2006年甘肃省兰州市初中毕业暨高中阶段招生考试

遂宁市2006年初中毕业暨高中阶段学校招生统一考试数学试卷说明：1、本试卷分为第Ⅰ卷和第Ⅱ卷第Ⅰ卷1—2页，第Ⅱ卷3—8页，请将第Ⅰ卷的正确选项填在第Ⅱ卷前面的第Ⅰ卷答题表内，第Ⅱ卷用钢笔或圆珠笔直接答在试卷上．2、本试卷满分150分,答题时间为120分钟.第Ⅰ卷（选择题 共40分）一、 选择题：（本大题共10个小题,每题4分,共40分）在每小题给出的四个选项中,有且仅有一项符合题目要求的,请选出并将其番号填写在第Ⅱ卷前面的第Ⅰ卷答题表内1、计算:︱-4︱=A 、0B 、-4C 、1/4D 、4 2、一组数据：1，3，2，3，1，0，2的中位数是A 、0B 、1C 、2D 、3 3、如图，已知AB 是⊙O 的直径， ⌒ = ⌒ = ⌒ ∠BOC=400，那么∠AOE =A 、400B 、 600C 、600D 、1200 4、下列计算正确的一个是A 、a 5+ a 5 =2B 、a 5+ a 5= a 10C 、a 5+ a 5 =aD 、x 2y+ xy 2=2 x 3 y 35、为了作三项调查：①了解炮弹的杀伤半径；②审查书稿有哪些科学性错误；③考查人们对环境的保护意识．其中不适合作普查而适合作抽样调查的个数是Ａ、０ Ｂ、１ Ｃ、２ Ｄ、３6、如果线段上一点P 把线段分割为两条线段PA 、PB 当PA 2=PB ·AB,即PA ≈0.618AB 时,则称点P 是线段AB 的黄金分割点,现已知线段AB=10,点P 是线段AB 的黄金分割点,如图AB CD DE _ O_ E _ D_ C_ B_ A所示,那么线段PB 的长约为A 、6.18B 、0.382 C 、0.618 D 、3.827、如图所示的物体,是由四个相同的小长方体堆砌而成的,那么这个物体的左视图是8、已知函数y = (x>0),那么 A 、函数图象在一象限内,且y 随x 的增大而减小 B 、函数图象在一象限内,且y 随x 的增大而增大 C 、函数图象在二象限内,且y 随x 的增大而减小 D 、函数图象在二象限内,且y 随x 的增大而增大9、某种手机卡的市话费上次已按原收费标准降低了m 元/分钟,现在两次下调20％，使收费标准为n 元/分钟,那么原收费标准为A 、( n – m)元/分钟B 、( n + m)元/分钟C 、( n – m)元/分钟D 、( n + m)元/分钟 10、已知△ABC 的三边长分别为20cm,50cm,60cm,现要利用长度分别为30cm 和60cm 的细木条各一根,做一个三角形木架与△ABC 相似,要求以其中一根为一边,将另一根截成两段(允许有余料)作为另外两边,那么另外两边的长度(单位: cm)分别为A 、10,25B 、10,36或12,36C 、12,36D 、10,25或12,36遂宁市2006年初中毕业暨高中阶段学校招生统一考试_ B_ P_ AABCDx34 5 5 45 51 1试题答案(非选择题 共110分)二、填空题:(本大题共5个小题,每小题4分,共20分) A11、分解因式: m 2- 1 = ____________________.D12、如图,已知,线段DE 由线段AB 平移而得, AB=DC=4cm,EC=5cm,则△DCE 的周长是_______ cm. C13、不等式组 的解集是__________________ 14、已知样本:3,4,0,-2,6,1,那么这个样本的方差是_____________.15、如图在梯形ABCD 中, ∠DCB=90 0;AB∥CD, AB=25,BC=24.将该梯形折叠,点A 愉好与点D 重合,BE 为折痕,那么AD 的长度为_________. 三、解答题(本大题共10个小题,共90分)16、(本题满分8分)计算:(-2)2+ (3-1)0-3tan60017、(本题满分8分)计算:⎩ ⎨ ⎧ －2 x < 0 x – 2 < 0_ E_ B_ B_ x (x－3) 3x －3 118、(本题满分9分)如图,已知点M 、N 分别是平行四边形ABCD 的边AB 、、DC 的中点,求证: ∠DAN=∠BCM.19、(本题满分8分)某班一次数学测验,全班学生的成绩频数分布表如下:（1） 问:59.5分以上的学生人数占全班学生人数百分之几? （2） 把右边的频数分布直方图补充完整?20、(本题满分9分)地表以下岩层的温度t(0C)随时着所处的深度h(千米)的变化而变化,t 与h 在一寂静范围内近似成一次函数关系(1)根据下表,求t(0C)与 h(千米)之间的函数关系式; (2)求当岩层温度达到7700C 时,岩层所处的深度为多少千米?_ D_ N_ C_ M_ A_ B_ 成绩 _ 学生数_ 14_ 13_ 12_ 11 _ 10 _9 _8_7_6 _5 _4 _3 _2 _1_o 49．5 59．5 69．5 79．5 89．5 10021、(本题满分9分)如图,PA 、PB 是⊙O 的两条切线,切点分别为A 、B 若直径AC=12cm, ∠P=600,求弦AB 的长.22、(本题满分9分)已知二次函数y = x 2 + 4x(1)用配方法把该函数化为y=a(x-h)2 + k(其中a 、h 、k 都是常数且a ≠0)的形式,并指出函数图象的对称轴和顶点坐标. (2) 函数图象与x 轴的交点坐标.23、(本题满分9分)将分别标有数字2,3,5的三张质地,大小完全一样的卡片背面朝上放在桌面上___ P①.随机抽取一张,求抽到奇数的概率②.随机抽取一张作为个位上的数字(不放回),再抽取一张作为十位上的数字,能组成哪些两位数?并求出抽取到的两位数恰好是35的概率.24、(本题满分10分)如图,把正方形ACFG 与Rt △ACB 按如图(甲)所示重叠在一起,其中AC=2, ∠BAC=600,若把Rt △ACB 绕直角顶点C 按顺时针方向旋转,使斜边AB 恰好经过正方形ACFG 的顶点F,得△A ′B ′C ′，A B 分别与A ′C, A ′B ′相交于D 、E,如图(乙)所示.①. △ACB 至少旋转多少度才能得到△A ′B ′C ′?说明理由.②.求△ACB 与△A ′B ′C ′的重叠部分(即四边形CDEF)的面积(若取近似值,则精确到0.1)25、(本题满分11分)有一种笔记本原售价为每8元,甲商场用如下办法促梢,每次购买1～8本打九折、9～16本打八五折、17～25本打八折、超过25本打七五折.乙商场用如下办法促销:①.请仿照乙商场的促销列表,列出甲商场促销笔记本的购买本数与本价格的对照表 ②.某学校有A 、B 两个班都需要买这种笔记本,A 班需要8本,B 班需要15本,问他们到哪家商场购买花钱较少?③设某班需要购买这种笔记本本数为x 且9 ≤x ≤40,总花费为y 元,从最省钱的角度出发,写出y 与x 的函数关系式._ G_ B_ F_ C _ A(甲)_ E_ D_ G_ B_ F_ C_ A(乙)北清桥东中培训中心。

2006年甘肃省普通高校招生考生报名登记表市（州）县（市、区）考试类型科类：1、“考试类型”栏。

指全国统考、三职生两项。

全国统考指参加3+文/理科综合、民考民、民考汉考试的考生。

三职生指参加全省统考3+专业课综合考试的职业高中、普通中专、技工学校的毕业生。

2、“科类”栏。

分为文史类、体育文、音乐文、美术文；理工类、体育理、音乐理、美术理八类。

考生应根据自己的情况选报其中一类，三职生不填写此栏。

考生报考的艺术专业不能明确音乐、美术类时，按音乐文、音乐理选报。

3、“考生号”、“身份证号”、“邮政编码”栏。

考生将自己的学生号、身份证号、邮政编码按左对齐（依照从左至右的顺序，数字之间不允许有空格出现）填写在方格内。

4、“学校代码”、“学校名称”栏。

先填写毕业学校名称，然后从《甘肃省高级中等教育学校名称代码表》中查出该校的代码并填写在“学校代码”栏内。

如查不到学校名称及代码时，只填写毕业学校的名称即可，往届生一律填写原毕业学校。

5、“通知书邮寄地址”栏。

应届考生一律填写毕业学校通讯地址，往届考生一律填写报名县（市、区）招办的通讯地址。

此栏专为发放考生《录取通知书》设置，考生务必按要求准确填写。

否则造成《录取通知书》投递错误，由此产生的后果自负。

6、“外语语种”栏。

设英、俄、日、德、法、西班牙六种外语。

考生应根据自己所学的语种，选择其中一种语种填写。

民考汉、民考民考生不填写此栏。

7、“考生类别”栏。

考生户口是城镇的，应届生填城市应届，往届生填城市往届；户口是农村的，应届生填写农村应届，往届生填写农村往届。

8、“毕业类别”栏。

分为普通中学、中等师范、职业高中、技工学校、其他中等专业学校及中等学历教育毕业等六类。

9、“采用何种少数民族语种答卷”、“加试科目”栏。

用本民族语种答卷的考生可在藏文、蒙文、哈文三种语言试卷中选择所学的一种民族语言填写，同时在加试科目栏中填写汉语文；报考民族院校（班）且用汉语文答卷的藏族考生，须填写加试藏语文。

绝密★启用前[考试时间：2006年6月14日下午2：30—5：00]乐山市2006年初中毕业会考暨高中阶段招生统一考试政史地（旧课程版）本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(综合题)两部分。

第Ⅰ卷1至6页，第Ⅱ卷7至l 4页，共200分。

考试时间为150分钟。

考试结束后，第Ⅱ卷和答题卡按规定装袋上交。

第Ⅰ卷(选择题共101分)注意事项：1．答第Ⅰ卷前，考生务必将自己的姓名、准考证号、考试科目用铅笔涂写在答题卡上。

2．每小题选出答案后，用铅笔把答题卡上对应题目的答案标号涂黑。

如需改动，用橡皮擦干净后，再选涂其它答案，不能答在试题卷上。

3．试结束后，本试卷由考场统一收回，集中管理不上交。

本卷共43小题，其中l题至6题为地理，7题至28题为历史，每小题2分；29题至43题为政治，每小题3分。

地理、历史、政治共101分。

在下列各小题的四个选项中，只有一项是符合题目要求的。

1.右图是一幅经纬网示意图。

有关该图的叙述，正确的是A.南回归线的纬度是23.5ºNB.纬线中，数码①所示纬线最长C.数码②所示经线具有指示东西方向的作用D.点③位于北回归线与北极圈之间，属于寒带2.我们的家乡乐山市地形起伏较大，最大相对高度达3968米。

境内最低处位于岷江出市境处，海拔仅为320米，最高处位于西部的马鞍山。

马鞍山山顶的海拔应是A.4288米B.3648米C.4608米D.3968米农业是受自然条件影响很大的一个生产部门。

美国充分利用本国地形、气候条件，扬长避短发展农业，是农业高度发达的国家。

读美国年降水量分布示意图和美国农业带分布示意图，回答3－4题。

3.美国年降水量分布示意图中，从①地至②地沿线降水变化的规律是A.逐渐减少B.逐渐增加C.先增加后减少D.先减少后增加4.美国乳畜带、棉花带、玉米带、亚热带作物带按纬度由高到低排列正确的是A.亚热带作物带、棉花带、玉米带、乳畜带B.棉花带、玉米带、乳畜带、亚热带作物带C.玉米带、亚热带作物带、棉花带、乳畜带D.乳畜带、玉米带、棉花带、亚热带作物带我国北方地区与南方地区自然环境差异明显，对生产、生活影响深刻。

兰州市2010年初中毕业生学业考试试卷数学（A ）参考答案及评分标准一、选择题（本题15小题，每小题4分，共60分）16．54m ≤且1m ≠ 17．5 1819．6 20．12三、解答题（本题8小题，共70分．解答时写出必要的文字说明、证明过程或演算步骤.） 21．（本题满分10分） （1）（本小题满分4分）解：原式=34132++-- ··························································································· 2分=3332++- ·································································································· 3分5= ··························································································································· 4分 （2）（本小题满分6分）解：1y 与2x 成正比例，2y 与x 成反比例设211y k x =，22k y x=，221k y k x x =+················································································ 2分把1x =，3y =，1x =-，1y =分别代入上式得121231k k k k =+⎧⎨=-⎩ ········································· 3分∴1221k k =⎧⎨=⎩， 212y x x =+ ································································································· 5分当12x =-，211132212222y ⎛⎫=⨯-+=-=- ⎪⎝⎭- ································································ 6分 22．（本题满分6分） （1）（本小题满分4分）用尺规作出两边的垂直平分线 ······························································································· 2分 作出圆······································································································································ 3分O 即为所求做的花园的位置.（图略） ·············································································· 4分（2）（本小题满分2分）解：∵90BAC ∠=，8AB =米，6AC =米，∴10BC =米∴ABC △外接圆的半径为5米 ····························································································· 5分 ∴小明家圆形花坛的面积为25π平方米 ················································································ 6分 23．（本题满分6分）（1）所有可能的结果如下表：一共有16种结果，每种结果出现的可能性相同．·············· 2分 和为偶数的概率为83166= 所以小莉去上海看世博会的概率为38···················································································· 3分 （2）由（1）列表的结果可知：小莉去的概率为83，哥哥去的概率为85，所以游戏不公平，对哥哥有利． ·························································································································· 4分游戏规则改为：若和为偶数则小莉得5分，若和为奇数则哥哥得3分，则游戏是公平的．··········································································································· 6分（游戏规则的修改有多种多样，阅卷老师视情况给分） 24．（本题满分8分）（1）如图，作AD BC ⊥于点D ··················· 1分Rt ABD △中，sin 4542AD AB ==⨯=················· 2分在Rt ACD △中，∵30ACD ∠=∴2 5.6AC AD == ·································································································· 3分 即新传送带AC 的长度约为5.6米． ····················································································· 4分 （2）结论：货物MNQP 应挪走． ······················································································· 5分 解：在Rt ABD △中，cos 4542BD AB ==⨯=················································· 6分 B CND在Rt ACD △中，cos30CD AC ===∴ 2.1CB CD BD =-==≈∵ 2.1 1.92PC PB CB =--=<≈4 ·················································································· 7分 ∴货物MNQP 应挪走．········································································································· 8分 25．（本题满分9分）解：（1）11POA △的面积将逐渐减小． ················································································· 2分 （2）作11PC OA ⊥，垂足为C ，因为11POA △为等边三角形，所以11OC PC ==，1P ． ·········································································· 3分代入ky x=，得k =y x =． ······································ 4分作212P D A A ⊥，垂足为D ，设1A D a =，则22OD a P D =+=，，所以2(2)P a +． ············································································································· 6分代入y =(2)a +=2210a a +-=解得：1a =-··············································································································· 7分∵0a > ∴1a =-+··································································································· 8分所以点2A 的坐标为 ·································································································· 9分26．（本题满分10分）解：（1）∵OA OC =，∴A ACO ∠=∠ ∵22COB A COB PCB ∠=∠∠=∠， ∴A ACO PCB ∠=∠=∠ ······································································································ 1分 ∵AB 是O 的直径∴90ACO OCB ∠+∠=······································································································ 2分 ∴90PCB OCB ∠+∠= ，即OC CP ⊥ ············································································· 3分 ∵OC 是O 的半径 ∴PC 是O 的切线 ··············································································································· 4分 （2）∵PC AC = ∴A P ∠=∠ ∴A ACO PCB P ∠=∠=∠=∠∵COB A ACO CBO P PCB ∠=∠+∠∠=∠+∠， ∴CBO COB ∠=∠ ··············································································································· 5分 ∴BC OC =∴12BC AB =······························································ 6分 （3）连接MA MB ， ∵点M 是弧AB 的中点∴弧AM =弧BM ∴ACM BCM ∠=∠ ··············· 7分 ∵ACM ABM ∠=∠ ∴BCM ABM ∠=∠ ∵BMC BMN ∠=∠ ∴MBN MCB △∽△ ∴BMMN MC BM = ∴2BM MC MN = ··············································································································· 8分 ∵AB 是O 的直径，弧AM =弧BM ∴90AMB AM BM ∠==，∵4AB =∴BM =·································································································· 9分 ∴28MC MN BM == ······································································································· 10分 27．（本题满分10分） 解：（1）∵AC BD ⊥ ∴四边形ABCD 的面积12S AC BD =⨯ 11082=⨯⨯ 40= ·························································································· 2分 （2）过点A 作AE BD ⊥，垂足为E ·················································································· 3分 ∵四边形ABCD 为平行四边形152AO CO AC === 142B O D O B D ===在Rt AOE △中，sin AEAOE AO∠=∴sin sin 605AE AO AOE AO =∠=⨯==··················································· 4分∴11422AOD S OD AE ==⨯= △········································································ 5分 ∴四边形ABCD的面积4AOD S S ==△····································································· 6分 （3）如图所示过点A ，C 分别作AE BD ⊥，CF BD ⊥，垂足分别为E ，F ············ 7分PAM第26题图在Rt AOE △中，sin AEAOE AO∠=∴sin sin AE AO AOE AO θ=∠=⨯ 同理可得sin sin CF CO COF CO θ=∠=⨯ ·························· 8分 ∴四边形ABCD 的面积1122ABD CBD S S S BD AE BD CF =+=+ △△ 1sin ()2BD AO CO θ=+ 1sin 2BD AC θ= 1sin 2ab θ= ·············································································· 10分 28．（本题满分11分）解：（1）因抛物线c bx x y ++-=2经过坐标原点O （0，0）和点E （4，0） 故可得04c b ==，所以抛物线的解析式为24y x x =-+ ···················································································· 1分由xx y 42+-=()224y x =--+ 得当2x =时，该抛物线的最大值是4． ··············································································· 2分 （2）①点P 不在直线ME 上.已知M 点的坐标为（2，4），E 点的坐标为（4，0）， 设直线ME 的关系式为y kx b =+．于是得⎩⎨⎧=+=+4204b k b k ，解得⎩⎨⎧=-=82b k所以直线ME 的关系式为28y x =-+ ·················································································· 3分由已知条件易得，当411=t 时，OA=AP=411，111144P ⎛⎫⎪⎝⎭， ················································ 4分∵P 点的坐标不满足直线ME 的关系式28y x =-+． ∴ 当411=t 时，点P 不在直线ME 上． ·············································································· 5分 ②以P 、N 、C 、D 为顶点的多边形面积可能为5 ∵点A 在x 轴的非负半轴上，且N 在抛物线上， ∴OA=AP=t .O ABCD第27题图F E。