九年级上学期期末复习试题
九年级数学上册期末复习综合测试题(含答案)
(第4题)九年级数学上册期末复习综合测试题(含答案)一、选择题(本大题共6小题,每小题2分,共12分.) 1.一元二次方程 x 2=x 的根是( )A .x 1=0,x 2=1B .x 1=0,x 2=-1C .x 1=x 2=0D .x 1=x 2=12.一个不透明布袋中有2个红球,3个白球,这些球除颜色外无其他差别,摇匀后从中随机摸出一个小球,该小球是红色的概率为( )A .12B .23C .15D .253.若一组数据 2,3,4,5,x 的方差比另一组数据 5,6,7,8,9 的方差大,则 x 的值可能是( ) A .1B .4C .6D .84.如图,OA 、OB 是⊙O 的半径,C 是⊙O 上一点.若∠OAC =16°,∠OBC =54°,则 ∠AOB 的度数是( )A .70°B .72°C .74°D .76°5.若关于x 的一元二次方程ax 2+k =0的一个根为2,则二次函数y =a (x +1)2+k 与x 轴的交点坐标为( ) A .(-3,0)、(1,0) B .(-2,0)、(2,0) C .(-1,0)、(1,0)D .(-1,0)、(3,0)6.如图,在Rt △ABC ,∠ACB =90°,AC =4,BC =3,点D ,E 分别在AB ,AC 上,连接DE ,将△ADE 沿DE 翻折,使点A 的对应点F 落在BC 的延长线上,若FD 平分∠EFB ,则AD 的长为( ) A . 157B .207C .258D .259二、填空题(本大题共10小题,每小题2分,共20分.) 7(第12题)l 1 l 2l 3A BCEFD (第11题)8.若a b =43,则a -b b= .9.设x 1、x 2是方程x 2+mx -m +3=0的两个根,则x 1+x 2-x 1x 2= .10.把抛物线y =-x 2向左平移2个单位,然后向上平移3个单位,则平移后该抛物线相应的函数表达式为 .11.如图,l 1∥l 2∥l 3,若AD =1,BE =3,CF =6,则ABBC的值为 .12.如图,点A 、B 、C 在⊙O 上,⊙O 的半径为3,∠AOC =的长为 . 13.已知关于x 的函数y =x 2+2mx +1,若x >1时,y 随x 的增大而增大,则m 的取值范围是 .14.如图,弦AB 是⊙O 的内接正六边形的一边,弦AC 是⊙O 的内接正方形的一边,若 BC =2+23,则⊙O 的半径为 .15.如图,正方形ABCD 的边长是4,点E 在DC 上,点F 在AC 上,∠BFE =90°,若 CE =116.如图,在矩形ABCD 中,AB =2,AD =4,点E 、F 分别为AD 、CD 边上的点,且EF 的长为2,点G 为EF 的中点,点P 为BC 上一动点,则P A +PG 的最小值为 . 三、解答题(本大题共11小题,共88分.请在答题卡指定区域.......内作答,解答时应写出文字说明、证明过程或演算步骤)17.(8分)解方程:(1)x 2-4x -5=0; (2)x 2-4=2x (x -2).18.(8分)甲乙两人在相同条件下完成了5次射击训练,两人的成绩(单位:环)如下(1)甲射击成绩的中位数为 环,乙射击成绩的众数为 环;(2)计算两人射击成绩的方差;(3)根据训练成绩,你认为选派哪一名队员参赛更好,为什么?19.(8分)某校开展秋季运动会,需运动员代表进行发言,从甲、乙、丙、丁四名运动员中随机抽取.(1)若随机抽取1名,甲被抽中的概率为 ; (2)若随机抽取2名,求甲在其中的概率.20.(7分)如图,在△ABC 中,点D 、E 分别在AB 、AC 上,且∠BCE +∠BDE =180°. (1)求证:△ADE ∽△ACB ;(2)连接BE 、CD ,求证:△AEB ∽△ADC .21.(8分)如图是二次函数y =-x 2+bx +c 的图像. (1)求该二次函数的关系式及顶点坐标; (2)当y >0时 x 的取值范围是 ;(3)当m <x <m +4时,-5<y ≤4,则m 的值为 .22.(7分)在Rt △ABC ,∠BAC =90°,AB =AC ,D 、E、F 分别为BC 、AB 、AC 边上的点,且∠EDF =45°.(1)求证:△EBD ∽△DCF ;(2)当D 是BC 的中点时,连接EF ,若CF =5,DF =4,则EF 的长为 .23.(8分)某超市销售一种商品,成本为每千克50元.当每千克售价60元时,每天的销售量为60千克,经市场调查,当每千克售价增加1元,每天的销售量减少2千克. (1)为保证某天获得750元的销售利润,则该天的销售单价应定为多少? (2)当销售单价定为多少时,才能使当天的销售利润最大?最大利润是多少?24.(8分)如图,AB 为⊙O 的直径,弦CD ⊥AB 于点P ,连接BC ,过点D 作DE ⊥CD ,交⊙O 于点E ,连接AE ,F 是DE 延长线上一点,且∠BCD =∠F AE . (1)求证:AF 是⊙O 的切线;(2)若AF =2,EF =1,求⊙O 的半径.25.(8分)已知二次函数y =(x -2)(x -m )(m 为常数). (1)求证:不论m 为何值,该函数的图像与x 轴总有公共点;(2)若M (-1,0), N (3,0),该函数图像与线段MN 只有1个公共点,直接写出 m 的取值范围;(3)若点A (-1,a ),B (1,b ),C (3,c )在该函数的图像上,当abc <0时,结合函数图像,直接写出m 的取值范围.26.(8分)如图,四边形ABCD 内接于⊙O ,AB =AC ,BD ⊥AC ,垂足为E . (1)求证:∠BAC =2∠DAC ; (2)若AB =10,CD =5,求BC 的长.27.(10分)定义:圆心在三角形的一边上,与另一边相切,且经过三角形一个顶点(非切点)的圆,称为这个三角形圆心所在边上的“伴随圆”.(1) 如图①,在△ABC 中,∠C =90°,AB =5,AC =3,则BC 边上的伴随圆的半径为 . (2)如图②,△ABC 中,AB =AC =5,BC =6,直接写出它的所有伴随圆的半径. (3)如图③,△ABC 中,∠ACB =90°,点E 在边AB 上,AE =2BE ,D 为AC 的中点,且∠CED =90°.①求证:△CED 的外接圆是△ABC 的AC 边上的伴随圆; ②DE的值为 .参考答案说明:本评分标准每题给出了一种或几种解法供参考,如果考生的解法与本解答不同,参照本评分标准的精神给分.一、选择题(本大题共6小题,每小题2分,共12分)二、填空题(本大题共10小题,每小题2分,共20分)7.9 8.13 9.-3 10.y =-(x +2)2+3 11.2312.2π 13.m ≥-1 14. 2 2 15.322 16.4 2 -1三、解答题(本大题共11小题,共88分) 17.(8分)(1)解:x 2-4x -5=0 x 2-4x +4=5+4(x -2)2=9 ········································································································ 1分x -2=±3 ········································································································ 2分 ∴ x 1=5,x 2=-1. ··························································································· 4分 (2)解:x 2-4=2x (x -2) x 2-4=2x 2-4xx 2-4x +4=0 ··································································································· 5分 (x -2)2=0 ········································································································ 6分 ∴ x 1=x 2=2. ··································································································· 8分 18.(8分)(1)7;8 ········································································································ 2分 (2)s 2甲=(7-8)2+(7-8) 2+(10-8)2+(9-8)2+(7-8)25=1.6环2. ······························ 4分s 2乙=(8-8)2+(8-8) 2+ (7-8)2+(8-8)2+(9-8)25=0.4环2. ······································ 6分(3)选择乙.因为甲乙两人平均数相同均为8,说明两人实力相当,但s 2乙<s 2甲,乙的成绩更加稳定,所以选乙. ······················································································· 8分19.(8分)(1)14. ·········································································································· 2分(2)解:随机抽取两名运动员,共有6种等可能性结果:(甲,乙)、(甲,丙)、(甲,丁)、(乙,丙)、(乙,丁)、(丙,丁).其中满足“有甲运动员”(记为事件A )的结果只有3种,所以P (A )=12. ·································································································· 8分20.(7分)(1)证明:∵ ∠BCE +∠BDE =180°, ∠EDA +∠BDE =180°,∴ ∠EDA =∠BCE . ·························································································· 1分 又 ∠A =∠A , ································································································· 2分 ∴ △ADE ∽△ACB . ·························································································· 3分 (2)∵ △ADE ∽△ACB , ∴ AD AC =AE AB, ·········································· 4分 ∴AD AE =ACAB, ······································· 5分 又 ∠A =∠A , ········································ 6分 ∴ △AEB ∽△ADC . ································· 7分21.(8分)(1)将(0,3)、 (3,0)代入,得⎩⎨⎧3=c ,0=-9+3b +c································································································· 1分解得⎩⎨⎧c =3,b =2····································································································· 2分∴ y =-x 2+2x +3 ····························································································· 3分 ∴ 顶点坐标为(1,4) ························································································ 4分 (2)-1<x <3. ······························································································ 6分 (3)-2或0 ···································································································· 8分 22.(7分)(1)解:∵∠BAC =90°,AB =AC ,∴ ∠B =∠C =45°. ··························································································· 1分 ∴ 在△BDE 中,∠BED +∠BDE =180°-∠B =135°, ∵ ∠EDF =45°,∴ ∠BDE +∠CDF =135°,∴ ∠BED =∠CDF . ·························································································· 3分 ∵ ∠B =∠C ,∴ △EBD ∽△DCF . ·························································································· 5分 (2 ········································································································ 7分23.(8分)(1)解:设每千克的销售价增加x 元,根据题意,得(60+x -50) (60-2x )=750 ··················································································· 2分 ∴ x 1=5,x 2=15. ····························································································· 3分 60+5=65或60+15=75 ···················································································· 4分 答:销售单价为65或75元时获得利润750元. (2)解:每千克的销售价增加x 元,利润为w 元.w =(60+x -50) (60-2x ) ···················································································· 6分 =-2(x -10)2+800 ···························································································· 7分 ∵ a =-2<0,∴ 当x =10时,w 有最大值800. ········································································ 8分 60+10=70答:当销售单价为70元时获得最大利润,为800元. 24.(8分) (1)连接BD .∵ AB 为⊙O 的直径,CD ⊥AB ,∴ ⌒BC = ⌒BD , ························································· 1分 ∴ ∠BDC =∠BCD .∵ 四边形ABDE 为⊙O 的内接四边形,∴ ∠BDE +∠BAE =180°,即∠BDC +∠CDF +∠BAE ····· 2分∵ DE ⊥CD , ∴ ∠CDF =90°, ∴ ∠BDC +∠BAE =90°.∵ ∠BCD =∠F AE , ·························································································· 3分 ∴ ∠BAE +∠F AE =90°,即∠F AB =90°, ∴ AF ⊥AB . 又 点A 在⊙O 上,∴ AF 与⊙O 相切. ·························································································· 4分 (2)过点O 作OG ⊥DF 垂足为G . ∵ ∠F AB =∠D =∠APD =90°, ∴ 四边形APDF 是矩形, ∴ ∠F =90°.∵ ∠F AB =∠F =∠OGF =90°, ∴ 四边形AOGF 是矩形,∴ AF =OG ,AO =GF . ···················································· 5分 设OE =OA =r ,则GE =r -1.在Rt △OGE 中,由勾股定理得OG 2+GE 2=OE 2, ···················································· 6分 即4+(r -1)2=r 2, ···························································································· 7分 解得r =5 2 . ····································································································· 8分25.(8分)(1)令y =0,即(x -2)(x -m )=0 ········································································· 1分 ∴ x 1=2,x 2=m . ····························································································· 2分 当m =2时,x 1=x 2,方程有两个相等的实数根; 当m ≠2时,x 1≠x 2,方程有两个不等的实数根. ∴ 不论m 为何值,方程总有实数根;∴ 不论m 为何值,该函数的图像与x 轴总有公共点. ·············································· 3分 (2)m =2或m >3或m <-1. ··········································································· 6分 (3)-1<m <1或m >3. ·················································································· 8分 26.(8分)。
北京市西城区月坛中学2025届九年级数学第一学期期末复习检测试题含解析
北京市西城区月坛中学2025届九年级数学第一学期期末复习检测试题注意事项1.考生要认真填写考场号和座位序号。
2.试题所有答案必须填涂或书写在答题卡上,在试卷上作答无效。
第一部分必须用2B 铅笔作答;第二部分必须用黑色字迹的签字笔作答。
3.考试结束后,考生须将试卷和答题卡放在桌面上,待监考员收回。
一、选择题(每题4分,共48分)1.一个扇形的半径为4,弧长为2π,其圆心角度数是( ) A .45B .60C .90D .1802.如图,四边形ABCD 是菱形,∠A=60°,AB=2,扇形BEF 的半径为2,圆心角为60°,则图中阴影部分的面积是( )A .2332π-B .233π- C .32π-D .3π-3.如图,一段抛物线26 (0)6y x x x =-+≤≤,记为抛物线1C ,它与x 轴交于点1O A 、;将抛物线1C 绕点1A 旋转180︒得抛物线2C ,交x 轴于点2A ;将抛物线2C 绕点2A 旋转180︒得抛物线3C ,交x 轴于点3A .···如此进行下去,得到一条“波浪线”,若点()2020,M m 在此“波浪线”上,则m 的值为( )A .6-B .6C .8-D .84.如图,在莲花山滑雪场滑雪,需从山脚下乘缆车上山,缆车索道与水平线所成的角为31︒,缆车速度为每分钟40米,从山脚下A 到达山顶B 缆车需要15分钟,则山的高度BC 为( )米.A .60031tan ⋅︒B .60031tan ︒C .60031sin ⋅︒D .600sin 31︒5.下列语句中正确的是( )A .长度相等的两条弧是等弧B .平分弦的直径垂直于弦C .相等的圆心角所对的弧相等D .经过圆心的每一条直线都是圆的对称轴6.已知23a b=(a≠0,b≠0),下列变形错误的是( ) A .23a b = B .2a=3b C .32b a =D .3a=2b7.一条排水管的截面如图所示,已知排水管的半径5OB =,水面宽8AB =,则截面圆心O 到水面的距离OC 是( )A .2B .3C .23D .2.58.如图,正方形ABCD 中,点E 、F 分别在边CD ,AD 上,BE 与CF 交于点G .若4BC =,1DE AF ==,则GF 的长为( )A .135B .125C .195D .1659.如图,AB 切⊙O 于点B ,C 为⊙O 上一点,且OC ⊥OA ,CB 与OA 交于点D ,若∠OCB =15°,AB =3,则⊙O 的半径为( )A .3B .2C .3D .410.在△ABC 中,点D 、E 分别在边AB 、AC 上,DE ∥BC ,AD :DB =4:5,下列结论中正确的是 A .45DE BC = B .94BC DE = C .45AE AC = D .54EC AC = 11.如图,已知AB 、AC 都是⊙O 的弦,OM ⊥AB ,ON ⊥AC ,垂足分别为M ,N ,若MN =5,那么BC 等于( )A .5B 5C .5D 1012.一元二次方程x 2﹣x ﹣2=0的解是( ) A .x 1=﹣1,x 2=﹣2 B .x 1=1,x 2=﹣2 C .x 1=1,x 2=2 D .x 1=﹣1,x 2=2二、填空题(每题4分,共24分)13.以原点O 为位似中心,作△ABC 的位似图形△A ′B ′C ′,△ABC 与△A ′B ′C ′相似比为13,若点C 的坐标为(4,1),点C 的对应点为C ′,则点C ′的坐标为_____.14.已知二次函数2y ax bx c =++(0a ≠),y 与x 的部分对应值如下表所示:x-1 0 1 2 3 4y61-2-3-2m下面有四个论断:①抛物线2y ax bx c =++(0a ≠)的顶点为(2,3)-;②3m =-;③关于x 的方程22ax bx c ++=-的解为11x =,23x =;④当0.5x =-时,y 的值为正,其中正确的有_______.15.当_____时,11x在实数范围内有意义.16.已知P是线段AB的黄金分割点,PA>PB,AB=2cm,则PA为___cm.17.在一个不透明的袋子中有10个除颜色外均相同的小球,通过多次摸球试验后,发现摸到白球的概率约为30%,估计袋中白球有个.18.从1,2,3三个数字中任取两个不同的数字,其和是奇数的概率是_________.三、解答题(共78分)19.(8分)数学活动课上老师带领全班学生测量旗杆高度.如图垂直于地面的旗杆顶端A垂下一根绳子.小明同学将绳子拉直钉在地上,绳子末端恰好在点C处且测得旗杆顶端A的仰角为75°;小亮同学接着拿起绳子末端向前至D处,拉直绳子,此时测得绳子末端E距离地面1.5 m且与旗杆顶端A的仰角为60°根据两位同学的测量数据,求旗杆AB 的高度.(参考数据:sin75°≈0.97,cos75°≈0.26,sin60°≈0.87,结果精确到1米)20.(8分)如图,在□ABCD中,E是AD的中点,延长CB到点F,使BF=12BC,连接BE、AF.(1)求证:四边形AFBE是平行四边形;(2)若AB=6,AD=8,∠C=60°,求BE的长.21.(8分)如图,已知抛物线y=﹣x2+bx+c与x轴交于A、B两点,交y轴于点C,已知A(﹣1,0)对称轴是直线x =1.(1)求抛物线的解析式及点C的坐标;(2)动点M从点O出发,以每秒2个单位长度的速度向点B运动,过M作x轴的垂线交抛物线于点N,交线段BC 于点Q.设运动时间为t(t>0)秒.①若AOC与BMN相似,请求出t的值;②BOQ能否为等腰三角形?若能,求出t的值.22.(10分)如图,正方形ABCD 中,112, 4AB AE AB ==,点P 在BC 上运动(不与,B C 重台),过点P 作PQ EP ⊥,交CD 于点Q ,求P 运动到BP 多长时,CQ 有最大值,并求出最大值.23.(10分)如图,已知抛物线25y ax bx =+-()0a ≠与x 轴相交于A 、B 两点,与y 轴相交于C 点,对称轴为1x =-,直线3y x =-+与抛物线相交于A 、D 两点.(1)求此抛物线的解析式;(2)P 为抛物线上一动点,且位于3y x =-+的下方,求出ADP ∆面积的最大值及此时点P 的坐标; (3)设点Q 在y 轴上,且满足OQA OCA CBA ∠+∠=∠,求CQ 的长.24.(10分)为深化课改,落实立德树人目标,某学校设置了以下四门拓展性课程:A .数学思维,B .文学鉴赏,C .红船课程,D .3D 打印,规定每位学生选报一门.为了解学生的报名情况,随机抽取了部分学生进行调查,并制作成如下两幅不完整的统计图,请回答下列问题:(1)求这次被调查的学生人数; (2)请将条形统计图补充完整;(3)假如全校有学生1000人,请估计选报“红船课程”的学生人数.25.(12分)孝感商场计划在春节前50天里销售某品牌麻糖,其进价为18元/盒.设第x 天的销售价格为y (元/盒),销售量为m (盒).该商场根据以往的销售经验得出以下的销售规律:①当130x ≤≤时,38y =;当3150x ≤≤时,y 与x 满足一次函数关系,且当36x =时,37y =;40x =时,35y =.②m 与x 的关系为330m x =+.(1)当3150x ≤≤时,y 与x 的关系式为 ;(2)x 为多少时,当天的销售利润W (元)最大?最大利润为多少? 26.在平面直角坐标系xOy 中,抛物线y =x 2﹣2mx+m 2﹣1. (1)求抛物线顶点C 的坐标(用含m 的代数式表示);(2)已知点A (0,3),B (2,3),若该抛物线与线段AB 有公共点,结合函数图象,求出m 的取值范围.参考答案一、选择题(每题4分,共48分) 1、C【分析】根据弧长公式即可求出圆心角的度数.【详解】解:∵扇形的半径为4,弧长为2π, ∴42180n ππ⨯=解得:90n =,即其圆心角度数是90︒ 故选C . 【点睛】此题考查的是根据弧长和半径求圆心角的度数,掌握弧长公式是解决此题的关键. 2、B【分析】根据菱形的性质得出△DAB 是等边三角形,进而利用全等三角形的判定得出△ABG ≌△DBH ,得出四边形GBHD 的面积等于△ABD 的面积,进而求出即可. 【详解】连接BD ,∵四边形ABCD 是菱形,∠A=60°, ∴∠ADC=120°, ∴∠1=∠2=60°, ∴△DAB 是等边三角形, ∵AB=2,∴△ABD 3,∵扇形BEF 的半径为2,圆心角为60°, ∴∠4+∠5=60°,∠3+∠5=60°, ∴∠3=∠4,设AD 、BE 相交于点G ,设BF 、DC 相交于点H , 在△ABG 和△DBH 中,2{34A AB BD ∠=∠=∠=∠, ∴△ABG ≌△DBH (ASA ),∴四边形GBHD 的面积等于△ABD 的面积,∴图中阴影部分的面积是:S 扇形EBF -S △ABD =2602123602π⨯-⨯=23π故选B . 3、D【分析】根据图象的旋转变化规律以及二次函数的平移规律得出平移后解析式,进而求出m 的值. 【详解】∵一段抛物线:26 (0)6y x x x =-+≤≤, ∴图象与x 轴交点坐标为:(0,0),(6,0), ∵将C 1绕点A 1旋转180°得C 2,交x 轴于点A 2; 将C 2绕点A 2旋转180°得C 3,交x 轴于点A 3; ……如此进行下去,直至得C n .∴C n 的与x 轴的交点横坐标为(6n ,0),(6n+3,0), ∴()2020,M m 在C 337,且图象在x 轴上方, ∴C 337的解析式为:()()33720162022y x x =---, 当2020x =时,()()20202016202020228y =---=. 即8m =, 故答案为D. 【点睛】此题主要考查了二次函数的平移规律,根据已知得出二次函数旋转后解析式是解题关键. 4、C【分析】在Rt ABC ∆中,利用∠BAC 的正弦解答即可.【详解】解:在Rt ABC ∆中,90ACB ∠=︒,31BAC ∠=︒,4015600AB =⨯=(米), ∵sin BCBAC AB∠=,sin 600sin31BC BAC AB ∴=∠⋅=⋅︒(米). 故选C . 【点睛】本题考查了三角函数的应用,属于基础题型,熟练掌握三角函数的定义是解题的关键. 5、D【解析】分析:根据垂径定理及逆定理以及圆的性质来进行判定分析即可得出答案.详解:A 、在同圆或等圆中,长度相等的两条弧是等弧;B 、平分弦(不是直径)的直径垂直于弦;C 、在同圆或等圆中,相等的圆心角所对的弧相等;D 、经过圆心的每一条直线都是圆的对称轴;故选D .点睛:本题主要考查的是圆的一些基本性质,属于基础题型.理解圆的性质是解决这个问题的关键. 6、B【分析】根据两内项之积等于两外项之积对各选项分析判断即可得解. 【详解】解:由23a b=得,3a=2b , A 、由等式性质可得:3a=2b ,正确; B 、由等式性质可得2a=3b ,错误; C 、由等式性质可得:3a=2b ,正确; D 、由等式性质可得:3a=2b ,正确; 故选B . 【点睛】本题考查了比例的性质,主要利用了两内项之积等于两外项之积. 7、B【解析】根据垂径定理求出BC ,根据勾股定理求出OC 即可. 【详解】解:OC AB ⊥,OC 过圆心O 点,118422BC AC AB ∴===⨯=,在Rt OCB ∆中,由勾股定理得:3OC ==, 故选:B . 【点睛】本题考查了勾股定理和垂径定理的应用;由垂径定理求出BC 是解决问题的关键. 8、A【分析】根据正方形的性质以及勾股定理求得5BE CF ==,证明BCE CDF ∆≅∆,根据全等三角形的性质可得CBE DCF ∠=∠,继而根据cos cos BC CGCBE ECG BE CE∠=∠==,可求得CG 的长,进而根据GF CF CG =-即可求得答案.【详解】∵四边形ABCD 是正方形,4BC =, ∴4BC CD AD ===,90BCE CDF ∠=∠=︒, ∵1AF DE ==, ∴3DF CE ==,∴5BE CF ===, 在BCE ∆和CDF ∆中,BC CD BCE CDF CE DF =⎧⎪∠=∠⎨⎪=⎩, ∴()BCE CDF SAS ∆≅∆, ∴CBE DCF ∠=∠,∵90CBE CEB ECG CEB CGE ∠+∠=∠+∠=︒=∠,cos cos BC CGCBE ECG BE CE∠=∠==, ∴453CG =,125CG =, ∴1213555GF CF CG =-=-=, 故选A. 【点睛】本题考查了正方形的性质,勾股定理,全等三角形的判定与性质,三角函数等知识,综合性较强,熟练掌握和灵活运用相关知识是解题的关键.注意数形结合思想的运用. 9、B【分析】连接OB ,由切线的性质可得∠OBA=90°,结合已知条件可求出∠A=30°,因为AB 的长已知,所以⊙O 的半径可求出. 【详解】连接OB , ∵AB 切⊙O 于点B , ∴OB ⊥AB , ∴∠ABO =90°,∵OC ⊥OA ,∠OCB =15°, ∴∠CDO =∠ADO =75°, ∵OC =OB ,∴∠C =∠OBD =15°, ∴∠ABD =75°,∴∠ADB =∠ABD =75°, ∴∠A =30°,∴BO=12 AO,∵AB=23,∴BO2+AB2=4OB2,∴BO=2,∴⊙O的半径为2,故选:B.【点睛】本题考查了切线的性质、等腰三角形的判定和性质以及勾股定理的运用,求出∠A=30°,是解题的关键.10、B【分析】根据平行线分线段成比例,相似三角形性质,以及合比性质,分别对每个选项进行判断,即可得到答案. 【详解】解:如图,在△ABC中,DE∥BC,AD∶DB=4∶5,则∴△ADE∽△ABC,∴49DE AD ADBC AB AD DB===+,故A错误;则94BCDE=,故B正确;则49AE ADAC AB==,故C错误;则59EC DBAC AB==,故D错误.故选择:B.【点睛】本题考查了相似三角形的性质,平行线分线段成比例,合比性质,解题的关键是熟练掌握平行线分线段成比例的性质.11、C【解析】先根据垂径定理得出M 、N 分别是AB 与AC 的中点,故MN 是△ABC 的中位线,由三角形的中位线定理即可得出结论.【详解】解:∵OM ⊥AB ,ON ⊥AC ,垂足分别为M 、N ,∴M 、N 分别是AB 与AC 的中点,∴MN 是△ABC 的中位线,∴BC =2MN =故选:C .【点睛】本题考查垂径定理、三角形中位线定理;熟知平分弦的直径平分这条弦,并且平分弦所对的两条弧是解答此题的关键. 12、D【解析】试题分析:利用因式分解法解方程即可.解:(x ﹣2)(x+1)=0,x ﹣2=0或x+1=0,所以x 1=2,x 2=﹣1.故选D .考点:解一元二次方程-因式分解法.二、填空题(每题4分,共24分)13、()12,3或()12,3--【解析】根据位似变换的性质计算即可.【详解】解:∵△ABC 与△A'B'C'相似比为13,若点C 的坐标为(4,1), ∴点C′的坐标为()43,13⨯⨯或()()()43,13⨯-⨯-∴点C′的坐标为()12,3或()12,3--故答案为()12,3或()12,3--【点睛】本题考查的是位似变换,在平面直角坐标系中,如果位似变换是以原点为位似中心,相似比为k ,那么位似图形对应点的坐标的比等于k 或﹣k .14、①③④【分析】根据表格,即可判断出抛物线的对称轴,从而得到顶点坐标,即可判断①;根据抛物线的对称性即可判断②;根据表格中函数值为-2时,对应的x 的值,即可判断③;根据二次函数的增减性即可判断④.【详解】解:①根据表格可知:抛物线2y ax bx c =++(0a ≠)的对称轴为x=2,∴抛物线2y ax bx c =++(0a ≠)的顶点为(2,3)-,故①正确;②根据抛物线的对称性可知:当x=4和x=0时,对应的函数值相同,∴m=1,故②错误;③由表格可知:对于二次函数2y ax bx c =++,当y=-2时,对应的x 的值为1或3∴关于x 的方程22ax bx c ++=-的解为11x =,23x =,故③正确;④由表格可知:当x <2时,y 随x 的增大而减小∵0.50-<,抛物线过(0,1)∴当0.5x =-时,y >1>0∴当0.5x =-时,y 的值为正,故④正确.故答案为:①③④.【点睛】此题考查的是二次函数的图象及性质,掌握二次函数的对称性、顶点坐标与最值、二次函数与一元二次方程的关系和二次函数的增减性是解决此题的关键.15、x ≥1且x ≠1【分析】二次根式及分式有意义的条件:被开方数为非负数,分母不为1,据此解答即可.有意义,∴x≥11≠1,∴x≥1且x≠1在实数范围内有意义, 故答案为:x≥1且x≠1【点睛】本题考查二次根式和分式有意义的条件,要使二次根式有意义,被开方数为非负数;要使分式有意义分母不为1.161【分析】把一条线段分割为两部分,使较大部分与全长的比值等于较小部分与较大的比值,则这个比值即为黄金分割,【详解】∵P 为线段AB 的黄金分割点,且PA >PB ,AB=2cm , ∴()5151251.22PA AB cm --==⨯=- 故答案为51-.【点睛】分析题意可知,本题主要考查了黄金分割,弄清楚黄金分割的定义是解答此题的关键;17、1【分析】根据摸到白球的概率公式=40%,列出方程求解即可.【详解】解:不透明的布袋中的小球除颜色不同外,其余均相同,共有10个小球,其中白色小球x 个,根据古典型概率公式知:P (白色小球)==10%, 解得:x=1.故答案为1.考点:已知概率求数量. 18、23【分析】由1,2,3三个数字组成的无重复数字的两位数字共有6个,其中奇数有4个,由此求得所求事件的概率.【详解】解:由1,2,3三个数字组成的无重复数字的两位数字共有3×2=6个,其中奇数有2×2=4个, 故从中任取一个数,则恰为奇数的概率是4263=, 故答案为:23. 【点睛】本题考查古典概型及其概率计算公式的应用,属于基础题.解题的关键是掌握概率公式进行计算.三、解答题(共78分)19、15米.【分析】根据题意分别表示出AB 、AF 的长,进而得出等式求出答案.【详解】过E 作EF ⊥AB 于F ,设AC=AE=x∵AB ⊥CD ,ED ⊥CD ,∴四边形FBDE 为矩形,∴ 1.5BF ED ==,在Rt AEF ⊿中 ∵AF sin AEF AE∠= , ∴60?AF xsin =︒,∴AB=AF+BF 60 1.5xsin =︒+,在Rt ACB ⊿中, ∵AB sin ACB AC∠=, ∴75AB xsin =︒,∴75?60 1.5xsin xsin ︒=︒+,1.57560x sin sin =︒-︒, ∴ 1.5 1.5750.970.97151575600.970.87AB sin sin sin =︒⨯≈⨯=⨯=︒-︒-(米). ∴旗杆AB 的高度为15米.【点睛】本题主要考查了解直角三角形的应用,正确应用锐角三角函数关系是解题关键.20、(1)证明见解析;(2)7【分析】(1)根据平行四边形的性质证明AE BF =,再由一组对边平行而且相等的四边形是平行四边形判定即可判定;(2)过点A 作AG ⊥BF 于G ,构造30读直角三角形,利用平行四边形的性质和勾股定理解答即可.【详解】证明:(1)∵四边形ABCD 为平行四边形,∴//AD BC ,AD BC =,又∵E 是AD 的中点,12BF BC =, ∴AE BF =,又∵//AE BF ,∴四边形AFBE 是平行四边形.(2)过点A 作AG BF ⊥于G ,由ABCD 可知://AB DC ,∴60ABF C ∠=∠=,∴30BAG ∠=,又∵6AB =,8AD =,∴3BG =,4BF AE ==,∴1FG =,在Rt ABG ∆中,由勾股定理得:222226327AG AB BG =-=-=,在Rt AGF ∆中,由勾股定理得:22227128AF AG FG =+=+=, ∴27BE AF ==【点睛】本题考查了平行四边形的判定与性质、勾股定理.平行四边形的判定方法共有4种,应用时要认真领会它们之间的联系与区别,同时要根据条件合理、灵活地选择方法.21、(1)2y x 2x 3=-++;()0,3;(2)①t=1;②当3t 4=632-秒时,△BOQ 为等腰三角形. 【分析】(1)将A 、B 点的坐标代入y =﹣x 2+bx+c 中,即可求解;(2)①△AOC 与△BMN 相似,则MB OA MN OC =或OC OA,即可求解;②分OQ=BQ ,BO=BQ ,OQ=OB 三种情况,分别求解即可;【详解】(1)∵A(﹣1,0),函数对称轴是直线x =1,∴()3,0B ,把A 、B 两点代入y =﹣x 2+bx+c 中,得: 93010b c b c ⎧-++=⎨--+=⎩,解得23b c =⎧⎨=⎩,∴抛物线的解析式为2y x 2x 3=-++,∴C 点的坐标为()0,3.(3)①如下图2443MN t t =-++,32MB t =-,△AOC 与△BMN 相似,则MB OA MN OC =或OC OA , 即2323443t t t -=-++或13, 解得32t =或1-3或3或1(舍去32,1-3,3), 故t=1.②∵()2,0M t ,MN x ⊥轴,∴()2,32Q t t -,∵△BOQ 为等腰三角形,∴分三种情况讨论:第一种:当OQ=BQ 时,∵QM OB ⊥,∴OM=MB ,∴232t t =-,∴3t 4=; 第二种:当BO=BQ 时,在Rt △BMQ 中,∵45OBQ ∠=︒,∴BQ =,即()3-2t ,∴64t -=; 第三种:当OQ=OB 时,则点Q 、C 重合,此时t=0,而t >0,故不符合题意;综上所述,当3t 4=秒或64-秒时,△BOQ 为等腰三角形. 【点睛】本题主要考查了二次函数的综合,准确分析求解是做题的关键.22、当BP =6时,CQ 最大,且最大值为1.【分析】根据正方形的性质和余角的性质可得∠BEP =∠CPQ ,进而可证△BPE ∽△CQP ,设CQ =y ,BP =x ,根据相似三角形的性质可得y 与x 的函数关系式,然后利用二次函数的性质即可求出结果.【详解】解:∵四边形ABCD 是正方形,∴∠B =∠C =90°,∴∠BEP +∠BPE =90°,∵PQ EP ⊥,∴∠QPC +∠BPE =90°,∴∠BEP =∠CPQ . ∴△BPE ∽△CQP ,∴BE BP PC CQ=. 设CQ =y ,BP =x ,∵AB=BC =12,∴CP =12﹣x .∵AE =14AB ,AB =12,∴BE =9, ∴912x x y =-,化简得:y =﹣19(x 2﹣12x ),即y =﹣19(x ﹣6)2+1, 所以当x =6时,y 有最大值为1.即当BP =6时,CQ 有最大值,且最大值为1.【点睛】本题考查了正方形的性质、相似三角形的判定和性质和二次函数的性质等知识,属于常见题型,熟练掌握相似三角形的性质和二次函数的性质是解答的关键.23、(1)212533y x x =+-;(2)当52t =-时,ADP S ∆取最大值133124,此时P 点坐标为555,212⎛⎫-- ⎪⎝⎭. (3)7CQ =或17.【分析】(1)根据对称轴与点A 代入即可求解;(2)先求出()8,11D -,过P 点作y 轴的平行线,交直线AD 于点M ,设212,533P t t t ⎛⎫+- ⎪⎝⎭,得到(),3M t t -+,215833PM t t =--+,表示出21111582233ADP A D S PM x x t t ∆⎛⎫=-=--+ ⎪⎝⎭,根据二次函数的性质即可求解; (3)根据题意分①当Q 在y 轴正半轴上时, ②当Q 在y 轴负半轴上时利用相似三角形的性质即可求解.【详解】(1)∵对称轴为x =−1,∴−2b a=−1, ∴b =2a ,∴y =ax 2+2ax −5,∵y =−x +3与x 轴交于点A (3,0),将点A 代入y =ax 2+2ax−5可得a =13∴212533y x x =+-. (2)令2125333x x x +-=-+,解得:13x =,28x =-, ∴()8,11D -,过P 点作y 轴的平行线,交直线AD 于点M , 设212,533P t t t ⎛⎫+- ⎪⎝⎭,则(),3M t t -+, ∴215833PM t t =--+,83t -<<, 则21111582233ADP A D S PM x x t t ∆⎛⎫=-=--+ ⎪⎝⎭, ∵103-<, ∴当52t =-时,ADP S ∆取最大值133124,此时P 点坐标为555,212⎛⎫-- ⎪⎝⎭. (3)存在, 理由:①当Q 在y 轴正半轴上时,如图,过点Q 作QN AC ⊥于N ,根据三角形的外角的性质得,OQA OCA QAN ∠+∠=∠, 又∵45OQA OCA CBA ∠+∠=∠=︒,∴45QAN CBA ∠=∠=︒,∴AN QN =,∵3AO =,5CO =,∴AC =设AN QN m ==,则CN AC AN m =+=+又∵90QNA COA ∠=∠=︒,QCN ACO ∠=∠,∴COA CNQ ∆∆∽, ∴CO AO AC CN QN QC==,3m QC ==,∴17QC ==, ②当Q 在y 轴负半轴上时,记作'Q ,由①知,17512OQ QC CO =-=-=,取'12OQ OQ ==,如图,则由对称知:'OQ A OQA ∠=∠,∴'45OQ A OCA OQA OCA CBA ∠+∠=∠+=∠=︒,因此点'Q 也满足题目条件,∴''1257Q C OQ OC =-=-=, 综合以上得:7CQ =或17.【点睛】本题考查二次函数的综合;熟练掌握二次与一次函数的图象及性质,掌握三角形相似、直角三角形的性质是解题的关键.24、(1)80人 (2)见解析 (3)375【分析】(1)根据条形统计图和扇形统计图可知,选择文学鉴赏的学生16人,占总体的20%,从而可以求得调查的学生总人数;(2)根据 3D 打印的百分比和(1)中求得的调查的学生数,可以求得选择3D 打印的有多少人,进而可以求得选择数学思维的多少人,从而可以将条形统计图补充完整;(3)根据调查的选择红船课程的学生所占的百分比,即可估算出全校选择体育类的学生人数.【详解】解:(1)16÷20%=80人; (2)如图所示;(3)30100080⨯=375(人). 【点睛】本题考查了条形统计图、样本估计总体、扇形统计图,解题的关键是明确题意,找出所求问题需要的条件,利用数形结合的思想解答问题.25、(1)1552y x =-+;(2)32, 2646元. 【分析】(1)设一次函数关系式为(0)y kx b k =+≠,将“当36x =时,37y =;40x =时,35y =”代入计算即可;(2)根据利润等于单件利润乘以销售量分段列出函数关系式,再根据一次函数及二次函数的性质得出最大利润即可.【详解】解:(1)设一次函数关系式为(0)y kx b k =+≠∵当36x =时,37y =;40x =时,35y =,即37363540k b k b =+⎧⎨=+⎩,解得:1255k b ⎧=-⎪⎨⎪=⎩ ∴1552y x =-+ (2)(18)W y m =-∴当130x ≤≤时,(3818)(330)60600W x x =-+=+ ∵60>0∴当x=30时,W 最大=2400(元)当3150x ≤≤时1(5518)(330)2W x x =-+-+ 239611102x x =-++ 23(32)26462x =--+ ∴当x=32时,当天的销售利润W 最大,为2646元.2646>2400∴故当x=32时,当天的销售利润W 最大,为2646元.【点睛】本题考查了二次函数的实际应用,根据题意列出函数关系式并熟知函数的基本性质是解题关键.26、(1)C (m ,﹣1);(3)﹣3≤m≤0或3≤m≤3.【分析】(1)化成顶点式,即可求得顶点C 的坐标;(3)由顶点C 的坐标可知,抛物线的顶点C 在直线y =﹣1上移动.分别求出抛物线过点A 、点B 时,m 的值,画出此时函数的图象,结合图象即可求出m 的取值范围.【详解】(1)y =x 3﹣3mx+m 3﹣1=(x ﹣m )3﹣1,∴抛物线顶点为C (m ,﹣1).(3)把A (0,3)的坐标代入y =x 3﹣3mx+m 3﹣1,得3=m 3﹣1,解得m=±3.把B(3,3)的坐标代入y=x3﹣3mx+m3﹣1,得3=33﹣3m×3+m3﹣1,即m3﹣3m=0,解得m=0 或m=3.结合函数图象可知:﹣3≤m≤0或3≤m≤3.【点睛】本题考查了二次函数的图象与系数的关系,二次函数图象上点的坐标特征,提现了转化思想和数形结合思想的应用.。
期末复习练习 2022-2023学年苏科版九年级上册物理
2022-2023学年第一学期苏科版九年级物理期末复习班级______ 姓名_______ 学号_____一.选择题1.下列实例中,有力对物体做功的是()A.跳水运动员从跳台跳下B.吊车吊着集装箱水平缓慢移动C.举重运动员举着杠铃停在空中D.小球在光滑水平面上滚动2.下列做法中符合安全用电原则的是()A.家用电器意外着火时,第一时间用水灭火B.使用试电笔辨别火线时,用手接触笔尖金属体C.洗衣机、电冰箱等家用电器的金属外壳要接地D.遇到雷雨天气,在大树底下避雨3.某同学连接的电路如图所示,闭合开关后,无论怎样调节滑动变阻器,灯泡L1、L2都不亮。
他用电压表进行电路故障检测,把电压表先后并联在ab、af、ae两点之间,电压表都有示数,而把电压表并联在ad两点之间时电压表无示数。
则可以判定电路中的故障是()A.灯泡L1短路B.灯泡L2短路C.de间断路D.be间断路4.探究某物质熔化和沸腾的实验如图甲所示,把100g某种固体碾碎放入试管中,插入温度计,再将试管放在装有水的烧杯中加热(物质在相同时间内吸收的热量相等)。
根据实验数据画出的图像如图乙所示,从开始加热到实验结束的10min内消耗4g酒精,酒精的热值3.0×107J/kg,这种物质的液态比热容c=1.5×103 J/(kg·℃),下列选项错误的()A.该物质在第5min末的内能大于第4min末的内能B.在AB段和BC段,该物质吸收的热量相同C.该物质固态时的比热容为3.0×103 J/(kg·℃)D.该装置能量的转化效率为50%5.如图所示的电路中,电源电压恒定,R1为定值电阻,R2为滑动变阻器,闭合开关S后,滑动变阻器滑片P从一端向另一端滑动过程中,以下说法错误的是()A.若滑动变阻器滑片P自a向b移动,电流表A的示数变小,电压表V2的示数与电流表A的示数之比变大B.若滑动变阻器滑片P自b向a移动,电压表V1的示数与电压表V2的示数比值变大C.若滑动变阻器滑片P自b向a移动,电压V2的示数变化量∆U2与电流表A的示数变化量∆I的比值变小D.不管滑片P怎么移动,电压V1的示数变化量∆U1与电流表A的示数变化量∆I的比值不变6.关于热量、温度、内能之间的关系,下列说法中正确的是()A.物体的温度升高,它的内能会增大B.物体吸收热量,温度一定升高C.温度为﹣40℃的冰块没有内能D.物体温度越高,所含有的热量越多7.如图所示,用同样的器材进行实验,现仅根据加热时间长短即可比较两种不同物质的吸热本领,则以下说法中错误的是()A.应控制两种物质的质量相等B.应使用相同燃料对两种物质进行加热C.应该控制两种物质的温度增加量相等D.两次加热时需控制消耗燃料质量相等8.有一种专门烤肉串的烤箱,将肉串挂到箱内支架上,关上箱门(相当于闭合开关S1),支架在电动机的带动下旋转;再将温控开关S2闭合,开始加热;若只闭合开关S2,则不能加热。
九年级数学第一学期期末考试综合复习测试题(含答案)
九年级数学第一学期期末考试综合复习测试题(含答案)一.选择题(共10小题,每小题3分,共30分) 1.2022的相反数是( )A .2022B .2022-C .12022D .2022± 2.若代数式3125m x y -与822m nx y +-是同类项,则( )A .73m =,83n =-B .3m =,4n =C .73m =,4n =- D .3m =,4n =-3.下列四组线段中,能组成直角三角形的是( ) A .1a =,3b =,3c = B .2a =,3b =,4c = C .2a =,4b =,5c =D .3a =,4b =,5c = 4.如图所示,直线//a b ,231∠=︒,28A ∠=︒,则1(∠= )A .61︒B .60︒C .59︒D .58︒5.下列关于事件发生可能性的表述,正确的是( )A .“在地面向上抛石子后落在地上”是随机事件B .掷两枚硬币,朝上面是一正面一反面的概率为13C .在同批次10000件产品中抽取100件发现有5件次品,则这批产品中大约有500件左右的次品D .彩票的中奖率为10%,则买100张彩票必有10张中奖6.某校10名学生参加课外实践活动的时间分别为:3,3,6,4,3,7,5,7,4,9(单位:小时),这组数据的众数和中位数分别为( ) A .9和7 B .3和3 C .3和4.5 D .3和5 7.一个正多边形的每一个内角都是150︒,则它的边数为( ) A .6 B .9 C .12 D .158.若不等式组841x x x m +<-⎧⎨>⎩的解集是3x >,则m 的取值范围是( )A .3m <B .3mC .3m >D .3m9.已知关于x 的一元二次方程22(21)0x m x m --+=有实数根,则m 的取值范围是( ) A .14m 且0m ≠ B .14m C .14m < D .14m >10.如图1,一个扇形纸片的圆心角为90︒,半径为6.如图2,将这张扇形纸片折叠,使点A 与点O 恰好重合,折痕为CD ,图中阴影为重合部分,则阴影部分的面积为( )A .9632π-B .693π-C .91232π-D .94π二.填空题(共5小题,每小题3分,共15分) 11.将数据2022万用科学记数法表示为 .12.已知当3x =时,代数式35ax bx +-的值为20,则当3x =-时,代数式35ax bx +-的值是 .13.将抛物线229y x x =-+-向左平移2个单位,再向上平移1个单位后,得到的抛物线的解析式为 .14.已知ABC ∆中,点O 是ABC ∆的外心,140BOC ∠=︒,那么BAC ∠的度数为 .15.如图,在正方形ABCD 中,顶点(5,0)A -,(5,10)C ,点F 是BC 的中点,CD 与y 轴交于点E ,AF 与BE 交于点G ,将正方形ABCD 绕点O 顺时针旋转,每次旋转90︒,则第2023次旋转结束时,点G 的坐标为 .三.解答题(一)(共3小题,每小题8分,共24分) 16.计算(1)2()(2)x y x y x +--;(2)2219(1)244a a a a --÷--+.17.如图,90ACB ∠=︒,AC AD =.(1)过点D 作AB 的垂线DE 交BC 与点E ,连接AE .(尺规作图,并保留作图痕迹) (2)如果8BD =,10BE =,求BC 的长.18.如图,在四边形ABCD 中,AC 与BD 交于点O ,BE AC ⊥,DF AC ⊥,垂足分别为点E ,F ,且BE DF =,ABD BDC ∠=∠.求证:四边形ABCD 是平行四边形.四.解答题(二)(共3小题,每小题9分,共27分) 19.阳光中学为了丰富学生的课余生活,计划购买围棋和中国象棋供棋类兴趣小组活动使用,若购买3副围棋和5副中国象棋需要98元;若购买1副围棋和2副中国象棋需要36元.(1)求每副围棋和每副中国象棋各多少元;(2)阳光中学决定购买围棋和中国象棋共40副,总费用不超过538元,且围棋的副数不低于象棋的副数,问阳光中学有几种购买方案;(3)请求出最省钱的方案需要多少钱?20.我市某中学举行“中国梦⋅我的梦”的演讲比赛,赛后整理参赛学生的成绩,将学生的成绩分为A,B,C,D四个等级,并将结果绘制成如图所示的条形统计图和扇形统计图,但均不完整,请你根据统计图解答下列问题.(1)参加比赛的学生人数共有名,在扇形统计图中,表示“D等级”的扇形的圆心角为度,图中m的值为;(2)补全条形统计图;(3)组委会决定从本次比赛中获得A等级的学生中,选出两名去参加市中学生演讲比赛,已知A等级中男生只有1名,请用画树状图或列表的方法求出所选学生恰是一男一女的概率.21.22.某网店专售一款新型钢笔,其成本为20元/支,销售中发现,该商品每天的销售量y与销售单价x(元/支)之间存在如下关系:10400y x=-+,自武汉爆发了“新型冠状病毒”疫情该网店店主决定从每天获得的利润中抽出200元捐赠给武汉,同时又让顾客得到实惠,当销售单价定位多少元时,捐款后每天剩余利润为550元?五.解答题(三)(共2小题,每小题12分,共24分)22.如图,以点O为圆心,AB长为直径作圆,在O上取一点C,延长AB至点D,连接DC,过点A作O的切线交DC的延长线于点E,且DCB DAC∠=∠.(1)求证:CD是O的切线;(2)若6AD=,2:3BC CA=,求AE的长.23.如图,在平面直角坐标系中,直线33y x =--与x 轴交于点A ,与y 轴交于点C .抛物线2y x bx c =++经过A 、C 两点,且与x 轴交于另一点B (点B 在点A 右侧). (1)求抛物线的解析式;(2)若点M 是线段BC 上一动点,过点M 的直线ED 平行y 轴交x 轴于点D ,交抛物线于点E ,求ME 长的最大值及此时点M 的坐标; (3)在(2)的条件下:当ME 取得最大值时,在x 轴上是否存在这样的点P ,使得以点M 、点B 、点P 为顶点的三角形是等腰三角形?若存在,请直接写出所有点P 的坐标;若不存在,请说明理由.答案一.选择题1. B .2. D .3. D .4. C .5. C .6. C .7. C .8. B .9. B .10. C . 二.填空题11. 72.02210⨯.12. 30-.13. 228y x x =---.14. 70︒或110︒.15. (4,3)-. 三.解答题16.解:(1)2()(2)x y x y x +--22222x xy y xy x =++-- 2y =;(2)2219(1)244a a a a --÷--+ 23(3)(3)2(2)a a a a a ---+=÷-- 23(2)2(3)(3)a a a a a --=⋅---+ 23a a -=--. 17.解:(1)如图所示即为所求作的图形. (2)ED 垂直AB , 90ADE EDB ∴∠=∠=︒,在Rt BDE ∆中,22221086DE BE BD =-=-=, 在Rt ADE ∆和Rt ACE ∆中, AC ADAE AE =⎧⎨=⎩, Rt ADE Rt ACE(HL)∴∆≅∆, 6EC ED ∴==, 16BC BE EC ∴=+=.18.证明:ABD BDC ∠=∠, //AB CD ∴.BAE DCF ∴∠=∠.在ABE ∆与CDF ∆中, 90BAE DCF AEB CFD BE DF ∠=∠⎧⎪∠=∠=︒⎨⎪=⎩. ()ABE CDF AAS ∴∆≅∆. AB CD ∴=.∴四边形ABCD 是平行四边形.19.解:(1)设每副围棋x 元,每副中国象棋y 元,根据题意得:3598236x y x y +=⎧⎨+=⎩,∴1610x y =⎧⎨=⎩,∴每副围棋16元,每副中国象棋10元;(2)设购买围棋z 副,则购买象棋(40)z -副, 根据题意得:1610(40)538m m +-,40m z -,2023m ∴,m 可以取20、21、22、23则有:方案一:购买围棋20副,购买中国象棋20副方案二:购买围棋21副,购买中国象棋19副方案:购买围棋22副,购买中国象棋18副方案四:购买围棋23副,购买中国象棋17副由4种方案;(3)由上一问可知共有四种方案:方案一:购买围棋20副,购买中国象棋20副;方案二:购买围棋21副,购买中国象棋19副;方案三:购买围棋22副,购买中国象棋18副;方案四:购买围棋23副,购买中国象棋17副;方案一需要20162010520x x +=; 方案二需要21161910526x x +=; 方案三需要22161810532x x +=; 方案四需要23161710538x x +=; 所以最省钱是方案一,需要520元.20.(1)解:根据题意得:总人数为:315%20÷=(人), 表示“D 等级”的扇形的圆心角为43607220⨯︒=︒;C等级所占的百分比为8100%40% 20⨯=,所以40m=,故答案为:20,72,40.(2)解:等级B的人数为20(384)5-++=(人),补全统计图,如图所示:(3)解:根据题意,列出表格,如下:男女1女2男女1、男女2、男女1男、女1女2、女1女2男、女2女1、女2共有6种等可能结果,其中恰是一男一女的有4种,所以恰是一男一女的概率为42 63 =.21.解:由题意可得(20)(10400)200550x x--+-=解得125x=,235x=因为要让顾客得到实惠,所以25x=答:当销售单价定为25元时,捐款后每天剩余利润为550元.22.(1)证明:连接OC,OE,如图,AB为直径,90ACB∴∠=︒,即190BCO∠+∠=︒,又DCB CAD∠=∠,1CAD∠=∠,1DCB∴∠=∠,90DCB BCO ∴∠+∠=︒,即90DCO ∠=︒, CD ∴是O 的切线;(2)解:EC ,EA 为O 的切线, EC EA ∴=,AE AD ⊥, OC OA =, OE AC ∴⊥,90BAC EAC ∴∠+∠=︒,90AEO EAC ∠+∠=︒, BAC AEO ∴∠=∠, tan tan BAC AEO ∴∠=∠,∴23BC AO AC AE ==, Rt DCO Rt DAE ∆∆∽,∴23CD OC OA DA AE AE ===, 2643CD ∴=⨯=, 在Rt DAE ∆中,设AE x =,222(4)6x x ∴+=+, 解得52x =. 即AE 的长为52.23.解:(1)直线33y x =--与x 轴、y 轴分别交于点A 、C , (1,0)A ∴-,(0,3)C -抛物线2y x bx c =++经过点(1,0)A -,(0,3)C -, ∴103b c c -+=⎧⎨=-⎩,解得23b c =-⎧⎨=-⎩,∴抛物线的解析式为223y x x =--.(2)设(E x ,223)(03)x x x --<<,则(,3)M x x -, 222393(23)3()24ME x x x x x x ∴=----=-+=--+,∴当32x =时,94ME =最大,此时3(2M ,3)2-. (3)存在.如图3,由(2)得,当ME 最大时,则3(2D ,0),3(2M ,3)2-,32DO DB DM ∴===; 90BDM ∠=︒,223332()()222OM BM ∴==+=. 点1P 、2P 、3P 、4P 在x 轴上, 当点1P 与原点O 重合时,则1322PM BM ==,1(0,0)P ; 当2322BP BM ==时,则232632322OP -=-=, 2632(2P -∴,0); 当点3P 与点D 重合时,则3332P M P B ==,33(2P ,0); 当4322BP BM ==时,则432632322OP +=+=, 4632(2P +∴,0). 综上所述,1(0,0)P ,2632(2P -,0),33(2P ,0),4632(2P +,0).。
2023-2024年道德与法治九年级上册期末复习试题(含答案)
2023-2024年道德与法治九年级上册期末复习试题的是()①国庆期间,人们排队乘坐无人驾驶的光谷空轨旅游观光②9月30日,党和国家领导人出席向人民英雄敬献花篮仪式③10月20日,公安部交管局开展冬季突出违法行为专项整治行动④杭州亚运会期间,观众通过VR、多赛同看等科技手段可随时沉浸在比赛现场A.①②B.①④C.②③D.③④6.下列选项不属于我国全过程人民民主实践的是()A.2022年全国县先两级人大换届选举,有超过10亿选民参与B.歙县发改委通过互联网就《歙县“十四五”服务业发展规划》向市民公开征集意见C.某领导根据自身经验进行判断,对所在单位进行大刀阔斧的改革D.黄山市确定基层立法联系点,在群众家门口搭建反映立法工作意见建议的“直通车”7.小马同学整理了保障人民当家作主的政治制度,其中认识正确的是()A.基层群众自治制度是人民直接行使民主权利,直接掌握国家政权的根本途径B.中国共产党领导的多党合作和政治协商制度是发展基层民主,建设社会主义民主政治的基础C.民族区域自治制度是一项独具中国特色的实现民族平等、保障少数民族合法权利的基本政治制度D.全国人民代表大会制度是我国的一项根本政治制度,通过协商求同存异,找到最大公约数,画出最大同心圆8.从管理垃圾分类到杜绝噪音扰民,从社区监控全覆盖到缓解社区停车难……各地在社区治理方面的举措正让居民获得更好的社区服务。
居民的以下做法正确的是()①提升社会责任感,爱护社区绿化环境②培养主人翁意识,积极参与社区建设③提高监督意识,经常向社区投诉抱怨④增强民主意识,主动为社区建言献策A.②③④B.①②③C.②③④D.①②④9.厉行法治是对全体社会成员的共同要求。
下列做法不符合厉行法治要求的是()①某企业先提价后打折,搞虚假促销②刘某未经他人同意,私自拆开他人信件③某市交警使用执法记录仪上岗执勤④某市司法局召开立法听证会,公开征求意见A.①②B.①③C.②④D.③④10.毛某通过某网络平台,发布了一条时长1分钟左右的作品,使用侮辱性语言对马某及其家人进行辱骂。
期末复习题 2023—2024学年统编版语文九年级上册
2023-2024年度上学期期末复习质量检测九年级语文试题卷一、单选题(本大题共3小题,共6分)1.下列有关作家作品的连线,不正确的一项是()A. 《我爱这土地》——艾青——《九三年》B. 《沁园春•雪》——毛泽东——革命家C. 《你是人间四月天》——林徽因——建筑学家、文学家D. 《水浒传》——施耐庵——元末明初2.下列加下划线的成语使用有错误的一项是()。
A. 泰州举办老街旅游文化节,各地游客纷至沓来B. 洪宗礼先生说,没有实践,教材编写便缺了源头活水C. 诈骗分子推陈出新,利用最新的市场漏洞,想出了新的诈骗手法D. 《朗读者》《见字如面》《经典咏流传》等文化类节目如一股股清泉,沁人心脾3.下列语句中没有语病的一项是()。
A. 汪国真的诗作曾点燃了一代人的青春梦想。
他猝然长逝,怎不让人扼腕叹息B. 通过我市举办的“名师好课”系列送教活动,促进了全市城乡教育的均衡发展C. “赣剧进校园”的成效并不显著,原因是对地方文化的重要性认识不足造成的D. 实施“校园足球计划”,旨在普及足球运动,进一步培养青少年足球运动水平二、默写(本大题共1小题,共4分)4.填空。
(1) 小时候,______ ,我在这头,母亲在那头。
而现在,______ ,我在这头,大陆在那头。
(2) 为什么我的眼里常含泪水?______(3) 毛泽东《沁园春·雪》中由写景到论史,起承上启下作用的句子是“______,______ ”。
(4) “月”是古诗词中的常见意象。
温庭筠在《商山早行》中描写月清霜冷的诗句是“______ ,______ ”;杜甫在《月夜忆舍弟》中流露月夜思乡之情的诗句是“______ ,______ ”。
三、综合题(本大题共1小题,共9分)5.阅读下面的文字,完成题目。
①毛泽东诗词纵览天下风云,俯瞰(kàn)历史兴衰,感受时光飞逝,把握时代潮流,反映了毛泽东的历史观和人生观,浸(jìn)透着历史智慧,洋溢..着壮志豪情。
九年级语文上册期末复习测试题(含答案)
九年级语文上册期末复习测试题(含答案)一、积累与运用(26分)1.下列词语中加点字注音、字形全部正确的一项是()(2分)A.莅.临(wèi)余晖.(huī)迭.起(dié)合.颜悦色(hé)B.驾驭.(yù)亢.奋(kàng)元勋.(xūn)取义成仁.(rén)C.镐.头(gǎo)慢.游(màn)秀颀.(qín)风云变幻.(huàn)D.健.树(jiàn)旗帜.(zhì)妩.媚(fǔ)经.世奇才(jīng)2.下列词语书写完全正确的一项是()(2分)A.朗润矜持见义思迁家喻户晓B.遨游娴熟和颜悦色因地治宜C.斟酌荣膺人情世故相辅相成D.聪颖鼎盛穿流不息取义成仁3.下列句子中,加点词语使用最恰当...的一项是()A.车间一线的青年工人勤学苦练,对技术精益求精....,争当“青年岗位能手”。
B.志愿者在冬奥会闭幕式上向各国健儿挥手,言外之意....是:欢迎再来中国!C.果农们冒着炎热的天气奋战了好几个昼夜,轻而易举....地把水果销售一空。
D.市民陈先生参与毽子操、游泳、跳绳等多项全民云健身运动,各得其所....。
4.下列表述正确的一项是()(2分)A.“毛遂在平原君那里,三年也没有遇到施展才华的机会,于是他大胆自荐,并提出了一个观点。
”是一个表因果关系的复句B.《红楼梦》是我国古典小说的巅峰之作,叙述了以贾家为代表的四大家族的兴衰史,反映了封建社会广阔的社会现实。
C.明清科举制度,童生经过童试,录取人学后称为“进学”,也就成了秀才;秀才参加乡试,考中的是贡士。
D.《三国演义》是我国第一部章回体长篇历史演义小说,该书反映了三国时期政治腐败、生灵涂炭、农民起义、诸侯割据的社会现实。
5.下列句子没有语病的一项是()(2分)A.各地中小学要对延期开学期间学生居家学习的情况逐一进行诊断、摸底,精准掌握每个学生的学习情况。
2023年北京三帆中学英语九年级第一学期期末复习检测试题含解析
2023年北京三帆中学英语九年级第一学期期末复习检测试题注意事项1.考生要认真填写考场号和座位序号。
2.试题所有答案必须填涂或书写在答题卡上,在试卷上作答无效。
第一部分必须用2B 铅笔作答;第二部分必须用黑色字迹的签字笔作答。
3.考试结束后,考生须将试卷和答题卡放在桌面上,待监考员收回。
Ⅰ. 单项选择1、It’s said that many workers in our city have _________ some organizations to help the elderly and the disabled people. They are helpful and we should learn from them.A.stayed up B.put up C.took up D.set up2、—Whose hair band is this?—It _____ be Grace’s. She is the only girl at the meeting.A.might B.wouldn’tC.must D.can’t3、--- Tom, you won’t mak e much progress_________ you work really hard.---Ok, I will try my best.A.if B.though C.unless D.when4、---Ms. Wang,I'm afraid I can't finish the work in two days.---Don't worry. I'll give you ____________ days.A.two another B.two moreC.more two D.two many5、Don’t talk . Your grandmother is sleeping now.A.happily B.nearly C.loudly D.hardly6、—How many letters have you got from your online friends since last year?—_______. We’re just chatting with each other when we’re free.A.Nothing B.None C.No D.No one7、--Where’s your father? We haven’t seen each other .---____________.A.He has been to America B.He has gone to EnglandC.He is going to England D.He would visit my grandparents8、My twin brothers promised my parents to give up playing computer games, butof them made it.A.none B.either C.neither9、Many people are shy when they ________ English so before you begin, take a deep breath and smile.A.say B.speak C.tell D.talk10、--- Look at the lovely dog. Guess_______ at birth.--- I think he might be 100 grams.A.what could he do B.what his weight wasC.what was his weight D.what he could doⅡ. 完形填空11、第二节词语填空(共10小题;每小题1分,满分10分)阅读下面的短文,掌握其大意,然后从下列各小题所给的A、B、C三个选项中选出可以填入空白处的最佳答案,并在答题卡上将该编号字母涂黑。
语文九年级上学期期末复习文言文综合检测试题附解析1
语文九年级上学期期末复习文言文综合检测试题附解析1 一、文言文1.阅读下文,回答问题。
马说世有伯乐,然后有千里马。
千里马常有,而伯乐不常有。
故虽有名马,祗辱于奴隶人之手,骈死于槽枥之间,不以千里称也。
马之千里者,一食或尽粟一石。
食马者不知其能千里而食也。
是马也,虽有千里之能,食不饱,力不足,才美不外见,且欲与常马等不可得,安求其能千里也?策之不以其道,食之不能尽其材,鸣之而不能通其意,执策而临之,曰:“天下无马!”呜呼!其真无马邪?其真不知马也。
(1)解释下列加下划线词语在句子中的意思。
①不以千里称也________②一食或尽粟一石________③策之不以其道________(2)把文中画线的句子翻译成现代汉语。
①故虽有名马,只辱于奴隶人之手,骈死于槽枥之间,不以千里称也。
②食马者不知其能千里而食也(3)下列对文章的分析理解,不正确的一项是()A.本文用托物寓意的写法,把能赏识人才的人(贤君)比作伯乐,把人才比作千里马,把愚妄无知、不识人才的统治者比作食马者。
B.作者借千里马表达了“有了伯乐,才能发现人才”的观点,指出“千里马常有,而伯乐不常有”,寄寓了作者对人才遭屈辱,被埋没的感慨、悲愤之情。
C.“策之不以其道,食之不能尽其材,鸣之而不能通其意”运用了排比的修辞手法,增强语势,刻画食马者愚妄无知的形象,表达作者对“食马者”辛辣的讽刺和无情的嘲弄。
D.造成千里马“欲与常马等不可得”的直接原因是“食马者不知其能千里而食也”。
2.阅读下文,回答问题。
鱼,我所欲也;熊掌,亦我所欲也。
二者不可得兼,舍鱼而取熊掌者也。
生,亦我所欲也;义,亦我所欲也。
二者不可得兼,舍生而取义者也。
生亦我所欲,所欲有甚于生者,故不为苟得也;死亦我所恶,所恶有甚于死者,故患有所不辟也。
如使人之所欲莫甚于生,则凡可以得生者何不用也?使人之所恶莫甚于死者,则凡可以辟患者何不为也?由是则生而有不用也,由是则可以辟患而有不为也。
是故所欲有甚于生者,所恶有甚于死者。
人教版九年级数学上册期末综合复习测试题(含答案)
人教版九年级数学上册期末综合复习测试题(含答案)时间:100分钟 总分:120分一、 选择题(每题3分,共24分)1.已知关于x 的方程()222310---=m m x x +是一元二次方程,则m 的值为( ) A .2m =B .4m =C .2m =±D .2m =-2.如图,将AOB ∆绕点O 按逆时针方向旋转40°后得到A OB ''△,若15AOB ∠=︒,则AOB '∠的度数是 ( )A .25°B .30°C .35°D .40°3.顶点(2,1),且开口方向、形状与函数22y x =的图像相同的抛物线是 ( ) A .221y x =+ B .22(2)1y x =-+ C .22(2)1y x =++D .22(2)1y x =+-4.把方程2630x x +-=化成2)x m n (的形式,则m n += ( ) A .15-B .9C .15D .65.如图,ABC ∆内接于O ,直径8cm AD =,=60B ∠︒,则AC 的长度为 ( )A .5cmB .42C .43D .6cm6.在一个不透明的口袋中有红色、黄色和绿色球共60个,它们除颜色外,其余完全相同.在不倒出球的情况下,要估计袋中各种颜色球的个数.同学们通过大量的摸球试验后,发现摸到红球、黄球和绿球的频率分别稳定在20%,40%和40%.由此,推测口袋中黄色球的个数有( ) A .15个B .20个C .21个D .24个7.在同一坐标系中,一次函数y ax k =+与二次函数2y kx a =+的图象可能是 ( )A .B .C .D .8.二次函数2y ax bx c =++的图像如图所示,对称轴是直线1x =.下列结论:①0abc >;②30a c +>;③a c b +<-;④520a b c -+<.其中结论正确的个数为 ( )A .1个B .2个C .3个D .4个二、填空题(每题3分,共24分)9.若n 是方程2210x x --=的一个根,则代数式232n n -+-的值是________. 10.如图,AB 是半圆的直径,C 、D 是半圆上的两点,且20BAC =︒∠,点D 是AC 的中点,则BAD ∠=______.11.点()()1122,,,A x y B x y 在二次函数232y x x =-++的图像上,若122x x <<-,则1y 与2y 的大小关系是1y _______________2y .(用“>”、“<”、“=”填空)12.已知关于x 的一元二次方程2()0(,,a x h k a h k -+=都是常数,且0)a ≠的解为1213x x =-=,,则方程2(1)0(,,a x h k a h k --+=都是常数,且0)a ≠的解为___________.13.如图,正方形ABCD 的边长为3,点E 为AB 的中点,以E 为圆心,3为半径作圆,分别交AD 、BC 于M 、N 两点,与DC 切于P 点.则图中阴影部分的面积是______.14.如图,正方形OABC 的顶点B 在抛物线2y x 的第一象限的图象上,若点B 的纵坐标是横坐标的2倍,则对角线AC 的长为_________.15.如图,抛物线2y ax c =+与直线y mx n =+交于()1,A p -,()3,B q 两点,则不等式2ax mx c n ++<的解集是__________.16.如图,以(0,3)G 为圆心,半径为6的圆与x 轴交于A ,B 两点,与y 轴交于C ,D 两点,点E 为⊙G 上一动点,CF AE ⊥于F ,点E 在G 的运动过程中,线段FG 的长度的最小值为______.三、解答题(每题8分,共72分) 17.解方程: (1)(2)(3)12x x --= (2)23410x x -+=18.已知关于x 的一元二次方程24250x x m --+=有两个实数根. (1)求m 的取值范围;(2)若该方程的两个根都是符号相同的整数,直接写出它的根.19.已知二次函数图像与x 轴两个交点之间的距离是4个单位,且顶点M 为()14-,,求二次函数的解析式.20.如图,抛物线2(0)y ax bx c a =++≠与直线1y x =+相交于(-10)A ,,(4)B m ,两点,且抛物线经过点(50)C ,(1)求抛物线的解析式;(2)点P 是抛物线上的一个动点(不与点A .点B 重合),过点P 作直线PD ⊥x 轴于点D ,交直线AB 于点E.当PE =2ED 时,求P 点坐标;(3)点P 是直线上方的抛物线上的一个动点,求ABP ∆的面积最大时的P 点坐标.21.一个不透明的口袋中有四个完全相同的小球.把它们分别标记为1,2,3,4.(1)随机摸取一个小球的标号是偶数,该事件的概率为______;(2)小雨和小佳玩摸球游戏,两人各摸一个球,谁摸到的数字大谁获胜.小雨先从口袋中摸出一个小球,不放回,小佳再从口袋中摸出一个小球.用画树状图(或列表)的方法,分别求出小雨和小佳获胜的概率.22.如图,已知女排球场的长度OD 为20米,位于球场中线处的球网AB 的高度2.24米,一队员站在点O 处发球,排球从点O 的正上方2米的C 点向正前方飞去,排球的飞行路线是抛物线的一部分,当排球运行至离点O 的水平距离OE 为6米时,到达最高点G ,以O 为原点建立如图所示的平面直角坐标系.(1)写出C 点坐标___________;B 点坐标___________.(2)若排球运行的最大高度为3米,求排球飞行的高度p (单位:米)与水平距离x (单位:米)之间的函数关系式(不要求写自变量x 的取值范围);(3)在(2)的条件下,这次所发的球能够过网吗?如果能够过网,是否会出界?请说明理由.23.如图,在Rt ABC △中,90ACB ∠=︒,延长CA 到点D ,以AD 为直径作O ,交BA 的延长线于点E ,延长BC 到点F ,使BF EF =.(1)求证:EF 是O 的切线.(2)若9OC =,4AC =,8AE =,则BC =______,BE =______.24.如图,已知等边ABC ,直线AM BC ⊥,点M 为垂足,点D 是直线AM 上的一个动点,线段CD 绕点D 顺时针方向旋转60°得线段DE ,联结BE 、CE .(1)如图1,当点D 在线段AM 上时,说明BE AB ⊥的理由;(2)如图2,当点D 在线段MA 的延长线上时,设直线BE 与直线AM 交于点F ,求BFM ∠的度数;(3)定义:有一个内角是36︒的等腰三角形称作黄金三角形,联结DB ,当DBE 是黄金三角形吋,直接写出BEC ∠为______度.25.抛物线2y ax 2x c =++与x 轴交于(1,0)A -、B 两点.与y 轴交于点(0,3)C 、点(,3)D m 在抛物线上.(1)求抛物线的解析式.(2)如图1,连接BC 、BD ,点P 在对称轴左侧的抛物线上,若PBC DBC ∠=∠,求点P 的坐标.(3)如图2,过点A 的直线∥m BC ,点Q 是直线BC 上方抛物线上一动点,过点Q 作QE m ⊥,垂足为点E ,连接BE ,CE ,CQ ,QB .当四边形BECQ 的面积最大时,求点Q 的坐标及四边形BDCQ 面积的最大值。
2025届陕西省西安市爱知初级中学化学九年级第一学期期末复习检测试题含解析
2025届陕西省西安市爱知初级中学化学九年级第一学期期末复习检测试题注意事项:1.答题前,考生先将自己的姓名、准考证号填写清楚,将条形码准确粘贴在考生信息条形码粘贴区。
2.选择题必须使用2B铅笔填涂;非选择题必须使用0.5毫米黑色字迹的签字笔书写,字体工整、笔迹清楚。
3.请按照题号顺序在各题目的答题区域内作答,超出答题区域书写的答案无效;在草稿纸、试题卷上答题无效。
4.保持卡面清洁,不要折叠,不要弄破、弄皱,不准使用涂改液、修正带、刮纸刀。
一、单选题(本大题共10小题,共20分)1.下列四组物质遇到明火可能会发生爆炸的一组是A.氧气和氮气B.二氧化碳和氢气C.棉尘和空气D.氢气和一氧化碳2.深海鱼油中的DHA(化学式为C22H32O2)是一种特殊的不饱和脂肪酸,这种不饱和脂肪酸是大脑细胞形成、发育及运作不可缺少的物质基础。
下列关于DHA说法中错误的是A.一个DHA分子由56个原子构成B.DHA中C、H、O三种元素的质量比为33:4:4C.DHA的相对分子质量为328D.DHA中氢元素和氧元素的质量分数不相等3.下列叙述错误的有几项()①尘士飞扬,说明分子是运动的②硫在氧气中燃烧,发出明亮的蓝紫色火焰③水是由氢原子和氧原子组成的④地壳中含量最多的元素是铁⑤氢原子是保持氢气化学性质的最小微粒⑥CO2、CO、C60这三种物质都属于化合物⑦2Mg2+这个粒子中数字“2”表示意义相同⑧水从液态变成气态时,分子间隔变大A.4 B.5 C.6 D.74.检验、提纯等都是化学实验的基本方法。
下列实验方案中,合理的是A.检验氢气中是否含有氧气:用带火星的木条检验B.除去氧化铜粉末中的木炭粉:将混合物在空气中充分灼烧C.验证质量守恒定律:比较镁带燃烧前后石棉网上的固体质量D.验证酒精中含有碳、氢元素:将酒精点燃后在火焰上方罩个干燥冷却的烧杯5.食盐常添加碘酸钾(化学式为KIO3)用于补充碘成分,碘酸钾在化学的物质分类上属于A.混合物B.化合物C.氧化物D.单质6.3Cl2+6NaOH5NaCl+X+3H2O的反应中,X的化学式是A.O2B.NaClO C.HCl D.NaClO37.逻辑推理是学习化学常用的学习方法,下列推理正确的是()A.化合物由多种元素组成,因此由多种元素组成的纯净物一定是化合物B.置换反应中有单质生成,则有单质生成的反应一定是置换反应C.一氧化碳和二氧化碳都是碳的氧化物,所以两者的性质相同D.溶液是均一稳定的混合物,所以均一稳定的液体一定是溶液8.从环境保护的角度考虑,下列燃料中最理想的是()A.氢气B.煤炭C.汽油D.乙醇9.下列实验现象,描述错误的是( )A.将一氧化碳通入灼热的氧化铜,黑色粉末逐渐变为红色粉末B.铁丝在氧气中剧烈燃烧,火星四射,生成黑色固体C.向硫酸铜溶液中滴加氢氧化钠溶液,产生蓝色沉淀D.电解水时负极和正极产生气体的体积比为1: 210.二氧化硫是造成酸雨的主要成分之一,二氧化硫排放到空气中溶于水生成亚硫酸(H2SO3),其中亚硫酸中硫元素(S)的化合价为A.+2B.-2C.+4D.+6二、填空题(本大题共1小题,共4分)11.请运用化学知识回答以下问题:(1)在厨房炒菜时,在客厅都能闻到菜香味的主要原因是_______________________。
2025届成都市教科院附属学校英语九年级第一学期期末复习检测试题含解析
2025届成都市教科院附属学校英语九年级第一学期期末复习检测试题考生请注意:1.答题前请将考场、试室号、座位号、考生号、姓名写在试卷密封线内,不得在试卷上作任何标记。
2.第一部分选择题每小题选出答案后,需将答案写在试卷指定的括号内,第二部分非选择题答案写在试卷题目指定的位置上。
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考试结束后,请将本试卷和答题卡一并交回。
Ⅰ. 单项选择1、She________live alone. But she________living alone because she feels lonely.A.used to, doesn’t used to B.is used to, was used toC.used to, is not used to D.was used to, doesn’t used to2、The ____________ of printing helped to spread the knowledge.A.result B.experience C.invention D.reason3、—How is your head teacher?—Our head teacher is _________ with us and we usually feel nervous in his lesson.A.kind B.pleased C.strict D.good4、Look! Some _______ are on the farm..A.horse and cows B.monkey and cowC.sheep and cows D.sheep and cow5、—Sonia, is this your dictionary?—Oh, no, it’s not ______. Ask Li Lei. He is looking for ______.A.me; hers B.mine; him C.my; her D.mine; his6、-- Dad, about ________ of our classmates wear glasses.-- Oh, that's terrible. You all should take good care of your eyes.A.three fourthB.third fourthC.third fourthsD.three quarters7、—May we leave the classroom now? —No, you_____. You______to leave until the bell rings.A.can’t; are allowed B.mustn’t; aren’t allowedC.needn’t; are allowed D.needn’t; aren’t allowed8、— Dad, I am af raid I can’t pass the exam.—If you keep on studying hard, you will make it.A.succeed B.fail C.produce9、A new road __________ near my school next year.A.builds B.will buildC.is built D.will be built10、Frank had to take a ________/ breɪk/ and ate something after working all night.A.bread B.blow C.brown D.breakⅡ. 完形填空11、When the Spring Festival falls, red flowers bloom (盛开) on the windows of my home. Of course,these are not 1 flowers. But in the cold of 2 , these wonderful flowers bring warmth to us.They are made of paper by my3 ,who is now aged 70. With a few cuts, she4 a piece of paper into a beautiful work of art.When my grandma was young, she was considered clumsy(笨拙的) in the 5 . But on a coldwinter night, she saw some beautiful red paper 6 on her new neighbor's windows She was so interested in it and decided to 7from her neighbor.Tons of paper was wasted, and her fingers werecut many times. But she 8 practicing. It took 10 years 9 my grandma could make paper art skillfully (娴熟地). The red flowers, blooming like burning fire, light up her world. Now, everyone in the village 10 her.My grandma's story has shown me what it takes to realize one's dream.1.A.nice B.real C.full D.sweet2.A.spring B.summer C.autumn D.winter3.A.sister B.aunt C.mother D.grandma4.A.changes B.spreads C.places D.bums5.A.city B.town C.villages D.block6.A.art B.money C.walls D.books7.A.buy B.steal C.learn D.borrow8.A.ended up B.kept on C.put off D.thought about9.A.before B.after C.while D.since10.A.serves B.pardons C.respects D.warnsⅢ. 语法填空12、短文填空Other fail little it surprised they possible back serve be call toThousands of people 1.attracted(吸引) to a 2-meter-wide lane in Tongcheng, Anhui province this spring festival. Although it was 2.to count the exact number, at3.5,000 people visited it each day.This narrow lane4.Liuchi Lane in Chinese, won greater popularity after the story behind it was broadcast(播放)on a CCTV show. At first glance it may seem unimportant, but to those who know5.history,Liuchi Lane means much more than its appearance. In the early Qing Dynasty, it 6.as the boundary(边界) between Zhang family and Wu family. They quarreled(争论)fiercely over which family had the right to build a wall on the dividing line. After 7.to make acompromise(妥协), the Zhangs turned to Zhang Ying, a family member of Zhangs and also a top official in the emperor’s court. To their8., Zhang Ying did not provide the support they needed. He called for his family to move back 1 meter from the boundary. Seeing that, the Wus agreed to move 9.another meter, thus Liuchi Lane was born.In reality, the story of Liuchi Lane is not only about how to get along with10., but also a guideline for all the officials in ChinaⅣ. 阅读理解A13、Since 2005, office workers in Japan have been allowed to wear T-shirts and shorts to work, instead ofbusiness suits and ties. Why? The answer is simple. The government is trying to reduce the use of air conditioners.Although air conditioners make our lives easier in hot weather, they can be bad for the environment. The coolers use a great deal of electricity to function and add to the effects of global warming.But it is a vicious (恶性的) cycle. As temperatures rise, there will be more need for air conditioning.Pawel Wargocki, an expert at the Technical University of Denmark, told The New York Times: “The air conditioning causes a huge problem to global climate and energy use.” He added, “More and more air conditioners are used. It’s very dangerous.”We can’t live with air conditioning. But the thing is, we can’t live without it either. Scientific studies have shown that if indoor temperatures are high in hot weather, people will suffer from certain health problems and have a lower productivity.For example, when air conditioners were turned down in Japanese offices, the room temperature rose to 3 or higher. As a result, workers started getting headaches, and feeling sleepy.They couldn’t give all their attention to work and their productivity dropped by 2 percent, which means they did 30 minutes less work each day, according to Shinichi Tanabe, a professor of architecture at Waseda University in Tokyo, Japan.So the best way to balance our comfort with environmental awareness is to change our expectations. Most of the time, it’s not that the room is too hot, but that we are used to it being cooler.“The temperature many Americans find most comfortable indoors in summer—21C— feels uncomfortable to most Europeans, who find it too cold.” said Michael Sivak, professor at a University of Michigan, US. He said it is because Europeans care more about the environment and are willing to accept a bit more heat in summer.Also, we can try to change our daily behavior. Open the windows to let air in, for example. Turning on the air conditioner is not the only way to make us feel comfortable.1.What did the Japanese government do to reduce the use of air conditioners?A.To allow the workers to wear T-shirts to work.B.To stop the workers from using air conditioners.C.To let people accept a bit more heat in summer.D.To ask people to open windows to let air in.2.The writer probably agrees that ______.A.most people will be sick without air conditionersB.global warming is mainly caused by air conditionersC.people have to be used to living without air conditionersD.we should change our expectations to protect environment3.We can learn from the passage that ______.A.Japanese office workers are less productive in summerB.Americans care less about environment than EuropeansC.People in Denmark believe air conditioners are dangerousD.Europeans accept a bit cooler than Americans in summer4.Which of the following would be the best title for the passage?A.Protect Environment: Is it time to change business suits?B.Protect Environment: Is it time to focus on temperature?C.Protect Environment: Shall we change our expectations?D.Protect Environment: Shall we stop using air conditioners?B14、A sign is another kind of language. Here are some of them that you can see on the roads.Number One is a sign with the number forty on it. When driver see this sign, they must not go at more than forty kilometers one hour. We see this sign when we are getting near a town. Number Two is a sign that we’re near a crossing. We must drive carefully. Number Three is a sign that there is a bend (弯道) on the road. Again, we must drive slowly and carefully. It is not safe to go round a bend very fast. Number Four is a sign that there is another road coming infrom the right. There is a junction(交叉点) at this place.Number Five is a sign that there is a dangerous mountain road and Number Six is a sign that the road gets narrow. Drivers must go slowly and carefully when they see these signs. Number Seven is a sign that there is a school at the side of the street or the road. Perhaps there are children going to or leaving school. So drivers must look carefully and go slowly. Number Eight is a sign with a letter “P”on it. The letter “P” means “Parking”. At some places, there are signs “NO PARKING” or “NO WAITING”. If a driver leaves his car near one of these signs, a policeman may come and write down the number of his car.1.. At the place where you see Sign 1, ________.A.you are already out of the townB.you still have forty kilometers to goC.there must be a town forty kilometers awayD.there must be many houses and buildings not far away2.There stands Sign 2 near a place where ________.A.two roads cross B.the road gets narrowC.there are no traffic lights D.people can cross the road3.You have to drive not only slowly but also carefully when you find ________ according to the passage.A.either of signs 2 and 3 B.all of signs 3, 5, 6 and 7C.each of the eight signs D.any of signs 3, 6, 7 and 84.A driver can leave his car ________.A.near Sign 8 at any timeB.near a sign with “NO PARKING” on itC.near Sign 3 in the daytimeD.near a sign with “NO WAITING” if there’re no police there5.People put these signs on the roads to ________.A.show drivers the way B.stop cars going too fastC.make driving even safer D.learn another kind of languageC15、1.How much money can a tour guide get for working eight hours a day?A.$24. B.$48. C.$72. D.$96.2.If you find Lizzie the cat, you should ______.A.call 555-3826 B.call 555-4863C.********************D.keep it as your own pet3.According to the information above, which of the following is TRUE?A.A tour guide doesn’t have to speak English.B.Andrew can help you find a house with a yard.C.A single girl can rent a room for$35 a month.D.The owner will give you a large pet if you find Lizzie.D16、All the time you spent memorizing multiplication tables(乘法表)may have made you a better mathematician, according to a new study.A team of scientists from Stanford University, in California, have shown how the brain reorganizes itself as kids learn math.After a certain amount of time spent practicing math, kids can put away the calculator(计算器).They don’t even need to count on their fingers.They simply know the answers to subtraction(−), addition (+), and multiplication (×) facts.The quicker kids can recall basic math facts, the easier it is for them to solve more difficult math problems.The Stanford University researchers observed(观察)the brain activity of 28 students, ages 7 to 4, for the study.They took scans of the students’ brains as the students solved math calculations without the help of a calculator, pen or paper.A calculation—three plus four equals seven, for example—flashed on a screen.The students pushed a button to say if the answer was right or wrong.The scientists also recorded the response speed, and what parts of the brain became active as the kids pushed the button.These observations showed a process called_______.Rather than using their fingers to count, or scrawling outequations(列算式)on a piece of paper, the students pulled the answers from memory.It’s as if the answers to basic subtraction, addition, and multiplication problems are kept in a long-term storage part in the brain.The storage part was built from repetition.“Experience really does matter,” said Dr.Kathy Mann Koepeke of the National Institutes of Health.Children make the changefrom counting to fact retrieval when they are 8 to 4 years old, the study says.This is the time when most students are learning basic addition and subtraction.When kids have basic math facts memorized, the brain has more free space to learn more difficult math.This process has benefits for the future.The study showed, as kids got older, their answers relied more on memory and became quicker and more correct.Less brain activity was involved in counting.Some children make this change quicker than others.Scientists hope to use this research to develop new strategies(策略)to help kids learn math at all levels.One strategy the study suggests is for students to test themselves in different orders—solving five times three before five times two, for example.Mixing up the order keeps the brain active.Keep these trainings up, and you may be a math whiz in no time.1.In the study, kids put away calculators because they can ______.A.use calculators wellB.scrawl out equationsC.recalling basic math factsD.count on their fingers2.The process of “fact retrieval”in Paragraph 4probably means ______.A.picking out results by working together with their partnersB.finding out answers based on practicing carefully and patientlyC.trying out ways of dealing with problems by attending trainingD.working out problems according to what they have remembered3.What can we know from the study?A.Kids need to practice using calculators to keep their brains active.B.Math learners should change to counting practice at the age of 8 to 4.C.Training by mixing up the order of multiplication tables can help learn math better.D.Less brain activity may be involved in solving difficult problems when kids get older.5.What is the best title of the passage?A.Math from MemoryB.Strategy DevelopmentC.Brain ReorganizationD.Training by RepetitionE17、Learning vocabulary doesn’t have to be boringMany students have problems with vocabulary when preparing for tests like the SATs. Last week, TEENS offered advice on how to memorize new words. If you’re now familiar with those methods, w hy not try moving on to the next stage by using the following tips to build up your English vocabulary.Repeated encounters (反复遇见)The usual way of “memorizing” new material is through repeating it while it’s still in your mind. However, simply repeating a word seems to have little long-term effect.But one kind of repetition is effective: repetition of encounters with a word. Words stand a good chance of being remembered if they’ve been met at least seven times in your reading, at intervals (相隔一段时间). So, read English magazines, newspapers, and websites regularly.Spacing and reviewIt’s better to distribute (分配) memory work over a period of time rather than do it in a single session (一段时间). This is known as the “principle of distributed practice”, where new vocabulary introduced in one lesson is reviewed in the next, and then in later lessons.Use it or lose itPutting words to use in interesting ways helps add them to your long-term memory. In a study of vocabulary learning, students who made up their own sentences with new words and read them aloud remembered them better thanthose who learned the words separately and silently. Best of all, students were given the task of silently imagining a picture in their minds to go with a new word. So it might help if learners link words with a mental image.Mix it upLet’s face it —learning words can be really boring, but it doesn’t have to be. Thanks to the Internet, there are more interesting ways to get the job done. Try watching videos from English-speaking social media celebrities and copying the way they speak. You could even upload your own videos and try becoming a star yourself.1.According to the passage, a good way of memorizing new words is ________.A.simply repeating themB.learning the words silentlyC.reviewing them in one lessonD.reading English magazines regularly2.To make learning words more interesting, we can ________.A.try to become a famous starB.read new words on websitesC.make and upload our own English videosD.draw a picture on the paper with a new word3.What does the passage mainly talk about?A.Where to use new words. B.How to memorize new words.C.When to learn new words. D.What to write with new words.F18、Would you completely trust your friends if you couldn’t see anything? I f you are falling, do you believe they will catch you and help you stand up? Our teacher helped us to find the answer.Our teacher is named Morrie. One day, he said that he had a game for us to try. He divided us into two groups. One of the students from the first group, stands in front of us with his eyes closed, and then falls backwards(向后). The second group stand behind him. Before he falls on the floor, they should catch him and not let him get hurt. Most of us were uncomfortable with this. We were afra id that they wouldn’t catch us and we would hit the floor. So we didn’t move. We just stood there and laughed in embarrassment(尴尬).Finally, one student, thin, quiet, dark-haired girl called Sarah, decided to move. She crossed her arms in front of her body, closed her eyes, and slowly fell backwards. She looked quiet but brave.For a moment, I was sure she was going to fall on the floor. But before she hit the floor, the other students caught her, held her and finally helped her stand up.“Whoa!!” we were a ll surprised, and also encouraged. The girl was happy, too. Seeing this, Morrie smiled and said to the girl, “When you close your eyes, you can’t see anything. You won’t feel safe. When you are falling, the feeling gets stronger. You are not sure whether you will hit the cold, hard floor or the warm, soft hands of your friends. But if you are brave enough to trust them, they won’t let you down.”1.According to the passage, we can know Morrie is ___________.A.a student B.a teacher C.a manager D.a farmer2.The writer thought the girl was ___________.A.strong but shy B.active C.quiet but brave D.careless3.In the end, the girl ___________.A.got hurtB.fell on the floorC.was only helped by the writerD.was caught by the other students4.The students felt encouraged because ___________.A.they could stop the game finallyB.they wanted to do something elseC.they thought they had won the gameD.they felt that they could trust each otherⅤ.书面表达19、“没有规矩,不成方圆。
2025届四川省遂宁市安居区九年级数学第一学期期末复习检测模拟试题含解析
2025届四川省遂宁市安居区九年级数学第一学期期末复习检测模拟试题考生请注意:1.答题前请将考场、试室号、座位号、考生号、姓名写在试卷密封线内,不得在试卷上作任何标记。
2.第一部分选择题每小题选出答案后,需将答案写在试卷指定的括号内,第二部分非选择题答案写在试卷题目指定的位置上。
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考试结束后,请将本试卷和答题卡一并交回。
一、选择题(每小题3分,共30分)1.国家实施”精准扶贫“政策以来,很多贫困人口走向了致富的道路.某地区2016年底有贫困人口9万人,通过社会各界的努力,2018年底贫困人口减少至1万人.设2016年底至2018年底该地区贫困人口的年平均下降率为x ,根据题意列方程得( ) A .()9121x -=B .()2911x -=C .()9121x +=D .()2911x +=2.将半径为5的圆形纸片,按如图方式折叠,若AB 和BC 都经过圆心O ,则图中阴影部分的面积是( )A .256π B .253π C .232π D .25π3.已知关于x 的一元二次方程280x mx +-=的一个根为1,则m 的值为( ) A .1B .-8C .-7D .74.如果关于x 的一元二次方程k 2x 2-(2k+1)x+1=0有两个不相等的实数根,那么k 的取值范围是( ) A .k>-14B .k>-14且0k ≠ C .k<-14D .k ≥-14且0k ≠ 5.下列约分正确的是( )A .632x x x=B .0x yx y +=+ C .222142xy x y =D .1()a b x a b x+=+6.小明同学对数据26,36,46,5■,52进行统计分析,发现其中一个两位数的个位数字被墨水涂污看不到了,则分析结果与被涂污数字无关的是 ( ) A .平均数B .方差C .中位数D .众数7.如图,点P 是以AB 为直径的半圆上的动点,CA AB PD AC ⊥⊥,于点D ,连接AP ,设AP x PAPD y =,﹣=,则下列函数图象能反映y 与x 之间关系的是( )A .B .C .D .8.已知一次函数()10y kx m k =+≠和二次函数()220y ax bx c a =++≠部分自变量和对应的函数值如表:x … -1 0 2 4 5 … y 1 … 0 1 3 5 6 … y 2…-159…当y 2>y 1时,自变量x 的取值范围是 A .-1<x <2B .4<x <5C .x <-1或x >5D .x <-1或x >49.下列算式正确的是( ) A .110--=B .()33--=C .231-=D .|3|3--=10.如图,△ABC 中,∠A =65°,AB =6,AC =3,将△ABC 沿图中的虚线剪开,剪下的阴影三角形与原三角形不构成相似的是( )A .B .C .D .二、填空题(每小题3分,共24分)11.已知二次函数222(1)22y x m x m m =--+--(m 为常数),若对于一切实数m 和均有y ≥k ,则k 的最大值为____________.12.如图,已知圆锥的底面半径为3,高为4,则该圆锥的侧面积为______.13.已知关于x 的一元二次方程x 2+kx ﹣6=0有一个根为﹣3,则方程的另一个根为_____.14.二次函数26y x x =-+图象的开口向__________.15.方程x 2﹣9x =0的根是_____.16.设O 为△ABC 的内心,若∠A =48°,则∠BOC =____°. 17.某种商品每件进价为10元,调查表明:在某段时间内若以每件x 元(10≤x ≤20且x 为整数)出售,可卖出(20﹣x )件,若使利润最大,则每件商品的售价应为_____元.18.函数224y x x =-+沿直线1y =翻折所得函数解析式为_____________. 三、解答题(共66分)19.(10分)()1解方程:2230x x +-=.()2如图,在平面直角坐标系中,ABC 的顶点坐标分别为()()()1,1,2,3,4,2A B C .以点()1,1A 为位似中心画出ABC 的位似图形11AB C △,使得11AB C △与ABC 的位似比为2:1,并写出点11,B C 的坐标.20.(6分)如图,在矩形ABCD中,AB=10,动点E、F分别在边AB、AD上,且AF=12AE.将△AEF绕点E顺时针旋转10°得到△A'EF',设AE=x,△A'EF'与矩形ABCD重叠部分面积为S,S的最大值为1.(1)求AD的长;(2)求S关于x的函数解析式,并写出自变量x的取值范围.21.(6分)已知关于的一元二次方程:.(1)求证:对于任意实数,方程都有实数根;(2)当为何值时,方程的两个根互为相反数?请说明理由.22.(8分)如图,直线l的解析式为y=34x,反比例函数y=xk(x>0)的图象与l交于点N,且点N的横坐标为1.(1)求k 的值;(2)点A 、点B 分别是直线l 、x 轴上的两点,且OA =OB =10,线段AB 与反比例函数图象交于点M ,连接OM ,求△BOM 的面积.23.(8分)已知抛物线24y ax bx =+-经过点()2,0A ,()4,0B -,与y 轴交于点C .(1)求这条抛物线的解析式;(2)如图,点P 是第三象限内抛物线上的一个动点,求四边形ABPC 面积的最大值.24.(8分)如图,在平面直角坐标系中,OA ⊥OB ,AB ⊥x 轴于点C ,点A (3,1)在反比例函数ky x=的图象上.(1)求反比例函数ky x=的表达式; (2)在x 轴的负半轴上存在一点P ,使得S △AOP =12S △AOB ,求点P 的坐标; (3)若将△BOA 绕点B 按逆时针方向旋转60°得到△BDE ,直接写出点E 的坐标,并判断点E 是否在该反比例函数的图象上,说明理由.25.(10分)已知关于x 的一元二次方程x 2+(2m +3)x +m 2=1有两根α,β (1)求m 的取值范围;(2)若α+β+αβ=1.求m 的值.26.(10分)如图,在小山的东侧A 处有一一热气球,以每分钟28米的速度沿着与垂直方向夹角为30°的方向飞行,半小时后到达C 处,这时气球上的人发现,在A 处的正西方向有一处着火点B ,5分钟后,在D 处测得着火点B 的俯角是15°,求热气球升空点A 与着火点B 的距离.(结果保留根号,参考数据:626215,15,1523,cot152344sin cos tan -+︒=︒=︒=-︒=+)参考答案一、选择题(每小题3分,共30分) 1、B【分析】等量关系为:2016年贫困人口()212018⨯-=下降率年贫困人口,把相关数值代入计算即可.【详解】解:设这两年全省贫困人口的年平均下降率为x ,根据题意得:()2911x -=,故选B . 【点睛】本题考查由实际问题抽象出一元二次方程,得到2年内变化情况的等量关系是解决本题的关键. 2、B【解析】如图(见解析),先利用翻折的性质、直角三角形的性质求出AOD ∠的度数,再根据垂径定理、等腰三角形的性质得出AOB ∠度数,从而得出AOC ∠的度数,最后根据翻折的性质得出AOC S S =阴影扇形,利用扇形的面积公式即可得.【详解】如图,过点O 作⊥OD AB ,并延长OD 交圆O 与点E ,连接OA 、OB 、OCAD BD ∴=(垂径定理)由翻折的性质得11,22AOC OD ED OE OA S S ====阴影扇形 30,9060OAD AOD OAD ∴∠=︒∠=︒-∠=︒2120AOB AOD ∴∠=∠=︒(等腰三角形的三线合一)同理可得120BOC ∠=︒360120AOC AOB BOC ∴∠=︒-∠-∠=︒21205253603AOC S ππ⋅∴==扇形 253S π∴=阴影 故选:B .【点睛】本题考查了垂径定理、翻折的性质、扇形的面积公式等知识点,利用翻折的性质得出AOD ∠的度数是解题关键. 3、D【解析】直接利用一元二次方程的解的意义将x=1代入求出答案即可. 【详解】∵关于x 的一元二次方程x 2+mx −8=0的一个根是1, ∴1+m −8=0, 解得:m=7. 故答案选:D. 【点睛】本题考查的知识点是一元二次方程的解,解题的关键是熟练的掌握一元二次方程的解. 4、B【分析】在与一元二次方程有关的求值问题中,必须满足下列条件:(1)二次项系数不为零;(2)在有两个实数根下必须满足△=b 2-4ac≥1.【详解】由题意知,k≠1,方程有两个不相等的实数根,所以△>1,△=b 2-4ac=(2k+1)2-4k 2=4k+1>1. 因此可求得k >14-且k≠1. 故选B . 【点睛】本题考查根据根的情况求参数,熟记判别式与根的关系是解题的关键. 5、D【分析】根据约分的运算法则,以及分式的基本性质,分别进行判断,即可得到答案.【详解】解:A 、642x x x=,故A 错误;B 、1x yx y+=+,故B 错误; C 、22242=xy y x y x,故C 错误;D 、1()a b x a b x+=+,正确;故选:D . 【点睛】本题考查了分式的基本性质,以及约分的运算法则,解题的关键是熟练掌握分式的基本性质进行解题. 6、C【分析】利用平均数、中位数、方差和标准差的定义对各选项进行判断.【详解】解:这组数据的平均数、方差和标准差都与被涂污数字有关,而这组数据的中位数为46,与被涂污数字无关. 故选:C . 【点睛】本题考查了方差:它也描述了数据对平均数的离散程度.也考查了中位数、平均数和众数的概念.掌握以上知识是解题的关键. 7、C【解析】设圆的半径为R ,连接PB ,求出1sin 22AP ABP x R R∠==,根据CA ⊥AB,求出21122PD APsin x x R R α⨯===,即可求出函数的解析式为212y PA PD x x R-+==-.【详解】设:圆的半径为R ,连接PB ,则1sin 22AP ABP x R R∠==, CA AB ⊥,即AC 是圆的切线,则PDA PBA α∠∠==,则2122x PD APsin x x R Rα⨯=== 则212y PA PD x x R-+==-图象为开口向下的抛物线, 故选:C . 【点睛】本题考查了圆、三角函数的应用,熟练掌握函数图像是解题的关键. 8、D【分析】利用表中数据得到直线与抛物线的交点为(-1,0)和(1,5),-1<x <1时,y 1>y 2,从而得到当y 2>y 1时,自变量x 的取值范围.【详解】∵当x=0时,y 1=y 2=0;当x=1时,y 1=y 2=5; ∴直线与抛物线的交点为(-1,0)和(1,5), 而-1<x <1时,y 1>y 2,∴当y 2>y 1时,自变量x 的取值范围是x <-1或x >1. 故选D . 【点睛】本题考查了二次函数与不等式:对于二次函数y=ax 2+bx+c (a 、b 、c 是常数,a≠0)与不等式的关系,利用两个函数图象在直角坐标系中的上下位置关系求自变量的取值范围,可作图利用交点直观求解,也可把两个函数解析式列成不等式求解. 9、B【解析】根据有理数的减法、绝对值的意义、相反数的意义解答即可. 【详解】A. 112--=-,故不正确; B. ()33--=,正确; C. 231-=-,故不正确;D. |3|3--=-,故不正确; 故选B. 【点睛】本题考查了有理数的运算,熟练掌握有理数的减法法则、绝对值的意义、相反数的意义是解答本题的关键. 10、C【分析】根据相似三角形的判定定理对各选项进行逐一判定即可.【详解】A 、阴影部分的三角形与原三角形有两个角相等,故两三角形相似,故本选项不符合题意; B 、阴影部分的三角形与原三角形有两个角相等,故两三角形相似,故本选项不符合题意; C 、两三角形的对应角不一定相等,故两三角形不相似,故本选项符合题意; D 、两三角形对应边成比例且夹角相等,故两三角形相似,故本选项不符合题意. 故选:C . 【点睛】本题考查了相似三角形的判定,熟练掌握相似三角形的判定定理是解题的关键.二、填空题(每小题3分,共24分) 11、134-【分析】因为二次函数系数大于0,先用含有m 的代数式表示出函数y 的最小值,得出min y 23m m =+-,再求出于m 的函数23M m m =+-的最小值即可得出结果. 【详解】解: 222(1)22y x m x m m =--+--, 22min41(22)4(1)41m m m y ⨯⨯----=⨯23m m =+-,关于m 的函数为23M m m =+-, 2min 41(3)113=414M ⨯⨯--=-⨯,∴134k -≥,∴k 的最大值为134-. 【点睛】本题考查二次函数的最值问题,先将函数化为顶点式,即可得出最值. 12、15π【分析】根据圆锥的底面半径为3,高为4可得圆锥的母线长,根据圆锥的侧面积S=rl π即可得答案. 【详解】∵圆锥的底面半径为3,高为4,,∴该圆锥的侧面积为:π×3×5=15π,故答案为:15π【点睛】本题考查求圆锥的侧面积,如果圆锥的底面半径为r ,母线长为l ,则圆锥的侧面积S=rl π;熟练掌握圆锥的侧面积公式是解题关键.13、1【分析】设方程的另一个根为a ,根据根与系数的关系得出a+(﹣3)=﹣k ,﹣3a=﹣6,求出即可.【详解】设方程的另一个根为a ,则根据根与系数的关系得:a+(﹣3)=﹣k ,﹣3a=﹣6,解得:a=1,故答案为1.【点睛】本题考查了根与系数的关系和一元二次方程的解,能熟记根与系数的关系的内容是解此题的关键.14、下【分析】根据二次函数的二次项系数即可判断抛物线的开口方向.【详解】解:∵26y x x =-+,二次项系数a =-6,∴抛物线开口向下,故答案为:下.【点睛】本题考查二次函数的性质.对于二次函数y=ax 2+bx+c (a ≠0),当a >0时,抛物线开口向上,当a <0时,抛物线开口向下.15、x 1=0,x 2=1【分析】观察本题形式,用因式分解法比较简单,在提取x 后,左边将变成两个式子相乘为0的情况,让每个式子分别为0,即可求出x .【详解】解:x 2﹣1x =0即x (x ﹣1)=0,解得x 1=0,x 2=1.故答案为x 1=0,x 2=1.【点睛】此题主要考查一元二次方程的求解,解题的关键是熟知因式分解法的应用.16、1【详解】解:∵点O 是△ABC 的内切圆的圆心, 11,22OBC ABC OCB ACB ∴∠=∠∠=∠, 11()(180)6622OBC OCB ABC ACB A ∴∠+∠=∠+∠=-∠=, 180(0)18066114.BOC BC OCB ∴∠=-∠+∠=-=故答案为1.17、1【解析】本题是营销问题,基本等量关系:利润=每件利润×销售量,每件利润=每件售价﹣每件进价.再根据所列二次函数求最大值.【详解】解:设利润为w 元,则w =(20﹣x )(x ﹣10)=﹣(x ﹣1)2+25,∵10≤x ≤20,∴当x =1时,二次函数有最大值25,故答案是:1.【点睛】本题考查了二次函数的应用,此题为数学建模题,借助二次函数解决实际问题.18、2(1)1y x =---【解析】函数224y x x =-+沿直线1y =翻折所得函数图像开口向下,只要根据轴对称的性质求出对称后的顶点坐标即可.【详解】∵224y x x =-+=(x -1)2+3,∴其顶点坐标是(1,3),∵(1,3)关于直线1y =的点的坐标是(1,-1),∴所得函数解析式为y =-(x -1)2-1.故答案为:()211y x =---.本题考查了二次函数的轴对称变换,其形状不变,但开口方向相反,因此a 值为原来的相反数,顶点位置改变,只要根据轴对称的点坐标特征求出新的顶点坐标,即可确定解析式.三、解答题(共66分)19、(1)123,1x x =-=;(2)见解析,点1B 的坐标为()3,5;点1C 的坐标为()7,3.【分析】 ⑴根据配方法解出即可;⑵根据相似比找到对应的点1B ,1C 即可.【详解】解:()21230,x x +-= 223x x +=,22131x x ++=+,()214x +=,12,=-±.123,1x x ∴=-=.(解法不唯一)()2解:如图,11AB C △即为所求.点1B 的坐标为()3,5;点1C 的坐标为()7,3.此题主要考查了解一元二次方程的配方法及位似图形的性质,熟练掌握相关知识是解题的关键.20、(1)6AD =;(2)221(06)49(67)1440(710)xx S x x x x ⎧<≤⎪⎪=<≤⎨⎪-+-<<⎪⎩【分析】(1)根据题意,当A F ''在CD 上时,AD A E AE '==,则重叠的面积有最大值1,根据面积公式,即可求出AD 的长度(2)根据题意,需要对x 的值进行讨论分析,分成三种情况进行解题,分别求出S 与x 的关系式,即可得到答案.【详解】(1)如图,当A F ''在CD 上时,S 9=,∵A E AE x '==,1122A F AF AE x ''===, ∴2111192224S A F A E x x x ''''=•=⨯⋅==. 解方程,得:6x =或6x =-(舍去),∴6AD A E '==.(2)①当06x <≤时,如图,21124S A F A E x ''''∴=⨯⋅=. ②如图可知,EF '经过点C 时,A F CD ''∥,A EF GEC ''∴△∽△. CG EG A F A E ∴='''. 12CG A F EG A E ''∴=='. 11163222CG EG AD ∴===⨯=, 1037AE DG CD DG AB DG ==-=-=-=.当67x <≤时,如图,A F CD ''∥,A EF GEH ''∴△∽△.GH EG A F A E∴='''. 12GH A F EG A E ''∴=='. 11163222GH EG AD ∴===⨯=. 1163922S EG GH ∴=⋅⨯⨯=. ③当710x <≤时,如图,//EA BC ',A EF BME ''∠=∠,在A EF ''△和BME ∆中,A EF BME ''∠=∠,90A B '∠=∠=︒,A EF BME ''∴△∽△.12BE BM BM A F A F A E BE A E ''∴=•==''''. 22()2(10)BM EB AB AE x ∴==-=-.∵S S =矩形12BCGE BME BE S GE EB BM -=⋅-⋅△, 216(10)(10)2(10)14402S x x x x x ∴=---⋅-=-+-. 综上所述:221(06)49(67)1440(710)xx S x x x x ⎧<≤⎪⎪=<≤⎨⎪-+-<<⎪⎩. 【点睛】此题是四边形综合题,主要考查了旋转的性质,矩形的性质,相似三角形的判定和性质,三角形的面积公式,熟练运用分类讨论的思想进行解题是解本题的关键.21、(1)见解析;(2)1,理由见解析.【解析】试题分析:(1)根据方程的系数结合根的判别式,可得出△=(t ﹣3)2≥0,由此可证出:对于任意实数t ,方程都有实数根;(2)设方程的两根分别为m 、n ,由方程的两根为相反数结合根与系数的关系,即可得出m+n=t ﹣1=0,解之即可得出结论.试题解析:(1)证明:在方程x 2﹣(t ﹣1)x+t ﹣2=0中,△=[﹣(t ﹣1)]2﹣4×1×(t ﹣2)=t 2﹣6t+9=(t ﹣3)2≥0,∴对于任意实数t ,方程都有实数根;(2)解:设方程的两根分别为m 、n ,∵方程的两个根互为相反数,∴m+n=t ﹣1=0,解得:t=1.∴当t=1时,方程的两个根互为相反数.考点:根与系数的关系;根的判别式.22、(1)27;(2)2【分析】(1)把x =1代入y =34x ,求得N 的坐标,然后根据待定系数法即可求得k 的值; (2)根据勾股定理求得A 的坐标,然后利用待定系数法求得直线AB 的解析式,再和反比例函数的解析式联立,求得M 的坐标,然后根据三角形面积公式即可求得△BOM 的面积.【详解】解:(1)∵直线l 经过N 点,点N 的横坐标为1,∴y =34×1=92,∴N (1,92), ∵点N 在反比例函数y =x k (x >0)的图象上, ∴k =1×92=27;(2)∵点A 在直线l 上,∴设A (m ,34m ), ∵OA =10,∴m 2+(34m )2=102,解得m =8, ∴A (8,1),∵OA =OB =10,∴B (10,0),设直线AB 的解析式为y =ax +b ,∴8m n 610m n 0+=⎧⎨+=⎩,解得330m n =-⎧⎨=⎩, ∴直线AB 的解析式为y =﹣3x +30, 解33027y x y x =-+⎧⎪⎨=⎪⎩得127x y =⎧⎨=⎩或93x y =⎧⎨=⎩, ∴M (9,3),∴△BOM 的面积=11032⨯⨯=2. 【点睛】本题考查了反比例函数与一次函数的交点,一次函数图象上点的坐标特征,待定系数法求反比例函数的解析式和一次函数的解析式,求得A 、M 点的坐标是解题的关键.23、(1)2142y x x =+-;(2)1 【分析】(1)将()2,0A ,()4,0B -代入抛物线中求解即可;(2)利用分割法将四边形面积分成AOC OCP OBP S S S ∆∆∆++,假设P 点坐标,四边形面积可表示为二次函数解析式,再利用二次函数的图像和性质求得最值.【详解】解:(1)∵抛物线24y ax bx =+-经过点()2,0A ,()4,0B -,∴424016440a b a b +-=⎧⎨--=⎩,解得121a b ⎧=⎪⎨⎪=⎩, ∴抛物线的解析式为2142y x x =+-, (2)如图,连接OP ,设点21,42P x x x ⎛⎫+- ⎪⎝⎭, 40x -<<,四边形ABPC 的面积为S ,由题意得点()0,4C -,∴AOC OCP OBP S S S S ∆∆∆=++21111244()442222x x x ⎛⎫=⨯⨯+⨯⨯-+⨯⨯--+ ⎪⎝⎭24228x x x =---+2412x x =--+2(2)16x =-++,∵10-<,∴开口向下,S 有最大值,∴当2x =-时,四边形ABPC 的面积最大,最大值为1.【点睛】本题考查了待定系数法求二次函数解析式、分割法求面积、二次函数的图象及性质的应用,比较综合,是中考中的常考题型.24、(1)3y x=;(2)P (3-,0);(3)E (3-1),在.【分析】(1)将点A ,1)代入k y x=,利用待定系数法即可求出反比例函数的表达式;(2)先由射影定理求出BC=3,那么B ,﹣3),计算求出S △AOB =124=S △AOP =12S △AOB 点P 的坐标为(m ,0),列出方程求解即可;(3)先解△OAB ,得出∠ABO=30°,再根据旋转的性质求出E 1),即可求解.【详解】(1)∵点A ,1)在反比例函数k y x=的图象上,∴k=∴反比例函数的表达式为y =(2)∵A 1),AB ⊥x 轴于点C ,∴AC=1,由射影定理得2OC =AC•BC ,可得BC=3,B 3),S △AOB =12×4=∴S △AOP =12S △AOB 设点P 的坐标为(m ,0),∴12×|m|×,∴|m|=∵P 是x 轴的负半轴上的点,∴m=﹣∴点P 的坐标为(-,0);(3)点E 在该反比例函数的图象上,理由如下:∵OA ⊥OB ,OA=2,OB=AB=4,∴sin ∠ABO=OA AB =24=12, ∴∠ABO=30°,∵将△BOA 绕点B 按逆时针方向旋转60°得到△BDE ,∴△BOA ≌△BDE ,∠OBD=60°,∴BO=BD=OA=DE=2,∠BOA=∠BDE=90°,∠ABD=30°+60°=90°,而BD﹣OC=3,BC﹣DE=1,∴E(3-,﹣1),∵3-×(﹣1)=3,∴点E在该反比例函数的图象上.考点:待定系数法求反比例函数解析式;反比例函数系数k的几何意义;坐标与图形变化-旋转.25、(1)m≥﹣;(2)m的值为2.【解析】(1)根据方程有两个相等的实数根可知△>1,求出m的取值范围即可;(2)根据根与系数的关系得出α+β与αβ的值,代入代数式进行计算即可.【详解】(1)由题意知,(2m+2)2﹣4×1×m2≥1,解得:m≥﹣;(2)由根与系数的关系得:α+β=﹣(2m+2),αβ=m2,∵α+β+αβ=1,∴﹣(2m+2)+m2=1,解得:m1=﹣1,m1=2,由(1)知m≥﹣,所以m1=﹣1应舍去,m的值为2.【点睛】本题考查的是根与系数的关系,熟知x1,x2是一元二次方程ax2+bx+c=1(a≠1)的两根时,x1+x2=﹣,x1x2=是解答此题的关键.26、9803980.【分析】过D作DH⊥BA于H,在Rt△DAH中根据三角函数即可求得AH的长,然后在Rt△DBH中,求得BH的长,进而求得BA的长.【详解】解:由题意可知AD=(30+5)×28=980,过D作DH⊥BA于H.在Rt△DAH中,DH=AD•sin60°=98033,AH=AD×cos60°=980×12=490,在Rt △DBH 中,BH=tan15DH =4903×(2+3)=1470+9803, ∴BA=BH-AH=(1470+9803)-490=980(1+3)(米).答:热气球升空点A 与着火点B 的距离为980(1+3)(米).【点睛】本题主要考查了仰角和俯角的定义,一般三角形的计算可以通过作高线转化为直角三角形的计算.。
辽宁省大石桥市水源镇九一贯制学校2025届九年级化学第一学期期末复习检测试题含解析
辽宁省大石桥市水源镇九一贯制学校2025届九年级化学第一学期期末复习检测试题注意事项1.考试结束后,请将本试卷和答题卡一并交回.2.答题前,请务必将自己的姓名、准考证号用0.5毫米黑色墨水的签字笔填写在试卷及答题卡的规定位置.3.请认真核对监考员在答题卡上所粘贴的条形码上的姓名、准考证号与本人是否相符.4.作答选择题,必须用2B铅笔将答题卡上对应选项的方框涂满、涂黑;如需改动,请用橡皮擦干净后,再选涂其他答案.作答非选择题,必须用05毫米黑色墨水的签字笔在答题卡上的指定位置作答,在其他位置作答一律无效.5.如需作图,须用2B铅笔绘、写清楚,线条、符号等须加黑、加粗.一、单选题(本题包括12个小题,每小题3分,共36分.每小题只有一个选项符合题意)1.图示实验操作正确的是()A.加热液体B.振荡试管C.放回滴瓶D.检查气密性2.下列厨房常见物质中,能溶于水形成溶液的是A.食盐B.面粉C.菜籽油D.辣酱3.如图表示的基本反应类型是()A.化合反应B.分解反应C.置换反应D.还原反应4.小茜同学对部分物质的性质和用途做了如下归纳,错误的是()A.氮气的化学性颜稳定——填充食品包装袋防腐B.稀有气体通电时能发出不同颜色的光一一用于霓虹灯C.氦气的密度很小且稳定一一探空气球D.氮气高温时性质活泼——光合作用的原料5.下面关于水电解实验的叙述正确的是()A.实验说明水是由氢、氧两种元素组成的B.水电解的化学方程式:2H2O→2H2↑+O2↑C.实验说明水是由氢气和氧气组成的D.若a试管气体为5mL,则b试管气体为10mL6.小红在峨眉山旅游到达金顶时,感觉身体不适,可能发生了“高原反应”,主要是因为高海拔地区的空气稀薄,不能给人体提供足量的A.二氧化碳B.氧气C.氮气D.稀有气体7.下列金属不能与CuSO4溶液反应的是()A.Ag B.Fe C.Zn D.Al8.下列实验操作中,正确的是()A.点燃酒精灯B.过滤C.加热液体D.称量固体9.氨基钠(NaNH2)常用于医药制造,根据其性质,氨基钠应该用时制备,不要长时间贮存。
福建省厦门市双十中学2025届九年级英语第一学期期末复习检测试题含解析
福建省厦门市双十中学2025届九年级英语第一学期期末复习检测试题考生须知:1.全卷分选择题和非选择题两部分,全部在答题纸上作答。
选择题必须用2B铅笔填涂;非选择题的答案必须用黑色字迹的钢笔或答字笔写在“答题纸”相应位置上。
2.请用黑色字迹的钢笔或答字笔在“答题纸”上先填写姓名和准考证号。
3.保持卡面清洁,不要折叠,不要弄破、弄皱,在草稿纸、试题卷上答题无效。
Ⅰ. 单项选择1、-- I don't know _______ during the summer vacation. Any advice?-- How about doing some part-time jobs?A.what should I doB.where should I goC.what I should doD.where I should go2、--- Notre Dame de Paris, a symbol of European culture, was on fire last month.--- And this should be considered as a great to humans.A.past B.risk C.loss D.theft3、---Look! A book is on the floor .Whose is it?---It____be Rick’s.It has his name on it.A.mustn’t B.can’t C.must D.need4、—Hi, Anna. Are these your sister’s pencils?—Oh, no. They’re not ______.A.her B.him C.hers D.his5、There is not many differences between the two. I really don’t know ____.A.what should I choose B.which I should chooseC.which should I choose D.what I should choose6、Today, there are _____ chances to get to know people from other c ultures. Don’t just stare at your shoes. Go and say “Hi”.A.plenty of B.quite a lot C.too much7、The man tried several times to start the car, and he succeeded ____.A.in the past B.in the end C.at first D.at once8、I think ______ important to prepare more food for tomorrow’s party.A.that B.it C.this D.us9、There’s no ticker left for Lang Lang’s piano concert .you your sister can go to it .A.Both ; and B.Not only ; but alsoC.Either ; or D.Neither ; nor10、Lily Lucy may go with you because one of them must stay at home.A.Not only; but also B.Neither; norC.Both; and D.Either; orⅡ. 完形填空11、Mrs. Green 1 to work every day. She usually parks(停车) her car in the street 2 her office. One day after work, she got into her car and started driving 3 . Suddenly, she 4 a black car following her. The 5 was a big man. When she turned left, the black car turned left. When she turned right, the black car turned right. When she stopped 6 the traffic lights, the black car stopped 7 .She was 8 , so she drove quickly to the 9 . She was 10 to see that the black car also stopped behind her car. She jumped out of her car quickly and ran to a policeman. She asked for 11 and told him everything.The policeman quickly ran to the black car. Just then the driver in the black car came out and 12 the policeman, “I want 13 back this purse(钱包) to her. I think she 14 it on the street.”Mrs. Green’s face turned red. She thanked 15 man and the policeman.1.A.drove B.drives C.is driving D.drive2.A.in B.on C.outside D.inside3.A.home B.work C.to home D.to work4.A.thought B.knew C.found D.met5.A.driver B.policeman C.cleaner D.worker6.A.under B.over C.on D.at7.A.after B.before C.behind D.in front8.A.happy B.excited C.surprised D.afraid9.A.office B.police station C.market D.traffic lights10.A.happy B.excited C.surprised D.afraid11.A.her car B.money C.help D.her key12.A.asked B.spoke C.talked D.said to13.A.give B.to give C.giving D.gave14.A.lost B.lose C.put D.bought15.A.a B./ C.an D.theⅢ. 语法填空12、根据短文内容及首字母提示写出所缺单词,并将完整单词写在下面对应题号后的横线上。
江苏省如皋市常青初级中学2025届九年级英语第一学期期末复习检测试题含解析
江苏省如皋市常青初级中学2025届九年级英语第一学期期末复习检测试题注意事项:1.答题前,考生先将自己的姓名、准考证号填写清楚,将条形码准确粘贴在考生信息条形码粘贴区。
2.选择题必须使用2B铅笔填涂;非选择题必须使用0.5毫米黑色字迹的签字笔书写,字体工整、笔迹清楚。
3.请按照题号顺序在各题目的答题区域内作答,超出答题区域书写的答案无效;在草稿纸、试题卷上答题无效。
4.保持卡面清洁,不要折叠,不要弄破、弄皱,不准使用涂改液、修正带、刮纸刀。
Ⅰ. 单项选择1、On Nov 9,2016,Donald Trump _________ as the 45th president of the United States.A.chooses B.choseC.is chosen D.was chosen2、―The lake is said to be dry. Is that true?―It true. Look, some kids are swimming in it.A.must be B.can′t be C.may be D.could be3、Lucy got plenty of pocket money during Spring Festival, so she can _____ /əˈfɔ:rd/ the bike.A.avoid B.attend C.afford D.around4、When I was your age, I a lot of housework to do after school.A.will have B.am having C.have D.had5、Ice is not often seen here in winter as the temperature normally stays zero.A.up B.down C.above D.below6、Modern plastics can ______ very high and very low temperatures.A.stand B.hold C.keep D.take7、—Do you know ?—Let me see. I remember it was on March 18th.A.why did they move here B.why they moved hereC.when did they move here D.when they moved here8、We all think that no one can replace Tom who specializes in operating the computer in our company.A.take place B.instead C.take the place of D.instead of9、—It seems _____ you like to work with children.—Yes, they are so lovely.A.that B.what C.to D.as10、Every day, _______ people wait to watch the raising of the national flag at Tian’anmen Square.A.thousands B.two thousands C.thousand of D.thousands ofⅡ. 完形填空11、Alex was a giraffe who lived in a zoo with his mum, dad and other giraffe friends. As Alex grew, he 1 his neck was much shorter than the others’. He began to get 2 . But his mother told him that he was special. She said it did not matter how short or long his neck was.Alex tried not to think about his short 3 . But as his friends grew taller and taller, his neck 4 short. Sometimes the other giraffes 5 him, but his parents told him not to worry. They said that he should be happy with himself the way he was. He tried, but he could not even play in some giraffe sports. He was just too 6 .One day, one of the zoo workers came and took Alex away. Alex was scared. Would he have to leave the zoo just7 he didn't have a long neck? The zoo worker calmed (使平静)Alex down and gave him some food. He felt 8 Then he led Alex into another area of the zoo.Soon mum and dad came to see Alex. “I told you that you were special,” his mother said. “The zoo has made a place just for 9 because so many people want to see how special you are!” She was right. Visitors pointed to him and screamed(尖叫) 10 . They looked happy to see something so special at the zoo.1.A.wondered B.found C.hoped D.agreed2.A.powerful B.interested C.energetic D.worried3.A.leg B.tail C.neck D.nose4.A.stayed B.became C.turned D.felt5.A.looked after B.thought of C.laughed at D.agreed with6.A.tall B.short C.fat D.thin7.A.after B.unless C.because D.before8.A.good B.better C.bad D.worse9.A.him B.you C.us D.me10.A.excitedly B.angrily C.sadly D.heavilyⅢ. 语法填空12、请用适当的词完成下面的短文,并把所缺单词填写在答题卡指定的位置上。
2025届山东省聊城市城区英语九年级第一学期期末复习检测试题含解析
2025届山东省聊城市城区英语九年级第一学期期末复习检测试题注意事项:1.答卷前,考生务必将自己的姓名、准考证号、考场号和座位号填写在试题卷和答题卡上。
用2B铅笔将试卷类型(B)填涂在答题卡相应位置上。
将条形码粘贴在答题卡右上角"条形码粘贴处"。
2.作答选择题时,选出每小题答案后,用2B铅笔把答题卡上对应题目选项的答案信息点涂黑;如需改动,用橡皮擦干净后,再选涂其他答案。
答案不能答在试题卷上。
3.非选择题必须用黑色字迹的钢笔或签字笔作答,答案必须写在答题卡各题目指定区域内相应位置上;如需改动,先划掉原来的答案,然后再写上新答案;不准使用铅笔和涂改液。
不按以上要求作答无效。
4.考生必须保证答题卡的整洁。
考试结束后,请将本试卷和答题卡一并交回。
Ⅰ. 单项选择1、A notice was in order to remind the students of the changed speech time.A.sent up B.given up C.set up D.put up2、—How many friends of ________ can come to celebrate your birthday?—________, I think. They’re all preparing for the new term.A.you, No one B.yours, None C.you, None3、I always play basketball to relax myself Saturdays.A.on B.in C.at D.by4、—What will you do if you have two days_______?— I'd like to go_______ a trip to Suzhou.A.of; on ; B.off; inC.off; on D.of; for5、Wechat ____widely ____ by people all over the world. We can chat with our friends on it.A.is; using B.are; used C.is; used6、—I think this skirt is very nice.—I ag ree. But it’s 500 yuan. I don’t think I can it.A.afford B.find C.put D.see7、Mr Wang ________ because he is not in excellent condition.A.replaced B.will be replacedC.will replace D.has replaced8、—Shall we take a car?—No, we . It's only five minutes' walk.A.can't B.mustn't C.needn't D.couldn't9、You can’t imagine __________ good weather it was yesterday!A.how B.what C.what a D.how a10、—Look in the mirror. What happened_____ your face?—Oh, there's some ink on my face.A.to B.on C.in D.withⅡ. 完形填空11、Amy is 12 years old. She is in a middle school now. Today is her first day to go to 1 . She gets to school 2 7:30. She sees many boys and girls in the school. Amy gets her classroom quickly (很快地). It’s on 3 floor (层).The class begins at 8:00. 4 old woman comes into the classroom. She 5 very nice. “Good morning, I’m Mrs. Green. I’m your 6 teacher. Our English classes are on Monday and Wednesday. I think you’ll7 my classes,” says Mrs. Green. Af ter the English class, Amy has science and music. Music is her 8 subject. In the afternoon, Amy has art and 9 .This is Amy’s first day at school. Do you remember (记得) 10 ?1.A.room B.bed C.library D.school2.A.in B.on C.at D.for3.A.two B.the two C.second D.the second4.A.A B.An C.The D.不填5.A.sees B.looks C.meets D.watch6.A.P.E. B.music C.art D.English7.A.lose B.finish C.like D.do8.A.first B.next C.favorite D.dear9.A.apple B.milk C.history D.ruler10.A.mine B.yours C.his D.hersⅢ. 语法填空12、用括号内所给动词的适当形式填空,必要时可加助动词或情态动词.It’s four o’clock in the afternoon. Some students1.(talk) about the future. Here is one of their predictions.I believe we will be able to live in the space. There2.(be) many big clean hotels in the space. We can fly our rockets to the space hotel there. And also we 3.play) sports and enjoy all kinds of interesting food in the sky. — John Lisa Hawkins loves riding her BMX bike (极限单车)and her dream is to go to the Olympics. Lisa first got interested in BMX racing at the age of ten. Once she hired(租) a bike for a pound and immediately 4.(fall) in love with the sport. “It was the best pound I ever spent,” she says. Lisa is only 16 but she 5.already have) lots of success, and got first prize in a world BMX competition last y ear. BMX racing is quite a dangerous sport. But this didn’t keep herfrom6.(want) to do more competition. Lisa exercises every day. Like many sports people, she7.(believe) in luck as well as hard work.Ⅳ. 阅读理解A13、People drink alcohol(酒) in many parts of the world. Some medical scientists say light drinking may even be good for your health, especially for the heart. But they say such health benefits(好处) should be compared to the many health risks connected with alcohol use.Some research suggests that having one to two drinks of alcohol a day may be good for your health. Several large studies have shown that this type of drinking may lower the risk of heart disease and other diseases.A study last year suggested that drinking a little red wine may help lower the risk of breast cancer(乳腺癌) in women. Researchers at Medical Center in California studied 36 women. Each woman drank a glass of red or white wine every day for almost a month. Researchers collected blood samples(样本) from the women two times a month to test their hormone(激素) levels. The next month these women were told to exchange their wine. The researchers found that the women who drank red wine had lower levels of the female hormone than the white wine drinkers. They said red grapes have chemicals that may help to lower the risk of breast cancer.However, recent studies suggested that even a little alcohol may generally increase the risk of breast cancer in women. They found that women who drink four small glasses of wine a week increase their risk of breast cancer by 15 percent.Many studies have examined the harmful effects of alcohol use on the body. Alicia Ann Kowalchuk serves as medical director, saying “Healthy use for adults --- that's men under age 65 --- is no more than 4 drinks in a day and no more than 14 drinks in a week. And for women of all ages, it's no more than 7 drinks in a week and no more than 2 drinks in a day.” She says too much alcohol use has something to do with liver (肝脏) diseases, and many other cancers.1.How did the researchers carry out their study?A.By following an example. B.By listing facts.C.By collecting information worldwide. D.By comparing tested objects.2.Which of the following shows the structure of the passage?A.B.C.D.3.What is the best title of this passage?A.Alcohol drinking may do harm to your healthB.Alcohol drinking can be both good and bad for youC.Drinking alcohol is popular in many parts of the worldD.Drinking alcohol a day may give your health some benefits.B14、Being safe in your everyday life needs knowledge. If you remember the following information, your life will be much safer.●Always notice the environment around you. Y ou shouldn't walk alone outside. Make sure where the public phones are. If anything dangerous happens, you can find them easily.●Your bag should be carried towards the front of your body instead of putting on your back. When a bus is full of people, it is easy enough for a thief to take away the things in the bag on your back.●If you are follow ed by someone you don't know, cross the street and go to the other way, let the person understand that you know he or she is after you. Next, don't go home at once. You are safer in the street than you are alone in your home or in a lift (电梯).●If you have to take a bus to a place far away, try to get to the stop a few minutes earlier before the bus leaves. This stops other people from studying you. On the bus, don't sit alone. Sit behind the driver or with other people. Don't sleep. 1.Which of the following is Not safe when you are out?A.Go home alone late at night.B.Make sure where the public phones are.C.Don't get to the bus stop too early.D.Always notice the environment around you.2.You'd better put your bag _______ when there are too many people on a bus.A.on your back B.beside youC.in front of you D.behind the driver3.When you are followed by someone on your way home, you should _________to make yourself safe.A.run homeB.find a lift and go inC.turn back and walk towards him or her at onceD.cross the street and go to the other way4.When you take a bus alone, it's safe for you _____.A.to sit behind the driver or with other people, but not to sleepB.to talk with the driverC.to call your friendsD.to get off the bus at once5.What can you learn from the text?A.How to notice the environment around you.B.How to be safe in your everyday life.C.How to cross the street.D.How to use the public phones.C15、Fashion is a term commonly used to describe a style of clothing worn by most people of a country. A fashion usually keeps popular for 1-3 years and then is replaced by another fashion. Even though there are a lot of changes in fashion, most people do not easily hope for the changes.Nowadays more and more young people prefer following the fashion. On this phenomenon(现象), different people have different opinions. It is held that one should follow the fashion so as not to be looked down or seem strange. But it is also held that one should have his or her own judgement(判断).Those who hold the first opinion think that society is progessing and one should get used to the changes. In their opinions, if everyone always follows the old fashion, there won’t be any change and naturally there won’t be much progress. However, those who hold the second vi ew think that one shouldn’t follow the fashion without having his or her own judgement and characters.As for me, I agree with the second opinion. Of course, it’s good for people to keep up with the fashion, but this is not to say that people should follow the fashion blindly(盲目地). If all the people follow the same fashion, there will be no variety and the whole society will be boring.1.(小题1). The second paragraph mainly tells us ________.A.what the fashon is aboutB.different people have different opinions about the fashionC.whether one should follow the fashion or notD.more and more young people prefer following the fashion2.(小题2)According to those who hold the first view, why should one follow the fashion?A.One should follow the fashion so that they won’t be looked down.B.One should follow the fashion so that society is progressing.C.One should follow the fashion so that people will have their own judgement.D.One should follow the fashion so that society will be boring.3.(小题3).We can tell from the story that the writer thinks we should _________.A.keep up with the same fashionB.be against the fashionC.say nothing about the fashionD.follow the fashion by our own judgement4.(小题4)What is the best title for the passage?A.Hope For Fashion B.Have Your Own JudgementC.How to follow the fashion D.Two ideas about the fashionD16、Ammie Reddick from East Kilbride, Lanarkshire, was only 18 months old when she had the accident that had scarred(留下创伤)her for life. The curious child reached up to catch the wire(电线) of a hot kettle(壶) in the family kitchen and poured boiling water over her little body.Her mother Ruby turned round and, seeing Ammie terribly burnt, called an ambulance(救护车) which rushed her daughter to a nearby hospital. Twenty per(组织)tak en from unburned areas of Ammie’s body, surgeons(外科医生)performed difficult skin grafts(移植)to close her wounds and control her illness, an operation that took about six hours. Over the next 16 years, Ammie experienced 12 more operations to repair her body.When she started school at Maxwelton Primary at age 4, other pupils say cruel words about her or simply wouldn’t play with her. “ I was the only burned child in the street, the class and the school,” she recalled, “some children refused to become friends b ecause of that.”Today, age 17, Ammie can only ever remember being a burned person with scars(疤), suffering from ache is a lasting part of her body. She still has to have two further skin grafts. Yet she is a confident, outgoing teenager who offers encouragement and hope to other young burnt victims.She is a member of the Scottish Burned Chindren’s Club, a charity set up last year. This month, Ammie will join the younger children at the Graffham Water Center in Cambridgeshire for the charity’s first summer camp. “I’ll show them how to pay no attention to unkind stares(注视) from others,” she says. Ammie loves wearing fashionablesleeveless dresses, and she plans to show the youngsters at the summer camp that they can too. “I do not go togreat lengths to hide m y burnt scars,” she says, “I gave up wondering how other people would react years ago.”1.How many operations has Ammie already had?A.One B.TwelveC.Thirteen D.Fifteen2.What did other children do when Ammie first went to school?A.They laughed at herB.They were cruel to herC.They were friendly to herD.They were afraid of her3.Ammie will teach the younger children at the Graffham Water Center to ____.A.live a normal lifeB.recover quicklyC.face others’ unkindness bravelyD.hide their scars by proper dressing4.What does the underlined word “sleeveless” mean in Chinese?A.无袖的B.漂亮的C.昂贵的D.合身的5.Which of the following words is not proper to describe Ammie?A.Confident B.StubbornC.Helpful D.OutgoingE17、Big as the World is, Home is Where We Go Back (《世界再大,也要回家》) is a group of hand-drawn comics by a college student. It shows us parents’ deep love when their daughter is at home and away.But one day, you happen to find it hard for your parents to do things they used to be able to. Their memory and their hearing are getting worse and worse. You are going faster and faster without knowing that they are walking slower and slower— they are growing old. When you finally understand it is time to repay your parents, could there be enough time left? So, go home before it is too late! Show your love by spending more time with your parents.1.What would parents do when their daughter was at home?A.They would complain about her laziness to friends.B.They would prepare just one dish and let her eat first.C.They would talk much about her weakness.D.They would get her up when she wants to sleep late.2.What is the meaning of the underlined word“competent”in Picture 5 ?A.能干的B.自信的C.强势的D.有经验的3.According to the passage, which statement is TRUE?A.Eating simple and vegetarian is good for our health.B.The daughter in the comic didn’t listen to her father.C.There isn’t enough time to repay parents, so don’t go home late. D.Being with your parents is the longest confession of love to parents.F根据材料内容,选择最佳答案。
2025届重庆市璧山区青杠初级中学化学九年级第一学期期末复习检测试题含解析
2025届重庆市璧山区青杠初级中学化学九年级第一学期期末复习检测试题请考生注意:1.请用2B铅笔将选择题答案涂填在答题纸相应位置上,请用0.5毫米及以上黑色字迹的钢笔或签字笔将主观题的答案写在答题纸相应的答题区内。
写在试题卷、草稿纸上均无效。
2.答题前,认真阅读答题纸上的《注意事项》,按规定答题。
一、单选题(本大题共15小题,共30分)1.如图为某化学反应的微观示意图,图中“●”表示硫原子,“○”表示氧原子。
下列说法中错误..的是()A.甲是SO2B.该反应属于化合反应C.此反应中反应物都是化合物D.反应中,甲、乙、丙三种物质的分子个数比为2:1:22.下列食品、调味品制作过程中,没有发生化学变化的是()A.糯米酿甜酒B.水果榨果汁C.鲜奶制酸奶D.黄豆酿酱油3.在一个密闭容器中放入X,Y,Z,W四种物质,在一定条件下发生化学反应,一段时间后,测得相关数据如下表,下列关于此反应的认识,正确的是物质X Y Z W反应前的质量/g 10 m 8 7反应后的质量/g 2 3 30 25A.若W为水,则X或Y必定为酸或碱B.该反应可能为化合反应C.m的数值为32D.参加反应的X与Y的质量比为1∶44.化石能源在人类发展史上有着重大贡献,下列不属于化石能源的是A.石油B.天然气C.核能D.煤5.下列各图所表示的化学实验操作正确的是()A.点燃酒精灯B.闻气体气味C.塞进橡皮塞D.量筒读数6.将一枚洁净的铁钉浸入稀硫酸(H2SO4)中,下列叙述中正确的是。
①铁钉表面产生气泡;②液体由无色逐渐变为浅绿色;③铁钉的质量减轻;④液体的质量减轻。
A.②③B.①②④C.①②③D.①②③④7.将浑浊的红色污水进行了如下净化处理,下列说法正确的是()A.①中液体无色浑浊B.经上述处理后,②中液体为纯净物C.②中液体无色澄清D.若③中出现大量泡沫,则②中液体为硬水8.下列关于石油的叙述不正确的是A.石油是一种化工产品B.石油是一种混合物C.可利用石油产品发电D.将石油分馏可得到多种产品9.下列有关物质的变化描述正确的是( )A.食物腐败是缓慢氧化引起的B.酒精燃烧既是化合反应也是氧化反应.C.工业制取二氧化碳和实验室制取二氧化碳的方法都属于分解反应D.生石灰、硝酸铵溶于水都放热,发生了化学变化10.把X、Y、Z三种金属分别放入稀盐酸中,X、Y溶解并有气泡产生,Z不反应,取过量X放入Y(NO3)2和Z(NO3)2的混合溶液中充分反应后过滤,经检验,滤渣中只有X、Z。
2025届山东省临沂商城外国语学校九年级化学第一学期期末复习检测试题含解析
2025届山东省临沂商城外国语学校九年级化学第一学期期末复习检测试题注意事项:1.答卷前,考生务必将自己的姓名、准考证号填写在答题卡上。
2.回答选择题时,选出每小题答案后,用铅笔把答题卡上对应题目的答案标号涂黑,如需改动,用橡皮擦干净后,再选涂其它答案标号。
回答非选择题时,将答案写在答题卡上,写在本试卷上无效。
3.考试结束后,将本试卷和答题卡一并交回。
一、单选题(本大题共15小题,共30分)1.下列有关碳和碳的氧化物的说法中,错误的是A.古代字画长久不变色是因为常温下碳的化学性质不活泼B.用煤火取暖时应注意通风,防止一氧化碳中毒C.二氧化碳的排放是导致酸雨的原因D.金刚石和石墨物理性质不同,是因为它们的碳原子排列方式不同2.目前世界上最精确的钟是以锶做钟摆的钟,锶的原子结构示意图及元素周期表中某些信息如图所示,下列说法正确的是A.x的值为8B.锶属于非金属元素C.锶的相对原子质量为87.62gD.锶原子在化学反应中易失去电子3.如图所示是某反应前后的微观示意图,“”和“”表示两种不同的原子,则下列说法正确的是()A.该反应是化合反应B.该反应的生成物为两种C.反应前后分子和原子的种类都发生改变D.参加反应的两种分子个数比为4:1A.锌B.氢气C.铜D.锌和铜5.用分子的相关知识解释下列生活中的现象,其中错误的是()A.热胀冷缩,说明分子的大小随温度升降而改变B.“酒香不怕巷子深”,说明分子在不停地运动C.食物腐败,说明分子本身发生改变D.湿衣服在夏天比冬天容易晾干,说明分子的运动速率随温度升高而加快6.下列变化属于化学变化的是()A.金属拉丝B.冰雪融化C.酒精挥发D.甲烷燃烧7.化学与人类的生产、生活密切相关。
下列描述正确的是()A.引起贫血的主要原因是缺锌B.将垃圾分类并回收利用C.是电池回收标志D.用水喷淋燃着的酒精以降低着火点8.下列有关物质用途的说法中错误的是A.浓硫酸在实验室中作干燥剂B.用氢氧化钠改良酸性土壤C.氢氧化镁用于治疗胃酸过多D.用稀盐酸除去金属表面的锈9.把一定质量的甲、乙、丙、丁四种物质放入一密闭容器中,在一定条件下反应一段时间,测得反应后各物质的质量如下,则下列说法中错误的是()物质甲乙丙丁反应前的质量(g) 7.6 2.0 0.2 0反应后的质量(g) 3.2 X 5.8 0.4A.x=0.4B.该反应是化合反应C.甲和乙是反应物D.反应中的甲、丙发生改变的质量比是11:1410.下列物质的性质与应用对应关系正确的是( )A .一氧化碳有可燃性,可用于工业冶炼金属B .氧气能支持燃烧,可做发射火箭的助燃剂C .铝能与盐酸反应,可在铁栏杆的表面涂铝粉防止生锈D .氩气通电时能发光,可用作不锈钢焊接的保护气11.生活中常见的下列现象,发生化学变化的是( )A .冰雪融化B .生米煮成熟饭C .湿衣服晾干D .玻璃杯炸裂12.在一密闭容器内加入甲、乙、丙、丁四种物质,在一定条件下充分反应,测得反应前后各物质质量如图所示。
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九年级上学期期末复习试题一、选择题1、光合作用:消耗二氧化碳,产生氧气;制造有机物,贮存能量;维持生物圈的碳---氧平衡促进光合作用:延长光照时间,增加二氧化碳的浓度、合理密植等2、呼吸作用:消耗氧气,产生二氧化碳;分解有机物,释放能量;为生物生命活动提供能量。
促进:松土、排涝等;抑制:低温、低氧(如水果外的保鲜袋)等3、蒸腾作用:通过气孔散失水分。
促进生物圈的水循环。
应用:移栽植物去掉部分枝叶4、消化系统由消化道和消化腺组成,人体消化和吸收营养物质的主要器官是小肠。
因小肠较长,内表面有许多环形的皱襞和小肠绒毛,有丰富的毛细血管,小肠内有多种消化液。
5、呼吸系统由呼吸道和肺组成,肺是呼吸系统的主要器官,是气体交换的场所。
因肺泡数目多,肺泡外包绕着丰富的毛细血管,肺泡壁和毛细血管壁都只有一层上皮细胞构成,大大地增加了气体交换的面积。
肺泡里的氧进入血液、血液中的二氧化碳进入肺泡,是通过气体扩散作用实现的。
肺内气体扩散要经过两层细胞。
进入血液中的氧,通过血液循环输送到全身各处的组织细胞里,氧最后是在细胞的线粒体内被利用。
6、结构层次:植物:细胞→组织→器官→植物体动物:细胞→组织→器官→系统→动物体7、低倍镜转高倍镜:细胞体积大,数目少,视野变暗8、血液循环系统由心脏和血管组成,功能是运输体内物质基础。
血液循环的动力器官是心脏9泌尿系统由肾脏、输尿管、膀胱和尿道组成,肾脏是形成尿液的器官,肾脏的结构和功能的基本单位是肾单位。
尿的形成包括肾小球和肾小囊壁的过滤作用和肾小管的重吸收作用。
6、如下图按照构成生物体的结构层次排序,正确的是:A.④→②→①→③→⑤ B.①→②→③→④→⑤C.⑤→①→③→④→② D.②→①→④→③→⑤7、有人将某植株一片叶的主脉切断(右图中甲),另一片叶两面都贴上不透明的锡箔纸(右图中乙),黑暗中放置24小时后光照4小时。
取下两叶经脱色后用碘液处理,发现A部位里棕褐色,B部位呈蓝黑色,C部位不呈蓝黑色。
以上实验能够证明光合作用需要( )A、水和二氧化碳B、叶绿素和阳光C、水和阳光D、水和叶绿素二、填空1、器官的结构总具有与其功能相适应的特点。
比如,心脏主要由心肌构成,所以能有力的收缩和扩张。
肺是呼吸系统的主要器官,是人体进行气体交换的场所,它的结构中有许多与气体交换相适应的特点:(1)肺由许多肺泡组成,增大了进行气体交换的面积;(2)肺泡外包绕着毛细血管;(3)肺泡壁和毛细血管壁都很薄,只有一层上皮细胞构成,有利于物质进出,肺泡内的气体进入血液只需要穿过层细胞,充分体现了相适应的生物学观点。
2、食物中的蛋白质在消化道中必须转变成______________才能被消化道吸收到血液。
3、人体每天形成的原尿大约有150升,但最终排出的尿液约为 1.5升,原因是全部、绝大部分水分和部分无机盐等被重新吸收。
4、根吸收水份和无机盐的主要部位是区,因此区生有。
人体消化食物和吸收营养物质的主要器官是。
5、小小萝卜籽种下后发芽,并逐渐的长大,茎膨大成“萝卜”是因萝卜叶的作用,积累了有机物;萝卜贮存了就空心,是因萝卜的作用,消耗了有机物。
6、植物运输有机物的主要通道是管;人体运输营养物质的主要通道是管。
7、人体气体交换的主要器官是;人体形成尿液的主要器官是。
三、简答题:1、植物是生态系统中的生产者,结合植物各项生理活动知识回答:(1)植物进行光合作用的原料是,条件是,场所是,产物是,能量变化是将光能转变成能贮存在它所制造的中。
植物进行光合作用的主要器官是。
(2)茂密的森林中空气清新,凉爽宜人,与植物的等生理活动分不开。
(3)植物进行各项生命活动都需要能量。
能将有机物彻底氧化分解,将有机物中的能量释放出来,供各项生命活动利用的结构是。
(4)如果要探究植物进行光合作用是否需要光照,取两盆长势相似的盆栽同种植物,暗处理后,一盆放置在适宜的光照条件下,另一盆放置在环境中(其他条件都相同)。
(5)在农业生产上,要保证作物有效地进行光合作用的各种条件,尤其是光,,既不能使叶片互相遮挡,又不能使光照到地面上,因此在种植农作物时应该。
2、右上图是植物的一片叶。
①②③④代表进出叶的物质及方向。
(1)如果此图表示绿色植物进行光合作用,则光合作用的原料有,产物是。
(填标号和名称)(2)如果去掉箭头①和②,此图可以表示绿色植物体内物质运输,③表示沿着向上运输,④表示沿着向下运输。
(3)去掉箭头①和④,此图可表示绿色植物进行蒸腾作用,绿色植物蒸腾失水的“门户”是,此结构在植物叶的(正面或背面)分布的更多一些。
绿色植物的蒸腾作用不仅促进了生物圈的水循环,是植物吸收水分的动力,还促进了在植物体内的运输。
植物吸收水分的主要部位是。
(4)绿色植物的光合作用可以维持生物圈的平衡,制造的有机物不仅满足自身生长、发育、繁殖的需要,而且为生物圈中其他生物提供了来源。
(5)乙醇、沼气等都是可再生的清洁能源,乙醇、沼气等所含有的能量都是植物通过作用将能转化而来的。
3、分析简答题右图表示人体毛细血管网(球),请据图回答下列问题(1)若B为肺部的毛细血管网,则A结构内流动的是血;(2)若B为脑部的毛细血管网,则C结构内流动的是血;(3)若B为肾小球,则A结构内流动的血,肾小球在尿液的形成过程中起作用。
(4)毛细血管的管壁非常薄,只由层扁平的上皮细胞构成;(5)毛细血管的结构特点保证了其血管内血液与组织细胞充分地进行物质交换,这种现象体现了结构与相适应的生物学观点。
4、根据“人体血液循环路线图”回答有关的人体生理学问题(1)在心房和心室之间、心室和动脉之间都有结构,可以防止血液倒流。
(2)在肺里完成气体交换后,从肺部流出去的血液是含氧丰富的血。
(3)消化食物和吸收养料的主要场所是。
(4)通过血液循环为全身各器官、组织细胞运来营养物质和氧,组织细胞通过作用将有机物分解,释放出供给生命活动需要。
(5)血液流经肾脏时,会形成尿液,正常人的尿液里通常不含。
5、下图为淀粉消化后的终产物a进入血液循环,并进入组织细胞进行代谢及其代谢终产物排出体外的过程示意图。
图中a、b、c、d、e、f分别代表不同的物质。
请据图回答下列问题:(1)a是淀粉被彻底消化后形成的物质,那么a是,淀粉在消化道的中开始被消化,在中被彻底消化为a。
(2)图中d代表进入肾小囊的物质,e代表肾小管中回到血液中的物质,f代表经尿道排出的物质。
d物质与血液相比,d中没有。
正常情况下f中是不含a物质的,若某人f中含有a物质,可能是图中的功能出现了问题。
(3)图中肺泡壁、小肠绒毛壁、毛细血管壁以及肾小囊壁等结构上的共同特点是:__ ____ _____ ___,这些结构特点是与物质交换相适应的,充分体现了的生物学观点。
(4) 小肠吸收来的a被运至脑部细胞,在脑部细胞的中被利用。
(5)经过肺泡与包绕在肺泡周围的毛细血管之间的气体交换后,从肺部流出的血液成分变化为:。
6、右图表示发生在人体内的几种生理过程。
据图回答问题:(1)食物在消化道内经过消化,最终分解为葡萄糖、氨基酸等营养物质,这些营养物质主要在内被吸收。
当血液流经消化系统主要功能器官后,血液成分的主要变化是。
(2)外界的空气经过呼吸道的处理后进入肺,在与血液进行气体交换,血液由血变为血。
(3)人的心脏中,心室壁比心房壁厚,左心室壁比右心室壁厚,有利于血液循环,体现了相适应的生物学观点。
一人因肠炎住院,护士从他的左脚上的静脉打吊针,药液从他的心脏的四个腔中经过的顺序依次是:。
(4)医生在检验某人的尿液时发现了较多的葡萄糖,如果是肾有疾病,出现病症的可能是肾单位的。
如果是内分泌系统的疾病,出现病变的内分泌腺是胰岛。
(5)在肺循环中,血液流经肺部毛细血管网时,红细胞中的与来自肺泡中的氧结合,这样血液由静脉血变为动脉血。
红细胞所运输的氧最终在组织细胞的被利用。
7、根据下式回答(1)过程a和b分别表示植物的作用和作用;(2)植物进行a过程的主要器官是,其上面有半月形保卫细胞构成,有利于进行气体交换和蒸腾作用;(3)a和b两个过程,过程对于维持生物圈中的碳—氧平衡具有重要作用。
(4)温带落叶树木过冬时,仍持续进行过程。
8、右图是植物细胞结构图,请据图回答下列问题:(1)与外界进行物质交换并能控制物质进出且使每一个细胞都具有独立性的结构是。
(2)细胞进行生命活动的重要场所是细胞质,因为它内有能量转换器,其中的与光合作用有关,与呼吸作用有关的结构是。
(3)细胞中的“遗传信息库”是,因为细胞中的物质和能量变化,主要是由它来控制。
(4)综合来看,是生物体结构和功能的基本单位。
生物体各部分结构进化得相当完美,主要表现在结构与功能相适应。
如每个肾脏大约有100多万个肾单位,每个肾单位内含有一个由数十条毛细血管弯曲盘旋而成的肾小球。
这样的结构增加了滤过面积,与肾脏滤过功能相适应。
请依照以上方法另举一例说明:9、如图所示,曲线A表示某种蔬菜的一张叶片光合作用每小时合成有机物的质量与温度的关系(光照强度为E时),曲线B表示该叶片呼吸作用每小时分解有机物的质量与温度的关系,据图回答:(1)菜农在光照强度为E的温室内栽培该种蔬菜,温度最好控制在0C左右。
(2)当温度在350C时,该叶片内积累的有机物的质量是。