200年莆田市初中毕业、升学考试试卷
数 学
(满分:150分,考试时间:120分钟)
一、细心填一填(本大题共10小题,每小题4分,共40分.直接把答案填在题中的横线上.)
1.3-的相反数是 .
2.2009年莆田市参加初中毕业、升学考试的学生总人数约为43000人,将43000用科学记数法表示是___________.
3.在组成单词“Probability ”(概率)的所有字母中任意取出一个字母,则取到字母“b ”的概率是 .
4.如图,A B 、两处被池塘隔开,为了测量A B 、两处的距离,在AB 外选一适当的点C ,连接AC BC 、,并分别取线段AC BC 、的中点E F 、,测得EF =20m ,则AB =__________m .
5.一罐饮料净重500克,罐上注有“蛋白质含量≥0.4%”,则这罐饮料中蛋白质的含量至
少为__________克.
6.如图,菱形ABCD 的对角线相交于点O ,请你添加一个条件: ,使得该菱形为正方形.
7.甲、乙两位同学参加跳高训练,在相同条件下各跳10次,统计各自成绩的方差得2
2
S S <乙甲,则成绩较稳定的同学是___________.(填“甲”或“乙”)
8.已知1O ⊙和2O ⊙的半径分别是一元二次方程()()120x x --=的两根,且122O O =,则
1O ⊙和2O ⊙的位置关系是 .
9.出售某种文具盒,若每个获利x 元,一天可售出()6x -个,则当x = 元时,一天出售该种文具盒的总利润y 最大.
10.如图,在x 轴的正半轴上依次截取
112233445OA A A A A A A A A ====,过点12345
A A A A A 、、、、分别作x 轴的垂线与反比例函数()2
0y x x =≠的图象相交于点
12345P P P P P 、、、、,得直
角三角形
1
1122
3
O
P A A P A A P A A P
A A P A
2、、、、,并设其面积分别为12345S S S S S 、、、、,则5S 的值为 .
(第4题图) A B
D
D C B
A O (第
6题图)
O
(第10题图)
2
二、精心选一选(本大题共6小题,每小题4分,共24分,每小题给出的四个选项中有且只有一个是正确的,请把正确选项的代号写在题后的括号内,答对的得4分;答错、不答或答案超过一个的一律得0分). 11
有意义,则x 的取值范围是( )
A .x ≥0
B .0x <
C .0x ≠
D .0x > 12.下列各式运算正确的是( )
A .22
a a a ÷= B .()
2
224ab
a b =
C .248
a a a ·= D .55a
b b a -= 13.如图是一房子的示意图,则其左视图是( )
A .
B .
C
.
D. 14.某班5位同学参加“改革开放30周年”系列活动的次数依次为1
2333、、、
、,则这组数据的众数和中位数分别是( )
A .22、
B . 2.43、
C . 32、
D .33、 15.不等式组2410
x x ?
+>?,
的解集在数轴上表示正确的是( )
A B
C D
16.如图1,在矩形MNPQ 中,动点R 从点N 出发,沿N →P →Q →M 方向运动至点M 处停止.设点R 运动的路程为x ,MNR △的面积为y ,如果y 关于x 的函数图象如图2所示,则当9x =时,点R 应运动到( )
A .N 处
B .P 处
C .Q 处
D .M 处
三、耐心做一做(本大题共9小题,共86分.解答应写出必要的文字说明、证明过程或演
算步骤.)
(第16题图)
(图1)
17.(8
分)计算:0
133??
???
.
18.(8分)先化简,再求值:22
442
42
x x x x x x +++÷---,其中1x =. 19.(8分)已知:如图在ABCD
中,过对角线BD 的中点O 作直线EF 分别交DA 的延
长线、AB DC BC 、、的延长线于点E M N F 、、、.
(1)观察图形并找出一对全等三角形:△________≌△____________,请加以证明; (2)在(1)中你所找出的一对全等三角形,其中一个三角形可由另一个三角形经过怎样的
变换得到?
20.(8分)(1)根据下列步骤画图..并标明相应的字母:(直接在图1中画图) ①以已知线段AB (图1)为直径画半圆O ;
②在半圆O 上取不同于点A B 、的一点C ,连接AC BC 、; ③过点O 画OD BC ∥交半圆O 于点D . (2)尺规作图..
:(保留作图痕迹,不要求写作法、证明) 已知:AOB ∠(图2). 求作:AOB ∠的平分线.
21.(8分)某校课题研究小组对本校九年级全体同学体育测试情况进行调查,他们随机抽
查部分同学体育测试成绩(由高到低分A B C D 、、、四个等级),根据调查的数据绘制成如下的条形统计图和扇形统计图.
图2 O B A 图1 (第20题图) E B M O
D N C (第19题图)
A
(第21题图)
请根据以上不完整的统计图提供的信息,解答下列问题:
(1)该课题研究小组共抽查了__________名同学的体育测试成绩,扇形统计图中B 级所占的百分比b =___________; (2)补全条形统计图;
(3)若该校九年级共有400名同学,请估计该校九年级同学体育测试达标(测试成绩C 级以上,含C 级)约有___________名. 22.(10分)已知,如图,BC 是以线段AB 为直径的O ⊙的切线,AC 交O ⊙于点D ,过点D 作弦DE AB ⊥,垂足为点F ,连接BD BE 、.. (1)仔细观察图形并写出四个不同的正确结论:①________,②________ ,③________,④____________(不添加其它字母和辅助线,不必证明); (2)A ∠=30°,CD
=
3
,求O ⊙的半径r .
23.(10分)面对全球金融危机的挑战,我国政府毅然启动内需,改善民生.国务院决定从2009年2月1日起,“家电下乡”在全国范围内实施,农民购买人选产品,政府按原价购买..总额的...13%...
给予补贴返还.某村委会组织部分农民到商场购买人选的同一型号的冰箱、电视机两种家电,已知购买冰箱的数量是电视机的2倍,且按原价购买冰箱总额为40000元、电视机总额为15000元.根据“家电下乡”优惠政策,每台冰箱补贴返还的金额比每台电视机补贴返还的金额多65元,求冰箱、电视机各购买多少台?
(2)列出方程(组)并解答.
(第22题图)
24.(12分)已知:等边ABC △的边长为a . 探究(1):如图1,过等边ABC △的顶点A B C 、、依次作AB BC CA 、、的垂线围成MNG △,求证:MNG △是等边三角形且
.MN =;
探究(2):在等边ABC △内取一点O ,过点O 分别作OD AB OE BC OF CA ⊥⊥⊥、、,垂足分别为点D E F 、、.
①如图2,若点O 是ABC △的重心,我们可利用三角形面积公式及等边三角形性质得到两个正确结论(不必证明):结论1
.OD OE OF ++=;结论2.32
AD BE CF a ++=
; ②如图3,若点O 是等边ABC △内任意一点,则上述结论12、是否仍然成立?如果成立,
请给予证明;如果不成立,请说明理由.
N M A G C B A F C B D A F C
B
D (图1)
(图2)
(图3)
(第24题图)
O A
F C
B D
(图4)
O O
25.(14分)已知,如图1,过点()01E -,
作平行于x 轴的直线l ,抛物线2
14
y x =上的两点A B 、的横坐标分别为-1和4,直线AB 交y 轴于点F ,过点A B 、分别作直线l 的垂线,垂足分别为点C 、D ,连接CF DF 、. (1)求点A B F 、、的坐标; (2)求证:CF DF ⊥; (3)点P 是抛物线2
14
y x =
对称轴右侧图象上的一动点,过点P 作PQ PO ⊥交x 轴于点Q ,是否存在点P 使得OPQ △与CDF △相似?若存在,请求出所有符合条件的点P 的坐标;若不存在,请说明理由.
2009年莆田市初中毕业、升学考试试卷
数学试卷参考答案及评分标准
说明:
(一)考生的解法与“参考答案”不同时,可参照“答案的评分标准”的精神进行评分 (二)如解答的某一步计算出现错误,这一错误没有改变后续部分的考查目的,可酌情给分,
但原则上不超过后面应得分数的二分之一;如属严重的概念性错误,就不给分. (三)以下解答各行右端所注分数表示正确做完该步骤应得的累计分数. (四)评分的最小单位是1分,得分或扣分都不能出现小数. 一、细心填一填(本大题共10小题,每小题4分,共40分.)
1.3 2.4
4.310?(不必考虑有效数字) 3.
2
11
4.40 5.2 6.AB BC ⊥或AC BD =或AO BO =等 7.甲 8.相交 9.3 10.1
5
二、精心选一选(本大题共6小题,每小题4分,共24分.) 11.A 12.B 13.C 14.D 15.A 16.C 三、耐心做一做(本题共9小题,共86分)
17.(1)解:原式
=341+ ······························································································· 6分
(图1)
备用图
(第25题图)
=············································································································ 8分
注:33=(2分)
4=(2分),13?? ???
=1(2分)
18.解:原式=()()()
2
22222x x x
x x x +-?-+-+ ················································································· 6分
=1x -················································································································ 7分
当1x =时原式=110-= ···························································································· 8分 注:()()()2
2222
442422?22
x x x x x x x x x x +-++=+-=+-÷=?
-+、、?(各2分) 19. (1)DOE BOF ①△≌△; ···························· 2分
证明:∵四边形ABCD 是平行四边形
∴AD BC ∥ ·············································· 3分 ∴EDO FBO E F ∠=∠∠=∠, ································· 4分
又∵OD OB =
∴()DOE BOF AAS △≌△ ··················································································· 5分
BOM DON ②△≌△ ····························································································· 2分
证明:∵四边形ABCD 是平行四边形
∴AB CD ∥ ·············································································································· 3分
∴MBO NDO BMO DNO ∠=∠∠=∠, ······························································· 4分 又∵BO DO =
∴()BOM DON AAS △≌△ ················································································· 5分
ABD CDB ③△≌△; ··························································································· 2分
证明:∵四边形ABCD 是平行四边形
∴AD CB AB CD ==, ··························································································· 3分
又∵BD DB = ·········································································································· 4分
∴()ABD CDB SSS △≌△ ····················································································· 5分 (2)绕点O 旋转180°后得到或以点O 为中心作对称变换得到. ································ 8分 20.(1)正确完成步骤①、②、③,各得1分,字母标注完整得1分,满分4分.
(2)说明:①以点O 为圆心,以适当长为半径作弧交OA OB 、于两点C D 、 ·············· 5分
②分别以点C D 、为圆心,以大于
1
2
CD 长为半径作弧, 两弧相交于点E ·························································································· 7分
③作射线OE ·································································································· 8分
E
B M O D
N
C
(第19题图)
A
A E
D
C
C
21.(1)80 ························································································ 2分 40% ··································································································· 4分 (2)补全条形图(如右图) ··························································· 6分 (3)380 ······························································································· 8分
22.(1)BC AB AD BD ⊥⊥,,DF FE BD BE ==,,
BDF BEF △≌△, BDF △∽BAD △,BDF BEF ∠=∠,A E DE BC ∠=∠,∥等 (每写出一个正确结论得1分,满分4分.)
(2)解:AB 是O ⊙的直径90ADB ∴∠=° ·································· 5分 又30E ∠= °
30A ∴∠=° ·
····················································································· 6分 1
2
BD AB r ∴=
= ·
··········································································· 7分 又BC 是O ⊙的切线
90CBA ∴∠=° ·
·················································································· 8分 60C ∴∠=? 在Rt BCD △
中,CD =
tan 60BD r
DC ∴
==° ·
············································································································ 9分 2r ∴=·
········································································································································· 10分 23
(2)解:依题意得
2x -65x
= ·
······························································ 7分 解得10x = ··································································································································· 8分
经检验10x =是原分式方程的解 ·································································································
9分
220x ∴=. 答:冰箱、电视机分别购买20台、10台 ······································· 10分 24.证明:如图1,ABC △为等边三角形 60ABC ∴∠=
°
BC MN BA MG ⊥⊥ ,
∴90CBM BAM ∠=∠=° (第22题图)
N
M
A G C
B (图1)
(第21题图)
9030ABM ABC ∴∠=∠=?°- ····················································· 1分 9060M ABM ∴∠=?∠=?- ·
························································· 2分 同理:60N G ∠=∠=?
MNG ∴△为等边三角形.··········································································································· 3分 在Rt ABM △
中,sin sin 60AB a BM M =
==?
在Rt BCN △
中,tan tan 60BC a BN N =
==? ···································································· 4分
MN BM BN ∴=+= ····················································································································· 5分
(2)②:结论1成立.
证明;方法一:如图2,连接AO BO CO 、、 由ABC AOB BOC AOC S S S S =++△△△△=()1
2
a OD OE OF ++ ·
·············· 7分 作AH BC ⊥,垂足为H ,
则sin sin 602AH AC ACB a a =∠=??=
11222
ABC S BC AH a ∴=
=△·· (
)1122a OD OE OF a ∴++=
OD OE OF ∴++=
··········································································································· 8分 方法二:如图3,过点O 作GH BC ∥,分别交AB AC 、于点G H 、,过点 H 作HM BC ⊥于点M , 6060DGO B OHF C ∴∠=∠=∠=∠=°,° AGH ∴△是等边三角形
GH AH ∴= ·
········································································· 6分 OE BC ⊥ OE HM ∴∥
∴四边形OEMH 是矩形
HM OE ∴= ·
··········································································· 7分 在Rt ODG △
中,sin sin 60OD OG
DGO OG =∠=?=·· 在Rt OFH △
中,sin sin 60OF OH
OHF OH =∠=?=·· A F
C
B
D
(图2)
O
A
F C
B
D (图3)
O
H
G
在Rt HMC △
中,sin sin 602
HM HC
C HC HC ==?=··
222
OD OE OF OD HM OF HC ∴++=++=
++
)222
GH HC AC =+== ···························· 8分 (2)②:结论2成立.
证明:方法一:如图4,过顶点A B C 、、依次作边AB BC CA 、、的垂线围成MNG △,由(1)得M N G △为等边
三角形且MN = ··················································· 9分 过点O 分别作OD MN '⊥于D ',OE NG '⊥于NG 于点
E OF
MG ''⊥,于点F ' 由结论1得:
322OD OE OF MN a '+'+'===2 ·········································
······························· 10分 又OD AB AB MG OF MG ⊥⊥'⊥ ,,
90ADO DAF OF A ∴∠=∠'=∠'=? ∴四边形ADOF '为矩形 OF ∴'=AD
同理:OD BE '=,OE CF '= ··································································································· 11分
3
2
AD BE CF OD OE OF a ∴++='+'+'=
····················································································· 12分 方法二:(同结论1方法二的辅助线) 在Rt OFH △
中,tan OF FH OHF =
=∠
在Rt HMC △
中,sin HM HC C == ··························· 9分
33
CF HC FH OE OF ∴=+=
+
同理:3333
AD OF BE =
+=+, ··················································· 10分 AD BE CF ∴++
+++ A
F C
B
D
(图3)
O
H
G A F C
E
B
D
(图4)
O F '
D '
M
G
N
E '
)
OD OE OF
++ ················································································································ 11分由结论1
得:OD OE OF
++=
3
22
AD BE CF a a
∴++==························································································· 12分方法三:如图5,连接OA OB OC
、、,根据勾股定理得:
22222
BE OE OB BD OD
+==+①
22222
CF OF OC CE OE
+==+②
22222
AD OD AO AF OF
+==+③························································································· 9分①+②+③得:
222222
BE CF AD BD CE AF
++=++ ··················································································· 10分()()()
222
222
BE CF AD a AD a BE a CF
∴++=-+-+-
222222
222
a AD a AD a BE a BE a CF a CF
=-++-++-+
············································ 11分整理得:()2
23
a AD BE CF a
++=
3
2
AD BE CF a
∴++= ·············································································································· 12分25.(1)解:方法一,如图1,当1
x=-时,
1
4
y=
当4
x=时,4
y=
∴1
A
??
-
?
??
1
,
4
·····················································································1分
()
44 B,······························································································2分
设直线AB的解析式为y kx b
=+ ···················································3分
则
1
4
44
k b
k b
?
-+=
?
?
?+=
?
解得
3
4
1
k
b
?
=
?
?
?=
?
(图1)
A
F
C
B
D
(图5)
O
∴直线AB 的解析式为3
14
y x =+ ·
················································· 4分 当0x =时,1y =
()01F ∴, ······································································································································ 5分
方法二:求A B 、两点坐标同方法一,如图2,作FG BD ⊥,AH BD ⊥,垂足分别为G 、H ,交y 轴于点N ,则四边形FOMG 和四边形
NOMH 均为矩形,设FO x = ························································· 3分 BGF BHA △∽△
BG FG
BH AH ∴= 44
1544x -∴=-·
································································································································· 4分 解得1x =
()0F ∴,1 ··································································································································· 5分
(2)证明:方法一:在Rt CEF △中,1,2CE EF ==
22222125CF CE EF ∴=+=+=
CF ∴=·································································································································· 6分
在Rt DEF △中,42DE EF ==,
222224220DF DE EF ∴=+=+=
DF ∴=由(1)得()()1
141C D ---,,, 5CD ∴= 22525CD ∴==
222CF DF CD ∴+= ············································································································ 7分
90CFD ∴∠=°
∴CF DF ⊥ ·
························································································································· 8分 方法二:由 (1
)知5544AF AC ===,
AF AC ∴=·
··························································································································· 6分 同理:BF BD =
(图2)
ACF AFC ∴∠=∠ AC EF ∥
ACF CFO ∴∠=∠
AFC CFO ∴∠=∠ ·
··············································································································· 7分 同理:BFD OFD ∠=∠
90CFD OFC OFD ∴∠=∠+∠=° 即CF DF ⊥ ·························································································································· 8分
(3)存在.
解:如图3,作PM x ⊥轴,垂足为点M ············ 9分 又PQ OP ⊥
Rt Rt OPM OQP ∴△∽△
PM OM
PQ OP
∴
=
PQ PM
OP OM ∴
=
························································ 10分 设()2
104
P x x x ??> ???
,,则2
14
PM x OM x =
=, ①当Rt Rt QPO CFD △∽△时,
1
2
PQ CF OP DF === ········································································································ 11分 21142x
PM OM x ∴== 解得2x =
()121P ∴,
······························································································································ 12分 ②当Rt Rt OPQ CFD △∽△时,
2PQ DF OP CF ===········································································································· 13分 2
142x
PM OM x ∴== 解得8x = ()2816P ∴,
综上,存在点()121P ,
、()2816P ,使得OPQ △与CDF △相似. ······································ 14分
图3
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(I)求f(x)的定义域; (II)若f(m)+f(m-1)=1,求m 的值. 17.(本小题满分10分) 2012年春季期小学四年级数学质量检测 (考试时间90分钟) 一、口算。(18分) 0.86-0.26= 0.8-0.3= 10-0.8= 0.4+0.6= 6.4+6= 0.57+0.18= 4.6×10= 18.7÷10= 10÷100= 二、填空。(10 分) 1.三角形都有()条边,()个顶点和()角。 2.等腰直角三角形的每个锐角都是()度。 3.任何一个三角形的内角的和都是()度。 4.三条边都相等的三角形叫()三角形。 5.等边三角形底边上的高,把顶角分成()的两个角,每个角都是()度。 6.由()条线段()的图形叫三角形。 三、我是小判官。(对的打“√”,错的打“×”)(12分) 1.任何一个三角形都有3条高。() 2.直角三角形只有两条高。() 3.等腰三角形的两个内角是相等的。() 4.三角形具有稳定性。() 四、分别画出与已知的底边a、b、c相对应的高。(9分) a b 五、列式计算并且验算下面各题。(9分) 2.7+9.8= 3.54-3.04= 0.897-0.45= 验算:验算:验算:六、计算。(9分) 0.384+0.26+2.6 4.2-0.07-0.175 15.02+(18.3-0.375) 七、下面各题怎样算简便就怎样算。(12分) 1.2+9.6+3.8 5.8- 2.6-2.4 10+0.009+0.391 八、解决问题。(第1小题5分,第2小题6分,第3小题10分,共21分) 1、鸵鸟每小时走54.3千米,卡车每小时行45.7千米。鸵鸟的速度比卡车快多少千米? 2、已知一个直角三角形的一个内角是35°,另一个内角是多少度? 3、小明的妈妈去超市买水果。梨子每500克要1.18元,葡萄每500克要6.80元。(1)这两种水果,哪种便宜些?每500克便宜多少元? (2)小明妈妈买梨和葡萄各1000克,要付多少钱? 2018年上海市普通高等学校春季招生统一文化考试 数学试卷 一、填空题(本大题共有12题,满分54分,第1~6题每题4分,第7~12题每题5分) 1.不等式||1x >的解集为__________. 2.计算:31 lim 2 n n n →∞-=+__________. 3.设集合{|02}A x x =<<,{|11}B x x =-<<,则A B =__________. 4.若复数1z i =+(i 是虚数单位),则2 z z + =__________. 5.已知{}n a 是等差数列,若2810a a +=,则357a a a ++=__________. 6.已知平面上动点P 到两个定点(1,0)和(1,0)-的距离之和等于4,则动点P 的轨迹为 __________. 7.如图,在长方形1111B ABC A C D D -中,3AB =,4BC =,15AA =, O 是11AC 的 中点,则三棱锥11A AOB -的体积为__________. 第7题图 第12题图 8.某校组队参加辩论赛,从6名学生中选出4人分别担任一、二、三、 四辩.若其中学生 甲必须参赛且不担任四辩,则不同的安排方法种数为__________. 9.设a R ∈,若9 22x x ? ?+ ?? ?与9 2a x x ??+ ???的二项展开式中的常数项相等,则a =__________. 10.设m R ∈,若z 是关于x 的方程22 10x mx m -+=+的一个虚根,则||z 的取值范围 是__________. 11.设0a >,函数()2(1)sin()f x x x ax =+-,(0,1)x ∈,若函数21y x =-与() y f x = 学大教育对口升学考试数学模拟试卷(一) 一、单项选择题(每小题3分,共45分) 1.已知全集{1,2,3,4,5,6,7,8},{3,4,5},{1,3,6},{2,7,8}U A B ===则集合是( ) A .A B B .A B C .U U C A C B D .U U C A C B 2.若2(2)2,(2)f x x x f =-=则( ) A .0 B .1- C .3 D .2 3.已知点(,3),(5,2),(4,5),,A x B y AB x y -= 且则的值为( ) A .1,10x y =-= B .1,10x y == C .1,10x y ==- D .1,10x y =-=- 4.关于余弦函数cos y x =的图象,下列说法正确的是( ) A .通过点(1,0) B .关于x 轴对称 C .关于原点对称 D .由正弦函数sin 2 y x x π =的图象沿轴向左平移个单位而得到 5.6 2 20.5与的等比中项是( ) A .16 B .2± C .4 D .4± 6.2210,C x xy y C -++=如果曲线的方程为那么下列各点在曲线上的是( ) A .(1,2)- B .(1,2)- C .(2,3)- D .(3,6) 7.直线10x -+=的倾斜角是( ) A . 6 π B . 3 π C . 23 π D . 56 π 8.若40,,x x x x >+ 要使取最小值则必须等于( ) A .1 B .2± C .—2 D .2 9.若圆柱的轴截面的面积为S ,则圆柱的侧面积等于( ) A .S π B . 2 S C 2 S D .2S π 10.如图,在正方体11111,ABC D A B C D AC BD -中异面直线与所成的角是( ) A .90 B .60 C .45 D .30 友情提醒: 2015 苏州小学教师教学基本功大赛 小学数学试题 1. 本试卷满分 100 分,答题时间为 90 分钟。 2. 本试卷共 4 页,共 5 大题,59 小题。 3. 答案要求全部做在提供的答题纸上,在本试卷上答题无效。 一、选择题(第 1~20 题为单选题,每题 1 分;第 21~25 题为多选题,每题 2 分,多选、错选、漏选均不得分,合计 30 分) l (第 11 题图) 1. 一学生在测验时遇到某个难题,暂时跳过去,先做简单的,这表明他已经掌握了一些( )。 A. 组织策略 B. 问题解决的策略 C. 元认知策略 D. 精细加工策略 2. 在维纳的归因理论中,属于内部而稳定的因素是( )。 A. 努力 B. 能力 C. 难度 D. 运气 3. “君子一言,驷马难追”或“一诺千金”体现的是( )对从众行为的影响。 A. 道德感 B. 承诺感 C. 模糊性 D. 匿名 4. 数学教师在教解决实际问题时,一再强调要学生看清题目,必要时可以画一些示意图,这样做的目 的是为 了( )。 A. 牢记住题目内容 B. 很好地完成对心理问题的表征 C. 有效地监控解题过程 D. 熟练地使用计算技能 5. 学习了“分数”概念基础上,又学习了“真分数”、“假分数”的概念,这种概念同化的形式是 ( )。 A. 类属同化 B. 并列同化 C. 总结同化 D. 上位同化 6. 根据实施教学评价的时机不同,可以将教学评价分为( )。 A. 准备性评价、形成性评价和总结性评价 B. 常模参照评价与标准参照评价 C. 标准化学绩测验和教师自编测验 D. 发展性评价和过程性评价 7. “鸡兔同笼”问题是我国古代名题之一,它出自我国古代的一部算书,书名是( )。 A. 《孙子算经》 B. 《周髀算经》 C. 《九章算术》 D. 《海岛算经》 8. 为了布置教室,王晓用一张长 30 厘米、宽 15 厘米的彩纸,剪成直角边分别是 8 厘米和 5 厘米的 直角三角 形彩旗(不可以拼接),最多能剪( )面。 A. 9 B. 18 C. 20 D. 22 b a c 姓名 学校 考试号 ………………………………密…………………………………………封……………………………………线…………………………………………… 2019年江苏省高考数学试卷 一、填空题 1.已知集合123A ,,,245B ,,,则集合A B U 中元素的个数为_______. 2.已知一组数据4,6,5,8,7,6,那么这组数据的平均数为 ________. 3.设复数z 满足234z i (i 是虚数单位),则z 的模为_______. 4.根据如图所示的伪代码,可知输出的结果S 为________. 5.袋中有形状、大小都相同的 4只球,其中1只白球,1只红球,2只黄球,从中一次随机摸出2只球,则这2只球颜色不同的概率为________. 6.已知向量21a r ,,2a r 1,,若98ma nb mn R r r ,,则m-n 的值为______. 7.不等式 224x x 的解集为________. 8.已知tan 2,1 tan 7,则tan 的值为_______. 9.现有橡皮泥制作的底面半径为 5,高为4的圆锥和底面半径为2、高为8的圆柱各一个。若将它们重新制作成总体积与高均保持不变,但底面半径相同的新的圆锥与圆柱各一个,则新的底面半径为 。10.在平面直角坐标系 xOy 中,以点)0,1(为圆心且与直线)(012R m m y mx 相切的所有圆中,半径最大的圆的标准方程为 。11.数列}{n a 满足 11a ,且11n a a n n (*N n ),则数列}1{n a 的前10项和 为。12.在平面直角坐标系 xOy 中,P 为双曲线122y x 右支上的一个动点。若点P 到直线01y x 的距离对c 恒成立,则是实数c 的最大值为 。13.已知函数 |ln |)(x x f ,1,2|4|10,0)(2x x x x g ,则方程1|)()(|x g x f 实根的 个数为。14.设向量)12,,2,1,0)(6cos 6sin ,6(cos k k k k a k ,则1201)(k k k a a 的值 为。四年级数学试卷word
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