数 学
一、选择题
1.5-的相反数是( ▲ ).
A .
1
5
B .15
-
C .5
D .5-
2.下列运算正确的是( ▲ ).
A .2
3
6
·a a a = B .()3
28
a
a = C .()3
2
63a b a b = D .623a a a ÷=
3.为迎接2014年青奥会,在未来两到三年时间内,一条长53公里,总面积约11000亩的鸀色长廊将串起南京的观音门、仙鹤门、沧波门等8座老城门遗址.数据11000用科学记数法可表示为( ▲ ). A .3
1110? B .4
1.110? C .5
1.110? D .5
0.1110?
4.如图,不等式组???x +1>0,x -1≤0
的解集在数轴上表示正确的是( ▲ ).
5.如图,在12?网格的两个格点上任意摆放黑、白两个棋子,且两棋子不在同一条格线上.其中恰好如
图示位置摆放的概率是( ▲ ). A .
6
1
B .
91 C . 12
1 D .
18
1
6.如图,在扇形纸片AOB 中,OA =10,AOB =36,OB 在桌面内的直线l 上.现将此扇形沿l 按顺时针方向旋转(旋转过程中无滑动),当OA 落在l 上时,停止旋转.则点O 所经过的路线长为( ▲ ). A . π12 B .π11 C
55
l
(第6题图)
(第5题图)
A .
B . -1 1
C .
D .
二、填空题
7.数据3,5,5,1-,1,1,1的众数是 ▲ . 8.分解因式2
69x
x -+的结果是 ▲ .
9.如图,已知AB ∥CD ,80AEF ∠=°,则DCF ∠为 ▲ °.
10.观察:12341111111
13355779
a a a a =-=
-=-=-,,,,…,则n a = ▲ (n 为正整数)
. 11.如图,AB 是⊙O 直径,且AB =4cm ,弦CD ⊥AB ,∠COB =45°,则CD 为 ▲ cm . 12.如图,是水平放置的长方体,它的底面边长为2和4,左视图的面积为6,则该长方体的体积为 ▲ . 13.当分式
12x -与3x
的值相等时,x 的值为 ▲ .
14.如图,正比例函数1y x =和反比例函数2k
y x
=的图象都经过点A (1,1).则在第一象限内,当12y y >时,x 的取值范围是 ▲ .
15.如图,在梯形ABCD 中,AD ∥BC ,点E 、F 、G 、H 是两腰上的点,AE =EF =FB ,CG =GH =HD ,且四边形EFGH
的面积为6cm 2,则梯形ABCD 的面积为 ▲ cm 2.
16.一张矩形纸片经过折叠得到一个三角形(如图),则矩形的长与宽的比为 ▲ .
(第16题图)
D B
E F A G H (第15题图)
(第9题图)
A
C D F E
4
2
(第12题图)
(第11题图)
三、解答题
17.(5分)计算:182)31(0+---.
18.(5分)先化简,再求值:
2
2222a b b a b a b
+++-,其中a =-2,b =1.
19.(6分)如图,在△ABC 中,AB=AC ,AD ⊥BC ,垂足为D ,AE ∥BC , DE ∥AB .
证明:(1)AE =DC ;(2)四边形ADCE 为矩形.
20.(6分)某区为了解全区2800名九年级学生英语口语考试成绩的情况,从中随机抽取了部分学生的成
A
B
C
D
E
(第19题图)
绩(满分24分,得分均为整数),制成下表:
(1)填空:
①本次抽样调查共抽取了▲名学生;
②学生成绩的中位数落在▲分数段;
③若用扇形统计图表示统计结果,则分数段为x≤16的人数所对应扇形的圆心角为▲°;
(2)如果将21分以上(含21分)定为优秀,请估计该区九年级考生成绩为优秀的人数.
21.(6分)某初级中学准备随机选出七、八、九三个年级各1名学生担任领操员.现已知这三个年级分别选送一男、一女共6名学生为备选人.
(1)请你利用树状图或表格列出所有可能的选法;
(2)求选出“两男一女”三名领操员的概率.
22.(6分)受国际原油价格持续上涨影响,某市对出租车的收费标准进行调整.
.
(1)调整前出租车的起步价为 ▲ 元,超过3km 收费 ▲ 元/km ;
(2)求调整后的车费y (元)与行驶路程x (km )(x >3)之间的函数关系式,并在图中画出其函数图象.
23.(8分) 现有一张宽为12cm 练习纸,相邻两条格线间的距离均为.调皮的小聪在纸的左上角用印章印
出一个矩形卡通图案,图案的顶点恰好在四条格线上(如图),测得∠α=32°.
(1)求矩形图案的面积;
(2)若小聪在第一个图案的右边以同样的方式继续盖印(如图),最多能印几个完整的图案
(参考数据:sin32°≈,cos32°≈,tan32°≈)
……
12cm
α
(第23题图)
(第22题图) 调整前: 调整后:
调整方案: 加收1元燃油附加费,其它收费标准保持不变.
x /km 车费y/元
0 1 2 3 4 5 6 16
15 14
13
12
11 10 9
24.(8分)某手机专营店代理销售A 、B 两种型号手机.手机的进价、售价如下表:
(1)第一季度:用36000元购进 A 、B 两种型号的手机,全部售完后获利6300元,求第一季度购进A 、
B 两种型号手机的数量;
(2)第二季度:计划购进A 、B 两种型号手机共34部,且不超出第一季度的购机总费用,则A 型号手
机最多能购多少部
25. (8分)如图,在△ABC 中,AB=AC ,点O 为底边上的中点,以点O 为圆心,1为半径的半圆与边AB
相切于点D .
(1)判断直线AC 与⊙O 的位置关系,并说明理由;
(2)当∠A =60°时,求图中阴影部分的面积.
26.(9分)已知二次函数m x x y ++-=22
的图象与x 轴相交于A 、B 两点(A 左B 右),与y 轴相交于
点C ,顶点为D . (1)求m 的取值范围;
(2)当点A 的坐标为(3,0)-,求点B 的坐标; (3)当BC ⊥CD 时,求m 的值.
B
(第25题图)
27.(9分)
操作:小明准备制作棱长为1cm 的正方体纸盒,现选用一些废弃的圆形纸片进行如下设计:
纸片利用率=纸片被利用的面积
纸片的总面积
×100%
发现:(1)方案一中的点A 、B 恰好为该圆一直径的两个端点.
你认为小明的这个发现是否正确,请说明理由. (2)小明通过计算,发现方案一中纸片的利用率仅约为%.
请帮忙计算方案二的利用率,并写出求解过程.
探究:(3)小明感觉上面两个方案的利用率均偏低,又进行了新的设计(方案三),请直接写出方案
三的利用率.
说明: 方案一图形中的圆过点A 、B 、C ; 方案二直角三角形的两直角边与展开图左下角的正方形边重合,斜边经过两个正方形的顶点. 说明: 方案三中的每条边均过其中两个正方形的顶点. A B
C
方案一 方案三 方案二
28.(12分)如图,在Rt△ABC中,∠C=90°,AC=BC=4cm,点D为AC边上一点,且AD=3cm,动点E从点A出发,以1cm/s的速度沿线段AB向终点B运动,运动时间为x s.作∠DEF=45°,与边BC相交于点F.设BF长为y cm.
(1)当x= ▲s时,DE⊥AB;
(2)求在点E运动过程中,y与x之间的函数关系式及点F运动路线的长;
(3)当△BEF为等腰三角形时,求x的值.
(第28
题图)
(第28题备用图)
参考答案
一、选择题(每小题2分,共计12分)
二、填空题
7.1 8.()2
3-x 9.100 10.
1
21
121+-
-n n 11.2 2
12.24 13.3 14.x >1 15.18 16.2︰ 3 (或2 3 或2 3
3 )
三、解答题 17.(本题5分)
解:原式=1-2+3 2 ················································································································ 3分
=-1+3 2 ·················································································································· 5分 18.(本题5分)
解:原式)
)((2))(())(2(2
b a b a b b a b a b a b a -++-+-+= ·········································································· 2分
))((2b a b a ab
a -++=
········································································································· 3分 b
a a -=
························································································································ 4分 当a=-2,b=1时,原式=
-2 -2-1
= 2
3 ·········································································· 5分
19.(本题6分)
证明:
(1)在△ABC 中,∵AB=AC ,AD ⊥BC , ∴BD=DC ········································································································································ 1分 ∵AE ∥BC , DE ∥AB , ∴四边形ABDE 为平行四边形 ···································································································· 2分 ∴BD=AE , ···································································································································· 3分 ∵BD=DC ∴AE = DC . ··································································································································· 4分 (2)
解法一:∵AE ∥BC ,AE = DC , ∴四边形ADCE 为平行四边形. ································································································ 5分 又∵AD ⊥BC , ∴∠ADC=90°,
∴四边形ADCE 为矩形. ············································································································ 6分 解法二:
∵AE ∥BC ,AE = DC ,
A B C D E
∴四边形ADCE 为平行四边形 ···································································································· 5分 又∵四边形ABDE 为平行四边形 ∴AB=DE .∵AB=AC ,∴DE=AC . ∴四边形ADCE 为矩形. ············································································································ 6分 20.(本题6分) 解法一:(1)用表格列出所有可能结果:
······················································································································································· 3分
(2)从上表可知:共有8种结果,且每种结果都是等可能的,其中“两男一女”的结果有3种. 5分
所以,P (两男一女)=3
8 . ······································································································· 6分 解法二:(1)用树状图列出所有可能结果:
······················································································································································· 3分
(2)从上图可知:共有8种结果,且每种结果都是等可能的,其中“两男一女”的结果有3种. 5分
所以,P (两男一女)=3
8 . ······································································································· 6分 21.(本题6分)
(1)①300 ··································································································································· 1分
②21≤x ≤22 ······················································································································· 3分 ③12 ····································································································································· 4分 (2)2800×112+128
300 =2240(人) ··························································································· 5分 答:该区所有学生中口语成绩为满分的人数约为2240人. ················································· 6分 22.(本题6分) 解:(1)9;; ······························································································································· 2分
(男,男,男) (男,男,女) 男 女
男
(男,女,男) (男,女,女) 男 女 (女,男,男) (女,男,女)
男 女
男 (女,女,男) (女,女,女) 男 女 女
男
女
开始
七年级 八年级
九年级 结果
(2)y=10+(x -3)=+ ·········································································································· 5分
······················································································································································· 6分
23.(本题8分)
(1)如图,在Rt △BCE 中,
∵sin α=CE BC ,∴BC = CE sin α =5.08.0 = ························································································· 2分
∵矩形ABCD 中,∴∠BCD=90°,∴∠BCE +∠FCD=90°, 又∵在Rt △BCE 中,∴∠EBC +∠BCE=90°,∴∠FCD=32°.
在Rt △FCD 中,∵cos ∠FCD=FC CD ,∴CD=?32cos FC
=8.06.1=2 ···················································· 4分
∴橡皮的长和宽分别为2cm 和.
(2)如图,在Rt △ADH 中,易求得∠DAH=32°.∵cos ∠DAH=AD
AH , ∴AH=
?32cos AD =8
.06
.1=2 ················································································································ 5分
在Rt △CGH 中,∠GCH=32°.∵tan ∠GCH=GH
CG ,
∴GH=CG tan32°= × = ················································································································ 7分 又∵6×2+>12,5×2+<12,3×4+,∴最多能摆放5块橡皮. ··············································· 8分 24.(本题8分)
(1)解:设该专营店第一季度购进A 、B 两种型号手机的数量分别为x 部和y 部. ········ 1分
由题意可知: ???1200x +1000 y =36000,180x +200y =6300 ··················································································· 3分
解得:???x =15,
y =18
答:该专营店本次购进A 、B 两种型号手机的数分别为15部和18部. ····························· 4分
C
F G 调整后的图像如图: 调整前: 调整后:
x /km
B (2)解:设第二季度购进A 型号手机a 部. ········································································· 5分 由题意可知:1200a +1000(34-a )≤36000,············································································· 6分 解得:a ≤10 ································································································································· 7分 不等式的最大整数解为10
答:第二季度最多能购A 型号手机10部. ············································································· 8分
25.(本题8分)
解:(1)直线AC 与⊙O 相切. ·············································· 1分 理由是:
连接OD ,过点O 作OE ⊥AC ,垂足为点E . ∵⊙O 与边AB 相切于点D ,
∴OD ⊥AB . ·································································································································· 2分
∵AB=AC ,点O 为底边上的中点,
∴AO 平分∠BAC ··························································································································· 3分 又∵OD ⊥AB ,OE ⊥AC
∴OD= OE ······································································································································· 4分 ∴OE 是⊙O 的半径.
又∵OE ⊥AC ,∴直线AC 与⊙O 相切.····················································································· 5分 (2)∵AO 平分∠BAC ,且∠BAC=60°, ∴∠OAD=∠OAE=30°, ∴∠AOD=∠AOE=60°,
在Rt △OAD 中,∵tan ∠OAD = OD AD ,∴AD=OD
tan ∠OAD =3,同理可得AE=3
∴S 四边形ADOE =12 ×OD ×AD ×2=1
2 ×1×3×2=
3 ······························································· 6分 又∵S 扇形形ODE =120π×12 360 =13 π ····································································································· 7分 ∴S 阴影= S 四边形ADOE -S 扇形形ODE = 3 -1
3 π. ·············································································· 8分 26.(本题9分)
解:(1)∵二次函数m x x y ++-=22
的图象与x 轴相交于A 、B 两点
∴b 2-4ac >0,∴4+4m >0, ····································································································· 2分 解得:m >-1 ······························································································································ 3分 (2)解法一:
∵二次函数m x x y ++-=22
的图象的对称轴为直线x =-b 2a =1 ···································· 4分 ∴根据抛物线的对称性得点B 的坐标为(5,0) ··································································· 6分 解法二:
把x =-3,y =0代入m x x y ++-=22中得m=15 ··································································· 4分
∴二次函数的表达式为1522
++-=x x y
令y =0得01522
=++-x x ······································································································ 5分 解得x 1=-3,x 2=5
∴点B 的坐标为(5,0) ··········································································································· 6分
(3)如图,过D 作DE ⊥y 轴,垂足为E .∴∠DEC =∠COB =90°, 当BC ⊥CD 时,∠DCE +∠BCO =90°,
∵∠DEC =90°,∴∠DCE +∠EDC =90°,∴∠EDC =∠BCO .
∴△DEC ∽△COB ,∴EC OB =ED
OC .····························································································· 7分 由题意得:OE =m+1,OC =m ,DE =1,∴EC =1.∴ 1OB =1
m .
∴OB =m ,∴B 的坐标为(m ,0). ························································································· 8分 将(m ,0)代入m x x y ++-=22
得:-m 2+2 m + m =0.
解得:m 1=0(舍去), m 2=3. ······························································································ 9分 27.(本题9分) 发现:(1)小明的这个发现正确. ··························································································· 1分 理由:解法一:如图一:连接AC 、BC 、AB ,∵AC =BC = 5 ,AB =10
∴AC 2+BC 2=AB 2 ∴∠BAC =90°, ······················································· 2分 ∴AB 为该圆的直径. ·············································································· 3分
解法二:如图二:连接AC 、BC 、AB .易证△AMC ≌△BNC ,∴∠ACM =∠CBN .
又∵∠BCN +∠CBN =90°,∴∠BCN +∠ACM =90°,即∠BAC =90°, ······· 2分 ∴AB 为该圆的直径. ······················································································ 3分
(2)如图三:易证△ADE ≌△EHF ,∴AD =EH =1. ····························································· 4分 ∵DE ∥BC ,∴△ADE ∽△ACB ,∴AD AC =DE CB ∴14 =2
CB ,∴BC =8. ···································· 5分 ∴S △ACB =16. ······························································································································· 6分 ∴该方案纸片利用率=展开图的面积纸板的总面积 ×100%=616 ×100%=% ········································· 7分
探究:(3)180
361 ··························································································································· 9分 28.(本题12分)
O y x A B C D E 图一 M 图二 N C B A D E F
H 图三
解:(1)3
2
2 ···························································································································· 2分
(2)∵在△ABC 中,∠C =90°,AC =BC =4.
∴∠A =∠B =45°,AB =4 2 ,∴∠ADE +∠AED =135°;
又∵∠DEF =45°,∴∠BEF +∠AED =135°,∴∠ADE =∠BEF ;
∴△ADE ∽△BEF ··························································································································· 4分 ∴AD BE =AE BF , ∴
3 4 2 -x
=x y ,∴y =-13 x 2+4
3
2 x ················································································· 5分
∴y =-13 x 2+4
3 2 x =-13 ( x -2 2 )2+8
3
∴当x =2 2 时,y 有最大值=8
3 ···························································································· 6分 ∴点F 运动路程为16
3 cm ············································································································ 7分
(3)这里有三种情况:
①如图,若EF =BF ,则∠B =∠BEF ;
又∵△ADE ∽△BEF ,∴∠A =∠ADE =45° ∴∠AED =90°,∴AE =DE =3
2
2 ,
∵动点E 的速度为1cm/s ,∴此时x =3
2
2 s ;
②如图,若EF =BE ,则∠B =∠EFB ;
又∵△ADE ∽△BEF ,∴∠A =∠AED =45° ∴∠ADE =90°,∴AE =3 2 , ∵动点E 的速度为1cm/s ∴此时x =3 2 s ;
A 第28题(1)(2)图 A
B
C
D
E F
第28题(3)①图
A
B
C
D E F
第28题(3)②图
A
B
C D
E F
第28题(3)③图
③如图,若BF =BE ,则∠FEB =∠EFB ; 又∵△ADE ∽△BEF ,∴∠ADE =∠AED ∴AE =AD =3,
∵动点E 的速度为1cm/s ∴此时x =3s ;
综上所述,当△BEF 为等腰三角形时,x 的值为3
2 2 s 或
3 2 s 或3s .
(注:求对一个结论得2分,求对两个结论得4分,求对三个结论得5分)
一、判断题 1、新课标提倡关注知识获得的过程,不提倡关注获得知识结果。【错】 2、要创造性地使用教材,积极开发、利用各种教学资源为学生提供丰富多彩的学习素材。【对】 3、不管这法那法只要能提高学生考试成绩就是好法。【错】 4、《基础教育课程改革纲要》指出:课程标准是教材编写、教学、评估和考试命题的依据,是国家管理和评价课程的基础。【对】 5、《纲要》提出要使学生“具有良好的心理素质”这一培养目标很有必要,不仅应该在心理健康教育课中培养,在数学课上也应该关注和培养学生的心理素质。【对】 1、教师即课程。(X) 2、教学是教师的教与学生的学的统一,这种统一的实质是交往。(V) 3、教学过程是忠实而有效地传递课程的过程,而不应当对课程做出任何变革。(X) 4、教师无权更动课程,也无须思考问题,教师的任务是教学。(X) 5、从横向角度看,情感、态度、价值观这三个要素具有层次递进性。(V) 6、从纵向角度看,情感、态度、价值观这三个要素具有相对贸易独立性。(V) 7、从推进素质教育的角度说,转变学习方式要以培养创新精神和实践能力为主要目的。(V) 8、课程改革核心环节是课程实施,而课程实施的基本途径是教学。(V) 9、对于求知的学生来说,教师就是知识宝库,是活的教科书,是有学问的人,没有教师对知识的传授,学生就无法学到知识。(X) 1.课程改革的焦点是协调国家发展需要和学生发展需要二者间的关系. (V) 2.素质教育就是把灌输式与启发式的教学策略相辅相成. (X) 3.全面推进素质教育的基础是基本普及九年义务教育. (X) 4.现代信息技术的应用能使师生致力于改变教与学的方式,有更多的精力投入现实的探索性的数学活动中去. (V) 5.新课程评价只是一种手段而不是目的,旨在促进学生全面发展. (V) 二、选择题(每小题3分,共24分) 1、新课程的核心理念是【为了每一位学生的发展】 2、教学的三维目标是【知识与技能、过程与方法、情感态度价值观】 3、初中数学课程为课标中规定的第几学段【第三】 4、《基础教育课程改革纲要》为本次课程改革明确了方向,基础教育课程改革的具体目标中共强调了几个改变【 6个】 5、课标中要求“会解一元一次方程、简单的二元一次方程组、可化为一元一次方程的分式方程”。这里要求方程中的分式不超过【两个】 6、对“平行四边形、矩形、菱形、正方形、梯形的概念和性质”,课标中知识技能的目标要求是【掌握】 7、七年级上册第七章《可能性》属于下面哪一部分内容【统计与概率】 8、课标中要求“掌握有理数的加、减、乘、除、乘方及简单的混合运算”,这里的运算步骤要【以三步为主】 9、《新课程标准》对“基本理念”进行了很大的修改,过去的基本理念说:“人人学有价值的数学,人人获得必须的数学,不同人在数学上得到不同的发展。”,现在的《新课标》改为:.“人人都能获得良好的数学教育,不同的人在数学教育中得到不同的发展。 10、什么叫良好的数学教育? 就是不仅懂得了知识,还懂得了基本思想,在学习过程中得到磨练。 11.旧的标准理念中,为了突破过去的东西,写的时候有一些偏重,非常强调学生的独立学习,强调
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初中数学综合练习题 一、选择题(每小题3分,共30分) 1.如果a a -=-,下列成立的是( ) A .0a < B .0a ≤ C .0a > D .0a ≥ 2.把2 3x x c ++分解因式得:2 3(1)(2)x x c x x ++=++,则c 的值为( ) A .2 B .3 C .2- D .3- 3.分别剪一些边长相同的①正三角形,②正方形,③正五边形,④正六边形,如果用其中一种正多边形镶嵌,可以镶嵌成一个平面图案的有( ) A .①②③ B .②③④ C .①②④ D .①②③④都可以 4.用 表示三种不同的物体,现放在天平上比较两次,情况如图所示,那么 这三种物体按质量从大到小的顺序排列应为( ) A . B . C . D . 通过图表,估计这个病人下午16:00时的体温是( ) A .38.0℃ B .39.1℃ C .37.6℃ D .38.6℃ 6.给定一列按规律排列的数:1111 1 3579,,,,,它的第10个数是( ) A . 1 15 B . 117 C .119 D .121 7.如图,1O ,2O ,3O 两两相外切,1O 的半径11r =,2O 的半径22r =,2O 的半径33r =,则123O O O △是( ) A .锐角三角形 B .直角三角形 a b c a b c a b c a b c a b c 体温/℃ O 2 a b c
C .钝角三角形 D .锐角三角形或钝角三角形 8.如图是光明中学乒乓球队队员年龄分布的条形图. 这些年龄的众数、中位数依次分别是( ) A .15,15 B .15,15.5 C .14.5,15 D .14.5,14.5 9.由棱长为1的小正方体组成新的大正方体,如果不允许切割,至少要几个小正方体( ) A .4个 B .8个 C .16个 D .27个 10.在Rt ABC △中,90C ∠=,5BC =,15AC =,则A ∠=( ) A .90 B .60 C .45 D .30 二、填空题(本大题共8个小题,每小题4分,共32分.请把答案填在题中横线上) 11.如图,是一块三角形木板的残余部分,量得100A ∠=,40B ∠=,这块三角形木板另外一个角是 度. 12.足球联赛得分规定如图,大地足球队在足球联赛的5场比赛中得8分,则这个队比赛的胜、平、负的情况是 . 13.星期天小华去书店买书时,从镜子内看到背后墙上普通时钟的时针(粗)与分针(细)的位置如图所示,此时时针表示的时间是 .(按12小时制填写) 14.已知一次函数的图象过点(03),与(21),,则这个一次函数y 随x 的 增大而 . 15.上小学五年级的小丽看见上初中的哥哥小勇用测树的影长和自己的影长的方法来测树高,她也学着哥哥的样子在同一时刻测得树的影长为5米,自己的影长为1米.要求得树高,还应测得 . 16.如图,已知AC 平分BAD ∠,12∠=∠,3AB DC ==, 则BC = . 17.如图,一块长方体大理石板的A B C ,,三个面上的边长如图 人数 10 8 6 4 2 0 13 14 15 16 17 18 年龄 C B A 第11题图 第12题图 1 2 A B C D 第13题图
相交线与平行线测试题 、精心选一选,慧眼识金!(每小题3分,共24分) 6 .某人在广场上练习驾驶汽车,两次拐弯后, 行驶方向与原来相同,这两次拐弯的角度可能 是 C. 7. 如图所示是“福娃欢欢”的五幅图案,②、 到( A .'② 8. (2009 . A.80° 如图AB// CD 可以得至U ) 2. A . 7 1=7 2 B 如图所示,7 1和72是对顶角的是( ) D 2 C. 7 1=7 4 D A. B . C. D. .7 3=7 4 如图, 同位角相等,两直线平行 内错角相等,两直线平行 同旁内角互补,两直线平行 两直线平行,同位角相等 给出了过直线外一点作已知直线的平行线的方法,其依据是( 4 . (2007 ?北京)如图,Rt △ ABC 中, 若7 BCE=3&则7 A 的度数为( A . 35 B . 45 C 55 D 5 . (2009 .重庆)如图,直线 则7D 等于( ) A. 70° B. 80° 3 7 ACB=90,DE 过点C 且平行于AB, ) .65 AB CD 相交于点 E, DF// AB.若7 AEC=1O0, A C. 90° D. 100° s B D 第一次左拐30°,第二次右拐30° B .第一次右拐50°,第二次左拐130° 第一次右拐50°,第二次右拐130° 笔.③占.④D 四川遂宁)如图,已知7仁7 2,7 3=80°,则7 4=() B. 70 ° C. 60 ° D. 50 ° 锤定音!(每小题3分,共24分) 二、耐心填一填, 9 . (2009 .上海)如图,已知a / b ,7 1=40°,那么7 2的度数等于 10 .如图,计划把河水引到水池 A 中,先引AB 丄CD ------------------
初中数学期中考试反思精选 反思一 在刚刚结束的期中考试中,我们学校初三年级的数学考试整个年级组的成绩不是很理想,平均分为85分左右。其实看起来数学试卷的难度并不是很难,而每一个小题都不是一眼就能看出答案的,都有一定的技巧在里面,所以从学生答题情况来看,基础知识掌握得较好,概念理解得较透彻,解分式方程的准确率较高,但部分学生理解能力较差,应用题审题不清,导致出现不少错误。几何证明题分析问题的思路上不去,分析问题的方法掌握得不够好。另外,学生的解决问题的能力不同,部分发展不理想的学生学习习惯较差,接受能力较差,碰到思维力度较强的题目就无法解答。在今后的教学中,要特别注重对发展不理想学生的辅导,注重对学生理解能力、解决实际问题等方面的能力培养。 在今后的教学中,我们要在以下几个方面多下功夫: 一、引导学生逐渐认识实际生活中的问题。 如结合信息科技,为学生创设熟悉的教学情境,让学生认识到生活中处处存在数学问题,数学来源于生活又应用于生活,激发学生学习数学的兴趣和认识学习数学的必要性,调动学生学习数学的主观能动性。 二、指导学生解决实际问题时,要留给学生思考的余地。 学生用数学不是靠教师教会的,而是学生想懂的。古人云授之以鱼不如授之以渔。在解决实际生活问题中充分发挥学生灵活运用数学知识解决问题的能力,使学生的思维得到充分的发展。教学过程当中教师要注意让学生亲身感受数学的由来及关注知识的生成,既要有提前的预设,更的灵活处理教学过程中随时可能出现的意外,要有全盘掌握课堂的能力。 三、因材施教、分层实施差异教学 在这次考试中,原本一些不及格的学生,数学成绩却考到了60分以上,主要的原因:其一是他们自身的努力,其二是考前曾降低对他们的要求,每一阶段对他们提出他们能做到的目标,其三是树立他们以及家长的自信心,密切做到家长与老师的配合。他们的进步,我们做老师的从内心深处为他们高兴。从他们的身上也给了我们很大的启示:1、要对每一位学生切切实实做到分层教学分层练习,在每周的练习中让不同的学生做不同的练习。2、对于中下游的学生要及时了解他们薄弱环节,有针对性的对他们进行必要的练习。3、树立每一位学生学习的自信心。不是锤的敲打,而是水的抚摸,才使鹅卵石这般光滑剔透。作为一个老师,如果在威严中不失宽容,多总结教学中的得与失,多找找自身的原因,我想,教育学生才会真正有效。 反思二 考试后,同学们最为关心的莫过于各门功课的分数了。其实分数只不过是对你这一阶段
初中数学综合测试题 一、选择题 1.对任意三个实数c b a ,,,用{}c b a M ,,表示这三个数的平均数,用{}c b a ,,m in 表示这三个数中最小的数,若{}{}y x y x y x y x y x y x M -+++=-+++2,2,22m in 2,2,22,则=+y x ( ) A. ﹣4 B.﹣2 C.2 D.4 2.如图,ABC Rt ?的斜边AB 与圆O 相切与点B ,直角顶点C 在圆O 上,若,则圆O 的半径是( ) A.3 B.32 C.4 D.62 3.现有一张圆心角为108°,半径为40cm 的扇形纸片,小红剪去圆心角为θ的部分扇形纸片后,将剩下的纸片制作成一个底面半径为10cm 的圆锥形纸帽(接缝处不重叠), 则剪去的扇形纸片的圆心角θ的度数为( ) A.18° B.36° C.54° D.72° 4.已知二次函数y=ax 2+bx+c(a ≠0)的图象如图所示,给出以下结论①b 2<4ac;②abc>0;③2a+b=0;④8a+c<0;⑤9a+3b+c<0;其中正确的结论有() A.1个 B.2个 C.3个 D.4个 10.如图,直线都与直线l 垂直,垂足分别为M ,N ,MN=1,正方形ABCD 的边长为,对角线AC 在直线l 上,且点C 位于点M 处,将正方形ABCD 沿l 向右平移,直到点A 与点N 重合为止,记点C 平移的距离为x ,正方形ABCD 的边位于 之间分的长度和为y ,则y 关于x 的函数图象大致为( ) A. B. C. D. 二.填空题 6.如图,点A (m ,2),B (5,n )在函数y=(k >0,x >0)的图象上,将该函数图象向上平移2个单位长度得到一条新的曲线,点A 、B 的对应点分别为A′、B′.图中阴影部分的面积为8,则k 的值为___. 7.如图,正方形ABCD 的边长是16,点E 在边AB 上,AE=3,点F 是边BC 上不与点B ,C 重合的一个动点,把△EBF 沿EF 折叠,点B 落在B′处.若△CDB′恰为等腰三角形,则DB′的长为______.