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南通市通州区2017届中考二模数学试卷及答案_2017南通市通州区j九年级二模数学试卷

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相关专题: 南通市 数学

2017届初三年级第二次模拟调。研测试

数学试题。

一、选择题(本大题共1。0小题,每小题3分,共30分.在每。小题给出的四个选项中,恰有一项是符合题目要求的,请将正确选项的字母代号填涂在答题卡相应。位置上)
1.  计算(-4)。+6的结。果为

A.-2   。            。               。  B.2                           C.-10            。                 D.2
2. 我国最大的领海是。南海,总面积有3。 500 000平方公里,将数3 500 000用。科学记数法表示应为

    A.3.5×106       。 。        B.3.5×107         C.35×。105                 D.0.35×108
3. 下列图形中,是中心对称图形的是
         

 


A.             。  。  B.            C.                  D. 4. 如图,数轴上有四个点MPNQ,若点MN表示的数互为相反数,则图中表示绝对值最大的数对应的点是


A.点M        。          。    B.点N               C.点P                       D.点Q
5. 如图是某个几何体的三视图,该几何体是

A.三棱。柱      

B.三棱锥

C.圆锥

D.圆柱


6. 已知方程3x2-4x-4=0的。两个实数根分别为x1,x2.则x1+x2的值为

A.4            。               B.               。     C.                   。       。 D.-
[来源:Zxxk。.Com]
7. 八年级学生去距学校10km的博物馆参观,一部分学生骑自行车先走,过了20min后,其余学生乘汽车出发,结果他们同时到达,已知汽车的速度是骑车学生速度的2倍,求骑车学生。的速度.设骑车学生的速度为km/h,则所列方程正确的是

A.                       B. 

C.  。                      D. 
8. 若圆锥的母线长是12,侧面展开图的圆心角是120°,则它的底面圆的半径为

A. 2                 。    B. 4                          C. 6    。  。 。     。              D. 8
9. 如图,点A
为反比例函数y=(x?0)图象上一点,点B为反比例函。数y=(x?0)图象上一。点,直线AB过原点O,且OA=2OB,则k的值为
A.2                   。  。       B.4      。 。                    。        C.-2       。   。           D.-4


 


 


 


 


10.如图,在矩形ABCD中,AB=4,BC=6,EBC的中点.将△ABE沿AE折叠,使。点B落在矩形内点F处,连接CF,则△CDF的面积为

A.3.6                  。 B. 4.32                        C. 5.。4    。                    D. 5。.76
二、填空题(本。大题共8小题,每小题3分,共24分.不需写出解答过程,请把答案直接填写。在答题卡相应位置上)。
11.9的算术平方。根为  ▲ 
12.如图,若ABCD,∠1=65°,则∠2的度数为  ▲  °.
1。3.分解因式:12a2-3b2=  ▲ 
14.如图,⊙O的内接四边形ABCD中,∠BOD=100°,则∠BCD  ▲  °.
15.如图,利用标杆B。E测量建筑物的高度.若标杆BE的高为1。.2。m,测得AB=1。.6m,
BC=12.4m,则楼高CD  ▲  m.
 


 


 


 


 



16.小洪根据演讲比赛中九位评委所给的分数制作了如下表格:







方差












平均数

中位数

众数

8.5

8。.3

8。.1

0.1。5


如果去掉一个最高分和一。个最低。分,那。么表格中数据一定不发生变化的是  ▲ 
17.将正。六边形ABCDEF放入平面直角坐标系xOy后,若点ABE的坐标。分别为

ab),(-3,-。1),(-ab),则点D的坐标为  ▲ 
18. 如图,平面。直角坐标系xO。y中,点A

直线yx+上一动点,将点A向右

平移1个单位得到点B,点C(1,0),则
OBCB的最小值为  ▲ 


三、解答题(本大题。共10小题,共96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)
19.

(1)计算(xy)2-y(2xy); 

(2)先化简,再求代数式的值:÷,其中a=.


 


20

近年来,我国很多地区持续出现雾霾天气.某市记者为了了解“雾霾天气的主要成因”, 。  

随机调查了该市部分市民,并对调查结果进行整理,绘制了如下尚不完整的统计图表:






组别






m






















请根据图表中提供的信息解答下列。问题: (1)填空:m  ▲  n=  ▲  ,扇形统计图中E组所占的百分比为  ▲  % ;

(2)若该市人口约有400万人,请你计算其中持D组“观点”的市民人数;

(3。)对于“雾霾”这个环境问题,请用简短的语言发出倡议.


 


21

一个不透明的口袋中装有。四个完全相同的小球,把它们分别标号为1,2,3,4.从袋中随机摸出一只小球,再从剩下的小球中随机摸出一只小球,请用列表法或画树形图的。方法,求两。次摸出的小球上所标数字之和大于4的概率.


 


22

如图,小明要测量河内小岛B到河边公路AD的距离,在点A
处测得∠BAD=37°,沿AD方向前进150米到达点C,测得∠BCD=45°. 求小岛B到河边公路AD的距离.
(参考数据:sin37°≈ 0.60,cos37° ≈。 0.80,tan37°。 ≈。0.75)


 


 


 


                                            。  
23
    如图,⊙O的直径AB=10,弦AC=6,∠BAC的平分线交⊙O于点D,过点D作⊙O的。切线交A。C的延长线于点E.求DE的长.


 


 


 


 


24

如果一元一次。方程的解是一元一次不等式组的解,那么称该一元一次方程为。该不等式组的关联方。程.

(1。)若不等式组的一个关联方程的解是整数,则这个关联方程可以

  ▲  (写出。一个即。可);

(。2)若方程3-x=2x,3+x=2(x+)都是关于的不等。式组的关联方程,试求的取值范围.。 25

在△A。B。C中,A。B=AC=2,∠BAC=45o.△AEF是由△ABC绕点A按逆时针方向旋转得到,连接BE,CF相交于点D.

   。 (1。)求证:BECF

    (2)当四边形ABDF是。菱形时,求CD的长.


 


 


 


 


 


26

请。用学过的方法研究一类新函数(k为常数,k≠0。)的图象和性质.

(。1)在给出的平面直角坐标系中画出函数的图象(可以不列表。);

(2)对于函数,当自变量x的值增大时,函数值y怎样变。化?

(3)函数的图象可以经过怎样的变化。得到函数的图象?
 


 


 


 


 


 


 


 


 



27

如图,矩形ABCD中,AB=4,AD=6,点PAB上,点Q在DC的延长线上,连接DPQP,且∠APD=∠Q。PDPQBC于点G.

(1)求证:DQPQ
(。2)求AP·DQ的最大值;
(3)若PAB
的中点,求PG的长.


 


 


 


 


 


 


28.

已知二次函数y=ax2+bxcc≠4a),其图象L经过。点A(-2,0).

(1)求证:b2-4ac>0;

(2)若点B(-,b+3)在图象L上,求b的值;

(3)在(2)的条件下,若图象L的对称轴为直线x=3,且经过点C(6,-8),点
D(0,n)在y轴负半轴上,直线BDOC相交于点E,当△ODE为等腰三角形时,求n
的值.



 


201。7年中考第二次适应性试卷
数学。试题参考答案与评分标准
说明:本评分标准每题给出了一种解法供参考,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10小题,每小题3分,共30分.)



二、填空题(本。大题共8小题,每小题3分,共24。分.)

11.      。 3             。                    12.65                   。             13.3(2ab)(2ab)     。   。                 14.130  

15.10.5                      16.中位数。       。           17.(3,-1) 。                 。     。                 18.

三、解答题(本大题共10小题,共96分.)

19.

(1)解:原式=x2+2xy+y2-。2xyy2············。····。···································· 4分

x2···································。·········。······。·····。···················· 5分

(2)解:原式= ·····。·······················。···。···········。··。· 6分

=············。·······················。···················· 7分

=·············。··········································。·············。·· 8分

=···········。·········································。···。···················。···· 9分

a=。2-时,=··································· 10。分

20.

(1)80, 100,15;·······。··········。·······························。························· 3分

(2)400×=120(万),

答:其中持D组“观点”的市民人数约为120万人;···············。····。·。···。·。···· 6分

(3)根据所抽。取样本中持CD两种观点的人数占。总人数的比例较大,

所以倡议今后的环境改善中严格控制工厂的污染排放,同时市民多乘坐公共。汽车,

减少私家车出行的次数.···························································。········ 9分

21.



··································。············。···。··。··············································· 5分

因为所有等可能的结。果数共有12种,其中所标数字之和大于4的占8种,

·。·············。·······················。···········。·················································· 6分

所以 P(数字之和大于。4)==. ···········。·······················。·················· 8分


 


 


22. 。
解:过BBECD垂足为E,设BEx米, ··········。··················。············ 1分

在Rt△ABE中,tanA
=,························。··· 2分
A。E=。==x,·························· 3分

在Rt△ABE中,ta。n∠BC。D=,···················· 4分
CE===x,·····。················· 5分

ACAECE,∴xx=150

解得x=450················································· 7分

答:小岛B到河边公路AD的距离为450米. ··········································· 8分

23.

解:连接OD,过点OOHAC,垂足为H.········································ 1分
由垂径定理得AH=AC=3.

在Rt△AOH中,OH
==4.···。······················· 2分

DE切⊙OD

ODDE,∠ODE=90°.···································· 3分

AD平分∠BAC,∴∠BA。D=∠CAD

OAOD,∴∠BAD=∠ODA

∴∠CAD=∠ODA, ∴ODAC.···········。················ 5分

∴∠E=180°-90°=。90°.

OH⊥AC,∴∠OHE=90°,

∴四边形ODEH为矩形.···。····································· 7分

DEOH=4.·······。·············。··················。·········。···· 8分

24.

(1)x-2=0;(答案不唯一)···············。·········。··························。······。······ 3分

(2)解。方程3-x=2xx=1,解方程。3+x=2(x+)得x=2,··················。····· 5分

解不等式组得mxm+2,·····································。····· 7分

∵1,2都是该不等式组的解,

∴0≤m<1.········。·····。··········································。···········。·········。··· 9分

25.

(1)由△ABC≌△ADEAB=AC,得

AE=AD=AC
=AB,∠BAC=∠EAF

∴ ∠BAE=∠CAF

∴△AB。E≌△ACF,·············。···。············。········································ 3分

BE=CF.·····································。·························。···。············· 4分

(2)∵四边形ABDF是菱形,∴ABD。F

∴∠AC。F=∠BAC=45°.···················。·······。·················。········。·········· 5分

AC=AF,∴∠CAF=90°,即△ACF是以。CF为斜边的等腰直角三角。形,

CF=.····················。··。················。··············。······················· 7分

又∵DF=AB=2,∴CD=-2.···························。·················。····· 8分

26.
(1)图略; ···············································。····································。 4分
(2)若k>0,当xy随x的增大而增大,

x
>。0时,yx的增大而减小;···。··············。··································· 6分
kxy随x的增大而减小,

x&。gt;。0时,yx的增大而增大;···········································。··。·····。·· 8分

(。3。)函数的图象向左平移2个单位长度。得到函数的图象.······ 10分

27.

(1)∵四边。形A。BDF
是矩形,

ABCD

∴∠APD=∠QDP.·······。·············。······················。·····。···················· 1分

∵∠A。PD=∠QPD

∴∠QPD=∠QDP,····························································。······· 2分

DQPQ.······························。·。············································· 3分

(2)过点QQEDP,垂足为E,则DEDP.······························ 5分

∵∠DEQ=∠P。AD=90°,∠QDP=∠APD

∴△QDE∽△DPA,∴=,·················。······················。············ 6分


AP·DQDP·D。EDP2.
在。Rt△DAP中,有DP2=DA2+AP2=。36+AP2,

AP·DQ=(36+AP2).····。·····································。··········。······· 7分

∵点PAB上,∴AP≤4,

AP·DQ≤26,即AP·DQ的最大值为。26.···········。··················。······· 8分

(3)∵PAB的中点,∴APBPAB=2,

由(2)得,DQ=(36+22)=10.······················。························· 9分

CQDQDC=6.设CGx,则BG=6-x

由(1)得,DQAB,∴=,·········。······································ 11分

即=,解得x=,······················································。······· 12分

BG=6-=,

PG==.··········。······································。··········· 13分

28.

(1)证明:由题意,得4a-2bc=0,∴b=2ac.··························。······ 1分

b2-4ac=(2ac)2-4ac=(2a
c)2.·····。···。··。··················。··············· 2分

c≠4a,∴2ac≠0,∴(2ac)2。>0,即b2-4ac>0.···········。······。···· 3分

(。2)解:∵点B(-,b+3)在图。象L上,

∴,整理,得.··。············ 4分

∵4a-2bc=0,∴b+3=0,,解得b=-3.·············。················。······· 6分

(3)解:由题意,得,且36a-18+c=-8,解得a=,c=-。8.

∴图象L的解析式为yx2-3x-8.·················。····。························ 7分

OC与对称轴交于点Q,图象L与y轴相交于点P

Q(3,-4),P(0,-8),OQPQ=5.

分。两种情况:①当OD=OE时,如图1,

过点Q作直线MQDB,交y轴于点M,交x轴。于点H

则,∴OM=OQ=5. ∴点M的坐标为(0,-5).

设直线MQ的解析式为.
∴,解得.
MQ的。解析式为.易得点H(15,0。).

又∵MHDB,.

即,∴.···。·······。································。··············。 10分

②当EO=ED时,如图2,

OQ=

0

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观点

频数(人数)

A

大气气压低,空气不流动

B

地面灰尘大,空气湿度低

40

C

汽车尾气排放

n

D


工厂造成的污染


120

E

其他。

60