九年级第一次模拟考试08.3

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二、填空题（本大题 6 小题，每小题 4 分，共 24 分）请将下列各题的正确答案填写
在答题卡相应的位置上．
11、已知反比例函数 y k 的图象经过(3，－2)，则 k ____ x
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7 元，8 元，9 元，10 元，则这组数据的中位数与众数分别为 （ ）
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9、在一个不透明的口袋中，装有 5 个红球 3 个白球，它们除颜色外都相同，从中任
意摸出一个球，摸到红球的概率为
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B
7．据广东信息网消息，2011 年第一季度，全省经济运行呈现平稳增长态势．初
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）
A． 5.206 102 亿元

翔宇教育集团江苏省淮安外国语学校2007-2008学年度第二学期第一次模拟初三数学参考答案一、选择题（每题3分，共30分）⒈D⒉C⒊B⒋C⒌B⒍D⒎A⒏A⒐B⒑D二、填空题（每题3分，共24分）⒒略⒓40°⒔4π ⒕553 ⒖2 ⒗16⒘y=27x 3 ⒙40162009 三、解答题 ⒚原式=1-x x …………………………………………………………………………………4分 =212+………………………………………………………………………………………8分⒛1≤x ＜3………………………………………………………………………………6分表示略.………………………………………………………………………………………8分21.⑴略；………………………………………………………………………………………5分 ⑵仍成立,理由略.……………………………………………………………………………9分22.⑴相切,理由略；……………………………………………………………………………5分 ⑵22.…………………………………………………………………………………………9分23.⑴京；………………………………………………………………………………………2分 ⑵①心心相扣的心形，象征志愿者与运动员、奥林匹克大家庭和所有宾客心连着心、用心服务、奉献爱心，为奥林匹克运动增添光彩；②欢快舞动的人形，展现了志愿者奉献为乐的志愿精神，志愿者真挚的笑容、出色的服务、友善的行为将唤起每一位奥运会参与者的心灵共鸣．注：答对“心心相扣”、“欢快舞动的人形”其中之一大概意义的即可给分……………4分 ⑶作图…………………………………………………………………………………………8分 含义【应包含“淮外”（曙光）、“校友”】 ………………………………………………10分24. ⑴作图略；…………………………………………………………………………………3分 ⑵45π…………………………………………………………………………………………8分25.⑴y=x －1；y=x3……………………………………………………………………………4分 ⑵（0,5）、………………………………………………………………………………………7分 （0,－7）………………………………………………………………………………………10分26.⑴0.60或0.6；………………………………………………………………………………2分 ⑵黑球有8只，白球有12只；………………………………………………………………4分 ⑶树状图或列表正确 …………………………………………………………………………6分 P （两只球颜色不同）=9548.…………………………………………………………………10分 27. ⑴240；……………………………………………………………………………………2分⑵300－a ，……………………………………………………………………………………3分0.8+0.05a ，……………………………………………………………………………………4分P=－0.14(a-150)2+3290；……………………………………………………………………8分⑶5m ，168.75万元(295m ，409.75舍去) (12)分 四、综合与实践28．⑴圆心角α=90°，蚂蚁所走的最短路程为205cm ；……………………………………4分 ⑵圆心角α=180°，蚂蚁由A 沿母线AS 到达顶点S ，绕点S 旋转一周后按原路返回到母线的中点M ．……………………………………………………………………………………………8分⑶①分类不周全；………………………………………………………………………………9分 ②分三种情形：（Ⅰ）当扇形的圆心角0°＜α＜180°时(如图①)，A M ′是最短路径；………………10分 （Ⅱ）当圆锥的锥角等于60°时,圆锥的侧面展开图是一个半圆面（α=180°）(如图②)，对于这种情形，最短路径还可以认为是图中线段AM ′=AS+SM ′．……………………………11分（Ⅲ）但是，当圆锥的锥角大于60°时,圆锥的侧面展开图是一个大于半圆面的扇形(如图③)，此时，蚂蚁的最短路径应该是折线AS+S M ′，也可以认为是AS+SM ，我们可以这样理解，这种路径就是蚂蚁由A 沿母线AS 到达顶点S ，绕点S 旋转一周后按原路返回到母线的中点M ．……………………………………………………………………………………………………12分图② 图③ 图①。

2008年初中毕业生学业考试模拟（一）数学参考答案三、解答题：（本大题共9小题，共102分.）注：下面只是给出各题的一般解法，其余解法对应给相应的分数17、（9分）解：原式2(1)(1)1aaa-=+--·······················································································4分(1)(1)a a=+--·························································································6分2 =··············································································································9分18、（9分）解：设每个小组原先每天生产x件产品 ·································································1分根据题意可得310500310(1)500xx⎧⎪⎨⎪⎩⨯<⨯+>·············································································6分解得22151633x<< ····································································································7分x∵的值应是整数，16x=∴． ···················································································································8分答：每个小组原先每天生产16件产品． ······························································9分证明：（1）由折叠可知：D D '∠=∠，CD AD '=，C D AE '∠=∠． ∵四边形ABCD 是平行四边形，B D ∠=∠∴，AB CD =，C BAD ∠=∠． ······························································ 2分 ∴B D '∠=∠，AB AD '=，D AE BAD '∠=∠， 即1223∠+∠=∠+∠． 13∠=∠∴．ABE AD F '∴△≌△． ············································· ················································ 4分 （2）四边形AECF 是菱形． ·················································································· 5分 由折叠可知：AE EC =，45∠=∠．∵四边形ABCD 是平行四边形， AD BC ∴∥． 56∠=∠∴．46∠=∠∴． AF AE =∴． AE EC =∵，AF EC =∴．又AF EC ∵∥，∴四边形AECF 是平行四边形． ············································································ 8分 AF AE =∵，∴四边形AECF 是菱形． ······················································································ 10分20、（10分）解：（1）由两个统计图可知该校报名总人数是16016040040%0.4==（人）． ········· 2分 （2）选羽毛球的人数是40025%100⨯=（人）． ············································· 4分因为选排球的人数是100人，所以10025%400=， ··············································· 6分 因为选篮球的人数是40人，所以4010%400=， 即选排球、篮球的人数占报名的总人数分别是25%和10%． ···························· 8分 （3）如图． ············································································································ 10分A F D C EB D '12 34 5 6解：(1)因为正方形也是菱形，在6张中，抽一张为菱形的概率为26=13. ········· 3分 (2)∵在6张卡片中，中心对称图形有：平行四边形、矩形、菱形、正方形4种， ···························································································································· 4分∴随机抽取，则抽一张为菱形的概率为4263=. ····················································· 6分(3)在6张卡片中，轴对称图形有矩形、菱形、正方形,分别记为a 、b 、c ，其··························· 9分 由上表知，随机抽取两张共有5×6=30种（含先后顺序）可能，其中符合条件的有2×3=6种， ···································································································· 10分∴两张都是轴对称图形的概率为Ｐ＝61305=. ··························································· 12分 22、（12分） 解：(1)∵点122B ⎛⎫-- ⎪⎝⎭，在反比例函数12k y x=图象上， 12122k -=⎛⎫⨯- ⎪⎝⎭∴ 12k =∴ ∴反比例函数的解析式为1y x=． ············································································ 2分 又(1)A n ∵，在反比例函数图象上，11n =∴ 1n =∴A ∴点坐标为(11)，． ··································································································· 4分 ∴一次函数2y k x b =+的图象经过点1(11)22A B ⎛⎫-- ⎪⎝⎭，，， 221122k b k b +=⎧⎪⎨-+=-⎪⎩∴ 221k b =⎧⎨=-⎩∴ ∴一次函数的解析式为21y x =-． ·········································································· 8分(2)存在符合条件的点P ···························································································· 9分 可求出点P 的坐标为((20)(10)， ·························································· 12分(1) 证明：BE CD ∥，AB CD ⊥， AB BE ∴⊥． 又AB 为直径， BE ∴为⊙O 的切线． ······························································································· 4分(2)AB 为直径，AB CD ⊥，116322CM CD ∴==⨯=． ······················································································· 6分⌒ BC = ⌒ BD．BAC BCD ∴∠=∠．1tan 2BCD ∠=， 12BM CM ∴=． 1322BM CM ∴==． ································································································· 8分1tan tan 2CM BAC BCD AM ∴=∠=∠=， 6AM ∴=． ············································································································· 10分∴⊙O 的直径315622AB AM BM =+=+=． ······················································· 12分E第23题图解：（1）回答不唯一，如平行四边形、等腰梯形等． ·············································· 2分（2）答：与A ∠相等的角是BOD ∠（或COE ∠）． ············································· 3分四边形DBCE 是等对边四边形． ························································ 5分（3）答：此时存在等对边四边形，是四边形DBCE ． ······································ 7分证法一：如图1，作CG BE ⊥于G 点，作BF CD ⊥交CD 延长线于F 点． ····· 8分∵12DCB EBC A ∠=∠=∠，BC 为公共边，∴BCF CBG △≌△．………………………………… ……10分 ∴BF CG =．∵BDF ABE EBC DCB ∠=∠+∠+∠，BEC ABE A ∠=∠+∠， ∴BDF BEC ∠=∠． 可证BDF CEG △≌△． ························································································· 12分 ∴BD CE =．∴四边形DBCE 是等边四边形． ·········································································· 14分证法二：如图2，以C 为顶点作FCB DBC ∠=∠，CF 交BE 于F 点． ············· 8分∵12DCB EBC A ∠=∠=∠，BC 为公共边，∴BDC CFB △≌△． ····························································································· 10分 ∴BD CF =，BDC CFB ∠=∠．∴ADC CFE ∠=∠．∵ADC DCB EBC ABE ∠=∠+∠+∠，FEC A ABE ∠=∠+∠， ∴ADC FEC ∠=∠．∴FEC CFE ∠=∠．∴CF CE =．∴BD CE =．∴四边形DBCE 是等边四边形． ·········································································· 14分说明：当AB AC =时，BD CE =仍成立．只有此证法，只给1分．B OADE CF 图2B O AD E C F 图1 G（1）当12p =时，1(100)2y x x =+-，即1502y x =+． ∴y 随着x 的增大而增大，即12p =时，满足条件（ii ）． ··································· 3分 又当20x =时，12050602y =⨯+=；当100x =时，1100501002y =⨯+=．而原数据都在20~100之间，所以新数据都在60~100之间，即满足条件（i ）．综上可知，当12p =时，这种变换满足要求； ······················································ 6分（2）本题是开放性问题，答案不唯一．若所给出的关系式满足：()20a h ≤；()b 若20100x =，时，y 的对应值m n ，都能落在60~100之间，则这样的关系式都符合要求．如取20h =，2(20)y a x k =-+， ············································································ 8分0a >，∴当20100x ≤≤时，y 随x 的增大而增大． ···································· 10分令20x =，60y =，得60k = ① 令100x =，100y =，得280100a k ⨯+= ②由①②解得116060a k ⎧=⎪⎨⎪=⎩，． 21(20)60160y x =-+． ················································ 14分。

2008年中考第一次模拟考试九年级历史试题（考试时间：50分钟试卷分值：60分）同学们，经过半学期对中考综合复习，相信大家一定有很大的收获，那就一起来检验一下吧，祝你成功！注意：要细心，不要把答案写错地方哦！请将所有答案写在答题纸上！选择题部分（24分）一、选择题：在下列每小题所给出的四个选项中选出最符合题目要求的一项。

（第1—10题，每小题1分；第11—17题，每小题2分。

共24分）专题一：农业·农村·农民1．张骞通西域后，不少西域的物种传到中原，丰富了中原人民的生活。

下列物种，哪一种不是从西域传人的A.小麦B.核桃C.石榴D.葡萄2．东晋南朝时期，江南地区得以开发的最主要原因是A．江南统治者调整统治政策B．北方农民南迁带来先进技术C．南方战乱少，社会安定D．南方的自然条件优越3.十六届五中全会通过的《中共中央关于制定国民经济和社会发展第十一个五年规划的建议》，把社会主义新农村建设作为农村工作的重要目标。

标志着社会主义制度在农村初步建立的事件是A.1952年，土地改革的完成B. 1958年，开展人民公社化运动C. 1956年，全国农民加入农业生产合作社D.1978年，实行家庭联产承包责任制4．从“大跃进”运动、人民公社化运动中，应吸取的教训是①经济建设要遵循客观规律，不能急于求成②经济建设要立足于本国国情⑧坚持实事求是，一切从实际出发④生产关系一定要适应生产力的发展A. ①③④B.①②③C.②③④D.①②③④专题二：科技·生产·生活5.连云港田湾核电站的建成，将为经济的进一步发展提供更丰富的电力资源。

这一重大成就得益于A. 哥白尼的太阳中心说B.达尔文的生物进化论C. 牛顿的经典力学理论D.爱因斯坦的“相对论”6.下列现实生活中涉及到的科技发明，不属于第三次科技革命的是A.访问互联网B. 在电灯下看书C.用数码相机拍照D. 使用“克隆”技术专题三：民主·人权·法制.7. 依据古代雅典民主，以下哪些人具有公民大会的选举权和被选举权A．侨居雅典的斯巴达人B．雅典贵族布拉达斯未满十岁的儿子C．雅典30岁的男子昂纳多D．雅典贵族布拉达斯的妻子8.某男子生于1821年，40岁时，他从农奴身份成为法律上的“自由人”，并自己出钱购买了一块“份地”。