教材课后习题参考答案新版_学生版(1-4)章_介绍

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统编版小学语文四年级下册精品上课课件 8.第八单元 26 宝葫芦的秘密(节选)

统编版小学语文四年级下册精品上课课件  8.第八单元 26 宝葫芦的秘密(节选)

课后作业
1. 当王葆真的得到一个宝葫芦,会发 生什么故事呢?课外阅读《宝葫芦 的秘密》这本书。
2. 读一读王葆和奶奶的对话,运用语 言描写和动作描写,写一写你和一 位亲人相处的日常细节。
课后习题参考答案
呢?可这葫芦好像知道李四的心思似的,说:“哈哈, 你觉得很奇怪是吧?我可是一只宝葫芦,是这里最大的 宝贝!”李四不信,说:“这里有这么多宝贝,每一样 都价值连城,你凭什么说你是最大的宝贝?”宝葫芦说: “谁要是拥有了我,就等于拥有了一切。你想要什么, 我都能够变出来。你看,你想要氧气罐,我就给你变出 了一个氧气罐。你说我是不是最大的宝贝?”李四一听, 心想:“这葫芦可真是最大的宝贝。”宝葫芦说:“我
课后习题参考答案
东西的时候,就想有一个宝葫芦来帮助自己。终于 有一天,他在河边钓鱼时遇到了一个宝葫芦。这个 宝葫芦什么都能变出来,帮王葆钓到鱼、做成高级 的模型、完成家庭作业……真是要什么有什么,想 什么就办成什么。文中的宝葫芦可以说是一个让人 又爱又恨的捣蛋鬼,每次帮助王葆实现心愿,结果 却总是害得王葆狼狈不堪,心惊胆战。拥有了这个
不过我得提从出王我葆的总条是件要:求“听那故,事非,得我讲看 故事。”于出是王奶葆奶是又一讲个了顽一皮个又—淘—气又、是充宝满葫好 芦的故事。奇心,并且特别爱听故事的孩子。
品读课文
我听了这些故事,常常就联系到自己: “我要是有了一个宝葫芦,我该怎么办?我 该要些什么?”
一直到我长大了,有时候还想起它来。 我有几次对着一道算术题发愣,不知道要怎 么列式子,就由“8”字想到了宝葫芦—— 假如我有这么一个,那可就省心了。
丫拽福舔 葵瘦棒罢
右半部分第一笔是横, 最后两点要并排。
第六、七、八笔分 别为撇、撇、捺。

化学方程式的书写及应用(学生版) 2024--2025学年沪教版(上海)九年级化学上册第4章同步练习

化学方程式的书写及应用(学生版) 2024--2025学年沪教版(上海)九年级化学上册第4章同步练习

第4章认识化学反应第2节化学方程式的书写及应用(1)板块导航01/学习目标明确内容要求,落实学习任务02/思维导图构建知识体系,加强学习记忆03/知识导学梳理教材内容,掌握基础知识04/效果检测课堂自我检测,发现知识盲点05/问题探究探究重点难点,突破学习任务06/分层训练课后训练巩固,提升能力素养一、化学方程式概念用来表示化学反应的式子,叫做化学方程式。

二、化学方程式的书写1.书写原则:①以客观事实为依据。

②符合2.书写步骤3.配平方法(1)最小公倍数法(2)奇数配偶法+一、选择题1.(23-24八年级下·山东泰安·期中)下列关于书写化学方程式的说法中,不正确的是A .以客观事实为基础B .若反应条件为加热,可用“D ”表示C .遵循质量守恒定律D .生成物中的气体一定要标注“↑”2.下列化学方程式书写正确的是( )A .P +O 2P 2O 5B .H 2O 2H 2↑+O 2↑C .2H 2O2H 2↑+O 2↑D .Mg +O 2MgO 23.(23-24八年级下·山东济宁·期中)化学方程式为38222aC H bO cCO dH O +=+,配平后化学计量数之和为A .13B .12C .11D .10二、非选择题4.化学方程式的书写原则(1)必须以 为依据,不能凭空臆造事实上不存在的物质和化学反应。

(2)必须遵循 ,要使反应前后各 的种类与数目保持不变。

5.配平下列方程式:(1) Al+ H 2SO 4═Al 2(SO 4)3+ H 2↑(2) Mg+O (3)Fe+ HCl ═FeCl 2+H 2↑(4)CH 4+ O 2+ H 2O6.(24-25九年级上·天津红桥·期中)完成下列化学方程式。

(1)通电分解水。

(2)镁带在氮气中点燃生成二氮化三镁 。

(3)铁在氧气中燃烧。

(4)白色固体粉末氯酸钾(3KClO )在二氧化锰作催化剂并加热的条件下能较快地生成氯化钾和氧气:。

第4章 基础实验3 燃烧条件的探究(学生版) 2025学年沪教版(上海)九年级化学上册第4章同步练习

第4章 基础实验3 燃烧条件的探究(学生版) 2025学年沪教版(上海)九年级化学上册第4章同步练习

第4章认识化学反应基础实验3 燃烧条件的探究板块导航01/学习目标明确内容要求,落实学习任务02/思维导图构建知识体系,加强学习记忆03/知识导学梳理教材内容,掌握基础知识04/效果检测课堂自我检测,发现知识盲点05/问题探究探究重点难点,突破学习任务06/分层训练课后训练巩固,提升能力素养探究燃烧条件【问题讨论】1.结合你在上述实验中所观察到的现象和得到的结论,试着总结燃烧所需要的条件。

答:物质要具有可燃性,跟氧气充分接触,温度要达到着火点2.列举日常生活中常用的灭火方法,并试着说明其灭火原理。

A.灭火的方法:做饭时关闭液化气瓶阀门,使火焰熄灭依据的原理:将可燃物撤离燃烧区B.灭火的方法: 盖上灯帽,熄灭燃烧的酒精灯依据的原理:将可燃物与空气(或氧气)隔绝C.灭火的方法:柴草发生火灾时,用水浇灭依据的原理:将可燃物的温度降低到着火点以下+1.(2024·重庆·中考真题)利用红磷和白磷进行燃烧条件探究,实验装置如下图所示。

下列说法不正确的是A .a 处物质为白磷B .红磷的着火点高于80℃C .a 、b 两点对比说明燃烧需要氧气D .c 处通入氧气白磷能燃烧2.(2024·辽宁丹东·模拟预测)通过比较、推理等方法可总结归纳出燃烧与灭火的规律并加以应用,请回答:(1)如图所示的“火三角”可表示燃烧条件,其中缺少的一个条件是。

(2)图中熄灭酒精灯的原理是。

(3)消防队员在火灾现场扑灭明火后,还要继续喷水防止复燃,利用的原理是。

(4)物质燃烧的剧烈程序与氧气的接触面积有关,如(举一日常生活实例)。

3.如图是“燃烧的条件”实验活动的部分操作示意图:(1)实验一:加热片刻观察到A中棉球上的酒精燃烧产生蓝色火焰,B中棉球上的水不燃烧,由此得出燃烧的条件之一是:燃烧需要。

(2)实验二:观察到先燃烧的物质是乒乓球碎片;若将铜片上的物质换成等量的木屑和煤粉进行实验,观察到木屑先燃烧,说明木屑的着火点比煤粉。

部编版语文一年级上册教科书(教材)课后习题参考答案

部编版语文一年级上册教科书(教材)课后习题参考答案
背诵指导:借助课本上的插图,边看画面, 边按顺序背诵。
你喜欢哪个季节?仿照课文说一说。
提示:课文前三节都运用了叠词,说 出了事物的特征;课文第一、二节的句式 是“什么对什么说”,而第三、四节的句 式是“什么怎么样地说”。这些我们都可 以学着用一用。
示例:我喜欢夏天和秋天。
▷ 西瓜圆圆,他鼓着肚皮说:“我是夏天。” ▷ 柿子红红,他仰着小脸儿,对南飞的大雁
背诵指导:课文先写我们的国旗是五星红 旗,然后描述升国旗时的庄严情景,接着写我 们在升国旗时注视、立正、敬礼。结合自己参 加升国旗仪式的感受,熟读成诵。
课后习题 课文
5 影子
朗读课文,读准字音。
朗读指导:朗读时,先读准字音,“常、 着、是”是翘舌音,“子、在、左”是平舌 音,“子、着、的、友”读轻声。要重读表 示影子出现的位置的词和表现影子形状的词, 语速稍快。
读一读,说一说,看谁说得多。
金色的
雪白的
快活的
示例: 金色的(太阳/麦穗/沙漠) 雪白的(羽毛/棉花/梨花) 快活的(孩子/小鸟/星星)
课后习题 课文 12 雪地里的小画家
朗读课文。背诵课文。
朗读指导:第一句读得大声、响亮,读 出欢呼的语气。“啦”字用在句末,加强了 感叹语气,读出兴奋激动的感觉。想象下大 雪的时候,“小画家们”和我们一样兴奋地 跑向雪地,边跑边欢呼,朗读时要把自己的
写字 写作业
课后习题 课文 8 雨点儿
分角色朗读课文。
朗读指导:读小雨点儿的话时声音可以 天真一些,读大雨点儿的话时声音可以稳重 一些。问句要读出疑问语气,语速稍慢;回 答时要表现小雨点儿和大雨点儿都急着去自 己想去的地方的情态,语速稍快。
读下面的句子,注意读好停顿。
不久,有花有草的地方,花 更·红·了,

小学语文教材课后习题参考答案

小学语文教材课后习题参考答案

教材课后习题参考答案1绿叶的梦2.因为“我们"在采集绿叶时能爬大树钻丛林,嬉笑,打闹,给"我们"带来了许多欢乐。

"我们”还能用元宝树叶串成项链在旷野的课堂里,绿叶和野花谱写成"我们"生活的乐章。

秋天一到“我们"背着筐,扛着筢,奔跑着,欢呼着搂树叶,堆成垛,躺在上面打滚,翻跟头挑选出那些漂亮的叶子装饰教室老师让"我们”观察并讲述这些叶子丰富"我们”的生活知识。

★说说自己童年有趣的事,跟同学一起分享快乐。

回忆自己的童年,从有趣入手。

说话要有条理口齿要清楚。

说起自己的童年,我就觉得好笑又好玩,它给我留下了永久的记忆。

那是去年暑假的一天妈妈让我午休,我偏偏不听,和亚强一起到邻居家的枣树上去偷枣,枣摘得不多,却摸着了马蜂窝重得我俩脸都像馒头一样。

从那以后,我再也不敢东跑西颠了。

2芦叶船2."我们"在竖河里放芦叶船时产生过这样的联想有的找不到了,联想到它已经远航了;有的靠在岸边,联想到它进港了。

联想到的原因是他们见过真正的船是怎样航行的又是怎样远航和进港的,这与他们的生活经历是分不开的。

"我们"联想芦叶船开到长江里开到东海里,这是对芦叶船充满了希望,也是"我们”对生活的希望。

3.在第一组里,第二句"那里的河道特别多,横的、竖的,像蜘蛛网一样"较第一句多加了“像蜘蛛网一样",就把那里的河道不仅多而且纵横交错的情景形象地展现在人们眼前了。

在第二组里,第二句"只见这只芦叶船顺着风顶着浪,越开越远,一会儿就看不见了”。

这里的"顺着风,顶着浪",不仅把芦叶船是在怎样的情况下开走的交代清楚了,还能使人想到"我”的表哥确实"更会"做芦叶船,进而能理解"我"那"又惊又喜”的心情。

新版高等数学教材答案

新版高等数学教材答案

新版高等数学教材答案在新版高等数学教材中,有很多难题和复杂的数学理论需要学生深入理解和掌握。

为了帮助学生更好地学习和掌握高等数学知识,答案是必不可少的工具之一。

下面是我整理的一些高等数学教材答案,希望对广大学生有所帮助。

第一章:函数与极限1. 函数极限的概念和性质2. 常见函数的极限计算3. 极限的四则运算法则4. 极限存在准则第二章:导数与微分1. 导数的定义和性质2. 常见函数的导数计算3. 高阶导数与隐函数导数计算4. 微分的概念及其应用第三章:中值定理与导数应用1. 罗尔中值定理和拉格朗日中值定理2. 柯西中值定理和达布中值定理3. 费马引理和罗尔定理的推广4. 高阶导数的应用第四章:不定积分与定积分1. 基本积分表与不定积分2. 常用换元法和分部积分法3. 定积分的概念和性质4. 牛顿—莱布尼茨公式和换元法在定积分中的应用第五章:曲线与曲面积分1. 曲线积分的概念和性质2. Green公式和高斯公式3. 曲面积分的概念和性质4. 斯托克斯公式和格林公式的推广第六章:无穷级数1. 数列极限的概念和性质2. 数列极限判定法3. 无穷级数的概念和性质4. 常见无穷级数的求和计算第七章:多元函数微分学1. 多元函数的极限和连续性2. 偏导数和全微分3. 方向导数和梯度4. 隐函数存在定理第八章:多元函数积分学1. 二重积分的概念和性质2. 二重积分的计算方法3. 三重积分的概念和性质4. 三重积分的计算方法和应用第九章:常微分方程1. 常微分方程的基本概念和分类2. 一阶常微分方程的解法3. n阶常微分方程的解法4. 常微分方程的应用这些是新版高等数学教材中的一些题目的答案,每章的内容都有所涉及。

希望对学生们在学习高等数学过程中有所帮助,更好地理解和掌握数学知识。

当然,答案只是辅助学习工具,学生还需进行适当的思考和练习,从而更好地提升自己的数学能力。

通过对高等数学教材答案的整理和提供,希望能够为广大学生提供一个辅助学习的工具,使学习高等数学变得更加轻松和高效。

教材各章习题参考答案.doc

教材各章习题参考答案.doc

教材各章习题参考答案一、判断题1、×2、×3、×4、√5、×6、√7、×8、×9、√10、√二、选择题三、双选题四、填空题1、42、机器机语言3、字节4、声音、图像5、0.625 、0.A6、源程序7、内存8、外内存器9、1A.D 10、控制器、运算器、11冯.诺依曼12、61 13、阴极射线管(显示器)14、速度15、ROM 16、信息17、二进制18、7 128 1 0 19、132.C20、编译21、运算器22、4FFFH 23、11100.001 34.1 1C.224、7AH FFEFH FFEEH (说明:此题,设计算机字长为16位)25、内存正在运行的程序和数据RAM ROM 26、16K27、单片机28、源程序目标程序第二章Windows 2000操作系统习题答案一、选择题1、D2、A3、B4、A5、B6、D7、C8、D9、B 10、C11、B 12、A 13、A 14、D 15、C 16、D 17、B 18、BD 19、AD 20、AD 21、AC 22、CD二、填空题1、实时2、对话框3、存档4、活动窗口5、EXIT6、.bmp7、F18、刷新9、网上邻居10、回收站11、添加/删除硬件12、不封闭13、控制面板14、软(A)15、自动排列16、关闭窗口三、判断题1、√2、×3、×4、√5、√6、√7、√8、×9、√10、×11、×12、×第三章文稿处理软件Word 2000习题答案一、判断题1、×2、×3、√4、×5、×6、×7、√8、×9、×10、√11、×12、√13、×二、选择题1、A2、A3、D4、A5、D6、C7、D8、B9、D 10、A11、B 12、A 13、C 14、B 15、B 16、A 17、B 18、C 19、A 20、C21、C 22、A,B,E 23、B,C 24、C,D 25、A,C,E,F,G 26、B,C27、A,D 28、B,E,F三、填空题1、显示/隐藏编辑标记2、23、文档1.doc4、标尺5、自动图文集6、分隔符7、宋体8、文本层9、页边距+缩进距10、表格自动套用格式11、大纲视图12、最小化13、符号14、保存第四章电子表格软件EXCEL 2000习题答案一、判断题1、×2、√3、×4、×5、×6、√7、√8、×9、×10、√11、×12、×13、×14、×15、×二、选择题1、A2、D3、C4、C5、B6、A7、A8、D9、A 10、D11、D 12、B 13、A 14、A 15、A 16、A 17、A 18、C 19、A 20、ABC 21、CD 22、AC三、填空题1、.XLS2、双击3、折分窗口4、3,2555、双击6、表示数据清单中有25条记录,当前在第10条7、复制工作表8、满足条件的数据9、=MAX(A2:E20)10、填充柄11、36 12、自动筛选13、$ 14、’15、条件格式16、公式17、A3:C618、排序19、保护工作表20、不合格第五章演示文稿制作软件PowerPoint 2000习题答案一、选择题1、C2、C3、C4、B5、B6、D7、C8、A9、C 10、C11、C 12、B 13、A 14、B 15、A 16、BC 17、BC 18、CD 19、AB 20、AC 21、AD 22、AB二、填空题1、内容提示向导,模板,空演示文稿2、不能3、占位符4、PowerPoint放映,.PPS5、幻灯片版面设置,应用设计模板6、演讲者放映,观众自行浏览,在展台浏览7、幻灯片浏览,大纲8、背景音乐9、常用,新建10、剪辑库11、隐藏幻灯片12、对象13、图片14、层次,顺序15、放映设置16、应用设计模板17、效果,换页方式,声音18、动画效果19、时间,效果20、录制旁白21、彩色投影机,Web演示文稿三、判断题1、√2、×3、×4、√5、×6、√7、×8、×9、√10、×11、√12、×13、×14、√15、√16、×17、√第六章计算机网络基础习题答案一、选择题1、D2、A3、A4、A5、D6、A7、D8、B9、BC 10、A 11、B 12、B 13、B 14、B 15、B 16、B17、D 18、D 19、A 20、D二、判断题1、×2、√3、√4、×5、√6、√7、√8、√9、×10、×三、填空题1、计算机技术、通信技术2、网卡、传输介质3、集线器4、局域网、广域网5、网络层6、层、协议7、WWW、FTP、E-MAIL 8、主页9、帐号10、ARPANet 11、ISP邮件服务器12、HTTP13、另存为14、同轴电缆、光纤15、工具、Internet选项第七章网页制作与网站建设基础习题答案一、选择题1、A2、B3、A4、D5、B6、B7、C8、C9、A 10、A 11、B 12、C二、判断题1、×2、√3、×4、√5、×6、√7、×8、×9、√10、×三、填空题1、HTML、主页2、打开…、打开站点3、WEB站点、本地计算机硬盘中的文件4、须保存网页5、JPEG、GIF第八章信息安全与职业道德习题答案一、略……二.单选题1、D2、A3、B4、D5、D6、B7、A8、A9、B 10、C三、填空题1、信息保密性、完整性、可用性、可控性的破坏2、COM3、网络安全、病毒防护安全4、信息处理环节的脆弱性、计算机信息系统自身的脆弱性5、通信双方身份6、1991。

全新版大学英语(第二版)综合教程4学生用书_课后习题答案(后附testyourself重要词翻译)

全新版大学英语(第二版)综合教程4学生用书_课后习题答案(后附testyourself重要词翻译)

全新版大学英语(第二版)综合教程4学生用书_课后习题答案(后附testyourself重要词翻译)UNIT1VocabularyI.1. alliance;2. at the cost of;3. stroke;4. limp;5. minus;6. regions;7. declarations;8. siege;9. raw; 10. bide his time; 11. have taken their toll;12. in the case of1. is faced with;2. get bogged down;3. is pressing on/pressed on;4. drag on5. get by;6. dine out;7. have cut back; 8. get through;1.lead to the conquest of cancer in the near feature2.has been brought to a halt by the delayed arrival of raw materials due to the dock workers’strike3.will/should never get in the way of her career4.caught the foreign minister off guard5.of the electronic calculator has rendered the slide rule outof date/obsolete1. Being faced with; the occupation of; regions;2. crucial to; efficient; to reckon with; weaken; be brought toa halt;3. a heroic; the decisive; turned the tide; siege; by launchingII.1.During the First World War, battles occurred here and there over vast areas. Some of the mostdramatic fighting took place in the gloomy trenches of France and Belgium.2.Elizabeth made careful preparations for the interview and her efforts/homework paid off.3.I spent hours trying to talk him into accepting the settlement, but he turned a deaf ear to all mywords.4.Pneumonia had severely weakened her body, and I wondered how her fragile body couldwithstand the harsh weather.III.1. fall ill;2. lay dead at home for two weeks;3. dropped dead from a heart attack;4. fell asleep;5. marrying young;6. to sit still for longer than a few minutes;Comprehensive ExercisesI.1. invasion;2. stand in the way;3. Conquest;4. catching… off his guard;5. launching;6. declaration;7. campaign;8. drag on;9. reckon with; 10. bringing… to a halt1. allow;2. reckoned;3. highly;4. forecasts;5. rapidly;6. instant;7. delivery;8. advantage;9. observing; 10. powerful;II.1.1)Mr. Doherty and his family are currently engaged in getting the autumn harvest in on thefarm.2)We must not underestimate the enemy. They are equipped with the most sophisticatedweapons.3)Having been out of a job/Not having had a job for 3 months, Phil is getting increasinglydesperate.4)Sam, as the project manager, is decisive, efficient, and accurate in his judgment.5)Since the chemical plant was identified as the source of pollution, the village neighborhoodcommittee decided to close it down at the cost of 100 jobs.2.The offensive had already lasted three days, but we had not gained much ground. The division commande r instruc ted our battalion to get around to the rear of the enemy at night and launch a surprise attack. T o do so, however, we had to cross marshland and many of us were afraid we might get bogged down in the mud. Our battalion commander decided to take a gamble. Lucky enough, thanks to the severely cold weather which made the marchland freeze over, we arrived at our destination目的地before dawn and began attacking the enemy from the rear. This t urned the tide of the battle. The enemy, caught off guard, soon surrendered.UNIT2VocabularyI.1. expansion;2. automated;3. vapor;4. take control of;5. hazards;6. satellite;7. vibrated;8. magnetic;9. bunched; 10. in the air;11. got/was stuck in; 12. approximately;1. send out;2. stand up for;3. pass for;4. were closing in on;5. starting up;6. went through;7. fill out;8. fall into1.incorporates all the latest safety features2.two trees ten feet apart3.awarding lucrative contracts to his construction firm4.the prototype of a new model before they set up a factory to make the cars5.are correlated in all racial groups1. the application; remote; has turned into a reality; are poised to2. that vibrate; can detect; frequency3. lanes; are mounted in; alert a; hazard;II.1.kilogram;2. memorandum;3. gymnasium;4. liberation;5. doctor;6. veterinarian;7. preparatory;8. automobile;9. influenza;10. medical care; 11. electronic mail; 12. communications satellite;13. news broadcast; 14. sky hijack; 15. European dollar;16. breakfast and lunch; 17. television broadcast; 18. Oxford and CambridgeIII.1. swimming pool;2. drawing board;3. enriched Middle English;4. disturbing change;5. fully developed prototype;6. Canned foods7. working population; 8. puzzling differences;Comprehensive ExercisesI.1. computerized;2. start up;3. be poised to;4. alert;5. hazards;6. monotonous;7. take control of;8. steer;9. lane; 10. decrease;11. calculate; 12. eliminate; 13. getting stuck in; 14. mounted;15. detect; 16. vapor;1. generates;2. related;3. revolutionized;4. enabled;5. opportunities;6. overall;7. manufacturing;8. dependent;9. interact; 10. fatalities;II.1.6)There was an unusual quietness in the air, except for the sound of artillery in the distance.7)The expansion of urban areas in some African countries has been causing a significant fall inliving standards and an increase in social problems.8)The research shows that atmospheric carbon dioxide levels are closely correlated with globaltemperatures.9)The frequency of the bus service has been improved from 15 to 12 minute recently.10)The diver stood on the edge of the diving board, poised to jump at the signal from the coach.2.Automobiles have, since their invention, revolutionized transportation, changing forever the way people live, travel, and do business. On the other hand, they have brought hazards,危害especially highway fatalities. However, today the application of computer technology and electronic sensors in designing and manufacturing cars makes it possible to eli minate most of traffic accidents. For example, electronic sensors mounted 安装in your car can detect alcohol v apor 蒸汽in the air and refuse to start up the engine. They can also monitor road conditions by receiving radio signals s ent o ut from orbiting satellites and greatly reduce your chances of g etting stuck in traffic jams. UNIT3VocabularyI.1. rude;2. physically;3. structure;4. made a difference;5. blurted (out);6. chuckling;7. measurable;8. prospective;9. preparations; 10. sparkled; 11. took a crack at; 12. partner;1. go after;2. look back at/on;3. be put up;4. stood for;5. build in;6. follow up;7. be hooked up to;8. closed up6.grilled her about where she had been all night7.beyond Cinderella’s wildest dreams that she could one day dance in the King’s palace8.will be in readers’ hands soon9.do your homework before going on an interview10.was in the neighborhood of 150 dollars1. applicants; veteran; the prospective;2. from his standpoint; has made every endeavor to go after;3. as the saying goes; to have a crack at; barely;II.1. behave;2. keep (used to avoid repetition);3. clean;4. get along;5. perform/complete;6. perform/complete;7. study;8. be enough;9. be acceptable;III.1.There is so much to say and it is hard to know where to begin. Ok, I’ll talk about myself first.2.Thank you very much, John, for your beautiful Christmas card. By the way, I have somethinghere for you.3.The new computer language can be quite easily understood by anyone who can read the dailynewspaper. Now, why is this an advantage?4.I’m going to work out the outline and will let you know how it goes. By the way, I will seeyou in February, as I plan to attend your seminar in Shanghai.5.OK, you got the job. Now, how to maximize your profits with as little effort as possible?6.Chris is back from Australia. Incidentally, those pictures you sent me are wonderful.Comprehensive ExercisesI.1. prospective;2. As I see it;3. done your homework;4. beforehand;5. endeavor;6. structure;7. partners;8. Respond;9. take a crack; 10. from the standpoint; 11. make a difference;12. follow up1. encouraging;2. inquiry;3. relevant;4. samples;5. references;6. advice;7. preparing;8. seriously;9. probably; 10. exhibit;II.1.11)Despite the inadequate length of the airstrip in this emergency landing, the veteran pilotmanaged to stop the plane after taxiing for only o short while.12)Grilled by the reporters, the movie star eventually blurted (out) that she had undergone towplastic surgeries.13)We have the technology and our partner has the capital. Working together, we’ll have thefuture in our hands.14)If I had known beforehand that you would bring so many friends home, I would have madebetter preparations. You see, I have barely enough food and drinks for a snack.15)People gave generously upon learning that new school rooms with stronger structures were tobe built in the earthquake-stricken area.2.Well begun, half done, as the saying goes. It is extremelyimportant for a job applicant to do his homework while seeking employment. From my standpoint, whether or not one has done his homework clearly makes a difference in his chance of success.I have a friend who is earning somewhere in the neighborhood of 100,000 dollars a years in a large computer software company. He told me that from his own experience the decision makers who interview prospective employees like people who are well prepared. Those who make no endeavor to learn as much about his prospectiv e employer as possible don’t have much of a chance of success.UNIT5VocabularyI.1. In a way;2. in accordance with;3. vacancy;4. in good condition;5. transparent;6. rub;7. spicy;8. hitherto;9. with (a) bad grace; 10. instinct; 11. pawned; 12. current;1. turned up;2. will stick to;3. brought back;4. go about;5. driving at;6. put away;7. turning over;8. took aback;11.has a very weak constitution—she may not be able to survive the operation12.was taken aback by the insurance company’s rejection of my compensation claim13.was something of a surprise when we ran into each other in a place like that14.needs trimming/to be trimmed—it’s getting too long15.are often deceptive1. Oddly enough; went broke; wrinkled; he had gone all to pieces2. definite; is capable of; her vanity;3. too mild; sipping; stroke;II.1. Except for;2. except that;3. except;4. except when;5. except to;6. except what;7. except where;8. except that;III.1. flat;2. can;3. candy;4. lift;5. faculty;6. ground floor;7. petrol;8. mail;9. railroad; 10. trousers;11. tube, underground; 12. maize; 13. store; 14. favorable; 15. paralyse; 16. labour; 17. meter; 18. catalogue; 19. levelling ; 20. theatre ;21. defense ; 22. plow ; 23. programme ; 24. practise;25.characterize; 26. tire;Comprehensive ExercisesI.1. insane;2. current;3. candid;4. capable;5. was taken aback;6. in good condition;7. constitution;8. go all to pieces;9. gone broke; 10. vacancy; 11. mild; 12. deceptive;1. suspected;2. pleading;3. confirmed;4. stunned;5. lucrative;6. jewellery;7. wealthy;8. urge;9. spell; 10. arrested;II.1.16)I have an instinct that Henry will seek to join the expedition, because he is something of anadventurer.17)He is c apable of sticking to the task at hand, even if he is exposed to noises.18)The trademark was registered i n accordance with与一致the laws h itherto到目前为止inforce.19)Oddly enough, many people volunteered to help organize the meeting, but only a few turnedup.20)The teacher’s affectio nate words, along with his candid comments, changed the way Mikeperceived 感知the society and himself.2.For my own part, I find that appearances are all too often deceptive. For instance, you might be wrong if you judged by appearances only people like Edward Hyde Burton. In appearance, he seemed a man all of a piece. He was a tiny little fellow with white hair and mild blue eyes, gentle and candid. Nevertheless然而, he turned out to be very ,sane venture荒唐的冒险. What was still more surprising was that he was completely indifferent 冷漠to Lenny’s death. Without doubt, Burton was a man with a heart of stone.UNIT6VocabularyI.1. appliances;2. comparative;3. multiply;4. distribution;5. prosperity;6. decorate;7. famine;8. large quantities of/a large quantity of9. streamline; 10. fax; 11. pointed the way to; 12. bewildered;1. eat into;2. cling to;3. stand out/stood out;4. wears away;5. set about;6. switch off;7. will be turned loose;8. poured in;16.is forecast to be below average next year, which at the moment is 4 percent17.to enter the building and find the baby girl proved futile as rescuers were driven out by theheat and flames18.was urged to divert some of its attention from expanding production and get more involvedwith issues of market demand.19.can really eat heavily into your profits when you are selling suits at£900 and dresses at£2,00020.has toiled endlessly over the exercise machine for the last twenty years in order to keep herbody in shape1. reaction to; discontent; provoked;2. Convention; evading tax; the confines of;3. a burden; are always on the go/seem forever on the go; to cope;II.1. nervousness;2. tension;3. stress; stress;4. tension;1. honorary;2. Honorable;3. honorable;4. honorary;5. honorable;6. Honorary;III.1.Dealing with the extinction crisis is no simple matter. Is it sensible, we may ask, to spendlarge sums of money to save some pieces—be it an elephant or an orchid—in a nation in which a large proportion of the population is living below the poverty line?2.This new technology could be used anywhere large numbers of people need to be quicklyscreened—at airports, train stations, bus terminals or border crossings. However, expert suspect, there is also the risk that people will learn to fool the machine the same way they try to fool polygraph readings by controlling their breath or taking drugs to relax themselves.3.With a high percentage of marriages ending in divorce, often due to financial difficulties, youwould say that money is a big factor in making a good marriage. But, believe it or not, it isn’t money that ensures youa happy marriage; it is your philosophy of life that does.4.Not all the risks on the Internet are sexual, you know. Sitespromoting violence are just a clickaway, and may include instructions for making bombs and other destructive devices.Comprehensive ExercisesI.1. switch off ;2. obliged ;3. on the go ;4. cope;5. shortage;6. large quantity of;7. pouring in;8. by nature;9. fraction; 10. futile;1. advantage;2. wisely;3. faithfully;4. waking;5. includes;6. schedule;7. sticking;8. priorities;9. set; 10. respect;II.1.21)They are exploring the new frontiers of medical science in an attempt to find remedies forincurable diseases/cures for diseases that are beyond remedy治疗so far.22)Her unique teaching methods apart, Ms Wilson, my mathteacher, never tried to cramknowledge into my head.23)The regular weather forecast by the Central TV Station keeps us up with the changes ofweather wherever we go on a trip.24)The appalling 骇人的explosion started a big fire and caused the partial collapse坍塌of thebuilding.25)In the modern world, there are more ways than ever to w aste awa y time, and all kinds ofdistractions are eating into our precious time.2.Today we are under constant 持续的pressure to work longer hours, to produce more, and to possess more. Lots of people hold the wrong perception 感觉that happiness lies in working hard and earning well/good money.Many women today feel the same stress to work and get ahead and, at the same time, to nurture their offspring 下一代and shoulder the burden of domestic 家庭的responsibilities.Research shows that workaholism tends to distance us from our immediate families. It forces us to toil辛苦longer and longer hours, leaving a minute fraction 小部分of our time to be physically and emotionally available to our loved ones. Intimacy 亲密among family members i s doomed to die in the process.UNIT7VocabularyI.1. divined;2. nerves;3. solidarity;4. sacred; mourn5. coated6. perish;7. hijack;8. grief;9. farewell; 10. take revenge on;11. revolves; revolves; 12. denounced;1. drop off;2. applied for;3. went off;4. are gaining on;5. bring down;6. blotted out;7. think back on;8. picking at;21.brought down the American housing market in 200822.what will happen after his son steps into his shoes?23.not in the mood to go out24.long before the market began to show signs of weakness25.mourn the loss of the tranquil life we had in the countryside1. in the aftermath of; to blot out; the tragic;2. armed; at dusk; accomplices; explosive;3. in the space of; no illusion;II.1. a little of;2. a few; most of/many of;3. much;4. few;5. many;6. many of;7. much of;8. little;9. few of; 10. Some;III.1. As the boy grew older;2. she signs as beautifully as a nightingale/sings like a nightingale;3. they don’t see themselves as servants of the people;4. As she had left her key in the office;5. Just do as you are told;6. Areas once regarded as rural;7. as they do in China8. As he was brave and loyal as wellComprehensive ExercisesI.1. mood;2. tragic;3. in the…aftermath of;4. chaos;5. toppling;6. solidarity;7. take revenge on;8. thinking back on;9. mourning; 10. perished1. crashed;2. horrible;3. harsh;4. protect;5. remove;6. utterly;7. truly;8. justify;9. rewarded; 10. devastating;II.1.26)Some high-ranking officers of the armed forces started toa coup, toppling the governmentand throwing the country into chaos.27)The falling market shattered her illusion about getting rich quickly.28)Thinking back on the history of World War II, we can see that the formation of the Allies wasthe natural product of the development of political and military circumstances then.29)Paul felt stung when Jim called him a religious fanatic. But as he was in no mood for aquarrel/not in a quarreling mood, he simply pretended not to bear it.30)People say that time heals all wounds. But for those who have lost their loved ones in theevent, will time fill up the void in their hearts?2.Today, long after the earthquake shook/hit my home town, I can still recall, in crystal detail, what I saw as I ran out of my home with my parents. The building just across the street toppled right before our eyes, debris flew everywhere and a cloud of choking dust blotted out the sun. Horror-stricken people ran in all directions, crying and screaming.Now, many years after that tragic event, a new town has risenon the wreckage of the old one. In the town square, a memorial has been built to remember those killed in the disaster. It seems the wounds in people’s hearts have healed, but the memory will linger.UNIT8VocabularyI.1. heap;2. was smeared;3. warmed;4. dissolve;5. thrash;6. out of the way;7. hollows;8. tangled;9. get his hands on;10. opaque; 11. at the edge of; illumined; 12. hop;1. take apart;2. result from;3. run out of;4. feed on;5. come forward;6. woke up;7. focused on;26.was dying to see the movie based on it27.as a rule, the sheer distance mutes all sounds from the ground28.fuss too much over details29.slumped into an armchair, (feeling) completely exhausted30.was reputed to be the wittiest woman of her time1.swarmed to; the spectacular; paddled;2.in the heart of; out of the range of; trailing over; gliding in;3.Day after day; strip; the heaving; slap;II.1. worth;2. worthy;3. worthwhile;4. worth; worth;5. worthwhile;6. worthy;1. lone;2. alone;3. alone;4. lonely;5. lonely;6. alone; lonely;III.1.ice-cream;2. teas;3. wines;4. cloth;5. soap;6. beer;7. fuels; 8. soils; 9. sugars; 10. grass;Comprehensive ExercisesI.1. are dying to ;2. in the heart of ;3. tangled;4. paddles;5. loop;6. out of sight;7. in flocks;8. hopping;9. gliding; 10. opaque; 11. thrashing; 12. darting;13. swarms; 14. spectacular;1. running;2. fancy;3. incorporate;4. exploring;5. guide;6. adventure;7. rarely;8. diverse;9. survive; 10. lucky;II.1.31)Janet was just the kind of girl Mike knew he could trust, so he bared his heart to her on theirfirst date.32)At first the girls played on the fringe of the dark forest, now laughing, now screaming, butbefore long they were out of sight.33)The moment the football players disembarked from the plane, they saw a fleet of cars waitingfor their arrival.34)Carson condemned his opponent for using misleading information to smear his character.35)Alex gave the policeman a wallet stuffed with banknote. He said he had found it on the curbwhen he hopped off his school bus.2.Last Wednesday, my classmate Caroline and I visited Zhouzhuang, a well-known town looped all around by streams. When we arrived at the town, Caroline was so excited that she darted towards the first bridge she saw and began singing loudly there. Suddenly her voice hushed when she found that she had startled a flock of ducks not too far from us. Now as Caroline was dying for a boat ride, we decide to tour the town by boat. Now loud, now soft, Caroline talked to all the creatures in the stream and was fussing about everything while I looked at the boats gliding over the water in silence. Though we did not see anything spectacular, we enjoyed every minute in the town that lies out of the range of the heavy traffic and noise of the large city.Zhouzhuang is worth visiting and, time permitting, I’d like to go there again.Test yourself u1 to u4Globallization by the bookGlobalization is actually quite easy to define. It is simply an extension of economic freedoms beyond national boundaries. Many years ago I got a chance to enjoy the freedom that globalization brings when on my 20th birthday. I boarded a plane that brought me from Bombay 孟买to New York. And when I landed in New York it was still my 20th birthday and I celebrated it by eating Chinese food that came in funny little white containers I‘d never seen before。

英语Unit 1-4课后答案

英语Unit 1-4课后答案

Unit 1Language Points1.Translate the following sentences into English.1) 大家认为有必要对青年学生进行传统教育。

It is considered necessary to give the young students traditional education.2)我认为放弃这个面试的机会太遗憾了。

I consider it a great pity to give up the chance of the interview.2. Translate the following sentences into English.1) 我以前通常很晚才睡觉。

I used to go to bed very late.2) 那里以前有一座石桥。

There used to be a stone bridge there.3. Translate the following sentences into English.1)朋友之间往往只用“嗨”来打招呼。

Friends tend to say “Hi” as a greeting to each other.2) 书信有可能被电子邮件所替代。

Written letters tend to be replaced by E-mails.4. Translate the following sentences into English.1) 他已经非常不习惯没有网络的生活。

He hasn’t been quite used to the life without network.2)我不习惯在大庭广众之下被人表扬。

I am not used to being praised in public.rather than / rather…than 在更大程度上; 而不是; 比较起来倒不如……Translate the following sentences into English.1) 我愿意工作而不是闲着。

最新教材高中数学课后习题答案大全2019人A版

最新教材高中数学课后习题答案大全2019人A版
(3) ⫋;⫌
2.解析 D⫋C⫋B⫋A,
Venn 图如图所示.

1.1 集合的概念

A∪( B∩C) = {1,2,3,4,5,6,7,8} .
3.解析 “ 每位同学最多只能参加两项比赛”
表示为 A∩B∩C = ⌀.
(1) A∪B 表示参加 100 m 或参加 200 m 跑
的同学.
(2) A∩C 表示既参加 100 m 又参加 400 m
跑的同学.
综合运用
4. 解 析 因 为 A = { x | 3 ≤ x < 7 }, B =
8} ;A∪B = {3,5,6,8} ∪{ 4,5,7,8} = { 3,4,
5,6,7,8} .
2.解析 A = { x | x 2 -4x-5 = 0} = { x | ( x- 5) ·( x
+1) = 0} = {5,-1} ,B = { x | x 2 = 1} = { - 1,1} ,
3.解析 充 分 条 件: ( 1) ∠1 = ∠4, ( 2) ∠1 =
∠2,(3) ∠1+∠3 = 180°.
必要条件:( 1) ∠1 = ∠4,( 2) ∠1 = ∠2,( 3)
∠1+∠3 = 180°.
1.4.2 充要条件
练习
1.解析 ( 1) p 是 q 的充要条件. ( 2) p 是 q 的
(2) { 1,2,3,12,21,13,31,23,32,123,132,
213,231,312,321} .
(3) {4,5,6} .
(3) { 造纸术,印刷术,指南针,火药} .
4.解析 (1) { y | y = x 2 -4,x∈R}
= { y | y≥-4} .
(2) x≠0 时,函数 y =

人教版高一数学课后答案精品

人教版高一数学课后答案精品

人教版高一数学课后答案第一章集合与函数概念1.1集合1.1.1集合的含义与表示练习(第5页)1.(1)中国A ,美国A ,印度A ,英国A ;中国和印度是属于亚洲的国家,美国在北美洲,英国在欧洲.(2)1A 2{|}{0,1}A x x x .(3)3B 2{|60}{3,2}Bx xx.(4)8C ,9.1C9.1N .2.解:(1)因为方程290x的实数根为123,3x x ,所以由方程290x的所有实数根组成的集合为{3,3};(2)因为小于8的素数为2,3,5,7,所以由小于8的所有素数组成的集合为{2,3,5,7};(3)由326y x yx ,得14x y,即一次函数3y x与26yx的图象的交点为(1,4),所以一次函数3yx与26yx 的图象的交点组成的集合为{(1,4)};(4)由453x ,得2x ,所以不等式453x 的解集为{|2}x x.1.1.2集合间的基本关系练习(第7页)1.解:按子集元素个数来分类,不取任何元素,得;取一个元素,得{},{},{}a b c ;取两个元素,得{,},{,},{,}a b a c b c ;取三个元素,得{,,}a b c ,即集合{,,}a b c 的所有子集为,{},{},{},{,},{,},{,},{,,}a b c a b a c b c a b c .2.(1){,,}a a b c a 是集合{,,}a b c 中的一个元素;(2)20{|0}x x 2{|0}{0}x x ;(3)2{|10}xR x 方程210x无实数根,2{|10}x R x;(4){0,1}N(或{0,1}N ){0,1}是自然数集合N 的子集,也是真子集;(5){0}2{|}x xx (或2{0}{|}x xx )2{|}{0,1}x xx ;(6)2{2,1}{|320}x x x方程2320xx 两根为121,2x x .3.解:(1)因为{|8}{1,2,4,8}Bx x 是的约数,所以A B ;(2)当2kz 时,36k z ;当21kz 时,363k z ,即B 是A 的真子集,BA ;(3)因为4与10的最小公倍数是20,所以A B .1.1.3集合的基本运算练习(第11页)1.解:{3,5,6,8}{4,5,7,8}{5,8}A BI I ,{3,5,6,8}{4,5,7,8}{3,4,5,6,7,8}A BU U .2.解:方程2450xx 的两根为121,5x x ,方程210x的两根为121,1x x ,得{1,5},{1,1}A B ,即{1},{1,1,5}A BA BI U .3.解:{|}A Bx x I 是等腰直角三角形,{|}A Bx x U 是等腰三角形或直角三角形.4.解:显然{2,4,6}U Be ,{1,3,6,7}U Ae ,则(){2,4}U A B I e ,()(){6}U UA B I 痧.1.1集合习题1.1 (第11页) A组1.(1)237Q 237是有理数;(2)23N 239是个自然数;(3)Q 是个无理数,不是有理数;(4)2R 2是实数;(5)9Z93是个整数;(6)2(5)N2(5)5是个自然数.2.(1)5A ;(2)7A ;(3)10A .当2k 时,315k ;当3k 时,3110k ;3.解:(1)大于1且小于6的整数为2,3,4,5,即{2,3,4,5}为所求;(2)方程(1)(2)0x x 的两个实根为122,1x x ,即{2,1}为所求;(3)由不等式3213x ,得12x,且xZ ,即{0,1,2}为所求.4.解:(1)显然有20x,得244x,即4y,得二次函数24yx的函数值组成的集合为{|4}y y;(2)显然有0x,得反比例函数2yx的自变量的值组成的集合为{|0}x x;(3)由不等式342x x ,得45x ,即不等式342x x 的解集为4{|}5x x.5.(1)4B ;3A ;{2}B ;BA ;2333x x x,即{|3},{|2}A x x Bx x;(2)1A ;{1}A ;A ;{1,1}=A ;2{|10}{1,1}Ax x;(3){|}x x 是菱形{|}x x 是平行四边形;菱形一定是平行四边形,是特殊的平行四边形,但是平行四边形不一定是菱形;{|}x x 是等边三角形{|}x x 是等腰三角形.等边三角形一定是等腰三角形,但是等腰三角形不一定是等边三角形.6.解:3782x x ,即3x,得{|24},{|3}A x x B x x,则{|2}A Bx xU ,{|34}A B x xI .7.解:{|9}{1,2,3,4,5,6,7,8}A x x 是小于的正整数,则{1,2,3}A B I ,{3,4,5,6}A CI ,而{1,2,3,4,5,6}B CU ,{3}B CI ,则(){1,2,3,4,5,6}A B C I U ,(){1,2,3,4,5,6,7,8}A B C U I .8.解:用集合的语言说明这项规定:每个参加上述的同学最多只能参加两项,即为()A B C I I .(1){|}A B x x U 是参加一百米跑或参加二百米跑的同学;(2){|}A Cx x I 是既参加一百米跑又参加四百米跑的同学.9.解:同时满足菱形和矩形特征的是正方形,即{|}B Cx x I 是正方形,平行四边形按照邻边是否相等可以分为两类,而邻边相等的平行四边形就是菱形,即{|}A B x x 是邻边不相等的平行四边形e ,{|}S A x x 是梯形e .10.解:{|210}A Bx xU ,{|37}A Bx xI ,{|3,7}R Ax xx 或e ,{|2,10}R Bx xx或e ,得(){|2,10}R A B x x x U 或e ,(){|3,7}R A B x x x I 或e ,(){|23,710}R A B x xxI 或e ,(){|2,3710}R A B x xxxU 或或e .B 组1.4集合B 满足A BA U ,则B A ,即集合B 是集合A 的子集,得4个子集.2.解:集合21(,)|45x y D x y xy 表示两条直线21,45xyxy的交点的集合,即21(,)|{(1,1)}45x y Dx y xy,点(1,1)D 显然在直线y x 上,得D C .3.解:显然有集合{|(4)(1)0}{1,4}Bx x x ,当3a时,集合{3}A ,则{1,3,4},A B A BU I ;当1a 时,集合{1,3}A ,则{1,3,4},{1}A B A B U I ;当4a时,集合{3,4}A ,则{1,3,4},{4}A BA BU I ;当1a ,且3a ,且4a时,集合{3,}A a ,则{1,3,4,},A Ba A BU I .4.解:显然{0,1,2,3,4,5,6,7,8,9,10}U,由U A B U ,得U B A e ,即()U UA B B I 痧,而(){1,3,5,7}U A B I e ,得{1,3,5,7}U B e ,而()U UB B 痧,即{0,2,4,6,8.9,10}B.第一章集合与函数概念1.2函数及其表示1.2.1函数的概念练习(第19页)1.解:(1)要使原式有意义,则470x ,即74x,得该函数的定义域为7{|}4x x ;(2)要使原式有意义,则1030xx ,即31x ,得该函数的定义域为{|31}x x.2.解:(1)由2()32f x xx ,得2(2)322218f ,同理得2(2)3(2)2(2)8f ,则(2)(2)18826f f ,即(2)18,(2)8,(2)(2)26f f f f ;(2)由2()32f x xx ,得22()3232f a aa aa ,同理得22()3()2()32f a a a aa ,则222()()(32)(32)6f a f a aa aa a ,即222()32,()32,()()6f a aa f a aa f a f a a .3.解:(1)不相等,因为定义域不同,时间0t ;(2)不相等,因为定义域不同,()(0)g x x x.1.2.2函数的表示法练习(第23页)1.解:显然矩形的另一边长为2250x cm ,222502500y x xx x ,且050x,即22500(050)yx x x .2.解:图象(A )对应事件(2),在途中遇到一次交通堵塞表示离开家的距离不发生变化;图象(B )对应事件(3),刚刚开始缓缓行进,后来为了赶时间开始加速;图象(D )对应事件(1),返回家里的时刻,离开家的距离又为零;图象(C )我出发后,以为要迟到,赶时间开始加速,后来心情轻松,缓缓行进.3.解:2,2|2|2,2x x y x x x,图象如下所示.4.解:因为3sin 602o,所以与A 中元素60o相对应的B 中的元素是32;因为2sin 452o,所以与B 中的元素22相对应的A 中元素是45o.1.2函数及其表示习题1.2(第23页)1.解:(1)要使原式有意义,则40x ,即4x ,得该函数的定义域为{|4}x x ;(2)xR ,2()f x x 都有意义,即该函数的定义域为R ;(3)要使原式有意义,则2320x x ,即1x 且2x ,得该函数的定义域为{|12}x x x且;(4)要使原式有意义,则4010xx ,即4x 且1x ,得该函数的定义域为{|41}x x x 且.2.解:(1)()1f x x 的定义域为R ,而2()1xg x x的定义域为{|0}x x ,即两函数的定义域不同,得函数()f x 与()g x 不相等;(2)2()f x x 的定义域为R ,而4()()g x x 的定义域为{|0}x x ,即两函数的定义域不同,得函数()f x 与()g x 不相等;(3)对于任何实数,都有362xx ,即这两函数的定义域相同,切对应法则相同,得函数()f x 与()g x 相等.3.解:(1)定义域是(,),值域是(,);(2)定义域是(,0)(0,)U ,值域是(,0)(0,)U ;(3)定义域是(,),值域是(,);(4)定义域是(,),值域是[2,).2()352f x x x ,所以4.解:因为2(2)3(2)5(2)2852f ,(2)852f ;即同理,22()3()5()2352f a a a aa,即2()352f a aa ;22(3)3(3)5(3)231314f aa a aa ,即2(3)31314f aaa ;22()(3)352(3)3516f a f a a f aa ,即2()(3)3516f a f aa .5.解:(1)当3x时,325(3)14363f ,即点(3,14)不在()f x 的图象上;(2)当4x时,42(4)346f ,即当4x 时,求()f x 的值为3;(3)2()26x f x x,得22(6)xx,即14x .6.解:由(1)0,(3)0f f ,得1,3是方程20x bxc的两个实数根,即13,13b c ,得4,3bc,即2()43f x xx ,得2(1)(1)4(1)38f ,即(1)f 的值为87.图象如下:8.解:由矩形的面积为10,即10xy,得10(0)yxx,10(0)xyy,由对角线为d ,即22d xy ,得22100(0)dx xx,由周长为l ,即22l xy ,得202(0)l x xx,另外2()l xy ,而22210,xydxy ,得22222()22220(0)l x y xyxy dd ,即2220(0)ldd.9.解:依题意,有2()2dxvt ,即24v x t d,显然0x h ,即240v th d,得24h d tv,得函数的定义域为2[0,]4h dv和值域为[0,]h.10.解:从A到B的映射共有8个.分别是()0()0()0f af bf c,()0()0()1f af bf c,()0()1()0f af bf c,()0()0()1f af bf c,()1()0()0f af bf c,()1()0()1f af bf c,()1()1()0f af bf c,()1()0()1f af bf c.B组1.解:(1)函数()r f p的定义域是[5,0][2,6)U;(2)函数()r f p的值域是[0,);(3)当5r,或02r时,只有唯一的p值与之对应.2.解:图象如下,(1)点(,0)x和点(5,)y不能在图象上;(2)省略.3.解:3, 2.522,211,10 ()[]0,011,122,233,3xxxf x x xxxx图象如下4.解:(1)驾驶小船的路程为222x,步行的路程为12x,得2221235x xt,(012)x,即241235x xt,(012)x.(2)当4x时,2441242583()3535t h.第一章集合与函数概念1.3函数的基本性质1.3.1单调性与最大(小)值1.答:在一定的范围内,生产效率随着工人数量的增加而提高,当工人数量达到某个数量时,生产效率达到最大值,而超过这个数量时,生产效率随着工人数量的增加而降低.由此可见,并非是工人越多,生产效率就越高.2.解:图象如下[8,12]是递增区间,[12,13]是递减区间,[13,18]是递增区间,[18,20]是递减区间.3.解:该函数在[1,0]上是减函数,在[0,2]上是增函数,在[2,4]上是减函数,在[4,5]上是增函数.4.证明:设12,x x R ,且12x x ,因为121221()()2()2()0f x f x x x x x ,即12()()f x f x ,所以函数()21f x x 在R 上是减函数. 5.最小值.1.3.2单调性与最大(小)值练习(第36页)1.解:(1)对于函数42()23f x xx ,其定义域为(,),因为对定义域内每一个x 都有4242()2()3()23()f x x x x xf x ,所以函数42()23f x x x 为偶函数;(2)对于函数3()2f x xx ,其定义域为(,),因为对定义域内每一个x 都有33()()2()(2)()f x x x xx f x ,所以函数3()2f x xx 为奇函数;(3)对于函数21()xf x x,其定义域为(,0)(0,)U ,因为对定义域内每一个x 都有22()11()()x xf x f x xx,所以函数21()xf x x 为奇函数;(4)对于函数2()1f x x,其定义域为(,),因为对定义域内每一个x 都有22()()11()f x x xf x ,所以函数2()1f x x为偶函数.2.解:()f x 是偶函数,其图象是关于y 轴对称的;()g x 是奇函数,其图象是关于原点对称的.习题1.3A 组1.解:(1)5(,)2上递减;函数在5[,)2函数在上递增;(2)(,0)上递增;函数在[0,)上函数在递减.12x x ,而2.证明:(1)设2212121212()()()()f x f x xx x x x x ,由12120,0x x x x ,得12()()0f x f x ,即12()()f x f x ,所以函数2()1f x x 在(,0)上是减函数;(2)设120x x ,而1212211211()()x x f x f x x x x x ,由12120,0x x x x ,得12()()0f x f x ,即12()()f x f x ,所以函数1()1f x x在(,0)上是增函数.3.解:当0m时,一次函数y mxb 在(,)上是增函数;当0m 时,一次函数ymx b 在(,)上是减函数,令()f x mx b ,设12x x ,而1212()()()f x f x m x x ,当0m时,12()0m x x ,即12()()f x f x ,得一次函数y mxb 在(,)上是增函数;当0m时,12()0m x x ,即12()()f x f x ,得一次函数y mxb 在(,)上是减函数.4.解:自服药那一刻起,心率关于时间的一个可能的图象为5.解:对于函数21622100050xyx ,当162405012()50x 时,max 307050y (元),即每辆车的月租金为4050元时,租赁公司最大月收益为307050元.6.解:当0x时,0x,而当0x时,()(1)f x x x ,即()(1)f x x x ,而由已知函数是奇函数,得()()f x f x ,得()(1)f x x x ,即()(1)f x x x ,所以函数的解析式为(1),0()(1),0x x x f x x x x.B 组1.解:(1)二次函数2()2f x xx 的对称轴为1x,则函数()f x 的单调区间为(,1),[1,),且函数()f x 在(,1)上为减函数,在[1,)上为增函数,函数()g x 的单调区间为[2,4],且函数()g x 在[2,4]上为增函数;(2)当1x 时,min()1f x ,因为函数()g x 在[2,4]上为增函数,所以2min()(2)2220g x g .2.解:由矩形的宽为x m ,得矩形的长为3032x m ,设矩形的面积为S ,则23033(10)22x xx S x,当5x 时,2max 37.5S m ,即宽5xm 才能使建造的每间熊猫居室面积最大,且每间熊猫居室的最大面积是237.5m .3.判断()f x 在(,0)上是增函数,证明如下:设120x x ,则120x x ,因为函数()f x 在(0,)上是减函数,得12()()f x f x ,又因为函数()f x 是偶函数,得12()()f x f x ,所以()f x 在(,0)上是增函数.复习参考题A 组1.解:(1)方程29x的解为123,3x x ,即集合{3,3}A ;(2)12x,且xN ,则1,2x ,即集合{1,2}B ;(3)方程2320xx 的解为121,2x x ,即集合{1,2}C.2.解:(1)由PAPB ,得点P 到线段AB 的两个端点的距离相等,即{|}P PA PB 表示的点组成线段AB 的垂直平分线;(2){|3}P POcm 表示的点组成以定点O 为圆心,半径为3cm 的圆.3.解:集合{|}P PAPB 表示的点组成线段AB 的垂直平分线,集合{|}P PA PC 表示的点组成线段AC 的垂直平分线,得{|}{|}P PAPB P PAPC I 的点是线段AB 的垂直平分线与线段AC 的垂直平分线的交点,即ABC 的外心.4.解:显然集合{1,1}A ,对于集合{|1}B x ax ,当0a 时,集合B,满足BA ,即0a ;当0a时,集合1{}B a,而BA ,则11a,或11a,得1a,或1a ,综上得:实数a 的值为1,0,或1.5.解:集合20(,)|{(0,0)}30x y A Bx y x y I ,即{(0,0)}A B I ;集合20(,)|23x y A Cx y x y I ,即A C I ;集合3039(,)|{(,)}2355x y B Cx y xyI ;则39()(){(0,0),(,)}55A B B C I U I .6.解:(1)要使原式有意义,则2050x x ,即2x ,得函数的定义域为[2,);(2)要使原式有意义,则40||5x x ,即4x ,且5x ,得函数的定义域为[4,5)(5,)U .7.解:(1)因为1()1x f x x,所以1()1a f a a ,得12()1111a f a aa,即2()11f a a ;(2)因为1()1x f x x,所以1(1)(1)112a a f a a a,即(1)2a f a a.8.证明:(1)因为221()1x f x x,所以22221()1()()1()1x x f x f x x x,即()()f x f x ;(2)因为221()1x f x x,所以222211()11()()111()x x f f x x xx,即1()()f f x x .9.解:该二次函数的对称轴为8k x,函数2()48f x xkx 在[5,20]上具有单调性,则208k ,或58k ,得160k,或40k ,即实数k 的取值范围为160k,或40k.10.解:(1)令2()f x x ,而22()()()f x x xf x ,即函数2y x 是偶函数;(2)函数2y x 的图象关于y 轴对称;(3)函数2y x 在(0,)上是减函数;(4)函数2yx 在(,0)上是增函数.B 组1.解:设同时参加田径和球类比赛的有x 人,则158143328x ,得3x ,只参加游泳一项比赛的有15339(人),即同时参加田径和球类比赛的有3人,只参加游泳一项比赛的有9人.2.解:因为集合A ,且20x,所以0a .3.解:由(){1,3}U A B U e ,得{2,4,5,6,7,8,9}A BU ,集合A B U 里除去()U A B I e ,得集合B ,所以集合{5,6,7,8,9}B.4.解:当0x时,()(4)f x x x ,得(1)1(14)5f ;当0x时,()(4)f x x x ,得(3)3(34)21f ;(1)(5),1(1)(1)(3),1a a a f a a a a.5.证明:(1)因为()f x axb ,得121212()()222x x x x a f a b x x b ,121212()()()222f x f x ax b ax b a x x b ,所以1212()()()22x x f x f x f ;(2)因为2()g x x ax b ,得22121212121()(2)()242x x x x g xx x x a b ,2212121()()22x x x x a b ,因为2222212121212111(2)()()0424xx x x xx x x ,即222212121211(2)()42x x x x xx ,所以1212()()()22x x g x g x g .6.解:(1)函数()f x 在[,]b a 上也是减函数,证明如下:设12bx x a ,则21ax x b ,因为函数()f x 在[,]a b 上是减函数,则21()()f x f x ,又因为函数()f x 是奇函数,则21()()f x f x ,即12()()f x f x ,所以函数()f x 在[,]b a 上也是减函数;(2)函数()g x 在[,]b a 上是减函数,证明如下:设12bx x a ,则21ax x b ,因为函数()g x 在[,]a b 上是增函数,则21()()g x g x ,又因为函数()g x 是偶函数,则21()()g x g x ,即12()()g x g x ,所以函数()g x 在[,]b a 上是减函数.7.解:设某人的全月工资、薪金所得为x元,应纳此项税款为y元,则x,由该人一月份应交纳此项税款为26.78元,得25004000 x,得2517.825(2500)10%26.78x,所以该人当月的工资、薪金所得是2517.8元.。

新教材第四章答案修正版

新教材第四章答案修正版

4-18 图4.1所示系统中,发电机以发电机-变压器组方式接入系统,最大开机方式为4台机全开,最小开机方式为两侧各开1台机,变压器T5和T6可能2台也可能1台运行。

其参数为:图4.1 系统示意图115/3kV E ϕ=; 1.1 2.1 1.2 2.215G G G G X X X X ====Ω; 1.3 2.3 1.4 2.410G G G G X X X X ====Ω,1.1 1.4~10T T X X =Ω,0.10.4~30T T X X =Ω, 1.5 1.620T T X X ==Ω,0.50.640T T X X ==Ω;60km A B L -=,40km B C L -=;线路阻抗120.4/km Z Z ==Ω,0 1.2/km Z =Ω,线路阻抗角均为075,.max .max 300A A B L C B L I I --==,负荷功率因数角为030; 1.2ss K =, 1.2re K =,0.85rel K I =,0.75rel K II =,变压器均装有快速差动保护。

试解答:(1) 为了快速切除线路上的各种短路,线路A-B 、B-C 应在何处配备三段式距离保护,各选用何种接线方式?各选用何种动作特性?答:应该在1,2,3,4处配备三段式距离保护;选用接地距离保护接线方式和相间距离保护接线方式;它们的I ,II 段选择具有方向特性的距离保护,III 选用具有偏移特性的距离保护。

(2) 整定保护1~4的距离I 段,并按照你选定的动作特性,在一个阻抗复平面上画出各保护的动作区域。

答:线路AB 的正序阻抗 10.46024AB A B Z Z L -==⨯=Ω线路BC 的正序阻抗 10.44016BC B C Z Z L -==⨯=Ω 保护1,2的距离I 段 .1,20.852420.4IIset rel AB Z K Z ==⨯=Ω 保护3,4的距离I 段 .3,40.851613.6IIset rel BC Z K Z ==⨯=Ω保护1~4距离I 段在阻抗复平面上的动作区域如图4.1.1所示,圆周1、2、3、4分别对应保护1、2、3、4距离I 段的动作特性:图4.1.1 保护1~4距离I 段的动作特性(3) 分别求出保护1、4接地距离II 段的最大、最小分支系数; 答:对保护1:(I )当与相邻下级线路距离保护I 段相配合时1max 2.88b K =,1min 1.59b K =(II )当与相邻变压器的快速保护相配合时1max 2.88b K =,1min 2.01b K =对保护4:(I )当与相邻下级线路距离保护I 段相配合时4max 2.26b K =,4min 1.41b K =(II )当与相邻变压器的快速保护相配合时4max 1.99b K =,4min 1.53b K =(4) 分别求出保护1、4接地距离II 、III 段的定值及时限,并校验灵敏度。

2020年最新版初一数学教材课后习题参考答案

2020年最新版初一数学教材课后习题参考答案

习题1.1第2题答案(1)0.08m表示水面高于标准水位0.08m;-0.2m表示水面低于标准水位0.2m(2)水面低于标准水位0.1m,记作-0.1m;高于标准水位0.23m,记作+0.23m(或0.23m)习题1.1第3题答案不对O既不是正数,也不是负数习题1.1第4题答案表示向前移动5m,这时物体离它两次移动前的位置为Om,即回到了它两次移动前的位置习题1.1第5题答案这七次测量的平均值为(79.4+80.6+80.8+79.1+80+79.6+80.5)/7=80(m)以平均值为标准,七次测量的数据用正数、负数表示分别为:-0.6m,+0.6m,+0.8m,-0.9m,Om,-0.4m十0.5m习题1.1第6题答案氢原子中的原子核所带电荷可以用+1表示,氢原子中的电子所带电荷以用-1表示习题1.1第7题答案由题意得7-4-4=-1(℃)习题1.1第8题答案中国、意大利服务出口额增长了;美国、德国、英国、日本服务出日额减少了;意大利增长率最高;日本增长率最低习题1.2第1题答案正数:{15,0.15,22/5,+20,…)负数:{-3/8,-30,-12.8,-60,…}习题1.2第2题答案如下图所示:习题1.2第3题答案当沿数轴正方向移动4个单位长时,点B表示的数是1当沿数轴反方向移动4个单位长时,点B表示的数是-7习题1.2第4题答案各数的相反数分别为4,-2,1.5,0,-1/3,9/4在数轴上表示如下图所示:习题1.2第5题答案丨-125丨=125,丨+23丨=23,丨-3.5丨=3.5,丨0丨=0,丨2/3丨=2/3,丨-3/2丨=3/2,丨-0.05丨=0.05-125的绝对值最大,0的绝对值最小习题1.2第6题答案-3/2<-2/3<-1/2<-0.25<-0.15<0<0.05<+2.3习题1.2第7题答案各城市某年一月份的平均气温(℃)按从高到低的顺序排列为:13.1;3.8;2.4;-4.6;-19.4习题1.2第8题答案因为丨+5丨=5,丨-3.5丨=3.5,丨+0.7丨=0.7,丨-2.5丨=2.5,丨-0.6丨=0.6所以从左向右数,第五个排球的质量最接近标准习题1.2第9题答案-9.6%最小;增幅是负数说明人均水资源占有量在下降习题1.2第10题答案表示数1的点与表示-2和4的点的距离相等,都是3习题1.2第11题答案(1)有;如-0.1,-0.12,-0.57,…有;如-0.15,-0.42,-0.48,…(2)有,-2;-1,0,1(3)没有(4)如:-101,-102,-102.5习题1.2第12题答案不一定,x还可能是-2;x=0;x=0习题1.3第1题答案(1)-4(2)8(3)-12(4)-3(5)-3.6(6)-1/5(7)1/15(8)-41/3(1)3(2)0(3)1.9(4)-1/5习题1.3第3题答案(1)-16(2)0(3)16(4)0(5)-6(6)6(7)-31(8)102(9)-10.8(10)0.2(1)1(2)1/5(3)1/6(4)-5/6(5)-1/2(6)3/4(7)-8/3(8)-8习题1.3第5题答案(1)3.1(2)3/4(3)8(4)0.1(5)-63/4(6)0两处高度相差:8844.43-(-415)=9259.43(m)习题1.3第7题答案半夜的气温为:-7+11-9=-5(℃)习题1.3第8题答案解:132-12.5-10.5+127-87+136.5+98=383.5(元)答:一周总的盈亏情况是盈利383.5元习题1.3第9题答案解:25×8+1.5-3+2-0.5+1-2-2-2.5=200-5.5=194.5(kg)答:这8筐白菜一共194.5kg习题1.3第10题答案解:各天的温差如下:星期一:10-2=8(℃)星期二:12-1=11(℃)星期三:11-0=11(℃)星期四:9-(-1)=10(℃)星期五:7-(-4)=11(℃)星期六:5-(-5)=10(℃)星期日:7-(-5)=12(℃)答:星期日的温差最大,星期一的温差最小习题1.3第11题答案(1)16(2)(-3)(3)18(4)(-12)(5)(-7)(6)7习题1.3第12题答案解:(-2)+(-2)=-4(-2)+(-2)+(-2)=-6(-2)+(-2)+(-2)+(-2)=-8(-2)+(-2)+(-2)+(-2)+(-2)=-10(-2)×2=4,(-2)×3=-6(-2)×4=8,(-2)×5=-10法则:负数乘正数积为负,积的绝对值等于两个数的绝对值的积习题1.3第13题答案解:第一天:0.3-(-0.2)=0.5(元)第二天:0.2-(-0.1)=0.3(元)第三天:0-(-0.13)=0.13(元)平均值:(0.5+0.3+0.13)÷3=0.31(元)题1.4第1题答案(1)(-8)×(-7)=56(2)12×(-5)=-60(3)2.9×(-0.4)=-1.16(4)-30.5×0.2=-6.1(5)100×(-0.001)=-0.1(6)-4.8×(-1.25)=6习题1.4第2题答案(1)1/4×(-8/9)=-2/9(2)(-5/6)×(-3/10)=1/4(3)-34/15×25=-170/3(4)(-0.3)×(-10/7)=3/7习题1.4第3题答案(1)-1/15(2)-9/5(3)-4(4)100/17(5)4/17(6)-5/27习题1.4第4题答案(1)-91÷13=-7(2)-56÷(-14)=4(3)16÷(-3)=-16/3(4)(-48)÷(-16)=3(5)4/5÷(-1)=-4/5(6)-0.25÷3/8=-2/3习题1.4第5题答案-5,-1/5,-4,6,5,1/5,-6,4习题1.4第6题答案(1)(-21)/7=-3(2)3/(-36)=-1/12(3)(-54)/(-8)=27/4(4)(-6)/(-0.3)=20习题1.4第7题答案(1)-2×3×(-4)=2×3×4=24(2)-6×(-5)×(-7)=-6×5×7=-210(3)(-8/25)×1.25×(-8)=8/25×8×5/4=16/5(4)0.1÷(-0.001)÷(-1)=1/10×1000×1=100(5)(-3/4)×(-11/2)÷(-21/4)=-3/4×3/2×4/9=-1/2(6)-6×(-0.25)×11/14=6×1/4×11/14=33/28(7)(7)×(-56)×0÷(-13)=0(8)-9×(-11)÷3÷(-3)=-9×11×1/3×1/3=-11习题1.4第8题答案(1)23×(-5)-(-3)÷3/128=-115+3×128/3=-115+128=13(2)-7×(-3)×(-0.5)+(-12)×(-2.6)=-7×3×0.5+12×2.6=-10.5+31.2=20.7(3)(13/4-7/8-7/12)÷(-7/8)+(-7/8)÷(13/4-7/8-7/12)=(7/4-7/8-7/12)×(-8/7)+(-7/8)÷7/24=7/24×(-8/7)-3=-31/3(4)-丨-2/3丨-丨-1/2×2/3丨-丨1/3-1/4丨-丨-3丨=-2/3-1/3-1/12-3=-49/12习题1.4第9题答案(1)(-36)×128÷(-74)≈62.27(2)-6.23÷(-0.25)×940=23424.80(3)-4.325×(-0.012)-2.31÷(-5.315)≈0.49(4)180.65-(-32)×47.8÷(-15.5)≈81.97习题1.4第10题答案(1)7500(2)-140(3)200习题1.4第11题答案解:450+20×60-12×120=210(m)答:这时直升机所在高度是210m习题1.4第12题答案(1)<,<(2)<,<(3)>,>(4)=,=习题1.4第13题答案2,1,-2,-1一个非0有理数不一定小于它的2倍,因为一个负数比它的2倍大习题1.4第14题答案(-2+3)a习题1.4第15题答案(1)(2)均成立,从它们可以总结出:分子、分母以及分数这三者的符号,改变其中两个,分教的值不变复习题1第1题答案如下图所示:-3.5<-2<-1.6<-1/3<0<0.5<2<3.5复习题1第2题答案将整数x的值在数轴上表示如下图所示:复习题1第3题答案a=-2的绝对值、相反数和倒数分别为:2,2,-1/2b=-2/3的绝对值、相反数和倒数分别为:2/3,2/3,-3/2c=5.5的绝对值、相反数和倒数分别为:5.5、-5.5,2/11复习题1第4题答案互为相反数的两数的和是0;互为倒数的两数的积是1复习题1第5题答案(1)100(2)-38(3)-70(4)-11(5)96(6)-9(7)-1/2(8)75/2(9)(-0.02)×(-20)×(-5)×4.5=-0.02×4.5×20×5=-0.09×100=-9(10)(-6.5)×(-2)÷(-1/3)÷(-5)=6.5×2×3×1/5=7.8(11)6+(-1/5)-2-(-1.5)=6-0.2-2+1.5=5.3(12)-66×4-(-2.5)÷(-0.1)=-264-25=-289(13)(-2)2×5-(-2)3÷4=4×5-(-8)÷4=20-(-2)=22(14)-(3-5)+32×(1-3)=-(-2)+9×(-2)=2+(-18)=-16复习题1第6题答案(1)245.635≈245.6(2)175.65≈176(3)12.004≈12.00(4)6.5378≈6.54复习题1第7题答案(1)100000000=1×108(2)-4500000=-4.5×106(3)692400000000=6.924×1011复习题1第8题答案(1)-2-丨-3丨=-2-3=-5(2)丨-2-(-3)丨=丨-2+3丨=1复习题1第9题答案(82+83+78+66+95+75+56+93+82+81)÷10=791÷10=79.1复习题1第10题答案C复习题1第11题答案解:星期六的收入情况表示为:458-[-27.8+(-70.3)+200+138.1+(-8)+188]=458-420=38因为38>0所以星期六是盈余的,盈佘了38元复习题1第12题答案解:(60-15)×0.002=0.09(mm)(5-60)×0.002=-0.11(mm)0.09-0.11=-0.02(mm)答:金属丝的长度先伸长了0.09mm,又缩短了0.11mm,最后的长度比原长度伸长了-0.02mm复习题1第13题答案解:1.4960亿km=1.4960×108km答:1个天文单位是1.4960×108km复习题1第14题答案(1)当a=1/2时,a的平方为1/4,a的立方为1/8,所以a大于a的平方大于a的立方,即a>a2>a3(0<a<1)(2)当b=-1/2时,b的平方为1/4,b的立方为-1/8,所以b的平方大于b的立方大于b,即b2>b3>b(-1<b<o)复习题1第15题答案特例归纳略(1)错,如:0的相反数是0(2)对,因为任何互为相反数的两个数的同—偶数次方符号相同,绝对值相等(3)错,对于一个正数和一个负数来说,正数大于负数,正数的倒数仍大于这个负数的倒数,如2和-3,2>-3,1/2>-1/3复习题1第16题答案1;121;12321;1234321(1)它们有一个共同特点:积的结果各数位上的数字从左到右由1开始依次增大1,当增大到乘式中一个乘数中1的个数后,再依次减小1,直到1(2)12345678987654321习题2.1第2题答案(1)(t+5)℃(2)3(x-y)km或(3x-3y)km(3)(100-5x)(4)(πR2a-πr2a)cm3习题2.1第3题答案习题2.1第4题答案(1)年数每增加一年,树高增加5cm (2)(100+5n)cm习题2.1第5题答案第2排有(a+1)个座位第3排有(a+2)个座位第n排的座位数为(a+n-1)20+19-1=38(个)习题2.1第6题答案解:V=(1/2a2-πr2)h(cm3)当a=6cm,r=0.5cm,h=0.2cm时V≈(1/2×62-3×0.52)×0.2=3.45(cm3)习题2.1第7题答案(1)2n(2)2n+1或(2n-1)习题2.1第8题答案3个球队比赛,总的比赛场数是[3(3-1)]/2=34个球队比赛,总的比赛场数是[4(4-1)]/2=65个球队比赛,总的比赛场数是[5(5-1)]/2=10n个球队比赛,总的比赛场数是[n(n-1)]/2习题2.1第9题答案密码L dp d jlou,破译它的“钥匙”x-3密码的意思是“I am a girl”(注:此题答案不唯一,合理即可)习题2.2第1题答案(1)2x-10.3x=(2-10.3)x=-8.3x(2)3x-x-5x=(3-1-5)x=-3x(3)-b+0.6b-2.6b=(-1+0.6-2.6)b=-3b(4)m-n2+m-n2=(1+1)m+(-1-1)n2=2m-2n2习题2.2第2题答案(1)2(4x-0.5)=8x-1(2)-3(1-1/6x)=-3+1/2x(3)-x+(2x-2)-(3x+5)=-x+2x-2-3x-5=-2x-7(4)3a2+a2-(2a2-2a)+(3a-a2)=3a2+a2-2a2+2a+3a-a2=a2+5a 习题2.2第3题答案(1)原式=5a+4c+7b+5c-3b-6a=-a+4b+9c(2)原式=8xy-x2+y2-x2+y2-8xy=-2x2+2y2(3)原式=2x2-1/2+3x-4x+4x2-2=6x2-x-5/2(4)原式=3x2-(7x-4x+3-2x2)=3x2-7x+4x-3+2x2=5x2-3x-3习题2.2第4题答案(-x2+5+4x)+(5x-4+2x2)=-x2+5+4x+5x-4+2x2=x2+9x+1当x=-2时,原式=(-2)2+9×(-2)+1=4-18+1=-13(1)比a的5倍大4的数为5a+4,比a的2倍小3的数是2a-3(5a+4)+(2a-3)=5a+4+2a-3=7a+1(2)比x的7倍大3的数为7x+3,比x的6倍小5的数是6x-5(7x+3)-(6x-5)=7x+3-6x+5=x+8习题2.2第6题答案解:水稻种植面积为3ah m2,玉米种植面积为(a-5)h m23a-(a-5)=3a-a+5=(2a+5)(h m2)习题2.2第7题答案(1)πa2/2+4a2=(π+8)/2a2(cm2)(2)πa+2a×3=πa+6a=(π+6)a(cm)习题2.2第8题答案3(a+y)+1.5(a-y)=3a+3y+1.5a-1.5y=4.5a+1.5y17a,20a,…,(3n+2)a习题2.2第10题答案S=3+3(n-2)=3n-3当n=5时,S=3×5-3=12当n=7时,S=3×7-3=18当n=11时,S=3×11-3=30习题2.2第11题答案(1)10b+a(2)10(10b+a)(3)10b+a+10(10b+a)=11(10b+a)这个和是11的倍数,因为它含有11这个因数习题2.2第12题答案36a2;cm2复习题2第1题答案(1)(t+15)°C(2)nc元,(100-nc)元(3)0.8b元,(0.8b-10)元(4)a/30m,1500m,(a/30-1500)m 复习题2第2题答案复习题2第3题答案(1)-2x2y(2)10.5y2(3)0(4)-1/12mn+7(5)8ab2+4(6)3x3-2x2复习题2第4题答案(1)原式=4a3b-10b3-3a2b2+10b3=4a3b-3a22b2(2)原式=4x2y-5xy2-3x2y+4xy2=x2y-xy2(3)原式=5a2-(a2+5a2-2a-2a2+6a)=5a2-a2-5a2+2a+2a2-6a=a2-4a(4)原式=15+3-3a-1+a+a2+1-a+a2-a3=18-3a+2a2-a3(5)原式=4a2b-3ab-5a2b+2ab=a2b-ab(6)原式=6m2-4m-3+2m2-4m+1=8m2-8m-2(7)原式=5a2+2a-1-12+32a-8a2=-3a2+34a-13(8)原式=3x2-(5x-1/2x+3+2x2)=3x2-5x+1/2x-3-2x2=x2-9/2x-3复习题2第5题答案解:原式=(5-3-2)x2+(-5+6)x-1=x-1当x=-3时原式=-3-1=-4复习题2第6题答案(1)5/2(2)(x+y)/10复习题2第7题答案(h+20)m(h-30)m(h+20)-(h-30)=h+20-h+30=50(m)复习题2第8题答案S长方形=2x×4=8x(cm2)S梯形=1/2(x+3x)×5=10x(cm2)S梯形>S长方形S梯形-S长方形=10x-8x-2x(cm2)复习题2第9题答案解:2πr×2-(2πr+2π×r/2+2π×r/6+2π×r/3)=0因此所需材料一样多复习题2第10题答案解:a×(1+22%)=1.22a(元)1.22a×85%=1.037a(元)1.037a-a=0.037a(元)答:按成本增加22%定出价格,每件售价1.22a元;按原价的85%出售,现售价1.037a元;每件还能盈利0.037a元复习题2第11题答案解:10a+b;10b+a;(10b+a)+(10a+b)=11(a+b)答:这个数能被11整除复习题2第12题答案(1)原式=(4+2-1)(0+6)=5(a+b)=5a+5b(2)原式=(3+8)(z+y)2+(-7+6)·(x+y)=11(x+y)2-(x+y)习题3.1第1题答案(1)a+5=8(2)1/3b=9(3)2x+10=18(4)1/3x-y=6(5)3a+5=4a(6)1/2b-7=a+b习题3.1第2题答案(1)a+b=b+a(2)a·b=b·a(3)a·(b+c)=a.b+a·c(4)(a+b)+c=a+(b+c)习题3.1第3题答案x=3是方程(3)3x-2=4+x的解x=0是方程(1)5x+7=7-2x的解x=-2是方程(2)6x-8=8x-4的解习题3.1第4题答案(1)x=33(2)x=8(3)x=1(4)x=1习题3.1第5题答案解:设七年级1班有男生x人,有女生(4/5x+3)人,则x+(4/5x+3)=48习题3.1第6题答案解:设获得一等奖的学生有x人,则200x+50(22-x)=1400习题3.1第7题答案解:设去年同期这项收入为x元,则x·(1+8.3%)=5109习题3.1第8题答案解:设x个月后这辆汽车将行驶20800km,则12000+800x=20800习题3.1第9题答案解:设内沿小圆的半径为x cm,则102π-πx2=200习题3.1第10题答案解:设每班有x人,则10x=428+2210x+1-(10+x)=18,x=3习题3.2第1题答案(1)x=2(2)x=3(3)y=-1(4)b=18/5习题3.2第2题答案例如:解方程5x+3=2x,把2x改变符号后移到方程左边,同时把3改变符号后移到方程右边,即5x-2x=-3,移项的根据是等式的性质1习题3.2第3题答案(1)合并同类项,得4x=-16.系数化为1,得x=-4(2)合并同类项,得6y=5.系数化为1,得y=5/6(3)移项,得3x-4x=1-5.合并同类项,得-x=-4.系数化为1,得x=4(4)移项,得-3y-5y=5-9.合并同类项,得-8y=-4.系数化为1,得y=1/2(1)根据题意,可列方程5x+2=3x-4.移项,得5x-3x=-4-2.合并同类项,得2x=-6.系数化为1,得x=-3(2)根据题意,可列方程-5y=y+5.移项,得-5y-y=5.合并同类项,得-6y=5.系数化为1,得y=-5/6习题3.2第5题答案解:设现在小新的年龄为x.根据题意,得:3x=28+x移项,得2x=28系数化为1,得x=14答:现在小新的年龄是14习题3.2第6题答案解:设计划生产I型洗衣机x台,则计划生产Ⅱ型洗衣机2x台,计划生产Ⅲ型洗衣机14x台.根据题意得:x+2x+14x=25500合并同类项,得17x=25500系数化为1,得x=1500因此2x=3000,14x=21000答:这三种型号洗衣机计划分别生产1500台、3000台、21000台习题3.2第7题答案解:设宽为xm,则长为1.5xm根据题意,得2x+2×1.5x=60合并同类项,得5x=60系数化为1,得x=12所以1.5x=18答:长是18m,宽是12m习题3.2第8题答案(1)设第一块实验田用水xt,则第二块实验田用水25%xt,第三块实验田用水15%xt(2)根据(1),并由题意得:x+25%x+15%x=420合并同类项,得1.4x=420系数化为1,得x=300.所以25%x=75,15%x=45答:第一块实验田用水300t,第二块实验田用水75t,第三块实验田用水45t 习题3.2第9题答案解:设它前年10月生产再生纸xt,则去年10月生产再生纸(2x+150)t.根据题意得:2x+150=2050移项,合并同类项,得2x=1900系数化为1,得x=950答:它前年10月生产再生纸950t习题3.2第10题答案在距一端35cm处锯开习题3.2第11题答案解:设参与种树的人数是x.根据题意得:10x+6=12x-6移项,得10x-12x=-6-6合并同类项,得-2x=-12系数化为1,得x=6答:参与种树的人数是6习题3.2第12题答案解:设相邻三行里同一列的三个日期数分别为x-7,x,x+7根据题意,假设三个日期数之和能为30,则(x-7)+x+(x+7)=30去括号,合并同类项,得3x=30系数化为1,得x=10x=10符合题意,假设成立x-7=10-7=3,x+7=10+7=17所以相邻三行里同一列的三个日期数之和能为30.这三个数分别是3,10,17习题3.2第13题答案方法1:设这个两位数的个位上的数为x,则十位上的数为(3x+1),这个两位数为:10(3x+1)+x根据题意,得x+(3x+1)=9解这个方程,得x=23x+1=3×2+1=7这个两位数为10(3x+1)+x=10×7+2=72答:这个两位数是72方法2:设这个两位数的个位上的数为x,则十位上的数为(9-x),这个两位数为10(9-x)+x根据题意,得3x+1=9-x解这个方程,得x=2这个两位数为10(9-x)+x=10×(9-2)+2=72答:这个两位数是72习题3.3第1题答案(1)a=-2(2)b-1(3)x=2(4)y=-12习题3.3第2题答案(1)去括号,得2x+16=3x-3.移项、合并同类项,得-x=-19.系数化为1,得x=19(2)去括号,得8x=-2x-8.移项、合并同类项,得10x=-8.系数化为1,得x=-4/5(3)去括号,得2x-2/3x-2=-x+3.移项、合并同类项,得7/3x=5.系数化为1,得x=15/7(4)去括号,得20-y=-1.5y-2.移项、合并同类项,得0.5y=-22.系数化为1,得y=-44习题3.3第3题答案(1)去分母,得3(3x+5)=2(2x-1).去括号,得9x+15=4x-2.移项、合并同类项,得5x=-17.系数化为1,得x=-17/5.(2)去分母,得-3(x-3)=3x+4.去括号,得-3x+9=3x+4.移项、合并同类项,得6x=5.系数化为1,得x=5/6.(3)去分母,得3(3y-1)-12=2(5y-7).去括号,得9y-3-12=10y-14.移项、合并同类项,得y=-1.(4)去分母,得4(5y+4)+3(y-1)=24-(5y-5).去括号,得20y+16+3y-3=24-5y+5.移项、合并同类项,得28y=16.系数化为1,得y=4/7习题3.3第4题答案(1)根据题意得:1.2(x+4)=3.6(x-14)去括号得:1.2x+4.8=3.6x-50.4移项,得1.2x-3.6x=-50.4-4.8合并同类项,得-2.4x=-55.2系数化为1,得x=23(2)根据题意得:1/2(3y+1.5)=1/4(y-1)去分母(方程两边乘4)得:2(3y+1.5)=y-1去括号,得6y+3=y-1移项,得6y-y=-1-3合并同类项,得5y=-4系数化为1,得y=-4/5习题3.3第5题答案解:设张华登山用了x min,则李明登山所用时间为(x-30)min 根据题意得:10x=15(x-30)解得x=90山高10x=10×90=900(m)答:这座山高为900m习题3.3第6题答案解:设乙车的速度为xkm/h,甲车的速度为(x+20)km/h根据题意得:1/2x+1/2(x+20)=84解得x=74x+20=74+20=94答:甲车的速度是94km/h,乙车的速度是74km/h习题3.3第7题答案(1)解:设无风时这架飞机在这一航线的平均航速为x km/h,则这架飞机顺风时的航速为(x+24)km/h,这架飞机逆风时的航速为(x-24)km/h根据题意,得2.8(x+24)=3(x-24)解这个方程,得x=696(2)两机场之间的航程为2.8(x+24)km或3(x-24)km所以3(x-24)=3×(696-24)=2016(km)答:无风时这架飞机在这一航线的平均航速为696km/h两机场之间的航程是2 016km习题3.3第8题答案蓝布料买了75m,黑布料买了63m习题3.3第9题答案解:设每个房间需要粉刷的墙面面积为x m2,则(8x-50)/3=(10x+40)/5+10,解得x=52答:每个房间需要刷粉的墙面面积为52m2习题3.3第10题答案解:从10时到12时王力、陈平两人共行驶36+36=72(km),用时2h,所以从8时到10时王力、陈平用时2h也行驶72km,设A,B两地间的路程为z km,则x-72=36,得x=108答:A,B两地间的路程为108km解:设两地间的路程为x km,上午10时,两人走的路程为(x-36)km,速度和为(x-36)/2km/h,中午12时,两人走的路程为(x+36)km,速度和为(x+36)/4km/h,根据速度和相等列方程,得(x-36)/2=(x+36)/4,得x=108答:A,B两地之间的路程为108km习题3.3第11题答案(1)设火车的长度为xm,从车头经过灯下到车尾经过灯下火车所走的路程为xm,这段时间内火车的平均速度为x/10m/s(2)设火车的长度为xm,从车头进入隧道到车尾离开隧道火车所走的路程为(300+x)m,这段时间内火车的平均速度为((300+x)/20)m/s(3)在这个问题中火车的平均速度没有发生变化(4)根据题意,可列x/10=(300+x)/20,解得x=300,所以这列火车的长度为300m 习题3.4第1题答案略习题3.4第2题答案解:设计划用x m3的木材制作桌面,(12-x)m3的木材制作桌腿,才能制作尽可能多的桌子根据题意得:4×20x=400(12-x)解得x=10,12–x=12-10=2答:计划用10m3的木材制作桌面,2m3的木材制作桌腿才能制作尽可能多的桌子习题3.4第3题答案解:设甲种零件应制作x天,乙种零件应削作(30-x)天根据题意得:500x=250(30-x)解得x=10,30-x=30-10=20答:甲种零件应制作10天,乙种零件应制作20天习题3.4第4题答案解:设共需要x h完成,则(1/7.5+1/5)+1/5(x-1)=1解得x=13/3,13/3h=4h20min答:如果让七、八年级学生一起工作1h,再由八年级学生单独完成剩余部分,共需4h20min习题3.4第5题答案解:设先由x人做2h,则x/80×2+(x+5)/80×8=3/4解得x=2,x+5=7(人)答:先安排2人做2h,再由7人做8h,就可以完成这项工作的3/4习题3.4第6题答案解:设这件衣服值x枚银币,则(x+10)/12=(x+2)/7,解得x=9.2答:这件衣服值9.2枚银币习题3.4第7题答案解法1:设每台B型机器一天生产x个产品,则每台A型机器一天生产(x+1)个产品根据题意,得(5(x+1)-4)/8=(7x-1)/11,解得x=19,因此(7×19-1)/11=12(个)答:每箱装12个产品解法2:设每箱装x个产品,根据“每台A型机器一天生产的产品=每台B型机器一天生产的产品+1”根据题意列方程,得(8x+4)/5=(11x+1)/7+1.解得x=12答:每箱装12个产品习题3.4第8题答案(1)由题意知时间增加5min,温度升高15℃,所以每增加1min,温度升高3℃,则21min时的温度为10+21X3=73(℃)(2)设时间为x min,列方程3x+10=34,解得x=8习题3.4第9题答案解:设制作大月饼用x kg面粉,制作小月饼用(4500-x)kg面粉,才能生产最多的盒装月饼根据题意得:(x/0.05)/2=((4500-x)/0.02)/4化简,得8x=10(4500-x)解得x=25004500-x=4500-2500=2000答:制作大月饼应用2500kg面粉,制作小月饼用2000kg面粉,才能生产最的盒装月饼习题3.4第10题答案解:设相遇时小强行进的路程为x km,小刚行进的路程为(x+24)km,小强行进的速度为x/2km/h,小刚行进的速度为(x+24)/2km/h根据题意得:(x+24)/2×0.5=x解得x=8所以x/2=8/2=4,(x+24)/2=(8+24)/2=16相遇后小强到达A地所用的时间为:(x+24)/4=(8+24)/4=8答:小强行进的速度为4km/h.小刚行进的速度为16km/h.相遇后经过8h小强到达A地习题3.4第11题答案解:设销售量要比按原价销售时增加x%.根据题意得:(1-20%)(1+x%)=1解得x=25答:销售量要比按原价销售时增加25%习题3.4第12题答案(1)设此月人均定额是x件,则(4x+20)/4=(6x-20)/5,解得x=45答:此月人均定额是45件(2)设此月人均定额为y件,则(4y+20)/4=(6y-20)/5+2,解得y=35答:此月人均定额是35件(3)设此月人均定额为z件,则(4z+20)/4=(6z-20)/5-2,解得z=55.答:此月人均定额是55件习题3.4第13题答案(1)设丢番图的寿命为x岁,则1/6x+1/12x+1/7x+5+1/2x+4=x,解得x=84所以丢番图的寿命为84岁(2)1/6x+1/12x+1/7x+5=38(岁),所以丢番图开始当爸爸时的年龄为38岁(3)x-4=80,所以儿子死时丢番图的年龄为80岁复习题3第1题答案(1)t-2/3t=10(2)(n-110)/n×100%=45%或(1-45%)n=110(3)1.1a-10=210(4)60/5-x/5=2复习题3第2题答案(1)移项,得-8x+11/2x=3-4/3.合并同类项,-5/2x=5/3.系数化为1,得x=-2/3(2)移项,得0.5x+1.3x=6.5+0.7.合并同类项,得1.8x=7.2.系数化为1,得x=4(3)去括号,得1/2x-1=2/5x-3.移项,得1/2x-2/5x=-3+1.合并同类项,得1/10x=-2.系数化为1,得x=-20(4)去分母,得7(1-2x)=3(3x+1)-63.去括号,得7-14x=9x+3-63.移项、合并同类项,得-23x=-67.系数化为1,得x=67/23复习题3第3题答案(1)根据题意得:x-(x-1)/3=7+(x+3)/5去分母得:15x-5(x-1)=105-3(x+3)去括号得:15x-5x+5=105-3x-9移项、合并同类项,得13x=91系数化为1,得x=7∴当x=7时,x-(x-1)/3的值与7-(x+3)/5的值相等(2)根据题意得:2/5x+(-1)/2=(3(x-1))/2-8/5x,去分母(方程两边同乘10)得:4x+5(x-1)=15(x-1)-16x去括号得:4x+5x-5=15x-15-16x移项得:4x+5x-15x+16x=-15+5合并同类项,得10x=-10系数化为1,得x=-1复习题3第4题答案解:梯形面积公式s=1/2(n+6)h(1)当S=30,a=6,h=4时,30=1/2(6+b)×4去括号,得12十2b=30移项、合并同类项,得2b=18系数化为1,得b=9(2)当S=60,b=4,h=12时,60=1/2(a+4)×12,去括号,得6a+24=60移项、合并同类项,得6a=36系数化为1,得a=6(3)当S=50,a=6,b=5/3a时,b=5/3a=5/3×6=10.50=1/2(6+10)×h去括号,得8h=50系数化为1,得h=25/4复习题3第5题答案解:设快马x天可以追上慢马,根据题意得:240x=150(12+x),解得x=20.。

大学计算机教材课后习题参考答案新版1-4_章

大学计算机教材课后习题参考答案新版1-4_章

《Visual Basic程序设计基础》教材习题第1章 Visual Basic 6.0程序设计概述习题一、判断题参考答案:1.√2.√3.√4.×5.√6.×7.√8.×9.√10.√11.×12.√13.×14.√15.√二、选择题参考答案:1. C2. B 3.D 4.C 5.B 6.A 7.D 8.A 9.B 10.D三、填空题参考答案:1.对象、事件2.属性、方法、事件3.控件、屏幕(Screen)4.Left 5. Top 6.属性、<对象名>.<属性名>=<表达式>7.网格8.F o rm19.Load或Initialize或Resize或Activate(自动执行的先后顺序也按此排列)10.Activate、Deactivate四、程序设计题参考代码:程序1.界面设计参看表1。

表1 各控件的主要属性设置过程设计如下:Private Sub Command1_Click()Label2.Visible = False ‘使控件Label2看不见Label1.Caption = Text1.Text + ": 欢迎使用" + Label1.Caption Text1.Visible = False ‘使控件Text1看不见Command1.Enabled = False ‘使控件Command1不能用Command2.Enabled = True ‘使控件Command2能用End SubPrivate Sub Command2_Click()End ‘结束程序运行End Sub程序2.界面设计,过程设计如下:Private Sub Command1_Click() '字体变大Form1.FontSize = Form1.FontSize + 3Form1.Cls: Print "欢迎使用VB"End SubPrivate Sub Command2_Click() '字体变小Form1.FontSize = Form1.FontSize - 3Form1.ClsPrint "欢迎使用VB"End SubPrivate Sub Command3_Click() '字体加粗Form1.FontBold = TrueForm1.Cls: Print "欢迎使用VB"End SubPrivate Sub Command4_Click() '字体不加粗Form1.FontBold = FalseForm1.Cls: Print "欢迎使用VB"End Sub运行时,不得连续、过多次单击“字体变小”按钮,因为当FontSize属性值小于0会产生适时错误。

课后习题答案.doc

课后习题答案.doc

,轮 2 与轮 1 的转向相反。
对运
自由度为零,为一刚性桁架,所以构件之间不能产生相
动。
( 2)图 b 中的 CD 杆是虚约束,去掉与否不影响机构的运动。故图
b 中机构的自由度为:
所以构件之间能产生相对运动。
谢谢聆听
1-13 解 该导杆机构的全部瞬心如图所示,构件
1 、3 的角速比为:
1-14 解 该正切机构的全部瞬心如图所示,构件
3 的速度为:


谢谢聆听
要求轮 1 与轮 2 的角速度之比,首先确定轮 1 、轮 2 和机架 4 三个构件的三个瞬心,即


,如图所示。则:
1-16 解 ( 1 )图 a 中的构件组合的自由度为:
1-1 至 1-4 解 机构运动简图如下图所示。
第 1 章 课后习题答案
谢谢聆听
图 1.11 题 1-1 解图
图 1.12 题 1-2 解图
1-5 解 1-6 解 1-7 解 1-8 解 1-9 解 1-10 解 1-11 解 1-12 解
图 1.13 题 1-3 解图
图 1.14 题 1-4 解图
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《Visual Basic程序设计基础》教材习题第1章Visual Basic 6.0程序设计概述习题一、判断题参考答案:1.√2.√3.√4.×5.√6.×7.√8.×9.√10.√11.×12.√13.×14.√15.√二、选择题参考答案:1. C2. B 3.D 4.C 5.B 6.A 7.D 8.A 9.B 10.D三、填空题参考答案:1.对象、事件2.属性、方法、事件3.控件、屏幕(Screen)4.Left 5. Top 6.属性、<对象名>.<属性名>=<表达式>7.网格8.F o rm19.Load或Initialize或Resize或Activate(自动执行的先后顺序也按此排列)10.Activate、Deactivate四、程序设计题参考代码:程序1.界面设计参看表1。

表1 各控件的主要属性设置过程设计如下:Private Sub Command1_Click()Label2.Visible = False ‘使控件Label2看不见Label1.Caption = Text1.Text + ": 欢迎使用" + Label1.Caption Text1.Visible = False ‘使控件Text1看不见Command1.Enabled = False ‘使控件Command1不能用Command2.Enabled = True ‘使控件Command2能用End SubPrivate Sub Command2_Click()End ‘结束程序运行End Sub程序2.界面设计,过程设计如下:Private Sub Command1_Click() '字体变大Form1.FontSize = Form1.FontSize + 3Form1.Cls: Print "欢迎使用VB"End SubPrivate Sub Command2_Click() '字体变小Form1.FontSize = Form1.FontSize - 3Form1.ClsPrint "欢迎使用VB"End SubPrivate Sub Command3_Click() '字体加粗Form1.FontBold = TrueForm1.Cls: Print "欢迎使用VB"End SubPrivate Sub Command4_Click() '字体不加粗Form1.FontBold = FalseForm1.Cls: Print "欢迎使用VB"End Sub运行时,不得连续、过多次单击“字体变小”按钮,因为当FontSize属性值小于0会产生适时错误。

程序3.界面设计运行效果图过程设计如下(窗体Height值的计量包括标题、边框,而ScaleHeight值仅与窗体有效绘图区域的高度有关):Private Sub Form_Load()Text1.Text = "Visual Basic 程序设计"End SubPrivate Sub Form_Resize()Text1.Top = 0: Text1.Left = 0Text1.Width = Form1.ScaleWidth/2 '用Form1.ScaleWidth效果优于用Form1. WidthText1.Height = Form1.ScaleHeight/2 '用Form1.ScaleHeight优于用Form1. HeightCommand1.Top = Form1.ScaleHeight - Command1.HeightCommand1.Left = Form1.ScaleWidth - Command1.WidthEnd SubPrivate Sub Command1_Click()EndEnd Sub第2章程序设计基础习题一、判断题1.√2.×3.×4.×5.√6.×7.√8.√9.√10.×二、选择题1.B 2.D 3.A 4.C 5.C 6.B 7.A 8.D 9. C 10.A三、填空题参考答案1.String 2.4 3."aaa" 4.145 5.6 6.3 7.18 8.False9.(x Mod 10) * 10 + x \ 10 10.10 + Int( Rnd * 90 ) 11.a * b Mod c12.Log(x)+Sin(3.141593/6) 13.Const PI=3.1415926 14.日期15.Int(x)+1四、程序设计题参考答案程序1、Private Sub Command1_Click()Dim x As Double, y#, s#x = Val(InputBox("请输入x"))y = Val(InputBox("请输入y"))s = Sqr( (x ^ 3 + Exp(-6) * Log(y)) * (Sin(x)* Cos(y)) / (x * x + y * y) _+ (2 * Sin(90 * 3.14156 / 180) + 2 * x * Exp(y)) / Sqr(Abs(x * y)))Print "当x="; x; "当y="; y; "时,计算结果:"; sEnd Sub程序2、Private Sub Command1_Click()Dim s As Double, l As DoubleDim r As DoubleConst pi = 3.1415926r = Val(Text1.Text)l = 2 * 3.1415926 * rs = r * r * 3.1415926Label2.Caption = Str(s)Label5.Caption = Str(l)End Sub'删除文本框后Private Sub Command2_Click()Dim s As Double, l As DoubleDim r As DoubleConst pi = 3.1415926r = Val(InputBox("请输入半径:"))l = 2 * 3.1415926 * rs = r * r * 3.1415926Label2.Caption = Str(s)Label5.Caption = Str(l)End Sub程序3、Private Sub Command1_MouseMove(Button As Integer, Shift As Integer, X As Single, Y As Single)Command1.Move Int(Rnd * (Form1.ScaleWidth - Command1.Width)), Int(Rnd * (Form1.ScaleHeight - Command1.Height))End Sub程序4、Private Sub Command1_Click()Command1.Move Int(Rnd * (Form1.ScaleWidth - Command1.Width)), Int(Rnd * (Form1.ScaleHeight - Command1.Height))End Sub程序5、Private Sub Command1_Click()Text1.Text = Int(Rnd * (6)) + 1Text2.Text = Int(Rnd * (6)) + 1Text3.Text = Val(Text1.Text) + Val(Text2.Text)Form1.BackColor = RGB(Int(Rnd * 255) + 1, Int(255 * Rnd) + 1, Int(255 * Rnd) + 1)End Sub第3章结构化程序设计与数组习题一、判断题参考答案1.√2.×3.√4.×5.√6.×7.×8.√9.×10.×二、选择题参考答案1.A 2.A 3.D 4.C 5.C 6.D 7. D 8.B(提示:因为If m Mod i=0 Then Exit For :Next i 在同一行上,系统认为是if语句)三、填空题参考答案1.If x>y Then t=y: y=x: x=t 2.IS3.For i = 0 to 9: B(i) = InputBox("B(" & i & ")="): Next i4.ReDim c(n) as single 5.String 6.8 (4个元素,每个元素2个字节)四、程序阅读题(写出下列程序的运行结果)参考答案:程序1. 55 11 程序2. 20 程序3. 2 4 7 11 16程序4. 1 4 9 16 25程序5. 程序6.五、程序填空题参考答案:1.(1)s=1 (2)n (3)s=s+p2.(1)While r<>0 (2)r=m Mod n (3) print n3.(1)v = 0 (2)v=v+a(i) (3) v=v/n (4) a(i)>v4.(1)m (2)n (3) exit for (4) j<=n5.(1)k=i (2)k=j (3)a(k)=t (4)a(i) Mod 2 = 16.(1)j+i-1 (2)6-i+1 (3) print六、程序设计题参考答案:程序1.Private Sub Form_Click()Dim x As Integer, y As Integer, z As IntegerDim max As Integer, midd As Integer, min As Integerx = InputBox("请输入x的值:")y = InputBox("请输入y的值:")z = InputBox("请输入z的值:")If x < y Then t = x: x = y: y = tIf x < z Then t = x: x = z: z = tIf y < z Then t = y: y = z: z = tPrint x; y; zEnd Sub程序2.界面设计略,过程设计如下:Private Sub Form_Click()Dim x As Single, y As Singlex = InputBox("请输入x的值:")If x > 3 Theny = x + 3ElseIf x >= 1 Theny = x * xElseIf x > 0 Theny = Sqr(x)Elsey = 0End IfPrint yEnd Sub程序3.Private Sub Form_Click()Dim i As Byte, j As ByteFor i = 1 To 9For j = 1 To iPrint Tab(8 * (j - 1)); Trim(Str(i)); "*"; Trim(Str(j)); "="; Trim(Str(i * j)); Next jPrintNext iEnd Sub程序4.Private Sub Form_Click()Dim x As Integer, n As Integer, i As Integer, p As Single, s As Single n = Val(InputBox("请输入n"))x = Val(InputBox("请输入x"))s = 1 + x: p = 1For i = 2 To (n + 1) Step 1p = p * x / is = s + pNext iPrint sEnd Sub程序5.Private Sub Form_Click()Dim e As Single, a As Single, n As Integera = 1 '级数第1项为1e = 1 '将第1项存入en = 0 '变量n记录当前已累加的项数While a > (0.0001)n = n + 1: a = a / n: e = e + aWendPrint "e="; e; "n="; nEnd Sub程序6.方法一Public Sub Form_Click()'同构数是指一个数出现在它的平方数的右端'如25在25平方625的右端,则25为同构数'利用数字转字符再取出右端字符的方法进行判断Dim i As LongDim x1 As String, x2 As StringFor i = 1 To 1000x1 = Trim(Str(i)) '将i转字符型x2 = Trim(Str(i ^ 2)) '将i^2转字符型If Right(x2, Len(x1)) = x1 ThenForm1.Print i; "是同构数"End IfNext iEnd Sub方法二Private Sub Form_Click()Dim i&, ws As IntegerFor i = 1 To 1000ws = Len(Str(i)) - 1If (i * i - i) Mod 10 & ^ ws = 0 Then Print i; NextEnd Sub程序7.Private Sub Form_Click()Dim i%, j%, s As Double, x(10) As Single, y(10) As Single RandomizeFor i = 1 To 10 '生成10个坐标点x(i) = Int(Rnd * 90) + 10Print "(" & x(i) & ",";y(i) = Int(Rnd * 90) + 10Print y(i) & ")";NextFor i = 1 To 9 '按公式计算For j = i + 1 To 10s = s + Sqr((x(i) - x(j)) ^ 2 + (y(i) - y(j)) ^ 2)NextNextPrintPrint "各点距离之和:" & sEnd Sub程序8.Private Sub Form_Click()Dim n As Integer, m As Integer, i As Integer, j As Integerm = Val(InputBox("请输入a数组元素个数: "))n = Val(InputBox("请输入b数组元素个数: "))ReDim a(m) As Integer, b(n) As IntegerFor i = 1 To ma(i) = Val(InputBox("a(" & i & ")="))Next iFor i = 1 To nb(i) = Val(InputBox("b(" & i & ")="))Next iFor i = 1 To mFor j = 1 To nIf a(i) = b(j) Then Exit ForNext jIf j > n Then Print a(i);Next iFor j = 1 To nFor i = 1 To mIf a(i) = b(j) Then Exit ForNext iIf i > m Then Print b(j);Next jEnd Sub程序9.Private Sub Form_Click()Dim a(5, 5) As Integer, i As Integer, j As Integer For i = 1 To 5For j = 1 To 5 - i + 1a(i, j) = 7 - i - jNext jNext iFor i = 2 To 5For j = 5 - i + 1 To 5a(i, j) = j + i - 5Next jNext i'打印二维数组For i = 1 To 5For j = 1 To 5Print a(i, j);Next jPrintNext iEnd Sub程序10.(1)Option Base 1Private Sub Form_Click()Dim n As Integer, i As Integer, j As Integer, k As Integer, c%() n = Val(InputBox("请输入行数n:"))n = n + 1ReDim c(n, n)For i = 1 To nc(i, 1) = 1c(i, i) = 1Next iFor i = 3 To nFor j = 2 To i - 1c(i, j) = c(i - 1, j - 1) + c(i - 1, j)Next jNext iFor i = 1 To nFor j = 1 To iPrint c(i, j);Next jPrintNext iEnd Sub(2)Private Sub Form_Click()Dim n As Integer, i As Integer, j As Integer, k As Integer, c%() Do ‘假设n范围在1-9n = Val(InputBox("请输入行数n:"))Loop Until n > 0 And n < 10n = n + 1ReDim c(n, n)'产生杨辉三角形For i = 1 To nc(i, 1) = 1c(i, i) = 1Next iFor i = 3 To nFor j = 2 To i - 1c(i, j) = c(i - 1, j - 1) + c(i - 1, j)Next jNext i'打印杨辉三角形,每个数字占4列For i = 1 To n '打印n 行Print Tab((4 * n - 4 * i) / 2); '确定打印起始列For j = 1 To i '打印i 个数字Print Space(4 - Len(Trim(Str(c(i, j))))) & Trim(Str(c(i, j)));Next jPrintNext iEnd Sub第4章函数与过程习题一、判断题1.×2.√3.×4.×5.√6.×7.×8.√9. √10. √二、填空题参考答案1.按地址传送2.b() As Long 3.6 4.按值传递5.按地址传递6.Public x As Single 7.Static x As Integer 8.Form2.y9. function f9(a() as single,n) as single 10.sub f10(a() as single ,n as integer)11. sub f10(a() as single ,m as integer,n as integer ,max as single,min as single)12. c=form1.f12(a,b)三、程序阅读题(写出下列程序的运行结果)程序1. s = 2 程序2. command1 command2s = 5 1 1s = 9 1 21 31 4程序3. Byref时分别输出0 101 ,0 110 ,0 1011 ,0 10001(注意:语句Print n; f(n)在一行上结果如上)Byval时分别输出5 101 ,6 110 ,11 1011 ,17 10001程序4. 分别输出:0 3 2 3程序5. 结果(杨辉三角形)四、程序填空题参考答案:1.(1)ByeVal (2)n Mod k=0 (3)n = n \ i (4)Call pp(i) 2.(1)as string (2)n>0 (3)f16=chr(65+k)+f16 (4)n=n\16 3.(1)a() As Double, n As Integer (2)t = t * x (3)f = s4. (1)a() Aa Double,n as integer (2)n-1 (3)a(j) < a(k)五、程序设计题参考答案程序1.Function f1(x%, y%, z%) As IntegerDim max As Integermax = xIf max < y Then max = yIf max < z Then max = zf1 = maxEnd Function‘验证Private Sub Command1_Click()Dim a%, b%, c%a = 40:b = 120:c = 30Print f1(a, b, c)End Sub程序2.Private Function fsum(x() As Double, n As Integer) As DoubleDim i As IntegerFsum=0For i = 1 To nfsum = fsum + x(i)Next ifsum = fsum / nEnd Function程序3.界面设计略,过程设计如下:Private Sub ff(a() As Single, n As Integer)Dim i As Integer, x As SingleFor i = 1 To n \ 2x = a(n - i + 1): a(n - i + 1) = a(i): a(i) = x‘(a(1)⇔a(n),a(2)⇔a(n-1),...,a(i)⇔a(n-i+1))Next iEnd Sub程序4.Private Sub find(x() As Single,m As byte,n As Byte,ki As Byte,kj As Byte)xmax = x(1,1)ki = 1: kj = 1For i = 1 To mFor j = 1 To nIf Abs(x(i, j)) > Abs(xmax) Then xmax = x(i, j): ki = i: kj = j Next j, iEnd Sub程序5.Option ExplicitPublic Function g1(a%(), n%) As DoubleDim i%For i = 1 To ng1 = g1 + a(i)Nextg1 = g1 / nEnd FunctionPublic Function g2(a%(), n%) As Double Dim i%, t#t = g1(a, n)For i = 1 To ng2 = g2 + (a(i) - t) ^ 2Next ig2 = Sqr(g2) / nEnd Function'调用Private Sub Command1_Click()Dim x%(10), g#, i%For i = 1 To 10x(i) = iNext iPrint g2(x, 10) '计算g2End Sub。

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