2010-2011学年下期郑州四十七中学区七八年级期中联合考
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河南省郑州市第四十七中学2010-2011学年高一第二次月考(历史)试题本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,共100分,考试时间50分钟。
第Ⅰ卷(选择题,共60分)一、选择题(共30小题,每小题2分,共计60分。
在每小题给出的四个选项中,只有一项是最符合题目要求的)1.下面是某学校高一年级同学对分封制、宗法制的认识,请你对此判断正误:①甲同学认为分封制是把包括镐京、洛邑在内的广大地区土地和人民授予王族、功臣和古代帝王的后代②乙同学认为从“礼乐征伐自天子出”这一现象到“礼乐征伐自诸侯出”,反映了分封制的瓦解③丙同学认为宗法制的核心是嫡长子继承制④丁同学认为在周王室中周王正妻所生之子一定能够成为大宗A.①②正确③④错误B.①③正确②④错误C.②③正确①④错误D.①④正确②③错误2.《姓氏起源》一书对“宋”姓起源的解释:周武王克商灭纣,建立周朝,封微子启(商王后裔)于商丘,建立宋国,共传36代,亡于楚国。
宋亡国后,原王公之族散居各地,以原国“宋”为姓,乃成宋姓。
由此不能得出的历史信息是()A.周朝实行分封制和宗法制B.姓氏是维护中央集权统治的有力工具C.诸侯争霸是宋亡国的主要原因D.反映了我国古代早期政治制度的基本特点3.亚里士多德在《雅典政制》中说,雅典议事会的成员由400人改为500人,每(地区)部落出50人,而在以前,每(血缘)部落则出100人。
上述变化发生于A.梭伦改革前B.梭伦改革时期C.克利斯提尼改革时期D.伯利克里任首席将军期间4.假如你是古罗马一位执政官,当你遇到以下案子时,你判为合法的是①主人拒绝为他工作了20年而要求给予自由的奴隶②一个夜里行窃的人被人当场杀死,但杀人者自称其行为合法③某平民将他大儿子财产的一部分分给其小儿子。
④拥有100个奴隶的罗马贵族,在急需劳动力时为了赚钱卖了20个奴隶A.①②④B.①②③C.①③④D.②③④5.一位与伯利克里同时代的人曾经自豪地说:“假如你未见过雅典,你是个笨蛋;假如你见到雅典而不狂喜,你是一头蠢驴;假如你自愿把雅典抛弃,你是一头骆驼。
河南省郑州市第47中学10-11学年高一下学期第二次月考(语文)一、本大题共7小题,每题3分,共21分。
1.下列词语中加点的字,注音有两处错误的的一项是( )A.窥伺(cì)延宕(dàng)撚断(niǎn)繁文缛节(rù)B.撺掇(cuān)连累(leì)盗跖(zhì)自怨自艾(aì)C.罪愆(qiān)宿愿(sù)暴虐(nuè)前合后偃(yàn)D.披靡(mí)揾泪(wèn)倨傲(jū)献愁供恨(gòng)2.下列词语字形无误的一项是()A.寒暄烦躁各行其事良辰美景B.羸弱规距翻云覆雨出奇制胜C.暮霭摈弃欢度春节责无旁待D.天谴肉袒如法炮制莼羹鲈脍3.下列句子中,加点的成语使用不恰当的一项是()A.长期以来,一些商业电视广告“打造优等生”“不能让孩子输在起跑线上”的蛊惑之词不绝于耳....,对一些家长的教育观念产生了负面影响。
B.新修订的《老年人权益保障法》增加了给予老年人生活关照和精神慰藉的内容,这对那些虐待老人的不肖子孙....起到了震慑作用。
C.中华文理学院中文系孔立远教授博览群书、学养深厚、才气横溢,他的诗文犹如天马行空....,令他的学生十分佩服。
D.肖仁福的小说《权规则》是一本表现权利背景下各色人等心态的作品,对官场上尔虞我诈现象的讽刺批判入木三分....。
4.下列句子中,没有语病、语意明确的一项是()A.京剧是融唱、念、做、打于一体的戏剧表演艺术,表现了中国传统的戏剧美学理想,它是人类共同的宝贵文化遗产。
B.畜牧局下发通知,要排查并打击在饲料原料和产品中添加有害物质及在畜禽饲养、贩运过程中使用“瘦肉精”等违禁药物。
C.甲状腺减低症简称“甲减”,发病率高达5%,但在人群中知晓率却很低,多数人不但没听说过这种病,甚至还很不了解这种病。
D.以色列军方一位发言人说,以色列军队几天前就开始占领了靠近以色列北部边境的两个黎巴嫩村庄中的阵地。
郑州市第四十七中2010—2011学年高一年级第二次月考试题英语命题人:樊淑婷一.单项填空(共30题;每小题1分,满分30分)从所给的四个选项中,选出可以填入空白处的最佳选项。
1. ––Do you mind my opening the window? It’s a bit hot in here.––_________, as a matter of fact.A. Go aheadB. Yes, my pleasureC. Yes, I doD. Come on2. I don’t think the experiment is_______ failure. At least we have gained ________ experience for future success.A. /; theB. a; theC. a; /D. /; /3. As the busiest woman in Norton, she made _____ her duty to look after all the other people’s affairs in that town.A. thisB. thatC. oneD. it4. I would walk to school every day _______ ride a bicycle.A. rather thanB. more thanC. other thanD. less than5. ––I’d like to go to the cinema with you, Dad.––Sorry, my darling, but the film is ______ for adults only.A. promisedB. permittedC. admittedD. designed6. They are letting us use their computers, and _______ we are giving them the results of our research.A. in turnB. in theoryC. in returnD. in general7. Mary never does her reading in the evening, __________.A. so does JohnB. John does tooC. John doesn’t tooD. nor does John8. Jenny hopes that Mr. Smith will suggest a good way to have her written English ________ in a short period.A. improvedB. improvingC. to improveD. improve9. ––Is the meeting held in Room 302 or 303?––It should be 302. But I hear that it ______ till tomorrow.A. was put offB. will put offC. has been put offD. is put off10. A new _______ bus service to the railway station started some days ago.A. normalB. usualC. commonD. regular11. Do you remember the chicken farm ________ we visited three months ago?A. whereB. whenC. thatD. what12. My sister always tells me her __________ opinions and forces me to live in her way.A. publicB. personalC. singleD. similar13. It’s easy to take a machine ________ but difficult to put it together.A. awayB. apartC. outD. off14. She brought with her three friends, none of ______ I had ever met before.A. themB. whoC. whomD. these15. —Can I help you?—I’d like a room with a bath. How much do you ______?A. offerB. affordC. chargeD. want16. Peter was so excited ______ he received an invitation from his friend to visit Chongqing.A. whereB. thatC. whyD. when17. Why are you so anxious? It isn’t your problem ________.A. on purposeB. in allC. on timeD. after all18. You _______ Bob at this meeting yesterday; he has been on holiday in Paris for a week.A. can’t have seenB. shouldn’t seeC. mustn’t have beenD. couldn’t see19. ––Why is Tom so worried?––Because he doesn’t know ________ his shop.A. what to deal withB. how to do withC. what can he do withD. how to deal with20. ______ is known to everybody that Chinese football has to improve in the future.A. ItB. AsC. ThatD. What21. ––Do you think it’s a good film?––Oh, yes. It’s ______worth ______.A. very; watchingB. well; seeingC. very; to be watchedD. well; being seen22. She watches TV _______ evening and changes channels(频道) ____ few minutes.A. in; everyB. every; everyC. every; eachD. every; in23. Lucy advised him not to buy the knife, but he bought it _______.A. anyhowB. howeverC. thoughD. totally24. I’d like to study law at university, _____ my cousin prefer s geography.A. thoughB. asC. whileD. for25. ––Haven’t you solved the problem yet?––Sorry, but I promise it _________ in an hour.A. will be settledB. will settleC. has been settledD. has settled26. Mary is a much younger teacher but she has the great _________ of knowledge.A. interestB. successC. honorD. advantage27. There is no doubt _______ John will come on time.A. ifB. whyC. thatD. whether28. It was in the office _________ my sister works ______ I met the new manager.A. that; thatB. where; thatC. where; whereD. that; where29. The class club has decided to send me to Korea to _________ a friendly chess game.A. make use ofB. go throughC. take part inD. come up30. ––How about your journey to Shanghai?––Every thing was wonderful except that our car _______ twice on the way.A. slowed downB. broke downC. got downD. put down二.完形填空(共20题;每小题1.5分,满分30分)阅读下列短文,从每题所给的四个选项中,选出最佳选项,并在答题卡上将该项涂黑。
2023-2024学年河南省郑州四十七中八年级(下)期中物理试卷一、单选题:本大题共5小题,共10分。
1.下列有关运动和力的说法中,正确的是()A.力是维持物体运动的原因B.物体受到平衡力作用,它一定是静止的C.力是改变物体运动状态的原因D.物体运动的速度越大,受到的力也越大2.下列有关事例的描述正确的是()A.三峡船闸利用了连通器的原理B.医生用注射器将药液注入到肌肉时利用了大气压强C.马德堡半球实验测出了大气压的值D.高原边防哨所战士用压力锅煮面条是因为气压越高,液体沸点越低3.如图所示为小明测量体重时的场景,下列属于平衡力的是()A.小明对体重计的压力与体重计对他的支持力B.小明的重力与他对体重计的压力C.体重计对小明的支持力与小明的重力D.体重计对小明的支持力与地面对体重计的支持力4.校运动会爬杆比赛,小明第一个爬到杆顶。
如图所示,小明紧紧握住杆子保持静止,此时他受到的摩擦力为;片刻后,小明适度减小握杆子的力量使自己匀速滑下,此时他受到摩擦力为,则()A.、均竖直向上,且B.、均竖直向上,且C.竖直向上,竖直向下,且D.竖直向上,竖直向下,且5.如图所示,是我国自主研发的C919商用大飞机。
关于客机的相关物理知识,下列说法不正确的是()A.客机所在高空的大气压强比海平面附近的大气压强小B.客机采用密度小的材料,可以减轻它的重力C.飞机在跑道上加速滑行时惯性逐渐增大D.机翼被设计成“上凸下平”的形状,利用空气流速越大压强越小的原理获得升力二、多选题:本大题共3小题,共6分。
6.如图所示是“验证阿基米德原理”实验时的情形。
下列说法正确的是()A.实验最合理的顺序是甲、乙、丙、丁B.重物受到的浮力可表示为C.溢水杯中装满水的目的是使重物排开的水多于排出到烧杯中的水D.若,则可验证阿基米德原理7.在生物实验操作考试中,学生正在制作“黄瓜表层果肉细胞临时装片”,下列关于在此制作过程中涉及的物理知识说法正确的是()A.用胶头滴管从试剂瓶中吸取清水,利用了大气压B.用单面刀片刮取黄瓜表层果肉,刀片做的锋利是为了增大压力C.手拿镊子夹取盖玻片时,盖玻片受到的力的施力物体是手D.当用纱布向左擦拭载玻片时,纱布受到玻璃片对它向右的摩擦力8.放在水平桌面上的甲、乙两个相同容器中,盛有同种液体,体积相等的a、b两个物体在液体中静止时,两液面相平,如图所示,则下列说法正确的是()A.物体的密度B.物体所受浮力C.容器底部所受液体压力D.容器对桌面压强三、填空题:本大题共6小题,共14分。
学必求其心得,业必贵于专精郑州市第47中学2010-2011学年下期高一年级第一次月考试题英语命题人:刘红本试题卷分选择题和非选择题两部分,共8页,六大题,满分120分,考试时间100分钟。
选择题(共四大题,满分75分)一、单项选择(共15题每小题1分,满分15分)1. My neighour asked me to go for ________ walk, but I don’t think I've got _________ enery.A. a,,/B. the ,the C。
/ ,theD. a, the2. Why are you so anxious?It isn’t your problem 。
A. on purpose B。
in all C. on timeD. after all3.The country ,_______ cherry tree flowers ,looks _____ it is a sea of pink snow .A. covers with… as though B。
covered with… as though C. covering with…as if D。
is covered with …。
as though4. He had a wonderful childhood,_____ with his mother to all corners of the world。
A. travelB. to travelC. traveling D。
traveled5.。
It took me a long time before I was able to fully appreciate what they _____ for me。
A. had done B。
did C. would do D。
were doing学必求其心得,业必贵于专精6。
河南省郑州市第四十七中学2024届英语八年级第二学期期末经典模拟试题满分120分,时间90分钟一、完形填空(10分)1、阅读下面短文,从短文后所给的A、B、C、D 四个选项中选出能填入相应空白处的最佳选项,并在答题卡上将相应字母编号涂黑。
I once volunteered (做志愿) at the pet rescue centre. A dog named Minnie caught my eye. Its body ____1____ a sausage (香肠). Its hair was too thin to cover its body. Its eyes were so ____2____ that I didn’t know whether they were open or closed.It looked so strange that its owner (主人) didn’t want it, so it was left in the centre. But I thought Minnie was ____3____ and sweet. No one should laugh at its look. Every ____4____ should be loved by people. I gave Minnie a good bath. Then I put its pictures on the Internet and wrote, “Funny-looking dog, kind and sweet, needs a loving ____5____.” Later, a boy called me. He told me that his grandfather wanted to ____6____ Minnie very much.The next day, an old car came. Two kids got off the car and ran to me ____7____. They put Minnie into their arms and rushed to their grandfather excitedly. The grandfather put Minnie in his arms and ____8____ its ha ir gently with his hands. “It’s perfect, and I like it!” the old man said. I was thankful that Minnie had found a good owner. When they were going to leave, I saw the two kids helping the grandfather into the car. At that time, I knew the old man couldn’t see anything—he was ____9____.One day, I missed Minnie so much. So I visited it. The grandfather was walking in the garden with Minnie _____10_____ the way. Both of them were happy.1. A. looked after B. looked for C. looked at D. looked like2. A. big B. small C. perfect D. beautiful3. A. interesting B. brave C. clever D. famous4. A. bird B. plant C. animal D. flower5. A. school B. family C. hospital D. zoo6. A. act B. clean C. keep D. sell7. A. angrily B. worriedly C. sadly D. happily8. A. cut B. cleaned C. touched D. washed9. A. hurt B. blind C. deaf D. sick10. A. leading B. building C. giving D. making二、阅读理解(40分)2、Jenny is a bright-eyed, pretty five-year-old girl. One day when she and her mother were checking out at the grocery store (杂货店), Jenny saw a plastic pearl necklace priced at $2.50. How Jenny loved it! Her mother bought it for her and she wore it everywhere.Jenny had a very loving daddy. One night Jenny’s father asked her, “Jenny, do you love me?”“Oh yes, Daddy, you know I love you,” Jenny said.“Well, then, give me your pearl necklace,” her father joked. “I love it, too.”“Oh! Daddy, not my pearl necklace!” Jenny said to her father, “But you can have Rosy, my toy bear. Or you can have Ribbons, my favorite toy horse.”The next morning, when Jenny’s father left home for his office, Jenny was sitting on the sofa and her lips were trembling (颤抖). “Here you are, Daddy.” she said. “Here is the pearl necklace you like.” She put it in her father’s hand.Dad took the pearl necklace with tears in his eyes. Because of her love for both the necklace and her father, how hard it was for her to make this final decision.That afternoon, Jenny’s father went back home very early. He took out a blue box from his pocket. In the box was a real, beautiful pearl necklace.“Jenny, here is the pearl necklace you like. I love you.”1. What was Jenny’s first neck lace made of?A. Plastic.B. Paper.C. Real pearl.2. Why didn’t Jenny want to give Daddy her pearl necklace?A. She wanted to give him her toy horse.B. She didn’t love her father.C. She loved it.3. How did the father feel when Jenny gave him the necklace?A. Surprised.B. Moved.C. Interested.4. What did Jenny’s father give her that afternoon?A. A real pearl necklace.B. A beautiful toy.C. A pencil box.3、In Singapore, many middle school students spend a lot of their time on their studies. Good education is often regarded asa ticket to success in their future. So, many of these students try their best to get a good mark (成绩) in their examinations. They have a lot of homework every day and exams are a big headache.Sometimes, some of them are even made to go to remedial (补习的) classes after school. Schools run programs outsideschool hours. The students can take part in sports and games, music and dance, hiking and rock-climbing, etc. They are also very active in community service (社区服务). In their spare time, most students like to listen to pop music. Hollywood blockbusters (大片), Hong Kong and Singapore movies are very popular among them. They understand IT very well. Some of them also spend their free time surfing the Internet, e-mailing their friends, playing computer and video games. They sometimes go to cafes, fast-food restaurants, shopping centers and big bookstores. So, it looks like life as a middle school student in Singapore is not easy but it is rich and colorful.1. Many students want to get good education to be ________ in the future.A. popularB. successfulC. comfortableD. generous2. ________ go to remedial classes after school in Singapore.A. All of the studentsB. Few of the studentsC. Not all the studentsD. Most of the students3. I n the passage, “spare” here means ________.A. freeB. workingC. busyD. favorite4. According to the passage, most students in Singapore like ________.A. Hollywood blockbustersB. writing to their friendsC. talking to their friendsD. having exams5. The meaning of the underlined sentence in the passage is ________.A. the middle school students’ life in Singapore is easyB. the middle school students’ life in Singapore is amazing (疯狂的)C. the middle school students’ life in Singapore is boringD. the middle school students’ life in Singapore is hard, but it’s interesting4、Wuyi Rock Tea is a kind of Oolong tea. It got its name from the place where it’s produced—Wuyi Mountain in Fujian Province. The tea trees usually grow on rock cracks(缝隙). The traditional home of Rock Tea is only from a small area in the mountain. Wuyi Rock Tea is famous for its rich smell and good flavour. There are different kinds. The best is Dahongpao.With its long history, Wuyi Rock Tea dates back to the Shang Dynasty. People used to send it to the emperor as a tribute (贡品). In 2006, the skills of making Wuyi Rock Tea were added to China’s national intangible cultural heritage list.Zhang Huichun, 65, is a master of making Wuyi Rock Tea. In 1980, he learned to make the tea and opened a tea factory. To his surprise, the skills were very complex. After 10 years of learning, he could finally make the tea by himself. Zhang said there were 13 steps to make it, including drying, shaking and roasting(焙火). The most challenging for him is shaking. It takes strength and patience to shake the tea leaves by holding the round bamboo trays. “You have to decide when the best time is to stop based on the color,” Zhang said. “Some times you stop after shaking dozens of(几十)times, while sometimes it can be hundreds of times. Only years of experience can help you figure it out(弄明白).”“The last step—roasting—is also important. With three stages, it usually takes more than 20 hours! Each stage needs temperatures from 125℃ to 155℃. Tea makers have to keep a close eye on the process. Every 50 minutes, you have to turn over the leaves, or it might burn,” Zhang said.Zhang is already a master, but he is still learning. “The most fascinating p art is that every step is full of changes—different tea trees, weather and temperatures. Every time is an adventure.”In recent years, Zhang has often worked with schools to train tea makers. “It’s my responsibility to pass on the traditional skills,” he s aid.1. Wuyi Rock Tea got its name from ________.A. a mountainB. rock cracksC. its smellD. tea trees2. The underlined word “complex” in paragraph 3 most probably means ________.A. specialB. difficultC. strangeD. clever3. Why is shaking the most challenging step according to Zhang Huichun?A. Tea makers need years of experience in order to tell when to stop shaking.B. Tea makers have to shake hundreds of times in order to get the right colour.C. Tea makers need to spend more than 20 hours shaking the tree leaves.D. Tea makers have to keep a close eye on the leaves, or it will burn.4. Which can be the best title of this passage?A. Wuyi Mountain is callingB. The history of the famous teaC. Wuyi Rock Tea—traditional treasureD. Schools try to pass on the traditional tea5、The summer sun is out, and we feel the heat. Before you go outside to have fun, make sure you put on the right clothes!Do you know what kind of clothes can make you look hot and keep you cool? Here are some ideas for clothes to help you look great.Fun colors and cartoons are back this year. People will look for bright colors this summer. Girls in pink look cute and sweet. And green makes everyone look more lively. Some of the best colors are pink, orange and green. These colors go well with jeans or just about anything else. As soon as you put on one of these colors, you’ll feel like a star!Some girls like something more interesting than all solid (单一的) colors. For them, there are lots of floral patterns (花卉图案) to choose from shirts, dresses or skirts. A floral shirt looks nice with a light-colored skirt or trousers.Of course, jeans are in for boys. They look great. But if you feel too hot, try something lighter, like khaki cargo pants. They’ll have enough pockets (口袋) to keep everything you need for studying, playing and keeping cool! Army green is the most popular this year.1. This passage mainly tells us ________.A. what color to choose for summer clothesB. how to make summer clothesC. right colors for boysD. how to have a fashion show2. The underlined word “in” in Para 5 probably means ________.A. 在里面B. 流行的C. 参加D. 喜欢3. Which of the following is True?A. Boys in pink and green look cute and sweet.B. A floral shirt looks nice with a light-colored skirt.C. Girls like all solid colors.D. Army green is out this year.4. Khaki cargo pants are special because they ________.A. go well with floral patterns in shirtsB. make people want to danceC. have enough pocketsD. are cheap5. Which of the following colors may not look good in summer?A. Orange.B. Pink.C. Army green.D. Black.三、补全对话(10分)6、根据下面对话内容,在空白处填上一个适当的句子,使对话通顺、完整。
河南省郑州市第47中学2010-2011学年高一下学期第一月考试题(数学)(共三大题,22道小题,满分150分,时间120分钟)一.选择题(每题只有一个正确答案,请把你认为正确的答案的题号写在答题卡相应位置,每题5分,共60分.) 1.613sin π的值为 ( )A.21-B.21C.23-D. 232. 对于函数y =sin(132π-x ),下面说法中正确的是 ( ) A. 函数是最小正周期为π的奇函数 B.函数是最小正周期为π的偶函数 C. 函数是最小正周期为2π的奇函数 D.函数是最小正周期为2π的偶函数3. .设是第二象限角,则sin cos αα=( )(A) 1 (B)tan 2α (C) - tan 2α (D) 1-4设A 是第三象限角,且|sin2A |= -sin 2A ,则2A是 ( ) A.第一象限角 B. 第二象限角 C.第三象限 D 第四象限角 5.x y tan ,2x 2=<<-函数时ππ的图像 ( )A.关于原点对称B.关于x 轴对称C.关于y 轴对称D.不是对称图形 6.若α满足sin 2cos 2sin 3cos αααα-=+,则sin cos αα的值等于( )A.±865B. 865C. - 865D 以上都不对7.将函数x y sin =的图象上的每个点的纵坐标不变,横坐标缩小为原来的21,然后沿y 轴正方向平移2个单位,再沿x 轴正方向平移6π个单位,得到( ) A .22sin +=x y B .232sin +⎪⎭⎫⎝⎛-=πx y C .232sin +⎪⎭⎫⎝⎛+=πx y D .262sin +⎪⎭⎫ ⎝⎛-=πx y8.若θ是△ABC 的一个内角,且81cos sin -=θθ,则θθcos sin -的值为( ) A .23-B .23C .25-D .25 9.()ϕω+=x A y sin 的图象的一段如图所示,它的解析式是( )A .⎪⎭⎫⎝⎛+=322sin 32πx y B .⎪⎭⎫⎝⎛+=32sin 32πx yC .⎪⎭⎫ ⎝⎛-=32sin 32πx y D .⎪⎭⎫ ⎝⎛+=42sin 32πx y 10.同时具有以下性质:“①最小正周期是π;②图象关于直线3x π=对称;③在,63ππ⎡⎤-⎢⎥⎣⎦上是增函数”的一个函数是( ) A.)62sin(y π+=x B.)32cos(π+=x y C.)62sin(π-=x y D.)62cos(π-=x y11.定义运算:⎩⎨⎧>≤=*.,,b a b ba ab a 例如121=*,则函数x x x f cos sin )(*=的值域为 ( )A .⎥⎦⎤⎢⎣⎡-22,1B .[]1,1-C .⎥⎦⎤⎢⎣⎡1,22D .⎥⎦⎤⎢⎣⎡-22,2212.已知函数)0(sin 2>=ωωx y 在区间⎥⎦⎤⎢⎣⎡-4,3ππ上的最小值是-2,则ω的最小值等于( ) A.32 .B 23C.2D.3二、填空题:(本大题共4小题,将答案写在横线上,每小题5分,共20分)13. 函数y=5tan(2x-)的最小正周期是__________________;14.= ____________________;15. 函数[]的单调递增区πππ2,2),321sin(y -∈+=x x 是________________________;16.关于函数f(x)=4sin(2x+π3) (x ∈R),有下列命题:(1)y=f(x )的表达式可改写为y=4cos (2x — π6);(2)y=f(x )是以2π为最小正周期的周期函数;(3)y=f(x)的图象关于点(— π6 ,0)对称;(4)y=f(x)的图象关于直线x= — π6对称;其中正确的命题序号是________________________.三.解答题(本大题共6小题,解答应写出必要文字说明.演算步骤或推证过程)17(10分).已知角α的终边经过点P(1,3), (1)求sin(π-α)-sin()2απ+的值; (2)写出角α的集合S.18.(12分)已知tanα=2,sinα+cosα<0,求.)cos()3sin()cos()sin()2(sin 的值απαπαπαπαπ+∙-+-∙+∙-19.(12分)已知曲线该上最高点为),2,2()2,0,0)(sin(y πϕωϕω≤>>+=A x A 最高点到相邻的最低点间曲线与x 轴交于一点(6,0),求函数解析式,并求函数在[]0,6-∈x 上的值域.20.(12分)(14分)已知函数)42(cos 21)(π-+=x x f 。
一、选择题1.(0分)[ID:9905]如图,在矩形ABCD中,AB=2,BC=3.若点E是边CD的中点,连接AE,过点B作BF⊥AE交AE于点F,则BF的长为()A.3102B.3105C.105D.3552.(0分)[ID:9900]如图,在菱形ABCD中,AB=6,∠ABC=60°,M为AD中点,P为对角线BD上一动点,连接PA和PM,则PA+PM的最小值是( )A.3 B.2√3C.3√3D.63.(0分)[ID:9892]正方形具有而菱形不具有的性质是()A.四边相等 B.四角相等C.对角线互相平分 D.对角线互相垂直4.(0分)[ID:9888]为了让市民享受到更多的优惠,相关部门拟确定一个折扣线,计划使50%左右的人获得折扣优惠.某市针对乘坐地铁的人群进行了调查.调查小组在各地铁站随机调查了该市1000人上一年乘坐地铁的月均花费(单位:元),绘制了频数分布直方图,如图所示.下列说法正确的是()①每人乘坐地铁的月均花费最集中的区域在80~100元范围内;②每人乘坐地铁的月均花费的平均数范围是40~60元范围内;③每人乘坐地铁的月均花费的中位数在60~100元范围内;④乘坐地铁的月均花费达到80元以上的人可以享受折扣.A.①②④B.①③④C.③④D.①②5.(0分)[ID:9867]如图,在矩形ABCD中,E,F分别是边AB,CD上的点,AE=CF,连接EF,BF,EF与对角线AC交于点O,且BE=BF,∠BEF=2∠BAC,FC=2,则AB的长为()A.83B.8C.43D.66.(0分)[ID:9860]有一个直角三角形的两边长分别为3和4,则第三边的长为()A.5B.7C.5D.5或77.(0分)[ID:9857]如图,矩形纸片ABCD,3AB=,点E在BC上,且AE EC=.若将纸片沿AE折叠,点B恰好落在AC上,则矩形ABCD的面积是()A.12B.63C.93D.158.(0分)[ID:9854]如图,已知圆柱底面的周长为4dm,圆柱的高为2dm,在圆柱的侧面上,过点A和点C嵌有一圈金属丝,则这圈金属丝的周长最小为()A.42dm B.22dm C.25dm D.45dm9.(0分)[ID:9850]如图,在菱形ABCD中,AB=5,对角线AC=6.若过点A作AE⊥BC,垂足为E,则AE的长为( )A.4B.2.4C.4.8D.510.(0分)[ID:9843]下列二次根式:3418,,125,0.4823-12合并的有()A.1个B.2个C.3个D.4个11.(0分)[ID:9916]如图,点E F G H、、、分别是四边形ABCD边AB、BC、CD、DA的中点.则下列说法:①若AC BD=,则四边形EFGH为矩形;②若AC BD⊥,则四边形EFGH为菱形;③若四边形EFGH是平行四边形,则AC与BD 互相平分;④若四边形EFGH是正方形,则AC与BD互相垂直且相等.其中正确的个数是( )A .1B .2C .3D .4 12.(0分)[ID :9898]下列结论中,矩形具有而菱形不一定具有的性质是( )A .内角和为360°B .对角线互相平分C .对角线相等D .对角线互相垂直 13.(0分)[ID :9869]如图,菱形ABCD 的对角线AC ,BD 相交于点O ,E ,F 分别是AB ,BC 边上的中点,连接EF.若3EF =,BD=4,则菱形ABCD 的周长为( )A .4B .46C .47D .28 14.(0分)[ID :9851]下列各组数据中,不可以构成直角三角形的是( ) A .7,24,25B .2223,4,5C .53,1,44D .1.5,2,2.5 15.(0分)[ID :9925]已知一次函数y =﹣x +m 和y =2x +n 的图象都经过A (﹣4,0),且与y 轴分别交于B 、C 两点,则△ABC 的面积为( )A .48B .36C .24D .18二、填空题16.(0分)[ID :10028]使二次根式1x -有意义的x 的取值范围是 _____.17.(0分)[ID :10025]如图,在矩形ABCD 中,2AB =,对角线AC ,BD 相交于点O ,AE 垂直平分OB 于点E ,则AD 的长为__________.18.(0分)[ID :10013]如图,点E 在正方形ABCD 的边AB 上,若1EB ,2EC =,那么正方形ABCD 的面积为_.19.(0分)[ID :10010]若由你选择一个喜欢的数值m ,使一次函数()2y m x m =-+的图象经过第一、二、四象限,则m 的值可以是___________.20.(0分)[ID :9988]如图,正方形ABCD 的边长为3,点E 在BC 上,且CE=1,P 是对角线AC 上的一个动点,则PB+PE 的最小值为______.21.(0分)[ID :9986]若菱形的两条对角线长分别是6㎝和8㎝,则该菱形的面积是 ㎝2.22.(0分)[ID :9984]如图,△ABC 中,∠ACB =90°,CD 是斜边上的高,AC =4,BC =3,则CD =______.23.(0分)[ID :9952]在△ABC 中,∠C=90°,AC=1,BC=2,则AB 边上的中线CD=______.24.(0分)[ID :9935]如图,在菱形ABCD 中,AC 与BD 相交于点O ,点P 是AB 的中点,PO =2,则菱形ABCD 的周长是_________.25.(0分)[ID :10011]将一个矩形 纸片折叠成如图所示的图形,若∠ABC=26°,则∠ACD=____.三、解答题26.(0分)[ID :10098]星期五小颖放学步行从学校回家,当她走了一段路后,想起要去买彩笔做画报,于是原路返回到刚经过的文具用品店,买到彩笔后继续往家走.如图是她离家的距离与所用时间的关系示意图,请根据图中提供的信息回答下列问题:(1)小颖家与学校的距离是 米;(2)AB 表示的实际意义是 ;(3)小颖本次从学校回家的整个过程中,走的路程是多少米?(4)买到彩笔后,小颖从文具用品店回到家步行的速度是多少米/分?27.(0分)[ID :10076]如图平面直角坐标系中,已知三点 A (0,7),B (8,1),C (x ,0)且 0<x <8.(1)求线段 AB 的长;(2)请用含 x 的代数式表示 AC+BC 的值; (3)求 AC+BC 的最小值.28.(0分)[ID :10050]观察下列各式及验证过程:11122323-=211121223232323-===⨯⨯ 1111323438⎛⎫-= ⎪⎝⎭2111131323423423438⎛⎫-=== ⎪⨯⨯⨯⨯⎝⎭ 11114345415⎛⎫-= ⎪⎝⎭21111414345345345415⎛⎫-=== ⎪⨯⨯⨯⨯⎝⎭(1)按照上述三个等式及其验证过程中的基本思想,猜想111456⎛⎫-⎪⎝⎭的变形结果并进行验证.(2)针对上述各式反映的规律,写出用n(n为自然数,且n≥2)表示的等式,不需要证明.29.(0分)[ID:10045]某学校为改善办学条件,计划采购A、B两种型号的空调,已知采购3台A型空调和2台B型空调,需费用39000元;4台A型空调比5台B型空调的费用多6000元.(1)求A型空调和B型空调每台各需多少元;(2)若学校计划采购A、B两种型号空调共30台,且A型空调的台数不少于B型空调的一半,两种型号空调的采购总费用不超过217000元,该校共有哪几种采购方案?(3)在(2)的条件下,采用哪一种采购方案可使总费用最低,最低费用是多少元?30.(0分)[ID:10043]一天李师傅骑车上班途中因车发生故障,修车耽误了一段时间后继续骑行,按时赶到了单位,如图描述了他上班途中的情景,回答下列问题:(1)李师傅修车用了多时间;(2)修车后李师傅骑车速度是修车前的几倍.【参考答案】2016-2017年度第*次考试试卷参考答案**科目模拟测试一、选择题1.B2.C3.B4.C5.D6.D7.C8.A9.C10.B11.A12.C13.C14.B15.C二、填空题16.x≤1【解析】由题意得:1-x≥0解得x≤1故答案为x≤1点睛:二次根式有意义的条件是:a≥017.【解析】【分析】由矩形的性质和线段垂直平分线的性质证出OA=OB=AB=2得出BD=2OB=4由勾股定理求出AD即可【详解】解:∵四边形ABCD是矩形∴OB=ODOA=OCAC=BD∴OA=OB∵A18.【解析】【分析】根据勾股定理求出BC根据正方形的面积公式计算即可【详解】解:由勾股定理得正方形的面积故答案为:【点睛】本题考查了勾股定理如果直角三角形的两条直角边长分别是ab斜边长为c那么a2+b219.(答案不唯一满足均可)【解析】【分析】一次函数的图象经过第一二四象限列出不等式组求解即可【详解】解:一次函数的图象经过第一二四象限解得:m的值可以是1故答案为:1(答案不唯一满足均可)【点睛】此题主20.【解析】【分析】已知ABCD是正方形根据正方形性质可知点B与点D关于AC对称DE=PB+PE求出DE长即是PB+PE最小值【详解】∵四边形ABCD是正方形∴点B与点D关于AC对称连接DE交AC于点P21.24【解析】已知对角线的长度根据菱形的面积计算公式即可计算菱形的面积解:根据对角线的长可以求得菱形的面积根据S=ab=×6×8=24cm2故答案为2422.4【解析】【分析】在Rt中由勾股定理可求得AB的长进而可根据三角形面积的不同表示方法求出CD的长【详解】解:Rt中AC=4mBC=3mAB=m∵∴m=24m故答案为24m【点睛】本题考查勾股定理掌握23.【解析】【分析】先运用勾股定理求出斜边AB然后再利用直角三角形斜边上的中线等于斜边的一半解答即可【详解】解:由勾股定理得AB∵∠C=90°CD为AB边上的中线∴CD=AB=故答案为【点睛】本题考查的24.16【解析】【分析】根据菱形的性质可得AC⊥BDAB=BC=CD=AD再根据直角三角形的性质可得AB=2OP进而得到AB长然后可算出菱形ABCD的周长【详解】∵四边形ABCD是菱形∴AC⊥BDAB=25.128°【解析】【分析】如图延长DC到F根据折叠的性质可得∠ACB=∠BCF继而根据平行线的性质可得∠BCF=∠ABC=26°从而可得∠ACF=52°再根据平角的定义即可求得答案【详解】如图延长DC三、解答题26.27.28.29.30.2016-2017年度第*次考试试卷参考解析【参考解析】**科目模拟测试一、选择题1.B解析:B【解析】【分析】根据S △ABE =12S 矩形ABCD =3=12•AE•BF ,先求出AE ,再求出BF 即可. 【详解】如图,连接BE .∵四边形ABCD 是矩形,∴AB=CD=2,BC=AD=3,∠D=90°,在Rt △ADE 中,AE=22AD DE +=2231+=10, ∵S △ABE =12S 矩形ABCD =3=12•AE•BF , ∴BF=3105. 故选:B .【点睛】本题考查矩形的性质、勾股定理、三角形的面积公式等知识,解题的关键是灵活运用所学知识解决问题,学会用面积法解决有关线段问题,属于中考常考题型.2.C解析:C【解析】【分析】首先连接AC ,交BD 于点O ,连接CM ,则CM 与BD 交于点P ,此时PA+PM 的值最小,由在菱形ABCD 中,AB=6,∠ABC=60°,易得△ACD 是等边三角形,BD 垂直平分AC ,继而可得CM ⊥AD ,则可求得CM 的值,继而求得PA+PM 的最小值.【详解】解:连接AC ,交BD 于点O ,连接CM ,则CM 与BD 交于点P ,此时PA+PM 的值最小,∵在菱形ABCD 中,AB=6,∠ABC=60°,∴∠ADC=∠ABC=60°,AD=CD=6,BD 垂直平分AC ,∴△ACD 是等边三角形,PA=PC ,∵M为AD中点,∴DM=12AD=3,CM⊥AD,∴CM=√CD2−DM2=3√3,∴PA+PM=PC+PM=CM=3√3.故选:C.【点睛】此题考查了最短路径问题、等边三角形的判定与性质、勾股定理以及菱形的性质.注意准确找到点P的位置是解此题的关键.3.B解析:B【解析】解:正方形和菱形都满足:四条边都相等,对角线平分一组对角,对角线垂直且互相平分;菱形的四个角不一定相等,而正方形的四个角一定相等.故选B.4.C解析:C【解析】【分析】根据频数分布直方图中的数据,求得众数,平均数,中位数,即可得出结论.【详解】解:①根据频数分布直方图,可得众数为60−80元范围,故每人乘坐地铁的月均花费最集中的区域在60−80元范围内,故①不正确;②每人乘坐地铁的月均花费的平均数=876001000=87.6=87.6元,所以每人乘坐地铁的月均花费的平均数范围是80~100元,故②错误;③每人乘坐地铁的月均花费的中位数约为80元,在60~100元范围内,故③正确;④为了让市民享受到更多的优惠,若使50%左右的人获得折扣优惠,则乘坐地铁的月均花费达到80元以上的人可以享受折扣,故④正确.故选:C【点睛】本题主要考查了频数分布直方图,平均数以及中位数的应用,将一组数据按照从小到大(或从大到小)的顺序排列,如果数据的个数是奇数,则处于中间位置的数就是这组数据的中位数.如果这组数据的个数是偶数,则中间两个数据的平均数就是这组数据的中位数.5.D解析:D【解析】【分析】连接OB,根据等腰三角形三线合一的性质可得BO⊥EF,再根据矩形的性质可得OA=OB ,根据等边对等角的性质可得∠BAC=∠ABO ,再根据三角形的内角和定理列式求出∠ABO=30°,即∠BAC=30°,根据直角三角形30°角所对的直角边等于斜边的一半求出AC ,再利用勾股定理列式计算即可求出AB .【详解】解:如图,连接OB ,∵BE=BF ,OE=OF ,∴BO ⊥EF ,∴在Rt △BEO 中,∠BEF+∠ABO=90°,由直角三角形斜边上的中线等于斜边上的一半可知:OA=OB=OC ,∴∠BAC=∠ABO ,又∵∠BEF=2∠BAC ,即2∠BAC+∠BAC=90°,解得∠BAC=30°,∴∠FCA=30°,∴∠FBC=30°,∵FC=2,∴3∴3,∴22AC BC -22(43)(23)-6,故选D .【点睛】本题考查了矩形的性质,全等三角形的判定与性质,等腰三角形三线合一的性质,直角三角形30°角所对的直角边等于斜边的一半,综合题,但难度不大,(2)作辅助线并求出∠BAC=30°是解题的关键.6.D解析:D【解析】【分析】分4是直角边、4是斜边,根据勾股定理计算即可.【详解】当4是直角边时,斜边2234+,当4是斜边时,另一条直角边=22473-=,故选:D .【点睛】本题考查的是勾股定理,如果直角三角形的两条直角边长分别是a ,b ,斜边长为c ,那么a 2+b 2=c 2.7.C解析:C【解析】【分析】证明30BAEEAC ACE ,求出BC 即可解决问题.【详解】解:四边形ABCD 是矩形,90B ∴∠=︒, EA=EC ,EAC ECA ∴∠=∠,EACBAE , 又∵将纸片沿AE 折叠,点B 恰好落在AC 上,30BAE EAC ACE , 3AB =,333BC AB ,∴矩形ABCD 的面积是33393AB BC. 故选:C .【点睛】本题考查矩形的性质,翻折变换,直角三角形30角性质等知识,解题的关键是灵活运用所学知识解决问题.8.A解析:A【解析】【分析】要求丝线的长,需将圆柱的侧面展开,进而根据“两点之间线段最短”得出结果,在求线段长时,根据勾股定理计算即可.【详解】解:如图,把圆柱的侧面展开,得到矩形,则这圈金属丝的周长最小为2AC 的长度,圆柱底面的周长为4dm ,圆柱高为2dm ,2AB dm ,2BC BC dm , 22222448AC , 22AC dm ,∴这圈金属丝的周长最小为242ACdm . 故选:A .【点睛】本题考查了平面展开-最短路径问题,圆柱的侧面展开图是一个矩形,此矩形的长等于圆柱底面周长,高等于圆柱的高,本题就是把圆柱的侧面展开成矩形,“化曲面为平面”,用勾股定理解决.9.C解析:C【解析】【分析】连接BD ,根据菱形的性质可得AC ⊥BD ,AO=12AC ,然后根据勾股定理计算出BO 长,再算出菱形的面积,然后再根据面积公式BC•AE=12AC•BD 可得答案. 【详解】连接BD ,交AC 于O 点,∵四边形ABCD 是菱形,∴AB =BC =CD =AD =5,∴1,22AC BD AO AC BD BO ⊥==,, ∴90AOB ∠=,∵AC =6,∴AO =3,∴2594BO =-=, ∴DB =8,∴菱形ABCD 的面积是11682422AC DB ⨯⋅=⨯⨯=, ∴BC ⋅AE =24,24AE=,5故选C.10.B解析:B【解析】【分析】先将各二次根式进行化简,再根据同类二次根式的概念求解即可.【详解】=;=-=.=3=,合并的是故选:B.【点睛】本题考查了同类二次根式,解答本题的关键在于熟练掌握二次根式的化简及同类二次根式的概念.11.A解析:A【解析】【分析】因为一般四边形的中点四边形是平行四边形,当对角线BD=AC时,中点四边形是菱形,当对角线AC⊥BD时,中点四边形是矩形,当对角线AC=BD,且AC⊥BD时,中点四边形是正方形.【详解】因为一般四边形的中点四边形是平行四边形,当对角线BD=AC时,中点四边形是菱形,当对角线AC⊥BD时,中点四边形是矩形,当对角线AC=BD,且AC⊥BD时,中点四边形是正方形,故④选项正确,故选A.【点睛】本题考查中点四边形、平行四边形、矩形、菱形的判定等知识,解题的关键是记住一般四边形的中点四边形是平行四边形,当对角线BD=AC时,中点四边形是菱形,当对角线AC⊥BD时,中点四边形是矩形,当对角线AC=BD,且AC⊥BD时,中点四边形是正方形.12.C解析:C【解析】矩形与菱形相比,菱形的四条边相等、对角线互相垂直;矩形四个角是直角,对角线相等,由此结合选项即可得出答案.【详解】A 、菱形、矩形的内角和都为360°,故本选项错误;B 、对角互相平分,菱形、矩形都具有,故本选项错误;C 、对角线相等菱形不具有,而矩形具有,故本选项正确D 、对角线互相垂直,菱形具有而矩形不具有,故本选项错误,故选C .【点睛】本题考查了菱形的性质及矩形的性质,熟练掌握矩形的性质与菱形的性质是解题的关键.13.C解析:C【解析】【分析】首先利用三角形的中位线定理得出AC ,进一步利用菱形的性质和勾股定理求得边长,得出周长即可.【详解】解:∵E ,F 分别是AB ,BC 边上的中点,∴∵四边形ABCD 是菱形,∴AC ⊥BD ,OA=12OB=12BD=2,∴,∴菱形ABCD 的周长为.故选C .14.B解析:B【解析】【分析】由勾股定理的逆定理,只要验证两小边的平方和是否等于最长边的平方即可.【详解】解:A 、72+242=625=252,故是直角三角形,不符合题意;B 、222222(3)(4)81256337(5)+=+=≠,故不是直角三角形,符合题意;C 、12+(34)2=2516=(54)2,故是直角三角形,不符合题意; D 、1.52+22=6.25=2.52,故是直角三角形,不符合题意;故选:B .本题考查勾股定理的逆定理的应用.判断三角形是否为直角三角形,已知三角形三边的长,只要利用勾股定理的逆定理加以判断即可.15.C解析:C【解析】【分析】把A(﹣4,0)分别代入一次函数y=﹣x+m和y=2x+n中,求得m和n的值,根据所得的两个解析式,求得点B和点C的坐标,以BC为底,点A到BC的垂线段为高,求出△ABC的面积即可.【详解】把点A(﹣4,0)代入一次函数y=﹣x+m得:4+m=0,解得:m=﹣4,即该函数的解析式为:y=﹣x﹣4,把点A(﹣4,0)代入一次函数y=2x+n得:﹣8+n=0,解得:n=8,即该函数的解析式为:y=2x+8,把x=0代入y=﹣x﹣4得:y=0﹣4=﹣4,即B(0,﹣4),把x=0代入y=2x+8得:y=0+8=8,即C(0,8),则边BC的长为8﹣(﹣4)=12,点A到BC的垂线段的长为4,S△ABC11242=⨯⨯=24.故选C.【点睛】本题考查了一次函数图象上点的坐标特征,正确掌握代入法求一次函数的解析式是解题的关键.二、填空题16.x≤1【解析】由题意得:1-x≥0解得x≤1故答案为x≤1点睛:二次根式有意义的条件是:a≥0解析:x≤1【解析】由题意得:1-x≥0,解得x≤1.故答案为x≤1.a≥0.17.【解析】【分析】由矩形的性质和线段垂直平分线的性质证出OA=OB=AB=2得出BD=2OB=4由勾股定理求出AD即可【详解】解:∵四边形ABCD是矩形∴OB= ODOA=OCAC=BD∴OA=OB∵A解析:【分析】由矩形的性质和线段垂直平分线的性质证出OA =OB =AB =2,得出BD =2OB =4,由勾股定理求出AD 即可.【详解】解:∵四边形ABCD 是矩形,∴OB =OD ,OA =OC ,AC =BD ,∴OA =OB ,∵AE 垂直平分OB ,∴AB =AO ,∴OA =OB =AB =2,∴BD =2OB =4,∴AD故答案为:【点睛】此题考查了矩形的性质、线段垂直平分线的性质、勾股定理;熟练掌握矩形的性质,证明三角形是等边三角形是解决问题的关键.18.【解析】【分析】根据勾股定理求出BC 根据正方形的面积公式计算即可【详解】解:由勾股定理得正方形的面积故答案为:【点睛】本题考查了勾股定理如果直角三角形的两条直角边长分别是ab 斜边长为c 那么a2+b2解析:3.【解析】【分析】根据勾股定理求出BC ,根据正方形的面积公式计算即可.【详解】解:由勾股定理得,BC == ∴正方形ABCD 的面积23BC ==,故答案为:3.【点睛】本题考查了勾股定理,如果直角三角形的两条直角边长分别是a ,b ,斜边长为c ,那么a 2+b 2=c 2. 19.(答案不唯一满足均可)【解析】【分析】一次函数的图象经过第一二四象限列出不等式组求解即可【详解】解:一次函数的图象经过第一二四象限解得:m 的值可以是1故答案为:1(答案不唯一满足均可)【点睛】此题主 解析:(答案不唯一,满足02m <<均可)【解析】【分析】一次函数()2y m x m =-+的图象经过第一、二、四象限,列出不等式组200,m m -<⎧⎨>⎩求解即可.【详解】解:一次函数()2y m x m =-+的图象经过第一、二、四象限,200m m -<⎧⎨>⎩解得:02m <<m 的值可以是1.故答案为:1(答案不唯一,满足02m <<均可).【点睛】此题主要考查了一次函数图象,一次函数y kx b =+的图象有四种情况:①当0,0k b >>时,函数y kx b =+的图象经过第一、二、三象限;②当0,0k b ><时,函数y kx b =+的图象经过第一、三、四象限;③当0,0k b <>时,函数y kx b =+的图象经过第一、二、四象限;④当0,0k b <<时,函数y kx b =+的图象经过第二、三、四象限.20.【解析】【分析】已知ABCD 是正方形根据正方形性质可知点B 与点D 关于AC 对称DE=PB+PE 求出DE 长即是PB+PE 最小值【详解】∵四边形ABCD 是正方形∴点B 与点D 关于AC 对称连接DE 交AC 于点P 解析:10【解析】【分析】已知ABCD 是正方形,根据正方形性质可知点B 与点D 关于AC 对称,DE=PB+PE ,求出DE 长即是PB+PE 最小值.【详解】∵四边形ABCD 是正方形∴点B 与点D 关于AC 对称,连接DE ,交AC 于点P ,连接PB ,则PB+PE=DE 的值最小 ∵CE=1,CD=3,∠ECD=90° ∴22221310=++=DE CE CD ∴PB+PE 1010【点睛】本题考查正方形性质,作对称点,再连接,根据两点之间直线最短得结论.21.24【解析】已知对角线的长度根据菱形的面积计算公式即可计算菱形的面积解:根据对角线的长可以求得菱形的面积根据S=ab=×6×8=24cm2故答案为24 解析:24【解析】已知对角线的长度,根据菱形的面积计算公式即可计算菱形的面积.解:根据对角线的长可以求得菱形的面积,根据S=12ab=12×6×8=24cm2,故答案为24.22.4【解析】【分析】在Rt中由勾股定理可求得AB的长进而可根据三角形面积的不同表示方法求出CD的长【详解】解:Rt中AC=4mBC=3mAB=m∵∴m=24m故答案为24m【点睛】本题考查勾股定理掌握解析:4【解析】【分析】在Rt ABC中,由勾股定理可求得AB的长,进而可根据三角形面积的不同表示方法求出CD的长.【详解】解:Rt ABC中,AC=4m,BC=3m5=m∵1122ABCS AC BC AB CD =⋅=⋅∴125AC BCCDAB⋅==m=2.4m故答案为2.4 m【点睛】本题考查勾股定理,掌握勾股定理的公式结合利用面积法是解题关键.23.【解析】【分析】先运用勾股定理求出斜边AB然后再利用直角三角形斜边上的中线等于斜边的一半解答即可【详解】解:由勾股定理得AB∵∠C=90°CD 为AB边上的中线∴CD=AB=故答案为【点睛】本题考查的【解析】【分析】先运用勾股定理求出斜边AB,然后再利用直角三角形斜边上的中线等于斜边的一半解答即可.【详解】解:由勾股定理得,=∵∠C=90°,CD 为AB 边上的中线,∴CD=12 . 【点睛】 本题考查的是勾股定理和直角三角形的性质,掌握直角三角形斜边上的中线是斜边的一半是解答本题的关键.24.16【解析】【分析】根据菱形的性质可得AC⊥BDAB=BC=CD=AD 再根据直角三角形的性质可得AB=2OP 进而得到AB 长然后可算出菱形ABCD 的周长【详解】∵四边形ABCD 是菱形∴AC⊥BDAB=解析:16【解析】【分析】根据菱形的性质可得AC ⊥BD ,AB=BC=CD=AD ,再根据直角三角形的性质可得AB=2OP ,进而得到AB 长,然后可算出菱形ABCD 的周长.【详解】∵四边形ABCD 是菱形,∴AC ⊥BD ,AB=BC=CD=AD ,∵点P 是AB 的中点,∴AB=2OP ,∵PO=2,∴AB=4,∴菱形ABCD 的周长是:4×4=16, 故答案为:16.【点睛】此题主要考查了菱形的性质,关键是掌握菱形的两条对角线互相垂直,四边相等,此题难度不大.25.128°【解析】【分析】如图延长DC 到F 根据折叠的性质可得∠ACB=∠BCF 继而根据平行线的性质可得∠BCF=∠ABC=26°从而可得∠ACF=52°再根据平角的定义即可求得答案【详解】如图延长DC解析:128°.【解析】【分析】如图,延长DC 到F ,根据折叠的性质可得∠ACB=∠BCF ,继而根据平行线的性质可得∠BCF=∠ABC=26°,从而可得∠ACF=52°,再根据平角的定义即可求得答案.【详解】如图,延长DC 到F ,∵矩形纸条折叠,∴∠ACB=∠BCF ,∵AB ∥CD ,∴∠BCF=∠ABC=26°,∴∠ACF=52°,∵∠ACF+∠ACD=180°,∴∠ACD=128°,故答案为128°. 【点睛】本题考查了折叠的性质,平行线的性质,熟练掌握相关知识是解题的关键.三、解答题26.(1)2600;(2)小颖在文具用品店停留了10分钟;(3)小颖本次在从学校回家的整个过程中,走的路程是3400米;(4)小颖从文具用品店回到家步行的速度是90米/分.【解析】【分析】(1)根据函数图象,可知小颖家与学校的距离是2600米;(2)由函数图象可知,20~30分钟的路程没变,所以AB 表示的实际意义是小颖在文具用品店停留了10分钟;(3)小颖本次从学校回家的整个过程中,走的路程为26002180014003400+-=()(米). (4)用小颖从文具用品店回到家的路程除以所用时间即可.【详解】(1)根据函数图象,可知小颖家与学校的距离是2600米;(2)AB 表示的实际意义是小颖在文具用品店停留了10分钟;(3)26002180014003400+-=()(米).(列的式子只要合理都可) ∴小颖本次在从学校回家的整个过程中,走的路程是3400米.(4)1800503090/()(米分)÷-=. ∴小颖从文具用品店回到家步行的速度是90米/分.【点睛】考查一次函数的应用,读懂函数的图象,明确每一段图象所表示的实际意义是解题的关键. 27.(1)AB =10;(2249x +281x ()-+;(3)AC +BC 最小值为2.【解析】【分析】(1)根据两点间的距离公式可求线段AB 的长;(2)根据两点间的距离公式可求线段AC ,BC 的值,再相加即可求解;(3)作B 点关于x 轴对称点F 点,连接AF ,与x 轴相交于点C .此时AC +BC 最短.根据两点间的距离公式即可求解.【详解】(1)22807110AB =-+-=()();(2)AC +BC 2222070810x x =-+-+=-+-()()()()224981x x =++-+();(3)如图,作B 点关于x 轴对称点F 点,连接AF ,与x 轴相交于点C .此时AC +BC 最短.∵B (8,1),∴F (8,-1),∴AC +BC =AC +CF =AF =2222(80)(17)8882-+--=+=.即AC +BC 最小值为82.【点睛】本题考查了最短路线问题,利用了数形结合的思想,构造出符合题意的直角三角形是解题的关键.28.(1)见解析;(2)见解析.【解析】【分析】(1)类比题目中所给的运算方法即可解答;(2)观察题目所给的算式,根据算式总结出一般规律即可求解.【详解】(111115456524⎛⎫-= ⎪⎝⎭ 21111515456456456524⎛⎫-=== ⎪⨯⨯⨯⨯⎝⎭;(2=n 为自然数,且n ≥2) . 【点睛】本题是阅读理解题,能够从所给的案例中找出相应的规律是解决该类题型的关键. 29.(1)A 型空调和B 型空调每台各需9000元、6000元;(2)共有三种采购方案,方案一:采购A 型空调10台,B 型空调20台,方案二:采购A 型空调11台,B 型空调19台,案三:采购A 型空调12台,B 型空调18台;(3)采购A 型空调10台,B 型空调20台可使总费用最低,最低费用是210000元.【解析】分析:(1)根据题意可以列出相应的方程组,从而可以解答本题;(2)根据题意可以列出相应的不等式组,从而可以求得有几种采购方案;(3)根据题意和(2)中的结果,可以解答本题.详解:(1)设A 型空调和B 型空调每台各需x 元、y 元,3239000456000x y x y +⎧⎨-⎩==,解得,90006000x y ⎧⎨⎩==, 答:A 型空调和B 型空调每台各需9000元、6000元;(2)设购买A 型空调a 台,则购买B 型空调(30-a )台,()()13029000600030217000a a a a ⎧≥-⎪⎨⎪+-≤⎩, 解得,10≤a≤1213, ∴a=10、11、12,共有三种采购方案,方案一:采购A 型空调10台,B 型空调20台,方案二:采购A 型空调11台,B 型空调19台,方案三:采购A 型空调12台,B 型空调18台;(3)设总费用为w 元,w=9000a+6000(30-a )=3000a+180000,∴当a=10时,w 取得最小值,此时w=210000,即采购A 型空调10台,B 型空调20台可使总费用最低,最低费用是210000元.点睛:本题考查一次函数的应用、一元一次不等式组的应用、二元一次方程组的应用,解答本题的关键是明确题意,找出所求问题需要的条件,利用函数和不等式的思想解答. 30.(1)5分钟;(2)2倍【解析】【分析】(1)观察图象可得李师傅离家10分钟时开始修车、离家15分钟修完车,两数相减即可得解;(2)观察图象可得李师傅修车前后行驶的路程和时间,即可求得相应的行驶速度,两速度相除即可得解.【详解】解:(1)由图可得,李师傅修车用了15105-=(分钟);(2)∵修车后李师傅骑车速度是200010002002015-=-(米/分钟),修车前速度为100010010=(米/分钟)∴2001002÷=∴修车后李师傅骑车速度是修车前的2倍.【点睛】本题考查了从图象中读取信息的数形结合的能力,需要注意分析其中的“关键点”,还要善于分析各部分图象的变化趋势.。
郑州市第四十七中学人教版八年级生物上册期中期中试卷及答案一、选择题1.下列关于水螅的描述,错误的是()A.通常生活在水流缓慢、水草繁茂的清洁海水中B.它的身体几乎透明,长约1cmC.可附着在水草等物体上D.口周围伸展着柔软细长的触手2.下列都属于环节动物的是A.蚯蚓、鼠妇、水蛭 B.沙蚕、蚯蚓、水蛭 C.蝗虫、蚯蚓、沙蚕D.蛔虫、沙蚕、水蛭3.环节动物和节肢动物共有的特征是()A.有足B.体表有外骨骼C.有环带D.身体分节4.图是鱼的形态结构示意图,有关说法错误的是()A.②下方为鳃,鳃丝内有丰富的毛细血管B.①③④⑥在鱼的游泳运动中起协调作用C.⑤是侧线,能减少水中游泳时遇到的阻力D.体表常覆盖鳞片,表面有黏液,起保护作用5.如图是家鸽体内部分结构分布示意图,以下描述正确的是()A.家鸽呼吸时只在c处进行气体交换B.b和c都能进行气体交换C.a是气管b是肺c是气囊D.气体进入体内的途径是a→b→c6.受精的鸟卵在雌鸟体内开始发育,但鸟卵产出后就暂停发育,原因是外界()A.温度太低B.温度太高C.具有空气D.具有阳光7.壁虎、蜥蜴、海龟等动物的身体表面覆盖着角质鳞片或甲,其意义是()①保护身体②辅助呼吸③防止体内水分的散失④保温,有助于维持恒定体温A.①③B.①②③C.①③④D.①②③④8.海豹、海豚、鲸等都是生活中海洋中哺乳动物,它们的呼吸方式是()A.用皮肤呼吸B.用肺呼吸C.用鳃呼吸D.用气管呼吸9.下列各项叙述中,正确的是A.骨的运动要靠骨骼肌的牵引B.骨骼肌由中间的肌腱和两端的肌腹两部分组成。
C.动物的运动,只靠运动系统和神经系统的控制调节来完成D.所有动物体内都有骨骼10.生物的结构总是同功能相适应的,如人体运动依赖于一定的结构基础.如图表示人体部分运动器,下列有关叙述错误的是()A.关节由图乙中的⑥⑦⑨构成B.⑨内的滑液和⑩能使关节灵活C.每块完整的骨骼肌是一个器官D.产生丙图的动作时,甲图中③舒张和⑪收缩11.如图为人的屈肘动作示意图,与此有关的叙述正确的是A.肱二头肌收缩,肱三头肌舒张 B.肱三头肌收缩,肱二头肌舒张C.都收缩 D.都舒张12.关于人体运动系统的组成,不正确的叙述是()A.骨206块 B.骨之间连接一定是关节C.骨骼肌600多块 D.关节主要分布在躯干13.生物课上王老师指着自己受伤的左臂,幽默地说是伤了“支点”。
郑州市第47中学2010—2011学年下期高二年级第一次月考试题英语命题人:李志合第一部分语言知识运用(共两节,满分60)第一节单项选择(共20小题;每小题1。
5分,满分30分)从A、B、C、D四个选项中,选出可以填入空白处的最佳选项。
1。
I earn 10 dollars________hour as________supermarket cashier on Saturday。
A. an;theB. the;a C。
an;a D。
a;an2. —I'm sorry,sir.I didn't mean to hurt you by showing you those pictures?—________It just reminded me of my old friends。
A.Not at all.B. All the best。
C。
No sorry,please。
D。
Never mind.3. You can turn to the Internet for more information whenyou________the headline you are interested in while looking through the papers.A. cut out B。
turn out C. meet with D。
come with4. Many foreign leaders sent telegrams and letters,________ourgovernment on having successfully hosted the 29th Olympic Games in Beijing.A. congratulating B。
celebrating C. congratulated D. to congratulate5。
Some fans felt disappointed that few of them had direct________to the pop star after his performance on the stage.A. method B。
姓名得分: 考生注意:本试卷共32题,总分100 分,考试时间90分钟。
一、选择题(每题2分,共24分)
1.9的平方根是()
(A)3 (B)3(C)3
±(D)3
±
2.下列能构成直角三角形三边长的是()
(A)1、2、3 (B)2、3、4 (C)3、4、5 (D)4、5、6
3.以下五家银行行标中,既是中心对称图形又是轴对称图形的有()
(A)1个(B)2个(C)3个(D)4个
4.下列说法正确的是( )
(A)有理数只是有限小数(B)无理数是无限小数
(C)无限小数是无理数(D)
3
π
是分数
5.下列说法错误的是( )
(A)1
)1
(2=
-(B)()1
1
33-
=
-
(C)2的平方根是2
±(D)()2
3
2
)3
(-
⨯
-
=
-
⨯
-
6.一个多边形每个外角都等于72︒,则此多边形是()
(A)五边形(B)六边形(C)七边形(D)八边形
7.如图,三个正方形围成一个直角三角形,64、400
则图中字母所代表的正方形面积是()
(A)400+64 (B)2
264
400-(C)400-64 (D)2
264
400-
8.下列条件中能判断四边形ABCD为平行四边形的是()
(A)AB=BC CD=DA (B)AB∥CD AB=CD
(C)AD∥BC AB=CD (D)AD∥BC ∠B=∠C
9.将图形按顺时针方向旋转900后的图形是 ( )
(A)(B)(C)(D)
10.以下图形中,不能用来密铺的是()
11.矩形、菱形、正方形都具有的性质是( ) (A )一组邻边相等,对角线互相垂直平分 (B )一组邻角相等,对角线也相等 (C )一组对边平行且相等,对角线互相平分 (D )对角线相等,且互相垂直平分
12.如图,过圆心O 和圆上一点A 连一条曲线,将曲线OA 绕O 点按同一方向连续旋转三次,每次旋
转900,把圆分成四部分,则
(A ) 这四部分不一定相等 (B ) 这四部分相等
(C ) 前一部分小于后一部分 (D )不能确定
二、填空题(每题3分,共24分)
13.6的相反数是
;绝对值等于2的数是 . 14.□ABCD 中,∠A=60°,则∠B= ,∠C= . 15.化简()
=-2
3
2 ;1449⨯= .
16.估算比较大小:(填“>”、“<”或“=”) 3;
2
13-
2
1
17.用长4cm,宽3cm 的邮票300枚不重不漏摆成一个正方形,这个正方形的边长等于________cm.
18.边长为2的正方形对角线长为 ;以该正方形对角线为边长的新正方形的面积是 .
19.右图可以看作是由基本图形 经 得到的.
20.已知直角三角形的两条直角边分别是4和5,这个直角三角形的斜边的长度在两个相邻的整数之间,这两个整数是_______和________.
三、计算与化简(每题4分,共12分)
21.()
1525- 22.123
127+-
23.
(
)(
)
131
316
72-++
-
B
四、操作与探索(24题6分,25、26题各5分,共16分) 24.(1)如图,经过平移,小船上的点A 移到了点B ,作出平移后的小船。
A
B
(2)如图,正△ABC ,将此三角形绕点C 顺时针旋转,使CB 与CA 重合,得△ACD ①作出△ACD ②四边形ABCD 是什么四边形?
25.如图,正方形网格中的每个小正方形边长都是1,任意连结这些小正方形的顶点,可得到一些线段。
请在图中画出1352===EF CD AB 、、这样的线段,并选择其中的一个说明这
样画的道理.
26.图中字母表示为四边形、平行四边形,矩形、菱形、正方形、梯形、
等腰梯形、直角梯形从属关系,则字母所代表的图形为:
A为,B为,C为,
D为,E为,F为,
G为,H为.
五、解答下列各题(每题4分,共16分).
27.小文房间的面积为10.8㎡,房间地面恰巧由120块相同的正方形地砖铺成,每块地砖的边长是多少?
28.已知:如图,在平行四边形ABCD中,E、F分别在AB、CD上,且BE=DF 问:AF∥EC吗?试说明理由.
A
D
F
E
C
B
29.如图,已知:□ABCD 的对角线AC 、BD 相交于O 点,△AOB 为等边三角形,AB=4cm 。
(1)□ABCD 为矩形吗?请说明理由.
(2)求四边形ABCD 的面积.
30.已知:如图,菱形ABCD 的周长为8cm ,∠ABC :∠BAD =2:1,对角线AC 、BD 相交于点O ,求AC 的长及菱形的面积.
D C B A O
六、观察与思考(31题4分,32题4分,共8分)
31.如图,一个梯子AB长2.5 米,顶端A靠在墙AC上,这时梯子下端B与墙角C距离为1.5米,梯子滑动后停在DE的位置上,测得BD长为0.5米,求梯子顶端A下落了多少米?
32.某人欲从A点横渡一条河,由于水流的影响,实际上岸地点C偏离欲到达点B240米,结果他在水中实际游了510米,求该河的宽度。
七、附加题(本题5分)
31.如图,河两边有甲、乙两条村庄,现准备建一座桥,桥必须与河岸垂直,问桥应建在何处才能使由甲到乙的路程最短?请作出图形,并说说理由.
甲•
乙•。