2020年长沙市
位置: 首页 >范文大全 > 公文范文 > 文章内容

2020年长沙市

2022-07-01 09:20:04 来源:网友投稿

下面是小编为大家整理的2020年长沙市,供大家参考。

2020年长沙市

 

  2018 年 年 长沙市 初中毕业学业考试试卷 数 数

 学

 考生注意:共 本试卷共 26 道小题,时量 120 分钟,满分 120 分. 题次 次 一 一 二 二 三 三 四 四 五 五 总分 合分人 复分人 17- -19

 20- -22

 23 24 25 26 得分 分

  得

 分 分 评卷人 复评人 一、填 空题 题 (本题共 共 8 个小题,每小题 3 分,满分 24 )

 分)

 1、-8 的绝对值是

  . 2、函数 y= 2  x 中的自变量 x 的取值范围是

  . 3、△ABC 中,∠A=55,∠B=25,则∠C=

  . 4、方程 112 x的解为 x =

  . 5、如图,P 为菱形 ABCD 的对角线上一点,PE⊥AB 于点 E,PF⊥AD 于点 F,PF=m,则 P 点到 AB 的距离是

 cm.

 (第 5 题)

  (第 6 题)

 6、如图,在 Rt△ABC 中,∠C=90,AB=10cm,D 为 AB 的中点,则 CD=

 cm. 7、已知 a、b为两个连续整数,且a< 7 <b,则 b a  =

 . 8、在一次捐款活动中,某班 50 名同学人人拿出自己的零花钱,有捐 5 元、10 元、20 元的,还有捐 50 元和 100 元的。右边的统计图反映了不同捐款数的人数比例,那么该班同学平均每人捐款

 元.

  A B C D A B C D E F P 第 (第 8 题)

 20 元 44% 10 元 20% 50 元 16% 100元 12% 5元 8%

  得 分 评卷人 复评人 二、选择题(本题共 8 个小题,每小题 3 分,满分 24 分)

 请将你认为正确的选项的代号填在下面的表格里:

 题 号 9 10 11 12 13 14 15 16 答 案

  9、下面计算正确的是(

 )

 A、 2 21 

 B、 2 4  

  C、(3n m  ) 2 =6n m 

 D、4 2 6m m m  

 10、要反映长沙市一周内每天的最高气温的变化情况,宜采用(

 )

 A、条形统计图 B、扇形统计图 C、折线统计图

 D、频数分布直方图 11、若点 P( a , a  4 )是第二象限的点,则 a 必须满足(

 )

 A、 a <4

  B、 a >4

 C、 a <0

  D、0< a <4 12、如图是每个面上都有一个汉字的正方体的一种展开图,那么在正方体的表面,与“迎”相对的面上的汉字是(

 )

 A、文

  B、明

  C、奥

  D、运 13、在同一平面直角坐标系中,函数xy1  与函数 x y  的图象交点个数是(

 )

  A、0 个

  B、1 个

  C、2 个

  D、3 个 14、在同一时刻,身高 1.6 米的小强在阳光下的影长为 0.8 米,一棵大树的影长为 4.8 米,则树的高度为(

 )

 A、4.8 米

 B、6.4 米

 C、9.6 米

 D、10 米 15、如图,P 为⊙O 外一点,PA 切⊙O 于点 A,且 OP=5,PA=4,则 sin∠APO 等于(

 )

 A、54

  B、53

  C、34

  D、43

  (第 15 题)

 (第 16 题)

 16、二次函数 c bx ax y   2的图象如图所示,则下列关系式不正确的是(

 )

 A、 a <0

 B、 abc >0

 C、 c b a   >0

 D、 ac b 42 >0

  讲 文 明 迎 奥 运 第 (第 12 题)

 P O A · . .

  得

 分 分 评卷人 复评人 三、解答题共 (本题共 6 个小题,每小题 6 分,满分 36 )

 分)

 17、计算:0)1 51( 30 sin 2 273   .

  18、先化简,再求值:a aa 21422,其中21 a .

 19、在下面的格点图中,每个小正方形的边长均为 1 个单位,请按下列要求画出图形:

 (1)画出图①中阴影部分关于 O 点的中心对称图形;

 (2)画出图②中阴影部分向右平移 9 个单位后的图形;

 (3)画出图③中阴影部分关于直线 AB 的轴对称图形.

  (图①)

 (图②)

  (图③)

 20、解不等式组:   x xx14 3 40 121,并将其解集在数轴上表示出来.

 0 1 2 3 -1 -2 -3 -4 -5 -6

  21、当 m 为何值时,关于 x 的一元二次方程 02142    m x x 有两个相等的实数根?此时这两个实数根是多少?

  22、某商场开展购物抽奖活动,抽奖箱中有 4 个标号分别为 1、2、3、4 的质地、大小相同的小球,顾客任意摸取一个小球,然后放回,再摸取一个小球,若两次摸出的数字之和为“8”是一等奖,数字之和为“6”是二等奖,数字之和为其它数字则是三等奖,请分别求出顾客抽中一、二、三等奖的概率.

 得

 分 分 评卷人 复评人 四、解答题共 (本题共 2 个小题,每小题 8 分,满分 16 )

 分)

 23、(本题满分 8 分)

 “5·12”汶川大地震后,灾区急需大量帐篷。某服装厂原有 4 条成衣生产线和 5 条童装生产线,工厂决定转产,计划用 3 天时间赶制 1000 顶帐篷支援灾区。若启用 1 条成衣生产线和 2 条童装生产线,一天可以生产帐篷 105 顶;若启用 2 条成衣生产线和 3 条童装生产线,一天可以生产帐篷 178 顶. (1)每条成衣生产线和童装生产线平均每天生产帐篷各多少顶? (2)工厂满负荷全面转产,是否可以如期完成任务?如果你是厂长,你会怎样体现你的社会责任感?

  得

 分 分 评卷人 复评人 24 、(本题满分 8 分)

 如图,在 □ ABCD 中,BC=2AB=4,点 E、F 分别是 BC、AD 的中点. (1)求证:△ABE≌△CDF; (2)当四边形 AECF 为菱形时,求出该菱形的面积.

 A B C D E F

 得 分 评卷人 复评人 五、解答题共 (本题共2 个小题,每小题10 分,满分20 )

 分)

 25、(本题满分 10 分)

 在平面直角坐标系中,一动点 P( x ,y)从 M(1,0)出发,沿由 A(-1,1),B(-1,-1),C(1,-1),D(1,1)四点组成的正方形边线(如图①)按一定方向运动。图②是 P点运动的路程 s(个单位)与运动时间 t (秒)之间的函数图象,图③是 P 点的纵坐标 y 与 P点运动的路程 s 之间的函数图象的一部分.

  (图①)

 (图②)

 (图③)

 (1)s 与 t 之间的函数关系式是:

  ; (2)与图③相对应的 P 点的运动路径是:

 ;P 点出发

  秒首次到达点 B; (3)写出当 3≤s≤8 时,y 与 s 之间的函数关系式,并在图③中补全函数图象.

  得

 分 分 评卷人 复评人 26 、(本题满分 10 分)

 如图,六边形 ABCDEF 内接于半径为 r(常数)的⊙O,其中 AD 为直径,且AB=CD=DE=FA. (1)当∠BAD=75时,求BC⌒ 的长; (2)求证:BC∥AD∥FE; (3)设 AB= x ,求六边形 ABCDEF 的周长 L 关于 x 的函数关系式,并指出 x 为何值时,L 取得最大值.

  ·P A B C D E F O ·

  2018 年 年 长沙市 初中毕业学业考试试卷 数学参考答案及评分标准

 一、填空题

 1 、8

  2 、x ≥2

  3 、100 

  4 、3 5 、3

 6 、5

  7 、5

  8 、31 .2 二、选择题

 题号 9 10 11 12 13 14 15 16 答案 D C C A A C B C

 三、解答题

 17.原式=3+2× 21-1 ·················································································· (3 分)

 =3+1-1 ······················································································· (4 分)

 =3 ····························································································· (6 分)

 18.原式=   212 22  a a aa ···································································· (2 分)

 =    2 22 2  a aa a=   2 22 a aa ······································································· (3 分)

 =21 a ····························································································· (4 分)

 当21 a 时,原式=52. ········································································ (6 分)

 19.图略(“2018”字样),三部分图形各 2 分,共 6 分. 20.由11 024 3 14xx x  ≤得 52xx, ····························································· (4 分)

 不等式组的解集为-5<x≤2. ······························································· (5 分)

 解集在数轴上表示略. ········································································ (6 分)

 21.由题意,△=(-4) 2 -4(m-21)=0 ··································································· (2 分)

 即 16-4m+2=0,m=29. ············································································· (4 分)

 当 m=29时,方程有两个相等的实数根 x 1 =x 2 =2. ······································· (6 分)

 22.抽中一等奖的概率为161, ···································································· (2 分)

 抽中二等奖的概率为163, ···································································· (4 分)

 抽中三等奖的概率为43. ····································································· (6 分)

 四、解答题 23.(1)设每条成衣生产线和童装生产线平均每天生产帐篷各 x、y 顶,则 ············· (1 分)

   178 3 2105 2y xy x, ······················································································ (3 分)

 解得 x=41,y=32.

  答:每条成衣生产线平均每天生产帐篷 41 顶,每条童装生产线平均每天生产帐篷 32 顶.

 ············································································································· (5 分)

 (2)由 3(4×41+5×32)=972<1000 知,即使工厂满负荷全面转产,还不能如期完成任务.

 ······································································································· (7 分)

 可以从加班生产、改进技术等方面进一步挖掘生产潜力,或者动员其它厂家支援等,想法尽早完成生产任务,为灾区人民多做贡献. ················································ (8 分)

 24.(1)证明略; ······················································································· (4 分)

 (2)当四边形 AECF 为菱形时,△ABE 为等边三角形, ···································· (6 分)

 四边形 ABCD 的高为 3 , ········································································· (7 分)

 ∴菱形 AECF 的面积为 2 3 . ···································································· (8 分)

 五、解答题 25.(1)S= t21(t≥0) ···················································································· (2 分)

 (2)M→D→A→N, ·················································································· (4 分)

 10 ·········································································································· (5 分)

 (3)当 3≤s<5,即 P 从 A 到 B 时,y=4-s; ··················································· (6 分)

 当 5≤s<7,即 P 从 B 到 C 时,y=-1; ························································· (7 分)

 当 7≤s≤8,即 P 从 C 到 M 时,y=s-8. ······················································· (8 分)

 补全图象略. ························································································· (10 分)

 26.(1)连结 OB、OC,由∠BAD=75,OA=OB 知∠AOB=30, ························ (1 分)

 ∵AB=CD,∴∠COD=∠AOB=30,∴∠BOC=120, ····································· (2 分)

 故BC⌒ 的长为3r 2.···················································································· (3 分)

 (2)连结 BD,∵AB=CD,∴∠ADB=∠CBD,∴BC∥AD, ························· (5 分)

 同理 EF∥AD,从而 BC∥AD∥FE. ······················································ (6 分)

 (3) 过 点 B 作 BM⊥AD 于 M , 由 (2) 知 四 边 形 ABCD 为 等 腰 梯 形 , 从 而BC=AD-2AM=2r-2AM. ············································································ (7 分)

 ∵AD 为直径,∴∠ABD=90,易得△ BAM∽△DAB ∴AM=ADAB 2=rx22,∴BC=2r-rx 2,同理 EF=2r-rx 2 ···································· (8 分)

 ∴L=4x+2(2r-rx 2)= r x xr4 422   =   r r xr622   ,其中 0<x< r 2

  ······· (9 分)

 ∴当 x=r 时,L 取得最大值 6r. (10 分)


推荐访问:2020年长沙市 年长 沙市

猜你喜欢