第17周错题(有答案)

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17参考答案与解析

17参考答案与解析

参考答案与解析一、选择题(本大题共7小题,每小题3分,共21分.每小题只有一项符台题意,不选、多选、错选均得0分)1.“公共场所禁止大声喧哗”,其中“大声“是指声音的()A.音色B.音调C.响度D.频率【知识考点】音调、响度与音色的区分.【思路分析】音调是指声音的高低,响度是指声音的强弱,声音的响度与声源振动的幅度有关,振动幅度越大,响度越大。

【解答过程】解:在公共场所禁止大声喧哗,这里的“大声”是指声音的响度大。

故选:C。

【总结归纳】本题考查了音调和响度、音色的区分,理解各自的含义和特点是正确判断的关键。

2.下列几个实例中描述正确的是()A.夏天,给饮料里加冰块来降温,是利用了冰块熔化放出热量的特点B.给发烧的病人身上擦酒精降温,是利用了酒精蒸发有致冷作用的特点C.冬天,北方楼房中的“暖气”用水作介质,是利用水的比热容较小的属性D.高压锅更容易把食物煮熟,是利用了液体沸点随气压增大而降低的规律【知识考点】蒸发及其现象.【思路分析】(1)物质由固态变成液态的现象叫做熔化,熔化吸热;(2)物质由液态变成气态的现象叫汽化,汽化吸热,汽化包括蒸发和沸腾两种方式;(3)相同质量的水和其它物质比较,吸收或放出相同的热量,水的温度升高或降低的少;升高或降低相同的温度,水吸收或放出的热量多;(4)液态沸点与气压有关,气压越大沸点越高。

【解答过程】解:A、给饮料里加冰块来降温,是利用了冰块熔化吸收热量的特点。

故A错误;B、用酒精来给高温病人降温,利用了酒精蒸发吸热。

故B正确;C、用水来冷却发动机,利用了水的比热容大。

根据公式Q=cm△t可知,当物体的质量m和升高的温度△t相同时,物体吸收的热量Q和物质的比热成正比,水的比热容c大,因此利用水吸收的热量多,冷却效果好。

故C错误;D、高压锅更容易把食物煮熟,是利用了液体沸点随气压增大而降低的规律。

故D错误。

故选:B。

【总结归纳】这样的题目对学生的知识点要求比较高,考查了学生的综合能力。

17周易错题训练

17周易错题训练

1、把一根54米的绳子平均分成4段,每段长( )米,每段占全长的( )。

2、米比( )米多它的60%;( )比32少30% 。

3、钟面上时针的长1dm,一昼夜时针扫过的面积是( )。

4、把一个比的前项扩大2倍,后项缩小2倍,比值就( )。

5、一台碾米机65小时碾米127吨,1小时可碾米( )吨,碾1吨米要( )小时。

6、大小两个正方体的棱长比是3∶2;大小正方体的表面积比是( );大小正方体的体积比是( )。

7、一种油菜籽的出油率为35%,400千克油菜籽可以榨出( )千克油,要榨1400千克油需( )千克油菜籽。

8、7÷( )=8( )=25%=4∶( )=( )(填小数) 9、往30千克盐中加入( )千克水,可得到含盐率为30%的盐水。

10、甲乙两圆的周长比是2:3,其中一个圆的面积是18,另一个圆的面积可能是( ),也可能是( )。

11、用一个长10厘米,宽4厘米的长方形,剪一个最大的半圆,这个半圆的面积是( )。

12、原价90元的领带降价20%后是( )元,原价( )元的衬衫降价20%后是120元。

13、甲数除以乙数的商是 2.5,甲数与乙数的比是( )。

14、一种大豆的出油率是42%,2.1吨这样的大豆可榨油( )千克,( )千克的大豆可榨油2.1吨。

15、修一条20千米的路,若每天修它的101,要( )天修完,若每天修101千米,( )天修完。

16、直角三角形中两个锐角的度数比是1 :2,那么较大的锐角是( )度。

17、一块长方形地的周长是120米,其中宽比长短13,这块地的面积是( )平方米。

18、大圆的半径相当于小圆的直径,这两个圆的面积和是100平方厘米,大圆的面积是( )平方厘米。

19、A 的41与B 的61相等(A 不等于0),则A ∶B=( )。

20、因为甲×34 =乙×56,所以甲∶乙=( )。

21、一根绳子用去一半,再用去余下的一半,还剩下全长的( )22、甲数的53等于乙数的32(甲、乙 0),甲数与乙数相差10,甲数是( ),乙数是( )。

信丰一中2022-2023学年高二第一学期物理第17周周练试卷与参考答案讲评

信丰一中2022-2023学年高二第一学期物理第17周周练试卷与参考答案讲评

信丰一中2022-2023学年高二物理第17周周练试卷命题人:高二物理备课组审题人:高二物理备课组一、单选题(共7小题28分,每小题4分)1.关于简谐运动的回复力,下列说法正确的是()A .简谐运动的回复力可能是恒力B .做简谐运动的物体的加速度方向与位移方向可能相同C .简谐运动中回复力的公式F kx =-中k 是弹簧的劲度系数,x 是弹簧的长度D .做简谐运动的物体每次经过平衡位置回复力一定为零2.有两个简谐运动,其表达式分别是14sin(100)cm 3x t ππ=+,25sin(100)cm 6x t ππ=+,下列说法正确的是()A .它们的振幅相同B .它们的周期不相同C .它们的相位差恒定D .它们的振动步调一致3.弹簧振子做简谐振动,若从平衡位置O 开始计时,如图,经过0.2s (0.2s 小于振子的四分之一振动周期)时,振子第一次经过P 点,又经过了0.2s ,振子第二次经过P 点,则振子的振动周期为()A .0.4sB .0.8sC .1.0sD .1.2s4.如图所示,一个单摆在做简谐运动,关于摆球的运动,下列说法正确的是()A .单摆摆动到O 点时,回复力为零,加速度不为零B .摆球从A 到O 的过程中,机械能增大C .摆球向右经过O 点和向左经过O 点时,速度相同D .减小单摆的振幅,单摆的周期会减小5.冰雹(Hai )也叫“雹”,俗称雹子,是一种天气现象。

若一个100g 的冰雹形成掉落之后由于空气阻力,到达地面的速度相当于冰雹从7层楼高自由落体到地面的速度,落地时与地面碰撞的时间为2ms ,则该冰雹对地面产生的冲击力约为()A .500NB .1000NC .1500ND .2000N6.如图所示,质量相等的A 、B 两个球,原来在光滑水平面上沿同一直线相向做匀速直线运动,A 球的速度是6m/s ,B 球的速度是-2m/s ,A 、B 两球发生对心碰撞。

对于该碰撞之后的A 、B 两球的速度可能值,某实验小组的同学们做了很多种猜测,下面的猜测结果一定无法实现的是()A .'A 2m/s v =-,'B 6m/s v =B .'A 2m/s v =,'B 2m/sv =C .'A 1m/s v =,'B 3m/s v =D .'A 3m/s v =-,'B 7m/sv =7.最近,我国为“长征九号”研制的大推力新型火箭发动机联试成功,这标志着我国重型运载火箭的研发取得突破性进展。

科学第17周作业---声音、呼吸与消化复习 (1)

科学第17周作业---声音、呼吸与消化复习 (1)

14四上科学第17周作业---声音、呼吸与消化复习您的姓名: [填空题] *_________________________________您的班级: [单选题] *○四(1)班○四(2)班○四(3)班○四(4)班○四(5)班○四(6)班○四(7)班○四(8)班1. 声音是由物体振动引起的,物体振动停止后发声不会停止 [判断题] *对错(正确答案)答案解析:声音由物体振动引起的,物体振动停止后发声也会停止2. 我们拍球时,虽然能听见声音,但是看不见球面或地面的振动,因此拍球发声时不振动。

[判断题] *对错(正确答案)答案解析:它们发声了轻微振动,只是这种振动我们不容易看到,就像我们拍击桌面一样3. 只要我们对物体用力,物体就能发出声音。

[判断题] *对错(正确答案)答案解析:按压桌面就不会发出声音4. 使橡皮筋发出声音只能拉直拨动。

[判断题] *对错(正确答案)5. 弯曲钢尺是可以使钢尺发声的方法之一。

[判断题] *对错(正确答案)6. 陨石撞击月球,我们能听到响声。

[判断题] *对错(正确答案)7. 将振动的音叉放入水槽内的水中,我们用耳朵贴在水槽边可以听到音叉振动的声音,其传播路径是 [单选题] *音叉-水-水槽-人耳(正确答案)音叉-人耳音叉-空气-人耳8. 我们改变电视机的音量时,改变的是声音的强弱。

[判断题] *对(正确答案)错9. 用不同的力敲击同一个音叉,声音的高低会发生变化。

[判断题] *对错(正确答案)答案解析:用不同的力去敲击同一个音叉,声音的强弱会发生变化。

10. 用力拨动钢尺,发出的声音强,是因为钢尺的振动幅度大。

[判断题] *对(正确答案)错11. 声音的强弱可以用音量来描述。

[判断题] *对(正确答案)错12. 用力拍打篮球,篮球发出的声音越强,这是因为篮球的振动幅度越大。

[判断题] *对(正确答案)错13. 敲击小鼓的鼓面时,纸屏上吊着的泡沫塑料小球会被弹起。

二年级第十七周错题汇总

二年级第十七周错题汇总

第十七周周末乐园错题汇总一、填空题。

’1、在○里填上“>”“<”或“=”。

4米○40厘米 1米○45厘米+55厘米2米○85米-23米 70厘米○1米3米○30厘米 1米-50厘米○50厘米2、将下面的长度按从长到短的顺序排一排。

30厘米 10米 70 厘米 7米 3米99厘米 4米()>()>()>()>()>()3、16名同学排成一排。

从左数小红是第5个,从右数小明是第8个。

小红和小明之间有()名同学。

4、4名运动员,每两人握一次手,一共要握()次手。

5、把一根绳子对折,再对折,沿折痕剪开,每段绳子刚好长5厘米,这根绳子原来长()厘米。

6、一条红色彩带长48厘米,它剪去12厘米后,与一条绿色彩带同样长。

原来这条红色彩带比这条绿色彩带长()厘米。

二、解决问题1、商店有48个玩具,上午卖掉16个,下午卖掉17个。

现在比原来少了多少个?2、男生折了37只千纸鹤,女生折了48只干纸鹤,送给幼儿园一些后还剩下26只。

他们送给幼儿园多少只?3、美术组有36人,航模组有18人,美术组调多少人到航模组,两组人数就一样多了?4、把一根竹竿竖直插入水中,水面以上的部分长69厘米,水面以下的部分长31厘米。

这根竹竿长多少米?5、一间教室的长为8米,王老师沿着这间教室的长走了4个来回,他一共走了多少米?6、有一把断了的米尺,这把米尺上最小的刻度是36厘米,最大的刻度是90厘米,这把米尺一次最长可以量出多少厘米的长度?7、小红有一根铁丝,她用这根铁丝围了一个每条边的长都为8厘米的五边形后,这根铁丝还剩下12厘米,这根铁丝原来长多少厘米?8、小明家、小东家和学校在同一条笔直的道路的同一边,小明家离学校56米,小东家离学校38米,小明家和小东家相距多少米?9、把30米的绳子剪成相同的几段,一共剪了5下,每段长多少米?10、把一根绳对折两次后每段长6米,这根绳原来长多少米?如果把这根绳对折三次,那么每段长多少米?。

山东省淄博市六中2021-2022学年高二上学期第17周周末自主学习材料英语试题 Word版含答案

山东省淄博市六中2021-2022学年高二上学期第17周周末自主学习材料英语试题 Word版含答案

2022级高二英语自主学习材料文理通用第17周命题人:冯祚新任教班级:25班审核人:隽丕东任教班级: 7、8班姓名:班级:第一部分:阅读理解(共2节,满分70分)第一节(共30小题;每小题2分,满分60分)AProfessor Martin’s report says that children who attend a number of different schools, because their parents have to move around the country, probably make slow progress in their studies. There are also signs, says Professor Martin, that an unusually large number of such children are mentally affected.The Profes sor says, “It’s true, my personal feeling is that children should stay in one school. However, our findings are based on research and not on any personal feelings that I or my assistants may have on the subject.”Captain Thomas James, an Army lecturer for the past 20 years and himself a father of 2 children, said, “I’ve never heard such rubbish. Taking me for example, no harm is done to the education of my children who change school regularly---if they keep to the same system. In my experience----and I’ve known quite a few of them---- Army children are as well-adjusted(调整) as any others, if not more so. What the professor doesn’t appear to notice is the fact that in such situations children will adapt much better than grown-ups.”When this was put to Professor Martin, he said that at no time had his team suggested that all such children were mentally affected in some way, but simply that in their experience there was a clear tendency (倾向). “Our finding shows that while the very bright child can deal with regular change without harming his or her general progress in studies, the majority of children suffer from constantly having to enter a new learning situation.”1. Professor Martin’s report suggests that _______.A. it may not be good for children to change schools oftenB. parents should not move oftenC. more and more children are mentally affectedD. children will make rapid progress if they stay in one school.2. According to the passage, Professor Martin’s personal feeling____ .A. is the opposite of what his report has shownB. is in a way supported by his researchC. has played a big part in his researchD. is based on his own experience as a child3. From the passage, we can conclude that Captain Thomas’ children____.A. have been affected by changing schoolsB. have not been affected by changing schoolsC. like every army school they have been inD. are the brightest among the children who often change school4. According to Professor Martin, ____ suffer from changing schools regularly.A. non-army childrenB. bright childrenC. the majority of childrenD. few children5. Captain Thomas believes _____A. army children are generally better—adjusted than any othersB. army children are usually less experienced than any othersC. children can adapt more easily and quickly than grown-upsD. children can adapt as easily and quickly as grown-upsBDuring the 20th century there has been a great change in the lives of women. A woman marrying at the end of the 19th century probably have been in her middle 20’s and would be likely to have seven or eight children. By the time the youngest was 15, the mot her would have been in her early 50’s and would expect to live a further 20 years, during which chance and health made it hard for her to get paid work. Today women marry younger and have fewer children.Usually a woman’s youngest child will be 15 when sh e is 45 and she can be expected to live another 35years and is likely to get paid work until 60.This important change in women’s life has only recently begun to have its full effect on women’s economic position. Even a few years ago most girls left school and took a fulltime job. However, when they married they usually left work at once and never returned to it. Today the school-leaving age is 16, many girls stay at school after that age, and though women marry younger, more married women stay at work at least until shortly before their first child is born. Very many more return to work later. Such changes have led to a new relationship in marriage, with the husband accepting a greater share of the duties of family life.6. We learn from this passage that in the 19th century_____ .A. there were more children in the world than there are todayB. women spent a greater part of their lives raising children than they do todayC. there were more women in poor health than there are todayD. women married younger than they do today7. One reason why a married woman of today may take a job is that ____.A. she is usually younger when her children are old enough to look after themselvesB. she is obliged to help her husband support the familyC. she feels lonely at home when her children grow upD. she hopes to change the world a bit by doing up8. Many girls are now likely to ____.A. give up their jobs after they get marriedB. leave school as soon as they canC. marry early so that they can get better jobsD. continue working until they are going to have a baby9. According to the passage, it is now quite usual for women to ____ .A. stay at home after leaving schoolB. find jobs when they reach the age of 16C. start working again later in lifeD. marry and have children while still at school10. Now a husband probably ____ .A. plays a greater part in looking after the childrenB. does almost all of the houseworkC. feels unhappy about his wife’s going out to workD. takes a part-time job so as to help at homeCIn 1896, Einstein went to Zurich to study physics. There he met a girl from Hungary. They studied in the same class and the same interest in physics brought them together and they became good friends. Before long they fell in love. In 1903 when Einstein was 24, he married Marits, who was 4 years older than he.After their marriage, Einstein devoted himself to the research of the great theory. To give her husband more help, Marits gave up her own work, and became a good wife and assistant. She tried her best to encourage him whenever possible. She was sure that her husband would succeed. They often discussed the theory while walking outside or sitting together in the room. They even did that in their letters when one of them was away from their home.In1914, Einstein moved to Berlin and settled down there. At that time his theory proved to be correct and he had become famous all over the world. But it was not long before the First Would War broke out. Marits as well as her two sons, who was on holiday in Switzerland, couldn’t come back to Berlin any more. The war not only stopped Einstein’s work but also broke up the warm, happy family. In 1919, Einstein and Marits had to get divorced(离婚).11. Form this passage we can see that Marits was born in _____.A.1896B. 1879C. 1883D.187512. We can learn from the passage that Marits ______.A. greatly helped Einstein with his research workB. supported Einstein but helped him littleC. was also a great scientistD. was the real discoverer of the theory13. The last paragraph mainly tells us _____.A. why Einstein and Marits got divorcedB. why Einstein moved to BerlinC. how Einstein became famous all over the worldD. when the First World War broke out14. Why did Einstein and Marits get divorced?A. Because Marits didn’t love Einstein after the war broke out.B. Because the war stopped Marits from coming back to Berlin.C. Because Einstein only cared about his research.D. Because Einstein was famous all over the world.15. The writer wanted to tell us that _____.A. Marits should be respected as greatly as her husbandB. Einstein suffered a great deal in his lifeC. The bitter (苦难的) suffering was the mother of successD. We should remember Marits when talking about Einstein’ theoryDAfter giving a talk at a high school, I was asked to pay a visit to a special student. An illness had kept the boy home, but he had expressed an interest in meeting me, and it would mean a great deal to him. I agreed.During the nine-mile drive to his home, I found out something about Matthew. He had muscular dystrophy(肌肉萎缩症). When he was born, the doctor told his parents that he would not live to see five, and then they were told he would not make it to ten. Now he was thirteen. He wanted to meet me because I was a gold-medal power lifter, and I knew about overcoming obstacles (障碍) and going for my dreams.I spent over an hour talking to Matthew. Never once did he complain or ask, “Why me?” He spoke about winning and succeeding and going for his dreams. Obviously, he knew what he was talking about. He didn’t mention that his classmates had made fun of him because he was different. He just talked about his hopes for the future, and how one day he wanted to lift weight with me.When we finished talking, I went to my briefcase and pulled out the first gold medal I won and put it around his neck. I told him he was more of a winner and knew more about success and overcoming obstacles than I ever would. He looked at it for a moment, then took it off and handed it back to me. He said, “You are a champion. You earned that medal. Someday when I get to the Olympics and win my own medal, I will show it to you.”Last summer I received a letter from Matthew’s parents telling me that Matthew had passed away. They wanted me to have a letter he had written to me a few days before:Dear Rick,My mom said I should send you a thank-you letter for the picture you sent me. I also want to let you know that th e doctors tell me that I don’t have long to live anymore. But I still smile as much as I can. I told you someday I was going to the Olympics and win a gold medal. But I know now I will never get to do that. But I know I’m a champion, and God knows that too. When I get to Heaven, God will give me my medal and when you get there, I will show it to you. Thank you for loving me.Your friend,Mathew16. The boy wanted to meet the author because ______.A. he was interested in what the author was doingB. he wanted to get a gold medal himselfC. he admired the author very muchD. he wanted the author to know him too17. The underlined part in the third paragraph probably means “______”.A. Why do you come to see me?B. Why do I have to stay at home?C. Why does the disease fall on me?D. Why not give a gold medal to me?18. We can infer from the passage that ______.A. Matthew was a determined boy and considered himself as normalB. Rick used to have the same disease and later became a power lifterC. Matthew was to become a champion before he diedD. After meeting Matthew, Rick regarded him as normal19. The boy refused the author’s medal because ______.A. he wanted the picture insteadB. he would not be pitied by othersC. he did not know he would die soonD. he himself could earn one in the future20. The author wrote the passage with the purpose of ______.A. describing his unusual friendship with a disabled childB. showing his admiration towards the disabled childC. telling an experience of meeting a disabled childD. expressing his pity to all the disabled childrenENo one knows why we sleep, but it's certain that we need to. People who are prevented from sleeping begin to suffer obvious effects after a few days--they think less clearly, and they fall asleep during the working hours; some may have hallucinations(幻觉).There are no rules about sleep. Generally speaking, grown-ups sleep about 7 and a half hours each night and probably more than 60 percent get between seven and eight hours. But perhaps eight per cent are quite happy with 5 hours or less, and 4 per cent or so find that they want ten hours or more. If you feel all right, you're probably getting enough sleep. The important thing is not to worry how much other people get.Their needs may be different. Exercise doesn't seem to increase the need for sleep--office workers, for example, sleep for about as long as people doing physically active work.Children sleep more than grown-ups--perhaps 14 to 18 hours soon after birth, going down to grown-up levels by early teen-age. Sleep patterns also tend(倾向) to be different in the elderly, who may sleepless at night than they did when younger, find sleep getting more broken, and often make it a rule to sleep during the daytime.21. According to the passage, some people are unable to think clearly because ________.A. they have hallucinationsB. they feel sleepy during the working hoursC. they don't have enough sleepD. they are certain to be kept from going to bed22. Which of the following is true?A. All grown-ups must have at least eight-hour sleep.B. Most of grown-ups sleep for seven or eight hours.C. Quite a few people need only 5 hours or less for them to sleep.D. No grown-ups sleep more than ten hours.23. Whether you have got enough sleep is judged by ________.A. how many hours you have sleptB. how many hours you need to sleepC. whether you do exercise and physical workD. whether you feel fresh and energetic24. According to the passage, a boy of 14 years old sleeps ________.A. as long hours as a grown-upB. much longer hours than a grown-upC. for 14 hours each nightD. for less than 8 hours per night25. Who finds they sleep less during the night and must take a nap during the day?A. ChildrenB. Grown-upsC. Young peopleD. Old peopleFBill Javis took over our village news-agency at a time of life when most of us only want to relax. He just thought he would like something but not too much to do, and the news-agency was ready-made. The business produced little enough for him, but Bill was a man who only wanted the simplicity and order and regularity of the job. He had been a long-serving sailor, and all his life had done everything by the clock.Every day he opened his shop at six a. m. to catch the early trade; the papers arrived on his door-step before that. Many of Bill’s customers were city workers, and the shop was convenient for the station. Business was tailing off by 10 o’clock, so at eleven sharp Bill closed for lunch. It was hard luck on anybody who wanted a paper or magazine in the afternoon, for most likely Bill would be down on the river bank, fishing, and his nearest competitor was five kilometers away. Sometime in the afternoon, the evening paper landed on the doormat, and at 4 o’clock Bill reopened. The evening rush lasted t ill seven, and it was worthwhile.He lived in a flat above the shop, alone. Except in the very bad weather, you always knew where to find him in the afternoon, as I have said. Once, on a sunny afternoon, I walked home along the river bank from a shopping trip to the village. By my watch it was three minutes past four, so I was astonished to see Bill sitting there on his little chair with a line in the water. He had had no luck, I could see, but he was making no effort to move.“What’s wrong, Bill?” I call ed out from the path.For answer, he put a hand in his jacket and took out a big, golden object. For a moment I had no idea what it could be, and then it suddenly went off with a noise like a fire engine. Stopping the bell, Bill held the thing up and call ed back, “Ten to four, you see, and this is dead right.”I had never known anyone carrying a brass alarm clock round with him before.26. Bill Javis became a news-agent when ______.A. he needed the moneyB. he decided to take things easyC. he was quite an old manD. he gave up clock-repairing27. Bill opened the shop so early in the day because ______.A. he liked to do as much as possible before he went to workB. the shop had to be open when the morning papers cameC. he was never sure of the timeD. it was then that he did a lot of business28. You might say “Hard luck” to someone who ______.A. has just heard some very good newsB. is less fortunate than he or she ought to beC. puts great effort into whatever he or she triesD. fails through his or her own fault entirely29. On that sunny afternoon, the writer was surprised when he saw Bill because ______.A. he thought it was late for Bill to be still fishingB. he thought Bill was ill, since he was not moving at allC. Bill had not caught anything, and that seemed strangeD. Bill stayed in his flat30. From the information given in the passage, who or what do you think was wrong?A. The bell was; it must have gone off at the wrong time.B. Bill was; he had dropped off to sleep.C. The writer’s watch was fast.D. Bill’s clock was wrong; it was very old.其次节(共5小题;每小题2分,满分10分)依据短文内容,从短文后的选项中选出能填入空白处的最佳选项。

高考英语一轮复习 周末培优(第17周)七选五(含解析)新人教版-新人教版高三全册英语试题

高考英语一轮复习 周末培优(第17周)七选五(含解析)新人教版-新人教版高三全册英语试题

七选五〔测试时间:40分钟,总分:60分〕班级:____________ 姓名:____________ 座号:____________ 得分:____________I.单项填空(每一小题1分,共10分)从A、B、C、D四个选项中,选出可以填入空白处的最优选项。

1.—Where is Peter? I can’t find him anywhere.—He went to the library after breakfast and__________his essay there ever since.A.wroteB.had writtenC.has been writingD.is writing【答案】C【易错点拨】解答此题的关键是要抓住时间状语ever since,否如此会想当然地认为空处应和and 前面的went to the library的时态保持一致,误选一般过去时。

2.A common memory they all have__________ their school days is the school uniform.A.ofB.onC.toD.with【答案】A【解析】考查介词的用法。

句意:他们对校园岁月的共同记忆就是校服。

本句使用了have a memory of 这一搭配,含义为"对……有记忆"。

3.Many of the things we now benefit from would not be around __________Thomas Edison.A.thanks toB.regardless ofC.aside fromD.but for【答案】D【解析】考查介词短语辨析。

句意:要不是因为托马斯·爱迪生,很多现在让我们受益的东西都不会出现。

but for "要不是,倘假设没有",符合句意。

4.It might have saved me some trouble __________the schedule.A.did I knowB.have I knownC.do I knowD.had I known【答案】D【解析】考查虚拟语气的倒装。

二年级语文下册试题-能力训练周周练 ( 十七)(含答案)部编版

二年级语文下册试题-能力训练周周练 ( 十七)(含答案)部编版

二下语文周周练(十七)班级姓名成绩一、(北京市第一实验小学期末)多音字组词。

省()()称()()纤()()()()()()()()量()()杆()()剥()()()()()()()()二、照样子写词语。

亮晶晶家家户户小心翼翼三、加标点。

(1)小蝌蚪是怎样找到妈妈的(2)这朵花真漂亮啊(3)妈妈说小红你要做一个听话的孩子四、选择正确词语填空。

灯会登高粽子月饼这个学期,我们学到了很多节日,我们知道了立夏节要,端午节吃,中秋节吃,元宵节闹。

五、阅读分析。

这个办法真好一个晴朗的日子,小鸡和小鸭打算去探望小白鹅,于是,它们出发了。

当它们朝小白鹅家赶的时候,一道路障——河流挡住了它们的去路。

两个人面对宽宽的河,奔流的水显出一脸无奈。

这时,小鸭想:上次不是刚学会游泳吗?我就试试我的本领吧!小鸭转忧为喜,小鸡则急得直跺脚,像热锅上的蚂蚁。

是呀,小鸡怎么过河呢?想到小鸡,小鸭不禁又烦恼了起来。

嘿,有办法了!只见小鸭急急忙忙地向森林跑去。

过了一会儿,小鸭从森林里跑出来,上气不接下气地对小鸡说:“这对你过河有帮助。

”小鸡疑惑不解地看着小鸭把一块木板放在水里,小鸭让小鸡站在木板上,小鸭摆动脚上的蹼,推着小木板,开始过河。

小鸡过了河。

河边的花朵都在向小鸭微笑,像在夸奖小鸭呢!1、“探望”是一个带有“看”意思的词语,请再写出四个带有“看”意思的词语。

2、理解句子:小鸭转忧为喜,小鸡则急得直跺脚,像热锅上的蚂蚁。

(1)小鸭“转忧为喜”,“忧”是因为,“喜”是因为。

小鸡急是因为。

(2)“小鸡则急得直跺脚,像热锅上的蚂蚁。

”写出了小鸡着急的程度,请你再写一句同样表现着急的句子。

小鸡急得。

3、根据短文内容填空。

(1)小鸭子想的好办法是:小鸭找来,把它放在。

让小鸡,小鸭在水里。

(2)“小鸡疑惑不解地看着小鸭把一块木板放在水里”,“疑惑不解”的意思是。

4、想象写话。

“河边的花朵都在向小鸭微笑,像在夸奖小鸭呢!”花朵们会夸奖一些什么呢?六、看图写话。

人教版八年级数学下册第十七章勾股定理周周测及答案4(17.2).doc

人教版八年级数学下册第十七章勾股定理周周测及答案4(17.2).doc

【若缺失公式、图片现象属于系统读取不成功,文档内容齐全完整,请放心下载。

】第十七章勾股定理周周测4一选择题1.下列各组数中不能作为直角三角形的三边长的是( )A.6,8,10 B.5,12,13 C.1,2,3 D.9,12,152.五根小木棒,其长度分别为7,15,20,24,25,现将他们摆成两个直角三角形,其中正确的是()3.三角形的三边长为a,b,c,且满足(a+b)2=c2+2ab,则这个三角形是( )A.等边三角形B.钝角三角形C.直角三角形D.锐角三角形4.若△ABC的三边a.b.c,满足(a-b)(a2+b2-c2)=0,则△ABC是()A.等腰三角形B.直角三角形C.等腰三角形或直角三角形D.等腰直角三角形5.下列说法中, 不正确的是( )A. 三个角的度数之比为1:3:4的三角形是直角三角形B. 三个角的度数之比为3:4:5的三角形是直角三角形C. 三边长度之比为3:4:5的三角形是直角三角形D. 三边长度之比为5:12:13的三角形是直角三角形6.有长度为9cm,12cm,15cm,36cm,39cm的五根木棒,可搭成(首尾连接)直角三角形的个数为()A.1个B.2个C.3个D.4个7.有下列判断:①△ABC中,,则△ABC不是直角三角形;②若△ABC是直角三角形,,则;③若△ABC中,,则△ABC是直角三角形;④若△ABC是直角三角形,则(,正确的有()A.4个B.3个C.2个D.1个8.如图,矩形ABCD中,AB=3,AD=1,AB在数轴上,若以点A为圆心,对角线AC的长为半径作弧交数轴于点M,则点M表示的数为()A.2 B. C. D.第8题图第9题图9. 如图,有一块地,已知AD=4米,CD=3米,∠ADC=90°,AB=13米,BC=12米,则这块地的面积为( )A. 24平方米B. 26平方米C. 28平方米D. 30平方米10.在下列条件中:①在△ABC中,∠A:∠B:∠C=1:2:3;②三角形三边长分别为32,42,52;③在△ABC中,三边a,b,c满足(a+b)(a-b)=c2;④三角形三边长分别为m-1,2m,m+1(m为大于1的整数),能确定△ABC是直角三角形的条件有()A.1个B.2个C.3个D.4个二填空题11.在△ABC中,如果(a+b)(a﹣b)=c2,那么∠=90°.12.若三角形三边分别为6,8,10,那么它最长边上的中线长是.13.某住宅小区有一块草坪如图所示,已知AB=3米,BC=4米,CD=12米,DA=13米,且AB⊥BC,这块草坪的面积是.14.若一个三角形的三边长分别为1.a.8(其中a为正整数),则以a-2,a,a+2为边的三角形面积为.15.在△ABC中,若其三条边的长度分别为9,12,15,则以两个这样的三角形所拼成的长方形的面积是________.16.如图,Rt△ABC中,∠ACB=90°,∠ABC=60°,BC=2cm,D为BC的中点,若动点E以1cm/s的速度从A点出发,沿着A→B→A的方向运动,设E点的运动时间为t秒,连接DE,当△BDE是直角三角形时,t的值.三解答题17.如图,一块地,已知AD=4m,CD=3m,∠ADC=90°,AB=13m,BC=12m.求这块地的面积.18.如图,已知∠ADC=90°,AD=8,CD=6,AB=26,BC=24.(1)证明:△ABC是直角三角形.(2)请求图中阴影部分的面积.19.如图,在△ABC中,∠ABC=45°,CD⊥AB,BE⊥AC,垂足分别为D.E,F为BC中点,BE与DF,DC分别交于点G,H,∠ABE=∠CBE.(1)求证:BH=AC;(2)求证:BG2-GE2=EA2.20.已知a.b.c为△ABC的三边,且满足a2c2﹣b2c2=a4﹣b4,试判断△ABC的形状.解:∵a2c2﹣b2c2=a4﹣b4,①∴c2(a2﹣b2)=(a2+b2)(a2﹣b2).②∴c2=a2+b2.③∴△ABC是直角三角形.问:(1)在上述解题过程中,从哪一步开始出现错误?请写出该步的代号:;(2)错误的原因为;(3)写出正确的解题过程.第十七章勾股定理周周测4试题答案1.C2.C3.C4.C5.B6.B7.C8.C9.A 10.B11.A12.5 13.36 14.24提示:7<a<9,∴a=8.15. 10816.2,6,3.5,4.5解析:∵∠ACB=90°,∠ABC=60°,BC=2cm,∴AB=BC÷cos60°=2÷=4.①∠BDE=90°时,∵D为BC的中点,∴DE是△ABC的中位线,∴AE=AB=×4=2(cm),点E在AB上时,t=2÷1=2(秒),点E在BA上时,点E运动的路程为4×2-2=6(cm),∴t=6÷1=6(秒);②∠BED=90°时,BE=BD•cos60°=×2×=0.5.点E在AB上时,t=(4-0.5)÷1=3.5(秒),点E在BA上时,点E运动的路程为4+0.5=4.5(cm),t=4.5÷1=4.5(秒),综上所述,t的值为2或6或3.5或4.5.17.2418.(1)证明:∵在Rt△ADC中,∠ADC=90°,AD=8,CD=6,∴AC2=AD2+CD2=82+62=100,∴AC=10.在△ABC中,∵AC2+BC2=102+242=676,AB2=262=676,∴AC2+BC2=AB2,∴△ABC 为直角三角形.(2)解:S阴影=S Rt△ABC﹣S Rt△ACD=×10×24﹣×8×6=96.19.证明:(1)∵CD⊥AB,BE⊥AC,∴∠BDH=∠BEC=∠CDA=90°,∵∠ABC=45°,∴∠BCD=180°-90°-45°=45°=∠ABC∴DB=DC.∵∠BDH=∠BEC=∠CDA=90°,∴∠A+∠ACD=90°,∠A+∠HBD=90°,∴∠HBD=∠ACD. ∵在△DBH和△DCA中,∠BDH =∠CDA,BD=CD,∠HBD=∠ACD,∴△DBH≌△DCA(ASA),∴BH=AC.(2)连接CG,由(1)知DB=CD.∵F为BC的中点,∴DF垂直平分BC,∴BG=CG.∵点E为AC中点,BE⊥AC,∴EC=EA.在Rt△CGE中,由勾股定理得CG2-GE2=CE2.∵CE=AE,BG=CG,∴BG2-GE2=EA2.20.解:(1)③(2)除式可能为零;(3)∵a2c2﹣b2c2=a4﹣b4,∴c2(a2﹣b2)=(a2+b2)(a2﹣b2),∴a2﹣b2=0或c2=a2+b2,当a2﹣b2=0时,a=b;当c2=a2+b2时,∠C=90°,∴△ABC是等腰三角形或直角三角形.中考数学知识点代数式一、重要概念分类:1.代数式与有理式用运算符号把数或表示数的字母连结而成的式子,叫做代数式。

第17周《西游记》导读及周测第93-100回(含答案)

第17周《西游记》导读及周测第93-100回(含答案)

《西游记》阅读周计划 第17周(93-100回)第93回 给孤园问古谈因 天竺国朝王遇偶1. 在禅寺被锁女子的真实身份是__天竺国公__。

2. 假公主搭起彩楼,__抛绣球__招亲,故意打中__唐僧_悟空定下了__倚婚降怪__之计。

3. 假公主招亲的目的是什么?采取唐僧元阳真气,,以成太乙上仙4. 阅读选段,说说各表现了唐僧、悟空、八戒的什么特点A.三藏急还礼,扶起众人,回头埋怨行者道:“你这猴头,又是撮弄我也“绣球儿打在你头上,滚在你袖里,干我何事?埋怨怎么?”三藏道:“似此怎道:“师父,你且放心。

便入朝见驾,我回驿报与八戒、沙僧等候。

若是公主不换了关文就行;如必欲招你,你对国王说,召我徒弟来,我要分付他一声。

’那朝,我其间自能辨别真假。

此是‘倚婚降怪’之计。

”唐僧无已从言,行者转身回驿。

B.八戒听说,跌脚捶胸道:“早知我去好来!都是那沙僧惫懒!——你不阻我啊,我径奔彩楼之下,一绣球打着我老猪,那公主招了我,却不美哉,妙哉!俊刮标致、停当,大家造化耍儿子,何等有趣!”唐僧胆小,没主意。

悟空机智聪明,思维灵活。

八戒美慕富贵,贪恋美色。

第94回 四僧宴乐御花园 一怪空怀情欲喜1.悟空为何引唐僧前去彩楼?为圆师父慕古之意,为查探给孤园长老所托之事,验明天竺国公主真伪 。

2.在本章节中,三个徒弟有没有如愿见到天竺国“公主“真容?为什么?没有见到; 因为“公主”借口徒弟长相丑恶,恐其惊伤弱体,示意国王迅速打发他们出城。

第95回 假合真形擒玉兔 真阴归正会灵元1. 假公主原来是广寒宫里的__玉兔__,使用的兵器短棍实是___捣药杵___。

2. 如何评价下面两人的做法?(1) 正此观看处,猪八戒动了欲心,忍不住,跳在空中,把霓裳仙子抱住道:“姐姐,我与你是旧相识,我和你耍子儿去也”。

(2) 行者道:“他这山,名为白脚山。

进来说有蜈蚣成精,黑夜伤人,往来行旅,甚为不便。

我思蜈蚣惟鸡可以降服,可选绝大雄鸡千只,撒放山中,除此毒虫。

2023年秋八上语文第17周周练:期末复习——记叙文阅读测试卷(含答案)

2023年秋八上语文第17周周练:期末复习——记叙文阅读测试卷(含答案)

第17周周测卷:期末专题复习——记叙文阅读测试卷满分:120分注意事项:1.答卷前,一定要保持稳定呼吸,不要焦躁,要自信,相信自己能找到答案!2.拿到试卷后,考生务必将自己的姓名、准考证号填写在试卷上。

3.不要舍不得放弃,一道题无思路做不出来,你在那里耗时间,就是在减少试卷分数!4.越狂躁,越容易粗心,和答案失之交臂!一、(16分)(2023·四川成都·八年级统考期末)阅读下面的文章,完成下面小题。

家里的老水缸吴新星①我家有两只缸。

一只黄褐色,外面上了釉,摸起来非常光滑,缸口外壁还有一圈草绳纹,这只缸放在天井里。

还有一只是灰青色的,与灶台为邻,一家人的平常日子就从这里开始。

②父亲常说,水缸里不能缺水,缺了水的日子就像长在墙头上的草,撑不了几天就会蔫头巴脑。

③水缸不会说话,但水缸里的水就是水缸的心思,清澈明亮,像极了水缸边淳朴勤劳的父母。

每天早上,父亲用瘦削的肩膀从远在几里外的村头挑回一担担井水,哗啦啦地倒入缸内。

母亲从地里忙完,又忙着生火点燃灶膛,从水缸里取水烧饭。

只要水缸里有水,灶腔里的火就不会熄。

即使粗茶淡饭,也是世间至真至淳的美味。

母亲怕缸里落了灰,平常总是把一块半月形的木板搁在上面,又一再叮嘱我们要爱惜缸里的水,这朴实的话里蕴含着的传统美德,影响了我们的一生。

那只放在天井里的缸呢,相形之下,我们就跟它亲近多了。

④即便是空空的一只放在天井里,在那个贫瘠的年代对于我们也是一件特别的玩具。

我们踮起脚尖,俯身向缸内喊上一声,缸内嗡嗡作响,恍若空谷回音。

下雨了,雨水自檐间下来,经过一个长长的锡皮管,落在承接的缸中。

雨水时而淅沥时而哗哗。

雨停了,水缸满满地注着透亮的水,像______。

我喜欢把头探到缸面上,痴痴地看个半天,和缸水又像大眼瞪小眼。

其实缸里没什么东西,只是清水。

可也有趣呀,可以看到微凸的缸底、褐色的缸壁和水里的折光。

⑤寒冬腊月,水缸表面结了一层厚厚的冰,小孩子们蜂拥而至,围住水缸,伸出两个手指在冰面上“走”,滑溜溜的。

高一语文“每周一练”系列试题及答案(17)

高一语文“每周一练”系列试题及答案(17)

高一语文“每周一练”系列试题及答案(17)高一语文“每周一练”系列试题及答案(17)?1.下列词语中加点的字,读音全都正确的一组是( )A.菁华(qīng) 宁可(nìng) 冠心病(guān) 翘首回望(qiáo)B.吐蕃(fān) 庇护(bǐ) 歼击机(jiān) 呱呱坠地(gū)C.请帖(tiě) 梵文(fán) 发横财(hèng) 按捺不住(nà)D.链接(liàn) 创口(chuāng) 倒春寒(dào) 拈花惹草(niān)2.下列各项中,加点的成语运用不恰当的一项是( )A.邻里之间的是非大多是由日常生活中的一些琐屑小事引起的,不必寻根究底,你们还是大事化小,小事化了吧。

B.身处春秋鼎盛的时代,我们这些身强体壮的青年应当奋勉有为,主动向上,刻苦学习,为国家和社会多作贡献。

C.这位代表说的虽不是什么崇论宏议,但他的话发自肺腑,句句实在,没有套话和假话,因此我们更加重视。

D.今年有四到六成的作品流拍,成交总额同比削减一半,这说明以往超过底价数十倍成交的火爆场面已成明日黄花。

3.下列各句中,加点的词语运用不恰当的一句是( )A.我们要重整山河,建设更加美妙的新北川!此时此刻,北川的复原重建正在紧锣密鼓进行,对这个全国唯一的羌族自治县文化的爱护也已悄然绽开。

D.本次中国古巴女排对抗赛上,因为赛前发烧,赵蕊蕊、冯坤昨天整场竞赛作壁上观,代替她们首发的是马蕴雯和魏秋月C.这儿的路很陡峭,空气很挑剔,耸人云霄的大山标记人生的一种高度,风雪垭口不容许生命中的弱者通过。

D.伴随经济的飞速发展,中国传统文化正在全面回来和复兴,过去境遇冷落的原生态民族民间歌舞,起先大模大样登上舞台。

4.下列各句中,没有语病的一句是( )A.中国就东海问题重申:春晓油气田完全在中国的主权权利范围内,与共同开发无关;在东海划界问题上,中方不承认所谓“中间线”的立场没有改变。

2022-2023学年第一学期北师大版九年级数学第十七周周末综合作业题(附答案)

2022-2023学年第一学期北师大版九年级数学第十七周周末综合作业题(附答案)

2022-2023学年第一学期北师大版九年级数学第十七周周末综合作业题(附答案)一.选择题1.如图,点A、B、C在⊙O上,∠ABO=30°,∠ACO=45°,则∠BOC等于()A.60°B.90°C.150°D.160°2.关于抛物线y=x2﹣2x﹣1,下列说法中错误的是()A.开口方向向上B.对称轴是直线x=1C.当x>1时,y随x的增大而减小D.顶点坐标为(1,﹣2)3.如图,在Rt△ABC中,∠ACB=90°,CD⊥AB于D,下列式子正确的是()A.sin A=B.cos A=C.tan A=D.cos B=4.如图,⊙O是△ABC的外接圆,∠B=60°,OP⊥AC于点P,OP=2,则⊙O的半径为()A.4B.6C.8D.125.如图,某厂生产一种扇形折扇,OB=10cm,AB=20cm,其中裱花的部分是用纸糊的,若扇子完全打开摊平时纸面面积为πcm2,则扇形圆心角的度数为()A.120°B.140°C.150°D.160°6.抛物线y=ax2+bx+c的对称轴为直线x=﹣1,图象过(1,0)点,部分图象如图所示,下列判断中:其中正确的个数是()①abc>0;②b2﹣4ac>0;③9a﹣3b+c=0;④若点(﹣2.5,y1),(﹣0.5,y2)均在抛物线上,则y1>y2;⑤5a﹣2b+c<0.A.2个B.3个C.4个D.5个7.如图,抛物线y=ax2+bx+c与x轴交于点A(﹣1,0),顶点坐标(1,n)与y轴的交点在(0,2),(0,3)之间(包含端点),则下列结论:①3a+b<0;②﹣1≤a≤﹣;③对于任意实数m,a+b≥am2+bm总成立;④关于x的方程ax2+bx+c=n﹣1有两个不相等的实数根.其中结论正确的个数为()A.1个B.2个C.3个D.4个8.如图,以点P为圆心,以为半径的圆弧与x轴交于A,B两点,点A的坐标为(2,0),点B的坐标为(6,0),则圆心P的坐标为()A.(4,)B.(4,2)C.(4,4)D.(2,)9.如图,将边长为1cm的等边三角形ABC沿直线l向右翻动(不滑动),点B从开始到结束,所经过路径的长度为()A.cm B.(2+π)cm C.cm D.3cm二.填空题10.如图,是半圆,点O为圆心,C、D两点在上,且AD∥OC,连接BC、BD.若=65°,则∠ABD的度数为.11.如图,一渔船由西往东航行,在A点测得海岛C位于北偏东60°的方向,前进20海里到达B点,此时,测得海岛C位于北偏东30°的方向,则海岛C到航线AB的距离CD 等于海里.12.如图,平行四边形ABCD的一边AB在x轴上,长为5,且∠DAB=60°,反比例函数y=和y=分别经过点C,D,则AD=.13.有三张正面分别写有数字﹣1,1,2的卡片,它们背面完全相同,现将这三张卡片背面朝上洗匀后随机抽取一张,以其正面数字作为a的值,然后再从剩余的两张卡片随机抽一张,以其正面的数字作为b的值,则点(a,b)在第二象限的概率为.14.《孙子算经》是我国古代重要的数学著作,成书于约一千五百年前,其中有道歌谣算题:“今有竿不知其长,量得影长一丈五尺,立一标杆,长一尺五寸,影长五寸,问杆长几何?”歌谣的意思是:有一根竹竿不知道有多长,量出它在太阳下的影子长一丈五,同时立一根一尺五的小标杆,它的影长五寸(提示:仗和尺是古代的长度单位,1丈=10尺,1尺=10寸),可以求出竹竿的长为尺.15.如图是某几何体的三视图及相关数据,则该几何体的侧面积是.16.如图,某地修建高速公路,要从A地向B地修一条隧道(点A,B在同一水平面上).为了测量A,B两地之间的距离,一架直升飞机从A地出发,垂直上升900米到达C处,在C处观察B地的俯角为30°,则A,B两地之间的距离为.17.若函数y=16x与y=的图象有一个交点是,则另一个交点坐标是.18.一艘货轮由西向东航行,在A处测得灯塔P在它的北偏东60°方向,继续航行到达B 处,测得灯塔P在正南方向10海里的C处是港口,点A、B、C在一条直线上,则这艘货轮由A处到B处航行的路程为海里(结果保留根号).19.如图,已知AM为⊙O的直径,直线BC经过点M,且AB=AC,∠BAM=∠CAM,线段AB和AC分别交⊙O于点D、E,∠BMD=40°,则∠EOM=.20.已知边长为6cm的等边三角形ABC,以AB为直径画半圆(如图),则阴影部分的面积是(结果保留π)21.如图所示,一动点从半径为2的⊙O上的A0点出发,沿着射线A0O方向运动到⊙O上的点A1处,再向左沿着与射线A1O夹角为60°的方向运动到⊙O上的点A2处;接着又从A2点出发,沿着射线A2O方向运动到⊙O上的点A3处,再向左沿着与射线A3O夹角为60°的方向运动到⊙O上的点A4处.……按此规律运动到点A2020处,则点A2020与点A0间的距离是.22.如图,两圆相交于A,B两点,小圆经过大圆的圆心O,点C,D分别在两圆上,若∠ADB=100°,则∠ACB的度数为.三.解答题23.计算题:(1)计算:sin45°+cos230°•tan60°﹣tan45°;(2)已知α是锐角,2sin(α﹣15°)=,求﹣|cosα﹣tan|的值.24.在一次数学兴趣小组活动中,阳光和乐观两位同学设计了如图所示的两个转盘做游戏(每个转盘被分成面积相等的几个扇形,并在每个扇形区域内标上数字).游戏规则如下:两人分别同时转动甲、乙转盘,转盘停止后,若指针所指区域内两数和小于12,则阳光获胜,反之则乐观获胜(若指针停在等分线上,重转一次,直到指针指向某一份内为止).(1)请用列表或画树状图的方法表示出上述游戏中两数和的所有可能的结果;(2)游戏对双方公平吗?请说明理由.25.如图,在平面直角坐标系中,将一块等腰直角三角板ABC放在第二象限,点C坐标为(﹣1,0),点A坐标为(0,2).一次函数y=kx+b的图象经过点B、C,反比例函数y =的图象经过点B.(1)求一次函数和反比例函数的关系式;(2)直接写出当x<0时,kx+b﹣<0的解集;(3)在x轴上找一点M,使得AM+BM的值最小,直接写出点M的坐标和AM+BM的最小值.26.如图,某仓储中心有一斜坡AB,其坡度为i=1:2,顶部A处的高AC为4m,B、C在同一水平地面上.(1)求斜坡AB的水平宽度BC;(2)矩形DEFG为长方体货柜的侧面图,其中DE=2.5m,EF=2m,将该货柜沿斜坡向上运送,当BF =3.5m 时,求点D 离地面的高.(结果保留根号)27.汽车租赁公司拥有某种型号的汽车100辆.公司在经营中发现每辆车的月租金x (元)与每月租出的车辆数(y )有如下关系:x (元)3000 3200 3500 4000 y (辆) 100 96 90 80(1)观察表格,用所学过的一次函数、反比例函数或二次函数的有关知识,求按照表格呈现的规律,每月租出的车辆数y (辆)与每辆车的月租金x (元)之间的关系式.(2)已知租出的车每辆每月需要维护费150元,未租出的车每辆每月需要维护费50元.用含x (x ≥3000)的代数式填表:租出的车辆数(辆)未租出的车辆数(辆) 租出每辆车的月收益(元) 所有未租出的车辆每月的维护费(元)(3)若你是该公司的经理,你会将每辆车的月租金定为多少元,才能使公司获得最大月收益?请说明理由.28.如图,点M 在函数y =(x >0)的图象上,过点M 分别作x 轴和y 轴的平行线交函数y =(x >0)的图象于点B ,C .(1)若点M 的坐标为(1,3),求B ,C 两点的坐标;(2)若点M 是y =(x >0)的图象上任意一点,求△BMC 的面积.29.为了身体健康,越来越多的人喜欢上了行走健身,为方便群众步行健身,某地政府决定对一段如图1所示的坡路进行改造.如图2所示,改造前的斜坡AB =260米,坡度为1:;将斜坡AB 的高度AE 降低AC =30米后,斜坡AB 改造为斜坡CD ,其坡度为1:4.求斜坡CD的长.(结果保留根号)30.某校组织代表队参加市“与经典同行”吟诵大赛,初赛后对选手成绩进行了整理,分成5个小组(x表示成绩,单位:分).A组:75≤x<80;B组:80≤x<85;C组:85≤x <90;D组:90≤x<95;E组:95≤x<100,并绘制如下两幅不完整的统计图:请根据图中信息,解答下列问题:(1)参加初赛的选手共有名,请补全频数分布直方图;(2)扇形统计图中,E组人数占参赛选手的百分比是多少?它对应的圆心角是多少度?(3)学校准备组成8人的代表队参加市级决赛,E组6名选手直接进入代表队,现要从D组中的两名男生和两名女生中,随机选取两名选手进入代表队,请用列表或画树状图的方法,求恰好选中两名女生的概率.31.某广告公司设计一幅周长为16米的矩形广告牌,广告设计费为每平方米2000元,设矩形一边长为x,面积为S平方米.(1)求S与x之间的函数关系式,并写出自变量x的取值范围;(2)设计费能可以达到30000元吗?为什么?(3)当x是多少米时,设计费最多?最多是多少元?32.如图,一次函数y=kx+b与反比例函数y=(x>0)交于A(2,4),B(a,1)两点,与x轴、y轴分别交于点C,D.(1)求一次函数和反比例函数的表达式;(2)求证:AD=BC.33.如图,AB是⊙O的直径,C是⊙O上一点,D是的中点,E为OD延长线上一点,且∠CAE=2∠C,AC与BD交于点H,与OE交于点F.(1)求证:AE是⊙O的切线;(2)若DH=9,sin C=,求直径AB的长.34.如图1,已知抛物线y=ax2+bx+3(a≠0)与x轴交于点A(1,0)和点B(﹣3,0),与y轴交于点C.(1)求抛物线的表达式;(2)如图1,若点E为第二象限抛物线上一动点,连接BE,CE,求四边形BOCE面积的最大值,并求此时E点的坐标;(3)如图2,在x轴上是否存在一点D使得△ACD为等腰三角形?若存在,请求出所有符合条件的点D的坐标;若不存在,请说明理由.35.如图,AB是⊙O的直径,C是的中点,CE⊥AB于E,BD交CE于点F,(1)求证:CF=BF;(2)若CD=12,AC=16,求⊙O的半径和CE的长.36.某宾馆有30个房间供游客住宿,当每个房间的房价为每天160元时,房间会全部住满.当每个房间每天的房价每增加10元时,就会有一个房间空闲.宾馆需对游客居住的每个房间每天支出20元的各种费用.根据规定,每个房间每天的房价不得高于260元.设每个房间的房价每天增加x元(x为10的整数倍).(1)设一天订住的房间数为y,直接写出y与x的函数关系式及自变量x的取值范围;(2)设宾馆一天的利润为w元,求w与x的函数关系式;(3)一天订住多少个房间时,宾馆的利润最大?最大利润是多少元?37.如图,已知抛物线与x轴交于A(﹣1,0)、E(3,0)两点,与y轴交于点B(0,3).(1)求抛物线的解析式;(2)设抛物线顶点为D,求四边形AEDB的面积;(3)△AOB与△DBE是否相似?如果相似,请给以证明;如果不相似,请说明理由.38.如图,BC是⊙O的直径,A是⊙O上一点,过点C作⊙O的切线,交BA的延长线于点D,取CD的中点E,AE的延长线与BC的延长线交于点P.(1)说明:AP是⊙O的切线;(2)若OC=CP,AB=6,求CD的长.39.把一副三角板如图甲放置,其中∠ACB=∠DEC=90°,∠A=45°,∠D=30°,BC =+1.斜边AB、DC相交于点O.(1)求CO的长;(2)若把三角板DCE绕点C顺时针旋转15°得到△D1CE1(如图乙),这时AB与CD1相交于点O1,此时,求:CO1的长;(3)若把三角板D1CE1绕着点C顺时针再旋转15°得△D2CE2(如图丙),这时AB与CD2相交于点O2,此时,求:CO2的长.40.如图,抛物线y=﹣x2+bx+c交x轴于A,B两点,交y轴于点C,直线y=﹣x+2经过点B,C.(1)求抛物线的解析式;(2)点P是直线BC上方抛物线上一动点,设点P的横坐标为m.求△PBC面积最大值和此时m的值;(3)Q是抛物线上一点,若∠ABC=∠CBQ,直线BQ与y轴交于点M,请直接写出M 的坐标.参考答案一.选择题1.解:过A作⊙O的直径,交⊙O于D;在△OAB中,OA=OB,则∠BOD=∠ABO+∠OAB=2×30°=60°,同理可得:∠COD=∠ACO+∠OAC=2×45°=90°,故∠BOC=∠BOD+∠COD=150°.故选:C.2.解:抛物线y=x2﹣2x﹣1,∵a=1>0,∴开口方向向上,故选项A不合题意;对称轴是直线x=﹣=﹣=1,故选项B不合题意;当x>1时,y随x的增大而增大,故选项C符合题意;y=x2﹣2x﹣1=(x﹣1)2﹣2,顶点坐标为(1,﹣2),故选项D不合题意.故选:C.3.解:∵∠ACB=90°,CD⊥AB,∴∠A+∠DCA=90°,∠DCA+∠BCD=90°,∴∠A=∠BCD,∴sin A=sin∠BCD=,故选:A.4.解:∵圆心角∠AOC与圆周角∠B所对的弧都为,且∠B=60°,∴∠AOC=2∠B=120°,又OA=OC,∴∠OAC=∠OCA=30°,∵OP⊥AC,∴∠APO=90°,在Rt△AOP中,OP=2,∠OAC=30°,∴OA=2OP=4,则圆O的半径4.故选:A.5.解:∵OB=10cm,AB=20cm,∴OA=OB+AB=30cm,设扇形圆心角的度数为α,∵纸面面积为πcm2,∴﹣=π,∴α=150°,故选:C.6.解:①由图象开口向上,则a>0,故b>0,∵c<0,∴abc<0,故①错误.②∵抛物线与x轴有两个交点,∴b2﹣4ac>0,故②正确.③∵抛物线与x轴的一个交点是(1,0),对称轴是直线x=﹣1,∴抛物线与x轴的另一个交点是(﹣3,0),∴9a﹣3b+c=0,故③正确.④∵点(﹣0.5,y2)在抛物线上,对称轴为直线x=﹣1,∴(﹣1.5,y2)也在抛物线上,∵﹣1.5>﹣2.5,且(﹣1.5,y2),(﹣2.5,y1)都在对称轴的左侧,∴y1>y2,故④正确.⑤∵抛物线对称轴x=﹣1,经过(1,0),∴﹣=﹣1,a+b+c=0,∴b=2a,c=﹣3a,∴5a﹣2b+c=5a﹣4a﹣3a=﹣2a<0,∴⑤正确.故正确的判断是②③④⑤共4个.故选:C.7.解:∵抛物线开口向下,∴a<0,而抛物线的对称轴为直线x=﹣=1,即b=﹣2a,∴3a+b=3a﹣2a=a<0,所以①正确;∵2≤c≤3,把x=﹣1,y=0代入y=ax2+bx+c,得a﹣b+c=0,∴c=﹣3a,∴2≤﹣3a≤3,∴﹣1≤a≤﹣,所以②正确;∵抛物线的顶点坐标(1,n),∴x=1时,二次函数值有最大值n,∴a+b+c≥am2+bm+c,即a+b≥am2+bm,所以③正确;∵抛物线的顶点坐标(1,n),∴抛物线y=ax2+bx+c与直线y=n﹣1有两个交点,∴关于x的方程ax2+bx+c=n﹣1有两个不相等的实数根,所以④正确.故选:D.8.解:过点P作PC⊥AB于点C;即点C为AB的中点,又点A的坐标为(2,0),点B的坐标为(6,0),故点C(4,0)在Rt△P AC中,P A=,AC=2,即有PC=4,即P(4,4).故选:C.9.解:∵△ABC是等边三角形,∴∠ACB=60°,∴∠AC(A)=120°,点B两次翻动划过的弧长相等,则点B经过的路径长=2×=π(cm).故选:C.二.填空题10.解:∵是半圆,即AB是直径,∴∠ADB=90°,又∵AD∥OC,∴OC⊥BD,∴=65°,∴=180°﹣65°﹣65°=50°,∴∠ABD=.故答案为:25°.11.解:根据题意可知∠CAD=30°,∠CBD=60°,∵∠CBD=∠CAD+∠ACB,∴∠CAD=30°=∠ACB,∴AB=BC=20海里,在Rt△CBD中,∠BDC=90°,∠DBC=60°,sin∠DBC=,∴sin60°=,∴CD=20×sin60°=20×=10海里,故答案为:10.12.解:设点C(x,),则点D(﹣x,),∴CD=x﹣(﹣x)=x,∵四边形ABCD是平行四边形,∴CD=AB=5,∴x=5,解得x=2,∴D(﹣3,),作DE⊥AB于E,则DE=,∵∠DAB=60°,∴AD===2,故答案为2.13.解:画树状图图得:∵共有6种等可能的结果,点(a,b)在第二象限的有2种情况,∴点(a,b)在第二象限的概率为:=.故答案为:.14.解:设竹竿的长度为x尺,∵竹竿的影长=一丈五尺=15尺,标杆长=一尺五寸=1.5尺,影长五寸=0.5尺,∴=,解得x=45(尺).故答案为:45.15.解:根据图形可知圆锥的高为6,母线长为8,则底面半径为2,圆锥侧面积公式=底面周长×母线长×,圆锥侧面积=×π×2×2×8=16π.故答案为:16π.16.解:由题意知∠BAC=90°,∠ABC=30°,AC=900米,∵tan∠ABC=,∴AB===900(米),故答案为:900米.17.解:∵两函数图象关于原点对称,∴两函数图象交点关于原点对称,∴的对称点为(﹣,﹣4).故答案为(﹣,﹣4).18.解:根据题意得:PC=10海里,∠PBC=90°﹣45°=45°,∠P AC=90°﹣60°=30°,在直角三角形APC中,∵∠P AC=30°,∠C=90°,∴AC=PC=10(海里),在直角三角形BPC中,∵∠PBC=45°,∠C=90°,∴BC=PC=10海里,∴AB=AC=BC=(10﹣10)(海里);故答案为:(10﹣10).19.解:连接EM,∵AB=AC,∠BAM=∠CAM,∴AM⊥BC,∵AM为⊙O的直径,∴∠ADM=∠AEM=90°,∴∠AME=∠AMD=90°﹣∠BMD=50°∴∠EAM=40°,∴∠EOM=2∠EAM=80°,故答案为:80°.20.解:如图,根据等边三角形和圆的对称性,阴影部分的面积就是扇形OMN的面积,由题意得,扇形OMN的半径为3cm,圆心角的度数为60°,S阴影部分=S扇形OMN==,故答案为:.21.解:如图,∵⊙O的半径=2,由题意得,A0A1=4,A0A2=2,A0A3=2,A0A4=2,A0A5=2,A0A6=0,A0A7=4,…∵2020÷6=336…4,∴按此规律运动到点A2020处,A2020与A4重合,∴A0A2020=A0A4=2.22.解:如图:连接OA,OB,∵四边形AOBD是圆内接四边形,∴∠AOB+∠D=180°,∵∠ADB=100°,∴∠AOB=80°,∴∠ACB=40°.故答案为:40°.三.解答题23.解:(1)原式==1+﹣1=;(2)∵2sin(α﹣15°)=,∴,∴α﹣15°=45°,∴α=60°,∴原式=====1﹣24.解:(1)根据题意列表如下:6789 39101112410111213511121314可见,两数和共有12种等可能结果;(2)游戏对双方公平,理由如下:∵两数和共有12种等可能的情况,其中和小于12的情况有6种,∴阳光获胜的概率为=,∴乐观获胜的概率是,∵=,∴游戏对双方公平.25.解:(1)过点B作BF⊥x轴于点F,∵点C坐标为(﹣1,0),点A坐标为(0,2).∴OA=2,OC=1,∵∠BCA=90°,∴∠BCF+∠ACO=90°,又∵∠CAO+∠ACO=90°,∴∠BCF=∠CAO,在△AOC和△CFB中∴△AOC≌△CFB(AAS),∴FC=OA=2,BF=OC=1,∴点B的坐标为(﹣3,1),将点B的坐标代入反比例函数解析式可得:1=,解得:k=﹣3,故可得反比例函数解析式为y=﹣;将点B、C的坐标代入一次函数解析式可得:,解得:.故可得一次函数解析式为y=﹣x﹣.(2)结合点B的坐标及图象,可得当x<0时,kx+b﹣<0的解集为:﹣3<x<0;(3)作点A关于x轴的对称点A′,连接B A′与x轴的交点即为点M,∵A(0,2),∴A′(0,﹣2),设直线BA′的解析式为y=ax+b,将点A′及点B的坐标代入可得:,解得:.故直线BA′的解析式为y=﹣x﹣2,令y=0,可得﹣x﹣2=0,解得:x=﹣2,故点M的坐标为(﹣2,0),AM+BM=BM+MA′=BA′==3.综上可得:点M的坐标为(﹣2,0),AM+BM的最小值为3.26.解:(1)∵坡度为i=1:2,AC=4m,∴BC=4×2=8m.(2)作DS⊥BC,垂足为S,且与AB相交于H.∵∠DGH=∠BSH,∠DHG=∠BHS,∴∠GDH=∠SBH,∴=,∵DG=EF=2m,∴GH=1m,∴DH==m,BH=BF+FH=3.5+(2.5﹣1)=5m,设HS=xm,则BS=2xm,∴x2+(2x)2=52,∴x=m∴DS=+=2m.27.解:(1)由表格数据可知y与x是一次函数关系,设其解析式为y=kx+b.由题:,解之得:,∴y与x间的函数关系是y=﹣x+160.(2)如下表:租出的车辆数﹣x+160未租出的车辆数x﹣60租出的车每辆的月收益x﹣150所有未租出的车辆每月的维护费x﹣3000(3)设租赁公司获得的月收益为W元,依题意可得:W=(﹣x+160)(x﹣150)﹣(x﹣3000)=(﹣x2+163x﹣24000)﹣(x﹣3000)=﹣x2+162x﹣21000=﹣(x﹣4050)2+307050当x=4050时,Wmax=307050,即:当每辆车的月租金为4050元时,公司获得最大月收益307050元.故答案为:﹣x+160,x﹣60,x﹣150,x﹣3000.28.解:(1)∵点M的坐标为(1,3),MC⊥x轴,MB⊥y轴,且B,C在函数的图象上,∴当x=1时,y=1,∴点C的坐标为(1,1)∴当y=3时,,∴点B的坐标为;(2)设点M的坐标为(a,b)∵点M在函数的图象上,∴ab=3∵点B,C在函数的图象上,∴点C的坐标为,B点坐标为∴,,∴S△BMC=•BM•MC=••=•=.29.解:在Rt△ABE中,∵,∴∠ABE=30°,∵AB=260,∴,∵AC=30,∴CE=130﹣30=100,在Rt△CDE中,∵tan D=1:4,∴,∴,∴(米),答:斜坡CD的长是米.30.解:(1)参加初赛的选手共有:8÷20%=40(人),B组有:40×25%=10(人),频数分布直方图补充如下:故答案为:40;(2)E组人数占参赛选手的百分比是:×100%=15%;E组对应的圆心角度数是:360°×15%=54°;(3)根据题意画树状图如下:由上图可以看出,所有可能出现的结果有l2种,这些结果出现的可能性相等,选中两名女生的结果有2种,则选中两名女生的概率是==.31.解:(1)矩形的一边长为x米,周长为16米.另一边长为(8﹣x)米,∴S=x(8﹣x)=﹣x2+8x,其中0<x<8;(2)能.理由是:∵设计费为每平方米2000元,∴当设计费为30000元时,面积为:30000÷2000=15(平方米)即﹣x2+8x=15,解得x1=3,x2=5;∴设计费能达到30000元;(3)∵S=﹣x2+8x=﹣(x﹣4)2+16,∴当x=4时,S最大值=16,∴16×2000=32000.∴当x是4米时,矩形的最大面积为16平方米,设计费最多,最多是32000元.32.解:(1)将A(2,4)代入,得m=8,∴反比例函数的表达式为∴又B(a,1)在反比例函数的图象上,∴,解得,a=8∴B(8,1)将A(2,4),B(8,1)代入y=kx+b中,得,解得:∴一次函数的表达式为.(2)由(1)可知,一次函数的表达式为当x=0时,y=5;当y=0时,x=10;∴C(10,0),D(0,5),如图,过点A作y轴的垂线与y轴交于点E,过B作x轴的垂线与x轴交于点F,∴E(0,4),F(8,0),∴AE=2,DE=1,BF=1,CF=2∴在Rt△ADE中,由勾股定理得:在Rt△BCF中,由勾股定理得:∴AD=BC.33.证明:(1)连接OC,∵D是的中点,∴∠AOD=∠COD∵OA=OC,∴OE⊥AC,即∠AFE=90°,∴∠E+∠EAF=90°∵∠AOE=2∠C,∠CAE=2∠C,∴∠CAE=∠AOE∴∠E+∠AOE=90°,∴∠EAO=90°∴AE是⊙O的切线(2)∵∠C=∠B∵OD=OB,∴∠B=∠ODB,∴∠ODB=∠C∴,∴由勾股定理得:∵∠C=∠FDH,∠DFH=∠CFD∴△DFH∽△CFD∴∴∴设OA=OD=x,∴∵AF2+OF2=OA2∴,解得:x=10∴OA=10∴直径AB的长为20.34.解:(1)将点A(1,0),B(﹣3,0)代入y=ax2+bx+3,得,,解得,,∴抛物线表达式为y=﹣x2﹣2x+3;(2)如图1,过点E作EF⊥x轴于点F,设E(a,﹣a2﹣2a+3)(﹣3<a<0),∴EF=﹣a2﹣2a+3,BF=a+3,OF=﹣a,∴===,∴当时,S四边形BOCE最大,且最大值为;当时,,此时,点E坐标为;(3)如图2,连接AC,①当CA=CD时,此时CO为底边的垂直平分线,满足条件的点D1,与点A关于y轴对称,点D1坐标为(﹣1,0);②当AD=AC时,在Rt△ACO中,∵OA=1,OC=3,由勾股定理得,AC==,以点A为圆心,AC的长为半径作弧,交x轴于两点D2,D3,即为满足条件的点,此时它们的坐标分别为,;③当DA=DC时,线段AC的垂直平分线与x轴的交点D4,即为满足条件的点,设垂直AC的垂直平分线交y轴于点P,过AC中点Q,∵∠AOC=∠BOC=∠PQC=∠PQA=90°,∠D4PO=∠CPQ,∴∠ACO=∠OD4P,∴△D4AQ∽△CAO,∴=,即=,∴D4A=5,∴OD4=D4A﹣OA=4,∴点D4的坐标为(﹣4,0);综上所述,存在符合条件的点D,其坐标为D1(﹣1,0)或或或D4(﹣4,0).35.解:(1)证明:∵AB是⊙O的直径,∴∠ACB=90°,又∵CE⊥AB,∴∠CEB=90°,∴∠2=90°﹣∠ABC=∠A,又∵C是弧BD的中点,∴∠1=∠A,∴∠1=∠2,∴CF=BF;(2)∵C是弧BD的中点,∴=,∴BC=CD=12,又∵在Rt△ABC中,AC=16,∴由勾股定理可得:AB=20,∴⊙O的半径为10,∵S△ABC=AC•BC=AB•CE,∴CE==9.6.36.解:(1)y=30﹣x(0≤x≤100,且x是10的整数倍);(2)w=(30﹣x)(160+x﹣20)=﹣x2+16x+4200;(3)w=﹣x2+16x+4200=﹣(x﹣80)2+4840∴当x=80时,w最大为10600.当x=80时,y=30﹣x=22.答:一天订住22个房间时,宾馆每天利润最大,最大利润是4840元.37.解:(1)∵抛物线与y轴交于点(0,3),∴设抛物线解析式为y=ax2+bx+3(a≠0)根据题意,得,解得.∴抛物线的解析式为y=﹣x2+2x+3;(2)如图,设该抛物线对称轴是DF,连接DE、BD.过点B作BG⊥DF于点G.由顶点坐标公式得顶点坐标为D(1,4)设对称轴与x轴的交点为F∴四边形ABDE的面积=S△ABO+S梯形BOFD+S△DFE=AO•BO+(BO+DF)•OF+EF•DF=×1×3+×(3+4)×1+×2×4=9;(3)相似,如图,BD=;∴BE=DE=∴BD2+BE2=20,DE2=20即:BD2+BE2=DE2,所以△BDE是直角三角形∴∠AOB=∠DBE=90°,且,∴△AOB∽△DBE.38.(1)证明:连接AO,AC(如图).∵BC是⊙O的直径,∴∠BAC=∠CAD=90°.∵E是CD的中点,∴CE=DE=AE.∴∠ECA=∠EAC.∵OA=OC,∴∠OAC=∠OCA.∵CD是⊙O的切线,∴CD⊥OC.∴∠ECA+∠OCA=90°.∴∠EAC+∠OAC=90°.∴OA⊥AP.∵A是⊙O上一点,∴AP是⊙O的切线;(2)解:由(1)知OA⊥AP.在Rt△OAP中,∵∠OAP=90°,OC=CP=OA,即OP=2OA,∴sin P=.∴∠P=30°.∴∠AOP=60°.∵OC=OA,∴∠ACO=60°.在Rt△BAC中,∵∠BAC=90°,AB=6,∠ACO=60°,∴.又∵在Rt△ACD中,∠CAD=90°,∠ACD=90°﹣∠ACO=30°,∴CD====4.39.解:(1)过点O作OH⊥BC于点H.在Rt△OHB中,∠HOB=90°﹣∠B=45°=∠B∴OH=HB.∵在Rt△DCE中,∠DCE=90°﹣∠D=60°∴在Rt△OHC中,∠COH=90°﹣∠OCH=90°﹣60°=30°∴OC=2CH.又∵OH=CH•tan∠OCH=,∴HB=OH=.又∵CH+HB=CB,∴CH+=.∴CH=1.∴CO=2CH=2;(2)∵∠BCE1=15°∴∠O1CB=60°﹣15°=45°=∠B.∴∠CO1B=180°﹣(45°+45°)=90°∴CO1=BC•sin∠B==;(3)从甲图到丙图的过程中,由于旋转角均为15°,且在乙图中CO1⊥AB,所以CO2与CO在这个旋转过程中关于直线CO1成轴对称.所以CO2=CO=2.40.解:(1)针对于直线y=﹣x+2,令x=0,则y=2,∴C(0,2),令y=0,则﹣x+2=0,∴x=4,∴B(4,0),∵抛物线y=﹣x2+bx+c过点B,点C,∴,∴,∴抛物线的解析式为y=﹣x2+x+2;(2)如图1,过点P作PD∥y轴交直线BC于D,∵点P的横坐标为m,∴P(m,﹣m2+m+2),D(m,﹣m+2),∴PD=﹣m2+m+2﹣(﹣m+2)=﹣m2+4m,∴S△PBC=PD(x B﹣x C)=(﹣m2+4m)×4=﹣2(m﹣2)2+8,当m=2时,S△PBC最大,其值为8.(3)如图2,过点C作CN⊥BM于N,∴∠MNC=90°=∠BOC,∵∠ABC=∠CBQ,∴CN=OC=2,∵∠CMN=∠BMO,∠CNM=∠BOM=90°,∴△MNC∽△MOB,∴,∴,∴OM=2MN,∴CM=OM﹣OC=2MN﹣2,在Rt△CNM中,根据勾股定理得,MN2+CN2=CM2,∴MN2+4=(2MN﹣2)2,∴MN=0(舍)或MN=,∴OM=2MN=,∴M(0,),当点Q与点A重合时,点M和点O重合,此时M(0,0),即点M的坐标为(0,0)或(0,).。

2021-2022年高二上学期第十七次周练英语试题 含答案

2021-2022年高二上学期第十七次周练英语试题 含答案

Ⅰ.单项填空(每小题1分,共15分)从题中所给的A、B、C、D四个选项中选出一个最佳答案。

1.She's developing into someone who's going to bee very poor and require ________.A.looked after B.being looked afterC.looking after D.to look after2.The survey shows that a large number of teenagers are ________ cigarettes nowadays.A.devoted to B.related toC.connected to D.addicted to3.I searched the website. Unfortunately, the movie that I had seen many times was no longer ________.A.valuable B.reliableC.available D.incredible4.—Have you got the news about Jack? I haven't seen him for many years.—He ________ in a pany for two years. Now he is a teacher.A.worked B.has workedC.had worked D.would work5.Poor Tony has ________ a bad headache for a long time, so he didn't passthe mid­term examination.A.suffered from B.sufferedC.stood D.borne6.The racers ________ their positions on the starting line at the sound of the whistle.A.took up B.made upC.help up D.set up7.I ________ to give her a hand to carry the suitcase but she refused.A.afforded B.managedC.tried D.offered8.I took ________ for granted that the Chinese Women Volleyball Team would lose the final match, but to our astonishement, we won it.A.what B.thatC.one D.it9.A button ________ as I was climbing over the wall in a hurry.A.came away B.came outC.came down D.came off10.Mike ________ have stolen the money from the bank, ________ he appeared to be rich overnight.A.can't; for B.might; forC.might; since D.can't; since11.He was ________ with the results of the exam, which can be seen from his ________ smile.A.satisfied; satisfyingB.satisfied; satisfiedC.satisfying; satisfiedD.satisfying; satisfying12.He ________ to lend me his dictionary but I refused, for I don't think it of any use.A.managed B.volunteeredC.considered D.afford13.Your suggestion sounds fine ________, but would it work ________?A.in practice; in theoryB.in theory; in practiceC.in advance; in theoryD.in general: in practice14.—Hello, is that Tom speaking?—Sorry, you ________ the wrong number.A.may have dialed B.must have dialedC.could dial D.might dial15.________ time going on, we are growing up and e to understand our parents.A.When B.WithC.As D.WhileⅡ.完形填空(每小题1分,共20分)阅读下面短文,理解大意,从题中所给的A、B、C、D四个选项中选出一个最佳答案。

八年级现代文阅读全解全练第5部分第17周

八年级现代文阅读全解全练第5部分第17周

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【文学类文本阅读】 一号套餐 8 块钱
①有一年暑假,妈妈给我找了一份暑期工——在麦当劳做 售餐员。我每天要在柜台前站 8 个小时,挂着机械化的微笑, 把鸡翅、汉堡、薯条快速分类摆放在餐托上,再甜甜地说:“谢 谢光临,请您慢用。”对这份一天挣不了多少钱的工作,我始 终抱着应付的态度。
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⑧在这方面,美国女孩丽兹也给我们上了深刻的一课。丽 兹因为患有罕见病而外表怪异,被媒体粗暴地评为“世界最丑 的女人”。她在经历了噩梦般的精神崩溃之后,最终战胜自我, 勇敢地站了出来,举办演讲、建立个人网站和摄制纪录片,以 此来与偏见斗争,对舆论的霸凌说不。她找到了勇气和自信, 也赢得了尊重和尊严。
第五部分 读写素养及能力提 升组合训练•周周练
第十七周
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实用类文本 第十七周
阅读(10 分) 建议用时 限时训练 10 分钟 实际用时 得分情况 失误分析
文学类文本 阅读(17 分) 建议用时 20 分钟 实际用时
作文(50 分)
建议用时 40 分钟 实际用时
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【实用类文本阅读】 青春别被“颜值”裹挟
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(2)隐瞒,不一定就是欺骗,而是因为太爱对方;坦白,也不 一定是坦诚,有时候恰恰是冷漠。(从修辞手法的角度)
示例:运用对比的修辞手法,将“隐瞒”和“坦白”两种对 待亲人的态度进行对比,突出强调了二者背后所代表的深刻 含义和辩证关系,赞美亲情的珍贵,呼唤人间温情。
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3.请结合文章内容,分析文中“我”的人物形象。(4 分) ①“我”是一个涉世不深的学生,心地善良又善解人意。 “我”愿意帮着那个女大学生隐瞒价格,成全她的一片孝心。 ②同时“我”又是一个富有理性思考、善于发现生活、懂得 珍惜幸福的人。“我”从这件事中很受触动和启发,看到最 朴素的人间亲情并提醒自己要珍惜,认识到亲情是无法用金 钱衡量的。

2024年【每周一测】第十七周语文八年级上册基础练习题(含答案)

2024年【每周一测】第十七周语文八年级上册基础练习题(含答案)

2024年【每周一测】第十七周语文八年级上册基础练习题(含答案)试题部分一、选择题(每题2分,共20分)1. 下列词语中,加点字的注音全都正确的一项是()A. 摇曳(yè)惬意(qiè)纤夫(qiàn)B. 徜徉(cháng)感慨(kǎi)惊骇(hài)C. 沉溺(nì)呕心沥血(lì)豁达(huò)D. 稠密(chóu)氛围(fēn)惊悚(sǒng)2. 下列句子中,加点成语使用不正确的一项是()A. 他做事总是小心翼翼,生怕出错。

B. 这部电影情节跌宕起伏,引人入胜。

C. 在危急关头,他挺身而出,舍己为人。

D. 老师语重心长地教导我们,要珍惜时间,努力学习。

3. 下列句子中,没有语病的一项是()A. 通过这次活动,使同学们认识到了保护环境的重要性。

B. 春天的校园里,万物复苏,鸟语花香,充满了生机。

C. 为了防止疫情不再蔓延,我们要做好个人防护。

D. 他穿着一件蓝色外套,一顶帽子,在人群中格外显眼。

4. 下列诗句中,表达了作者对朋友依依不舍之情的是()A. 劝君更尽一杯酒,西出阳关无故人。

B. 莫愁前路无知己,天下谁人不识君。

C. 海内存知己,天涯若比邻。

D. 故人西辞黄鹤楼,烟花三月下扬州。

5. 下列文学常识,表述正确的一项是()A. 《背影》是朱自清的一篇散文,讲述了他离开家乡的故事。

B. 《春》是老舍的一篇散文,描绘了春天的美好景色。

C. 《济南的冬天》是鲁迅的一篇散文,赞美了济南冬天的美景。

D. 《记念刘和珍君》是茅盾为纪念刘和珍而写的散文。

6. 下列句子中,标点符号使用正确的一项是()A. “你今天怎么没来上课?”老师问。

B. “这道题怎么做?”他疑惑地看着我。

C. “我要去图书馆还书,”她说,“你一起去吗?”D. “这个书包真好看!”她说:“多少钱一个?”7. 下列词语中,没有错别字的一项是()A. 班门弄斧振聋发聩B. 气喘吁吁眼疾手快C. 和颜悦色和蔼可亲D. 情不自禁意味深长8. 下列句子中,加点词语使用不恰当的一项是()A. 他在这场比赛中表现英勇,被誉为赛场上的勇士。

七年级科学17周双休日作业

七年级科学17周双休日作业

七年级科学17周双休日作业班级_____________姓名____________家长签名_________一、选择题1、下列不属于科学问题的是 ( )A .千岛湖中哪种鱼脂肪含量最低B .杭州植物园内哪种花卉最美观C .杭州西溪国家湿地公园内哪几种植物存冬季开花D .天目山自然保护区内哪种植物自然生长速度最快2、有着“植物大熊猫”和世界上最长寿树种美称的红豆杉,成为世博会中国馆珍稀植物展出品种。

红豆杉因其种子成熟时假皮呈红色得名。

从植物分类上看,红豆杉属于( )A .被子植物B .裸子植物C .蕨类植物D .苔藓植物3、洋葱表皮细胞和口腔上皮细胞都具有的结构是 ( )A .细胞壁、细胞质、细胞核B .细胞膜、细胞质、细胞核C .细胞膜、叶绿体、液泡D .细胞壁、叶绿体、液泡4、小丽同学在做“观察人体口腔上皮细胞”实验时,通过低倍显微镜发现视野的上方有一个口腔上皮细胞,若要在高倍物镜下观察该细胞,应 ( )A .向下移动载玻片使细胞位于视野中央,转动转换器,调节细准焦螺旋B .向下移动载玻片使细胞位于视野中央,转动转换器,调节粗准焦螺旋C .向上移动载玻片使细胞位于视野中央,转动转换器,调节粗准焦螺旋D .向上移动载玻片使细胞位于视野中央,转动转换器,调节细准焦螺旋5、下列描述中,属于金属化学性质的是 ( )A .铜是紫红色固体B .汞在常温下呈液态C .铝能与酸反应D .钠的质地较软6、在“显微镜观察洋葱表皮细胞”的实验中,下列有关说法正确的是 ( )A.镜筒下降时,眼睛应注视目镜内的物像B.由低倍镜换用高倍镜后,视野变亮C.换用高倍镜后,应使用粗准焦螺旋调节D.可观察到细胞壁、细胞核、液泡等细胞结构7、下列以农历确定的节日中,可以观测到满月的是 ( )A 春节B 端午节C .中秋节D .重阳节 8、右图是一张日历,据此判断这一天晚上的月相是9、2010年春晚,台湾著名魔术师刘谦表演的魔术,再次成为人们热议的话题,现有下列猜想:①果汁消失是因为纸杯中有夹层并有吸水物质。

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