已知抛物线23y x mx =++对称轴为直线230x mx n ++-=(n 为实数)在1-<()A .26n ≤<B .26n <<二、填空题11.点()2,4P -关于原点对称点1P 的坐标为12.下列说法:①直径是弦;②弦是直径;③大于半圆的弧是优弧;④长度相等的弧是14.参加足球联赛的每两队之间都进行两场比赛,共要比赛参加比赛.15.二次函数y ①0abc <;②16.问题背景:如图,在正方形ABG ,接着证明△EM CD 交AF 于M 则四边形MNEF 的面积为三、解答题17.若关于x 的一元二次方程240x bx +-=有一个根是1x =,求b 的值及方程的另一个根.18.如图,利用函数243y x x =-+的图象,直接回答:(1)方程2430x x -+=的解是__________;(2)当x __________2时,y 随x 的增大而增大(填“≤、≥”);(3)当函数值y 小于0时,x 的取值范围是__________;(4)当14x -<<时,y 的取值范围是__________.19.如图,在四边形ABCD 中,已知AB AC =,90BAC ∠=︒,6BD =,4=AD ,将ABD △绕点A 顺时针旋转90︒得到ACE △,连接DE .(1)DAE ∠=__________°出结果);(2)若=45ADC ∠︒,求20.已知关于x 的一元二次方程(1)求m 的取值范围;(2)若1x ,2x 是一元二次方程21.已知在正方形的网格中,点图.(1)作图:画出ABC 关于直线AP 成轴对称的(2)作图:在AD 上找一点E ,使得PE ⊥(3)作图:若PE 交CD 于点F ,在线段BC 22.如图,小朋友燃放--种手持烟花,这种烟花每隔的飞行路径,爆炸时的高度均相同.小朋友发射出的第一发花弹的飞行高度飞行时间t (秒)变化的规律如下表.t/秒00.51 1.52 2.5h /米1.87.311.815.317.819.3(1)根据这些数据选择适当的函数表示写出相应的函数表达式__________;(2)当第一发花弹发射2.2秒后,第二发花弹达到的高度为多少米?(3)为了安全,要求花弹爆炸时的高度不低于19米.小朋友发现在第一发花弹爆炸的同时,第二发花弹与它处于同一高度,请分析花弹的爆炸高度是否符合安全要求?23.在等腰ABC 中,AB AC =,BAC α∠=,点P 为平面内一点.(1)如图1,已知60α=︒,点P 为边BC 上一点,以AP 为边向外作等边APD △,连接CD ,求证:AC CD PC =+;(2)如图2,已知120α=︒,当点P 在ABC 的外部,且满足60APC ∠=︒,连接BP .试判断AP 、BP 、CP 存在何种数量关系,并说明理由;(3)在(2)的条件下,若4AP PC +=,当BP 的长取最小值时,APC △的面积为__________.(直接写出结果)24.如图1,抛物线2y ax bx c =++与x 轴交于A 、B 两点,与y 轴交于C 点.已知()1,0A -,()3,0B ,点C 的纵坐标为3.(1)求抛物线的解析式;(2)若在抛物线上仅存在三个点到直线BC 的距离为m ,求m 的值;(3)如图2,直线y kx b =+交抛物线于P 、Q 两点,当APQ △的内心在x 轴上时,此时直线PQ 一定和经过原点的某条直线平行吗?若是,请求出这条过原点的直线解析式:若不是,请说明理由.。
湖北省武汉市江岸区武汉二中广雅中学2023-2024学年九年级9月月考化试题(解析版)

武汉二中2023-2024学年度上学期九年级9月理化试题一、选择题(每小题的四个选项中只有一个答案符合题意)1. 1836年科学家用电解氟化物的方法制出氟气(F2),工业上氟气可用于金属的焊接和切割,电镀玻璃加工等,氟气属于A. 纯净物B. 混合物C. 氧化物D. 稀有气体2. 下列叙述正确的是A. 氧气应用于焊接或切割金属,体现了氧气的可燃性B. 鱼、虾能在水中生存是由于氧气易溶于水C. 硫在氧气中燃烧的现象的发明亮的蓝色紫色火焰,生成二氧化硫D. 实验室加热高锰酸钾以及加热氯酸钾与二氧化锰制取氧气,反应结束后,反应容器中的剩余物都为混合物。
3. 下列实验操作及对现象的判断,正确的是()A. 检查气密性:导管口有气泡冒出,则说明装置漏气B. 滴瓶上的滴管用后不用洗涤就放回原瓶C. 读取液体的体积D. 向试管中倾倒液体4. 化学概念间有包含、并列、交叉等不同关系。
下列选项符合如图关系的是A B C DX化学反应混合物物理变化化合反应Y分解反应纯净物化学变化氧化反应A. AB. BC. CD. D5. 同学们在做氧气性质实验时,将点燃的木炭伸入集气瓶内,有的现象明显,有的却不明显。
导致现象不明显的原因可能是A. 排水法收集前未将集气瓶灌满水B. 导管口均匀连续放出气泡时开始收集氧气C排水法收集满后盖上毛玻璃片拿出水面D. 排水法收集结束时集气瓶内的水有残留6. 下列说法不正确的是( )①催化剂能改变化学反应速率,但不增加生成物的质量②氧气的化学性质活泼,能与所有物质发生化学反应③固体物质受热变为气体,可能是物理变化,也可能是化学变化④将一定量的氯酸钾和少量的高锰酸钾混合加热一段时间后,试管中只剩余三种固体物质⑤燃烧和缓慢氧化的过程都放出热量⑥高锰酸钾、氯酸钾、双氧水、分离液态空气,这四种制取氧气的方法都属于分解反应⑦有氧气参加的反应都是化合反应A. ①③⑤B. ②⑥⑦C. ②④⑥⑦D. ③④⑥⑦7. 下列图像正确的是A. 用高锰酸钾制取氧气中二氧化锰质量变化B. 足量红磷在密封的空气瓶中燃烧,瓶内气体压强与时间的关系C. 等质量的氯酸钾两份,其中一份加二氧化锰,另外一份不加二氧化锰,产生氧气质量与时间关系D. 用等质量、等浓度的过氧化氢溶液分别制取氧气8. 如图所示,集气瓶容积为200mL,量筒的量程为250mL,实验步骤如下(装置气密性良好,部分操作已略去,因导管产生的误差忽略不计),I.打开止水夹a和b,向集气瓶中缓慢鼓入一定量空气,测得进入量筒中水的体积为V1,关闭止水夹a、b;Ⅱ.强光照射引燃白磷;III.白磷熄灭并冷却至室温,打开止水夹b,最终测得量筒中水的体积变为V2;IV.计算空气中氧气的体积分数。
2021-2022学年湖北省武汉二中广雅中学九年级(上)元调综合测试数学试卷(3)

2021-2022学年湖北省武汉二中广雅中学九年级(上)元调综合测试数学试卷(3)一、选择题(本大题共10小题,共30分)1.将一元二次方程3x2=﹣2x+5化为一般形式后,二次项系数、一次项系数、常数项分别是()A.3,﹣2,5B.3,2,﹣5C.3,﹣2,﹣5D.3,5,﹣2 2.围棋起源于中国,古代称之为“弈”,至今已有4000多年的历史,流传下来很多经典棋局.现取其棋谱中的四个部分,由黑白棋子摆成的图案是中心对称的是()A.B.C.D.3.下列事件中,是必然事件的是()A.掷一枚质地均匀的硬币,一定正面向上B.车辆随机到达一个路口,遇到红灯C.如果a2=b2,那么a=bD.将花生油滴在水中,油会浮在水面上4.如图(1)是博物馆展出的古代车轮实物.为测量车轮半径,如图(2)所示,在车轮上̂所在圆的圆心为O,作弦AB的垂线OC,D为垂足,则D是AB的取A、B两点,设AB中点.经测量:AB=90cm,CD=15cm,则OA的长度是()A .60cmB .65cmC .70cmD .75cm5.利用配方法解方程x 2+4x ﹣5=0,经过配方,得到( )A .(x +2)2=9B .(x ﹣2)2=9C .(x +4)2=9D .(x ﹣4)2=96.将抛物线y =x 2+3x +2向右平移a 单位正好经过原点,则a 的值是( )A .a =1B .a =2C .a =﹣1或a =1D .a =1或a =27.如图所示,将Rt △ABC 绕点A 按顺时针旋转一定角度得到Rt △ADE ,点B 的对应点D 恰好落在BC 边上.若AB =1,∠B =60°,则CD 的长是( )A .0.5B .1.5C .√2D .18.有两把不同的锁和三把不同的钥匙,其中两把钥匙恰好分别能打开这两把锁,其余的钥匙不能打开这两把锁.现在任意取出一把钥匙去开任意一把锁,则一次打开锁的概率是( )A .12B .13C .14D .34 9.如图,AB 为⊙O 的直径,点C 为AB̂的中点,D 、E 为圆上动点,且D 、E 关于AB 对称,将AD̂沿AD 翻折交AE 于点F ,使点C 恰好落在直径AB 上点C ′处,若⊙O 的周长为10,则AF̂的长为( )A .1B .1.25C .1.5D .2 10.抛物线y =ax 2+bx +c 过点(x 1,t )和(x 2,t ),若点(5x 1−x 24+t ,y 1)和(5x 2−x 14−t ,y 2)均在抛物线上,关于y 1,y 2的关系描述正确的是( )A.y1>y2B.y1=y2C.y1<y2D.y1,y2的大小无法确定二、填空题(本大题共6小题,共18分)11.在平面直角坐标系xOy中,将点(﹣2,3)绕原点O旋转180°,所得到的对应点的坐标为.12.如图,激光打靶游戏板中每一块小正方形除颜色外都相同.若某人用激光枪向打靶游戏发射激光一次(光点落在游戏板上),则光点落在涂色部分的概率是.13.为了保障医护人员在抗击疫情期间的个人防护安全,我市不断增加一线医疗工作者的医疗防护保障资金,2019年我市一线医疗工作者年人均医疗防护费用为20000元,2021年年人均医疗防护费用为24200元.则2019年到2021年我市一线医疗工作者年人均医疗防护费用的年平均增长率是.14.如图,⊙O是△ABC的外接圆,直径BC=4,点E是△ABC的内心.连接AE并延长交⊙O于点D,则DE的大小是.15.下列关于二次函数y=x2﹣2ax+4a(a为常数)的结论:①该函数的图象与x轴有两个交点时,a必大于4;②该函数的图象必过一定点;③该函数的图象随着a的取值变化时,其顶点会两次落在x轴上;④A(x1,y1)与点B(x2,y2)在该函数的图象上,若a>﹣1且﹣a<x1<x2时,y1<y2.其中正确的结论是(填写序号).16.如图,直线MN过正方形ABCD的顶点A,且∠NAD=30°,AB=2√2,P为直线MN 上的动点,连BP,将BP绕点B顺时针旋转60°至BQ,连接CQ,则CQ的最小值是.三、解答题(木大题共8小题,共72分)17.已知关于x的一元二次方程x2﹣4x+c=0有一个根是x=3,求c与另一个根.18.如图,将△ABC绕点C顺时针旋转90°得到△EDC.若点A,D,E在同一条直线上,且∠ACB=20°,求∠CAE及∠B的大小.19.防疫期间,全市所有学校都严格落实测体温进校园的防控要求.某校开设了A,B,C 三个测温通道.某天早晨,小明和小丽两位同学随机通过测温通道进入校园.(1)小明从A测温通道通过的概率是;(2)利用画树状图或列表的方法,求小明和小丽通过同一个测温通道的概率.20.如图,由小正方形构成的10×10网格,每个小正方形的顶点叫做格点.⊙O经过A,B,C三个格点,仅用无刻度的直尺在给定网格中按要求画图.(保留连线痕迹)(1)在图(1)中作线段AB的垂直平分线;̂=BÊ;(2)在图(2)中的⊙O上画一点E,使AE(3)在图(3)中过A,B,C的圆上找一点F,使AF平分∠CAB.21.如图(1),⊙O与矩形ABCD的边AB相切于点H,与边AD,BC分别交于点G,E,̂=HF̂.F,K,EH(1)求证:∠AEH=∠BFH;(2)如图(2),连接GF,连接DF交⊙O于点M,且GM平分∠DGF,若半径r=5,ED=4,求BK.22.近年来我国无人机设备发展迅猛,新型号无人机不断面世,科研单位为保障无人机设备能安全投产,现针对某种型号的无人机的降落情况进行测试,该型号无人机在跑道起点处着陆后滑行的距离y(单位:m)与滑行时间x(单位:s)之间满足二次函数关系,其部分函数图象如图所示.(1)求y关于x的函数关系式;(2)若跑道长度为900(m),是否够此无人机安全着陆?请说明理由.(3)现对该无人机使用减整伞进行短距离着陆实验,要求无人机触地同时打开减速伞(开伞时间忽略不计),若减速伞的制动效果为开伞后每秒钟减少滑行距离2a(单位:m),无人机必须在200(单位:m)的短距跑道降落,请直接写出a的取值范围为.23.问题背景如图(1),△ABC为等腰直角三角形,∠BAC=90°,直线l绕着点A顺时针旋转,过B,C两点分别向直线l作垂线BD,CE,垂足为D,E,此时△ABD可以由△CAE通过旋转变换得到,请写出旋转中心、旋转方向及旋转角的大小(取最小转角度)尝试应用如图(2),△ABC为等边三角形,直线l绕着点A顺时针旋转,D、E为直线l上两点,∠BDA=∠AEC=60°,△ABD可以由△CAE旋转变换得到吗?若可以,请指出旋转中心O的位置并说明理由.拓展创新 如图(3)在问题背景的条件下,若AB =2,连接DC ,直接写出CD 的长的取值范围.24.抛物线y =ax 2+bx +c 的顶点坐标为(m ,n ).(1)若抛物线y =ax 2+bx +c 过原点,m =2,n =﹣4,求其解析式;(2)如图(1),在(1)的条件下,直线l :y =﹣x +4与抛物线交于A 、B 两点(A 在B 的左侧),M 、N 为线段AB 上的两个点,MN =2√2,在直线l 下方的抛物线上是否存在点P ,使得△PMN 为等腰直角三角形?若存在,求出M 点横坐标;若不存在,请说明理由;(3)如图(2),抛物线y =ax 2+bx +c 与x 轴负半轴交于点C ,与y 轴交于点G ,P 点在点C 左侧抛物线上,Q 点在y 轴右侧抛物线上,直线CQ 交y 轴于点F ,直线PC 交y 轴于点H ,设直线PQ 解析式为y =kx +t ,当S △HCQ =2S △GCQ ,试证明b k 是否为一个定值.。
武汉二中广雅学校2023年九年级上学期月考数学试题(原卷版)

九年级(上)数学课堂作业9.16(试卷满分:120分 练习时间:120分钟)一、选择题(共10小题,每小题3分,共30分)1. 已知一元二次方程2230x x −+=的二次项系数是2,则一次项系数和常数项分别是( ) A. 1,3 B. 1,3− C. 1−,3 D. 1−,3− 2. 一元二次方程2210x x −−=的根的情况为( )A 有两不相等实根B. 有两相等实根C. 无实根D. 不能确定3. 1x ,2x 是方程260x x +−=的两根,则12x x +和12x x ⋅的值分别是( )A. 1,6B. 1,6−C. 1−,6D. 1−,6− 4. 抛物线()2324y x =+−的顶点坐标是( )A. ()2,4−−B. ()2,4−C. ()2,4−D. ()2,4 5. 二次函数2y x ,当12x <<时,y 的取值范围是( )A. 14y −<<B. 14y <<C. 01y ≤<D. 04y ≤< 6. 将抛物线2y x =−向左平移2个单位后,得到的抛物线的解析式是( ).A. 2(2)y x =−+B. 22y x =−+C. 2(2)y x =−−D. 22y x =−− 7. 已知抛物线23y x =−上两点()11A x y ,,()22B x y ,,若211x x >>,则下列结论成立的是( )A. 12y y >B. 12y y <C. 12y y ≥D. 12y y ≤ 8. 如图,有一长为12cm ,宽为8cm 的矩形纸片,在它的四角各剪去一个同样的小正方形,然后折叠成一个无盖的长方形纸盒,若纸盒的底面(图中阴影部分)的面积为236cm ,求剪去的小正方形的边长,设剪去的小正方形的边长为cm x ,根据题意可列方程为( )A. 1284836x ×−×=B. ()()1228236x x −−=C. ()()12836x x −−=D. 2128436x ×−=.9. 四边形ABCD 的对角线AC BD ⊥,且16AC BD +=,则四边形ABCD 的面积( )A. 有最大值64B. 有最小值64C. 有最大值32D. 有最小值3210. 当04x <≤时,直线2y x m =+与抛物线222y x x −−有两个不同交点,则m 的取值范围是( )A 62m −<<− B. 62m −≤<− C. 62m −<≤− D. 62m −≤≤−二、填空题(共6小题,每小题3分,共18分)11. 把二次三项式2610x x −+化成()x p q ++2的形式应为______.12. 已知抛物线()()220y a x k a =++>,当x ≥______时,y 随x 的增大而增大. 13. 已知关于x 的一元二次方程()22120kx k x k −−+−=有两个实数根,则实数k 的取值范围是___. 14. 抛物线()2224y m x mx n =−−+的对称轴是2x =,且它的最高点在直线4y x =+上,则m =______,n =______.15. 若一元二次方程()200ax bx c ac ++=≠有两个不相等实根,则下列结论: ①240b ac −>;②方程20cx bx a ++=一定有两个不相等实根;③设2b m a=−,当0a >时,一定有22am bm ax bx +≤+;④s ,()t s t <是关于x 的方程()()10x p x q +−−=的两根,且p q <,则q t s p >>>,一定成立的结论序号是______.16. 如图,在ABC 中,以AC 为斜边作等腰Rt ADC ,45DBC ∠=°,BC =,则ABC S = ______.三、解答题(共8题,共72分)17 解方程:228=0x x −−.18. 二次函数()20y ax bx c a ++≠的图象的顶点C 的坐标为()14−,,与x 轴交于()30A −,,()10B ,两点...(1)求二次函数的解析式;(2)直接写出关于x 的不等式20ax bx c ++>的解集为______.19. 已知矩形周长24cm ,矩形绕它的一边旋转形成一个圆柱,矩形之长、宽各为多少时,旋转形成的圆柱侧面积最大?最大的侧面积是多少?20. 图中是抛物线形拱桥,当拱顶离水面2m 时,水面宽4m ,建立如图所示的平面直角坐标系: (1)求拱桥所在抛物线的解析式;(2)当水面下降1m 时,则水面宽度为多少?21. 如图,抛物线()211144y x =--+经过原点,与x 轴交于另一点A ,点()4,E m 在此抛物线上,连接OE ,作45OEF ∠=°,交抛物线于点F ,求点F 的坐标.22. 经市场调查,商场某种运动服成本80元/件,月销量y (件)是售价x (元)的一次函数2400y x =−+,(1)当售价是______元/件时,月销售利润最大,最大利润是______元;(2)由于某种原因,该商品进价降低了n 元/件()0n >,商家规定该运动服售价不得低于150元/件,该商场在今后的售价中,月销量与售价仍满足上述一次函数关系,若月销售最大利润是8000元,求n 的值. 23. 如图,在正方形ABCD 中,DF EB =.的图1 图2 图3 (1)求证:ADE FBC ∠=∠;(2)如图2,点P 、Q 分别是线段DE 、FB 上的动点,45PCQ ∠=°,连接PQ ,探究三条线段DP 、PQ 、BQ 之间满足的数量关系,并证明你的结论; (3)如图3,在(2)的条件下,8DE =,在P 、Q 运动过程中,若PQ CD ∥,当PQ 最小时,AD =______.24. 抛物线22y ax ax m =−+与x 轴交于()1,0A −和B 两点,与y 轴交于点()0,3C −.图1 图2(1)求此抛物线解析式;(2)如图1,E 为线段CB 上一点,作DE OC ∥交抛物线于D ,DE 的长度最大时,求点E 的坐标; (3)如图2,E 射线CB 上一点,D 在抛物线上,D 、E 均位于x 轴上方,且使DAB OCB ∠=∠,当ADE 为等腰三角形时,求E 点坐标.为。
湖北省武汉二中广雅中学2019-2020学年九年级(上)月考数学试卷(9月份)解析版

湖北省武汉二中广雅中学2019-2020学年九年级(上)月考数学试卷(9月份)一、选择题(共10小题,每小题3分,共30分)1.(3分)下列四个图案中,是中心对称图案的是( )A .B .C .D . 2.(3分)点P (2,3)关于原点的对称点Q 的坐标是( )A .(﹣2,3)B .(2,﹣3)C .(3,2)D .(﹣2,﹣3)3.(3分)抛物线y =﹣(x +)2﹣3的顶点坐标是( )A .(,﹣3)B .(﹣,﹣3)C .(,3)D .(﹣,3) 4.(3分)用配方法解方程x 2+2x ﹣1=0时,配方结果正确的是( )A .(x +2)2=2B .(x +1)2=2C .(x +2)2=3D .(x +1)2=3 5.(3分)如图,已知△OAB 是正三角形,OC ⊥OA ,OC =OA .将△OAB 绕点O 按逆时针方向旋转,使得OB 与OC 重合,得到△OCD ,则旋转的角度是( )A .150°B .120°C .90°D .60°6.(3分)如图所示的Rt △ABC 向右翻滚,下列说法正确的有( )(1)①⇒②是旋转(2)①⇒③是平移(3)①⇒④是平移(4)②⇒③是旋转.A .1种B .2种C .3种D .4种7.(3分)已知函数y =(k ﹣3)x 2+2x +1的图象与x 轴有交点,则k 的取值范围是( )A.k<4B.k≤4C.k<4且k≠3D.k≤4且k≠38.(3分)已知A(x1,﹣1)、B(x2,﹣2)两点都在抛物线y=﹣x2+2x+3上,且x1>1,x2>1,则x1、x2的大小关系为()A.x1>x2B.x1<x2C.x1=x2D.无法确定9.(3分)宾馆有50间房供游客居住,当毎间房每天定价为180元时,宾馆会住满;当毎间房每天的定价每增加10元时,就会空闲一间房.如果有游客居住,宾馆需对居住的毎间房每天支出20元的费用.当房价定为多少元时,宾馆当天的利润为10890元?设房价定为x元.则有()A.(180+x﹣20)(50﹣)=10890B.(x﹣20)(50﹣)=10890C.x(50﹣)﹣50×20=10890D.(x+180)(50﹣)﹣50×20=1089010.(3分)如图,二次函数y=ax2+bx+c的图象经过点A(﹣1,0)、点B(3,0)、点C(4,y1),若点D(x2,y2)是抛物线上任意一点,有下列结论:①二次函数y=ax2+bx+c的最小值为﹣4a;②若﹣1≤x2≤4,则0≤y2≤5a;③若y2>y1,则x2>4;④一元二次方程cx2+bx+a=0的两个根为﹣1和其中正确结论的个数是()A.1B.2C.3D.4二、填空题(本大题共6个小题,每小题3分,共18分)11.(3分)抛物线y=4x2﹣8x+3的对称轴是直线.12.(3分)x1、x2是方程x2+5x﹣3=0的两个根,则x1﹣x1x2+x2=.13.(3分)已知点A(a,b)绕着(0,﹣1)旋转180°得到B(﹣4,1),则A点坐标为.14.(3分)将抛物线y=(x﹣1)2+3向左平移1个单位,得到的抛物线与y轴的交点坐标是.15.(3分)将直角边长为5cm的等腰直角△ABC绕点A逆时针旋转15°后,得到△AB′C′,则图中阴影部分的面积是cm2.16.(3分)如图,△ABC中,∠BAC=30°且AB=AC,P是底边上的高AH上一点.若AP+BP+CP的最小值为2,则BC=.三、解答题(共8题,共72分)17.(8分)解方程:(1)x2﹣4x﹣7=0(用公式法)(2)x2﹣2x﹣24=018.(8分)如图,△AEC绕A点顺时针旋转60°得△APB,∠PAC=20°,求∠BAE.19.(8分)已知关于x的方程(k﹣1)x2﹣(k﹣1)x+=0有两个相等的实数根,求k的值,并求此时该方程的根.20.(8分)参加足球联赛的每两队之间都进行两场比赛.共要比赛90场.共有多少个队参加比赛?21.(8分)如图,在平面直角坐标系中,已知A(﹣2,﹣4)、B(0,﹣4)、C(1,﹣1)(1)画出△ABC绕O点逆时针旋转90°后的图形△A1B1C1,并写出C1的坐标;(2)将(1)中所得△A1B1C1先向左平移4个单位,再向上平移2个单位得到△A2B2C2,画出△A2B2C2,则C2(,)(3)若△A2B2C2可以看作△ABC绕某点旋转得来,则旋转中心的坐标为.22.(10分)如图,有长为24米的篱笆,一面利用长为10m的墙,围成中间隔有一道篱笆的长方形花圃,设花圃的宽AB为xm,面积为Sm2(1)设BC=y,求y与x的关系式,并写出自变量x的取值范围;(2)如果要围成面积为45m2的花圃,AB的长是多少?(3)能围成面积比45m2更大的花圃吗?如果能,请求出最大面积,并说明围法,如果不能,请说明理由.23.(10分)如图,点E是正方形ABCD中CD边上任意一点,AB=4,以点A为中心,把△ADE 顺时针旋转90°得到△AD′F(1)画出旋转后的图形,求证:点C、B、F三点共线;(2)AG平分∠EAF交BC于点G.①如图2,连接EF.若BG:CE=5:6,求△AEF的面积;②如图3,若BM、DN分别为正方形的两个外角角平分线,交AG、AE的延长线于点M、N.当MM∥DC时,直接写出DN的长.24.(12分)如图,已知直线y=x+2交x轴、y轴分别于点A、B,抛物线y=ax2+bx+c(a≠0)的对称轴为直线x=﹣,且抛物线经过A、B两点,交x轴于另一点C.(1)求抛物线的解析式;(2)点M是抛物线x轴上方一点,∠MBA=∠CBO,求点M的坐标;(4)过点A作AB的垂线交y轴于点D,平移直线AD交抛物线于点E、F两点,连结EO、FO.若△EFO为以EF为斜边的直角三角形,求平移后的直线的解析式.参考答案与试题解析一、选择题(共10小题,每小题3分,共30分)1.解:A、该图形不是中心对称图形,故本选项错误;B、该图形是中心对称图形,故本选项正确;C、该图形不是中心对称图形,故本选项错误;D、该图形旋转180度,阴影部分不能重合,故不是中心对称图形,故本选项错误;故选:B.2.解:根据中心对称的性质,可知:点P(2,3)关于原点O的对称点的坐标为(﹣2,﹣3).故选:D.3.解:y=﹣(x+)2﹣3是抛物线的顶点式,根据顶点式的坐标特点可知,顶点坐标为(﹣,﹣3).故选:B.4.解:∵x2+2x﹣1=0,∴x2+2x+1=2,∴(x+1)2=2.故选:B.5.解:∵△OAB是正三角形,∴∠BOA=60°,∵OC⊥OA,∴∠AOC=90°,∴∠BOC=∠BOA+∠AOC=60°+90°=150°,即旋转角是150°,故选:A.6.解:观察图形可知,(1)(3)(4)说法正确;(2)①⇒③需要改变旋转中心,经过两次旋转得到,不属于平移,错误;正确的有三种,故选C.7.解:①当k﹣3≠0时,(k﹣3)x2+2x+1=0,△=b2﹣4ac=22﹣4(k﹣3)×1=﹣4k+16≥0,k≤4;②当k﹣3=0时,y=2x+1,与x轴有交点.故选:B.8.解:∵抛物线y=﹣x2+2x+3的对称轴x=1,x1>1,x2>1,∴A、B在对称轴的右侧,抛物线开口向下,∵﹣1>﹣2,∴x1<x2,故选:B.9.解:设房价定为x元,根据题意,得(x﹣20)(50﹣)=10890.故选:B.10.解:抛物线解析式为y=a(x+1)(x﹣3),即y=ax2﹣2ax﹣3a,∵y=a(x﹣1)2﹣4a,∴当x=1时,二次函数有最小值﹣4a,所以①正确;当x=4时,y=a•5•1=5a,∴当﹣1≤x2≤4,则﹣4a≤y2≤5a,所以②错误;∵点C(4,5a)关于直线x=1的对称点为(﹣2,5a),∴当y2>y1,则x2>4或x<﹣2,所以③错误;∵b=﹣2a,c=﹣3a,∴方程cx2+bx+a=0化为﹣3ax2﹣2ax+a=0,整理得3x2+2x﹣1=0,解得x1=﹣1,x2=,所以④正确.故选:B.二、填空题(本大题共6个小题,每小题3分,共18分)11.解:∵y=4x2﹣8x+3,∴抛物线对称轴为x=﹣=1,故答案为:x=1.12.解:∵x1、x2是方程x2+5x﹣3=0的两个根,∴x1+x2=﹣5,x1x2=﹣3,∴x1﹣x1x2+x2=x1+x2﹣x1x2=﹣5﹣(﹣3)=﹣2.故答案是:﹣2.13.解:∵点A(a,b)绕着(0,﹣1)旋转180°得到B(﹣4,1),∴点(0,﹣1)为AB的中点,∴0=,1=,解得a=4,b=﹣3,∴A点坐标为(4,﹣3).故答案为(4,﹣3).14.解:抛物线y=(x﹣1)2+3的顶点坐标为(1,3),把点(1,3)向左平移1个单位得到点的坐标为(0,3),所以平移后抛物线解析式为y=x2+3,所以得到的抛物线与y轴的交点坐标为(0,3).故答案为(0,3).15.解:∵等腰直角△ABC绕点A逆时针旋转15°后得到△AB′C′,∵∠CAC′=15°,∴∠C′AB=∠CAB﹣∠CAC′=45°﹣15°=30°,AC′=AC=5,∴阴影部分的面积=×5×tan30°×5=.16.解:如图将△ABP绕点A顺时针旋转60°得到△AMG.连接PG,CM.∵AB=AC,AH⊥BC,∴∠BAP=∠CAP,∵PA=PA,∴△BAP≌△CAP(SAS),∴PC=PB,∵MG=PB,AG=AP,∠GAP=60°,∴△GAP是等边三角形,∴PA=PG,∴PA+PB+PC=CP+PG+GM,∴当M,G,P,C共线时,PA+PB+PC的值最小,最小值为线段CM的长,∵AP+BP+CP的最小值为2,∴CM=2,∵∠BAM=60°,∠BAC=30°,∴∠MAC=90°,∴AM=AC=2,作BN⊥AC于N.则BN=AB=1,AN=,CN=2﹣,∴BC===﹣.故答案为﹣.三、解答题(共8题,共72分)17.解:(1)∵a=1,b=﹣4,c=﹣7,∴△=16﹣4×1×(﹣7)=44>0,则x==2±,∴x1=2+,x2=2﹣;(2)∵x2﹣2x﹣24=0,∴(x+4)(x﹣6)=0,则x+4=0或x﹣6=0,解得:x1=﹣4,x2=6.18.解:根据旋转的性质可得△ABP≌△ACE,AC与AB是对应边,∠BAC=∠BAP+∠PAC=60°,∵∠PAC=20°,∴∠CAE=∠BAP=40°,∴∠BAE=∠BAC+∠CAE=100°.19.解:∵关于x的方程(k﹣1)x2﹣(k﹣1)x+=0有两个相等的实数根,∴,即,解得:k=2.当k=2时,原方程为x2﹣x+==0,解得:x1=x2=.20.解:设共有x个队参加比赛,根据题意得:2×x(x﹣1)=90,整理得:x2﹣x﹣90=0,解得:x=10或x=﹣9(舍去).故共有10个队参加比赛.21.解:(1)△A1B1C1如图所示,C1(1,1);(2)△A2B2C2如图所示;故答案为:﹣3,3.(3)如图所示,旋转中心为P(﹣3,﹣1).故答案为:(﹣3,﹣1).22.解:(1)由题可知,花圃的宽AB为x米,则BC为(24﹣3x)米这时面积y=24﹣3x(0<x<8).(2)由条件﹣3x2+24x=45化为x2﹣8x+15=0解得x1=5,x2=3∵0<24﹣3x≤10得≤x<8∴x=3不合题意,舍去即花圃的宽为5米.(3)S=﹣3x2+24x=﹣3(x2﹣8x)=﹣3(x﹣4)2+48(≤x<8)∴当x=时,S有最大值48﹣3(﹣4)2=46故能围成面积比45米2更大的花圃.围法:24﹣3×=10,花圃的长为10米,宽为4米,这时有最大面积46平方米.23.(1)证明:旋转后的图形如图1中所示,∵四边形ABCD是正方形,∴AD=AB,∠D=∠ABC=90°,∵∴点D′与点B重合,∵∠AD′F=90°,∴∠AD′F+′AD′C=180°,∴C,B,F共线.(2)①解:如图2中,连接EG.∵∠BAF=∠DAE,∴∠EAF=∠DAB=90°,∵AG平分∠EAF,∴∠EAG=×90°=45°,∴∠FAG=∠FAB+∠BAG=∠BAG+∠DAE=45°,∴∠FAG=∠EAG,∵AG=AG,AF=AE,∴△GAE≌△GAF(SAS),∴FG=EG,∴EG=BF+BG=DE+BG,∵BG:CE=5:6,∴可以假设BG=5k,CE=6k,则DE=4﹣6k,CG=4﹣5k,EG=4﹣k,在Rt△EGC中,∵EG2=EC2+CG2,∴(4﹣k)2=(6k)2+(4﹣5k)2,∴k=,∴DE=,∴AE=AF==,=•AE•AF=.∴S△AEF②解:如图3中,连接EG,延长MN交AD的延长线于点P,作MQ⊥AB交AB的延长线于点Q.由题意可知:△PDN,△BMQ都是等腰直角三角形,设DP=PN=x,BG=a,DE=b.∵四边形AQMP是矩形,∴MQ=BQ=AP=4+x,∵DE∥PN,∴=,即=①,∵BG∥MQ,∴=,即=②在Rt△BCG中,∵EG2=EC2+CG2,∴(a+b)2=(4﹣a)2+(4﹣b)2③,由①②③可得x=2﹣2或﹣2﹣2(舍弃)∴DN=x=2﹣2.24.解:(1)∵直线y=x+2交x轴、y轴分别于点A、B,∴A(﹣2,0),B(0,2),∵抛物线的对称轴x=﹣,A,C关于对称轴对称,∴C(1,0),设抛物线的解析式为y=a(x+2)(x﹣1),把(0,2)代入得到a=﹣1,∴抛物线的解析式为y=﹣x2﹣x+2.(2)如图1中,作EA⊥AB交BM的延长线于E,作EF⊥x轴于F.∵∠ABE=∠OBC,∠BAE=∠BOC=90°,∴△BAE∽△BOC,∴=,∴=,∴AE=,∵∠EAF+∠BAO=90°,∠BAO=45°,∴∠EAF=45°,∴EF=AF=1,∴E(﹣3,1),∴直线BE的解析式为y=x+2,由,解得或,∴M(﹣,).(3)如图2中,当直线AD向下平移时,设E(x1,y1),F(x2,y2),作EH⊥x轴于H,FG ⊥x轴于G.∵∠EOF=90°=∠PHE=∠OGF,由△EHO∽△OGF得到:=,∴=,∴x1x2+y1y2=0,由,消去y得到:x2+b﹣2=0,∴x1x2=b﹣2,x1+x2=0,y1y2=(﹣x1+b)(﹣x2+b)=x1x2+b2,∴2(b﹣2)+b2=0,解得b=﹣1﹣或﹣1+(舍弃),当直线AD向上平移时,同法可得b=﹣1+,综上所述,平移后的解析式为y=﹣x﹣1+或y=﹣x﹣1﹣.。
2023-2024学年湖北省武汉市江岸区二中广雅中学九年级上学期10月月考化学试题

初中英语个性化教案数词专题一、教学目标1. 知识目标学生能够理解并掌握数词的用法,包括基数词和序数词。
学生能够运用数词进行基本的数学计算和表达。
2. 能力目标学生能够听懂、说对、会读、会写数词。
学生能够在实际情景中运用数词进行交流。
3. 情感目标学生能够培养对英语数词学习的兴趣,提高学习积极性。
二、教学重难点1. 教学重点数词的用法,包括基数词和序数词。
数词在实际情景中的运用。
2. 教学难点序数词的构成和用法。
在实际情景中灵活运用数词。
三、教学方法1. 情境教学法:通过设定各种真实的生活情境,让学生在实际语境中学习和运用数词。
2. 互动教学法:引导学生参与课堂互动,进行数词的听、说、读、写训练。
3. 任务型教学法:设计各种任务,让学生在完成任务的过程中运用数词。
1. 教学材料:数词相关教材、练习册。
2. 教学工具:PPT、黑板、实物等。
3. 教学环境:教室。
五、教学过程1. 导入:通过歌曲、游戏等方式引导学生进入数词学习主题。
2. 教学新课:讲解数词的用法,包括基数词和序数词。
3. 练习巩固:设计各种练习题,让学生进行听、说、读、写训练。
4. 情景交际:设定生活场景,让学生运用数词进行实际交流。
6. 课后作业:布置相关数词的练习题,让学生巩固所学知识。
六、教学评价1. 形成性评价:在教学过程中,通过观察、提问、互动等方式及时了解学生的学习情况,给予指导和帮助。
2. 终结性评价:在课程结束后,通过测试、作业、表演等方式评估学生的学习成果。
七、教学拓展1. 开展数词竞赛活动,激发学生的学习兴趣。
2. 组织数词主题的英语角活动,让学生在实际交流中运用数词。
3. 推荐数词相关的英语学习资源,帮助学生拓展知识。
八、教学反思在教学过程中,要不断反思教学方法、教学内容、教学效果等方面,以便及时调整和改进。
根据教学目标和教学内容,制定详细的教学计划,包括每个课时的教学内容、教学方法、教学时间等。
十、教学资源1. 网络资源:利用互联网查找数词相关的教学素材、教案、课件等。
湖北省武汉二中广雅中学届九年级数学上学期月考试卷(三)(含解析)新人教版【含解析】

2015-2016学年湖北省武汉二中广雅中学九年级(上)月考数学试卷(三)一、选择题(共10小题,每小题3分,共30分)1.方程4x(x﹣2)=25的一次项系数和常数项分别为()A.﹣2,25 B.﹣2,﹣25 C.8,﹣25 D.﹣8,﹣252.下列四个图形中,既是轴对称图形又是中心对称图形的是()A.B.C.D.3.下列事件是必然事件的是()A.地球绕着太阳转B.抛一枚硬币,正面朝上C.明天会下雨D.打开电视,正在播放新闻4.如图,弦AC∥OB,∠B=25°,则∠O=()A.20° B.30° C.40° D.50°5.方程5x﹣1=4x2的两根之和为()A.B.﹣ C.D.﹣6.袋子中装有2个红球、3个白球和3个黄球,这些球除颜色外无其他差别,从袋子中随机摸出一个球,是白球的概率为()A.B.C.D.7.二次函数y=x2﹣6x+21的图象顶点坐标为()A.(﹣6,3)B.(6,3)C.(﹣6,75)D.(6,75)8.如图,在矩形ABCD中,AB=4,AD=5,AD,AB,BC分别与⊙O相切于E,F,G三点,过点D作⊙O的切线BC于点M,切点为N,则DM的长为()A.B.C.D.29.如图,在直角坐标系中,已知点A(﹣3,0)、B(0,4),对△OAB连续作旋转变换,依次得到△1、△2、△3、△4、…,△16的直角顶点的坐标为()A.(60,0) B.(72,0) C.(67,)D.(79,)10.在Rt△ABC中,∠C=90°,AC=10,BC=12,点D为线段BC上一动点.以CD为⊙O直径,作AD交⊙O于点E,连BE,则BE的最小值为()A.6 B.8 C.10 D.12二、填空题(本大题共6个小题,每小题3分,共18分)11.方程x(x﹣4)=2(x﹣4)的解为.12.将抛物线y=2(x+3)2+5向右平移2个单位后的抛物线解析式为.13.已知点A(a,1)与点B(3,b)关于原点对称,则线段AB= .14.有两人患了流感,经过两轮传染后共有242人患了流感,求每轮传染中平均一个人传染了几个人?设每轮传染中平均一个人传染了x人,则可列方程为.15.边心距为2的正六边形的面积为.16.将边长为的正方形ABCD与边长为的正方形CEFG如图摆放,点G恰好落在线段DE上.连BE,则BE长为.三、解答题(共8题,共72分)17.解方程:3x2﹣6x﹣2=0.18.已知二次函数图象的顶点为(3,﹣1),与y轴交于点(0,﹣4).(1)求二次函数解析式;(2)求函数值y>﹣4时,自变量x的取值范围.19.如图,⊙O中,弦AD=BC.(1)求证:AC=BD.(2)若∠D=60°,⊙O的半径为2,求弧AB的长.20.如图,方格纸中的每个小正方形都是边长为1个单位长度的正方形,每个小正方形的顶点叫格点.△ABC的顶点都在格点上,结合所给的平面直角坐标系解答下列问题:(1)画出△ABC关于点P成中心对称的△A1B1C1,点A1的坐标为.(2)画出△ABC绕点P逆时针方向旋转90°后所得到的△A2B2C2,点B2的坐标为.(3)在(2)中线段AB绕点P按逆时针方向旋转90°后得到线段A2B2过程中所扫过的面积为.21.如图,在△ABC中,AB=AC,以AC为直径的⊙O交BC于点D,交AB于点E,过点D作DF⊥AB,垂足为点F,连接DE.(1)求证:直线DF与⊙O相切;(2)若CD=3,BF=1,求AE的长.22.某工厂为了对新研发的一种产品进行合理定价,将该产品按拟定的价格进行试销,通过(1)计算这5天销售额的平均数(销售额=单价×销量);(2)通过对上面表格中的数据进行分析,发现销量y(件)与单价x(元/件)之间存在一次函数关系,求y关于x的函数关系式(不需要写出函数自变量的取值范围);(3)预计在今后的销售中,销量与单价仍然存在(2)中的关系,且该产品的成本是20元/件.为使工厂获得最大利润,该产品的单价应定为多少?23.已知,正方形ABCD的边长为4,点E是对角线BD延长线上一点,AE=BD.将△ABE绕点A顺时针旋转α度(0°<α<360°)得到△AB′E′,点B、E的对应点分别为B′、E′.(1)如图1,当α=30°时,求证:B′C=DE;(2)连接B′E、DE′,当B′E=DE′时,请用图2求α的值;(3)如图3,点P为AB的中点,点Q为线段B′E′上任意一点,试探究,在此旋转过程中,线段PQ长度的取值范围为.24.已知抛物线y=x2与直线y=x+1交于A、B两点(A在B的左侧)(1)求A、B两点的坐标.(2)在直线AB的下方的抛物线上有一点D,使得ABD面积最大,求点D的坐标.(3)把抛物线向右平移2个单位,再向下平移m(m>0)个单位,平移后的抛物线与x轴交于E、F两点,直线AB与y轴交于点C.当m为何值时,过E、F、C三点的圆的面积最小,最小面积是多少?2015-2016学年湖北省武汉二中广雅中学九年级(上)月考数学试卷(三)参考答案与试题解析一、选择题(共10小题,每小题3分,共30分)1.方程4x(x﹣2)=25的一次项系数和常数项分别为()A.﹣2,25 B.﹣2,﹣25 C.8,﹣25 D.﹣8,﹣25【考点】一元二次方程的一般形式.【分析】根据去括号、移项,可得一元二次方程的一般形式,根据ax2+bx+c=0(a,b,c是常数且a≠0)b,c分别叫一次项系数,常数项,可得答案.【解答】解:去括号、移项,得4x2﹣8x﹣25=0.一次项系数和常数项分别为﹣8,﹣25,故选:D.2.下列四个图形中,既是轴对称图形又是中心对称图形的是()A.B.C.D.【考点】中心对称图形;轴对称图形.【分析】根据轴对称图形与中心对称图形的概念求解.【解答】解:A、不是轴对称图形,是中心对称图形.故错误;B、不是轴对称图形,是中心对称图形.故错误;C、不是轴对称图形,是中心对称图形.故错误;D、是轴对称图形,也是中心对称图形.故正确.故选D.3.下列事件是必然事件的是()A.地球绕着太阳转B.抛一枚硬币,正面朝上C.明天会下雨D.打开电视,正在播放新闻【考点】随机事件.【分析】根据必然事件、不可能事件、随机事件的概念可区别各类事件.【解答】解:A、地球绕着太阳转是必然事件,故A符合题意;B、抛一枚硬币,正面朝上是随机事件,故B不符合题意;C、明天会下雨是随机事件,故C不符合题意;D、打开电视,正在播放新闻是随机事件,故D不符合题意;故选:A.4.如图,弦AC∥OB,∠B=25°,则∠O=()A.20° B.30° C.40° D.50°【考点】圆周角定理.【分析】先根据平行线的性质求出∠A的度数,再由圆周角定理即可得出结论.【解答】解:∵AC∥OB,∠B=25°,∴∠A=∠B=25°.∵∠A与∠O是同弧所对的圆周角与圆心角,∴∠O=2∠A=50°.故选D.5.方程5x﹣1=4x2的两根之和为()A.B.﹣ C.D.﹣【考点】根与系数的关系.【分析】把方程化为一般形式后,根据根与系数的关系得到两根之和即可.【解答】解:∵5x﹣1=4x2,∴4x2﹣5x+1=0,设方程4x2﹣5x+1=0的两根设为:x1,x2,∴x1+x2=.故选:A.6.袋子中装有2个红球、3个白球和3个黄球,这些球除颜色外无其他差别,从袋子中随机摸出一个球,是白球的概率为()A.B.C.D.【考点】概率公式.【分析】让白球的个数除以球的总数即为摸到白球的概率.【解答】解:∵布袋中装有2个红球、3个白球和3个黄球,共8个球,从袋中任意摸出一个球共有10种结果,其中出现白球的情况有3种可能,∴是白球的概率是,故选B.7.二次函数y=x2﹣6x+21的图象顶点坐标为()A.(﹣6,3)B.(6,3)C.(﹣6,75)D.(6,75)【考点】二次函数的性质.【分析】把函数的一般式化成顶点式,即可求得顶点坐标.【解答】解:∵y=x2﹣6x+21=(x﹣6)2+3,∴二次函数y=x2﹣6x+21的图象顶点坐标为:(6,3).故选B.8.如图,在矩形ABCD中,AB=4,AD=5,AD,AB,BC分别与⊙O相切于E,F,G三点,过点D作⊙O的切线BC于点M,切点为N,则DM的长为()A.B.C.D.2【考点】切线的性质;矩形的性质.【分析】连接OE,OF,ON,OG,在矩形ABCD中,得到∠A=∠B=90°,CD=AB=4,由于AD,AB,BC分别与⊙O相切于E,F,G三点得到∠AEO=∠AFO=∠OFB=∠BGO=90°,推出四边形AFOE,FBGO是正方形,得到AF=BF=AE=BG=2,由勾股定理列方程即可求出结果.【解答】解:连接OE,OF,ON,OG,在矩形ABCD中,∵∠A=∠B=90°,CD=AB=4,∵AD,AB,BC分别与⊙O相切于E,F,G三点,∴∠AEO=∠AFO=∠OFB=∠BGO=90°,∴四边形AFOE,FBGO是正方形,∴AF=BF=AE=BG=2,∴DE=3,∵DM是⊙O的切线,∴DN=DE=3,MN=MG,∴CM=5﹣2﹣MN=3﹣MN,在R t△DMC中,DM2=CD2+CM2,∴(3+NM)2=(3﹣NM)2+42,∴NM=,∴DM=3=,故选A.9.如图,在直角坐标系中,已知点A(﹣3,0)、B(0,4),对△OAB连续作旋转变换,依次得到△1、△2、△3、△4、…,△16的直角顶点的坐标为()A.(60,0) B.(72,0) C.(67,)D.(79,)【考点】规律型:点的坐标.【分析】根据题目提供的信息,可知旋转三次为一个循环,图中第三次和第四次的直角顶点的坐标相同,由①→③时直角顶点的坐标可以求出来,从而可以解答本题.【解答】解:由题意可得,△OAB旋转三次和原来的相对位置一样,点A(﹣3,0)、B(0,4),∴OA=3,OB=4,∠BOA=90°,∴AB=∴旋转到第三次时的直角顶点的坐标为:(12,0),16÷3=5 (1)∴旋转第15次的直角顶点的坐标为:(60,0),又∵旋转第16次直角顶点的坐标与第15次一样,∴旋转第16次的直角顶点的坐标是(60,0).故选A.10.在Rt△ABC中,∠C=90°,AC=10,BC=12,点D为线段BC上一动点.以CD为⊙O直径,作AD交⊙O于点E,连BE,则BE的最小值为()A.6 B.8 C.10 D.12【考点】切线的性质.【分析】连接CE,可得∠CED=∠CEA=90°,从而知点E在以AC为直径的⊙Q上,继而知点Q、E、B共线时BE最小,根据勾股定理求得QB的长,即可得答案.【解答】解:如图,连接CE,∴∠CED=∠CEA=90°,∴点E在以AC为直径的⊙Q上,∵AC=10,∴QC=QE=5,当点Q、E、B共线时BE最小,∵BC=12,∴QB==13,∴BE=QB﹣QE=8,故选:B.二、填空题(本大题共6个小题,每小题3分,共18分)11.方程x(x﹣4)=2(x﹣4)的解为4或2 .【考点】解一元二次方程-因式分解法.【分析】方程移项变形后,利用因式分解法求出解即可.【解答】解:方程变形得:x(x﹣4)﹣2(x﹣4)=0,因式分解得:(x﹣4)(x﹣2)=0,解得:x1=4,x2=2.故答案为4或2.12.将抛物线y=2(x+3)2+5向右平移2个单位后的抛物线解析式为y=2(x+1)2+5 .【考点】二次函数图象与几何变换.【分析】根据函数图象向左平移加,向右平移减,可得答案.【解答】解:将抛物线y=2(x+3)2+5向右平移2个单位,所得抛物线的解析式为y=2(x+3﹣2)2+5,即y=2(x+1)2+5.故答案为:y=2(x+1)2+5.13.已知点A(a,1)与点B(3,b)关于原点对称,则线段AB= 2.【考点】关于原点对称的点的坐标.【分析】根据关于原点对称的点的横坐标互为相反数,纵坐标互为相反数,可得a、b的值,再根据勾股定理,可得答案.【解答】解:由点A(a,1)与点B(3,b)关于原点对称,得a=﹣3,b=﹣1.A(﹣3,1),B(3,﹣1).由勾股定理得AB===2,故答案为:2.14.有两人患了流感,经过两轮传染后共有242人患了流感,求每轮传染中平均一个人传染了几个人?设每轮传染中平均一个人传染了x人,则可列方程为2+2x+(2+2x) x=242 .【考点】由实际问题抽象出一元二次方程.【分析】先根据题意列出第一轮传染后患流感的人数,再根据题意列出第二轮传染后患流感的人数,而已知第二轮传染后患流感的人数,故可得方程.【解答】解:设每轮传染中平均一个人传染了x个人,第一轮传染后患流感的人数是:2+x,第二轮传染后患流感的人数是:2+x+x(1+x),而已知经过两轮传染后共有242人患了流感,则可得方程,2+2x+(2+2x) x=242.故答案为:2+2x+(2+2x) x=242.15.边心距为2的正六边形的面积为24.【考点】正多边形和圆.【分析】根据题意画出图形,先求出∠AOB的度数,证明△AOB是等边三角形,得出AB=OA,再根据直角三角形的性质求出OA的长,再根据S六边形=6S△AOB即可得出结论.【解答】解:如图所示,∵图中是正六边形,∴∠AOB═60°.∵OA=OB,∴△OAB是等边三角形.∴OA=OAB=AB,∵OD⊥AB,OD=2,∴OA==4.∴AB=4,∴S△AOB=AB×OD=×4×2=4,∴正六边形的面积=6S△AOB=6×4=24.故答案为:24.16.将边长为的正方形ABCD与边长为的正方形CEFG如图摆放,点G恰好落在线段DE上.连BE,则BE长为.【考点】正方形的性质;全等三角形的判定与性质;勾股定理.【分析】连接BD,BG,设DC和BG相较于点O,利用△BOD∽△COG求出线段BO、OC、OD、OG,在RT△BGE中利用勾股定理即可求BE.【解答】解:(1)如图1,∵四边形ABCD、四边形CGEF都是正方形,∴BC=CD=,CG=CE=,∠BCD=∠GCE=90°,∠DEC=∠CGE=45°,∠BDC=45°,∴BD=,GE=2,∴∠BCG=∠DCE,在△BCG和△DCE中,,∴△BCG≌△DCE,∴∠BGC=∠DEC=45°,∴∠BGE=∠BGC+∠CGE=90°,∵∠DOB=∠GOC,∠BDO=∠OGC,∴△BDO∽△CGO,∴,设OC=k,则BO=k,∵BO2=OC2+BC2,∴5k2=5+k2,∴k=,∴OC=OD=,BO=2.5,OG=0.5,∴BG=BO+OG=3,在RT△BGE中,BG=3,EG=2,∴BE==,故答案为.三、解答题(共8题,共72分)17.解方程:3x2﹣6x﹣2=0.【考点】解一元二次方程-公式法.【分析】先根确定a=3,b=﹣6,c=﹣2,算出b2﹣4ac=36+24=60>0,确定有解,最后代入求根公式计算就可以了.【解答】解:∵a=3,b=﹣6,c=﹣2,∴b2﹣4ac=36+24=60>0,∴x=,∴x1=,x2=18.已知二次函数图象的顶点为(3,﹣1),与y轴交于点(0,﹣4).(1)求二次函数解析式;(2)求函数值y>﹣4时,自变量x的取值范围.【考点】待定系数法求二次函数解析式.【分析】(1)已知函数的顶点坐标,就可设出函数的顶点式,利用待定系数法求解析式.(2)根据二次函数的开口方向,顶点坐标以及对称性即可求解.【解答】解:(1)设抛物线的解析式为y=a(x﹣3)2﹣1,把(0,﹣4)代入得9a﹣1=﹣4,解得a=﹣.所以二次函数解析式为y=﹣(x﹣3)2﹣1;(2)∵a=﹣<0,∴抛物线开口向下,∵顶点为(3,﹣1),∴点(0,﹣4)对称点为(6,﹣4),∴函数值y>﹣4时,自变量0<x<6.19.如图,⊙O中,弦AD=BC.(1)求证:AC=BD.(2)若∠D=60°,⊙O的半径为2,求弧AB的长.【考点】圆心角、弧、弦的关系;弧长的计算.【分析】(1)由在同圆中,弦相等,则所对的弧相等和等量加等量还是等量求解;(2)根据同弧所对的圆周角等于所对圆心角的一半,得出弧AB所对的圆心角,再由弧长公式计算即可.【解答】证明:(1)∵AD=BC,∴=.∴+=+.∴=.∴AC=BD;(2)∵∠D=60°,∴弧AB所对的圆心角=120°,∴l===π,∴弧AB的长为π.20.如图,方格纸中的每个小正方形都是边长为1个单位长度的正方形,每个小正方形的顶点叫格点.△ABC的顶点都在格点上,结合所给的平面直角坐标系解答下列问题:(1)画出△ABC关于点P成中心对称的△A1B1C1,点A1的坐标为(0,3).(2)画出△ABC绕点P逆时针方向旋转90°后所得到的△A2B2C2,点B2的坐标为(2,﹣1).(3)在(2)中线段AB绕点P按逆时针方向旋转90°后得到线段A2B2过程中所扫过的面积为π.【考点】作图-旋转变换.【分析】(1)利用网格特点和中心对称的性质画出点A、B、C的对称点A1、B1、C1,则可得到△A1B1C1,然后写出点A1的坐标;(2)利用网格特点和旋转的性质画出点A、B、C的对称点A2、B2、C2,则可得到△A2B2C2,然后写出点BH2的坐标;(3)根据扇形的面积公式,利用线段A2B2过程中所扫过的面积为=S扇形APA2﹣S扇形BPB2进行计算即可.【解答】解:(1)如图,△A1B1C1为所作,点A1的坐标为(0,3);(2)如图,△A2B2C2为所作,点B2的坐标为(2,﹣1);(3)PA==2,PB=4,线段A2B2过程中所扫过的面积为=S扇形APA2﹣S扇形BPB2=﹣=π.故答案为(0,3),(2,﹣1),π.21.如图,在△ABC中,AB=AC,以AC为直径的⊙O交BC于点D,交AB于点E,过点D作DF⊥AB,垂足为点F,连接DE.(1)求证:直线DF与⊙O相切;(2)若CD=3,BF=1,求AE的长.【考点】切线的判定.【分析】(1)连接OD,利用AB=AC,OD=OC,证得OD∥AB,易证DF⊥OD,故DF为⊙O的切线;(2)根据内接四边形的性质得到∠AED+∠ACD=180°,由于∠AED+∠BED=180°,得到∠BED=∠ACD,由于∠B=∠B,推出△BED∽△BCA,根据相似三角形的性质得到,∠DEB=∠ODC,得到∠B=∠DEB,求得BE=2BF=2,BD=CD=BC=3,BC=6,即可得到结论.【解答】(1)证明:如图,连接OD,∵AB=AC,∴∠B=∠C,∵OD=OC,∴∠ODC=∠C,∴∠ODC=∠B,∴OD∥AB,∵DF⊥AB,∴OD⊥DF,∵点D在⊙O上,∴直线DF与⊙O相切;(2)解:∵四边形ACDE是⊙O的内接四边形,∴∠AED+∠ACD=180°,∵∠AED+∠BED=180°,∴∠BED=∠ACD,∵∠B=∠B,∴△BED∽△BCA,∴,∠DEB=∠ODC,∴∠B=∠DEB,∴BD=DE,∴BE=2BF=2,∵OD∥AB,AO=CO,∴BD=CD=BC=3,∴BC=6,∴,∴AB=9,∴AE=AB﹣BE=7.22.某工厂为了对新研发的一种产品进行合理定价,将该产品按拟定的价格进行试销,通过(2)通过对上面表格中的数据进行分析,发现销量y(件)与单价x(元/件)之间存在一次函数关系,求y关于x的函数关系式(不需要写出函数自变量的取值范围);(3)预计在今后的销售中,销量与单价仍然存在(2)中的关系,且该产品的成本是20元/件.为使工厂获得最大利润,该产品的单价应定为多少?【考点】二次函数的应用.【分析】(1)根据题中表格中的数据列出算式,计算即可得到结果;(2)设y=kx+b,从表格中找出两对值代入求出k与b的值,即可确定出解析式;(3)设定价为x元时,工厂获得的利润为W,列出W与x的二次函数解析式,利用二次函数性质求出W最大时x的值即可.【解答】解:(1)根据题意得: =934.4(元);(2)根据题意设y=kx+b,把(30,40)与(40,20)代入得:,解得:k=﹣2,b=100,则y=﹣2x+100;(3)设定价为x元时,工厂获得的利润为W,根据题意得:W=(x﹣20)y=(x﹣20)(﹣2x+100)=﹣2x2+140x﹣2000=﹣2(x﹣35)2+450,∵当x=35时,W最大值为450,则为使工厂获得最大利润,该产品的单价应定为35元.23.已知,正方形ABCD的边长为4,点E是对角线BD延长线上一点,AE=BD.将△ABE绕点A顺时针旋转α度(0°<α<360°)得到△AB′E′,点B、E的对应点分别为B′、E′.(1)如图1,当α=30°时,求证:B′C=DE;(2)连接B′E、DE′,当B′E=DE′时,请用图2求α的值;(3)如图3,点P为AB的中点,点Q为线段B′E′上任意一点,试探究,在此旋转过程中,线段PQ长度的取值范围为≤PQ≤4+2 .【考点】四边形综合题.【分析】(1)先由正方形的性质得到直角三角形AOE,再经过简单计算求出角,判断出△ADE ≌△AB′C即可;(2)先判断出△AEB′≌△AE′D,再根据旋转角和图形,判断出∠BAB′=∠DAB′即可;(3)先判断出点Q的位置,PQ最小时和最大时的位置,进行计算即可.【解答】解:(1)如图1,连接AC,B′C,∵四边形ABCD是正方形,∴AB=AD,AC⊥BD,AC=BD=2OA,∠CAB=ADB=45°,∵AE=BD,∴AC=AE=2OA,在Rt△AOE中,∠AOE=90°,AE=2OA,∴∠E=30°,∴∠DAE=∠ADB﹣∠E=45°﹣30°=15°,由旋转有,AD=AB=AB′∠BAB′=30°,∴∠DAE=15°,在△ADE和△AB′C中,,∴△ADE≌△AB′C,∴DE=B′C,(2)如图2,由旋转得,AB′=AB=AD,AE′=AE,在△AEB′和△AE′D中,,∴△AEB′≌△AE′D,∴∠DAE′=∠EAB′,∴∠EAE′=∠DAB′,由旋转得,∠EAE′=∠BAB′,∴∠BAB′=∠DAB′,∵∠BAB′+∠DAB′=90°,∴α=∠BAB′=45°,(3)如图3,由点到直线的距离,过点P 作PM ⊥BE ,∵AB=4,点P 是AB 中点,∴BP=2,∴PM=,在旋转过程中,△ABE 在旋转到点E 在BA 的延长线时,点Q 和点E 重合,∴AQ=AE=BQ=4,∴PQ=AQ+AP=4+2,故答案为≤PQ ≤4+2.24.已知抛物线y=x 2与直线y=x+1交于A 、B 两点(A 在B 的左侧) (1)求A 、B 两点的坐标.(2)在直线AB 的下方的抛物线上有一点D ,使得ABD 面积最大,求点D 的坐标.(3)把抛物线向右平移2个单位,再向下平移m (m >0)个单位,平移后的抛物线与x 轴交于E 、F 两点,直线AB 与y 轴交于点C .当m 为何值时,过E 、F 、C 三点的圆的面积最小,最小面积是多少?【考点】二次函数综合题.【分析】(1)直线解析式与二次函数解析式组成方程组,求得点A ,B 的坐标;(1)根据自变量与函数值的对应关系,可得D 、E 点坐标,根据面积的和差,可得二次函数,根据二次函数的性质,可得答案;(3)由抛物线平移后为:y=(x ﹣2)2﹣m ,其对称轴是x=2.由于过E 、F 的圆的圆心必在对称轴上,要使圆的面积最小,则圆的半径要最小,即点C 到圆心的距离要最短,过C 作CG 垂直抛物线的对称轴,垂足为G ,则符合条件的圆是以G 为圆心,GC 长为半径的圆,求得圆的面积和m 的值.【解答】解:(1)联立直线与抛物线,得解得:x2+3x﹣4=0,解得x=﹣4或x=1.当x=﹣4时y=4,当x=1时,y=;A点坐标为(﹣4,4),B点坐标为(1,);(2)如图1,作DE⊥x轴于E,设D(m, m2),E(m,﹣ m+1),DE=﹣m2﹣m+1.S△ABD=S△ADE+S BDE=DE•|x B﹣x A|=(﹣m2﹣m+1)×[1﹣(﹣4)]=﹣(m+)2+,当m=﹣时,S最大=,当m=﹣时, m2=×(﹣)2=,ABD面积最大,点D的坐标(﹣,);(3)如图2,抛物线平移后为:y=(x﹣2)2﹣m.其对称轴是x=2.由于过E、F的圆的圆心必在对称轴上,要使圆的面积最小,则圆的半径要最小,即点C到圆心的距离要最短,过C作CG垂直抛物线的对称轴,垂足为G,则符合条件的圆是以E为圆心,EC=2长为半径的圆,其面积为4π,CG=EG=2,EH==,OE=OH﹣HE=2﹣,E点坐标为(2﹣,0)把E点坐标代入抛物线的解析式,得×(2﹣﹣2)2﹣m=0,解得m=0.75,m的值0.75.。
湖北省武汉二中广雅中学2023-2024学年九年级上学期月考数学试题

湖北省武汉二中广雅中学2023-2024学年九年级上学期月考数学试题学校:___________姓名:___________班级:___________考号:___________一、单选题1.一元二次方程2461x x -=化成一般式后,其常数项为1-,则二次项、一次项分别是( )A .4,6-B .24x ,6x -C .4,6D .24x ,6x 2.“守株待兔”这个事件是( )A .随机事件B .确定性事件C .必然事件D .不可能事件 3.下列各曲线是根据不同的函数绘制而成的,其中是中心对称图形的是( ) A . B .C .D .4.用配方法解一元 e 二次方程2680x x --=配方后得到的方程是( ) A .()2628x += B .()2628x -= C .()2317x += D .()2317x -= 5.已知O e 的半径为4,4PO =,则过P 点的直线l 与O e 的位置关系是( ) A .相离 B .相交 C .相切 D .相交或相切 6.某电影上映第一天票房收入约3亿元,以后每天票房收入按相同的增长率增长,三天后累计票房收入达到10亿元.若增长率为x ,则下列方程正确的是( )A .3(1)10x +=B .23(1)10x +=C .233(1)10x ++=D .233(1)3(1)10x x ++++= 7.平面直角坐标系中,抛物线22y x x =+经变换得到抛物线22y x x =-,则这个变换是( )A .向左平移2个单位B .向右平移2个单位C .向左平移4个单位D .向右平移4个单位8.如图,在平面直角坐标系中,矩形ABCO 的两边与坐标轴重合,21OA OC ==,.将矩形ABCO 绕点O 顺时针旋转,每次旋转90︒,则第2024次旋转结束时,点B 的坐标是( )A .()21,B .()12-,C .()21-,D .()12-,9.如图,在正方形ABCD 中,点,E F 分别在,BC CD 上,连接,,AE AF EF ,45EAF ∠=︒.若BAE α∠=,则FEC ∠一定等于( )A .2αB .902α︒-C .45α︒-D .90α︒-10.已知二次函数()20y ax bx a =+≠,经过点()2P m ,.当1y ≤-时,x 的取值范围为13n x n -≤≤--.则下列四个值中有可能为m 的是( )A .2-B .3-C .4-D .5-二、填空题11.在平面直角坐标系中,点()2,3P -关于原点对称的点的坐标是.12.某商品经过连续两次降价,售价由原来的25元/件降到16元/件,则平均每次降价的百分率为.三、解答题17.已知;关于x 的方程210x kx +-=,(1)求证;无论k 为何值时,方程始终有两个不相等的实数根;(2)若2k =,且方程的两个根分别是α与β,求αβαβ+-的值.18.如图,将ABC V 绕A 点逆时针旋转得到AEF △,点E 恰好落在BC 上,若70ABC ∠=︒,28ACB ∠=︒,求FGC ∠的度数.。
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武汉二中广雅中学九年级(上)英语月考(一)一、单项选择26. —Do the dishes, Tim, or I will tell mom! —Mind your own _______ Kate!A. dutyB. businessC. actionD. way27. —How is Jack’s grandma? —She is over eighty, but she is still _______.A. aliveB. patientC. busyD. active28. —Everyone has the right to good medical care, and it has nothing to do with the _______ to pay.—I can’t agree more.A. abilityB. knowledgeC. patternD. activity29. —_______ weather it is today! Let’s go out for a walk.—OK. Please wait me for a moment.A. What a fineB. How a fineC. What fineD. How fine30. —Hurry up. Or we will miss the sea lion show in the Ocean Park.—_______. We still have fifteen minutes before the show begins.A. Take it easyB. Good luckC. I guess soD. Take care31. —Do you know _______Dengchao is?—He is an actor. He is the team leader in the popular reality TV show Running Men now.A. whichB. whenC. whatD. how32. —Hi, my name is Terry. It’s my first time to be here. —_______.A. Very well, thank you!B. It’s all right.C. That would be very nice.D. Nice to meet you!33. —Why do those middle-aged women dance on the square every night?—Then just want to be healthier by _______ their arms and legs as much as possible.A. putting outB. reaching outC. spreading outD. coming out34. —Do you know if the old man is still living?—I am sorry. He _______. He _______ for two months.A. died; has diedB. died; has diedC. has been dead; diedD. died; has been dead35. —I’d like to know how to protect our environment.—The answer must _______ in saving resources(资源).A. existB. liveC. lieD. fall36. —He must _______ his diet because he doesn’t want to put on weight any more.—Yes. It’s good for his health.A. pay attention toB. deal withC. take careD. look forward to37. —What happened on Center Street?—Because of the rain, one of the pipes(水管) broke and the water _______ out.A. rushedB. becameC. laidD. set38. —You know what? Terry lost his wallet yesterday.—Yes, I heard of that and all the evidence(证据) _______ that it must be Andy, his deskmate.A. suggestedB. appearedC. presentedD. discovered39. —Did you have fun swimming? —No, they _______ us against swimming in the river.A. toldB. askedC. warnedD. kept40. —What did your teacher say to yhou just now? —He asked me _______.A. how could I work outB. when did I go to the libraryC. why I am late for schoolD. if I did my homework二、完形填空The family had just moved. The young woman was feeling a little 41. It was Mother’s Day and 800 miles separated her from her parents.She had called her mother that morning to wish her a happy Mother’s Day, and her mother had 42 how colorful their backyard was now that spring had arrived. Later, she told her husband how she 43 those lilacs(丁香花) in her parents’ yard. “I know where we can find some,” he said. “Get the 44 and come on.”So off they went.Some time later, they stopped at a hill and there were lilacs all around. The young woman rushed up to the nearest 45 and buried her face in the flowers. Carefully, she 46 some. Finally, they returned to their car for the 47 home. The woman sat smiling, surrounded by her flowers.When they were near home, she shouted “stop”, got off quickly and 48 to a nearby nursing home. She went to the end of the porch(门廊), where an elderly patient was sitting in her wheelchair, and put the flowers in her lap. The two 49, bursting into laughter now and then. Later the young woman turned and ran back to her 50. As the car pulled away, the woman in the wheelchair waved with a smile, and held the lilacs 51.“Mom,” the kids asked, “ Why did you give her our flowers?”“It is Mother’s Day, and she seems so 52 while I have all of you. And anyone would be 53 by flowers.”This satisfied the kids, but not the husband. The next day he 54 some young lilacs around their yard.I was the husband. Now, every May, our yard is full of lilacs. Every Mother’s Day our kids 55 purple lilacs. And every year I remember that smile of the lonely old woman.41. A. moved B. worried C. angry D. sad42. A. learned B. imagined C. mentioned D. realized43. A. missed B. grew C. watered D. showed44. A. cars B. kids C. clothes D. lilacs45. A. bush B. hill C. yard D. door46. A. bought B. picked C. set D. raised47. A. break B. holiday C. trip D. dinner48. A. responsed B. pointed C. drove D. hurried49. A. argued B. waited C. sat D. chatted50 A. family B. mother C. path D. home51. A. sadly B. politely C. quickly D. tightly52. A. quiet B. confused C. alone D. patient53. A. calmed B. troubled C. disappointed D. cheered54. A. picked B. dried C. planted D. hid55. A. find B. gather C. receive D. sell三、阅读理解AHave you ever worried about the cost of the Christmas shopping? Have you regetted for soemthing that had been sold out? Have you found what you bought at full price last week are now on sale? These suggestions may help you.Hit the salesTo fight for the better deal, you need to get up earlier and go shopping in the morning markets. Goods there are always at lower price and fresher, especially food and vegetables. Some retailers(零售商) offer certain specials all through the day. If you’re lucky to run into the sale and you can buy more to store.Pay close attention to advertisementsOnce the Christmas shopping season is coming, retailers hungry for business will do just about whatever to get you in the store. Pay very close attention to weekly advertisements of sales, and you may find different prices for the same thing in different shops.Shop onlineSome of us don’t like running from store to store in order to get the best price, and some of us don’t like to go out in the cold at all, we can do online shopping. Online shopping lets people visit all the stores as well as some specialty goods that could only be found on the Internet. Customers can find exactly what they are looking for at the best possible price without leaving their house.56. To fight for a better deal, you can _______.A. directly buy from producersB. fight with other customersC. do shopping in the morning marketsD. buy more to store at one time57. With the help of advertisements, you may find _______ in different shops.A. the same thing at the same priceB. the same thing at different pricesC. different things at different pricesD. everything at the same price58. We can get the best price on gifts easily _______.A. from store to storeB. during shopping festivalsC. in the morning marketsD. on the Internet59. Who might have the most interest in this passage?A. HousewivesB. BusinessmenC. Government officialsD. Shop asisstants60. In this passage the writer means to _______.A. tell us the best way to do shoppingB. attract more readers to do wise shoppingC. show us how hard it is to do Christmas shoppingD. give advice on how to get a good deal at Christmas timeBA nine-year-old kid was sitting at his desk when suddenly there was a puddle(some water or liquid) between his feet and the front of his trousers was wet. He thought his heart was going to stop because hecouldn’t possibly imagine how this had happened. It had never happened before, and he knew that when the boys found out he would never hear the end of it. When the girls found out, they would never speak to him again as long as he lived.He prayed this prayer, “Dear God, I need help now! Five minutes from now I’m dead meat!”He looked up from his prayer and here came the teacher with a look in her eyes that said he had been discovered. As the teacher was walking toward him, a classmate named Susie was carrying a goldfish bowl of water. Susie tripped(绊倒) in front of the teacher and dumped(倒) the bowl of water in the boy’s lap. The boy pretended to be angry, but all the while he was saying to himself, “Thank you. God!”Now all of a sudden, instead of being the object of ridicule(嘲笑), the boy was the object of sympathy(同情). The teacher rushed him downstairs and gave him gym shorts to put on while his trousers dried out. All the other children were on their hands and knees cleaning up around his desk. The sympathy was wonderful. But as life would have it, the ridicule that should have been his had been transferred(转移) to someone else—Susie. She tried to help, but they told her to get out.When school was over, the boy walked over to Susie and whispered, “You did that on purpose, didn’t you?” Susie whispered back, “I wet my trousers once, too!”61. The underlined sentence in Paragraph 1 means _______.A. the boys would never play with himB. the boys would treat him as usualC. he would hardly hear any praise from the boysD. he would be laughed at by the boys endlessly62. After Susie dumped water in his lap, the boy was in a state of _______.A. excitementB. reliefC. anxietyD. anger63. What did the other kids do after the incident?A. They offered him dry clothesB. They laughed at the boy rudelyC. They helped the boy do the cleaningD. They urged the boy to get out angrily64. Why did Susie dump water in the boy’s lap?A. The boy asked her to do soB. She just did it by accidentC. The teacher tripped her on purposeD. She knew the boy’s embarrassment65. Which of the follow is true according to the text?A. God helped the boy to solve the problemB. Besides Susie, maybe the teacher also knows what had happened to the boyC. Susie wet the boy’s trousers because she was angry with himD. After the incident, the boy would never talk to Susie.CThe story is about a teacher who spent a whole week teaching her class about aerodynamics(空气动力学). To finish off the unit she organized a paper airplane contest. Each student was given a piece of paper. The plane that would fly the farthest won a prize.The students went to work immediately, carefully folding their paper. Before long, everyone was ready to go outside to start the contest except Jeff. He usually had his own view of life that was not always the same as the others in the class. Jeff didn’t make a paper plane. He just sat there looking out of thewindow-thinking.To give him more time, the teacher told Jeff he could go last. Some of the planes only flew two meters while others did quite well. Jeff was the last one to fly his plane. He stepped to the line and showed his masterpiece(杰作) ——a flat piece of paper. But just as the class began to laugh, Jeff confidently wadded up (将揉成团) the piece of piece into a tight ball, and then drew his hand back and threw it higher and farther than the leading plane that had landed.The rest of the class stood there in amazement. The silence was broken when the teacher began to clap her hands and gave Jeff the first-place prize. The rest of the students then cheered the champion.Jeff showed a new way of solving a problem. More importantly, he had the courage to act in his own way. The next time you come up with an idea that seems “outside the box”, have the courage to act on your ideas. If you fail, you fail. At least then you will know how far you can go or at the very least, what you must do to go farther next time. Remember, Christopher Columbus wanted to discover America; then he had to have the courage to lose sight of the beach.66. At the end of the courage, the teacher _______.A. had a paper cutting contestB. had a paper plane showC. gave each student a prizeD. had a paper airplane contest67. From the reading, we can infer that _______.A. Jeff was a slow studentB. Jeff was a hard-working boyC. Jeff was a little bit quietD. Jeff was a creative student68. Which of the following is TRUE according to the reading?A. The teacher was very patient with the students.B. The teacher liked Jeff best of all the students.C. Jeff finished his paper plane with the help of his teacher.D. Jeff was the bravest boy in his class.69. From the reading, we can know that _______.A. we don’t need to be afraid of failure.B. we can succeed only if we are self-confident.C. we should have the courage to act on our own ideas.D. we should try to do something unusual more often.70. The best title for the reading is _______.A. A Paper Airplane ContestB. An Interesting ClassC. A Clever BoyD. A New Way to Solve Problems四、选择正确的词或词组,用适当的形式完成句子(12选10,一个词或词组只能用一次)71. Try to guess a word’s meaning by _______ the sentences before and after it.72. I was afraid to ask questions because of my poor _______.73. My father hates waiting in long lines, so I think he is just _______.74. Do parents take their children to ask for candies and _______.75. After this, people started the tradition of _______ the moon and sharing mooncakes with their familes.76. I heard that it’s becoming more and more popular _______ Mother’s Day and Father’s Day in China.77. Not only do people spread them around in different _______ places for an egg hunt, but they also give out these treats as gifts.78. The number of phone users is on the _______.79. He is a complete _______ to me, I don’t know him at all.80. It’s about an old man _______ Scrooge who never laughs or smiles.五、阅读填词AHave you ever been to a special town in Thailand? In this town, monkeys are e 81 in the town. They run around the streets. They climb on cars, houses, and other buildings. They can go anywhere they want, and they get into everything!The monkeys are very noisy in this town and they often make a lot of t 82 . But how do the people of this town feel about these little animals? They know that sometimes the monkeys can make life d 83 . People also feel terrible when the monkeys sit on their cars, run everywhere and take food. But people there don’t worry about the trouble c 84 by the monkeys. They know that it’s not just the monkeys. The people of the town also cause the problem. So they a 85 the monkeys to run around the streets. They don’t try to stop the monkeys. The people of this town do even m 86 for the monkeys. They have a monkey party once a year!But why do they have this big monkey party? In fact, monkeys are very important for Thai people. Monkeys a 87 in Thai stories. Because of one special story, some Thai people c 88 monkeys to be heroes. In the story, a demon(恶魔) takes the god’s wife away. The monkey Hanuman helps to s 89 the god’s wife. As a result, Hanuman becomes a hero and many people in Thailand are e 90 kind to monkeys today.81.e________ 82. t________ 83. d________ 84. c________ 85. a________86.m________ 87. a________ 88. c________ 89. s________ 90. e________BI worked as a music teacher for twenty-five years. I have always known that music touches the soul.I would like to share how music broke through a physical barrier(障碍) and made a c 91 with a young student. For a few years I was happy to have the c 92 to teach pre-school handicapped(残疾的) students one afternoon a week. One of my most memorable students was a young girl I will call Jessica. Jessica was six years old, had difficultly w 93 , and could not speak. We mostly sat on the floor for our music lessons and Jessica liked to sit on my lap. One of her favorite s 94 was “John the Rabbit”. It was a call and response song where I sang the call and the students clapped two times while singing the repeating phrase, “Oh, yes!”Jessica liked to put her hands t 95 with mine and clap with me. We probably performed that song during every class, Jessica and I clapping together. She never said or sang a word.One day later in the school year, when the song was f 96 Jessica turned around, looked me dead in the eye, clapped her tiny hands two times and said the words “Oh, yes!” I opened my mouth in surpriseand for that moment I was the one w 97 could not speak. When my heart finally started beating again, I looked over at the homeroom teacher to find her also speechless. Through music, we had made an amazing connection.Several years later, I p 98 Jessica on the street in town. I stopped my car and waved to say hello. She waved back with a big smile and then clapped her hands two times, coping the son we had performed so many times on our music class. This precious little girl, through her connection with music, left an impression on me that will last forever. Every child has the a 99 to learn and grow. It is up to us as educators to d 100 the way to reach each and every one of our students. We all must find each child’s light.91.c________ 92. c________ 93. w________ 94. s________ 95. t________96.f________ 97. w________ 98. p________ 99. a________ 100. d________六、书面表达中秋节即将来临,但是刚刚上初三的Mary并不开心,你作为她的好朋友很担心她,所以你写了一封信安慰她。