2020年九年级入学考试试卷

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2020年九年级入学考试物理试题(含答案)

2020年九年级入学考试物理试题(含答案)

2020年春季九年级入学考试理科综合试题(物理:60分时间:120分钟)物理试题一、选择题(每小题只有一个选项符合题意,每题3分,共21分)1.交警在查“酒驾”时,先请司机打开车窗,若闻到酒精气味,则要求司机必须接受酒精含量的测试。

交警能闻到酒精气味是因为酒精()A.分子间有空隙B.分子很小C.分子在不停的运动D.分子可分2.下列现象中,属于用热传递的方法改变物体内能的是()A.刚从蒸笼里拿出的馒头,放一阵子变凉了B.冬天天冷,通过搓手发热取暖C.用锤子敲打石头时,锤子发热D.给自行车车胎打气,打气筒壁变热了3. A、B、C三个轻质小球,已知A带负电,A和B互相吸引,C和A互相排斥,则()A.B一定带正电,C带负电B.B可能不带电,C带负电C.B可能带正电,C带正电D.B一定不带电,C带正电4.如图所示的电蚊拍,具有灭蚊和照明等功能。

当开关S l闭合、S2断开时,只有灭蚊网通电起到灭蚊作用;当开关S l和S2都闭合时,灭蚊网与灯都通电同时起到灭蚊和照明作用。

下列电路设计符合这种要求的是()A B C D5.关于家庭电路和安全用电,下列说法中正确的是()A.家中保险丝烧断了,一定是发生了短路B.用测电笔辨别火线与零线时,手不能接触测电笔上的任何金属C.只有高压线路才会发生触电D.有人触电时,用干燥的竹竿或木棍将电线拨开6.对图中甲、乙、丙、丁四幅图解释合理的是()A.甲:磁场能产生电流B.乙:闭合开关,小磁针N极向左偏转C.丙:麦克风应用了磁场对电流的作用D.丁:电流相同时,线圈的匝数越多电磁铁磁性越强九年级物理模考试题第 1 页共 6 页7.为避免汽车超速行驶带来的危害,某校科技创新小组设计了如图所示的一个装置来自动限制车速。

电路中电源电压为6V,保护电阻R0的阻值是5Ω,R x是一个阻值随车速增大而减小电阻元件(装置中其它的电阻忽略不计)。

当车速达到130km/h时,电路中电流恰好为1A,电磁铁刚好能够吸下衔铁B,使油门C喷油量受到限制而控制车速。

2020 年万州二中九年级(下)历史入学考试试题

2020 年万州二中九年级(下)历史入学考试试题

克思用他强有力的思想,哺育了世界西方和东方两个半球的无产阶级运动。”下列说法中,不能从材料直
接得出是( )
A.马克思主义的诞生具有历史进步意义
B.马克思主义产生了世界性影响
C.马克思主义批判继承了前人思想的精华
D.马克思主义是无产阶级运动的指导思想
二、判断题:每项 1 分,本大题共 5 分。
到 12 世纪,劳作者阶层不仅包括农民,也包括数量日益增长的商人、工匠、手工业者和医生、律师
23. 材料一 意大利文艺复兴时代最根本的因素并非古典文化的复兴,而是“近代精神”(即“个人主义”)
的觉醒和发展。此种“觉醒和发展”是中世纪与文艺复兴时代的根本分野,是欧洲的“近代性”的开端。
欧洲人在精神上获得觉醒并高度发展之后向外和向内的扩展:向外的发展即所谓“世界的发现”;向内的
发展即“人的发现”,也就是个性的发展。
——高德步等主编《世界经济史》
材料三 工业革命期间工人的集中式贫穷,恩格斯用“非人类”来形容,从首都伦敦到伯明翰、诺丁汉、
格拉斯哥等工业城市,工人无家可归、露宿街头的情况是常见的。与经济上的拮据相配的,是政治上的不
公待遇。
——张莉、郭雨祺《从《北极星报》到《新莱茵报》:早期工人阶级报刊新探》
(1)根据材料一,概括工业革命对英国城市化的影响。(2 分)
沉重打击了印度的城市手工业,特别是棉纺织业。英国的工业革命不仅摧毁了印度的城市手工业,还打击
了农村手工业,破坏了古老的农村公社制度以及社会经济结构,从而使印度自给自足的村庄变为生产鸦片、
棉花、黄麻、蓝靛和其他原料用以换取英国货的农场。这样,由英国生产工业品、由印度生产原料的国际
分工,在摧毁印度旧世界的基础上形成了。这种演变过程在其他落后国家也相继发生,如中国。

2020年秋入学检测九年级语文试卷及答案

2020年秋入学检测九年级语文试卷及答案

2020年秋入学检测九年级语文试卷(考试时间:120分钟,满分:120分)注意事项:1.全卷共8页,满分为120分。

考试用时为120分钟。

2.答卷前,考生务必用黑色字迹的签字笔或钢笔在答题卡填写自己的准考证号、姓名、考场号、座位号,用2B铅笔把对应该号码的标号涂黑。

3.请你将选择题的答案用2B铅笔填涂在答题卡上,非选择题的答案请用黑色的签字笔或钢笔填写在答题卡相应的横线上;如需改动,先划掉原来的答案,然后再写上新的答案;不准使用铅笔和涂改液。

不按以上要求作答的答案无效。

一、基础(24分)1.根据课文默写古诗文。

(10分)(1)□□□□,君子好逑。

(《关雎》)(1分)(2)蒹葭萋萋,□□□□。

所谓伊人,□□□□。

(《蒹葭》)(2分)(3)气蒸云梦泽,□□□□□。

(《送杜少府之任蜀州望洞庭湖赠张丞相》)(1分)(4)杜甫在《茅屋为秋风所破歌》中写自己身居茅屋,但是心怀天下,为天下寒士振臂高呼的句子是“□□□□□□□,□□□□□□□□□! (白居易《茅屋为秋风所破歌》) (2分)(5)请把李白的《送友人》默写完整。

(4分)青山横北郭,白水绕东城。

此地一为别,孤蓬万里征。

□□□□□,□□□□□。

□□□□□,□□□□□。

2.根据拼音写出相应的词语。

(4分)(1)十几个别的少年也大悟,立刻cuān duo( )起来,说可以坐了这航船和我一同去。

(2)明晰了又huìàn( ),尔后最终永远明晰了的大彻大悟!(3)但至柔至和的水一旦被压迫竟会这样nù bù kěè( )。

(4)除了扑面而来的水汽,zhèn ěr yù lóng( )的涛声,什么也看不见,什么也听不见。

3.下列情景中语言表达得体的一项是( )(3分)A.喇叭里传出了电工收电费的通知:各位村民注意了,明天交电费,晚六点以前要是交不清,就别怪我不客气了。

B.“这算得了什么,以后要是再有困难,可以继续到府上来找我。

九年级上册数学2020届九年级(下)入学考试数学试卷(解析版)

九年级上册数学2020届九年级(下)入学考试数学试卷(解析版)

2020-2020年九年级(下)入学考试数学试卷一、选择题:(本大题12个小题,每小题4分,共48分)每个小题都给出了代号为A、B、C、D的四个答案,其中只有一个是正确的,请将正确答案的代号填入答题卡上对应位置中.1.已知△ABC中,AC=4,BC=3,AB=5,则sinA=()A.B.C.D.2.用配方法解方程x2+4x﹣5=0,下列配方正确的是()A.(x+2)2=1 B.(x+2)2=5 C.(x+2)2=9 D.(x+4)2=93.下列式子,正确的是()A.3+=3B.(+1)(﹣1)=1C.2﹣1=﹣2 D.x2+2xy﹣y2=(x﹣y)24.在▱ABCD中,若∠A:∠B=1:2,则∠A的度数是()A.60°B.90°C.120°D.150°5.已知一个等腰三角形的两条边长分别为3和8,则这个等腰三角形的周长为()A.11 B.14 C.19 D.14或196.二次函数y=﹣2(x﹣4)2﹣5的开口方向、对称轴分别是()A.开口向上、直线x=﹣4 B.开口向上、直线x=4C.开口向下、直线x=﹣4 D.开口向下、直线x=47.如图,在⊙O中,∠AOB=50°,则∠ACB=()A.30°B.25°C.50°D.40°8.如图,在△ABC中,AB=BC,∠B=30°,DE垂直平分BC,则∠ACD的度数为()A.30°B.45°C.55°D.75°9.某校九年级(1)班有7个合作学习小组,各学习小组的人数分别为:5,6,6,x,7,8,9,已知这组数据的平均数是7,则这组数据的中位数是()A.6 B.7 C.8 D.910.下列图形都是由几个黑色和白色的正方形按一定规律组成,图①中有2个黑色正方形,图②中有5个黑色正方形,图③中有8个黑色正方形,图④中有11个黑色正方形,…,依次规律,图⑩中黑色正方形的个数是()A.32 B.29 C.28 D.2611.如图,在平面直角坐标系中,将矩形OABC沿对角线OB对折,使点A(,0)落在点A1处,已知点B的坐标是(,1),则点A1的坐标是()A.(,)B.(,)C.(,2)D.(,)12.如图,在平面直角坐标系系中,直线y=k1x+2与x轴交于点A,与y轴交于点C,与反比例函数y=在第一象限内的图象交于点B,连接B0.若S△OBC=1,tan∠BOC=,则k2的值是()A.﹣3 B.1 C.2 D.3二.填空(本大题6个小题,每小题4分共24分)13.方程(x﹣2)2=4的根是.14.计算:2cos60°﹣tan45°=.15.已知一个菱形的两条对角线长分别为6cm和8cm,则这个菱形的面积为cm2.16.在某时刻的阳光照耀下,身高160cm的阿美的影长为80cm,她身旁的旗杆影长5m,则旗杆高为m.17.从﹣1,0,1,2,3这五个数中,随机抽取一个数记为m,则使关于x的不等式组有解,并且使函数y=(m﹣1)x2+2mx+m+2与x轴有交点的概率为.18.在▱ABCD中,AB<BC,已知∠B=30°,AB=2,将△ABC沿AC翻折至△AB′C,使点B′落在▱ABCD所在的平面内,连接B′D.若△AB′D是直角三角形,则BC的长为.三.解答题(本大题2小题,每小题7分,共14分)解答时每小题必须给出必要的演算过程或推理步骤.19.解二元一次方程组.20.为有效开展阳光体育活动,云洱中学利用课外活动时间进行班级篮球比赛,每场比赛都要决出胜负,每队胜一场得2分,负一场得1分.已知九年级一班在8场比赛中得到13分,问九年级一班胜、负场数分别是多少?四、解答题:(本大题4个小题,每小题10分,共40分,解答题时每小题必须给出必要的演算过程或推理步骤,画出必要的图形(包括作辅助线).)21.先化简,再求值:(﹣)÷,其中x=tan60°+2.22.2020年1月,市教育局在全市中小学中选取了63所学校从学生的思想品德、学业水平、学业负担、身心发展和兴趣特长五个维度进行了综合评价.评价小组在选取的某中学七年级全体学生中随机抽取了若干名学生进行问卷调查,了解他们每天在课外用于学习的时间,并绘制成如下不完整的统计图.根据上述信息,解答下列问题:(1)本次抽取的学生人数是;扇形统计图中的圆心角α等于;补全统计直方图;(2)被抽取的学生还要进行一次50米跑测试,每5人一组进行.在随机分组时,小红、小花两名女生被分到同一个小组,请用列表法或画树状图求出她俩在抽道次时抽在相邻两道的概率.23.“村村通公路”工程是国家为支持新农村建设的一项重大举措,为了落实这一举措,重庆潼南县政府计划在南北方向的A、B两村之间建一条公路AB.已知公路AB的一侧有C村,在公路AB上的M处测得C村在M的南偏东37°方向上,从M向南走270米到达N处,测得C村在N的东南方向上,且C村周围800米范围内为油菜花田,那么计划修建的公路AB 是否会穿过油菜花田,请说明理由(参考数据:sin37°≈0.8,cos37°≈0.8,tan37°≈0.75)24.长宽比为(n为正整数)的矩形称为矩形.下面,我们通过折叠的方式折出一个矩形,如图①所示.操作1:将正方形ABCD沿过点B的直线折叠,使折叠后的点C落在对角线BD上的点G 处,折痕为BH.操作2:将AD沿过点G的直线折叠,使点A,点D分别落在边AB,CD上,折痕为EF.则四边形BCEF为矩形.证明:设正方形ABCD的边长为1,则BD=.由折叠性质可知BG=BC=1,∠AFE=∠BFE=90°,则四边形BCEF为矩形.∴∠A=∠BFE.∴EF∥AD.∴,即,∴.∴.∴四边形BCEF为矩形.阅读以上内容,回答下列问题:(1)在图①中,所有与CH相等的线段是,tan∠HBC的值是;(2)已知四边形BCEF为矩形,模仿上述操作,得到四边形BCMN,如图②,求证:四边形BCMN为矩形;(3)将图②中的矩形BCMN沿用(2)中的方式操作3次后,得到一个“矩形”,则n的值是.五、解答题(本大题2个小题,每小题12分,共24分)解答时每小题都必须写出必要的演算过程或推理步骤,请将解答过程书写在答卷中对应的位置上.25.已知:四边形ABCD中,AD∥BC,AD=AB=CD,∠BAD=120°,点E是射线CD上的一个动点(与C、D不重合),将△ADE绕点A顺时针旋转120°后,得到△ABE′,连接EE′.(1)如图1,∠AEE′=°;(2)如图2,如果将直线AE绕点A顺时针旋转30°后交直线BC于点F,过点E作EM∥AD 交直线AF于点M,写出线段DE、BF、ME之间的数量关系;(3)如图3,在(2)的条件下,如果CE=2,AE=,求ME的长.26.如图,在平面直角坐标系中,矩形OABC的边OA在y轴的正半轴上,OC在x轴的正半轴上,∠AOC的平分线交AB于点D,E为BC的中点,已知A(0,4)、C(5,0),二次函数y=x2+bx+c的图象抛物线经过A,C两点.(1)求该二次函数的表达式;(2)F、G分别为x轴,y轴上的动点,顺次连接D、E、F、G构成四边形DEFG,求四边形DEFG周长的最小值;(3)抛物线上是否在点P,使△ODP的面积为12?若存在,求出点P的坐标;若不存在,请说明理由.2020-2020年九年级(下)入学考试数学试卷参考答案与试题解析一、选择题:(本大题12个小题,每小题4分,共48分)每个小题都给出了代号为A、B、C、D的四个答案,其中只有一个是正确的,请将正确答案的代号填入答题卡上对应位置中.1.已知△ABC中,AC=4,BC=3,AB=5,则sinA=()A.B.C.D.【考点】锐角三角函数的定义;勾股定理的逆定理.【分析】先根据直角三角形的三边长判断出三角形的形状,再根据锐角三角函数的定义求解即可.【解答】解:∵△ABC中,AC=4,BC=3,AB=5,即42+32=52,∴△ABC是直角三角形,∠C=90°.sinA==.故选A.2.用配方法解方程x2+4x﹣5=0,下列配方正确的是()A.(x+2)2=1 B.(x+2)2=5 C.(x+2)2=9 D.(x+4)2=9【考点】解一元二次方程-配方法.【分析】先将原方程进行配方,然后选项进行对照,即可得到正确选项.【解答】解:x2+4x﹣5=0,配方,得(x+2)2=9.故选C.3.下列式子,正确的是()A.3+=3B.(+1)(﹣1)=1C.2﹣1=﹣2 D.x2+2xy﹣y2=(x﹣y)2【考点】二次根式的乘除法;负整数指数幂.【分析】根据二次根式的加减、负整数指数幂和完全平方公式判断.【解答】解:A、不是同类二次根式,不能相加,故错误;B、正确;C、原式=,故错误;D、与完全平方公式不符,故错误.故选B.4.在▱ABCD中,若∠A:∠B=1:2,则∠A的度数是()A.60°B.90°C.120°D.150°【考点】平行四边形的性质.【分析】根据平行四边形的基本性质可知,平行四边形的邻角互补,由已知可得,∠A、∠B 是邻角,故∠A可求解.【解答】解:∵▱ABCD,∴∠A+∠B=180°,而∠A:∠B=1:2∴∠A=60°,∠B=120°∴∠A=60°.故选A.5.已知一个等腰三角形的两条边长分别为3和8,则这个等腰三角形的周长为()A.11 B.14 C.19 D.14或19【考点】等腰三角形的性质;三角形三边关系.【分析】分3是腰长与底边长两种情况讨论求解即可.【解答】解:①3是腰长时,三角形的三边分别为3、3、8,∵3+3=6<8,∴此时不能组成三角形;②3是底边长时,三角形的三边分别为3、8、8,此时能组成三角形,所以,周长=3+8+8=19,综上所述,这个等腰三角形的周长是19.故选C.6.二次函数y=﹣2(x﹣4)2﹣5的开口方向、对称轴分别是()A.开口向上、直线x=﹣4 B.开口向上、直线x=4C.开口向下、直线x=﹣4 D.开口向下、直线x=4【考点】二次函数的性质.【分析】已知抛物线解析式为顶点式,可根据顶点式求抛物线的开口方向,对称轴.【解答】解:由y=﹣2(x﹣4)2﹣5可知,二次项系数为﹣2<0,∴抛物线开口向下,对称轴为直线x=4,故选D.7.如图,在⊙O中,∠AOB=50°,则∠ACB=()A.30°B.25°C.50°D.40°【考点】圆周角定理.【分析】直接根据圆周角定理求解即可.【解答】解:∠ACB=∠AOB=×50°=25°.故选:B.8.如图,在△ABC中,AB=BC,∠B=30°,DE垂直平分BC,则∠ACD的度数为()A.30°B.45°C.55°D.75°【考点】线段垂直平分线的性质;等腰三角形的性质.【分析】根据等腰三角形的性质得到∠A=∠ACB=75°,根据线段垂直平分线的性质得到BD=CD,求得∠DCE=∠B=30°,即可得到结论.【解答】解:∵AB=BC,∠B=30°,∴∠A=∠ACB=75°,∵DE垂直平分BC,∴BD=CD,∴∠DCE=∠B=30°,∴∠ACD=∠ACB=∠DCB=45°,故选B.9.某校九年级(1)班有7个合作学习小组,各学习小组的人数分别为:5,6,6,x,7,8,9,已知这组数据的平均数是7,则这组数据的中位数是()A.6 B.7 C.8 D.9【考点】中位数;算术平均数.【分析】根据题意首先求出x的值,再利用中位数的定义求出答案.【解答】解:∵5,6,6,x,7,8,9,这组数据的平均数是7,∴5+6+6+x+7+8+9=7×7,解得:x=8,故这组数据按从小到大排列:5,6,6,7,8,8,9,则这组数据的中位数是:7.故选:B.10.下列图形都是由几个黑色和白色的正方形按一定规律组成,图①中有2个黑色正方形,图②中有5个黑色正方形,图③中有8个黑色正方形,图④中有11个黑色正方形,…,依次规律,图⑩中黑色正方形的个数是()A.32 B.29 C.28 D.26【考点】规律型:图形的变化类.【分析】仔细观察图形,找到图形的个数与黑色正方形的个数的通项公式后代入n=11后即可求解.【解答】解:观察图形发现:图①中有2个黑色正方形,图②中有2+3×(2﹣1)=5个黑色正方形,图③中有2+3(3﹣1)=8个黑色正方形,图④中有2+3(4﹣1)=11个黑色正方形,…,图n中有2+3(n﹣1)=3n﹣1个黑色的正方形,当n=10时,2+3×(10﹣1)=29,故选B.11.如图,在平面直角坐标系中,将矩形OABC沿对角线OB对折,使点A(,0)落在点A1处,已知点B的坐标是(,1),则点A1的坐标是()A.(,)B.(,)C.(,2)D.(,)【考点】翻折变换(折叠问题);坐标与图形性质.【分析】由已知可得∠AOB=30°,翻折后找到相等的角及相等的边,在直角三角形中,利用勾股定理可求得答案.【解答】解:过A1作A1D⊥OA,∵A(,0),B的坐标是(,1),∴OA=,AB=1,在Rt△OAB中,OB==2,AB=1,∴AB=OB,∵△AOB是直角三角形,∴∠AOB=30°,OB为折痕,∴∠A1OB=∠AOB=30°,OA1=OA=,Rt△OA1D中,∠OA1D=30°,∴OD=×=,A1D=×=,∴点A1的坐标(,).故选B.12.如图,在平面直角坐标系系中,直线y=k1x+2与x轴交于点A,与y轴交于点C,与反比例函数y=在第一象限内的图象交于点B,连接B0.若S△OBC=1,tan∠BOC=,则k2的值是()A.﹣3 B.1 C.2 D.3【考点】反比例函数与一次函数的交点问题.【分析】首先根据直线求得点C的坐标,然后根据△BOC的面积求得BD的长,然后利用正切函数的定义求得OD的长,从而求得点B的坐标,求得结论.【解答】解:∵直线y=k1x+2与x轴交于点A,与y轴交于点C,∴点C的坐标为(0,2),∴OC=2,∵S△OBC=1,∴BD=1,∵tan∠BOC=,∴=,∴OD=3,∴点B的坐标为(1,3),∵反比例函数y=在第一象限内的图象交于点B,∴k2=1×3=3.故选D.二.填空(本大题6个小题,每小题4分共24分)13.方程(x﹣2)2=4的根是4,0.【考点】解一元二次方程-直接开平方法.【分析】根据方程的特点,用直接开平方法解一元二次方程即可.【解答】解:(x﹣2)2=4,x﹣2=±2,解得:x1=4,x2=0.故答案为:4,0.14.计算:2cos60°﹣tan45°=0.【考点】特殊角的三角函数值.【分析】将特殊角的三角函数值直接代入即可求解.【解答】解:2cos60°﹣tan45°=2×﹣1=0.15.已知一个菱形的两条对角线长分别为6cm和8cm,则这个菱形的面积为24cm2.【考点】菱形的性质.【分析】根据菱形的面积等于两对角线乘积的一半求得其面积即可.【解答】解:∵一个菱形的两条对角线长分别为6cm和8cm,∴这个菱形的面积=×6×8=24(cm2).故答案为:24.16.在某时刻的阳光照耀下,身高160cm的阿美的影长为80cm,她身旁的旗杆影长5m,则旗杆高为10m.【考点】相似三角形的应用.【分析】在同一时刻物高和影长成正比,即在同一时刻的两个物体,影子,经过物体顶部的太阳光线三者构成的两个直角三角形相似.【解答】解:根据相同时刻的物高与影长成比例,设旗杆的高度为x m,则160:80=x:5,解得x=10.故答案是:10.17.从﹣1,0,1,2,3这五个数中,随机抽取一个数记为m,则使关于x的不等式组有解,并且使函数y=(m﹣1)x2+2mx+m+2与x轴有交点的概率为.【考点】概率公式;解一元一次不等式组;抛物线与x轴的交点.【分析】首先解不等式以及利用二次函数与x轴交点个数和△的关系分别得出m的取值范围,进而利用概率公式求出即可.【解答】解:∵x+1≤m,解得;x≤m﹣1,2﹣x≤2m,解得:x≥2﹣2m,∴使关于x的不等式组有解,则m﹣1≥2﹣2m,解得:m≥1,∵使函数y=(m﹣1)x2+2mx+m+2与x轴有交点,∴b2﹣4ac4m2﹣4(m﹣1)(m+2)=﹣4m+8≥0,解得:m≤2,∴m的取值范围是:1≤m≤2,∴从﹣1,0,1,2,3这五个数中,随机抽取一个数记为m,符合题意的有1,2,故使关于x的不等式组有解,并且使函数y=(m﹣1)x2+2mx+m+2与x轴有交点的概率为.故答案为:.18.在▱ABCD中,AB<BC,已知∠B=30°,AB=2,将△ABC沿AC翻折至△AB′C,使点B′落在▱ABCD所在的平面内,连接B′D.若△AB′D是直角三角形,则BC的长为4或6.【考点】翻折变换(折叠问题);平行四边形的性质.【分析】在▱ABCD中,AB<BC,要使△AB′D是直角三角形,有两种情况:∠B′AD=90°或∠AB′D=90°,画出图形,分类讨论即可.【解答】解:当∠B′AD=90°AB<BC时,如图1,∵AD=BC,BC=B′C,∴AD=B′C,∵AD∥BC,∠B′AD=90°,∴∠B′GC=90°,∵∠B=30°,AB=2,∴∠AB′C=30°,∴GC=B′C=BC,∴G是BC的中点,在Rt△ABG中,BG=AB=×2=3,∴BC=6;当∠AB′D=90°时,如图2,∵AD=BC,BC=B′C,∴AD=B′C,∵由折叠的性质:∠BAC=90°,∴AC∥B′D,∴四边形ACDB′是等腰梯形,∵∠AB′D=90°,∴四边形ACDB′是矩形,∴∠BAC=90°,∵∠B=30°,AB=2,∴BC=AB÷=2×=4,∴当BC的长为4或6时,△AB′D是直角三角形.故答案为:4或6.三.解答题(本大题2小题,每小题7分,共14分)解答时每小题必须给出必要的演算过程或推理步骤.19.解二元一次方程组.【考点】解二元一次方程组.【分析】方程组利用加减消元法求出解即可.【解答】解:②﹣①得:5y=5,即y=1,把y=1代入①得:x=3,则方程组的解为.20.为有效开展阳光体育活动,云洱中学利用课外活动时间进行班级篮球比赛,每场比赛都要决出胜负,每队胜一场得2分,负一场得1分.已知九年级一班在8场比赛中得到13分,问九年级一班胜、负场数分别是多少?【考点】一元一次方程的应用.【分析】设胜了x场,那么负了(8﹣x)场,根据得分为13分可列方程求解.【解答】解:设胜了x场,那么负了(8﹣x)场,根据题意得:2x+1•(8﹣x)=13,x=5,8﹣5=3.答:九年级一班胜、负场数分别是5和3.四、解答题:(本大题4个小题,每小题10分,共40分,解答题时每小题必须给出必要的演算过程或推理步骤,画出必要的图形(包括作辅助线).)21.先化简,再求值:(﹣)÷,其中x=tan60°+2.【考点】分式的化简求值;特殊角的三角函数值.【分析】原式括号中两项通分并利用同分母分式的减法法则计算,同时利用除法法则变形,约分得到最简结果,把x的值代入计算即可求出值.【解答】解:原式=[﹣]•=•=•=,当x=tan60°+2=+2时,原式=.22.2020年1月,市教育局在全市中小学中选取了63所学校从学生的思想品德、学业水平、学业负担、身心发展和兴趣特长五个维度进行了综合评价.评价小组在选取的某中学七年级全体学生中随机抽取了若干名学生进行问卷调查,了解他们每天在课外用于学习的时间,并绘制成如下不完整的统计图.根据上述信息,解答下列问题:(1)本次抽取的学生人数是30;扇形统计图中的圆心角α等于144°;补全统计直方图;(2)被抽取的学生还要进行一次50米跑测试,每5人一组进行.在随机分组时,小红、小花两名女生被分到同一个小组,请用列表法或画树状图求出她俩在抽道次时抽在相邻两道的概率.【考点】列表法与树状图法;扇形统计图;利用频率估计概率.【分析】(1)根据题意列式求值,根据相应数据画图即可;(2)根据题意列表,然后根据表中数据求出概率即可.【解答】解:(1)6÷20%=30,(30﹣3﹣7﹣6﹣2)÷30×360=12÷30×26=144°,答:本次抽取的学生人数是30人;扇形统计图中的圆心角α等于144°;故答案为:30,144°;补全统计图如图所示:(2)根据题意列表如下:设竖列为小红抽取的跑道,横排为小花抽取的跑道,小红小花 1 2 3 4 51 (2,1)(3,1)(4,1)(5,1)2 (1,2)(3,2)(4,2)(5,2)3 (1,3)(2,3)(4,3)(5,3)4 (1,4)(2,4)(3,4)(5,4)5 (1,5)(2,5)(3,5)(4,5)记小红和小花抽在相邻两道这个事件为A,∴.23.“村村通公路”工程是国家为支持新农村建设的一项重大举措,为了落实这一举措,重庆潼南县政府计划在南北方向的A、B两村之间建一条公路AB.已知公路AB的一侧有C村,在公路AB上的M处测得C村在M的南偏东37°方向上,从M向南走270米到达N处,测得C村在N的东南方向上,且C村周围800米范围内为油菜花田,那么计划修建的公路AB 是否会穿过油菜花田,请说明理由(参考数据:sin37°≈0.8,cos37°≈0.8,tan37°≈0.75)【考点】解直角三角形的应用-方向角问题.【分析】本题要求的实际上是C到AB的距离,过C点作CD⊥AB,CD就是所求的线段,由于CD是条公共直角边,可用CD表示出MD,ND,然后根据MN的长,来求出CD的长.【解答】解:如图,过C点作CD⊥AB于D,由题可知:∠CND=45°,∠CMD=37°.设CD=x千米,tan∠CMD=,则MD=.tan∠CND=,则ND==x,∵MN=270米,∴MD﹣ND=MN,即tan37°x﹣x=270,∴﹣x=270,解得x=810.∵810米>800米,∴计划修建的公路AB是不会穿过油菜花田.答:计划修建的公路AB是不会穿过油菜花田.24.长宽比为(n为正整数)的矩形称为矩形.下面,我们通过折叠的方式折出一个矩形,如图①所示.操作1:将正方形ABCD沿过点B的直线折叠,使折叠后的点C落在对角线BD上的点G 处,折痕为BH.操作2:将AD沿过点G的直线折叠,使点A,点D分别落在边AB,CD上,折痕为EF.则四边形BCEF为矩形.证明:设正方形ABCD的边长为1,则BD=.由折叠性质可知BG=BC=1,∠AFE=∠BFE=90°,则四边形BCEF为矩形.∴∠A=∠BFE.∴EF∥AD.∴,即,∴.∴.∴四边形BCEF为矩形.阅读以上内容,回答下列问题:(1)在图①中,所有与CH相等的线段是GH、DG,tan∠HBC的值是﹣1;(2)已知四边形BCEF为矩形,模仿上述操作,得到四边形BCMN,如图②,求证:四边形BCMN为矩形;(3)将图②中的矩形BCMN沿用(2)中的方式操作3次后,得到一个“矩形”,则n的值是6.【考点】几何变换综合题.【分析】(1)设CH=GH=DG=x,根据DC=DH+CH=1,列出方程即可求出HC,然后运用三角函数的定义求出tan∠HBC的值.(2)只需借鉴阅读中证明“四边形BCEF为矩形”的方法就可解决问题.(3)利用(2)中结论,寻找规律可得到n的值.【解答】解:(1)如图①中,由折叠可得:DG=HG,GH=CH,∴DG=GH=CH.设HC=x,则DG=GH=x.∵∠DGH=90°,∴DH=x,∴DC=DH+CH=x+x=1,解得x=﹣1.∴tan∠HBC===﹣1.故答案为:GH、DG,;(2)如图②中,∵BC=1,EC=BF=,∴BE==由折叠可得BP=BC=1,∠FNM=∠BNM=90°,∠EMN=∠CMN=90°.∵四边形BCEF是矩形,∴∠F=∠FEC=∠C=∠FBC=90°,∴四边形BCMN是矩形,∠BNM=∠F=90°,∴MN∥EF,∴=,即BP•BF=BE•BN,∴1×=BN,∴BN=,∴BC:BN=1:=:1,∴四边形BCMN是的矩形;(3)同理可得:将矩形沿用(2)中的方式操作1次后,得到一个“矩形”,将矩形沿用(2)中的方式操作1次后,得到一个“矩形”,将矩形沿用(2)中的方式操作1次后,得到一个“矩形”,所以将图②中的矩形BCMN沿用(2)中的方式操作3次后,得到一个“矩形”.故答案为6.五、解答题(本大题2个小题,每小题12分,共24分)解答时每小题都必须写出必要的演算过程或推理步骤,请将解答过程书写在答卷中对应的位置上.25.已知:四边形ABCD中,AD∥BC,AD=AB=CD,∠BAD=120°,点E是射线CD上的一个动点(与C、D不重合),将△ADE绕点A顺时针旋转120°后,得到△ABE′,连接EE′.(1)如图1,∠AEE′=30°;(2)如图2,如果将直线AE绕点A顺时针旋转30°后交直线BC于点F,过点E作EM∥AD 交直线AF于点M,写出线段DE、BF、ME之间的数量关系;(3)如图3,在(2)的条件下,如果CE=2,AE=,求ME的长.【考点】几何变换综合题.【分析】(1)根据旋转性质以及三角形内角和定理即可解决.(2)根据EM∥FE′可以得==,再根据AN=NE,BE′=DE即可得到线段DE、BF、ME之间的关系.(3)通过辅助线求出线段E′F=7,E′Q=9,再由(2)的结论得到ME的长.【解答】解:(1)∵△ABE′是由△ADE绕点A顺时针旋转120°得到,∴∠EAE′=120°,AE=AE′,∴∠E′=∠AEE′==30°,故答案为30°.(2)①当点E在CD上时,DE+BF=2ME,理由如下:如图1,当点E在线段CD上,AF交EE′于N,∵∠EAF=30°,∠EAE′=120,∴∠E′AN=90°,∴E′N=2AN,∵∠NAE=∠NEA=30°,∴NA=NE,E′N=2EN,∵EM∥FE′,∴==,∵BE′=DE,∴E′F=2ME,∴DE+BF=2ME.②当点E在CD延长线上,0°<∠EAD∠30°时,BF﹣DE=2ME,理由如下:如图2,∵∠EAF=30°,∠EAE′=120,∴∠E′AN=90°,∴E′N=2AN,∵∠NAE=∠NEA=30°,∴NA=NE,E′N=2EN,∵EM∥FE′,∴==,∵BE′=DE,∴E′F=2ME,∴BF﹣DE=2ME.③当30°<∠EAD∠90°时,DE+BF=2ME,理由如下:如图3,∵∠EAM=30°,∠EAE′=120,∴∠E′AN=90°,∴E′N=2AN,∵∠NAE=∠NEA=30°,∴NA=NE,E′N=2EN,∵EM∥FE′,∴==,∵BE′=DE,∴E′F=2ME,∴BF+DE=2ME.④当90°<∠EAD<120°时,DE﹣BF=2ME,理由如下:如图4,∵∠EAM=30°,∠EAE′=120,∴∠E′AN=90°,∴E′N=2AN,∵∠NAE=∠NEA=30°,∴NA=NE,E′N=2EN,∵EM∥FE′,∴==,∵BE′=DE,∴E′F=2ME,∴DE﹣BF=2ME.(3)如图5,作AG⊥BC于点G,DH⊥BC于H,AP⊥EE′于P,EQ⊥BC于Q,∵AD∥BC,AD=AB=CD,∠BAD=120°,易知四边形AGHD是矩形,在△AGB和△DHC中,,∴△AGB≌△DHC,∴BG=HC,AD=GH,∵∠ABE′=∠ADC=120°,∴点E′、B、C共线,设AD=AB=CD=x,则GH=x,BG=CH=x,在RT△EQC中,CE=2,∠ECQ=60°,∴CQ=EC=1,EQ=,∴E′Q=BC+BE′﹣CQ=3x﹣3,在RT△APE中,AE=2,∠AEP=30°,∴AP=,PE=,∵AE=AE′,AP⊥EE′,∴PE=PE′=,∴EE′=2,在RT△E′EQ中,E′Q==9,∴3x﹣3=9,∴x=4,∴DE=BE′=2,BC=8,BG=2,∴E′G=4,∵∠AE′G=′AE′F,∠AGE′=∠FAE′,∴△AGE′∽△FAE′,∴,∴,∴E′F=7,∴BF=E′F﹣E′B=7﹣2=5,∵DE+BF=2ME ∴ME=.26.如图,在平面直角坐标系中,矩形OABC的边OA在y轴的正半轴上,OC在x轴的正半轴上,∠AOC的平分线交AB于点D,E为BC的中点,已知A(0,4)、C(5,0),二次函数y=x2+bx+c的图象抛物线经过A,C两点.(1)求该二次函数的表达式;(2)F、G分别为x轴,y轴上的动点,顺次连接D、E、F、G构成四边形DEFG,求四边形DEFG周长的最小值;(3)抛物线上是否在点P,使△ODP的面积为12?若存在,求出点P的坐标;若不存在,请说明理由.【考点】二次函数综合题.【分析】(1)根据待定系数法,可得函数解析式;(2)延长EC至E′,使E′C=EC,延长DA至D′,使D′A=DA,连接D′E′,交x轴于F点,交y轴于G点,则有:GD=GD′,EF=E′F,从而得:(DG+GF+EF+ED)的最小值=D′E′+DE,求出D′E′与DE的长即可得到答案.(3)根据三角形的面积,首先求得点P到OD的距离,然后过点O作OF⊥OD,使OF等于点P到OD的距离,过点F作FG∥OD,求得FG的解析式,然后再求直线FG与抛物线交点的坐标即可得到点P的坐标.【解答】解:(1)将A(0,4)、C(5,0)代入二次函数y=x2+bx+c,得,解得.故二次函数的表达式y=x2﹣x+4;(2)如图:延长EC至E′,使E′C=EC,延长DA至D′,使D′A=DA,连接D′E′,交x轴于F点,交y 轴于G点,GD=GD′EF=E′F,=D′E′+DE,(DG+GF+EF+ED)最小由E点坐标为(5,2),BC的中点;D(4,4),直角的角平分线上的点;得D′(﹣4,4),E(5,﹣2).由勾股定理,得DE==,D′E′==,=D′E′+DE=+;(DG+GF+EF+ED)最小(3)如下图:OD=.∵S△ODP的面积=12,∴点P到OD的距离==3.过点O作OF⊥OD,取OF=3,过点F作直线FG∥OD,交抛物线与点P1,P2,在Rt△OGF中,OG===6,∴直线GF的解析式为y=x﹣6.将y=x﹣6代入y=得:x﹣6=,解得:,,将x1、x2的值代入y=x﹣6得:y1=,y2=∴点P1(,),P2(,)如下图所示:过点O作OF⊥OD,取OF=3,过点F作直线FG交抛物线与P3,P4,在Rt△PFO中,OG==6∴直线FG的解析式为y=x+6,将y=x+6代入y=得:x+6=解得:,y1=x1+6=,y2=x2+6=∴p3(,),p4(,)综上所述:点P的坐标为:(,)或(,)或(,)或(,).2020年4月15日。

2020-2021学年新人教版九年级开学考试数学试题及解析

2020-2021学年新人教版九年级开学考试数学试题及解析

山东省东营市垦利县实验学校2020年秋季开学考试数 学 试 题(总分12020 考试时间12020)第Ⅰ卷(选择题 共30分)一、选择题:本大题共10小题,在每小题给出的四个选项中,只有一项是正确的,请把正确的选项选出来.每小题选对得3分,选错、不选或选出的答案超过一个均记零分. 1.下列银行标志中,既不是中心对称图形也不是轴对称图形的是( ) A .B .C .D .2.下列式子从左到右变形是因式分解的是( B )A .a 2+4a-21=a(a+4)-21B .a 2+4a-21=(a-3)(a+7)C .(a-3)(a+7)=a 2+4a-21D .a 2+4a-21=(a+2)2-253.如图,AD 是∠EAC 的平分线,AD ∥BC ,∠B =30°,则∠C 为()A . 30°B . 60°C .80° D .120204.分式方程﹣1=的解是( ) A . x =1B . x =﹣1+C . x =2D . 无解 5.若43=x ,79=y,则y x 23-的值为( )A .74B .47C .3-D .726.如图,在△ABC 中,∠ABC=50°,∠ACB=60°,点E 在BC 的延长线上,∠ABC 的平分线BD 与∠ACE 的平分线CD 相交于点D ,连接AD ,下列结论中不正确的是( )A. ∠BAC=70°B . ∠DOC=90°C . ∠BDC=35°D . ∠DAC=55°7.根据下图所示程序计算函数值,若输入的x 的值为52,则输出的函数值为( ) A .32B .25C .425D .2548.若一个正n 边形的每个内角为156°,则这个正n 边形的边数是( ) A . 13B . 14C . 15D . 169.如图,在四边形ABCD 中,AD ∥BC ,DE ⊥BC ,垂足为点E ,连接AC 交DE 于点F ,点G 为AF 的中点,∠ACD=2∠ACB .若DG=3,EC=1,则DE 的长为( )A . 2B .C . 2D .10.如图,已知正方形ABCD ,顶点A (1,3)、B (1,1)、C (3,1).规定“把正方形ABCD 先沿x 轴翻折,再向左平移1个单位”为一次变换.如此这样,连续经过2020次变换后,输入x 值 y =x -1 (-1≤x <0)1y x(2≤x ≤4)y =x 2(0≤x <2)输出y 值正方形ABCD 的对角线交点M 的坐标变为( )A .(—2020,2)B .(一2020,一2) C. (—2020,—2) D. (—2020,2)第Ⅱ卷(非选择题 共90分)二、填空题:本大题共8小题,其中11-14题每小题3分,15-18题每小题4分,共28分.只要求填写最后结果.11.若关于x 的方程x 2+(k ﹣2)x+k 2=0的两根互为倒数,则k= . 12.若y =﹣2,则(x +y )y = .13.若等腰三角形的两条边长分别为7cm 和14cm ,则它的周长为 cm .14.如图,将边长为12的正方形ABCD 是沿其对角线AC 剪开,再把ABC ∆沿着AD 方向平移,得到C B A '''∆,当两个三角形重叠的面积为32时,它移动的距离A A '等于______. 15.在平面直角坐标系中,已知一次函数y=2x+1的图象经过P 1(x 1,y 1)、P 2(x 2,y 2)两点,若x 1<x 2,则y 1 y 2.(填“>”“<”或“=”)16.如图,在Rt △ABC 中,∠B =90°,AB =3,BC =4,将△ABC 折叠,使点B 恰好落在边AC 上,与点B ′重合,AE 为折痕,则EB ′= .17.图①所示的正方体木块棱长为6cm ,沿其相邻三个面的对角线(图中虚线)剪掉一角,得到如图②的几何体,一只蚂蚁沿着图②的几何体表面从顶点A 爬行到顶点B 的最短距离为 cm .18.如图,正方形ABCD的边长为6,点O是对角线AC、BD的交点,点E在CD 上,且DE=2CE,过点C作CF⊥BE,垂足为F,连接OF,则OF的长为.三、解答题:本大题共7小题,共62分.解答要写出必要的文字说明、证明过程或演算步骤.19.(本题满分7分,第⑴题3分,第⑵题4分)(1) 计算:(2) 化简:(1﹣)÷(﹣2)(3)求不等式组的解集.2020本题满分8分)州教育局为了解我州八年级学生参加社会实践活动情况,随机抽查了某县部分八年级学生第一学期参加社会实践活动的天数,并用得到的数据检测了两幅统计图,下面给出了两幅不完整的统计图(如图)请根据图中提供的信息,回答下列问题:(1)a= %,并写出该扇形所对圆心角的度数为,请补全条形图.(2)在这次抽样调查中,众数和中位数分别是多少?(3)如果该县共有八年级学生2020人,请你估计“活动时间不少于7天”的学生人数大约有多少人?21.如图,在Rt△ABC中,∠ACB=90°,点D、F分别在AB、AC上,CF=CB,连接CD,将线段CD绕点C按顺时针方向旋转90°后得CE,连接EF.(1)求证:△BCD≌△FCE;(2)若EF∥CD,求∠BDC的度数.22.(本题满分8分) 甲、乙两车从A地驶向B地,并以各自的速度匀速行驶,甲车比乙车早行驶2h,并且甲车途中休息了0.5h,如图是甲乙两车行驶的距离y(km)与时间x(h)的函数图象.(1)求出图中m,a的值;(2)求出甲车行驶路程y(km)与时间x(h)的函数解析式,并写出相应的x的取值范围;(3)当乙车行驶多长时间时,两车恰好相距50km.23.(本题满分8分)某商店销售10台A型和2020型电脑的利润为4000元,销售2020型和10台B型电脑的利润为3500元.(1)求每台A型电脑和B型电脑的销售利润;(2)该商店计划一次购进两种型号的电脑共100台,其中B型电脑的进货量不超过A型电脑的2倍。

湖北省黄冈市2020届九年级下学期入学考试数学试题(图片版)

湖北省黄冈市2020届九年级下学期入学考试数学试题(图片版)

参考答案与试题解析一.选择题(共8小题)1.A .2.B.提示:∵n是一元二次方程x2+x=4的根,∴n2+n=4,即n2=﹣n+4,∵m、n是一元二次方程x2+x=4的两个实数根,∴m+n=﹣1,mn=﹣4,∴2n2﹣mn﹣2m=2(﹣n+4)﹣mn﹣2m=﹣2(m+n)﹣mn+8=2+4+8=14.故选:B.3.D.4.C.提示:∵∠AOB=90°,∠B=30°,∴∠A=60°,∵△AOB绕点O顺时针旋转角度得到△A′OB′,∴OA=OA′,∠AOA'=α,∴△OAA′是等边三角形,∴∠AOA′=60°,即旋转角α的大小可以是60°,故选:C.5.A.6.D.7.B.提示:∵∠A=∠C=35o,∴OA∥BC,∴∠B=∠AOB,∵∠AOB=2∠C=70°,∴∠B=70°.故选:B.8.C.提示:由抛物线的对称性可知:(0,m)与(2,m)是对称点,故对称轴为x=1,∴(﹣2,n)与(4,n)是对称点,∴4a﹣2b+c=n<0,故选:C.二.填空题(共8小题)9..提示:根据题意x1+x2=5,x1•x2=3,==.故答案为:.10.<.提示:∵抛物线y=ax2(a>0)∴抛物线开口向上,在对称轴右侧y随x的增大而增大,对称轴为y轴∵点A、B的坐标分别为(3,m)、(4,n),且3<4,∴m<n 故答案为:<.11..提示:∵△DEC与△ABC关于点C成中心对称,∴△ABC≌△DEC,∴AB=DE=3,AC=DC=1,∴AD=2,∵∠D=90°,∴AE===,故答案为.12.1.提示:设圆锥的底面圆的半径为rcm,根据题意得2π•r=,解得r=1.故答案为:1.13.9.提示:设口袋中有x个白球,因为摸了100次,其中60次摸到白球,则估计摸到白球的概率为=,所以=,解得x=9,即可估计口袋中的白球大约有9个.故答案为9.14.8.提示:设点的的坐标为(a,b),∵双曲线y=经过点D,∴ab=4,∵AD∥x轴,∴AD=a,AO=b,又∵点O为AC的中点,∴AC=2AO=2b,∴▱ABCD面积=2×AD×AC=a×2b=2ab=8,故答案为:8.15..提示:连接OB,∵AC是⊙O的直径,弦BD⊥AC,∴BE=BD=6cm,在Rt△OEB中,OB2=OE2+BE2,即OB2=(OB﹣4)2+62,解得,OB=,则EC=AC﹣AE=9,BC===3,∵OF⊥BC,∴CF=BC=,∴OF===(cm),故答案为.16..提示:设AB=CD=AD=BC=a,∵抛物线y=(x+1)2﹣5,∴顶点E(﹣1,﹣5),对称轴为直线x=﹣1,∴C的横坐标为﹣1,D的横坐标为﹣1﹣,∵点C在抛物线y=(x+1)2﹣5上,∴C点纵坐标为(﹣1+1)2﹣5=﹣5,∵E点坐标为(﹣1,﹣5),∴B点纵坐标为﹣5,∵BC=a,∴﹣5﹣a=﹣5,解得:a1=,a2=0(不合题意,舍去),故答案为:.三.解答题(共9小题)17.解:方程整理得:3x(2x+3)﹣2(2x+3)=0,分解因式得:(3x﹣2)(2x+3)=0,可得3x﹣2=0或2x+3=0,解得:x1=,x2=﹣.18.解:(1)如图所示,△A1B1C1即为所求.(2)如图所示,△A2B2C2即为所求,点A2的坐标为(5,﹣1).19.解:(1)根据题意得:(10+20+15)÷(1﹣10%)=50(人),50×10%=5(人),答:D小区的人数有5人,补图如下:(2)根据题意画图如下:共有12种等情况数,其中正好抽到甲和乙的有2种,则正好抽到甲和乙的概率是=.20.解:(1)当m﹣2≠0时,△=1+8(m﹣2)≥0,∴m≥且m≠2,当m﹣2=0时,x﹣2=0,符合题意,综上所述,m≥(2)由根与系数的关系可知:x1+x2=,x1x2=,∵x12+x22=5,∴(x1+x2)2﹣2x1x2=5,∴+=5,∴=1或=﹣5,∴m=3或m=(舍去).21.解:(1)∵将△BCD绕点C顺时针旋转得到△ACE∴△BCD'≌△ACE ∴AC=BC,又∵∠ABC=45°,∴∠ABC=∠BAC=45°∴∠ACB=90°故旋转角的度数为90°(2)AE⊥BD.理由如下:在Rt△BCM中,∠BCM=90°∴∠MBC+∠BMC=90°∵△BCD'≌△ACE ∴∠DBC=∠EAC 即∠MBC=∠NAM又∵∠BMC=∠AMN ∴∠AMN+∠CAE=90°∴∠AND=90°∴AE⊥BD(3)如图,连接DE,由旋转图形的性质可知CD=CE,BD=AE,旋转角∠DCE=90°∴∠EDC=∠CED=45°∵CD=3,∴CE=3在Rt△DCE中,∠DCE=90°∴DE===3∵∠ADC=45°∴∠ADE=∠ADC+∠EDC=90°在Rt△ADE中,∠ADE=90°∴EA===∴BD=22.解:(1)把A(﹣1,m),B(n,﹣1)分别代入y=得﹣m=﹣2,﹣n=﹣2,解得m=2,n=2,所以A点坐标为(﹣1,2),B点坐标为(2,﹣1),把A(﹣1,2),B(2,﹣1)代入y=kx+b得,解得,所以这个一次函数的表达式为y=﹣x+1;(2)设直线AB交y轴于P点,如图,当x=0时,y=1,所以P点坐标为(0,1),所以S△OAB=S△AOP+S△BOP=×1×1+×1×2=.23.(1)证明:如图1,连接ON,∵∠ACB=90°,D为斜边的中点,∴CD=DA=DB=AB,∴∠BCD=∠B,∵OC=ON,∴∠BCD=∠ONC,∴∠ONC=∠B,∴ON∥AB,∵NE⊥AB,∴ON⊥NE,∴NE为⊙O的切线;(2)解:如图2,连接DN,ON∵⊙O的半径为,∴CD=5∵∠ACB=90°,CD是斜边AB上的中线,∴BD=CD=AD=5,∴AB=10,∴BC==8,∵CD为直径,∴∠CND=90°,且BD=CD,∴BN=NC=4,24.解:(1)设线段AB的解析式为:y=kx+b,∵点A(1,16),点B(5,17),∴,解得:,∴线段AB的解析式为:y=x+;∵点B是抛物线的顶点,∴设抛物线BC的解析式为:y=a(x﹣5)2+17,把C(10,42)代入得,42=a(10﹣5)2+17,解得:a=1,∴抛物线BC的解析式为:y=(x﹣5)2+17;(2)当1≤x≤5时,w=x+﹣13=x+,故当x=1时,w有最小值为3;当5<x≤10时,w=(x﹣5)2+17﹣(3x﹣2)=(x﹣6.5)2+1.75,∵x为正整数,∴当x=6或7时,w有最小值2,综上所述,当x=6或7时,w有最小值2.25.解:(1)抛物线y=ax2+bx+c经过点A(0,﹣3)、C(2,﹣3),则函数的对称轴为:x=1,故点E(3,0),抛物线表达式为:y=a(x﹣3)(x+1)=a(x2﹣2x﹣3),故﹣3a=﹣3,解得:a=1,故抛物线的表达式为:y=x2﹣2x﹣3 ……①;(2)四边形MBEP恰好是平行四边形时,则MP=BE=3,故t=4,则点P(4,5);(3)过点C作y轴的平行线交AE于点H,由点A、E的坐标得直线AE的表达式为:y=x﹣3,设点P(t,t2﹣2t﹣3),则点H(t,t﹣3),△PAE的面积S=PH×OE=(t﹣3﹣t2+2t+3)=(﹣t2+3t),当t=时,S有最大值;(4)直线AE表达式中的k值为1,则与之垂直的直线表达式中的k为﹣1.①当∠PEA=90°时,直线PE的表达式为:y=﹣x+b,经点E的坐标代入并解得:直线PE的表达式为:y=﹣x+3 ……②,联立①②并解得:x=﹣2或3(舍去3),故点P(﹣2,5);②当∠PAE=90°时,同理可得:点P(1,﹣4);综上,点P的坐标为:(﹣2,5)或(1,﹣4).。

2020年春九年级英语开学入学考试试题(附答案)

2020年春九年级英语开学入学考试试题(附答案)

2020年春九年级英语入学考试(检测时间:90分钟满分:120分)第一部分:读(共二节,满分70分)第一节:阅读理解(共20个小题,每小题2分,满分40分)阅读下面短文,从每题所给的四个选项(A、B、C、D)中,选出最佳选项。

AThanks to the ancient Silk Road, the lands of Italy and China have been connected for more than 2,000 years. Tourists from China and every corner of the world come to visit Italy every year. Here are four famous cities of Italy.Rome ColiseumPisa is a small city. It is famous for the leaning tower of Pisa. Ifyou think that Pisa has only one leaning tower, you may just bewrong. This city has three leaning towers. Home of the famousscientist Galileo, Pisa is also known for its universities becausethe city is home of some of the most important universities inLeaning Tower DavidVenice is a city over a hundred islands. You can visit mostinteresting places of the city by walks or by the famousgondolas. Art exhibitions, film festivals and famous carnival(狂欢节)make Venice a place that can be visited any time ofthe year.gondolasnow.A. RomeB. PisaC. FlorenceD. Venice2. Pisa is also famous for its _____________besides leaning towers.A. universitiesB. art exhibitionsC. film festivalsD. museums3. We can visit Arnolfo Tower ____________.A. at 8 am on SundayB. at 9 am on TuesdayC. at 9 pm on SaturdayD. at 10 am on Monday4. According to the passage, which one of the following is NOT true?A. Pisa has three leaning towers.B. You’ll find David when you go shopping in Florence.C. You can visit most interesting places of Venice by walks or by bus.D. Every year, thousands of tourists all over the world come to visit Italy.5. What is the best title of the passage?A. Italy and China.B. The ancient Silk Road.C. All roads lead to Rome.D. Four famous cities of Italy.BHow to talk to someone you don't know? It depends on the situation (情况).If one of your friends introduces you to a new friend, it would be OK to start talking about the person who introduces you. You may ask him / her how they knew each other. Show you're interested in it, even if you aren't. Nod after he / she talks. This will give you more time to listen to what he / she says, and you think about nextquestion from his / her talking.Now if you go to a class on the first day where you do not know anyone, the first thing you should remember is not to be shy when talking to other people. You may remember when you were in the first or second grade, someone would just walk up to you and talk to you. Be open, but think before you talk.When it comes to meeting people face to face for the first time, find a situation which both of you are in. Maybe you are sitting next to each other. Maybe the other person says something you don't understand. Find a situation that seems proper (恰当的). Start by a simple conversation. It may seem strange at first. But the more you talk and take action with them, the more they will like you.6.What does the underlined part "the person" refer to?A.Yourself.B.A new friend.C.One of your friends.D.The weather.7. Why do you need more time according to the second paragraph?A.To know clearly what they say.B.To find funny things.C.To find what they are interested in.D.To think about next question.8.What does the third paragraph want to tell us?A.Not to be shy when talking to strangers.B.To find a good talk to start.C.To remember to greet others.D.To attract people to talk to us.9.Which is a good situation to meet people face to face for the first time?A.Saying something you both know well.B.Saying something you know.C.Sitting face to face.D.Sitting next to each other.10.Which of the following is the best title?A.The different situations about talking with your friends.B.Some good ways to talk to strangers.C.How to make friends in your new class.D.How to make a good impression on others.C.On January 27, in order to win the new pneumonia Resistance War(抗击新型肺炎战), the Ministry of Education issued(教育部公布) a notice on putting off the start time of school in spring 2020. It was said that school should not start before March. The exact start time of school will depend on situations and further notices.However, many parents are very worried that putting off the start of school may affect(影响)children’s studies. If they have such a long winter vacation, children will not learn any knowledge, read books or study at all. They may forget what they have learned before. Staying at home for a long time will make them keep eating, sleeping and playing, which seriously affect their physical andmental health.In short, the close of school influences learning and develops bad habits, so how should we avoid these problems? Local education departments have also issued a notice, that is, to organize online teaching and create a network platform(平台), where necessary courses will be shown every day, and online teaching will be given by excellent teachers of the subject, and students can learn at home through computer or mobile network.It should be said that such a method is very good. It can not only keep students indoors and not worry about the infection(感染) of the disease, but also make them study at home, improve their learning ability, and truly achieve“no suspension of classes”. Both the platform itself and the recorded lesson resources are more useful, and are welcomed by parents and teachers.Of course, except for learning at home, it is also necessary to teach children about health knowledge, especially how to keep away from touching virus and other knowledge popularization; at the same time, it is also necessary for children to do some physical exercises or some games, which is good for physical healthat home.11. Parents worried about putting off the start time of school because _______.A. children may be infected by virusB. they don’t like the idea of online learningC. children’s study may be affectedD. parents have no time to look after children12. Except for learning at home, what is also necessary according to the writer?A. To keep playing online games.B. To keep touching different people.C. To do some physical exercises outside.D. To teach children about health knowledge.13. The underlined word “suspension” in Paragraph 4 probably means _______ .A. planB. stopC. growthD. change14. Which is TRUE according to the passage?A. Students would rather go to school.B. The date of going to school is decided.C. Online teaching will help students study.D. Parents believe children can teach themselves.15. What’s the best title for this passage?A. A Special Way to Learn at HomeB. A New War with a New ChanceC. A Way to Keep Mentally HealthyD. A Method to Make Parents WorriedD.Many of us don’t pay much attention to the importance of eye care. It is said that if you take care of your body, then you can surely be healthy. That is why our eyes should be given a lot of care. Natural eye care should be put in a number one place.There are several causes leading to poor eyesight like not eating enough food, genes (基因) and aging (老化). Televisions,computers and reading are also the causes of having poor eyesight. If you happen to work in front of a computer, you’d better take a break every once in a while.Something dirty can cause redness and it will make you feel uncomfortable. It is bad for your eyes, too. If this happens, the best way is to clean your eyes by using cold water. Y ou must also try your best to protect your eyes from harmful things. For example,sunglasses are not just for fashion but they can also serve as a great way to protect your eyesight from UV rays.Eating healthy foods will do good to your eyesight. Remember that vitamins A, C and E are good for eyes. Try to eat food groups that have these vitamins. And you should do eye exercises because they protect your eyesight, too. If a person exercises regularly (定期地) and eats the right kind of food, hiseyes will stay in good condition for a long time.All above are natural ways of eye care that help us keep healthy eyes. Being happy all the time can be helpful to a person’s eyesight, too. In a word, eye care is very important, no matter how old a person is.16.is the most important way to protect our eyes.A.Taking medicineB. Natural eye careC.Seeing the doctorD. Being happy all the time17.All the following causes can lead to poor eyesight EXCEPT.A. computersB. agingC. readingD. height18.What should you do if you have to work in front of a computer according to the passage?A. Have a rest after working for a while.B. Eat healthy foods.C. Clean the eyes by using cold water.D. Wear a pair of sunglasses.19. What do the underlined phase “UV rays” mean in Chinese?A. 闪电B.沙尘.C.紫外线D. 超声波20.What is the best title for the passage?A.Ways of Doing Eye ExerciseB.Ways of Eye CareB.Ways of Being Happy D.Ways of Being Healthy第二节:完形填空(共15个小题,每小题2分,满分30分)阅读下面短文,从每题所给的四个选项(A、B、C、D)中,选出可以填入空白处的最佳选项。

浙江省杭州十三中浙教版2020届九年级上学期开学考试数学试题 含答案

浙江省杭州十三中浙教版2020届九年级上学期开学考试数学试题 含答案

浙江省杭州十三中浙教版2020届九年级上学期开学考试数学试题一、选择题(共10个小题,满分30分,每小题3分)1.下列地铁标志图形中,属于中心对称图形的是()A.B.C.D.2.下列函数中,y关于x的二次函数是()A.y=ax2+bx+c B.y=x(x﹣1)C.D.y=(x﹣1)2﹣x23.已知二次根式,则a的取值范围是()A.B.C.D.4.若关于x的一元二次方程(k+2)x2﹣3x+1=0有实数根,则k的取值范围是()A.k<且k≠﹣2 B.k C.k≤且k≠﹣2 D.k5.若不等式k<<k+1成立,则整数k的值为()A.6 B.7 C.8 D.96.在反比例函数y=﹣图象上有三个点A(x1,y1)B(x2,y2)C(x3,y3),若x1<0<x2<x3,则下列结论正确的是()A.y1<y3<y2B.y2<y3<y1C.y3<y1<y2D.y3<y2<y17.如图,在▱ABCD中,对角线AC,BD相交于点O,点E,F是对角线AC上的两点,当点E,F满足下列条件时,四边形DEBF不一定是平行四边形()A.AE=CF B.DE=BF C.∠ADE=∠CBF D.∠AED=∠CFB 8.已知ab<0,一次函数y=ax﹣b与反比例函数y=在同一直角坐标系中的图象可能()A.B.C.D.9.如图,在菱形纸片ABCD中,对角线AC、BD长分别为16、12,折叠纸片使点A落在DB 上,折痕交AC于点P,则DP的长为()A.3B.C.3D.310.规定:如果关于x的一元二次方程ax2+bx+c=0(a≠0)有两个实数根,且其中一个根是另一个根的2倍,则称这样的方程为“倍根方程”现有下列结论①方程x2+2x﹣8=0是倍根方程;②若关于x的方程x2+ax+2=0是倍根方程,则a=±3;③若(x﹣3)(mx﹣n)=0是倍根方程,则n=6m或3n=2m;④若点(m,n)在反比例函数y=的图象上,则关于x的方程mx2﹣3x+n=0是倍根方程.上述结论中正确的有()A.①②B.③④C.②③D.②④二、填空题(共6小题,满分24分,每小题4分)11.一个多边形的每个外角都等于72°,则这个多边形的边数为.12.已知一组数据5,8,10,x,9的众数是8,那么这组数据的方差是.13.若二次函数y=ax2﹣bx+5(a≠5)的图象与x轴交于(1,0),则b﹣a+2014的值是.14.如图,在△ABC中,CD平分∠ACB,AD⊥CD,垂足为D,E为AB的中点,连接DE,AC=15,BC=27,则DE=.15.如图,在平面直角坐标系中,点A在抛物线y=x2﹣2x+3上运动,过点A作AC⊥x轴于点C,以AC为对角线作矩形ABCD,连结BD,则对角线BD的最小值为.16.如图,在正方形ABCD中,AB=6,E为CD上一动点,AE交BD于F,过F作FH⊥AE交BC于点H,过H作HG⊥BD于G,连结AH.在以下四个结论中:①AF=HE;②∠HAE=45°;③FC=2;④△CEH的周长为12.其中正确的结论有.三、解答题(本题有8小题,共6分)17.计算(1)(2)18.随着移动互联网的快速发展,基于互联网的共享单车应运而生.为了解某单位使用共享单车的情况,该单位有200名员工,某研究小组随机采访10位员工,得到这10位员工一周内使用共享单车的次数分别为:17,12,15,20,17,0,7,26,17,9.(1)这组数据的中位数是,众数是(2)试用平均数估计该单位员工一周内使用共享单车的总次数.19.在平面直角坐标系中,二次函数y=ax2+bx+3的图象经过点A(3,0)和点B(4,3).(1)求二次函数的表达式(2)求二次函数图象的顶点坐标和对称轴.20.已知关于x的一元二次方程x2﹣(m+2)x+2m=0(1)求证:不论m为何值,该方程总有两个实数根;(2)若此方程的一个根是1,请求出方程的另一个根,并求出以此两根为边长的直角角形的周长21.如图,在△ABC中,D、E分别是AB,AC的中点,BE=2DE,延长DE到点F,使得EF=BE,连接CF.(1)求证:四边形BCFE是菱形;(2)若CE=2,∠BEF=120°,求菱形BCFE的面积.22.直线y=3x与反比例函数y=的图象交于A(1,m)和点B.(1)求m、k的值,并直接写出点B的坐标(2)过点P(t,0)(﹣1≤t≤1且t≠0)作x轴的垂线分别交直线y=3x与反比例函数y=的图象于点E,F.①当t=时,求线段EF的长;②若0<EF≤8,请根据图象直接写出t的取值范围.23.如图,在等腰△ABC中,AC=BC=3,AB=6,点E从点B沿着射线BA以每秒3个单位的速度运动,过点E作BC的平行线交∠ACB的外角平分线CF于点F.(1)求证:四边形BCFE是平行四边形;(2)当点E是边AB的中点时,连结AF,试判断四边形AECF的形状,并说明理由;(3)设运动时间为t秒,是否存在t的值,使得以△EFC的其中两边为边所构造的平行四边形恰好是菱形?若存在,请求出t的值;若不存在,试说明理由.参考答案与试题解析一.选择题(共10小题)1.下列地铁标志图形中,属于中心对称图形的是()A.B.C.D.【分析】根据中心对称图形的定义即可作出判断.【解答】解:A、不是中心对称图形,故选项错误;B、不是中心对称图形,故选项错误;C、不是中心对称图形,故选项错误;D、是中心对称图形,故选项正确.故选:D.2.下列函数中,y关于x的二次函数是()A.y=ax2+bx+c B.y=x(x﹣1)C.D.y=(x﹣1)2﹣x2【分析】根据二次函数的定义,逐一分析四个选项即可得出结论.【解答】解:A、当a=0时,y=bx+c不是二次函数;B、y=x(x﹣1)=x2﹣x是二次函数;C、y=不是二次函数;D、y=(x﹣1)2﹣x2=﹣2x+1为一次函数.故选:B.3.已知二次根式,则a的取值范围是()A.B.C.D.【分析】直接利用二次根式的性质得出a的取值范围.【解答】解:∵二次根式有意义,∴2a﹣1≥0,解得:a≥,则a的取值范围是:a≥.故选:D.4.若关于x的一元二次方程(k+2)x2﹣3x+1=0有实数根,则k的取值范围是()A.k<且k≠﹣2 B.k C.k≤且k≠﹣2 D.k【分析】根据一元二次方程的定义和根的判别式得出k+2≠0且△=(﹣3)2﹣4(k+2)•1≥0,求出即可.【解答】解:∵关于x的一元二次方程(k+2)x2﹣3x+1=0有实数根,∴k+2≠0且△=(﹣3)2﹣4(k+2)•1≥0,解得:k且k≠﹣2,故选:C.5.若不等式k<<k+1成立,则整数k的值为()A.6 B.7 C.8 D.9【分析】先估算出的范围,即可得出选项.【解答】解:∵9<<10,∴k=9,k+1=10,故选:D.6.在反比例函数y=﹣图象上有三个点A(x1,y1)B(x2,y2)C(x3,y3),若x1<0<x2<x3,则下列结论正确的是()A.y1<y3<y2B.y2<y3<y1C.y3<y1<y2D.y3<y2<y1【分析】k=﹣2019,故图象在二、四象限,x>0,y随x增大而增大,y2<y3,且均为负值,x<0时,y>0,即可求解.【解答】解:k=﹣2019,故图象在二、四象限,x>0,y随x增大而增大,y2<y3,且均为负值,x<0时,y>0,故选:B.7.如图,在▱ABCD中,对角线AC,BD相交于点O,点E,F是对角线AC上的两点,当点E,F满足下列条件时,四边形DEBF不一定是平行四边形()A.AE=CF B.DE=BF C.∠ADE=∠CBF D.∠AED=∠CFB 【分析】根据平行四边形的性质以及平行四边形的判定定理即可作出判断.【解答】解:A、∵在平行四边形ABCD中,OA=OC,OB=OD,若AE=CF,则OE=OF,∴四边形DEBF是平行四边形;B、若DE与AC不垂直,则满足AC上一定有一点DM=DE,同理有一点N使BF=BN,则四边形DEBF不一定是平行四边形,则选项错误;C、∵在平行四边形ABCD中,OB=OD,AD∥BC,∴∠ADB=∠CBD,若∠ADE=∠CBF,则∠EDB=∠FBO,∴DE∥BF,则△DOE和△BOF中,,∴△DOE≌△BOF,∴DE=BF,∴四边形DEBF是平行四边形.故选项正确;D、∵∠AED=∠CFB,∴∠DEO=∠BFO,∴DE∥BF,在△DOE和△BOF中,,∴△DOE≌△BOF,∴DE=BF,∴四边形DEBF是平行四边形.故选项正确.故选:B.8.已知ab<0,一次函数y=ax﹣b与反比例函数y=在同一直角坐标系中的图象可能()A.B.C.D.【分析】根据反比例函数图象确定b的符号,结合已知条件求得a的符号,由a、b的符号确定一次函数图象所经过的象限.【解答】解:若反比例函数y=经过第一、三象限,则a>0.所以b<0.则一次函数y=ax﹣b的图象应该经过第一、二、三象限;若反比例函数y=经过第二、四象限,则a<0.所以b>0.则一次函数y=ax﹣b的图象应该经过第二、三、四象限.故选项A正确;故选:A.9.如图,在菱形纸片ABCD中,对角线AC、BD长分别为16、12,折叠纸片使点A落在DB 上,折痕交AC于点P,则DP的长为()A.3B.C.3D.3【分析】首先设O点的对应点为E,连接PE,由菱形的性质,可求得OD,OA与AD的长,由折叠的性质,根据勾股定理可得方程:即(8﹣x)2=42+x2,可求x的值,由勾股定理可求DP的长.【解答】解:设O点的对应点为E,连接PE,由折叠的性质可得:PE=OP,DE=OD,∵四边形ABCD是菱形,∴AC⊥BD,OA=AC=×16=8,OB=BD=×12=6,∴AD===10,设OP=x,则PE=x,AE=AD﹣DE=10﹣6=4,AP=OA﹣OP=8﹣x,在Rt△APE中,AP2=AE2+PE2,即(8﹣x)2=42+x2,解得:x=3,即OP=3,∴DP===3,故选:A.10.规定:如果关于x的一元二次方程ax2+bx+c=0(a≠0)有两个实数根,且其中一个根是另一个根的2倍,则称这样的方程为“倍根方程”现有下列结论①方程x2+2x﹣8=0是倍根方程;②若关于x的方程x2+ax+2=0是倍根方程,则a=±3;③若(x﹣3)(mx﹣n)=0是倍根方程,则n=6m或3n=2m;④若点(m,n)在反比例函数y=的图象上,则关于x的方程mx2﹣3x+n=0是倍根方程.上述结论中正确的有()A.①②B.③④C.②③D.②④【分析】①通过解方程得到该方程的根,结合“倍根方程”的定义进行判断;②设x2=2x1,得到x1•x2=2x12=2,得到当x1=1时,x2=2,当x1=﹣1时,x2=﹣2,于是得到结论;③根据“倍根方程”的定义即可得到结论;④若点(m,n)在反比例函数y=的图象上,得到mn=2,然后解方程mx2﹣3x+n=0即可得到正确的结论;【解答】解:①∵方程x2+2x﹣8=0的两个根是x1=﹣4,x2=2,则2×2≠﹣4,∴方程x2+2x﹣8=0不是倍根方程,故①错误;②若关于x的方程x2+ax+2=0是倍根方程,则2x1=x2,∵x1+x2=﹣a,x1•x2=2,∴2x12=2,解得x1=±1,∴x=±2,∴a=±3,故②正确;③解方程(x﹣3)(mx﹣n)=0得,x1=3,x2=,若(x﹣3)(mx﹣n)=0是倍根方程,则=6或2×=3,∴n=6m或3m=2n,故③错误;④∵点(m,n)在反比例函数y=的图象上,∴mn=2,即n=,∴关于x的方程为mx2﹣3x+=0,解方程得x1=,x2=,∴x2=2x1,∴关于x的方程mx2﹣3x+n=0是倍根方程,故④正确;故选:D.二.填空题(共6小题)11.一个多边形的每个外角都等于72°,则这个多边形的边数为 5 .【分析】利用多边形的外角和360°,除以外角的度数,即可求得边数.【解答】解:多边形的边数是:360÷72=5.故答案为:5.12.已知一组数据5,8,10,x,9的众数是8,那么这组数据的方差是 2.8 .【分析】根据众数的定义求出x的值,再根据平均数的计算公式求出这组数据的平均数,再根据方差公式进行计算即可.【解答】解:∵一组数据5,8,10,x,9的众数是8,∴x是8,∴这组数据的平均数是(5+8+10+8+9)÷5=8,∴这组数据的方差是:[(5﹣8)2+(8﹣8)2+(10﹣8)2+(8﹣8)2+(9﹣8)2]=2.8.故答案为:2.8.13.若二次函数y=ax2﹣bx+5(a≠5)的图象与x轴交于(1,0),则b﹣a+2014的值是2019 .【分析】把把(1,0)代入y=ax2﹣bx+5得a﹣b+5=0,然后利用整体代入的方法计算b﹣a+2014的值.【解答】解:把(1,0)代入y=ax2﹣bx+5得a﹣b+5=0,所以b﹣a=5,所以b﹣a+2014=5+2014=2019.故答案为2019.14.如图,在△ABC中,CD平分∠ACB,AD⊥CD,垂足为D,E为AB的中点,连接DE,AC=15,BC=27,则DE= 6 .【分析】证明△CDA≌△CDF,根据全等三角形的性质得到AD=DF,CF=AC,根据三角形中位线定理解答.【解答】解:在△CDA和△CDF中,,∴△CDA≌△CDF,∴AD=DF,CF=AC=15,∴BF=BC﹣CF=12,∵AD=DF,AE=EB,∴DE=BF=6,故答案为:6.15.如图,在平面直角坐标系中,点A在抛物线y=x2﹣2x+3上运动,过点A作AC⊥x轴于点C,以AC为对角线作矩形ABCD,连结BD,则对角线BD的最小值为 2 .【分析】利用配方法求出抛物线的顶点坐标,根据矩形的性质解答.【解答】解:y=x2﹣2x+3=(x﹣1)2+2,则抛物线的顶点坐标为(1,2),∴当点A在抛物线的顶点时,AC最小,最小值为2,∵四边形ABCD是矩形,∴AC=BD,∴对角线BD的最小值为2,故答案为:2.16.如图,在正方形ABCD中,AB=6,E为CD上一动点,AE交BD于F,过F作FH⊥AE交BC于点H,过H作HG⊥BD于G,连结AH.在以下四个结论中:①AF=HE;②∠HAE=45°;③FC=2;④△CEH的周长为12.其中正确的结论有②④.【分析】①作辅助线,延长HF交AD于点L,连接CF,通过证明△ADF≌△CDF,可得:AF=CF,故需证明FC=FH,可证:AF=FH;②由FH⊥AE,AF=FH,可得:∠HAE=45°;③F是动点,CF的长度不是定值;④作辅助线,延长AD至点M,使AD=DM,过点C作CI∥HL,则IL=HC,可证AL=HE,再根据△MEC≌△MIC,可证:CE=IM,故△CEH的周长为边AM的长,为定值.【解答】解:①连接FC,延长HF交AD于点L,∵BD为正方形ABCD的对角线,∴∠ADB=∠CDF=45°.∵AD=CD,DF=DF,∴△ADF≌△CDF.∴FC=AF,∠ECF=∠DAF.∵∠ALH+∠LAF=90°,∴∠LHC+∠DAF=90°.∵∠ECF=∠DAF,∴∠FHC=∠FCH,∴FH=FC.∴FH=AF.故①错误,②∵FH⊥AE,FH=AF,∴∠HAE=45°.③∵F是动点,CF的长度不是定值,本选项错误;④延长AD至点M,使AD=DM,过点C作CI∥HL,则:LI=HC,根据△MEC≌△CIM,可得:CE=IM,同理,可得:AL=HE,∴HE+HC+EC=AL+LI+IM=AM=12.∴△CEH的周长为12,为定值.故②④结论都正确.故答案为②④.三.解答题(共7小题)17.计算(1)(2)【分析】(1)根据二次根式的乘除法则运算;(2)利用平方差公式和二次根式的性质计算.【解答】解:(1)原式=1+=1+4=5;(2)原式=4﹣5﹣3=﹣4.18.随着移动互联网的快速发展,基于互联网的共享单车应运而生.为了解某单位使用共享单车的情况,该单位有200名员工,某研究小组随机采访10位员工,得到这10位员工一周内使用共享单车的次数分别为:17,12,15,20,17,0,7,26,17,9.(1)这组数据的中位数是16 ,众数是17(2)试用平均数估计该单位员工一周内使用共享单车的总次数.【分析】(1)将数据按照大小顺序重新排列,计算出中间两个数的平均数即是中位数,出现次数最多的即为众数;(2)根据平均数的概念,将所有数的和除以10即可;【解答】解:(1)按照大小顺序重新排列后,第5、第6个数分别是15和17,所以中位数是(15+17)÷2=16,17出现3次最多,所以众数是17,故答案是16,17;(2)×(0+7+9+12+15+17×3+20+26)=14,答:这10位居民一周内使用共享单车的平均次数是14次;19.在平面直角坐标系中,二次函数y=ax2+bx+3的图象经过点A(3,0)和点B(4,3).(1)求二次函数的表达式(2)求二次函数图象的顶点坐标和对称轴.【分析】(1)把A点和B点坐标代入y=ax2+bx+3得关于a、b的方程组,然后解方程组即可;(2)先把一般式配成顶点式,然后根据二次函数的性质解决问题;【解答】解:(1)∵抛物线y=ax2+bx+3经过点A(3,0)和点B(4,3).∴,解得,∴这条抛物线所对应的二次函数的表达式为y=x2﹣4x+3;(2)∵y═x2﹣4x+3=(x﹣2)2﹣1,∴抛物线顶点坐标为(2,﹣1),对称轴为x=2.20.已知关于x的一元二次方程x2﹣(m+2)x+2m=0(1)求证:不论m为何值,该方程总有两个实数根;(2)若此方程的一个根是1,请求出方程的另一个根,并求出以此两根为边长的直角角形的周长【分析】(1)由根的判别式△=(m﹣2)2≥0,可证出:不论m为何值,该方程总有两个实数根;(2)将x=1代入原方程可求出m的值,利用两根之积等于可求出方程的另一个根,再利用勾股定理及三角形的周长公式即可求出围成直角三角形的周长.【解答】(1)证明:△=[﹣(m+2)]2﹣4×1×2m=m2﹣4m+4=(m﹣2)2.∵(m﹣2)2≥0,即△≥0,∴不论m为何值,该方程总有两个实数根.(2)解:将x=1代入原方程,得:1﹣(m+2)+2m=0,∴m=1,∴方程的另一根为=2.∵=,∴围成直角三角形的周长=1+2+=3+.21.如图,在△ABC中,D、E分别是AB,AC的中点,BE=2DE,延长DE到点F,使得EF=BE,连接CF.(1)求证:四边形BCFE是菱形;(2)若CE=2,∠BEF=120°,求菱形BCFE的面积.【分析】(1)从所给的条件可知,DE是△ABC中位线,所以DE∥BC且2DE=BC,所以BC 和EF平行且相等,所以四边形BCFE是平行四边形,又因为BE=FE,所以是菱形;(2)∠BEF=120°,所以∠EBC=60°,所以菱形的边长也为2,求出菱形的高面积就可求.【解答】(1)证明:∵D、E分别是AB、AC的中点,∴DE∥BC且2DE=BC,又∵BE=2DE,EF=BE,∴EF=BC,EF∥BC,∴四边形BCFE是平行四边形,又∵BE=FE,∴四边形BCFE是菱形;(2)解:∵∠BEF=120°,∴∠EBC=60°,∴△EBC是等边三角形,∴菱形的边长为2,高为,∴菱形的面积为2×=2.22.直线y=3x与反比例函数y=的图象交于A(1,m)和点B.(1)求m、k的值,并直接写出点B的坐标(2)过点P(t,0)(﹣1≤t≤1且t≠0)作x轴的垂线分别交直线y=3x与反比例函数y=的图象于点E,F.①当t=时,求线段EF的长;②若0<EF≤8,请根据图象直接写出t的取值范围.【分析】(1)把A(1,m)代入y=3x求出m=3,得出A的坐标是(1,3),把A的坐标代入y=求出k,由中心对称的性质求得B的坐标即可;(2)①求得E、F的坐标,即可求得EF的长,②由图象可知当t=±时,EF=8,当t=±1时EF=0,即可得出t的取值范围.【解答】解:(1)把A(1,m)代入y=3x得:m=3,即A(1,3),把A的坐标代入y=得:3=,解得k=3,∵点B与点A关于原点对称,∴B(﹣1,﹣3);(2)①点P(t,0),∴点E的坐标为(t,3t),点F的坐标为(t,).①当k=时,则E(,1),F(,9),∴EF=9﹣1=8;②由函数的对称性可知当t=±时,EF=8,当t=±1时EF=0,∴若0<EF≤8,则t的取值范围是﹣1<t≤﹣或≤t<1.23.如图,在等腰△ABC中,AC=BC=3,AB=6,点E从点B沿着射线BA以每秒3个单位的速度运动,过点E作BC的平行线交∠ACB的外角平分线CF于点F.(1)求证:四边形BCFE是平行四边形;(2)当点E是边AB的中点时,连结AF,试判断四边形AECF的形状,并说明理由;(3)设运动时间为t秒,是否存在t的值,使得以△EFC的其中两边为边所构造的平行四边形恰好是菱形?若存在,请求出t的值;若不存在,试说明理由.【分析】(1)由等腰三角形的性质得:∠B=∠BAC,再由角平分线定义和三角形外角的性质可解答;(2)由有一个角是直角的平行四边形是矩形可解答;(3)分三种情况:①EF=CF;②CE=CF;②CE=EF;分别列方程可解答.【解答】证明:(1)如图1,∵AC=BC,∴∠B=∠BAC,∵CF平分∠ACH,∴∠ACF=∠FCH,∵∠ACH=∠B+∠BAC=∠ACF+∠FCH,∴∠FCH=∠B,∴BE∥CF,∵EF∥BC,∴四边形BCFE是平行四边形;(2)四边形AECF是矩形,理由是:∵E是AB的中点,AC=BC,∴CE⊥AB,∴∠AEC=90°,由(1)知:四边形BCFE是平行四边形,∴CF=BE=AE,∵AE∥CF,AE=CF,∴四边形AECF是平行四边形,且∠AEC=90°,∴四边形AECF是矩形;(3)①以EF和CF两边为邻边所构造的平行四边形恰好是菱形时,如图2,∴BE=BC,即3t=3,∴t=;②以CE和CF两边为邻边所构造的平行四边形恰好是菱形时,如图3,过C作CD⊥AB于D,连接GC,∵AC=BC=3,AB=6,∴BD=AD=3,由勾股定理得:CD===6,∵四边形CEGF是菱形,∴EF⊥GC,且EF∥BC,∴GC⊥BC,且∠EGC=∠ECG,∴∠EBC=∠ECB,∴BE=CE=3t,∵(3t)2=62+(3t﹣3)2,∴t=;③以CE和EF两边为邻边所构造的平行四边形恰好是菱形时,如图4,CA=AF=BC,此时E与A重合,∴t=2,综上所述,t的值为秒或秒或2秒;。

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学校 八年级( )班 姓名_______________ 座号__________……………………………装…………………………订…………………………线…………………………… 广元市利州中学2020年秋入学检测九年级 英语试题时间:120分钟 满分:150分 一.单项选择 从A 、B 、C 三个选项中选择最佳答案。

(每题1.5分,共30分) ( ) 1.—What did your mother say just now? —Sorry. I ______ to the weather report on TV . A. listen B. listened C. was listening D. have listened ( ) 2.— ________ have you learned to play the piano? — ________ five years ago. A. How long, For B. How often, For C. How long, Since D. How soon, Since ( ) 3.---He’s never been to an aquarium. ---_______. A. Neither have I B. Neither do I C. I never too D. I don’t too ( ) 4.He is very angry with his mother, because his mother always him with the best student. A. considers B. competes C. continues D. compares ( ) 5. It is a(n) ________ for a young man like you not to study hard. A. feeling B. shame C. deal D. opinion ( ) 6.--Could you please _____ in the room? --Oh, I’m sorry. A. not smoking B. not smoke C. not to smoke D. don’t smoke ( ) 7. I have __ _ decided when to leave, but my sister hasn’t decided __ _. A. yet, yet B. already, already C. already, yet D. yet, already ( ) 8.This is ______ music I ’ve ever heard. I enjoy it very much. A. beautiful B. more beautiful C. the most beautifulD. the least beautiful ( ) 9. Bob is always polite to everyone, and he is lovely all of us like him very much.A. too, toB. enough, toC.so, thatD. such, that( ) 10.She _______ a lot of money on clothes.A. takesB. costsC. buysD. spends( ) 11.I will stay at home and read books if it _______ tomorrow.A. is rainingB. rainsC. rainD. will rain( ) 12.—I ________ a new house recently.—Great. When ________ you ________ it?A. have bought, did, buyB. have bought, have, boughtC. bought, did, buyD. bought, have, bought( ) 13. Every year ________ people take the train to go back home before the Spring Festival.A. five millionsB. five million ofC. million ofD. millions of( ) 14. My mother used to the song “Where’s time going?” . Now she is used to it.A. hating; listenB. hated; listened toC. hate; listening toD. hating; listening( ) 15. --- Could you please help me the exercise books to the class?--- No problem.A. take outB. give outC.work outD. put out( ) 16. Mr. Jack ________ China for several years.A. has been toB. has come toC. has been inD. has gone to( ) 17. —Jim takes _______ his father.—You mean he is similar _______ his father?A. for; asB. to; withC. after; toD. away; from( ) 18. It’s time for the meeting. Let’s _______ the meeting.A. stop havingB. stop to haveC. stoppingD. stopping to have( ) 19. Don’t talk about this to anyone, ________ not to your mother, or she will be worried.A. actuallyB. simplyC. suddenlyD. especially( ) 20. —I’m going to see the Terracotta Army in Xi’an.—Great!_____________.A. You’re welcomeB. That’s all rightC. Never mindD. Have a good trip二.完形填空选择最佳答案。

(每题1.5分,共30分)(A)Nick, an Australian man, was born without arms or legs, so life was not easy for him. At school many students laughed at him because he looked different 21 everyone else. His classmates refused to be his friends, so he always felt 22 . However, he faced that. He learnedto read and write at the age of six, and he could 23 surf(冲浪). He became very successful in college(大学) and was among the 24 students. And he decided on what to do later in his life —to encourage others to work hard for 25 dreams.Now Nick is one of the most popular 26 in the world. He traveled to many countries and give speeches about his 27 against difficulty. “Living a life fully is to look at what you 28 , not what you don’t have,”he said. His story encourages many people.“I tell people to keep on getting up 29 they fall every time”he said. “If I can 30 just one person, then my job in this life works.”( ) 21. A. as B. from C. for D. with( ) 22. A. tired B. relaxed C. excited D. lonely( ) 23. A. never B. even C. still D. just( ) 24. A. best B. funniest C. richest D. youngest( ) 25. A. his B. her C. our D. their( ) 26. A. actors B. singers C. speakers D. teachers( ) 27. A. family B. story C. plan D. travel( ) 28. A. have B. make C. forget D. remember( ) 29. A. because B. before C. until D. when( ) 30. A. change B. turn C. call D. excuse(B)Do you write your own blog(博客)? Do you often 31 other’s blogs on the Internet?All over the world, not only famous persons but also many students use blogs as diaries and places to share their experiences and 32 advice or help.“ I write my activities every day,33 funny things and worries in my life.” says Lili, a high school student in Washington. “ I like putting pictures on my blog because it can make my blog more 34 and more people will like it. It’s a good way 35 my life to the world.” She has had her blog for eight months.Students like writing blogs 36 they can write anything they want to, but don’t need to 37 any money.When you start to write your blog, you will find 38 necessary to open up your eyes to a new world. But what you write down may trouble you 39 . So use the blog carefully 40 you can keep yourself away from some trouble. The blog makes everyone’s life colorful.( ) 31. A. read B. write C. forget D. miss( ) 32. A. learn from B. hear of C. ask for D. depend on( ) 33. A. without B. including C. between D.however( ) 34. A. common B. modern C. famous D.interesting( ) 35. A. to show B. showing C. show D.showed( ) 36. A. unless B. while C. until D. because( ) 37. A. save B. spend C. take D. stole( ) 38. A. that B. it C. one D. those( ) 39. A. sometimes B. always C. ever D.never( ) 40. A. only when B. even though C. so that D. ever since三.阅读理解(每小题2分,共50分)(A)Rosa is an American school girl. She is very happy because her parents will take her to China for a visit. They are going to fly from New York on July 3 and get to Beijing on July 4. They will stay at Lido Holiday Inn. Their Chinese friends will show them around Beijing. They are going to visit the Palace Museum, the Summer Palace, the Temple of Heaven and the Great Wall.They will also visit some Chinese homes and try some Chinese food. They will spend a week in Beijing and then go to Xi’an, Guilin and some other places for sightseeing.The whole trip will take three weeks and she is sure she will have a great time in China.( )41. Rosa is very happy because she will _____ with her parents.A. visit ChinaB. go homeC. watch TVD. go to New York( )42. Their _____ friends will show them around Beijing.A. AustralianB. ChineseC. AmericanD. England( )43. From the text we can know that they will ______ in Beijing.A. cook Chinese foodB. learn to speak ChineseC. try some Chinese foodD. dance ( )44. After they visit Beijing, they will go to _______.A. Xi’an a nd GuilinB. The Summer PalaceC. ShanghaiD. the Great Wall. ( )45. How long will the whole trip take?A. One weekB. Two weeksC. Three weeksD. A month.(B)Today almost everyone knows computers and the Internet. If I ask you “What is the most important in your life?”maybe you will say “Computers and the Internet.”The first computer was made in 1946. it was very big but it worked slowly. Today computersare getting smaller and smaller. But they work faster and faster. What can computers do? A writer has said, “People can’t live without computers today.”The Internet came a little later than computers. It is about twenty-five years later than computers. But now it can be found almost everywhere. We can use it to read books, writer letters, do shopping, play computer games or make friends.Many students like the Internet very much. They often go into the Internet as soon as they are free. They make friends on the Internet and maybe they have never seen these friends. They don’t know their real names, ages, and even sex. They are so interested in making the “unreal friends”that they can’t put their hearts into study. Many of them can’t catch up with others on many subjects because of that.We can use computers and the Internet to learn more about the world. But at the same time, we should remember that not all the things can be done by computers and the Internet.( ) 46. The Internet can not be used for______.A. studyingB. shoppingC. thinkingD. playing( ) 47. When the computer was invented, it was______.A. large and worked quicklyB. small and worked slowlyC. large but worked slowlyD. small but worked quickly( ) 48. The Internet was born in about________.A. 1960B. 1970C. 1980D. 1985( ) 49. Which of the following is TRUE?A. Few students like going into the Internet.B. These “Unreal friends”often meet each otherC. Students use the Internet to make “unreal friends.”D. Students know the friends on the Internet very well( ) 50. What does the writer think of the Internet?A. It is wonderfulB. It can make students study harderC. It is not good for studentsD. It is helpful, but we can’t do everything on it.(C)Sometimes we feel good, such as when we get good grades or when something good happens. Sometimes we feel bad, such as when we argue with our classmates or when the teachers criticize (批评)us. Both kinds of feelings are common. Of course, we all enjoy good feelings, but when we feel unhappy, sometimes we don't know what to do about it. Here is a plan that will help you whenyour feelings are bothering(困扰)you.1. ACCEPT your feelings. Say to yourself, "I am sad. " "I am scared. " "I am angry. " "I am embarrassed. " It's all right to have these feelings.2. RELAX and have a rest before you act. Take slow deep breaths (呼吸) and relax your body. Imagine that you are in a safe place. You can listen to some music or sing a beautiful song !3. THINK about ways to help yourself. Thinking helps you do something smart instead of hurting yourself or making things worse.4. DO something to help yourself. Maybe you can talk to someone, or do something you enjoy. If it doesn't work, go back to Step 3.()51. We feel unhappy when .A. we get good gradesB. everything goes wellC. someone is proud of usD. we argue with our classmates()52. If you want to relax yourself before you act, you can .A. say to yourself, "I am sad."B. listen to some music or sing a beautiful songC. think about ways to help yourselfD. watch thrillers with your friends()53. The underlined part (画线部分) "If it doesn't work," in the passage probably means .A. If it is brokenB. If it can't interest youC. If it is wrongD. If it can't help you()54. Which of the following is NOT TRUE according to the passage?A. We felt bad when we lost our expensive things.B.We felt good when we won the prize in the competition.C. We can always say to ourselves that makes us happy when we feel bad.D. We can tell someone about our bad feelings when we are sad.()55. The passage tells us .A. what we can do when we feel unhappyB. we can do nothing when we feel badC. we all enjoy good feelings, and bad feelings are unusual to usD. what we can do when we feel happy(D)If you have no special plans for your holidays, why not spend your time helping others?Don’t listen to people who say the young people today often think only of themselves. It’s not true at all. More and more people in the USA volunteer to spend their school holidays working for others.A nd they don’t do it for money. Here are some of their stories.Alice Hamilton, 17I’m going to help the Forest Centre build new hiking paths in the mountains. It’s going to beterrific—I’ll spend the whole summer living in the tent and breathing the clean mountain air. I’m going to sleep under the stars. It’ll be a nice sport, and I’ll be able to do something good at the same time.Jason Moore,18This summer, I’m going to volunteer with Special House Program. They build good, low-cost houses and sell t hem to families that are not very rich. They’ll teach me what to do, so I’ll help people and also have a chance to learn how to build houses.Trish Anderson, 16I’m going to teach kids who have trouble reading. I’ll work for a program called Reading For Life.Every d ay, I’m going to help kids choose and read books that they like. I want to be a teacher and I love children and reading, so this is going to be a great experience for me.( )56. Young people in the USA volunteer to spend ___ helping others.A. school holidaysB. WeekdaysC. every morningD. every afternoon( )57. What does Alice volunteer to do in summer?A. To breathe the clean mountain airB. To spend the whole summer living in the tentC. To sleep under the starsD. To help build new hiking paths in the mountain ( )58. From the passage, we know that Special House Program helps .A. children in poor familiesB. workers who build housesC. poor families can’t afford ho usesD. build hiking paths( )59.______ wants to be a teacher.A.Alice HamiltonB. Jason MooreC. Trish AndersonD. none of them( )60. Which is the most possible title for this passage?A. Reading for lifeB. A Special House ProgramC. The Forest CentreD. Young American Volunteers(E)“Cool” is a word with many meanings. Its old meaning is used to express a temperature that is a little bit cold. As the world has changed, the word has had many different meanings. “Cool” can be used to express feelings of interest in almost anything.When you see a famous car in the street, maybe you will say, “It’s cool.” You may think, “He’s so cool,” wh en you see your favourite footballer. We all maximize (扩大) the meaning of “cool”. You can use it instead of many words such as “new” or “surprising”. Here’s an interesting story we can use to show the way the word is used. A teacher asked her students to write about the waterfall (瀑布) they had visited. On one student’s paper was just the one sentence, “It’s so cool.” Maybe he thought it was the best way to show what he saw and felt.But the story also shows a scarcity(缺乏) of words. Without “cool”, some people have no words to show the same meaning. So it is quite important to keep some credibility(可信性). Can you think of many other words that make your life as colourful as the word “cool”? I can. And I think they are also very cool.( ) 61. We know that the word “cool” has had ______.A. only one meaningB. no meaningsC. many different meaningsD. the same meaning( ) 62. In the passage, the word “express” means “______.”A. seeB. showC. knowD. feel( ) 63. If you are ______ somethi ng, you may say, “It’s cool.”A. interested inB. angry aboutC. afraid ofD. unhappy with( ) 64. The writer takes an example to show he is ______ the way the word is used.A. pleased withB. strange toC. worried aboutD. careful with( ) 65. In the passage, the writer suggests(暗示) that the word “cool” ______.A. can be used instead of many wordsB. usually means something interestingC. can make your life colourfulD. may not be as cool as it seems四.情景交际(每小题1分,共5分)根据对话内容,从方框中选择恰当的选项补全对话,其中有两项多余。

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