上海市杨浦、虹口、宝山、普陀、松江五区2021届九年级上学期期末(中考一模)英语分类汇编：缺词填空
上海市杨浦、虹口、宝山、普陀、松江五区2021届九年级上学期期末（一模）英语试题分类汇编缺词填空上海市杨浦区2020-2021学年度第一学期九年级期末质量调研英语试题B. Fill in the blanks with proper words.Unless you are one of those rare people with an excellent memory, you probably have trouble remembering the names of people you have just met. This problem can be frustrating （令人沮丧的）and embarrassing. F 71 , there are a number of ways you can overcome this human weakness.Before examining the methods to help remember people’s names, let’s look at the reasons why we are so f 72 in the first place. Researchers think it’s due to a lack of motivation. People are better at remembering things that they want too learn. To put it simply, you’re just not that i 73 . Another reason is that putting a new face to a name is more complicated than we think, which leads to another problem. We usually don’t make the effort to m 74 new names immediately. Not only that, but when meeting someone new, we may be thinking about other things and not be paying enough attention.All of this means that we need to take measures to avoid the embarrassment of forgetting a person’s name. One good method is to say a person’s name when you g 75 them and also repeat it silently to yourself a few times. This type of repetition is essential for learning any type of skill. Using a memory method, such as l 76 individual names with something else, al soh答案：71. Fortunately 72. forgetful 73. interested 74. memories 75. greet 76. linking 77. tips 上海市虹口区2021届九年级上学期期终学生学习能力诊断测试（一模）英语试题B.Read the passage and fill in the blanks with proper words.Scientific Ways to Improve Our MemoryExercise r 71Research found that sitting for a long time every day can lead to memory loss. So put on your trainers, go o 72 and start working up a good sweat. It’s healthy for both your mind and your body.Eat your vegetablesIt’s a well-known f 73 that eating vegetables can help keep your body healthy, and that includes our brain. Various research p 74 show that fruits and vegetables can help lower oxidative stress（氧化压力）in your brain as well as help healthy cognitive（认知）functions.R 75 stress levelsResearch found stress at high levels have a bad influence on our memory and cognitive skills. So slow down and take a walk from time to time.Play video gamesYes, you heard us. Get on that controller and s 76 a couple of hours playing your favorite video game. It’s good for your mind. According to a r 77 study which came out last week, these pastimes can help improve the functioning of various memory-associated areas in the brain. But don’t go overboard, too much of a good thing can be bad for you.答案：71. regularly 72. outside 73. fact 74. projects 75. Reduce 76. spend 77. recent上海市宝山区2021届九年级上学期期末（一模）英语试题A.Read the passage and fill in the blanks with proper words.When you’re bored, what will you do first? If you’re like many, the answer is quite simple: You reach for your phone. Five minutes of short videos or funny pictures seem like a p 81 break.If that sounds familiar, a new study has bad news. Reaching for your phone is likely to leaveyou feeling more bored. The problem isn’t taking a break. The problem is y ourphone. A research shows using phones can’t help people r82 from theirheavy work.To figure out the r 83 between phones and boredom, the research teaminstalled（安装）an app on the phones of 83 volunteers to find out how often they used their phones. They also asked these volunteers to keep diaries for three days, recording their level of boredom every hour“P hone breaks were frequent （频繁的）：I n the 20 minutes following each questionnaire, the volunteers picked up their phones 52 percent of the time. They spent an average of around 90 seconds on it each time, ” reports the research team. “ The more tired we are, the most l 84 we are to reach for our phones. While we look to our phones to relieve（减轻）boredom, screen time seemed to i 85 feelings of boredom. The volunteers actually reported higher levels of boredom after having used their smartphones.”The research team pointed out that phone breaks may end up being more tiring than stimulating. In other words, watching the funny videos is very nice, but it’s now w86 the cost to your brain in effort and concentration. In addition, picking up your phone might remind you that you have a better choice to do something else if you don’t have to finish the tasks immediately. Both your time and effort are easily wasted.What should we do i 87 ? Taking a walk is good exercise. Calling someone can make your mood cheerful. Reading a book gives your brain a fresh start. You can do many things like that. Just remember, don’t reach for your phone!答案：81. pleasant 82. relax 83. relation 84. likely 85. increase 86. worth 87. instead上海市普陀区2021届九年级上学期质量调研（一模）英语试题C. Fill in the blanks with proper words（在短文的空格内填入适当的词, 使其内容通顺, 每空格限填一词, 首字母已给）（14 分）Recently researchers in Brown University have found that it only takes a couple of minutes or a man to start making judgments about you when he first meets you. That is why you shouldfollow the following tips on how to create a good i_ 71 .Before you meet someone, stat thinking about the p_ 72 _of the meeting. Decide what image （形象）you want to present. For example, if you want to make new friends in some social activities, you will want to be friendly. And if you decide to run for class president, you need to be c_ 73 .It helps you a lot when you meet someone new if you like s 74 . I shows that you are friendly and makes people around you more comfortable. Please make sure your teeth are clean by brushing them every day.Before you begin speaking, you will be judged on your body language. Therefore, it's important to show trust in yourself by standing up tall and putting your shoulders back. Besides, if you don't cross your arms, you will look calm and r 75 .Your smell can a_ 76_ how other people judge on you. If you have a bad body smell, it will put people off. In short, smell clean and don't put on a lot of scented products.What you wear really matters. While you should look clean and tidy, it's also important to dress p_77_, whether you're going to a birthday party or a sporting event. You should think about what your clothes say about you.答案：松江区2020学年度九年级第一学期英语期末试卷C. Read the passage and fill in the blanks with proper words（在短文的空格内填入适当的词, 使其内容通顺。
上海市松江区2021—2022学年九年级上学期期末(中考一模)考试语文试卷(word版 含答案)
上海市松江区2021—2022学年九年级上学期期末（中考一模）考试语文试卷（满分：150分，完成时间：100分钟，在答题纸上完成）2022.1一、文言文（36分）（一）默写（12分）1.东风不与周郎便，（杜牧《赤壁》）2.知困，（《虽有嘉肴》）3. ，一任群芳妒。
（陆游《卜算子·咏梅》）4.问君何能尔？（陶渊明《饮酒》）5.深秋的早晨，同学们看到田野和房屋上薄雾缭绕，而东方已是一片绯红。
甲生说：这便是《答谢中书书》中描绘的“”的画面；乙生说：可惜没有文中早晨“”的热闹场景啊。
（二）阅读下列诗文，完成第6-12题（24分）【甲】石壕吏（节选）暮投石壕村，有吏夜捉人。
老翁逾墙走，老妇出门看。
吏呼一何怒！妇啼一何苦！听妇前致词：三男邺城戍。
一男附书至，二男新战死。
存者且偷生，死者长已矣！室中更无人，惟有乳下孙。
有孙母未去，出入无完裙。
【乙】（节选）见渔人，乃大惊，问所从来。
具答之。
便要还家，设酒杀鸡作食。
村中闻有此人，咸来问讯。
自云先世避秦时乱，率妻子邑人来此绝境，不复出焉，遂与外人间隔。
问今是何世，乃不知有汉，无论魏晋。
此人一一为具言所闻，皆叹惋。
余人各复延至其家，皆出酒食。
停数日，辞去。
此中人语云：“不足为外人道也。
”【丙】子奇治阿子奇年十六，齐君使治阿①.既行，齐君悔之，遣使追。
追者反曰：“子奇必能治阿，共载皆白首也。
夫以老者之智，以少者决之，必能治阿矣！”子奇至阿，铸库兵以作耕器，出仓廪以赈贫穷，阿县大治。
魏闻童子治邑，库无兵，仓无粟，乃起兵击之。
阿人父率子，兄率弟，以私兵战，遂败魏师。
【注】①阿：地名，今山东阿县。
②白首：老年人。
③决：决断（政事）。
6.甲诗的作者是（人名），乙节选语段出自课文《》。
（2分）7.解释下列句中加粗词（4分）（1)遣使追（）（2)铸库兵以作耕器（）8.用现代汉语翻译下面的句子（3分）余人各复延至其家9.对文中划线句翻译正确的一项是（）（3分）A 走出粮仓去救济贫穷的百姓，全面治理阿县。
上海杨浦、虹口、宝山、普陀、松江五区2021届九年级上学期期末(中考一模)英语试题分类汇编：选词填空
上海市杨浦、虹口、宝山、普陀、松江五区2021届九年级上学期期末（一模）英语试题分类汇编选词填空上海市杨浦区2020-2021学年度第一学期九年级期末质量调研英语试题Ⅲ. Complete the following passage with the words or phrases in the box. Each can only be used once.（A）A. majorB. processC. produceD. inventionE. as well asChinese animation has a history of over 80 years, starting with the “Wan Brothers” who started to 36 Chinese cartoon films in the1920s in many kinds of categories, such as puppet, paper-cut and so on.Ink-wash animation is a 37 part of it. When it first appeared in the 1960s, it was a breakthrough in the form of expression and aesthetic（美学的）conception in animated area. Two ink-wash films called “Tadpoles Searching for Mother” and “Cowherd’s Flute,”，with Te Wei as art director and Qian Jiajun as technical director developed a high reputation both at home and abroad.The former received the Best Animated Film Prize at the First Hundred Flower Awards38 winning five international prizes, while “Cowherd’s Flute” was awarded the Golden Prize at the Odense International Fairy Tale Film Festival in Denmark. People called ink-wash animation the “fifth Chinese 39 .”（B）A. rolesB. amazingC. amusingD. exploringE. influencedBut even before the birth of Tadpoles Scarching for Mother,another Chinese animated feature film called Uproar in Heavenhad started 40 a connection between Chinese traditional painting and animation. The two-hour- long film mixed five different colors in its 41 , which included the Monkey King and the Jade Emperor. It was considered an ideal combination of Disney-style films and Chinese traditional arts.Japanese Cartoonist Osamu Tezuka, the “Godfather of Anime”. Said he was greatly 42 by Chinese ink- wash animation. Another Japanese anime artist also commented, sa ying. “It is unbelievable that ink-wash pictures could move like films. It’s 43 !”答案：36-39 CAED 40-43 DAEB上海市虹口区2021届九年级上学期期终学生学习能力诊断测试（一模）英语试题Ⅲ. Complete the following passage with the words or phrases in the box. Each word or phrase can only be used once.On October 30, 2019, 3M announced the results of its 2019 Young Scientist Challenge. 14-year-old Kara Fan’s spray-on bandage（喷雾绑带）beat out several other creative 36 to some of the wor ld’s biggest problems.The contest, which is run by 3M and Discovery Education, is open to students in grades 5-8. Hundreds of kids sent in ideas 37 . In June 2019, 10 finalists were chosen 38 their knowledge and 39 to share their idea clearly.The finalists were matched with a scientist. Students were given several months to improve their projects with the help and advice of these scientists.In October 2019, the finalists 40 met their scientists and other students face to face at 3M’s special science center in St. Paul, Minnesota. The students 41 their views in the group, and the winners were announced.Kara Fan from California won the big 42 for a special liquid bandage she created. It can be sprayed on cuts and 43 . One important part of Kara’s bandage is nano -silver（纳米银）--silver in extremely any form. Kara had to do a lot of research to get that part just right. For winning, Kara got $25,000.答案：36. D 37. A 38. E 39. C40. B 41. D 42. A 43. E上海市宝山区2021届九年级上学期期末（一模）英语试题Ⅲ. Complete the following passage with the words or phrases in the box. Each one can only be used once.A. gainedB. charactersC. rulesD. adventuresE. madeComic books are an unusual kind of storybook. They tell a story using only pictures and speech bubbles or captions. The 46 of these stories are usually superheroeswith special abilities.The Spider-Man comic books follow the 47 of a young man named Peter Parker. Peter’s parents died when he was very young. Peter accidentally 48 his spider-like abilities when he was 15. After that, Peter 49 many clever tools himself, such as his web-shooters. He uses his skills to fight evil enemies. Peter is very different when he is Spider-Man—He becomes strong and brave.A. exceptB. alsoC. femaleD. because ofE. besidesBatman is one of the world’s first superheroes. He wears special costume to make him look like a bat. He 50 has many clever tools to help him fight against his enemies. Batman has a friend named Robin who always helps Batman when he is in danger. Very few people know who Batman really is 51 Robin.Storm is the name of a 52 superhero from the X-man comic books. The X-men are born with superhero abilities. They team together to fight their enemies. Storm got her name 53 her special ability to control the weather.答案：46-49 BDAE 50-53BACD上海市普陀区2021届九年级上学期质量调研（一模）英语试题II. Complete the following passages with the words in the box. Each can only be used once （将下列单词填入空格。
2021年上海中考英语一模阅读B篇完型填空汇编
上海市16区2021届九年级上学期期末（一模）英语试题分类汇编完形填空上海市杨浦区2020-2021学年度第一学期九年级期末质量调研英语试题B.Choose the best answer and complete the passage.Shawn and his class were working on a new assignment. They were writing persuasive（劝说的）articles. When Shaw first learned about the assignment, he was 65 . Shawn disliked writing articles, and he wasn’t sure how to go about planning and writing a powerful article that would persuade someone to do something. So he asked Mr. Matthews for extra help planning his article.Mr. Matthews suggested that Shawn begin by selecting a 66 that wasimportant to him. Shawn wanted to be on the football team, so he decided towrite a paper explaining why he should be on the team. 67 , Mr. Matthewssuggested that Shawn list the strongest and best reasons that would persuadepeople. When Shawn had listed his reasons, he was ready to start writing.Shawn began with his main point—that he should be allowed to be on the football team. Then, he explained why he should be on the team, using the reasons he had listed. After that, he wrote the 68 -end of the paper. After Shawn had his draft（草稿）finished, he was ready to revise what he had written.Shawn read the paper to himself and to his parents. His mom and dad suggested ways and ideas that he could revise the paper to make his arguments stronger. Then, Shawn read the paper again. This time, he 69 a few ideas himself. After the revision was finished, it was time to edit.Shawn read the paper 70 to be certain that he had spelled everything the way it should be spelled. He also checked his paper for grammatical mistakes.The paper was written well enough that Mr. Matthews gave him a very good grade. The football coach read Shawn’s paper, too, and was so convinced by Shawn’s reasons tha t he permitted Shawn to work with the team!65. A. relaxed B. unhappy C. confident D. embarrassed66. A. hobby B. team C. topic D. character67. A. Next B. In addition C. However D. As a result68. A. plots B. details C. title D. conclusion69. A. reported B. thought of C. arranged D. focused on70. A. aloud B. properly C. carefully D. immediately答案：65-70 BCADBC上海市虹口区2021届九年级上学期期终学生学习能力诊断测试（一模）英语试题B. Choose the words or expressions and complete the passage.Arrest Report Sheet NAME OF SUSPECT（嫌疑犯）：Mr. John A. Berry65 FOR ARREST：Use of bad language in a public place.●Wasting a police off icer’s time.●Damaging property and causing 66 to the public.●There was something I didn’t like about him.DETAILED REPORT OF THE EVENTS LEADING UP TO THE ARRESTThe suspect came into Mean Street Police Station at 4:37 p.m. on Wednesday, February 15th, claiming（声称）to have lost his car. Of course, this story did not satisfy a police officer like myself, and I 67 started to question the suspect. After about 45 minutes, he changed his story, claiming that the car had been stolen. Something about this story, too, 68 extremely unlikely and I refused to accept it.The suspect kept telling his story. Luckily, I took a useful and 69 training course last weekend. So I used the active questioning skills from the course to get the real truth out of him.When he finally 70 under my questioning, he admitted that he had parked his car in a no-parking zone, and then blown it up（爆炸）. I have not yet worked out why he did this; my theory is that he is a mentally ill terrorist who sees parking rules as a symbol of everything he hates most in modern society.SignedDetective—Inspector Philip E. Morris65. A. REASONS B. CAUSES C. SUGGESTIONS D. EXCUSES66. A. excitement B. surprise C. danger D. variety67. A. politely B. immediately C. suddenly D. correctly68. A. looked B. tasted C. felt D. sounded69. A. noisy B. boring C. comfortable D. enjoyable70. A. broke down B. woke up C. shut up D. calmed down答案：65-70ACADDA上海市宝山区2021届九年级上学期期末（一模）英语试题B.Choose the words or expressions to complete the passage.Most people hate change, which is sad since we often go throughchanges in life. And for some of us, even the Smallest changes can upsetour day. So, the question is: Why do most of us find making adjustments（调整）to our lives so hard?Why we dislike changeFear of change is 75 new. Over a century ago, people in Paris were unhappy over an unusual addition to their city: the Eiffel Tower. In fact, the citizens were so angry about the plans for the tower that they were against its construction（建造）. As strange as it may seem, their anger was completely76 . They were given no choice about the huge change that was going to be made, so they became angry. But we get upset over changes even when we do have to a say in the matter. Changes are brought about every day by the decisions we make: which school to attend, which job to take. V oluntary（出于自愿的）changes also make most of us uneasy because we don’t know how those changes will 77 our future.How we can overcome our fear of changePeople have discovered that the key to overcoming the fear of change is to be flexible. When people are flexible, they can 78 new situations more easily. Being flexible is especially important in the 21st century as technology makes change occur faster than ever before. Those who were against change, especially with technology in the workplace, may end up finding themselves out of a job.When change comes, and you have no choice but to 79 it. A positive（积极的）attitude helps a lot. In fact, the change may turn out to be the best thing for you. That new job you got may end up being much better than your old one. You may make the best friends of your life in the new city you moved to. 80 , don’t just focus on how you feel about change. You should learn to accept the change. The change is the reality and it’s up to you to decide whether the change will be a success or a failure. You never know—your next change may be your life’s Eiffel Tower!75. A. something B. anything C. everything D. nothing76. A. important B. natural C. unusual D. ordinary77. A. assist B. reward C. affect D. arrange78. A. be aware of B. get used to C. be interested in D. get rid of79. A. forget B. make C. face D. refuse80. A. However B. What’s more C. Besides D. Therefore答案：75-80DBCBCD上海市普陀区2021届九年级上学期质量调研（一模）英语试题B. Choose the best answer and complete the passage （选择最恰当的选项完成短文）（12分）If you live on the streets with little food and money, what will you do with a bag filled with snacks and nearly￡500 in cash in front of you? To most people. it would be a source of temptation （诱惑）Something that happened last month has made the actions of the homeless Roy Kaufman out of the ordinary.Roy saw a 65 on the front seat inside a parked car with its window down. He stood guard inthe rain for along time waiting for the owner to_66 . Because of the coldness, he reached inside and pulled the bag out. He hoped to find some cards with its owner's name or phone numbers so that he could 67_ the driver. But he only discovered it contained some bars of chocolates and￡435 in notes, with another￡50 in spare change beside it.He then took it to a nearby police station after leaving a note behind to let the owner know it was 68 . When the car's owner Mr. Anderson returned to the car, he was shocked to find two policemen standing next to it and he was told what Roy Kaufman did.Mr. Anderson was later able to thank Mr. Smith for his _69_ and said, "I couldn't believe that the man never took a penny. The man has nothing and yet he didn't take the money for himself. He thought about others instead of himself. It's unbelievable. It just proves there are _70_ men out there. "Roy Kaufman's act drew much of the public's attention. He also won praise from social media users after Mr. Anderson posted about the story on Facebook.Now Mr. Anderson has set up an organization to raise money for other homeless people in the area. For Roy Kaufman, this is a possible life-changingchance. The story once again ells us that one good turn deserves another.65. A) report B) bag C) list D) symbol66. A) apologize B) drop C) investigate D) return67. A) contact B) reward C) refuse D) greet68. A) strange B) lost C) safe D) empty69. A) encouragement B) support C) kindness D) opinion70. A) generous B) honest C) poor D) polite答案：65 B 66 D 67 A 68 C 69 C 70 B松江区2020学年度九年级第一学期英语期末试卷B. Choose the best words and complete the passage（选择最恰当的单词完成短文）（12分）Amy and Susan both live an average life. They have an average house, an average job and an average family. There is really nothing _75_ about their lives, but there is also nothing really negative （负面的）about their lives.However, when they are both happy, it always seems that Amy is happier than Susan. It looks like nothing can _76_ Amy. She is the life of the party and always has a big smile on her face.Susan can't stop thinking about this. How is Amy so happy when they both lead such _77_ lives? They go to the same stores, watch the same movies, and even have the same schedules! How can Amy be so carefree when life is so stressful?Between one's daily life and personal dreams, there is no way one can be so happy and stress free all the time. There are always a million things going on.But when Susan asks Amy about this, Amy says, "I just take life as it comes. It seems pretty easythen. "So what is the _78_ between these two friends? How is it that Amy's life seems so much better most of the time?The problem with Susan is that she spreads herself too thin（试图同时干太多的工作）. Not _79_, but in her own head. But although Amy tries to do everything at once and think about everything at once, she takes everything one step _80_. She is where she is, stays present in the moment and allows herself to focus on just one thing.When you live each moment to its fullest, when you deal with where you are and are present in the now, most of your stress will go away.75. A) boring B) common C) special D) disappointing76. A) attract B) trouble C) change D) encourage77. A) similar B)successful C) traditional D)wonderful78. A) link B)relationship C) contact D)difference79. A)physically B) rarely C) correctly D) suddenly80. A) on time B)for the time being C) at a time D) from time to time答案：75.C 76. B 77.A 78.D 79.A 80.C2021奉贤一模Choose the words and complete the passage（选择最恰当的单词，完成短文）Dogs bury bones. Squirrels gather nuts to last through the winter. Camels store food and water so that they can travel many days across deserts. But do pigs save anything? No, pigs save nothing. So why do we save coins in a piggy bank? Because someone made a mistakeDuring the Middle Ages（中世纪）, in about the fifteenth century, metal was 65 and seldom used for household wares（家用器具）. Instead, dishes and pots were made of a kind of economical clay called pygg. Whenever housewives could save an extra coin, they dropped it into one of their clay jars. They call this their pygg bank or pyggy bank.Around the eighteenth century, the spelling of pygg changed. The word 66 the same as the word for the animal pig. Over the following 200 to 300 years, people 67 that pygg meant the clay. And the term "pygg jar" gradually changed into pig bank, or piggy bank.In the nineteenth century, when English potters received requests for piggy banks, they misunderstood. They produced jars shaped like a pig. 68 , the pigs became popular among customers and delighted children.Another reason for the name "piggy bank" is 69 the idea that the coin given to the piggy bank represent the food fed to a pig by a farmer. It costs the farmer money to feed the pig and he doesn't 70 until the pig is slaughtered（屠宰）for meant, which is known as "breaking the piggy bank."Today, pigs are still one of the most popular forms of coin banks sold in gift shops.65. A. cheap B. expensive C. heavy D. light66. A. looked B. smelt C. sounded D. felt67. A. forgot B. forgave C. remembered D. realized68. A. Importantly B. Suddenly C. Sadly D. Certainly69. A. similar to B. based on C. famous for D. made of70. A. turn it over B. take it awayC. get it back D. work it out答案：65-70 BCADBC2021崇明一模B. Choose the words or expressions and complete the passage (选择最恰当的单词或词语完成短文) (12分)Animals can move from place to place, but plants cannot. When an animal is under attack, it can run away or 65. Plants certainly cannot run away, and they have no teeth or claws. But plants can protect themselves by using both physical and chemical means(方法)Some plants have their own ways to keep animals 66 For example, the leaves of the holly plant have sharp points that discourage (打消:阻止) grass eating animals. Holly leaves on lower branches have more sham points than leaves on upper branches. This is because the lower leaves are 67 for most animals to reach,Some plants, such as the oak tree, have special ways to discourage leaf-cating animals. They have thick and hard leaves that are 68 for animals to eat Some grasses may contain a sandy material; eating such grasses wears down the animal’s teeth.Many plants also have chemical defenses. Some plants produce chemicals that taste bitter or cause an unpleasant reaction Some plants may fight against a(n) 69 by increasing the production of these chemicals. When an insert bites a tobacco leaf, the leaf produces a chemical messenger(信使) This messenger sends to the roots the information to produce more nicotine The higher levels of nicotine discourage the insect.Many plants 70 both physical and chemical defenses A certain plant in China, for example, has sharp leaves, and each sharp leaf contains p oisonous venom (毒液) A single experience with this kind of plant will leach an animal not to eat the leaves of it in the future.65. A) calm down B) come back C) fall asleep D) fight back66. A) away B) warm C) safe D) crazy67. A) sweeter B) easier C) bigger D) greener68. A) important B) convenient C) difficult D) delicious69. A) attack B) illness C) storm D) flood70. A) pay for B) learn from C) search for D) rely onKey: DABCAD黄浦一模B. Choose the best answer and complete the passage(选择最恰当的选项完成短文)(1 分)The National Gallery in London has introduced an “art route” that people must follow round the gallery. Are set routes actually a better way to enjoy museums and galleries? Or should people be 75 to view exhibitions in any order they choose? What do you think - are set routes better?Yes - people will get more from their visitExperts spend years learning about the items in their museums and galleries. They know the best order in which people should see the exhibits, so that visitors can enjoy and learn about them. 76. if there was an exhibition of a painter, the works could be arranged to show visitors how their style developed through their career. If a person is on an art route, they won’t 77 any important pictures, and will be guaranteed to see everything. The experience will be enjoyable for the visitor, and they will have their own space in which to appreciate great works.No - freedom is part of the funMost exhibitions are already 78 in some kind of order. However, not everyone wants to look at the same things in the same way. Some visitors just want to enjoy a few pieces and then leave. Besides, if everyone follows the same route, some people might take a long time looking at one object and 79 everyone down. Equally, if a visitor wants to study something for more time, they may feel pressured by people behind them. One of the great joys of exhibitions is 80 noticing something. When a visitor is drawn to an object, it can feel special and they can learn about it by themselves. If everyone is following the same route, this is less likely to happen, so the experience will not be as much fun.75. A) ready B) able C) free D) thankful76. A) In fact B) For example C) In addition D) That is to say 77. A) forget B) fail C) lose D) miss78. A) laid out B) put away C) dealt with D) turned on 79. A) put B) slow C) take D) break 80. A) carefully B) silently C) suddenly D) peacefully Key:75. C 76. B 77. D 78. A 79. B 80. C2021嘉定一模B. Choose the words or expressions to complete the passage (选择最恰当的单词或词语，完成 短文)(12分)Some people volunteer to gain experience. Some people volunteer to make themselves feel better. Some people volunteer to help those who were (65)______. However; I volunteer to thank my community. I am thankful for the help I received from others.When I was in high school, I took part in the Upward Bound Program. Through Upward Bound, I was able to gain skills and opportunities that have helped me become the man that I am today. So I now often give back to enable kids to have the same(66)______ at life that I have had.V olunteering is the best way to thank those who helped me on the road to success. I remember asking a counselor(顾问)when I was in Upward Bound how I could repay him. He smiled and said, "Shawn, you can help me by helping others?* A sentence so simple, yet so (67)______in its meaning.When I got my first volunteer position as a junior counselor at my local Boys and Girls Chib, I was eager to be a great role model for kids. But things did not work out as I thought at first I got so (68)______ .By the time I was ready to (69)______ , the kids looked for me everywhere that day. They said they liked having me around.My desire to give back to the community came back. Now the feeling that I can shape someone's life for the better is amazing.So it is a good idea for everyone to find a(an)(70)______ and volunteer. In this way not only will you be helping others, you will be helping yourself as well. Key:65-70 CAB BCC 2021金山一模B. Choose the best answer and complete the passage(选择最恰当的单词或短语，完成短文)（12分）Most people have heard of the Great Wall of China, but not many foreigners know about the Grand Canal. This man-made waterway —known in China as DaYunhe —is 1,700 kilometers long and some parts of it are more than 2,000 years old. It (65) the north of China with Hangzhou, a city in the southern part of the country. It's the longest man-made waterway in the world.It was built as a(66) to transport grain from the rich agricultural land in China's south to cities in the north. In fact, it's not one canal, but a system of canals and rivers linked (67) . It's still an important part of the transport system in China.Thousands of boats use it every day to transport grain and many other types of cargo.65. A) active B) honest C) unfortunate D) helpful 66. A) chance B) courage C) decision D) stress 67. A) strange B) deep C) funny D) different 68. A) excited B) disappointed C) satisfied D) proud 69. A) turn up B) wake up C) give up D)put up 70. A) exampleB)causeC) excuseD)solutionNow the Chinese government is doing new work on the Grand Canal.It is making parts of it (68) , so bigger ships can use it. The canal will also help to move water. There is a lot of (69) in the south of China, but not (as much) in the north. The canal will carry millions of liters of water from the south to the north.The Grand Canal is much(70) than the Great Wall, and not very many tourists visit(it). But it's possible to go on a cruise along some of the oldest parts of the canal. Passengers on these cruises see beautiful parts of China that other visitors don't see.60.A. compares B. connects C. controls D. communicates61.A. result B. trick C. way D. sample62.A. together B. instead C. along D.forward63.A. cleaner B. prettier C. longer D. deeper64.A. food B. population C. rain D. electricity70. A. smaller B. narrower C.less modem D.less famousKey:65-70 BCADCD2021静安一模B. Choose the words or expressions and complete the passage (选择最恰当的单词或词语完成短文)(12分)You can think of a memory, as a photograph. Our brain is like an app on the phone that stores useful photographs lor a long time. But il 75 useless photographs. Some people have always been able to recall(回想起)the smallest details about their lives. The experts are now studying Brad Williams and a woman named A.J. The two share the same talent. The goal of the study is to gain a deeper 76 of memory and how it works.Dr. James found 51-year-old Williams’ memory to be 77 perfect. He can usually tell you what he did on a certain date 40 years ago. He can also recall what was in the news. He can even tell you what the weather was like.Dr. James had seen this level of ability before. In 2006, he studied a woman named A.J. She too was able to recall tiny details from her life. These details 78 notes she had written in journals decades earlier.However, they view the ability differently. Williams enjoys having his memory tested. A.J., however, said her memories often flood her mind in an unpleasant way. “Most have called it a gift, but I call it a burden,” A.J wrote. “I run my entire life through my head every day and it makes me 79 !!!”Now, doctors are testing Williams and A.J. They want to know what sets the two apart from others. Williams and A.J. both perform well. But they perform best on topics that interest them. This fact causes doubt among some scientists. They suggest that Williams and A.J. are not so special 80. Some scientists say that people with truly special memories should be able to remember all kinds of information, not just what interests them.Dr. James hopes that his study will provide answers for both those who doubt and those who surprise at what Williams and A.J. are able to do. “The human brain is the most important machinery in the known universe,” he said. “My aim is to decrease the mystery of this surprising machinery.”75. A) gives up B) gets rid of C) picks up D) keeps off76. A) confidence B) entrance C) speed D) understanding77. A) rarely B) exactly C) nearly D) mainly78. A) created B) recorded C) added D) matched79. A) crazy B) excited C) flexible D) pale80. A) after all B) at present C) in the end D) at a timeKey:75-80 BDCDAA2021闵行一模B. Choose the words or expressions and complete the passage (选择最恰当的单词或词语完成短文)(12分)Manners Around the WorldEvery culture has set rules about how people should act. Patterns of good behavior, or manners, show respect and care for others. Yet the details of 65 to express respect are so different.Greeting people cheerfully is almost always considered polite. But ifs more important in some cultures than others. For example, when you enter a store in France, you, should always greet the person working there. Other cultures also 66 greeting people. But of course the greetings vary as people speak different languages.When I was growing up in the United States, I once, took part in a performance at my church. Afterwards, someone gave me a compliment；but I felt like I had done a bad job. So I tried to refuse the compliment, saying, "No, I didn’t do that well." My mother stopped me and told me that was very 67. Later I learned that refusing a compliment is accepted in China.Some of the biggest cultural differences have to do with 68. In North America, ifs polite to eat as quietly as possible. That means chewing with your mouth closed and not slurping your soup. The same goes for burping (打嗝).But in parts of Asia, slurping shows that you are enjoying your meal. And burping is a sign of being full and content.There are also differences in how people，eat across cultures. 69, in North America and Europe people use forks to bring food to their mouths. But in Thailand, a fork is only used to push food onto a spoon. In India and the Middle East, people eat with their hands. But eating with the left hand is frowned upon. This is because the left hand is considered unclean.Yet there are also similar points across cultures. Saying “please” and “thank you” is almost always polite. If you show appreciation and try to 70 local customs, people will generally respond well. Showing kindness can bring people together, no matter what culture they come from.()65. A) what B) how C) when D) where()66. A) mind B) avoid C) imagine D) value()67. A) rude B) simple C) common D) accurate()68. A) family parties B) table manners C) fashion tastes D) instruction signs()69. A) What’s more B) In short C) For example D) Above all()70. A) change B) develop C) follow D) mentionKey: BDABCC2021浦东一模B. Choose the best answer and complete the passage(选择最恰当的选项完成短文)（12分）One night before bed, Christine Carter was sitting with her daughter in her bedroom. They were making a list, "Three Good Things of the Day."Her daughter said," Mom, chatting with you is one of my three good things."Making a list was their way to think of things they were (65) ______ for. It's a skill in the research of happiness.Carter is a professor at the University of California. She studies how schools and families build up positive attitudes and help kids lead (66) ______ lives. This has become especially important during the coronavirus pandemic (新冠病毒引发的流行病). Many kids have missed important events. Some even have lost people in their lives.Finding the positive things sometimes doesn’t mean people won’t be sad any more. You' ll find it natural to fee l sad, but it passes more (67) ______ if we notice the good things in life."This is really about mental health,"Carter says."We can practise bringing happiness to ourselves. It's like putting money in the bank. We can deal with hard times in the future."Expressing thanks is also important. Carter says,"Let’s say you often argue with a brother or a sister. Think about whether they did something good for you or with you. Remember the good times. This makes it likely you’ll (68) ______ better tomorrow."A key step toward happiness is learning how to describe emotions, even negative ones. It' better to face your emotions than to (69) ______ them. Ask yourself, "How do I feel? Where in my body am I feeling it? Does it have a color or a shape?"The surest way to happiness is (70) ______,"Helping others gives life purpose because you are changing the world" Carter says, "Besides, people have a sense of connection and satisfaction. We are comfortable when were connected with other people."65.A. sorry B. ready C. famous D. thankful66. A. richer B. healthier C. busier D. freer67. A. quickly B. patiently C. strongly D. heavily68. A. go out B. get along C. show off D. grow up69 A. disturb B. forgive C. operate D. ignore70. A. entertainment B. ambition C. kindness D. freedomB篇：65-70 DBABDC2021青浦一模B. Choose the words or expressions and complete the passage(选择最恰当的单词或词语，完成短文)(12分)When you think of a journalist, do you see a reporter with a pen and paper? Writing is one way to tell a story, but some stories are told without (65)__________.Photojournalists are reporters too. They use a camera to capture a story in photographs (66)__________ a pen and paper. You can see their work in the pictures that appear with an article. Pictures help readers visualize the things being described. Photojournalists also create photo essays, which tell a story with photosAlice Aedy is a photojournalist and documentary filmmaker based in London, England. She specializes in stories about social justice and human rights, but she also covers (67)_________issues. Aedy said that she's “passionate about climate change” because of how it affects peoples livesPhotojoumalism helps people (68)__________current events by giving context to important moments in history. So photojournalists aren't just photographers. Theyre storytellers. Like other reporters, they need to be on the scene and able to interview people about what's happeningAedy has traveled the world to do her reporting. While working, she spends as much time as she can in the country where a story is taking place and does lots of interviews to get the full story. “I want my photos to build empathy(共鸣),” she says，“and make people care.”Aedy says that photography, like all forms of media, should be viewed “with a (69)________ eye.” Viewers sh ould ask themselves questions like “What is the context of the photo? When and where was it taken? Was the photo taken with the permission of its subjects? Why was it taken?”It's also important for viewers to take time to think about how a photograph make s them (70)__________. “We’re bombarded(轰炸) with images every day, ” Aedy says，“So my hope is that whatever you're looking at, you just truly engage.”65. A)words B)pens C)pictures D) writers66. A)in addition to B)instead of C)except for D)because of67. A)business B)environment C)technology D) entertainment68.A)watch B)hear C) explain D)understand69.A)critical B) close C)friendly D)real70.A)judge B)decide C)feel D) observeB篇：65-70ABBDAC2021徐汇一模B.Choose the words or expressions and complete the passage.Dan and Michelle sat at a picnic table in the park on a beautiful fall afternoon. The air around them was filled。
上海市宝山区2020-2021学年九年级上学期期末(中考一模)数学试卷带讲解
上海市宝山区2021届初三一模数学试卷一、选择题1. 如果C 是线段AB 延长线上一点，且:3:1AC BC =，那么:AB BC 等于（ ）．A. 2:1B. 1:2C. 4:1D. 1:4 【答案】A【分析】先画出图形，设BC 为k ，然后用k 表示出AB ，最后求出:AB BC 即可．【详解】解：根据题意可画出下图：∵:3:1AC BC =，设BC 为k ，∴AC=3k ，∴AB=AC-BC=2k ，∴:AB BC =2k∴k=2∶1．故答案为A ．【点睛】本题主要考查了线段的和差，根据题意画出图形成为解答本题的关键．2. 在Rt ABC △中，90C ∠=︒，5AB =，3BC =，那么sin A 的值为（ ）． A. 35 B. 34 C. 45 D. 43【答案】A【分析】根据正弦的定义解答即可．【详解】解：在Rt △ABC 中，∠C=90°，AB=5，BC=3，则sinA=35BC AB =， 故选：A ．【点睛】本题考查了锐角三角函数的定义，掌握锐角A 的对边a 与斜边c 的比叫做∠A 的正弦是解题的关键．3. 如图，//AB DE ，//BC DF ，已知::AF FB m n =，BC a =，那么CE 等于（ ）．A. am nB. an mC. am m n +D. an m n+【答案】D【分析】先证明：四边形DEBF 是平行四边形，可得DF BE =，利用::AF FB m n =，再求解AF m AB m n=+，再证明ADF ACB ∽，利用相似三角形的性质求解BE ，再利用线段的和差可得答案． 【详解】解： //AB DE ，//BC DF ，∴ 四边形DEBF 是平行四边形， DF BE ∴=，::AF FB m n =，AF m AB m n∴=+， //DF BC ，ADF ACB ∴∽AF DF AD AB BC AC∴==, //AB DE ，BE AD m BC AC m n∴==+， BC a =，ma BE m n∴=+， .ma na CE a m n m n ∴=-=++ 故选：.D4. 已知点M 是线段AB 的中点，那么下列结论中，正确的是（ ）． A. AM BM = B. 12AM AB = C. 12BM AB = D. 0AM BM +=【答案】B【分析】根据题意画出图形，因为点M 是线段AB 的中点，所以根据线段中点的定义解答．【详解】解：A 、AM MB =，故本选项错误；B 、12AM AB =，故本选项正确；C 、12BM BA =，故本选项错误； D 、0AM BM +=，，故本选项错误．5. 若将抛物线2y x 先向右平移1个单位长度，再向上平移2个单位长度，就得到抛物线（ ） A. 2(1)2y x =-+B. 2(1)2y x =--C. 2(1)2y x =++D. 2(1)2y x =+- 【答案】A【分析】根据二次函数图象左加右减，上加下减的平移规律进行解答即可．【详解】解：将抛物线2y x 先向右平移1个单位长度，再向上平移2个单位长度，就得到抛物线：()212y x =-+ 故答案为：A ．【点睛】本题考查二次函数的图象与性质，图象平移规律“左加右减，上加下减”是解题关键．6. 如图所示是二次函数()20y ax bx c a =++≠图像的一部分，那么下列说法中不正确的是（ ）．A. 0ac <B. 抛物线的对称轴为直线1x =C. 0a b c -+=D. 点()12,y -和()22,y 在拋物线上，则12y y >【答案】B 【分析】根据图象分别求出a 、c 的符号，即可判断A ；根据抛物线与x 轴的两个交点可判断出该抛物线的对称轴不是x =1，即可判断B ；把x =-1代入二次函数的解析式，再根据图象即可判断C ；将x =-2与x =2带入二次函数，可得出y 1与y 2的值，即可判断D ．【详解】解：∴二次函数图象开口向上，∴a ＞0，∴二次函数的图象交y 轴的负半轴于一点，∴c ＜0，∴ac ＜0 选项A 正确；∴由图像可看出，抛物线与x 轴的交点一个为x=-1，另一个在x=2和x=3中间，不关于x=1对称，∴抛物线的对称轴不是x=1 选项B 错误；把x=-1代入y=ax 2+bx+c 得：y=a-b+c ，由图像可知，x=-1时y=0，∴a-b+c=0 选项C 正确；把x=-2和x=2代入y=ax 2+bx+c 中，由图像可知，y 1＞0，y 2＜0，∴y 1＞y 2 选项D 正确；故选：B ．【点睛】本题考查二次函数的性质，解题的关键时熟练运用抛物线的图像判断系数a 、b 、c 之间的关系，同时注意特殊点与对称轴之间的关系，属于中等题型．二、填空题7. 如果2x ＝3y ，那么x y y+=___． 【答案】52【分析】直接利用已知得出x =32y ，进而代入得出答案． 【详解】解：∵2x =3y ，∴x =32y ， ∴3522y y x y y y ++==． 故答案为：52． 【点睛】本题主要考查了比例的性质，正确将已知变形是解题关键．8. 已知线段2a =厘米，8c =厘米，那么线段a 和c 的比例中项b 的长度为______厘米．【答案】4【分析】根据线段的比例中项可直接进行列式求解．【详解】解：由题意可得：22816b ac ==⨯=，∴4b =cm ；故答案为4．【点睛】本题主要考查比例中项，熟练掌握比例中项是解题的关键．9. 如果线段AB 的长为2，点P 是线段AB 的黄金分割点，那么较短的线段AP =______．【答案】3【分析】设较短的线段AP x =，则BP AB AP =-，根据黄金分割点的性质列方程并求解，即可得到答案．【详解】设较短的线段AP x =∵AB 的长为2∴2BP AB AP x =-=- ∴BP AP AB BP= ∴222x x x-=- ∴()222x x -=∴3x =+3（经检验均为方程的根）32+>，故舍去∵(22310x -=-=≠∴3x =-∴较短的线段3AP =故答案为：3【点睛】本题考查了黄金分割点、分式方程、一元二次方程、二次根式的知识；解题的关键是熟练掌握黄金分割点、分式方程、一元二次方程、二次根式的性质，从而完成求解．10. 计算：32a ba b ______． 【答案】54a b -【分析】根据向量的表示方法可直接进行解答．【详解】解：326354a ba b a b a b a b ， 故答案为：54a b -．【点睛】本题考查的是平面向量的知识，熟悉相关性质是解题的关键．11. 已知等腰梯形上底为5，高为4，底角的余弦值为35，那么其周长为______． 【答案】26【分析】作DF ⊥BC 于F ，AE ⊥BC 于E ，根据等腰梯形的性质就可以得出△AEB ≌△DFC 就可以求出FC=BE ，然后根据底角的余弦值为35，求得BE ，AB ，从而求出周长． 【详解】解：如图示，作DF⊥BC 于F ，AE⊥BC 于E ，∵四边形ABCD 是等腰梯形，∴∠B=∠C ，AB=CD ，AD ∥BC ，∴∠ADF=∠DFC=90°，∴∠AEF=∠DFE=∠ADF=90°，∴四边形AEFD 是矩形，5EF AD ，△AEB 和△DFC 中BC AEBDFC AE DF ， ∴△AEB ≌△DFC （AAS ），∴BE=CF ； ∵35cos E ABB B ， 设3BE x =，则5AB x =， 根据勾股定理，有：2222534AE AB BE x x ，解之得：1x =（取正值），∴3BE =，5AB =，∴3FCBE ，5DC AB ==， ⊥周长AB BE EF FC CD AD 53535526，故答案是：26．【点睛】本题考查了等腰梯形的性质的运用，三角函数，矩形的判定及性质的运用，等腰三角形的性质的运用，全等三角形的判定及性质的运用，能熟练应用相关性质是解题的关键．12. 某公司10月份的产值是100万元，如果该公司第四季度每个月产值的增长率相同，都为0)x x >（，12月份的产值为y 万元，那么y 关于x 的函数解析式是______．【答案】()21001y x =+；【分析】根据：现有量=原有量×(1+增长率)n ,即可列方程求解．详解】依题意得：()21001y x =+故答案为：()21001y x =+【点睛】考查了一元二次方程的应用，可直接套公式：原有量×(1+增长率)n =现有量，n 表示增长的次数． 13. 如果抛物线()21y m x m =++（m 是常数）的顶点坐标在第二象限，那么它的开口方向______． 【答案】向上【分析】根据解析式写出顶点，根据顶点坐标在第二象限求出m 的取值故可求解．【详解】∵抛物线()21y m x m =++的得到为（-1，m ）又顶点坐标在第二象限∴m ＞0∴开口向上故答案为：向上．【点睛】此题主要考查二次函数的性质，解题的关键是熟知顶点式的特点．14. 已知一条抛物线具有以下特征：（1）经过原点；（2）在y 轴左侧的部分，图像上升，在y 轴右侧的部分，图像下降；试写出一个符合要求的抛物线的表达式：______．【答案】2y x =-（答案不唯一）【分析】设出符合条件的函数解析式，再根据二次函数的图象在y 轴左侧部分是上升的，在y 轴右侧部分是下降的可知该函数图象的开口向下，对称轴为y 轴，即0a <，0b =，再把()0,0A 代入，得出符合条件的函数解析式即可．【详解】解：设出符合条件的函数解析式为：()20y ax bx c a =++≠， ∵二次函数的图象在y 轴左侧部分是上升的，在y 轴右侧部分是下降的，∴该函数图象的开口向下，对称轴为y 轴，即0a <，0b =，∵函数图象经过()0,0A ，∴0c ，∴符合条件的二次函数解析式可以为：2y x =-（答案不唯一）．故答案为：2y x =-（答案不唯一）．【点睛】本题考查的是二次函数的性质，先根据题意设出函数解析式，再根据二次函数的性质判断出a 的符号及对称轴是解答此题的关键，此题属开放性题目，答案不唯一．15. 如图，已知ABC 中，//EF AB ，12AF FC =，如果四边形ABEF 的面积为25，那么ABC 的面积为______．【答案】45【分析】根据//EF AB ，易得∴CFE ∽△CAB ，再依据相似三角形的面积比等于相似比的平方，即可求出三角形ABC 的面积．【详解】解：∵//EF AB∴△CFE ∽△CAB 又∵12AF FC = ∴32ACFC=， ∴94ABC FEC S S =△△ 设∴ABC 的面积为x 则9254x x =-， 解得，x=45，经检验x=45是原方程的根故答案为：45【点睛】本题考查了相似三角形的判定与性质，依据相似三角形面积比是相似比的平方，构建方程，是解决问题关键．16. 在一块直角三角形铁皮上截一块正方形铁皮，如图，已有的铁皮是Rt ABC △，90C ∠=︒，要截得的正方形EFGD 的边FG 在AB 上，顶点E 、D 分别在边CA 、CB 上，如果4AF =，9GB =，那么正方形铁皮的边长为______．【答案】6【分析】设正方形铁皮的边长为x ，证明△AEF ∽△DBG ，得到EF AF BG DG =，49x x=，求解即可． 【详解】设正方形铁皮的边长为x ，∵90C ∠=︒，∴∠A+∠B=90︒，在正方形EFGD 中，EF=DG=FG=x ，∠EFG=∠DGF=90︒,∴∠AFE=∠BGD=90︒，∴∠A+∠AEF=90︒,∴∠AEF=∠B ，∴△AEF ∽△DBG ， ∴EF AF BG DG=， ∴49x x =， 解得x=6（负值舍去），故答案为：6．【点睛】此题考查正方形的性质，相似三角形的判定及性质，根据已知条件证明△AEF ∽△DBG 是解题的关键．17. 如图，某堤坝的坝高为12米，如果迎水坡的坡度为1:0.75，那么该大坝迎水坡AB 的长度为______米．【答案】15【分析】过点B 作BC ⊥AC 于C ，由迎水坡的坡度为1:0.75，得到tan ∠BAC=43=BC AC ，求出AC=9米，再利用勾股定理求出答案．【详解】过点B 作BC ⊥AC 于C ，∵迎水坡的坡度为1:0.75，∴tan ∠BAC=43=BC AC ， ∵BC=12米，∴AC=9米，∴米)，故答案为：15．．【点睛】此题考查坡度的定义，解直角三角形的实际应用，勾股定理，正确理解迎水坡的坡度为1:0.75得到tan ∠BAC=43=BC AC 是解题的关键． 18. 在Rt ABC △中，90ACB ∠=︒，AC BC =，点E 、F 分别是边CA 、CB 的中点，已知点P 在线段EF 上，联结AP ，将线段AP 绕点P 逆时针旋转90°得到线段DP ，如果点P 、D 、C 在同一直线上，那么tan CAP ∠=______．1．【分析】分两种情形：⊥当点D 在线段PC 上时，延长AD 交BC 的延长线于H ．证明AD ＝DC 即可解决问题．【详解】解：⊥如图2中，当点D 在线段PC 上时，延长AD 交BC 的延长线于H ．⊥CE ＝EA ，CF ＝FB ，∴EF ∥AB ，∵AC ＝AB ，∠ACB ＝90°⊥⊥CEF ＝⊥CAB ＝45°，∵PD ＝P A ，∠APD ＝90°⊥⊥PAD ＝⊥PDA ＝45°，⊥⊥HDC ＝⊥PDA ＝45°，∵点E 是边CA 的中点，⊥EA ＝EP ＝EC⊥⊥EPC ＝⊥CEP ，∵∠HDC ＝∠DCA+∠DAC ＝45°，∠CEF ＝∠DCA+∠EPC ＝45°，⊥⊥DAC ＝⊥EPC ＝⊥ECP ，∴DA ＝DC ，设AP ＝a ，则DA DC =,∴)1PC a =∴)1tan 1a PC CAP PAa∠===②如图3中，当点P 在线段CD 上时，由①可知，EF ∥AB ，∠CAB ＝∠PDA ＝45°， ∴∠CAD ＝180°-∠ACD-45°， ∠COA ＝180°-∠ACO-45° ∴∠CAD ＝∠COA ， ∵EF ∥AB ， ∴∠CPE ＝∠COA ， ∴∠CPE ＝∠CAD ， ∵点E 是边CA 的中点， ⊥EA ＝EP ＝EC ∴∠ECP ＝∠CPE ， ∴∠ECP ＝∠CAD ，∴DA ＝DC ，设AP ＝a ，则PD ＝a ，DA DC ==,∴)1PC a =∴)1tan 1a PC CAP PAa∠===:点P 在线段EF 上，情况⊥不满足条件,情况⊥满足条件，综上所述，tan CAP ∠1．【点睛】本题考查了旋转变换，等腰直角三角形的性质，中位线的性质，外角的性质，三角形内角和，勾股定理和三角函数等知识，熟悉相关性质是解题的关键．三、解答题19. 计算：21cos 45cot 30sin 60tan 30-︒︒+︒⋅︒．【分析】根据特殊角的三角函数值进行计算求解．【详解】解：原式21112121112⎛- -=====． 【点睛】本题考查特殊角的三角函数值，解题的关键是掌握特殊角的三角函数值．20. 如图，已知ABC 中，//DE BC ，且DE 经过ABC 的重心点G ，BD a =，BC b =．（1）试用向量a 、b 表示向量BE ； （2）求作向量()233a b -（不要求写作法，但要指出图中表示结论的向量）． 【答案】（1）23BE a b =+；（2）见解析 【分析】（1）根据重心到顶点距离是它到对边中点距离的2倍，分析得到DE=23BC ，再根据向量的加法法则，首尾顺次相连，由三角形法则即可求解；（2）取AD 的中点J ，延长CB 到I ，使BI=DE ，以BJ 、BI 为邻边作平行四边形BJKI ，边接BK ，则BK 即是所求作的向量．【详解】解：（1）如图，连接AG 并延长交BC 于点F ，则GF=12AG ，AG 2=AF 3∴，DE//BC ，BC b = ADE ABC ∴△△∽， DE AG 2==BC AF 3∴， 23b DE BC =＝， 2a 3BE BD DE b ∴=+=+（2）BD a =，3BA a ∴=，作AD 的中点J ，2J=3a 23B a ∴⨯=，延长CB 到I ，使得BI=DE ，23BI b ∴=-，以BJ 、BI 为邻边作平行四边形BJKI ，则()2223a 33BK BJ BI a b b =+=-=-， ∴BK 即是所求的求作的向量【点睛】本题考查了向量的知识，掌握法则向量的平行四边形法则，向量的三角形法则是解题的关键．21. 已知二次函数()20y ax ax a =-≠的图像经过点()1,2-．（1）求该二次函数的解析式和顶点坐标；（2）能否通过所求得的抛物线的平移得到抛物线2132y x x =++？如果能，请说明怎样平移，如果不能，请说明理由． 【答案】（1）2yx x ，顶点为11,24⎛⎫- ⎪⎝⎭；（2）可以，先向左平移2个单位，再向下平移32个单位【分析】（1）把点()1,2-代入函数解析式，求出a 的值即可得到解析式，再把一般式写成顶点式得到顶点坐标； （2）把所给的函数解析式化为顶点式，根据函数图象的平移法则进行求解． 【详解】解：（1）把点()1,2-代入函数解析式，得2a a +=，解得1a =， ∴2yx x ，写成顶点式：21124y x ⎛⎫=-- ⎪⎝⎭，∴顶点坐标是11,24⎛⎫-⎪⎝⎭； （2）将2132y x x =++也写成顶点式，得23724y x ⎛⎫=+- ⎪⎝⎭，31222⎛⎫--= ⎪⎝⎭，713442-=， ∴把原抛物线先向左平移2个单位，再向下平移32个单位． 【点睛】本题考查二次函数解析式的求解和图象的平移，解题的关键是掌握解析式的求解方法和函数图象的平移方法．22. 如图，点O 是菱形ABCD 的对角线BD 上一点，联结AO 并延长，交CD 于点E ，交BC 的延长线于点F ．（1）求证：2AB DE BF =⋅； （2）如果1OE =，2EF =，求CFBF的长．【答案】（1）见解析；（2）33CF BF -=【分析】（1）根据菱形的性质证明ABO EDO ，BFO DAO ，得到AB BFED DA=，再由AB DA =，即可证明结论；（2）连接OC ，先证明()ADO CDO SAS ≅得到DAO DCO ∠=∠，就可以证明OEC OCF ，根据对应边成比例求出OC 的长，再根据ADE FCE ~，利用对应边成比例求出结果． 【详解】解：（1）∵四边形ABCD 是菱形， ∴//AB CD ，//AD BC ，AB DA =， ∴ABO EDO ，BFO DAO ，∴AB BO ED DO =，BF BODA DO =， ∴AB BFED DA=， ∵AB DA =， ∴2AB DE BF =⋅； （2）如图，连接OC ，∵四边形ABCD 是菱形， ∴AD=DC ，ADO CDO ∠=∠， 在ADO △和CDO 中，AD CD ADO CDO DO DO =⎧⎪∠=∠⎨⎪=⎩， ∴()ADO CDO SAS ≅， ∴DAO DCO ∠=∠， ∵//AD BF ， ∴DAO OFC ∠=∠， ∴DCO OFC ∠=∠，∵COE FOC ∠=∠， ∴OEC OCF ，∴OE OCOC OF=，即2OC OE OF =⋅， ∵1OE =，2EF =， ∴123OF =+=，∴OC =∴AO OC == ∵//AD CF ， ∴ADE FCE ~，∴12AD AE FC FE ==，∴12BC AD FC +==，1322BF BC CF FC FC FC =+=+=，∴(236CF BF===． 【点睛】本题考查相似三角形，解题的关键是掌握相似三角形的性质和判定．23. 某校数学活动课上，开展测量学校教学大楼()AB 高度的实践活动，三个小组设计了不同方案，测量数据如下表：（2）请选择其中一个可行方案及其测量数据，求出教学大楼的高度． 【参考数据：sin370.60︒≈，cos370.80︒≈，tan370.75︒≈】 【答案】（1）二；（2）36米【分析】（1）根据第二组只测了角度，未给出距离相关信息即可判断； （2）由锐角三角函数可求tan ABBC C =，tan AB BD ADB=∠，由BC BD CD -=，列出方程可求解． 【详解】（1）∴第二组中没有线段长度的数据，所以无法测出AB 的高度， ∴填第二组， 故答案为：二．（2）可选第一组的方案， 设AB xm =，在Rt ABC 中，90B ∠=︒，tan =ABC BC， ∴4=tan tan 373AB x BC x C ==︒，在Rt ABD △中，90B ∠=︒，tan =ABADB BD∠， ∴tan tan 45AB xBD x ADB ===∠︒，∴BC BD CD -=， ∴4123x x -=， ∴36x =．答：教学大楼高36米．【点睛】本题考查了解直角三角形的应用，利用数学知识解决实际问题是中学数学的重要内容．解决此问题的关键在于正确理解题意的基础上建立数学模型，把实际问题转化为数学问题．24. 已知抛物线()20y ax bx a =+≠经过 ()4,0A ，()1,3B -两点，抛物线的对称轴与x 轴交于点C ，点 D 与点B 关于抛物线的对称轴对称，联结BC 、BD ．（1）求该抛物线的表达式以及对称轴；（2）点E 在线段BC 上，当CED OBD =∠∠时，求点 E 的坐标；（3）点M 在对称轴上，点N 在抛物线上，当以点O 、A 、M 、N 为顶点的四边形是平行四边形时，求这个平行四边形的面积． 【答案】（1）231255y x x =-，对称轴为2x =；（2）1,1E ；（3）当OA 为边时，1445S =；当OA 为对角线时，485S =． 【分析】（1）将()4,0A ，()1,3B -代入抛物线2y ax bx =+，求解即可；（2）过B 点作BF x ⊥轴叫x 轴与点F ，过E 点作EH x ⊥轴叫x 轴与点H ，根据B 点坐标是()1,3-，对称轴为2x =，易得BCF △是等腰直角三角形，ECH 也是等腰直角三角形，求出BC =CED OBD =∠∠，点D 与点B 关于抛物线的对称轴对称，可证得OBCEDB ，DBE BCO ，则DBEBCO ，有DBEBBCOC，可得EB =EC =（3）分两种情况讨论：当OA 对角线时，当OA 为边时，分别求出N 点坐标，然后求解即可．【详解】解：（1）将()4,0A ，()1,3B -代入抛物线 2y ax bx =+，得：16403a b a b +=⎧⎨-=⎩，解之得： 35125a b ⎧=⎪⎪⎨⎪=-⎪⎩，∴该抛物线的表达式是231255y x x =-， ∴22231233124255555y x x x xx ， ∴对称轴为2x =；（2）如图示：过B 点作BF x ⊥轴叫x 轴与点F ，过E 点作 EH x ⊥轴叫x 轴与点H ，∴B 点坐标是()1,3-，对称轴为2x =， ∴3BF CF ==，∴BCF △是等腰直角三角形，则ECH 也是等腰直角三角形， ∴22223332BCBF CF ,∴CED OBD =∠∠,CED EBD EDB ∠=∠+∠,OBDEBD OBC∴OBCEDB ,∴点D 与点B 关于抛物线的对称轴对称，则D 点坐标是()5,3， ∴//BD FA∴DBE BCO ∴DBE BCO ∴DB EBBCOC, ∴6BD =,2OC =,2EB，即有EB =∴32222ECBCEB,∴ECH 是等腰直角三角形， ∴1EHHC∴1OH =即点E 的坐标是()1,1； （3）∴4OA =∴当OA 是平行四边形的边长时，如图2所示，则MN 必定在y 轴的上方，并有4MN OA ，∴点M 在对称轴上， ∴点N 的横坐标是6或-2， 又∴点N 在抛物线上， ∴当6x =时，23123666555y， ∴平行四边形OANM 的面积36144455；当2x =-时，23123622555y ， 同理可得平行四边形OANM 的面积36144455； ∴当OA 是平行四边形的对角线时，如图3所示，∵点M 在对称轴上，并MONA ∴点N 也在对称轴2x =上，∴当2x =时，23121222555y， ∴112244255OAN S ∴平行四边形OANM 的面积24482255OAN S ． 综上所述，平行四边形的面积为1445或485． 【点睛】本题考查了用待定系数法求函数解析式，二次函数坐标轴上的点，三角形的相似的判定与性质，平行四边形的判定与性质，熟悉相关性质是解题的关键．25. 如图，已知△ABC 中，∠ACB =90°，AC =BC ，点D 、E 在边AB 上，∠DCE =45°，过点A 作AB 的垂线交CE 的延长线于点M ，联结MD ．（1）求证：2CE BE DE =⋅；（2）当AC =3，AD =2BD 时，求DE 的长；（3）过点M 作射线CD 的垂线，垂足为点F ，设BD x BC =，tan ∠FMD =y ，求y 关于x 的函数关系式，并写出定义域．【答案】（1）见解析； （2）4DE =； （3）1(02y x =<<． 【分析】（1）证明两个角相等证明△CDE ∽△BCE ，列比例式可得结论；（2）如图2，过D 作DN ⊥AC 于N ，根据△ADN 是等腰直角三角形，得AN =DN ，由平行线分线段成比例定理得23AD AN AB AC ==，计算DN 和CN 的长，利用勾股定理计算CD 和BD 的长，根据（1）中的相似三角形，列比例式得：DE CE DC CE BE BC ===，设DE ，CE =3x ，代入比例式可得结论； （3）如图3，作辅助线构建全等三角形，证明△AMC ≌△BPC （ASA ），得CM =CP ，证明△MCD ≌△PCD（SAS ），得∠MDC =∠PDC =∠BDC ，证明△BCD ∽△CMD ，列比例式得BD CD BC CM=，根据三角函数的定义和等量代换可得比例式，并根据D ，E 是AB 上一点，∠DCE =45°，可知当点E 与A 重合时，BD 最大为12AB ，可得x 的取值范围．【小问1详解】证明：如图1，∵∠ACB =90°，AC =BC ，∴∠B =∠CAB =45°，∵∠DCE =45°，∴∠B =∠DCE ，∵∠CED =∠CEB ，∴△CDE ∽△BCE ， ∴CE DE BE CE=， ∴2CE BE DE =⋅；【小问2详解】解：如图2，过D 作DN ⊥AC 于N ，∴∠AND =90°，∵∠DAN =45°，∴△ADN 是等腰直角三角形，∵DN ∥BC ，AD =2BD ， ∴23AD AN AB AC ==， ∵AC =3，∴AB AN =DN =2，CN =1，∵AD =2BD ，∴BD由勾股定理得：DC =由（1）知：△CDE ∽△BCE ，∴3DE CE DC CE BE BC ===，设DE ，CE =3x ，3=，∴x ，∴DE ； 【小问3详解】解：如图3，过点C 作CP ⊥CM ，交AB 的延长线于点P ，∵∠DCE =45°，∠ACB =90°，∴∠ACM +∠BCD =45°=∠BCD +∠BCP ，∴∠BCP =∠ACM ，∵∠CBP =180°-45°=135°=∠CAM ，AC =BC ，∴△AMC ≌△BPC （ASA ），∴CM =CP ，∵∠DCM =∠DCP =45°，CD =CD ，∴△MCD ≌△PCD （SAS ），∴∠MDC =∠PDC =∠BDC ，∵∠ABC =45°=∠MCD ，∴△BCD ∽△CMD ， ∴BD BC CD CM =，即BD CD BC CM=， ∵FM ⊥FC ，∠DCE =45°，∴△CFM 是等腰直角三角形，∴CM FM ，∴y =tan ∠FMDDF MF CM==)CF CD CM-=CM -=1BD BC=x ；Rt △ABC 中，AC =BC ，∴AB BC ，∵D ，E 是AB 上一点，∠DCE =45°，∴当点E与A重合时，BD最大为12 AB，∵BDBC=x，∴0＜x∴y（0＜x＜2）．【点睛】本题是相似形的综合题，考查了全等和相似三角形的判定和性质，等腰直角三角形的性质，三角函数的定义等知识，添加恰当辅助线构造全等三角形是本题的关键．。
2021-2022学年上海市普陀区九年级(上)期末数学试卷(一模)(含答案解析)
2021-2022学年上海市普陀区九年级（上）期末数学试卷（一模）1.下列抛物线经过原点的是( )A. y=x2−2xB. y=(x−2)2C. y=x2+2D. y=(x+2)(x−1)2.在Rt△ABC中，∠C=90∘，已知sinA=13，下列结论正确的是( )A. sinB=13B. cosB=13C. tanB=13D. cotB=133.如图，已知AD//BE//CF，它们依次交直线l1和l2于点A、B、C和点D、E、F，如果AB：BC=2：3，那么下列结论中错误的是( )A. DEEF =23B. DEDF =25C. BECF =25D. EFDF =354.如图，已知点B、D、C、F在同一条直线上，AB//EF，AB=EF，AC//DE，如果BF=6，DC=3，那么BD的长等于( )A. 1B. 32C. 2D. 35.已知a⃗与b⃗ 是非零向量，且|a⃗|=|3b⃗ |，那么下列说法中正确的是( )A. a⃗=3b⃗B. a⃗=−3b⃗C. a⃗//b⃗D. |ab⃗|=36.已知在△ABC中，∠C=90∘，AC=√3，BC=2，如果△DEF∽△ABC，且△DEF两条边的长分别为EF=4和DE=2√7，那么△DEF第三条边的长为( )A. 2B. √7C. 2√3D. 2√117.已知x5=y3，那么x+yy=______.8.已知反比例函数y=k+1，如果在这个函数图象所在的每一个象限内，y的值随着x的值的x增大而增大，那么k的取值范围是______.9.已知函数f(x)=x2−3x+1，如果x=3，那么f(x)=______.10.已知抛物线的开口方向向下，对称轴是直线x=0，那么这条抛物线的表达式可以是______(只要写出一个表达式).11.已知e⃗是单位向量，a⃗与e⃗方向相反，且长度为6，那么a⃗=______.(用向量e⃗表示)12.已知二次函数y=a(x+1)2+c(a≠0)的图象上有两点A(2,4)、B(m,4)，那么m的值等于______.13.如图，在△ABC中，AD平分∠BAC，如果∠B=80∘，∠C=40∘，那么∠ADC的度数等于______.14.如图，在四边形ABCD中，AD//BC，对角线AC、BD相交于点O，如果S△AOB=2a，S△BOC=4a，那么S△ADC=______.(用含有字母a的代数式表示)15.某芭蕾舞演员踮起脚尖起舞，腰部就成为整个身形的黄金分割点，给观众带来美感，如图，如果她踮起脚尖起舞时，那么她的腰部以下高度a与身形b之间的比值等于______.16.如图，在△ABC中，∠A=90∘，斜边BC的垂直平分线分别交，AB=7，那么CD的长等AB、BC交于点D、E，如果cosB=78于______.17.如图，已知点D、E分别在线段AB和AC上，点F是BE与CD的交点，∠B=∠C，如果DF=4EF，AB=6，AC=4，那么AD的长等于______.18. 如图，在△ABC 中，AB =AC =5，BC =4，AD 是边BC 上的高，将△ABC 绕点C 旋转，点B 落在线段AD 上的点E 处，点A 落在点F 处，那么cos∠FAD =______.19. 计算：4sin 260∘−2sin30∘−cot45∘tan60∘−2cos45∘.20. 如图，已知AB//CD ，AD 、BC 相交于点E ，过E 作EF//CD 交BD 于点F ，AB ：CD =1：3. (1)求EFCD的值； (2)设CD ⃗⃗⃗⃗⃗ =a ⃗ ，BF ⃗⃗⃗⃗⃗ =b ⃗ ，那么EF ⃗⃗⃗⃗⃗ =______，AE ⃗⃗⃗⃗⃗ =______(用向量a ⃗ ，b ⃗ 表示)21. 在平面直角坐标系xOy 中，反比例函数y =kx (k ≠0)的图象与正比例函数y =2x 的图象相交于横坐标为1的点A.(1)求这个反比例函数的解析式；(2)如图，已知B 是正比例函数图象在第一象限内的一点，过点B 作BC ⊥x 轴，垂足为点C ，BC 与反比例函数图象交于点D ，如果AB =AC ，求点D 的坐标．22.图(1)为钓鱼竿安置于湖边的示意图，钓鱼竿有两部分组成，一部分为支架，另一部分为钓竿，图(2)是钓鱼竿装置的平面图，NF//MB，NF⊥MN，支架中的MN=AM=20厘米，AC=50厘米，∠CAB=37∘，AB可以伸缩，长度调节范围为65cm≤AB≤180cm，钓竿EF 放在支架的支点B、C上，并使钓竿的一个端点F恰好碰到水面．(1)当AB的长度越______(填“长”或“短”)时，钓竿的端点F与点N之间的距离越远；(2)冬季的鱼喜欢远离岸边活动，为了提高钓鱼的成功率，可适当调节AB的长度，使钓竿的端点F与点N之间的距离最远，请直接写出你选择的AB的长度，并求出此时钓竿的端点F 与点N之间的距离(参考数据：sin37∘≈0.6，cos37∘≈0.8，tan37∘≈0.75)23.已知：如图，在△ABC中，点D、E分别在边AC、BC上，BD=DC，BD⋅BC=BE⋅AC.(1)求证：∠ABE=∠DEB；(2)延长BA、ED交于点F，求证：FDFE =ADDC.24.如图，在平面直角坐标系xOy中，已知抛物线y=13x2+bx+c与直线y=−13x+1交于点A(m,0)，B(−3,n)，与y轴交于点C，连接AC.(1)求m、n的值和抛物线的表达式；(2)点D在抛物线y=13x2+bx+c的对称轴上，当∠ACD=90∘时，求点D的坐标；(3)将△AOC平移，平移后点A仍在抛物线上，记作点P，此时点C恰好落在直线AB上，求点P的坐标．25.如图，在△ABC中，边BC上的高AD=2，tanB=2，直线l平行于BC，分别交线段AB，AC，AD于点E、F、G，直线l与直线BC之间的距离为m.(1)当EF=CD=3时，求m的值；(2)将△AEF沿着EF翻折，点A落在两平行直线l与BC之间的点P处，延长EP交线段CD 于点Q.①当点P恰好为△ABC的重心时，求此时CQ的长；②连接BP，在∠CBP>∠BAD的条件下，如果△BPQ与△AEF相似，试用m的代数式表示线段CD的长．答案和解析1.【答案】A【解析】解：A、将x=0代入，得y=0，所以该抛物线经过原点，本选项符合题意；B、将x=0代入，得y=4，所以该抛物线不经过原点，本选项不符合题意；C、将x=0代入，得y=2，所以该抛物线不经过原点，本选项不符合题意；D、将x=0代入，得y=−2，所以该抛物线不经过原点，本选项不符合题意.故选：A.本题考查了二次函数图象上点的坐标特征，抛物线经过一点，则该点的坐标满足函数的解析式．将x=0分别代入各抛物线的解析式，如果求出y=0，那么该抛物线经过原点，依次带入求解即可．2.【答案】B【解析】解：∵sinA=cos(90∘−A)，∴sinA=cosB，∵sinA=13，∴cosB=1 3.故选：B.本题考查的是锐角三角函数的定义，掌握互余两角的三角函数的关系是解题的关键．根据一个角的正弦等于这个角的余角的余弦解答即可．3.【答案】C【解析】解：∵AD//BE//CF，AB：BC=2：3，∴DE EF =ABBC=23，∴DE DF =25，EFDF=35，故选项A、B、D结论正确，不符合题意；连接AF，交BE于点H，∵BE//CF，∴△ABH∽△ACF，∴BH CF =ABAC=DEDF=25，∴BE CF >25，故选项C结论错误，符合题意. 故选：C.本题考查的是平行线分线段成比例定理，灵活运用定理、找准对应关系是解题的关键．根据平行线分线段成比例定理列出比例式判断A、B、D；连接AF，交BE于点H，根据相似三角形的性质判断C.4.【答案】B【解析】解：∵AB//EF，∴∠B=∠F，∵AC//DE，∴∠ACB=∠EDF，在△ABC和△EFD中，{∠ACB=∠EDF ∠B=∠FAB=EF，∴△ABC≌△EFD(AAS)，∴BC=FD，∴BC−DC=FD−DC，∴BD=FC，∴BD=12(BF−DC)=12(6−3)=32.故选：B.本题主要考查了全等三角形的判定与性质，证得△ABC≌△EFD是解决问题的关键．由平行线的性质得到∠B=∠F，∠ACB=∠EDF，证得△ABC≌△EFD，得到BC=FD，进而得到BD=FC，即可得出BD=12(BF−DC)=32.5.【答案】D【解析】解：A、由a⃗与b⃗ 是非零向量，且|a⃗|=|3b⃗ |知，a⃗与3b⃗ 只是模相等，方向不一定相同，a⃗=3b⃗ 不一定成立，故不符合题意；B、由a⃗与b⃗ 是非零向量，且|a⃗|=|3b⃗ |知，a⃗与3b⃗ 只是模相等，方向不一定相反，即a⃗=−3b⃗ 不一定成立，故不符合题意；C、由a⃗与b⃗ 是非零向量，且|a⃗|=|3b⃗ |知，a⃗与3b⃗ 只是模相等，不一定共线，故不符合题意；D、由a⃗与b⃗ 是非零向量，且|a⃗|=|3b⃗ |知，|ab⃗|=3，符合题意．故选：D.本题考查了平面向量，注意平面向量既有大小，又有方向．根据平面向量以及模的定义的知识求解即可求得答案.6.【答案】C【解析】解：在△ABC 中，∠C =90∘，AC =√3，BC =2， ∴AB =√AC 2+BC 2=√7， ∵△DEF ∽△ABC ， ∴AB DE =BC EF =ACDF ， ∴24=√72√7=√3DF ，∴DF =2√3，则△DEF 第三条边的长为2√3. 故选：C.本题考查了相似三角形的性质，熟练掌握相似三角形的性质是解题的关键． 根据勾股定理得到AB =√AC 2+BC 2=√7，根据相似三角形的性质得到结论．7.【答案】83【解析】解：设x5=y3=k ， 则x =5k ，y =3k ，∴x +y y =5k +3k 3k =8k 3k =83. 故答案为：83.本题考查了比例的性质，能选择适当的方法求解是解此题的关键．设x5=y3=k ，根据比例的性质求出x =5k ，y =3k ，把x =5k ，y =3k 代入x+yy ，即可求出答案．8.【答案】k <−1【解析】解：∵函数y =k+1x的图象在其所在的每一象限内，函数值y 随自变量x 的增大而增大，∴k +1<0， 解得k <−1. 故答案为：k <−1.本题考查了反比例函数的性质，关键掌握以下性质：反比例函数y =kx (k ≠0)，当k >0时，在每一个象限内，函数值y 随自变量x 的增大而减小；当k <0时，在每一个象限内，函数值y 随自变量x 增大而增大．根据反比例函数的性质可得k +1<0，再解不等式即可．9.【答案】1【解析】解：f(3)=32−3×3+1=1.故答案为：1.本题考查了二次函数图象上点的坐标特征，函数图象上点的坐标适合解析式．把x=3代入函数关系式即可求得．10.【答案】y=−x2+2(答案不唯一)【解析】解：满足题意的抛物线解析式为：y=−x2+2(本题答案不唯一).故答案为：y=−x2+2(答案不唯一).本题考查了待定系数法求二次函数的解析式，熟知二次函数的性质是解题的关键．可根据顶点式求抛物线解析式，只需要对称轴是直线x=0，图象开口向下即可．11.【答案】−6e⃗【解析】解：∵e⃗是单位向量，a⃗与e⃗方向相反，且长度为6，∴a⃗=−6e⃗ .故答案为：−6e⃗ .本题考查平面向量，解题的关键是熟练掌握基本知识，属于中考常考题型．根据平面向量的性质解决问题即可．12.【答案】−4【解析】解：∵二次函数y=a(x+1)2+c(a≠0)，∴抛物线的对称轴为直线x=−1，∵点A(2,4)、B(m,4)都在抛物线上，∴点A、B关于直线x=−1对称，∴2+m=−1，2∴m=−4.故答案为：−4.本题考查二次函数图象上点的坐标特征，二次函数的性质，熟知二次函数的对称性是解决问题的关键．根据点A(2,4)、B(m,4)坐标特点可知这两个点关于对称轴对称，可求出m的值．13.【答案】110∘【解析】解：∵∠B=80∘，∠C=40∘，∴∠BAC=180∘−∠B−∠C=60∘，∵AD平分∠BAC，∴∠BAD=12∠BAC=30∘，∴∠ADC=∠B+∠BAD=110∘.故答案为：110∘.本题主要考查三角形的外角性质，三角形的内角和定理，解答的关键是对相应的知识的掌握．由三角形的内角和可求得∠BAC=60∘，再由角平分线的定义得∠BAD=30∘，利用三角形的外角性质即可求∠ADC的度数．14.【答案】3a【解析】解：在梯形ABCD中，AD//BC，∵S△AOB=2a，S△BOC=4a，∴S△AOB：S△BOC=1：2，∴AO：OC=1：2，∵AD//BC，∴△AOD∽△COB，∵AO：OC=1：2，∴S△AOD：S△BOC=1：4，∴S△AOD=a，∴S△COD=2a，∴S△ADC=S△AOD+S△DOC=3a.故答案为：3a.本题主要考查了相似三角形的判定与性质的应用，以及梯形的特征和应用，要熟练掌握．首先根据S△AOB：S△BOC=1：2，可得AO：OC=1：2，然后根据相似三角形的面积的比的等于它们的相似比的平方和等高的三角形面积比是底与底的比，进而可以解决问题．15.【答案】√5−12【解析】解：∵某芭蕾舞演员踮起脚尖起舞，腰部就成为整个身形的黄金分割点，∴ab=√5−12.故答案为：√5−12.本题考查了黄金分割，解决本题的关键是熟记黄金分割的比值．由黄金分割的定义即可得出答案．16.【答案】327【解析】解：在△ABC中，∠A=90∘，cosB=78，AB=7，∴BC=AB÷cosB=7÷78=8，∵斜边BC的垂直平分线分别交AB、BC交于点D、E，∴BE=12BC=4，CD=BD，∠BED=90∘，∴CD=BD=BE÷cosB=4÷78=327.故答案为：327.本题主要考查解直角三角形，熟练掌握三角函数的定义是解题的关键．根据cosB=78，AB=7，求出BC=8，则BE=4，BD=327，再根据CD=BD，即可求出CD.17.【答案】2【解析】解：∵∠DFB=∠EFC，∠B=∠C，∴△DBF∽△ECF，∴DF EF =BFFC=BDEC=4，∵∠B=∠C，∠A=∠A，∴△ABE∽△ACD，∴AB AC =AEAD，∵AB=6，AC=4，∴AE AD =64=32，设CE=x，则BD=4x，∴AE=AC−CE=4−x，AD=AB−BD=6−4x，∴4−x 6−4x =32，∴x=1，∴AD=2.故答案为：2.本题考查了相似三角形的判定与性质，证明△ABE∽△ACD是解题的关键．证明△DBF∽△ECF，由相似三角形的性质，得出DFEF =BFFC=BDEC=4，证明△ABE∽△ACD，由相似三角形的性质得出ABAC =AEAD，设CE=x，则BD=4x，得出方程4−x6−4x=32，求出x=1，则可得出答案．18.【答案】√21−2√310【解析】解：如图，过点F 作FG ⊥AD 交AD 于点G ，∵将△ABC 绕点C 旋转，点B 落在线段AD 上的点E 处，点A 落在点F 处， ∴CE =BC =4，CF =EF =AB =AC =5， ∵AB =AC ，AD 是边BC 上的高， ∴BD =CD =2， ∴cos∠ECD =CD CE =24=12， ∴∠ECD =60∘，∴DE =CE ⋅sin∠ECD =4×sin60∘=2√3， 易知∠ACF =∠ECD =60∘， ∴△ACF 是等边三角形， ∴AF =EF =5，在Rt △ACD 中，AD =√AC 2−CD 2=√52−22=√21， ∴AE =AD −DE =√21−2√3， ∵AF =EF ，FG ⊥AD ， ∴AG =EG =√21−2√32，∴cos∠FAD =AGAF=√21−2√325=√21−2√310.故答案为：√21−2√310.本题考查了旋转的性质，等腰三角形的性质，等边三角形的判定和性质，勾股定理，三角函数定义，解题关键是要熟练运用等腰三角形性质．如图，过点F 作FG ⊥AD 交AD 于点G ，由旋转可知：CE =BC =4，CF =EF =AB =AC =5，利用三角函数可得∠ECD =60∘，进而可得DE =2√3，AF =EF =5，运用勾股定理可得AD =√21，AE =√21−2√3，由等腰三角形性质可得AG =EG =√21−2√32，再运用三角函数可得cos∠FAD =AG AF=√21−2√310.19.【答案】解：原式=4×(√32)2−2×12−1√3−2×√22=4×34−1−1√3−√2 =√3−√2 =√3−√2=√3+√2.【解析】本题考查了实数的运算，特殊角的三角函数值，熟练掌握运算法则是解本题的关键． 原式利用特殊角的三角函数值计算即可求出值．20.【答案】解：(1)∵AB//CD ，∴∠EAB =∠EDC ，∠ABE =∠DCE ， ∴△ABE ∽△DCE ， ∴ABCD =BECE =13， ∴CE =3BE ， ∵EF//CD ， ∴∠BEF =∠BCD ， ∵∠EBF =∠CBD ， ∴△BEF ∽△BCD ， ∴BE BC =EF CD， ∵BC =BE +CE =BE +3BE =4BE ， ∴EFCD =14； (2)14a ⃗ ；112a ⃗ +b ⃗ . 【解析】解：(1)见答案； (2)由(1)知：EF =14CD ， ∴EF ⃗⃗⃗⃗⃗ =14CD ⃗⃗⃗⃗⃗ =14a ⃗ ， ∵BE ⃗⃗⃗⃗⃗ +EF ⃗⃗⃗⃗⃗ =BF ⃗⃗⃗⃗⃗ ， ∴BE ⃗⃗⃗⃗⃗ =BF ⃗⃗⃗⃗⃗ −EF ⃗⃗⃗⃗⃗ ， ∵BF ⃗⃗⃗⃗⃗ =b ⃗ ，∴BE ⃗⃗⃗⃗⃗ =b ⃗ −14a ⃗ ，∵AB ：CD =1：3，∴AB =13CD ， ∴AB ⃗⃗⃗⃗⃗ =13CD ⃗⃗⃗⃗⃗ =13a ⃗ ，则AE ⃗⃗⃗⃗⃗ =AB ⃗⃗⃗⃗⃗ +BE ⃗⃗⃗⃗⃗ =13a ⃗ +b ⃗ −14a ⃗ =112a ⃗ +b ⃗ .故答案为：14a ⃗ ；112a ⃗ +b ⃗ . 本题考查相似三角形的判定和性质以及平面向量，熟练掌握平行线的性质和平面向量的加、减运算是解题的关键．(1)根据平行线的性质和相似三角形的判定，证明△ABE ∽△DCE 和△BEF ∽△BCD 即可得出结论； (2)根据(1)中结论和平面向量的加、减运算即可得出结论．21.【答案】解：(1)把x =1代入y =2x ，得y =2，∴A 的坐标为(1,2)，把A 的坐标代入y =kx (k ≠0)，得k =1×2=2， 即反比例函数的表达式为y =2x ；(2)如图，过点A 作AE ⊥BC 交BC 于点E ， ∵BC ⊥x 轴， ∴AE//x 轴， ∵A(1,2)， ∴CE =2，∵AC =AB ，AE ⊥BC ， ∴CE =BE =2， ∴B 点的纵坐标为4，把y =4代入y =2x ，得4=2x ，解得x =2， 即B 点的坐标为(2,4)， ∵D 点的横坐标为2， 把x =2代入y =2x ，得y =1， ∴点D 的坐标为(2,1).【解析】本题考查了一次函数与反比例函数的交点问题，用待定系数法求反比例函数的解析式，能求出各个点的坐标是解此题的关键．(1)把x =1代入y =2x ，求出A 的坐标，把A 的坐标代入y =kx (k ≠0)，求出k 即可； (2)过点A 作AE ⊥BC 交BC 于点E ，求出CE =2，根据等腰三角形的性质求出CE =BE =2，得出B 点的纵坐标为4，代入y =2x ，求出B 点的坐标，即可得出D 点的横坐标，进而即可求得纵坐标．22.【答案】解：(1)长；(2)AB=180cm，FN=293cm.如图(2)中，过点C作CK⊥AB交AB于点K，过点A作AH⊥FN交FN于点H，过点B作BJ⊥FN交FN于点J，则四边形MNHA，四边形AHJB都是矩形，∴MN=AH=BJ=20厘米，AM=NH=20厘米，AB=HJ=180厘米，在Rt△ACK中，CK=AC⋅sin37∘≈30(厘米)，AK=AC⋅cos37∘≈40(厘米)，∴BK=AB−AK=180−40=140(厘米)，∵BM//FN，∴∠CBK=∠F，∴tan∠CBK=tanF，∴CK BK =JBFJ，∴30 140=20FJ，∴FJ≈93(厘米)，∴FN=NH+HJ+FJ=20+180+93=293(厘米).答：AB的长度是180厘米，此时钓竿的端点F与点N之间的距离约为293厘米．【解析】解：(1)观察图象可知，当AB的长度越长时，钓竿的端点F与点N之间的距离越远，故答案为：长；(2)见答案.本题考查解直角三角形的应用，解题的关键是理解题意，学会添加常用辅助线，构造直角三角形解决问题，属于中考常考题型．(1)观察图象可知，当AB的长度越长时，钓竿的端点F与点N之间的距离越远，(2)如图(2)中，过点C作CK⊥AB交AB于点K，过点A作AH⊥FN交FN于点H，过点B作BJ⊥FN 交FN于点J，则四边形MNHA，四边形AHJB都是矩形．分别求出NH，HJ，JF，可得结论．23.【答案】解：(1)证明：∵BD=DC，∴∠DBC=∠C，∵BD⋅BC=BE⋅AC，∴BD AC =BEBC，∴△ABC∽△DEB，∴∠ABC=∠DEB，即∠ABE=∠DEB；(2)如图所示，由(1)得△ABC∽△DEB，∴∠CAB=∠BDE，∴∠FAD=∠FDB，∵∠F=∠F，∴△FAD∽△FDB，∴FD FB =ADDB，又∠ABE=∠DEB，∴FB=FE，又∵BD=DC，∴FD FE =ADDC.【解析】本题考查相似三角形的判定和性质，关键是找到相似的三角形．(1)由BD⋅BC=BE⋅AC，得出BDAC =BEBC，又BD=DC，得出∠DBC=∠C，从而得出结论；(2)根据(1)的结论和已知证明△FAD∽△FDB即可．24.【答案】解：(1)将A(m,0)代入y=−13x+1，解得m=3，∴A(3,0)，将B(−3,n)代入y=−13x+1，解得n=2，∴B(−3,2)，把A(3,0)，B(−3,2)代入y =13x 2+bx +c 中，得{13×9+3b +c =013×9−3b +c =2，解得{b =−13c =−2，∴抛物线的解析式为y =13x 2−13x −2； (2)如图1，过点D 作DH ⊥y 轴于点H ，∵抛物线的解析式为y =13x 2−13x −2， ∴抛物线的对称轴为直线x =−b 2a=12，∴DH =12， ∵∠ACD =90∘， ∴∠ACO +∠DCH =90∘， 又∵∠DCH +∠CDH =90∘， ∴∠ACO =∠CDH ， ∴tan∠ACO =tan∠CDH ， ∴AO CO =CHDH ，由(1)可知OA =3，OC =2， ∴32=CH12，∴CH =34，∴OH=OC+CH=114，∴D(12,−114)；(3)如图2，若平移后的三角形为△PMN，则MN=OC=2，PM=OA=3，设P(t,13t2−13t−2)，∴N(t−3,13t2−13t−2−2)，∵点N在直线y=−13x+1上，∴1 3t2−13t−4=−13(t−3)+1，∴t=3√2或t=−3√2，∴P(3√2,4−√2)或P(−3√2,4+√2).【解析】本题属于二次函数综合题，考查了二次函数的性质，一次函数的性质，直角三角形的性质，锐角三角函数的定义，平移的性质等知识，解题的关键是理解题意，学会用转化的思想思考问题，学会利用参数构建方程确定点的坐标．(1)利用待定系数法求出A，B两点坐标，再代入抛物线表达式即可解决问题；(2)过点D作DH⊥y轴于点H，由直角三角形的性质得出tan∠ACO=tan∠CDH，则AOCO =CHDH，可列出方程求出CH的长，则可得出答案；(3)设P(t,13t2−13t−2)，得出N(t−3,13t2−13t−2−2)，由点N在直线AB上可得出t的值，则可得出答案．25.【答案】解：(1)如图1，在△ABC中，边BC 上的高AD=2，tanB=2，∴ADBD=tanB=2，∴BD=1，∵EF=CD=3，DG=m，∴BC=BD+CD=4，AG=AD−DG=2−m，∵EF//BC，∴EF BC =AGAD，即34=2−m2，解得m=12，∴m的值为12；(2)①如图2，∵将△AEF沿着EF翻折，点A落在△ABC的重心点P处，∴BD=CD=1，AP=2PD，即PD=13AD=23，AP=23AD=43，∴AG=GP=12AP=23，∴DP=GP，∵EF//BC，∴∠PGE=∠PDQ=90∘，△AEG∽△ABD，∴EG BD =AGAD，即EG1=232，∴EG=13，在△PQD和△PEG中，{∠QPD=∠EPG DP=GP∠PDQ=∠PGE，∴△PQD≌△PEG(ASA)，∴DQ=EG=13，∴CQ=CD−DQ=1−13=23，∴此时CQ的长为23；②在Rt△ABD中，AB=√AD2+BD2=√5，∵将△AEF沿着EF翻折，点A落在两平行直线l与BC之间的点P处，∴∠PBQ<∠ABD，∵EF//BC，∴∠AEF=∠ABD，∴∠PBQ<∠AEF，∵∠CBP >∠BAD ，∴∠BAD <∠PBQ <∠AEF ，∵GP =AG =2−m ，DG =m ，∴DP =DG −GP =m −(2−m)=2m −2，∴m >1，∴1<m <2，∵∠AEF =∠ABD ， ∴AG EG =tan∠AEF =tan∠ABD =2， ∴2−m EG =2， ∴EG =2−m 2，∵EF//BC ，∴△PEG ∽△PQD ，∴DQ EG =DP GP ，即DQ2−m 2=2m−22−m ，∴DQ =m −1，∴BQ =BD +DQ =m ，PQ =√5(m −1)，∵∠AEF =∠PEG =∠BQP ，∠PBQ <∠AEF ，∴△BPQ 与△AEF 相似，则△BPQ ∽△FAE 或△BPQ ∽△AFE ，Ⅰ.当△BPQ ∽△FAE 时，∵△FAE ∽△CAB ，∴△BPQ ∽△CAB ，∴BQ PQ =BC AB ，即m√5(m−1)=BC√5，∴BC =mm−1，∴CD =BC −BD =m m−1−1=1m−1；Ⅰ.当△BPQ ∽△AFE 时，∵△AFE ∽△ACB ，∴△BPQ ∽△ACB ，∴PQBQ =BC AB ，即√5(m−1)m =BC√5，∴BC =5(m−1)m ，∴CD =BC −BD =5(m−1)m −1=4m−5m ，综上，线段CD 的长为1m−1或4m−5m .【解析】本题考查了全等三角形判定和性质，相似三角形的判定和性质，勾股定理，三角函数，翻转变换的性质等，熟练掌握全等三角形判定和性质、相似三角形的判定和性质等相关知识，运用分类讨论思想和方程思想思考解决问题是解题关键．(1)根据ADBD=tanB=2，可得BD=1，再由EF=CD=3，DG=m，可得BC=4，AG=2−m，利用EF//BC，可得EFBC =AGAD，建立方程求解即可；(2)①由翻折可得：BD=CD=1，AP=2PD，即PD=13AD=23，AP=23AD=43，进而得出AG=23，推出DP=GP，再由EF//BC，可得出EG=13，利用ASA证明△PQD≌△PEG，即可求得答案；②分两种情况：Ⅰ.当△BPQ∽△FAE时，由△FAE∽△CAB，推出△BPQ∽△CAB，建立方程求解即可；Ⅰ.当△BPQ∽△AFE时，由△AFE∽△ACB，推出△BPQ∽△ACB，建立方程求解即可．。
2021-2022学年上海市虹口区九年级上学期期末数学试卷(一模)(含答案解析)
2021-2022学年上海市虹口区九年级上学期期末数学试卷(一模)一、选择题（本大题共6小题，共24.0分）1.如图(1)所示，E为矩形ABCD的边AD上一边，动点P，Q同时从点B出发，点P沿折线BE−ED−DC运动到点C时停止，点Q沿BC运动到点C时停止，它们运动的速度都是1cm/秒，设P、Q同时出发t秒时，△BPQ的面积为ycm2.已知y与t的函数关系图象如图(2)(曲线OM为抛物线的一部分)则下列结论正确的是()A. AB：AD=3：4B. 当△BPQ是等边三角形时，t=5秒C. 当△ABE∽△QBP时，t=7秒D. 当△BPQ的面积为4cm2时，t的值是√10或47秒52.已知非零向量a⃗、b⃗ 和c⃗，下列条件中，不能判定a⃗//b⃗ 的是()A. a⃗=−2b⃗B. a⃗=c⃗，b⃗ =3c⃗C. a⃗+2b⃗ =c⃗，a⃗−b⃗ =−c⃗D. |a⃗|=2|b⃗ |3.若y=(3+m)x m2−9是开口向下的抛物线，则m的值为()A. 3B. −3C. √11D. −√114.把抛物线y=(x−2)2向左平移2个单位长度，再向上平移2个单位长度，所得到的抛物线是()A. y=x2+2B. y=x2−2C. y=(x+2)2−2D. y=(x+2)2+25.如图，小王在长江边某瞭望台D处，测得江面上的渔船A的俯角为40°，若DE=3米，CE=2米，CE平行于江面AB，迎水坡BC的坡度i=1：0.75，坡长BC=25米，则此时AB的长约为()(参考数据：sin40°≈0.64，cos40°≈0.77，tan40°≈0.84)A. 10.4米B. 12.4米C. 27.4米D. 22.4米6.如图，四边形ABCD内接于⊙O，AB为直径，AD=CD，过点D作DE⊥AB于点E.连接AC交DE于点F.若cos∠CBA=35，EF=3.则AB的长为()A. 10B. 12C. 16D. 20二、填空题（本大题共12小题，共48.0分）7.如果3x−5y=0，且y≠0，那么xy=______．8.计算：2(2a⃗+3b⃗ )−13a⃗+12b⃗ =______ ．9.如图，是抛物线y=ax2+bx+c(a≠0)图象的一部分．已知抛物线的对称轴为x=2，与x轴的一个交点是(−1,0).有下列结论：①abc>0；②4a−2b+c<0；③4a+b=0；④抛物线与x轴的另一个交点是(5,0)；⑤点(−3,y1)，(6,y2)都在抛物线上，则有y1<y2.其中正确的是。
2021学年度上海杨浦区九上期末质量检测 物理学科(中考一模)
2021学年度第一学期初三期末质量检测物理学科（满分100分，时间90分钟） 2022.1 考生注意：1．本试卷物理部分含四个大题。
2．答题时，考生务必按答题要求在答题纸规定的位置上作答，在草稿纸、本试卷上答题一律无效。
一、选择题（共20分）1.下列各物理量中，可鉴别物质的是A．质量B．电阻C．体积D．密度2.研究得出“导体中电流与电压的关系”的科学家是A．牛顿B．帕斯卡C．欧姆D．伽利略3.下列生活器件中，利用连通器原理的是A．密度计B．茶壶C．抽水机D．注射器4.下列实例中，为了增大压强的是A．尖锐的逃生锤尖B．较宽的书包带C．宽大的骆驼脚掌D．有很多轮子的拖车5.九年级第一学期物理练习册共有70页，平放在水平桌面中央时，它对桌面的压强约为A．1帕B．35帕C．70帕D．140帕6.小球漂浮在酒精中，排开酒精的质量为0.2千克。
若该小球漂浮在水中，则排开水的质量A．一定大于0.2千克B．可能小于0.2千克C．一定等于0.2千克D．一定小于0.2千克7.在图1(a)(b)所示电路中，电源电压相同且不变，电路元件均完好，无论开关S1、S2同时断开，还是同时闭合，观察到电流表A1与电流表A2示数始终相同。
关于电阻R1、R2、R3的判断，以下列式中正确的是A．R3=R1+R2B．R2=R1+R3C．R1>R2>R3D．R3>R1>R28.相同柱形容器甲、乙置于水平地面上，容器内装有完全相同．．．．物块A、B，物块A与容器甲底部用一根细绳相连。
同时往两容器内注水，直至水面相平，如图2所示。
下列判断正确的是A．此时物块受到的浮力F浮A=F浮BB．此时容器对地面的压力F甲=F乙C．此时容器对地面的压强p甲>p乙D．此时水对容器底部的压力F水甲=F水乙9.在图3所示的电路中，电源电压保持不变。
闭合开关S，当滑动变阻器的滑片P由最右端向中间移动时，变小的是A．电流表 A 示数与电流表 A1 示数的差值B．电流表 A 示数与电流表 A1 示数的比值C．电压表 V 示数与电流表 A 示数的比值D．电压表 V 示数与电流表 A1 示数的比值10.如图4所示，均匀正方体甲、乙放置在水平地面上，它们的质量分别为m甲、m乙，对地面的压强分别为p甲、p乙。
上海市松江区2020-2021学年九年级(上)期末数学试卷(一模) 解析版
2020-2021学年上海市松江区九年级（上）期末数学试卷（一模）一、选择题（本大题共6题，每题4分，满分24分【下列各题的四个选项中，有且只有一个选项是正确的，选择正确项的代号并填涂在答题纸的相应位置上】1．如果两个相似三角形对应边的比为1：4，那么它们的周长比是（）A．1：2B．1：4C．1：8D．1：162．在Rt△ABC中，∠C＝90°，∠A＝α，BC＝2，那么AC的长为（）A．2sinαB．2cosαC．2tanαD．2cotα3．将抛物线y＝2x2向右平移3个单位，能得到的抛物线是（）A．y＝2x2+3B．y＝2x2﹣3C．y＝2（x+3）2 D．y＝2（x﹣3）24．已知＝2，下列说法中不正确的是（）A．﹣2＝0B．与方向相同C．∥D．||＝2||5．如图，一艘船从A处向北偏东30°的方向行驶10千米到B处，再从B处向正西方向行驶20千米到C处，这时这艘船与A的距离（）A．15千米B．10千米C．10千米D．5千米6．如图，已知在Rt△ABC中，∠C＝90°，点G是△ABC的重心，GE⊥AC，垂足为E，如果CB＝8，则线段GE的长为（）A．B．C．D．二、填空题（本大题共12题，每题4分，满分48分）【请将结果直接填入答题纸的相应位置上】7．已知，则＝．8．已知线段MN的长是4cm，点P是线段MN的黄金分割点，则较长线段MP的长是cm．9．计算：sin30°•cot60°＝．10．在Rt△ABC中，∠C＝90°，AC＝6，cos A＝，那么AB的长为．11．一个边长为2厘米的正方形，如果它的边长增加x（x＞0）厘米，则面积随之增加y平方厘米，那么y关于x的函数解析式为．12．已知点A（2，y1）、B（3，y2）在抛物线y＝x2﹣2x+c（c为常数）上，则y1y2（填“＞”、“＝”或“＜”）．13．如图，已知直线l1、l2、l3分别交直线l4于点A、B、C，交直线l5于点D、E、F，且l1∥l2∥l3，AB＝4，AC＝6，DF＝10，则DE＝．14．如图，△ABC在边长为1个单位的方格纸中，△ABC的顶点在小正方形顶点位置，那么∠ABC的正弦值为．15．如图，已知点D、E分别在△ABC的边AB和AC上，DE∥BC，＝，四边形DBCE 的面积等于7，则△ADE的面积为．16．如图，在梯形ABCD中，AD∥BC，BC＝2AD，设向量＝，＝，用向量、表示为．17．如图，正方形DEFG的边EF在△ABC的边BC上，顶点D、G分别在边AB、AC上．已知△ABC的边BC＝16cm，高AH为10cm，则正方形DEFG的边长为cm．18．如图，已知矩形纸片ABCD，点E在边AB上，且BE＝1，将△CBE沿直线CE翻折，使点B落在对角线AC上的点F处，联结DF，如果点D、F、E在同一直线上，则线段AE的长为．三、解答题（本大题共7题，满分78分）19．（10分）用配方法把二次函数y＝3x2﹣6x+5化为y＝a（x+m）2+k的形式，并指出这个函数图象的开口方向、对称轴和顶点坐标．20．（10分）如图，已知AB∥CD，AD、BC相交于点E，AB＝6，BE＝4，BC＝9，联结AC．（1）求线段CD的长；（2）如果AE＝3，求线段AC的长．21．（10分）如图，已知在Rt△ABC中，∠C＝90°，sin∠ABC＝，点D在边BC上，BD ＝4，联结AD，tan∠DAC＝．（1）求边AC的长；（2）求cot∠BAD的值．22．（10分）如图，垂直于水平面的5G信号塔AB建在垂直于水平面的悬崖边B点处（点A、B、C在同一直线上）．某测量员从悬崖底C点出发沿水平方向前行60米到D点，再沿斜坡DE方向前行65米到E点（点A、B、C、D、E在同一平面内），在点E处测得5G信号塔顶端A的仰角为37°，悬崖BC的高为92米，斜坡DE的坡度i＝1：2.4．（1）求斜坡DE的高EH的长；（2）求信号塔AB的高度．（参考数据：sin37°≈0.60，cos37°≈0.80，tan37°≈0.75．）23．（12分）如图，已知在▱ABCD中，E是边AD上一点，联结BE、CE，延长BA、CE相交于点F，CE2＝DE•BC．（1）求证：∠EBC＝∠DCE；（2）求证：BE•EF＝BF•AE．24．（12分）如图，在平面直角坐标系xOy中，抛物线y＝ax2+bx﹣2经过点A（2，0）和B （﹣1，﹣1），与y轴交于点C．（1）求这个抛物线的表达式；（2）如果点P是抛物线位于第二象限上一点，PC交x轴于点D，．①求P点坐标；②点Q在x轴上，如果∠QCA＝∠PCB，求点Q的坐标．25．（14分）如图，已知在等腰△ABC中，AB＝AC＝5，tan∠ABC＝2，BF⊥AC，垂足为F，点D是边AB上一点（不与A，B重合）．（1）求边BC的长；（2）如图2，延长DF交BC的延长线于点G，如果CG＝4，求线段AD的长；（3）过点D作DE⊥BC，垂足为E，DE交BF于点Q，联结DF，如果△DQF和△ABC 相似，求线段BD的长．2020-2021学年上海市松江区九年级（上）期末数学试卷（一模）参考答案与试题解析一、选择题（本大题共6题，每题4分，满分24分【下列各题的四个选项中，有且只有一个选项是正确的，选择正确项的代号并填涂在答题纸的相应位置上】1．如果两个相似三角形对应边的比为1：4，那么它们的周长比是（）A．1：2B．1：4C．1：8D．1：16【分析】直接利用相似三角形的性质得出答案．【解答】解：∵两个相似三角形对应边的比为1：4，∴它们的周长比是：1：4．故选：B．2．在Rt△ABC中，∠C＝90°，∠A＝α，BC＝2，那么AC的长为（）A．2sinαB．2cosαC．2tanαD．2cotα【分析】根据锐角三角函数的意义求解后，再做出判断即可．【解答】解：∵cot A＝，BC＝2，∴AC＝BC•cotα＝2cotα，故选：D．3．将抛物线y＝2x2向右平移3个单位，能得到的抛物线是（）A．y＝2x2+3B．y＝2x2﹣3C．y＝2（x+3）2 D．y＝2（x﹣3）2【分析】根据“左加右减、上加下减”的原则进行解答即可．【解答】解：由“左加右减”的原则可知，抛物线y＝2x2向右平移3个单位，能得到的抛物线是y＝2（x﹣3）2．故选：D．4．已知＝2，下列说法中不正确的是（）A．﹣2＝0B．与方向相同C．∥D．||＝2||【分析】根据平面向量的性质进行一一判断．【解答】解：A、由＝2得到：﹣2＝，故本选项说法不正确．B、由＝2知，与方向相同，故本选项说法正确．C、由＝2知，与方向相同，则∥，故本选项说法正确．D、由＝2知，||＝2||，故本选项说法正确．故选：A．5．如图，一艘船从A处向北偏东30°的方向行驶10千米到B处，再从B处向正西方向行驶20千米到C处，这时这艘船与A的距离（）A．15千米B．10千米C．10千米D．5千米【分析】根据直角三角形的三角函数得出AE，BE，进而得出CE，利用勾股定理得出AC 即可．【解答】解：如图，∵BC⊥AE，∴∠AEB＝90°，∵∠EAB＝30°，AB＝10米，∴BE＝5米，AE＝5米，∴CE＝BC﹣CE＝20﹣5＝15（米），∴AC＝（米），故选：C．6．如图，已知在Rt△ABC中，∠C＝90°，点G是△ABC的重心，GE⊥AC，垂足为E，如果CB＝8，则线段GE的长为（）A．B．C．D．【分析】延长AG交BC于D，如图，利用三角形重心的性质得到CD＝BD＝4，AG＝2GD，再证明GE∥CD，则可判断△AEG∽△ACD，然后利用相似比可求出EG的长．【解答】解：延长AG交BC于D，如图，∵点G是△ABC的重心，∴CD＝BD＝BC＝4，AG＝2GD，∵GE⊥AC，∴∠AEG＝90°，而∠C＝90°，∴GE∥CD，∴△AEG∽△ACD，∴＝＝＝，∴EG＝CD＝×4＝．故选：C．二、填空题（本大题共12题，每题4分，满分48分）【请将结果直接填入答题纸的相应位置上】7．已知，则＝．【分析】根据题意，设x＝5k，y＝3k，代入即可求得的值．【解答】解：由题意，设x＝5k，y＝3k，∴＝＝．故答案为：．8．已知线段MN的长是4cm，点P是线段MN的黄金分割点，则较长线段MP的长是（2﹣2）cm．【分析】根据黄金分割的概念得到MP＝MN，把MN＝4cm代入计算即可．【解答】解：∵P是线段MN的黄金分割点，∴MP＝MN，而MN＝4cm，∴MP＝4×＝（2﹣2）cm．故答案为（2﹣2）．9．计算：sin30°•cot60°＝．【分析】直接利用特殊角的三角函数值化简得出答案．【解答】解：原式＝×＝．故答案为：．10．在Rt△ABC中，∠C＝90°，AC＝6，cos A＝，那么AB的长为8．【分析】根据锐角三角函数的意义求解后，再做出判断即可．【解答】解：∵cos A＝＝，AC＝6，∴AB＝＝8，故答案为：8．11．一个边长为2厘米的正方形，如果它的边长增加x（x＞0）厘米，则面积随之增加y平方厘米，那么y关于x的函数解析式为y＝x2+4x．【分析】根据“面积的增加量就是边长增加前后的两个正方形的面积差”可得答案．【解答】解：由题意得，y＝（2+x）2﹣22＝x2+4x，故答案为：y＝x2+4x．12．已知点A（2，y1）、B（3，y2）在抛物线y＝x2﹣2x+c（c为常数）上，则y1＜y2（填“＞”、“＝”或“＜”）．【分析】先求得开口方向和对称轴，再根据二次函数的性质进行判断即可．【解答】解：∵y＝x2﹣2x+c，∴抛物线的开口向上，对称轴是直线x＝﹣＝1，∴在对称轴的右侧，y随x的增大而增大，∵1＜2＜3，∴y1＜y2，故答案为：＜．13．如图，已知直线l1、l2、l3分别交直线l4于点A、B、C，交直线l5于点D、E、F，且l1∥l2∥l3，AB＝4，AC＝6，DF＝10，则DE＝．【分析】直接根据平行线分线段成比例定理得到＝，然后根据比例的性质可计算出DE的长．【解答】解：∵l1∥l2∥l3，∴＝，即＝，∴DE＝．故答案为．14．如图，△ABC在边长为1个单位的方格纸中，△ABC的顶点在小正方形顶点位置，那么∠ABC的正弦值为．【分析】根据题意和图形，可以求得AC、BC和AB的长，然后根据勾股定理的逆定理可以判断△ACB的形状，然后即可求得∠ABC的正弦值．【解答】解：由图可得，AC＝＝，AB＝＝，BC＝＝2，∴AC2+BC2＝AB2，∴△ACB是直角三角形，∴sin∠ABC＝＝，故答案为：．15．如图，已知点D、E分别在△ABC的边AB和AC上，DE∥BC，＝，四边形DBCE 的面积等于7，则△ADE的面积为9．【分析】由DE∥BC可判定△ADE∽△ABC，根据相似三角形的面积比等于相似比的平方，可得＝（）2＝，从而求得＝，即可求得△ADE的面积为9．【解答】解：∵DE∥BC，∴△ADE∽△ABC，∴＝（）2＝，∴＝，∵四边形DBCE的面积等于7，∴S△ADE＝9．故答案为：9．16．如图，在梯形ABCD中，AD∥BC，BC＝2AD，设向量＝，＝，用向量、表示为+2．【分析】根据梯形的性质和三角形法则解答．【解答】解：如图，在梯形ABCD中，∵AD∥BC，BC＝2AD，＝，∴＝2＝2，∴＝+＝+2，故答案是：+2．17．如图，正方形DEFG的边EF在△ABC的边BC上，顶点D、G分别在边AB、AC上．已知△ABC的边BC＝16cm，高AH为10cm，则正方形DEFG的边长为cm．【分析】设正方形DEFG的边长为xcm，则DE＝PH＝xcm，所以AP＝（10﹣x）cm，再证明△ADG∽△ABC，则利用相似比得到＝，然后根据比例的性质求出x．【解答】解：如图，设正方形DEFG的边长为xcm，则DE＝PH＝xcm，∴AP＝AH﹣PH＝（10﹣x）cm，∵DG∥BC，∴△ADG∽△ABC，∴＝，即＝，∴x＝（cm），故答案为．18．如图，已知矩形纸片ABCD，点E在边AB上，且BE＝1，将△CBE沿直线CE翻折，使点B落在对角线AC上的点F处，联结DF，如果点D、F、E在同一直线上，则线段AE的长为．【分析】根据矩形的性质得到AD＝BC，∠ADC＝∠B＝∠DAE＝90°，根据折叠的性质得到CF＝BC，∠CFE＝∠B＝90°，EF＝BE＝1，DC＝DE，证明△AEF∽△DEA，根据相似三角形的性质即可得到结论．【解答】解：∵四边形ABCD是矩形，∴AD＝BC，AB＝CD，∠ADC＝∠B＝∠DAE＝90°，∵把△BCE沿直线CE对折，使点B落在对角线AC上的点F处，∴CF＝BC，∠CFE＝∠B＝90°，EF＝BE＝1，∠CEB＝∠CEF，∵矩形ABCD中，DC∥AB，∴∠DCE＝∠CEB，∴∠CEF＝∠DCE，∴DC＝DE，设AE＝x，则AB＝CD＝DE＝x+1，∵∠AFE＝∠CFD＝90°，∴∠AFE＝∠DAE＝90°，∵∠AEF＝∠DEA，∴△AEF∽△DEA，∴，解得x＝或x＝（舍去），∴AE＝．故答案为：．三、解答题（本大题共7题，满分78分）19．（10分）用配方法把二次函数y＝3x2﹣6x+5化为y＝a（x+m）2+k的形式，并指出这个函数图象的开口方向、对称轴和顶点坐标．【分析】利用配方法把一般式化为顶点式，根据二次函数的性质解答．【解答】解：y＝3x2﹣6x+5＝3（x2﹣2x）+5＝3（x2﹣2x+1﹣1）+5＝3（x﹣1）2+2，开口向上，对称轴为直线x＝1，顶点（1，2）．20．（10分）如图，已知AB∥CD，AD、BC相交于点E，AB＝6，BE＝4，BC＝9，联结AC．（1）求线段CD的长；（2）如果AE＝3，求线段AC的长．【分析】（1）证明△ABE∽△DCE，由相似三角形的性质得出，则可得出答案；（2）由相似三角形的性质求出DE＝，证明△ABC∽△ECD，由相似三角形的性质得出，则可求出答案．【解答】解：（1）∵AB∥CD，∴△ABE∽△DCE，∴，∵AB＝6，BE＝4，BC＝9，∴CD＝；（2）∵AE＝3，△ABE∽△DCE，∴，∴，∴DE＝，∵，＝，∴，∵AB∥DC，∴∠ECD＝∠ABC，∴△ABC∽△ECD，∴，∴，∴AC＝．21．（10分）如图，已知在Rt△ABC中，∠C＝90°，sin∠ABC＝，点D在边BC上，BD ＝4，联结AD，tan∠DAC＝．（1）求边AC的长；（2）求cot∠BAD的值．【分析】（1）根据题意和锐角三角函数，可以求得AC的长；（2）根据（1）中的结果，可以得到AC、CD的长，然后根据勾股定理可以得到AD的长，再根据等面积法可以求得DE的长，从而可以求得AE的长，然后即可得到cot∠BAD 的值．【解答】解：（1）设AC＝3x，∵∠C＝90°，sin∠ABC＝，∴AB＝5x，BC＝4x，∵tan∠DAC＝，∴CD＝2x，∵BD＝4，BC＝CD+BD，∴4x＝2x+4，解得x＝2，∴AC＝3x＝6；（2）作DE⊥AB于点E，由（1）知，AB＝5x＝10，AC＝6，BD＝4，∵，∴，解得DE＝，∵AC＝6，CD＝2x＝4，∠C＝90°，∴AD＝＝2，∴AE＝＝＝，∴cot∠BAD＝＝＝，即cot∠BAD的值是．22．（10分）如图，垂直于水平面的5G信号塔AB建在垂直于水平面的悬崖边B点处（点A、B、C在同一直线上）．某测量员从悬崖底C点出发沿水平方向前行60米到D点，再沿斜坡DE方向前行65米到E点（点A、B、C、D、E在同一平面内），在点E处测得5G信号塔顶端A的仰角为37°，悬崖BC的高为92米，斜坡DE的坡度i＝1：2.4．（1）求斜坡DE的高EH的长；（2）求信号塔AB的高度．（参考数据：sin37°≈0.60，cos37°≈0.80，tan37°≈0.75．）【分析】（1）过点E作EM⊥DC交DC的延长线于点M，根据斜坡DE的坡度（或坡比）i＝1：2.4可设EH＝x，则DH＝2.4x，利用勾股定理求出x的值，进而可得出EH；（2）结合（1）得DH的长，故可得出CH的长．由矩形的判定定理得出四边形EHCM 是矩形，故可得出EM＝HC，CM＝EH，再由锐角三角函数的定义求出AM的长，进而可得出答案．【解答】解：（1）过点E作EM⊥AC于点M，∵斜坡DE的坡度（或坡比）i＝1：2.4，DE＝65米，CD＝60米，∴设EH＝x，则DH＝2.4x．在Rt△DEH中，∵EH2+DH2＝DE2，即x2+（2.4x）2＝652，解得，x＝25（米）（负值舍去），∴EH＝25米；答：斜坡DE的高EH的长为25米；（2）∵DH＝2.4x＝60（米），∴CH＝DH+DC＝60+60＝120（米）．∵EM⊥AC，AC⊥CD，EH⊥CD，∴四边形EHCM是矩形，∴EM＝CH＝120米，CM＝EH＝25米．在Rt△AEM中，∵∠AEM＝37°，∴AM＝EM•tan37°≈120×0.75＝90（米），∴AC＝AM+CM＝90+25＝115（米）．∴AB＝AC﹣BC＝115﹣92＝23（米）．答：信号塔AB的高度为23米．23．（12分）如图，已知在▱ABCD中，E是边AD上一点，联结BE、CE，延长BA、CE相交于点F，CE2＝DE•BC．（1）求证：∠EBC＝∠DCE；（2）求证：BE•EF＝BF•AE．【分析】（1）通过证明△DEC∽△ECB，可得结论；（2）通过证明△ABE∽△EBF，可得△ABE∽△EBF，可得结论．【解答】证明：（1）∵四边形ABCD是平行四边形，∴AD∥BC，∴∠DEC＝∠BCE，∵CE2＝DE•BC，∴，∴△DEC∽△ECB，∴∠EBC＝∠DCE；（2）∵AD∥BC，AB∥CD，∴∠AEB＝∠EBC，∠F＝∠ECD，∴∠AEB＝∠F，又∵∠ABE＝∠EBF，∴△ABE∽△EBF，∴，∴BE•EF＝BE•AE．24．（12分）如图，在平面直角坐标系xOy中，抛物线y＝ax2+bx﹣2经过点A（2，0）和B （﹣1，﹣1），与y轴交于点C．（1）求这个抛物线的表达式；（2）如果点P是抛物线位于第二象限上一点，PC交x轴于点D，．①求P点坐标；②点Q在x轴上，如果∠QCA＝∠PCB，求点Q的坐标．【分析】（1）由待定系数法可求解析式；（2）①过点P作PE⊥x轴于E，由平行线分线段成比例可求PE的长，代入解析式可求解；②分两种情况讨论，利用全等三角形的性质和相似三角形的性质可求解．【解答】解：（1）∵抛物线y＝ax2+bx﹣2经过点A（2，0）和B（﹣1，﹣1），∴，解得：，∴抛物线解析式为：y＝x2﹣x﹣2；（2）①如图1，过点P作PE⊥x轴于E，∵抛物线y＝ax2+bx﹣2与y轴交于点C，∴点C（0，﹣2），∴OC＝2，∵PE∥OC，∴＝，∴PE＝，∴＝x2﹣x﹣2，∴x＝﹣2或x＝（不合题意舍去），∴点P（﹣2，）；②如图2，过点B作BH⊥CO于H，由①可知DO＝＝，∵B（﹣1，﹣1），点C（0，﹣2），A（2，0）∴OA＝OC＝2，BH＝CH＝1，∴∠BCH＝45°＝∠OCA，∴∠BCA＝90°，当点Q在线段AO上时，∵∠QCA＝∠PCB，∴∠DCO＝∠QCO，又∵CO＝CO，∠DOC＝∠QOC＝90°，∴△DOC≌△QOC（ASA），∴DO＝QO＝，∴点Q坐标为（，0），当点Q'在射线OA上时，∵∠Q'CA＝∠PCB，∴∠DCQ'＝90°，∴∠CDO+∠DQ'C＝90°，∠DCO+∠CDO＝90°，∴∠DQ'C＝∠DCO，又∵∠DOC＝∠Q'OC＝90°，∴△DOC∽△COQ'，∴，∴4＝×Q'O，∴Q'O＝，∴点Q'（，0），综上所述：点Q坐标为（，0）或（，0）．25．（14分）如图，已知在等腰△ABC中，AB＝AC＝5，tan∠ABC＝2，BF⊥AC，垂足为F，点D是边AB上一点（不与A，B重合）．（1）求边BC的长；（2）如图2，延长DF交BC的延长线于点G，如果CG＝4，求线段AD的长；（3）过点D作DE⊥BC，垂足为E，DE交BF于点Q，联结DF，如果△DQF和△ABC 相似，求线段BD的长．【分析】（1）先利用等腰三角形的性质判断出BC＝2BH，再用三角函数和勾股定理求出BH，即可得出结论；（2）先利用勾股定理和三角函数求出CF，再判断出△CFK∽△AFD和△CGK∽△BGD，得出比例式，即可得出结论；（3）先求出BF＝4，再判断出△BEQ∽△BFC，得出，设EQ＝m，则BQ ＝5m，BE＝2m，进而表示出BD＝10m，DQ＝3m，∠DQF＝∠C，再分两种情况，利用相似得出比例式表示出FQ，最后用BF＝4建立方程求出m，即可得出结论．【解答】解（1）如图1，过点A作DH⊥BC于H，∴∠AHB＝90°，∵AB＝AC＝5，∴BC＝2BH，在Rt△AHB中，tan∠ABC＝＝2，∴AH＝2BH，根据勾股定理得，AH2+BH2＝AB2，∴（2BH）2+BH2＝（5）2，∴BH＝5，∴BC＝2BH＝10；（2）∵AB＝AC，∴∠ABC＝∠ACB，∵tan∠ABC＝2，∴tan∠ACB＝2，由（1）知，BC＝10，∵BF⊥AC，∴∠BFC＝90°，在Rt△BFC中，tan∠ACB＝＝2，∴BF＝2CF，根据勾股定理得，BF2+CF2＝BC2，∴（2CF）2+CF2＝102，∴CF＝2，∴AF＝AC﹣CF＝5﹣2＝3，如图2，过点C作CK∥AB交FG于K，∴△CFK∽△AFD，∴，∴＝，∴△CGK∽△BGD，∴，∴CG＝4，∴＝，∴，∴，∴AD＝AB＝×5＝；（3）如备用图，在Rt△BFC中，根据勾股定理得，BF＝＝＝4，∵DE⊥BC，∴∠BEQ＝90°＝∠BFC，∵∠EBQ＝∠FBC，∴△BEQ∽△BFC，∵CF＝2，BC＝10，∴，∴，∴设EQ＝m，则BQ＝5m，根据勾股定理得，BE＝2m，在Rt△BEQ中，tan∠ABC＝＝2，∴DE＝2BE＝4m，根据勾股定理得，BD＝10m，∴DQ＝DE﹣EQ＝3m，∵DE⊥BC，∴∠BEQ＝90°，∴∠CBF+∠BQE＝90°，∵∠BQE＝∠DQF，∴∠CBF+∠DQF＝90°，∵∠BFC＝90°，∴∠CBF+∠C＝90°，∴∠DQF＝∠C，∵AB＝AC，∴∠ABC＝∠C＝∠DQF，∵△DQF和△ABC相似，∴①当△DQF∽△ACB时，∴，∴，∴QF＝6m，∵BF＝4，∴5m+6m＝4，∴BD＝10m＝，②当△DQF∽△BCA时，，∴，∴FQ＝m，∴m+5m＝4，∴m＝，∴BD＝10m＝，即BD的长为或．。
上海杨浦、虹口、宝山、普陀、松江五区2021届九年级上学期期末(中考一模)英语试题分类汇编：词形转换
上海市杨浦、虹口、宝山、普陀、松江五区2021届九年级上学期期末（一模）英语试题分类汇编词形转换上海市杨浦区2020-2021学年度第一学期九年级期末质量调研英语试题Ⅳ. Complete the sentences with the given words in their proper forms.44.The thing to do is to give the stray dogs food and shelter.（one）45.His were well written but his delivery was hopeless.（speech）46.It’s a new story full of , adventure and heroic characters.（act）47.The yard doors usually have weak locks and you can open them .（easy）48.So what’s going on, you sound on the phone?（worry）49.A lot of people think that happiness is about being rich, but I .（agree）50.They tried to the police into thinking they had left the country, but failed.（foolish）51.Rita found on a desert island, in a room with no doors or windows.（she）答案：44. first 45. speeches 46. action 48. worried 49. disagree 50. fool 51. herself上海市虹口区2021届九年级上学期期终学生学习能力诊断测试（一模）英语试题Ⅳ. Complete the sentences with the given words in their proper forms.44.The hotel offers different of amusements to attract the guests.（variety）45.I chose the flat on the floor because I think “9” stands for “long lasting”.（nine）46.The local government is planning to more jobs for young people.（creative）47.It’s wrong of you to make fun of the boy. Y ou should apologize to him.（able）48.I think one of the ways to improve memory is the link method.（effect）49.Y ou need the knowledge of I.T. to apply for this job.（base）50.W e all know that success in study depends on one’s own effort.（main）51.The movie card was no available, so I threw it away.（length）答案：44. varieties 45. ninth 46. create 47. disabled 48. effective 49. basic 50. mainly51. longer上海市宝山区2021届九年级上学期期末（一模）英语试题Ⅳ. Complete the sentences with the given words in their proper forms.54.Two armed took place on the same street on Monday afternoon.（robbery）55.I have already checked the bag , but there is no sign of my wallet.（two）56.The boy was so clever that he could untie the knots and fool the kidnappers by .（he）57.To tell you the , I have worked very hard to prepare for the final exam.（true）58.The comic strip h as a dramatic plot to keep the readers in reading it.（interest）59.The cities in this country have been damaged after the war.（serious）puters were considered as one of the greatest in the 1920s though theywere very huge at that time.（invent）61.Eating food with strong smell on the underground usually makes people .（comfort）答案：54. robberies 55. twice 56. himself 57. truth 58. interested 59. seriously60. invention 61. uncomfortable上海市普陀区2021届九年级上学期质量调研（一模）英语试题IV. Complete the sentences with the given words In their proper forms（用括号中所给单词的适当形式完成下列句子。
2021-2022学年上海市松江区九年级上学期期末数学试题
∴这个二次函数的二次项系数为负数，
∴符合条件的函数有y＝﹣x2+4x+5，
答案为：y＝﹣x2+4x+5，答案不唯一．
【点睛】此题主要考查了二次函数的性质，解题的关键是会利用函数的性质确定解析式的各项系数．
13.一位运动员投掷铅球，如果铅球运行时离地面高度为y（米）关于水平距离x（米）的函数解析式为 ，那么铅球运动过程中最高点离地面的距离为______m．
2.已知在Rt ABC中，∠C＝90°，AB＝c，AC＝b，那么下列结论一定成立的是（ ）
A.b＝ctanAB.b＝ccotAC.b＝csinAD.b＝ccosA
【答案】D
【分析】根据余弦的定义解答即可．
【详解】解： Rt△ABC中，∠C＝90°，AB＝c，AC＝b，
则cosA＝ ，
∴b＝ccosA，
8.把抛物线y＝x2+1向右平移1个单位，所得新抛物线的表达式是___．
【答案】
【分析】根据平移规律得到新抛物线顶点坐标，即可得 新抛物线的表达式．
【详解】∵抛物线 的顶点坐标为 ，
∴抛物线向右平移1个单位后，所得新抛物线的表达式为 ，即 ．
故答案为： ．
【点睛】本题考查的是二次函数的图象与几何变换，熟知“上加下减，左加右减”的原则是解答此题的关键．
【详解】解：∵抛物线开口向上，
∴a＞0，
∵抛物线对称轴在y轴右侧，
∴﹣ ＞0，
∴b＜0，
∵抛物线与y轴交点在x轴下方，
∴c＜0．
故选：D．
【点睛】本题考查了二次函数的图象，解题关键是掌握二次函数的图象与系数的关系．
4.已知 ＝2 ，那么下列判断错误的是（ ）