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  • 2005年上海高考数学真题(理科)试卷(word解析版).doc

    绝密★启用前2005年普通高等学校招生全国统一考试(上海卷)数学试卷(理工农医类)(满分150分,考试时间120分钟)考生注意1.本场考试时间120分钟,试卷共4页,满分150分,答题纸共2页.2.作答前,在答题纸正面填写姓名、准考证号,反面填写姓名,将核对后的条形码贴在答题纸指定位置.3.所有作答务必填涂或书写在答题纸上与试卷题号对应的区域,不得错位.在试卷上作答一律不得分.4.用2B铅笔作答选择题,用黑色字迹钢笔、水笔或圆珠笔作答非选择题.一、填空题()1.函数的反函数________________2.方程的解是___________________3.直角坐标平面中,若定点与动点满足,则点P的轨迹方程是______________4.在的展开式中,的系数是15,则实数______________5.若双曲线的渐近线方程为,它的一个焦点是,则双曲线的方程是____6.将参数方程(为参数)化为普通方程,所得方程是______7.计算:______________8.某班有50名学生,其15人选修A课程,另外35人选修B课程从班级中任选两名学生,他们是选修不同课程的学生的概率是____________(结果用分数表示)9.在中,若,,,则的面积S=_________10.函数的图像与直线又且仅有两个不同的交点,则的取值范围是____________4a5a3a2a4a5a3a2a11.有两个相同的直三棱柱,高为,底面三角形的三边长分别为、、用它们拼成一个三棱柱或四棱柱,在所有可能的情形中,全面积最小的一个是四棱柱,则的取值范围是_______12.用n个不同的实数可得到个不同的排列,每个排列为一行写成一个行的数阵对第行,记例如:用1,2,3可得数阵如下,由于此数阵中每一列各数之和都是12,所以,那么,在用1,2,3,4,5形成的数阵中,___________________二、选择题()13.若函数,则该函数在上是(A) 单调递减无最小值(B) 单调递减有最小值(C) 单调递增无最大值(D) 单调递增有最大值14.已知集合,,则等于(A)(B) (C) (D) 15.过抛物线的焦点作一条直线与抛物线相交于A、B两点,它们的横坐标之和等于5,则这样的直线(A) 又且仅有一条(B) 有且仅有两条(C) 有无穷多条(D)不存在16.设定义域为为R的函数,则关于的方程有7个不同的实数解得充要条件是(A)且 (B)且(C)且 (D)且三、解答题17.已知直四棱柱中,,底面123132213231312321D1C1B1A1DCBA是直角梯形,,,,,,求异面直线与所成的角的大小(结果用反三角函数表示)18.证明:在复数范围内,方程(为虚数单位)无解19.点A、B分别是椭圆长轴的左、右焦点,点F是椭圆的右焦点点P在椭圆上,且位于x轴上方,(1)求P点的坐标;(2)设M是椭圆长轴AB上的一点,M到直线AP的距离等于,求椭圆上的点到点M的距离d的最小值321-1-2-3FAPBMoyx20.假设某市2004年新建住房400万平方米,其中有250万平方米是中低价房预计在今后的若干年内,该市每年新建住房面积平均比上一年增长8%,另外,每年新建住房中,中低价房的面积均比上一年增加50万平方米那么,到那一年底,(1)该市历年所建中低价房的累计面积(以2004年为累计的第一年)将首次不少于4750万平方米?(2)当年建造的中低价房的面积占该年建造住房面积的比例首次大于85%?21.(本题满分18分)本题共有3个小题,第1小题满分4分,第2小题满分8分,第3小题满分6分对定义域是.的函数.,规定:函数(1)若函数,,写出函数的解析式;(2)求问题(1)中函数的值域;(3)若,其中是常数,且,请设计一个定义域为R的函数,及一个的值,使得,并予以证明22.在直角坐标平面中,已知点,,,,其中n是正整数对平面上任一点,记为关于点的对称点,为关于点的对称点,为关于点的对称点(1)求向量的坐标;(2)当点在曲线C上移动时,点的轨迹是函数的图像,其中是以3位周期的周期函数,且当时,求以曲线C为图像的函数在上的解析式;(3)对任意偶数n,用n表示向量的坐标2005年高考理科数学上海卷试题及答案参考答案1.2.x=03. x+2y-4=0 4. - 5.6.7. 38. 9.10. 11. 解析:①拼成一个三棱柱时,只有一种一种情况,就是将上下底面对接,其全面积为②拼成一个四棱柱,有三种情况,就是分别让边长为所在的侧面重合,其上下底面积之和都是,但侧面积分别为

    上传时间:2024-02-20 页数:9

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  • 北美儿童文学杂志红蜘蛛7. Spider Magazine Stories, Games, Activites and Puzzles for Children and Kids - September 2016.pdf

    北美儿童文学杂志红蜘蛛7. Spider Magazine Stories, Games, Activites and Puzzles for Children and Kids - September 2016.pdf

    上传时间:2024-02-06 页数:41

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  • 精品解析:2021年湖北省普通高中学业水平选择性考试化学试题(解析版).docx

    2021年湖北省普通高中学业水平选择性考试化学可能用到的相对原子质量:H-1C-12N-14O-16F-19S-32Cl-35.5K-39Ga-70In-115Sb-122一、选择题:本题共15小题,每小题3分,共45分。在每小题给出的四个选项中,只有一项是符合题目要求的。1. 碳中和有利于全球气候改善。下列有关CO2的说法错误的是A. CO2是Ⅴ形的极性分子B. CO2可催化还原为甲酸C. CO2晶体可用于人工增雨D. CO2是侯氏制碱法的原料之一【答案】A【解析】【分析】【详解】A.CO2的中心原子C的价层电子对数为2,无孤电子对,CO2的空间构型为直线形,其正负电荷中心重合,为非极性分子,A项错误;B.CO2可以被H2在催化剂作用下还原为甲酸,B项正确;C.固态二氧化碳(干冰)升华时吸热,可用于人工增雨,C项正确;D.侯氏制碱法中发生反应: NaC1 + H2O + NH3 +CO2= NaHCO3↓+NH4C1,CO2是生产原料之一,D项正确;故选A。2. 乌铜走银是我国非物质文化遗产之一。该工艺将部分氧化的银丝镶嵌于铜器表面,艺人用手边捂边揉搓铜器,铜表面逐渐变黑,银丝变得银光闪闪。下列叙述错误的是A. 铜的金属活动性大于银B. 通过揉搓可提供电解质溶液C. 银丝可长时间保持光亮D. 用铝丝代替银丝铜也会变黑【答案】D【解析】【分析】【详解】A.根据金属活动性顺序表,铜的金属活动性大于银,A项正确;B.通过揉搓,手上的汗水可提供电解质溶液,B项正确;C.银丝发生还原反应,氧化银转化为单质银,单质银活动性较弱,可长时间保持光亮,C项正确;学科网(北京)股份有限公司 D.铜的活动性比铝弱,氧化铝不能被铜还原为单质铝,因此铜不会变黑,D项错误;答案选D。3. 氢化可的松乙酸酯是一种糖皮质激素,具有抗炎、抗病毒作用,其结构简式如图所示。有关该化合物叙述正确的是A. 分子式为C23H33O6B. 能使溴水褪色C. 不能发生水解反应D. 不能发生消去反应【答案】B【解析】【分析】【详解】A.由图可知该化合物的分子式为,故A错误;B.由图可知该化合物含有碳碳双键,能与溴发生加成反应而使溴水褪色,故B项正确;C.由图可知该化合物含有酯基,能发生水解反应,故C错误;D.由图可知该化合物含有羟基,且与羟基相连碳原子的邻位碳原子上连有氢原子,能发生消去反应,故D错误;故答案:B。4. 对于下列实验,不能正确描述其反应的离子方程式是A. 向氢氧化钡溶液中加入盐酸:H++OH-=H2OB. 向硝酸银溶液中滴加少量碘化钾溶液:Ag++I-=AgI↓C. 向烧碱溶液中加入一小段铝片:2Al+2OH-+6H2O=2[Al(OH)4]-+3H2↑D. 向次氯酸钙溶液中通入少量二氧化碳气体:ClO-+CO2+H2O=HClO+HCO【答案】D【解析】【分析】此题考查离子方程式的书写。学科网(北京)股份有限公司 【详解】A.化学方程式为,强酸,强碱,可溶性盐拆开,离子方程式,A项正确;B.化学方程式为,钾盐,硝酸盐要拆开,离子方程式为,B项正确;C.化学方程式为,离子方程式,C项正确;D.向次氯酸钙溶液中通入少量气体,发生反应生成和,D项错误;故选D。5. 下列实验现象与实验操作不相匹配的是实验操作实验现象A.将盛有Fe(OH)3胶体的烧杯置于暗处,用红色激光笔照射烧杯中的液体在与光束垂直的方向看到一条光亮的通路B.向盛有少量Mg(OH)2沉淀的试管中加入适量饱和NH4Cl溶液,振荡白色沉淀溶解C.向盛有3.0mL无水乙醇的试管中加入一小块金属钠有气泡产生D.向盛有2.0mL甲苯的试管中加入3滴酸性KMnO4,溶液,用力振荡紫色不变A. AB. BC. CD. D【答案】D【解析】【分析】【详解】A.胶体具有丁达尔效应,A项正确;学科网(北京)股份有限公司 B.在溶液中存在溶解平衡:,可与结合为,使减小,溶解平衡正向移动,因此白色沉淀溶解,B项正确;C.向无水乙醇中加入金属钠可发生反应:,有气泡产生,C项正确;D.甲苯能被酸性高锰酸钾溶液氧化为苯甲酸,因此溶液紫色变浅或消失,D项错误。故选D。6. NA为阿伏加德罗常数的值。下列说法错误的是A. 23gCH3CH2OH中sp3杂化的原子数为NAB. 0.5molXeF4中氙的价层电子对数为3NAC. 1mol[Cu(H2O)4]2+中配位键的个数为4NAD. 标准状况下,11.2LCO和H2的混合气体中分子数为0.5NA【答案】A【解析】【分析】【详解】A.中C和O均为杂化,23g乙醇为0.5mol,杂化的原子数为,选项A错误;B.中氙的孤电子对数为,价层电子对数为,则中氙的价层电子对数为,选项B

    上传时间:2024-02-18 页数:21

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  • 中考数学冲刺:方案设计与决策型问题--巩固练习(基础).doc

    中考冲刺:方案设计与决策型问题—巩固练习(基础)【巩固练习】一、选择题1.小明中午放学回家自己煮面条吃,有下面几道工序:①洗锅盛水需2分钟;②洗菜需3分钟;③准备面条及佐料需2分钟;④用锅把水烧开需7分钟;⑤用烧开的水煮面条和菜需3分钟.以上各工序除(4)外,一次只能进行一道工序,小明要将面条煮好,最少用()A.14分钟B.13分钟 C.12分钟D.11分钟2.某学校组织340名师生进行长途考察活动,带有行李170件,计划租用甲、乙两种型号的汽车共10辆.经了解,甲车每辆最多能载40人和16件行李,乙车每辆最多能载30人和20件行李.请问可行的租车方案有()A.2种 B.3种C.4种 D.5种3.(2016•邯郸一模)如图是小李销售某种食品的总利润y元与销售量x千克的函数图象(总利润=总销售额﹣总成本).由于目前销售不佳,小李想了两个解决方案:方案(1)是不改变食品售价,减少总成本;方案(2)是不改变总成本,提高食品售价.下面给出的四个图象中虚线表示新的销售方式中利润与销售量的函数图象,则分别反映了方案(1)(2)的图象是()A.②,③B.①,③ C.①,④ D.④,②二、填空题4.(2016春•乳山市期中)某足球赛一个赛季共进行了26轮比赛(即每队均需26场),其中胜一场得3分,平一场得1分,负一场得0分,某队在这个赛季中平局的场数比负的场数多7场,结果共得34分,则这个队在第一赛季中胜、平、负的场数依次是 .5.开学初,小芳和小亮去学校商店购买学习用品,小芳用18元钱买了1支钢笔和3本笔记本;小亮用31元买了同样的钢笔2支和笔记本5本.(1)每支钢笔的价格为 ;每本笔记本的价格为 ;(2)校运会后,班主任拿出200元学校奖励基金交给班长,购买上述价格的钢笔和笔记本共48件作为奖品,奖给校运会中表现突出的同学,要求笔记本数不少于钢笔数,共有种购买方案?请你一一写出.6.五·一假期,梅河公司组织部分员工到A、B、C三地旅游,公司购买前往各地的车票种类、数量绘制成条形统计图,如图.根据统计图回答下列问题:1(1)前往A地的车票有_____张,前往C地的车票占全部车票的________%; (2)若公司决定采用随机抽取的方式把车票分配给100名员工,在看不到车票的条件下,每人抽取一张(所有车票的形状、大小、质地完全相同且充分洗匀),那么员工小王抽到去B地车票的概率为______. 三、解答题7.(2015春•高新区期末)为了实现区域教育均衡发展,我区计划对A,B两类学校分批进行改进,根据预算,改造一所A类学校和两所B类学校共需资金230万元,改造两所A类学校和一所B类学校共需资金205万元.(1)改造一所A类学校和一所B类学校所需的资金分别是多少万元?(2)我区计划今年对A、B两类学校共6所进行改造,改造资金由国家财政和地方财政共同承担.若今年国家财政拨付的改造资金不超过380万元,地方财政投入的改造资金不少于70万元,其中地方财政投入到A、B两类学校的改造资金分别为每所10万元和15万元,请你通过计算求出有几种改造方案?哪种改造方案所需资金最少,最少资金为多少?8.某商场将进价40元一个的某种商品按50元一个售出时,每月能卖出500个.商场想了两个方案来增加利润:方案一:提高价格,但这种商品每个售价涨价1元,销售量就减少10个;方案二:售价不变,但发资料做广告.已知这种商品每月的广告费用m(千元)与销售量倍数p关系为p=mm24.02;试通过计算,请你判断商场为赚得更大的利润应选择哪种方案?请说明你判断的理由!9.为执行中央节能减排,美化环境,建设美丽新农村的国策,我市某村计划建造A、B两种型号的沼气池共20个,以解决该村所有农户的燃料问题.两种型号沼气池的占地面积、使用农户数及造价见下表:型号占地面积(单位:m2/个 )使用农户数(单位:户/个)造价(单位: 万元/个)A15182B203032已知可供建造沼气池的占地面积不超过365m2,该村农户共有492户.(1)满足条件的方案共有几种?写出解答过程;(2)通过计算判断,哪种建造方案最省钱.10.阅读下列材料:小明遇到一个问题:5个同样大小的正方形纸片排列形式如图1所示,将它们分割后拼接成一个新的正方形.他的做法是:按图2所示的方法分割后,将三角形纸片①绕AB的中点O旋转至三角形纸片②处,依此方法继续操作,即可拼接成一个新的正方形DEFG.请你参考小明的做法解决下列问题:

    上传时间:2023-04-30 页数:6

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  • 高清彩绘英文原版绘本I Wear my Tutu Everywhere.pdf

    By Wendy Cheyette LewisonIllustrated by Mary MorganFor Beth with love—W.C.L.For my father—M.M.I Wear My Tutu Everywhere!Electronic book published by ipicturebooks.com24 W. 25th St.New York, NY 10010For more ebooks, visit us at:http://www.ipicturebooks.comAll rights reserved.Text copyright © 1996 by Wendy Cheyette LewisonIllustrations copyright © 1996 by Mary MorganOriginally published by Grosset & Dunlap, Inc. in 1996No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without permission in writing from the publisher.e-ISBN 1-59019-639-2Library of Congress Cataloging-in-Publication DataLewison, Wendy Cheyette.I wear my tutu everywhere! / by Wendy Cheyette Lewison ; illustrated by Mary Morgan.p.cm. — (All aboard books)Summary: When Tilly wants to wear her tutu all the time and rips it on the playground, Mama surprisesher with the best place to wear it—dancing class.[1. Ballet dancing—Fiction.]I. Morgan, Mary, 1957- ill.II. Title.III. Series.PZ7.L5884Iaad1995[E]—dc2094-36629CIPACBy Wendy Cheyette LewisonIllustrated by Mary MorganOnce there was a little girl named Tilly who lovedto dance.She danced while Mama was braiding her hair. Please hold still, said Mama.She danced while she was brushing her teeth.And she danced in her dreams.So, when Tillys birthday came around, Mamaand Papa knew just what to give her for a present.A tutu, a beautiful pink tutu to dance in—just like a real ballerina!It fit perfectly.Tilly loved her new tutu. She loved it so muchthat she wore it wherever she went!I WearMWyTWututE WvawhuW!MEu WMWyTWututuMrMTMMsang Tilly, as she danced down the aisles attthe supermarket.In the frozen-food section it was pretty chilly.tBut Tilly didnt mind.By Wend Ch tLtLwy iesot Iludwty rdlu Ch tLtLtdnduhradudbMwrW.v (WwWwTaWhraMWhrWhrWWT vMauEvuTWTaWv C'hWhh Wwv (W.v (D

    上传时间:2024-02-02 页数:34

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  • 72本贝贝熊早教英文绘本之Berenstain_Bears_and_Baby_Makes_Five.pdf

    a a a a a a a a a a a a a a a a The Berenstain Bears and Baby Makes Five Electronic book published by ipicturebooks.com 24 W. 25th St. New York, NY10010 aFor more ebooks, visit us at: http://www.ipicturebooks.com aAll rights reserved. aCopyright (c) 2000 by Stan and Jan Berenstain Originally published by Random House in 2000 No part of this book may be reproduced or transmitted in any form aor by any means, electronic or mechanical, including photocopying,arecording, or by any information storage and retrieval system,awithout permission in writing from the publisher. e-ISBN: 1-59019-223-0 Ebook conversion by wTree.comBig news in Bear Country!The Bear family, who lives in the bigtree house down a sunny dirt road, has anew member; a baby girl named Honey.What fun! What excitement! What a nuisance! Sometimes it seemed that it wascrying,feeding,burping,aspitting up,and diapering around the clock. And when it wasn't those things, it wascuddling,dandling,fussingand kitchy-cooing.a And when it wasnt those things, it wasshopping, shopping and more shopping for things that were needed for the new baby.At least that's how it seemed to Sister Bear.a Brother Bear understood that babies need a lot ofattention. He'd been through it before when Sister wasborn. But having a new baby in the house was a newexperience for Sister, and she wasn't enjoying it very much.It didn't help that when Papa came home fromwork every day, the first thing he did waspick up the new baby, make goo-goo eyes,and say, "How'smy darlin'little dumpling?"aDumpling. Good name for her,thought Sister-a fat little doughballthat was hard to swallow. It didn't help at all when Aunt Min andUncle Louie visited and made a big fuss overthe new baby."Well, Sis," said Uncle Louie, "I guessyou're not the big star around here anymor

    上传时间:2024-02-02 页数:33

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  • 初中8年级(下册)勾股定理的逆定理 (提高)巩固练习.doc

    【巩固练习】一.选择题1.(2015•诏安县校级模拟)下列几组数中不能作为直角三角形三边长度的是()A.a=7,b=24,c=25 B.a=1.5,b=2,c=2.5C. D.a=15,b=8,c=172. 下列三角形中,不是直角三角形的是( )A.三个内角之比为5∶6∶1 B. 一边上的中线等于这一边的一半C.三边之长为20、21、29D. 三边之比为1.5 : 2 : 33. 下列命题中,不正确的是()A. 三个角的度数之比为1:3:4的三角形是直角三角形;B. 三边之比为1: 3:2的三角形是直角三角形;C. 三个角的度数之比为1:2:2的三角形是直角三角形;D. 三边之比为2:2:2的三角形是直角三角形.4. 如图,在单位正方形组成的网格图中标有AB、CD、EF、GH四条线段,其中能构成一个直角三角形三边的线段是()A.CD、EF、GHB.AB、EF、GH C.AB、CF、EF D.GH、AB、CD5.五根小木棒,其长度分别为7,15,20,24,25,现将他们摆成两个直角三角形,其中正确的是()6. cba,,为直角三角形的三边,且c为斜边,h为斜边上的高,下列说法:①222,,cba能组成一个三角形②cba,,能组成三角形③hbahc,,能组成直角三角形 ④hba1,1,1能组成直角三角形1其中正确结论的个数是()A.1B.2 C.3 D.4二.填空题7.若△ABC中,2babac,则∠B=____________.8.如图,正方形网格中,每个小正方形的边长为1,则网格上的△ABC是______三角形.9.若一个三角形的三边长分别为1、a、8(其中a为正整数),则以2a、a、2a为边的三角形的面积为______.10.△ABC的两边ab,分别为5,12,另一边c为奇数,且abc是3的倍数,则c应为______,此三角形为______.11.(2015春•滑县期末)如果三角形的三边a,b,c满足a2+b2+c2+50=6a+8b+10c,则三角形为   三角形.12. 如果线段abc,,能组成一个直角三角形,那么2,2,2cba________组成直角三角形.(填能或不能).三.解答题13.已知abc、、是△ABC的三边,且222244acbcab,试判断三角形的形状.14.(2015春•江津区校级月考)如图所示,在正方形ABCD中,M为AB的中点,N为AD上的一点,且AN=AD,试猜测△CMN是什么三角形,请证明你的结论.(提示:正方形的四条边都相等,四个角都是直角)15.在等边△ABC内有一点P,已知PA=3,PB=4,PC=5.现将△APB绕A点逆时针旋转60°,使P点到达Q点,连PQ,猜想△PQC的形状,并论证你的猜想.2【答案与解析】一.选择题1.【答案】C;【解析】解:A、满足勾股定理:72+242=252,故A选项不符合题意;B、满足勾股定理:1.52+22=2.52,故B选项不符合题意;C、不满足勾股定理,不是勾股数,故C选项符合题意;D、满足勾股定理:152+82=172,故D选项不符合题意.故选:C.2.【答案】D;【解析】D选项不满足勾股定理的逆定理.3.【答案】C;【解析】度数之比为1:2:2,则三角形内角分别为36°:72°:72°.4.【答案】B;【解析】22222228,20,5,13,ABCDEFGHABEFGH,所以这三条线段能构成直角三角形.5.【答案】C;【解析】22222272425152025,.6.【答案】C;【解析】因为222abc,两边之和等于第三边,故222,,cba不能组成一个三角形,①错误;因为abc,所以cba,,能组成三角形,②正确;因为abch,所以2222222aabbhcchh,即222abhch,③正确;因为2222222222222111abccabababchh,所以④正确.二.填空题37.【答案】90°;【解析】由题意222bac,所以∠B=90°.8.【答案】直角;【解析】2AB=13,2BC=52,2AC=65,所以222ABBCAC.9.【答案】24;【解析】∵7<a<9,∴a=8. 10.【答案】13;直角三角形;【解析】7<c<17.11.【答案】直角;【解析】解:∵a2+b2+c2+50=6a+8b+10c∴a2+b2+c2﹣6a﹣8b

    上传时间:2023-04-30 页数:5

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  • 精品解析:2021年福建省普通高中学业水平选择性考试思想政治试题(解析版).docx

    2021年福建省普通高中学业水平选择性考试思想政治试题一、选择题1. 某国对食糖实施进口关税配额管理,即关税配额量内进口的食糖征收低关税,配额量外进口的食糖征收高关税。2021年,该国某地自由贸易港取消食糖进口关税配额管理,短期内会对当地以进口食糖为原料的甜品市场产生影响。若不考虑其他因素,能正确反映这种影响的图示是(图中D表示需求,S表示供给)( )A. B. C. D. 【答案】C【解析】【详解】该国某地自由贸易港取消食糖进口关税配额管理,有利于进口,增加供给。A:选项A表示需求增加,供给不变,与题意不符,故A不选。B:选项B表示供给减少,需求增加,与题意不符,故B不选。C:选项C表示供给增加,需求不变,与题意符合,故C入选。D:选项D表示供给和需求都增加,与题意不符,故D不选。故本题选C。2. 上世纪90年代以来,沙县小吃凭借风味独特和经济实惠逐步走向全国。几年前,为解决经营中出现的产品标准不一、产能供应不足等问题,沙县小吃从个体分散经营逐步走上集团化、品牌化、产业化的发展道路,实行总公司、子公司、终端店运作模式,现已发展成为在全球拥有8.8万家连锁店、年营业额近500亿元的大产业。小吃变成大产业主要得益于( )①调整产业定位,主动适应市场变化②扩大经营规模,增加市场占有量③创新管理模式,完善公司组织机构④树立品牌形象,提升产品竞争力A. ①②B. ①④C. ②③D. ③④【答案】B【解析】【详解】①④:为解决经营中出现的产品标准不一、产能供应不足等问题,沙县小吃从个体分散经营逐步走上集团化、品牌化、产业化的发展道路,实行总公司、子公司、终端店运作模式,现已发展成为在全球拥有8.8万家连锁店、年营业额近500亿元的大产业。小吃变成大产业主要得益于调整产业定位,主动适应市场变化,树立品牌形象,提升产品竞争力,故①④入选。②:增加市场占有量是结果,不是原因,故②不选。③:创新管理模式,完善公司组织机构虽然是企业成功经营的重要原因,并不是沙县小吃由小吃变成大产业的主要原因,故③不选。故本题选B。3. 2021年3月,国家开发银行发行首单碳中和债,这是目前全球金融市场发行金额最大的专项绿色债券,用于助力实现碳达峰、碳中和目标,获得境内外投资机构踊跃认购。碳中和债的发行有利于( )①拓宽绿色产业的融资渠道②满足金融市场投资多元化需要③增加绿色项目的投资收益④加快全球金融机构低碳化发展A. ①②B. ①④C. ②③D. ③④【答案】A【解析】【详解】①②:国家开发银行发行首单碳中和债,这是目前全球金融市场发行金额最大的专项绿色债券,用于助力实现碳达峰、碳中和目标,获得境内外投资机构踊跃认购。碳中和债的发行有利于拓宽绿色产业的融资渠道,满足金融市场投资多元化需要,故①②入选。③:碳中和债的发行不一定增加绿色项目的投资收益,故③不选。④:碳中和债的发行与全球金融机构低碳化发展无关,故④不选,故本题选A。4. 2020年中央经济工作会议将强化反垄断和防止资本无序扩张确定为2021年的重点任务之一;2021年1月,中央政法工作会议提出加强反垄断和反不正当竞争执法司法。国家重视反垄断和反不正当竞争是为了( )①维护公平竞争秩序,保障市场主体合法权益②优化自由竞争环境,促进经营者跨越式发展③完善市场法制体系,保护消费者自主选择权④激发全社会创新活力,推动经济高质量发展A. ①②B. ①④C. ②③D. ③④【答案】B【解析】【详解】①④:国家重视反垄断和反不正当竞争是为了维护公平竞争秩序,保障市场主体合法权益,激发全社会创新活力,推动经济高质量发展,故①④入选。②:国家重视反垄断和反不正当竞争有利于优化公平的竞争环境,促进经营者发展,但不一定是跨越式发展,故②不选。③:国家重视反垄断和反不正当竞争与完善市场法制体系无关,其目的不是完善市场法制体系,故③不选。故本题选B。5. 厦门市某社区对台服务工作法入选民政部百个优秀社区工作法。该社区立足台商多、台胞多的特点,吸纳台胞加入社区发展理事会,鼓励台胞参与社区志愿服务,开展闽台文化交流活动,为台胞开设医保、子女就学等绿色通道,赢得台胞赞誉。对台服务工作法( )①探索社区共建共治共享的治理新机制②增进两岸同胞情谊,改善台商的营商环境③调动台胞的积极性,主动参与社区民主选举④创新社区管理的组织形式,维护台胞合法权益A. ①②B. ①③C. ②④D. ③④【答案】A【解析】【详解】①②:吸纳台胞加入社区发展理事会,鼓励台胞参与社区志愿服务,开展闽台文化交流活动,为台胞开设医保、子女就学等绿色通道,赢得台胞赞誉,这意味着对台服务工作法在探索社区共建共治共享的治理新机制,有利于增进两岸同胞情谊,改善台商的营商环境,①②符合题意。③:对台服务工作法属于优秀

    上传时间:2024-02-20 页数:18

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    上传时间:2024-01-28 页数:32

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