广西柳州市中考数学试题(含答案)
2010年柳州市初中毕业升学考试试卷
数学
(考试时间共120分钟,全卷满分120分)
第Ⅰ卷(选择题,共36分)
注意事项:
1.答题前,考生务必先将自己的姓名、准考证号用蓝、黑色墨水笔或圆珠笔填写在试卷左边的密封线内.
2.第Ⅰ卷为第1页至第2页.答题时,请用2B 铅笔把各小题正确答案序号填涂在答题卡对应的题号内.如需改动,须用橡皮擦干净后,再填涂其它答案. 在第Ⅰ卷上答题无效.
一、选择题(本大题共12小题,每小题3分,满分36分.在每个小题给出的四个选项中,只有一项是正确的,每小题选对得3分,错选、不选或多选均得零分) 1.5-的相反数是
A .5
B.5- C.55-
D.55
2.如图1,点A B C 、、是直线l 上的三个点,图中共有线段条数是
A .1条 B.2条 C.3条 D.4条
3.三条直线a b c 、、,若a c ∥,b c ∥,则a 与b 的位置关系是
A .a b ⊥ B.a b ∥ C.a b a b ⊥或∥ D.无法确定 4.图2的几何体中,主视图、左视图、俯视图均相同的是
5.若分式2
3x
-有意义,则x 的取值范围是 A .3x ≠ B.3x = C.3x < D.3x > 6.不等式5x +≥8的解集在数轴上表示为
A . B. C. D.
7.一个正多边形的一个内角为120度,则这个正多边形的边数为 A .9 B.8 C.7 D.6
图1
图2
8.如图3,Rt ABC △中,90C ∠=°,ABC ∠的平分线BD 交AC 于D ,若3cm CD =,则点D 到AB 的距离DE 是
A .5cm B.4cm C.3cm D.2cm
9.如图4,在正方形ABCD 的外侧作等边ADE △,则AEB ∠的度数为 A .10° B.12.5° C.15° D.20°
10.上海“世界博览会”某展厅志愿者的年龄分布如图5,这些志愿者年龄的众数是 A .19岁 B.20岁 C.21岁 D.22岁
11.抛物线2
y x bx c =-++上部分点的横坐标x ,纵坐标y 的对应值如下表:
x … 2-
1-
0 1 2 … y
…
4
6
6
4
…
从上表可知,下列说法正确的个数是
①抛物线与x 轴的一个交点为(20)-, ②抛物线与y 轴的交点为(06), ③抛物线的对称轴是:1x = ④在对称轴左侧y 随x 增大而增大
A .1 B.2 C.3 D.4 12.如图6,四边形ABCD 是边长为9的正方形纸片,将其沿MN 折叠,使点
B 落在CD 边上的B '处,点A 对应点为A ',且3B
C '=,则AM 的长是
A .1.5 B.2 C.2.25 D.2.5
2010年柳州市初中毕业升学考试试卷
第Ⅱ卷(非选择题,共84分)
注意事项:
1.答题前,考生务必先将自己的姓名、准考证号用蓝、黑色墨水笔或圆珠笔填写在试卷左边的密封线内.
2.第Ⅱ卷为第3页至第10页.答题时,用蓝黑色墨水笔或圆珠笔直接将答案写在试卷上.
图3 图4 图5 图6
二、填空题(本大题共6小题,每小题3分,满分18分.请将答案直接填写在题中横线上的空白处)
13.计算:2
3·=
. 14.因式分解:2
9x -= .
15.写出一个经过点(11),的一次函数解析式 .
16.2010年广州亚运会吉祥物取名“乐羊羊”.图7中各图是按照一定规律排列的羊的组图,图①有1只羊,图②有3只羊,……,则图⑩有 只羊.
17.关于x 的一元二次方程(3)(1)0x x +-=的根是 . 18.如图8,AB 是O ⊙的直径,弦2cm BC =,F 是弦BC 的中点,60ABC ∠=°.若动点E 以2cm/s 的速度从A 点出发沿着A B A →→方向运动,设运动时间为()(03)t s t <≤,连结EF ,当t 值为 s
时,BEF △是直角三角形. 三、解答题(本大题8分,满分66分.解答应写出必要的文字说明、演算步骤或推理过程) 19.(本题满分6分)
计算:30
(2)(20103)tan 45-+--°.
20.(本题满分6分)
如图9,在88?的正方形网格中,ABC △的顶点和线段EF 的端点都在边长为1的小正方形的顶点上.
A
C
B
F
O
E 图8 图7
(1)填空:ABC ∠= .BC = ; (2)请你在图中找出一点D ,再连接DE DF 、,使以D E F 、、为顶点的三角形与ABC △全等,并加以证明. 21.(本题满分6分)
桌面上有4张背面相同的卡片,正面分别写着数字“1”、“2”、“3”“4”.先将卡片背面朝上洗匀.
(1)如果让小唐从中任意抽取一张,抽到奇数的概率是 ;
(2)如果让小唐从中同时抽取两张.游戏规则规定:抽到的两张卡片上的数字之和为奇数,则小唐胜,否则小谢胜.你认为这个游戏公平吗?说出你的理由. 22.(本题满分8分) 如图10,从热气球P 上测得两建筑物A B 、的底部的俯角分别为45°和30°,如果A B 、两建筑物的距离为90m ,P 点在地面上的正投影恰好落在线段AB 上,求热气球P 的高度.(结果精确到0.01m ,参考数据:3 1.732≈,2 1.414≈)
图9 45°
30°
图10
目前,“低碳”已成为保护地球环境的热门话题.风能是一种清洁能源,近几年我国风电装机容量迅速增长.图11是我国2003年-2009年部分年份的内力发电装机容量统计图(单位:万千瓦),观察统计图解答下列问题.
(1)2007年,我国风力发电装机容量已达万千瓦;从2003年到2009年,我国风力发电装机容量平均每年增长
......万千瓦;
(2)求2007~2009这两年装机容量的年平均增长率
......;(参考数据:5.04 2.24
≈,
1.26 1.12
≈,14 3.74
≈)
(3)按(2)的增长率,请你预测2010年我国风力发电装机容量.(结果保留到0.1万千瓦)
24.(本题满分10分)
某住宅小区计划购买并种植甲、乙两种树苗共300株.已知甲种树苗每株60元,乙种树苗每株90元.
(1)若购买树苗共用21000元,问甲、乙两种树苗应各买多少株?
(2)据统计,甲、乙两种树苗每株树苗对空气的净化指数分别为0.2和0.6,问如何购买甲、乙两种树苗才能保证该小区的空气净化指数之和不低于90而且费用最低?
图11
如图12,AB 为O ⊙直径,且弦CD AB ⊥于E ,过点B 的切线与AD 的延长线交于点F . (1)若M 是AD 的中点,连接ME 并延长ME 交BC 于N .求证:MN BC ⊥. (2)若4
cos 35
C DF ∠=
=,,求O ⊙的半径. 26.(本题满分12分)
如图13,过点(43)P -,作x 轴、y 轴的垂线,分别交x 轴、y 轴于A B 、两点,交双曲线
(2)k
y k x
=
≥于E F 、两点. (1)点E 的坐标是 ,点F 的坐标是 ;(均用含k 的式子表示) (2)判断EF 与AB 的位置关系,并证明你的结论; (3)记PEF OEF S S S =-△△,S 是否有最小值?若有,求出其最小值;若没有,请说明理由.
2010年柳州市初中毕业升学考试
数学参考答案及评分标准
题号
1
2
3
4
5
6
7
8
9
10
11
12
图12
图13
答案 A C B D A A D C C B C B
题号 13
14
15
16 17
18 答案
6
(3)(3)x x +-
如y x =,等等(答案不唯一,只要正确均可得分)
55
1x =或3x =-
1或1.75或2.25
(说明:第17题只写对一个结果给2分,两个结果都写对给3分;第18题每写对一个结果给1分) 三、解答题: 19.本题满分6分.
解:原式=811-+- ························································································ 3分
=8- ································································································ 6分
20.本题满分6分.
(1)135ABC ∠=°,22BC =, ·········································· 2分
(2)(说明:D 的位置有四处,分别是图中的
1234D D D D 、、、.此处画出D 在1D 处的位置及证明,
D 在其余位置的画法及证明参照此法给分) 解:EFD △的位置如图所示.
········································· 3分 证明:222222FD BC ==
+=Q
··············································· 4分
9045135EFD ABC ∠=∠==°+?° ·
································································· 5分 2EF AB ==
EFD ABC ∴△≌△ ·
······················································································ 6分 (说明:其他证法参照此法给分) 21.本题满分6分. 解:(1)
1
2
··································································································· 2分 (2)(方法一)
这个游戏不公平. ··························································································· 3分 理由如下:任意抽取两个数,共有6种不同的抽法,其中和为奇数的抽法共有4种.
P ∴(和为奇数)=
42
63
= ················································································ 4分 P (和为偶数)=13
························································································ 5分 (方法二)设2008年的风力发电装机容量为a 万千瓦.
5002520500a a
a
--=
······················································································· 4分 21260000a = ·
························································································ 0a >Q
1122a ∴≈ ·
···························································································· 5分 经检验,1122a ≈是所列方程的根.
则2007到2009这两年装机容量的年增长率为1122500
1.24124%500
-=≈ ·
················ 6分 答:2007到2009这两年装机容量的年平均增长率约为124%.
(3)(1 1.24)25205644.8+?=Q ····································································· 7分
∴2010年我国风力发电装机容量约为5644.8万千瓦. ··········································· 8分
24.本题满分10分.
解:(1)设甲种树苗买x 株,则乙种树苗买(300)x -株. ······································ 1分 6090(300)21000x x +-= ·
············································································· 3分 200x = ·
················································································· 4分 300200100-= ·
··············································································· 5分 答:甲种树苗买200株,乙种树苗买100株.
(2)设买x 株甲种树苗,(300)x -株乙种树苗时该小区的空气净化指数之和不低于90. 0.20.6(300)90x x +-≥ ·
··············································································· 6分 0.21800.690x x +-≥
0.490x --≥
225x ≤ ·
············································································· 7分 此时费用6090(300)y x x =+-
3027000y x =-+ ·
············································································ 8分 y Q 是x 的一次函数,y 随x 的增大而减少
∴当225x =最大时,302252700020250y =-?+=最小(元) ······························ 9分
即应买225株甲种树苗,75株乙种树苗时该小区的空气净化指数之和不低于90,费用最小为20250元. ······························································································· 10分 (说明:其他解法参照此法给分) 25.本题满分10分 (1)(方法一) 连接AC .
AB Q 为O ⊙的直径,且AB CD ⊥于E , 由垂径定理得:点E 是CD 的中点. ··························· 1分 又Q M 是AD 的中点
ME ∴是DAC △的中位线 ·
······································· 2分 MN AC ∴∥ ·
························································ 3分 AB Q 为O ⊙直径,90ACB ∴∠=°, ·
························ 4分 90MNB ∴∠=°即MN BC ⊥ ·
··································· 5分 (方法二)
AB CD ⊥Q ,90AED BEC ∴∠=∠=° ·
···················· 1分 M 是AD 的中点,ME AM ∴=,即有MEA A ∠=∠ ·
·········································· 2分 又MEA BEN ∠=∠Q ,由A ∠与C ∠同对?BD
知C A ∠=∠ C BEN ∴∠=∠ ·
···························································································· 3分 又90C CBE ∠+∠=Q °
90CBE BEN ∴∠+∠=° ·
················································································ 4分 90BNE ∴∠=°,即MN BC ⊥. ····································································· 5分 (方法三)
AB CD ⊥Q ,90AED ∴∠=° ·
········································································ 1分 由于M 是AD 的中点,ME MD ∴=,即有MED EDM ∠=∠
又CBE ∠Q 与EDA ∠同对?AC ,CBE EDA ∴∠=∠ ············································ 2分
又MED NEC ∠=∠Q
NEC CBE ∴∠=∠
························································································ 3分 又90C CBE ∠+∠=Q °
90NEC C ∴∠+∠=° ·
··················································································· 4分 即有90CNE ∠=°,MN BC ∴⊥ ···································································· 5分 (2)连接BD
BCD ∠Q 与BAF ∠同对?BD
,C A ∴∠=∠ 4
cos cos 5
A C ∴∠=∠=
······································ 6分 BF Q 为O ⊙的切线,90ABF ∴∠=°
在Rt ABF △中,4
cos 5
AB A AF ∠=
= 设4AB x =,则5AF x =,由勾股定理得:3BF x = ··········································································7分 又AB Q 为O ⊙直径,BD AD ∴⊥
ABF BDF ∴△∽△ BF DF
AF BF
∴=
································································································ 8分
即
33
53x x x
=
53
x = ··································································································· 9分 ∴直径520
4433AB x ==?=
则O ⊙的半径为10
3
······················································································· 10分
(说明:其他解法参照此法给分) 26.本题满分12分.
解:(1)44k E ??-- ???,
,33k F ??
???
, ······································································ 3分 (说明:只写对一个点的坐标给2分,写对两个点的坐标给3分)
(2)(证法一)结论:EF AB ∥ ······································································ 4分
证明:(43)P -Q ,44k E ??∴-- ???,
,33k F ??
???
,, 即得:3443
k k
PE PF =+
=+, ·
······································································ 5分 312412
12123443
PA PB k k PE k PF k ====
++++Q , APB EPF ∠=∠
PAB PEF ∴△∽△
PAB PEF ∴∠=∠ ························································································· 6分 EF AB ∴∥ ·
································································································ 7分 (证法二)结论:EF AB ∥ ············································································ 4分
证明:(43)P -Q ,44k E ?
?∴-- ???,
,33k F ??
???
,, 即得:3443
k k
PE PF =+
=+, ·
······································································ 5分 在Rt PAB △中,4
tan 3
PB PAB PA ∠=
= 在Rt PEF △中,4
4
3tan 3
34
k PF PEF k PE +∠===+
tan tan PAB PEF ∴∠=∠
PAB PEF ∴∠=∠ ························································································· 6分 EF AB ∴∥ ·
································································································ 7分
(3)(方法一)
S 有最小值 ·
·································································································· 8分 EAO FBO PAOB PEOF S S S =+Q △△矩形四边形+S
=12k +
12EOF PEF PEF PEOF S S S S k ∴=-=-+△△△四边形 ···················································· 9分
由(2)知,1
1342
243PEF k k S PE
PF ????==++ ???????
△·· 212PEF OEF PEF S S S S k ∴=---△△△= ···························································· 10分
221
(6)31212
k k k =+=+- ········································································ 11分 又2k Q ≥,此时S 的值随k 值增大而增大,
∴当2k =时,7
3
S =
最小 S ∴的最小值是73
························································································ 12分 (方法二)
S 有最小值 ·
·································································································· 8分 分别过点E F 、作PF PE 、的平行线,交点为P '
由(2)知,34k k P ??'- ???
,
Q 四边形PEP F '为矩形
P EF PEF S S '∴=△△ PEF OEF S S S ∴=-△△
=P EF OEF S S '-△△
=OME ONF OMP N S S S '++△△矩形 ······································································ 9分
=22122k k k
++ ························································································· 10分
=2
12
k k + =
21
(6)312
k +- ·
···················································································· 11分
又2k Q ≥,此时S 的值随k 值增大而增大,
∴当2k =时,73
S =
最小 S ∴的最小值是7
3
. ····················································································· 12分 (说明:其他解法参照此法给分)
广西柳州市中考数学试题与答案
数学试卷 一、选择题(各小题中只有一项是正确的,每小题4分,共40分) 1、一元二次方程2 230x x --=的两个根分别为( ). A 、X l =1, x 2=3 B 、X l =1, x 2=-3 C 、X 1=-1,X 2=3 D 、X I =-1, X 2=-3 2、下列图形中,既是中心对称图形,又是轴对称图形的是( )、 A 、等腰三角形 B 、等边三角形 C 、平行四边形 D 、菱形 3、某几何体的三种视图分别如下图所示,那么这个几何体可能是( )。 A 、 空心长方体 B 、圆柱 C 、 圆锥 D 、圆台 4、如图,E 、F 、G 、H 分别是四边形ABCD 四条边的中点,要使四边形EFGH 为矩形,四边形ABCD 应具备的条件是( ) A 、一组对边平行而另一组对边不平行 B 、对角线相等 C 、对角线互相垂直 D 、对角线互相平分 5、如图是某广告公司为某种商品设计的商标图案,若图中每个小长方形的面积都是1, 则阴影部分的面积是( ) A 、6 B 、6.5 C 、7 D 、7.5 6、若甲杆高1米,它在地面上的影长为0.8米,但在同一时刻去测量乙杆的影长时,因乙杆靠近墙壁,故其影子没有全落在地面上,有一部分留在了墙壁上,测得留在墙壁上的影高1.2米,又测得它留在地面上的影长为2.4米,则乙杆的长是( ) A 、3米 B 、4.2米 C 、4.5米 D 、不能确定 7、在匀速运动中,路程s(千米)一定时,速度v(千米/时)关于时间t(小时)的函数关系的大致图象是( ) v t v t v t v t
(第4题图) (第5题图) (第8题图) 8、某装饰公司要在如图所示的五角星型中,沿边每隔20厘米装一盏闪光灯。 若BC=(√5-1)米,则共需安装闪光灯( ) A 、100盏 B 、101盏 C 、102盏 D 、103盏 9、在平面直角坐标系中,已知A (2,-2),在y 轴上确定点P ,使△AOP 为等腰三角形,(O 为坐标原点)则符合条件的点P 共有( ) A 、2个 B 、3个 C 、4个 D 、5个 10、已知反比例函数 的图象上有两点A (x 1,y 1)和B (x 2,y 2),且x 1<x 2,那么下列结论正确的是( ) A 、y 1<y 2 B 、y 1>y 2 C 、y 1 = y 2 D 、y 1与y 2之间的大小关系不能确定 二、填空题(每小题5分,共30分) 11、已知反比例函数 的图象在第二、四象限内,则k 的值可以为 (写出满 足条件的一个k 的值即可) 12、已知:在等腰梯形ABCD 中,AD ∥BC ,对角线AC ⊥BD,AD=3cm,BC=7cm,则梯形的高是____ cm 。 13、如图是利用四边形不稳定性制作的可活动菱形晾衣架,已知其中每个菱形的边长为15cm ,∠1= 600,则在墙上悬挂晾衣架的两个铁钉A 和B 之间的距离为 cm 。 14、在ΔABC 中,AB= 4,AC= 2√2,∠B= 300,则∠BAC 的度数是 。 15、利用旧墙(旧墙长为7m)为一边,再用13米长的篱笆围成一个面积为20m 2的长方形场地,则长方形场地的长和宽分别是是 m 。 16、用边长为1cm 的小正方形搭如下的塔状图形,则第n 次所搭图形的周长 是 _ cm (用含n 的代数式表示) ··· 第1次 第2次 第3次 第4次 ··· 三、解答题(本部分共8大题,其中第17-20题每题8分,第21题10分,第22、23每 D C B A H G F E
2018年度中考数学压轴题
1、如图,在Rt△ABC中,∠C=90°,AB=10cm,AC:BC=4:3,点P从点A出发沿AB方向向点B运动,速度为1cm/s,同时点Q从点B出发沿B→C→A方向向点A运动,速度为2cm/s,当一个运动点到达终点时,另一个运动点也随之停止运动.(1)求AC、BC的长; (2)设点P的运动时间为x(秒),△PBQ的面积为y(cm2),当△PBQ存在时,求y与x的函数关系式,并写出自变量x的取值范围; (3)当点Q在CA上运动,使PQ⊥AB时,以点B、P、Q为定点的三角形与△ABC 是否相似,请说明理由; (4)当x=5秒时,在直线PQ上是否存在一点M,使△BCM得周长最小,若存在,求出最小周长,若不存在,请说明理由. 解:(1)设AC=4x,BC=3x,在Rt△ABC中,AC2+BC2=AB2, 即:(4x)2+(3x)2=102,解得:x=2,∴AC=8cm,BC=6cm; (2)①当点Q在边BC上运动时,过点Q作QH⊥AB于H,
∵AP=x ,∴BP=10﹣x ,BQ=2x ,∵△QHB ∽△ACB , ∴ QH QB AC AB = ,∴QH=错误!未找到引用源。x ,y=错误!未找到引用源。BP ?QH=1 2 (10﹣x )?错误!未找到引用源。x=﹣4 5 x 2+8x (0<x ≤3), ②当点Q 在边CA 上运动时,过点Q 作QH ′⊥AB 于H ′, ∵AP=x , ∴BP=10﹣x ,AQ=14﹣2x ,∵△AQH ′∽△ABC , ∴'AQ QH AB BC =,即:' 14106 x QH -=错误!未找到引用源。,解得:QH ′=错误!未找到引用源。(14﹣x ), ∴y= 12PB ?QH ′=12(10﹣x )?35(14﹣x )=310x 2﹣36 5 x+42(3<x <7); ∴y 与x 的函数关系式为:y=2 248(03)5 33642(37)10 5x x x x x x ?-+<≤????-+<?错误!未找到引用源。; (3)∵AP=x ,AQ=14﹣x ,
2020-2021学年广西南宁市中考数学第一次模拟试题及答案解析
广西南宁市最新中考数学一模试卷(含解析) 一、选择题(共12小题,每小题3分,满分36分) 1.﹣3的倒数是() A.﹣3 B.3 C.D.﹣ 【分析】根据乘积为的1两个数互为倒数,可得到一个数的倒数. 【解答】解:﹣3的倒数是﹣, 故选:D. 【点评】本题考查了倒数,分子分母交换位置是求一个数的倒数的关键. 2.如图所示的几何体是由一些小立方块搭成的,则这个几何体的俯视图是() A.B.C.D. 【分析】俯视图是从物体上面看所得到的图形.从几何体上面看,是左边2个,右边1个正方形. 【解答】解:从几何体上面看,是左边2个,右边1个正方形. 故选:D. 【点评】本题考查了三视图的知识,俯视图是从物体上面看所得到的图形,解答时学生易将三种视图混淆而错误的选其它选项. 3.下列运算正确的是() A.xx2=x2B.3=x6D.x2+x2=x4 【分析】根据同底数幂的除法,底数不变指数相减,合并同类项,系数相加字母和字母的指数不变,同底数幂的乘法,底数不变指数相加,幂的乘方,底数不变指数相乘,对各选项计算后利用排除法求解. 【解答】解:A、xx2=x3同底数幂的乘法,底数不变指数相加,故本选项错误; B、(xy)2=x2y2,幂的乘方,底数不变指数相乘,故本选项错误; C、(x2)3=x6,幂的乘方,底数不变指数相乘,故本选项正确;
D、x2+x2=2x2,故本选项错误. 故选C. 【点评】本题考查同底数幂的除法,合并同类项,同底数幂的乘法,幂的乘方很容易混淆,一定要记准法则才能做题,难度适中. 4.我市5月的某一周每天的最高气温(单位:℃)统计如下:19,20,25,22,25,26,27,则这组数据的中位数与众数分别是() A.25,25 B.25,22 C.20,22 D.22,24 【分析】根据众数的定义即众数是一组数据中出现次数最多的数;中位数的定义即中位数是将一组数据从小到大(或从大到小)重新排列后,最中间的那个数,即可得出答案. 【解答】解:25出现了2次,出现的次数最多, 则众数是25; 把这组数据从小到大排列19,20,22,25,25,26,27,最中间的数是25, 则中位数是25. 故选:A. 【点评】此题考查了众数和中位数,众数是一组数据中出现次数最多的数,中位数是将一组数据从小到大(或从大到小)重新排列后,最中间的那个数(最中间两个数的平均数),叫做这组数据的中位数. 5.如图所示,△ABC中,DE∥BC,若=,则下列结论中错误的是() A.=B.= C.=D.= 【分析】根据平行线的性质以及相似三角形的性质即可作出判断. 【解答】解:∵DE∥BC, ∴△ADE∽△ABC,==,故A正确, ∴==,
2017年广西省柳州市中考数学试卷(含解析版)
2017年广西省柳州市中考数学试卷 第I卷(选择题,共36分) 一、选择题(每小题3分,共12小题,共计36分) 1.(2017广西柳州,1,3分)计算:(-3)+(-3)=( ) A.-9 B.9 C.-6 D.6 2.(2017广西柳州,2,3分) 下列交通标志中,是轴对称图形的是( ) A.限制速度 B.禁止同行 C.禁止直行 D.禁止掉头 3.(2017广西柳州,3,3分)如图,这是一个机械模具,则它的主视图是( ) A.B.C. D. 4.(2017广西柳州,4,3分)现有四个看上去完全一样的纸团,每个纸团里面分别写着数字1,2,3,4,现任意抽取一个纸团,则抽到的数字是4的概率是( ) A.3 4B.1 2 C.1 4 D.1 5.(2017广西柳州,5,3分)如图,经过直线l外一点画l的垂线,能画出( )
A.1条B.2条C.3条D.4条 6.(2017广西柳州,6,3分)化简:2x-x=( ) A.2 B.1 C.2x D.x 7.(2017广西柳州,7,3分)如图,直线y=2x必过的点是( ) A.(2,1) B.(2,2) C.(-1,-1) D.(0,0) 8.(2017广西柳州,8,3分) 如图,这个五边形ABCDE的内角和等于( ) A.360°B.540°C.720°D.900°9.(2017广西柳州,9,3分)如图,在⊙O中与∠1一定相等的角是( )
A .∠2 B .∠3 C .∠4 D .∠5 10.(2017广西柳州,10,3分)计算5a ab g =( ). A .5ab B .26a b C .25a b D .10ab 11. (2017广西柳州,11,3分).化简:211()2x x x -=g ( ) A .-x . B .1x C .22x - D . 2 x 12. (2017广西柳州,12,3分).如果有一组数据为1,2,3,4,5,则这组数据的方差为( ) A .1 B .2 C .3 D .4 第II 卷(非选择题,共84分) 二、填空题(每小题3分,共18分). 13.(2017广西柳州,13,3分).如图,AB∥CD,若∠1=60°,则∠2= °. 14.(2017广西柳州,14,3分).计算35= . 15.(2017广西柳州,15,3分).若点A(2,2)在反比例函数k y x =(k≠0)的图像上,则k = . 16.(2017广西柳州,16,3分)某校为了了解本届初三学生体质健康情况,从全校初三学生中随进抽取46名学生进行调查,上述抽取的样本容量为 . 17.(2017广西柳州,17,3分)如图,把这个“十字星”形图绕其中心点O 旋转,
中考数学压轴题(共10题)
2010年中考数学压轴题10题精选 【1】如图,点P 是双曲线11( 00)k y k x x = <<,上一动点,过点P 作x 轴、y 轴的垂线,分别交x 轴、y 轴于A 、B 两点,交双曲线y = x k 2 (0<k 2<|k 1|)于E 、F 两点. (1)图1中,四边形PEOF 的面积S 1= ▲ (用含k 1、k 2的式子表示); (2)图2中,设P 点坐标为(-4,3). ①判断EF 与AB 的位置关系,并证明你的结论; ②记2PEF OEF S S S ??=-,S 2是否有最小值?若有,求出其最小值;若没有,请说明理由。 【2】一开口向上的抛物线与x 轴交于A (m -2,0),B (m +2,0)两点,记抛物线顶点为C ,且AC ⊥BC . (1)若m 为常数,求抛物线的解析式; (2)若m 为小于0的常数,那么(1)中的抛物线经过怎么样的平移可以使顶点在坐标原点? (3)设抛物线交y 轴正半轴于D 点,问是否存在实数m ,使得△BCD 为等腰三角形?若存在,求出m 的值;若不存在,请说明理由. 【3】如图,在梯形ABCD 中,24AD BC AD BC ==∥,,,点M 是AD 的中点,MBC △是等边三角形. (1)求证:梯形ABCD 是等腰梯形; (2)动点P 、Q 分别在线段BC 和MC 上运动,且60MPQ =?∠保持不变.设PC x MQ y ==,, 求y 与x 的函数关系式; (3)在(2)中:①当动点P 、Q 运动到何处时,以点P 、M 和点A 、B 、C 、D 中的两个点 B D A C O x y