外研高三广东9-12期解答案析

合集下载

广东省深圳市富源学校2023-2024学年高三上学期9月调研考试英语试题

广东省深圳市富源学校2023-2024学年高三上学期9月调研考试英语试题学校:___________姓名：___________班级：___________考号：___________一、阅读理解Since 2017, AR Together has presented over 200 workshops, camps, events, and gathering spaces for immigrants to meet, connect, and build lasting connections in the community. Below are some of ARTogether’s past and present community workshops and events.Water-coloring WorkshopOrganized in cooperation with Story Center, Water-coloring Workshop is a free online channel accessible to immigrants. These Zoom workshops are a creative space for partcipants to paint, chat, listen to music, and share stories.Y(our) Legacy: A Printmaking WorkshopThis is designed to give high school partcipants a creative practice towards their personal journey of self-acceptance and healing. Led by artist Sen Mendez, participants gather at Oakland International High School and are given ten small blocks to create a visual story about themselves.Women’s Craft CircleARTogether’s longest-running workshop series, the Women’s Craft Circle seeks to bring women together through the healing power of craft-making. In the Oakland Asian Cultural Center, participants are able to relax in a welcoming space, connecting through a love for art and shared experiences.Expressive Art Class for KidsAt West Oakland Branch Library, kids aged 6-11 from immigrant backgrounds explore fun interactive activities such as free drawing. Painting, and free dance that will let youth appreciate art without pressure or strict rules. In each class, activity leaders follow the principle of guiding, not telling, which creates a stress-free environment that encourages creativity.1．Which workshop is available for people living outside Oakland?A．Water-coloring Workshop B．Y(our) Legacy: A PrintmakingWorkshop.C．Women’s Craft Circle D．Expressive Art Class for Kids. 2．Who can sign up for Y(our) Legacy: A Printmaking Workshop?A．Visiting teacher interested in painting.B．A female artist specializing in craft-makingC．An exchange student in the international high school.D．An experienced librarian from immigrant backgrounds.3．What do the last two workshops have in common?A．They take place at local libraries.B．They require basic drawing skills.C．They offer instructions with strict rules.D．They encourage innovation or imagination.Sewing is an art. It is one that takes patience, time, and true talent. Julia,a woman from Chicago, is the definition of a talented tailor.Julia’s inspiration for her designs comes from her grandma’s sketches (草图) from the 1940s. Julia’s grandma went to fashion school in her late teens and dropped out but she kept many of her sketches. Grandma showed Julia the designs she made back in fashion school, and Julia’s mind was blown.When Grandma expressed her regret for not having the opportunity to see her sketches come to life, Julia decided not to have Grandma wait to witness it.After sewing her grandma’s designs, Julia has been posting the final products on social media one by one, which has made popular. One of the most excellent designs she made was a beautiful gold cocktail dress, which took Julia about two months to finish.Julia taught herself how to sew and has no professional training. Every month or so Julia continues to make her grandma’s designs a reality. Julia says,“She was very happy. I think she loves getting to connect with me and have something to do because she’s at a stage in life where, you know, many people don’t have many things to do. I think she is really proud and excited that I have completed it.”From the social media, Julia has even gotten offers from people willing to buy her dresses, Julia is very pleased, but she claims since she isn’t an expert, she doesn’t feel comfortable selling them. However, seeing Julia teach herself how to sew with social media videos and design these dresses is giving her viewers the confidence to attempt something oftheir own!4．How did Julia like Grandma’s sketches?A．She was excited about themB．She thought they were out of fashion.C．She felt regretful for their poor preservationD．She thought they were inspirational to other designers.5．Why did Julia do sewing?A．To prove her talent.B．To test her patience.C．To attract people’s attention.D．To help realize Grandma’s wish.6．How did Julia learn sewing?A．She consulted professionals.B．She attended a fashion school.C．She learned from social media videos.D．She followed Grandma’s instructions7．What is the viewers’ attitude towards Julia’s posts on social media?A．Doubtful.B．Admiring.C．Indifferent.D．Grateful.Anyone can be late a handful of times, but to be the person who is always late ―that’s an art, a frustrating art. Or, a side effect of your personality traits, scientists have found.So what is it that causes some people to constantly miss trains, make it to the wedding just after the bride’s shown up and regularly annoy their friends? And why is it so hard for us to fix it? “There are all sorts of punishments for being late, and the paradox is that we are late even when those punishments and consequences exist.” said Justin Kruger, a social psychologist at New York University.One of the commonest reasons why people are frequently late is that they fail to accurately judge how long a task will take―something known as the planning fallacy(谬误). Research has shown that people on average underestimate the time to complete a task by a significant 40 percent.Another trait is that forever-late-comers are more likely to be multitaskers. In a 2003 study run by Jeff Conte from San Diego State University found that out of 181 subwayoperators in New York City, those who preferred multitasking were more often late for their job. This is because multitasking makes it harder to have the awareness of what you’re doing. Conte also discovered there is a personality type that’s more likely to be late. While highly strung(紧张不安), achievement-oriented Type A individuals are more possible to be punctual. Type B individuals, however, who are more laid-back(漫不经心), have a higher chance to be late.Admittedly, knowing all of this doesn’t necessarily help fix the problem. But scientists are starting to work on strategies that can slowly improve our punctuality. For people who constantly underestimate tasks, breaking down an activity into detailed steps can help people estimate how long something will take more accurately. As for your personality type, unfortunately, there isn’t much you can do to change that. But accepting that you need to struggle for it may just help. Acceptance, after all, is the first step to change.8．What does the underlined word “paradox” mean in paragraph 2?A．Strategy.B．Argument.C．Solution.D．Puzzle. 9．What is a possible feature of forever-late-comers?A．They plan to spend more time on a task.B．They tackle more than one task at a time.C．They suffer from concentration difficulties.D．They have high expectations for achievements.10．Which advice can be given to people who are always late?A．Learn to accept who you are.B．Change your personality type.C．Divide a task into smaller ones.D．Keep to the timetable accurately. 11．What is the main idea of the text?A．Time management contributes to success.B．Late comers should be severely punished.C．One’s always being late is linked to personality.D．Changing personality helps improve punctuality.Every 40 days a language dies. This “catastrophic” loss is being intensified by the climate crisis, according to linguists. If nothing is done, conservative estimates suggest half of all the 7000 languages currently spoken will be extinct by the end of the century.Speakers of minority languages have experienced a long history of persecution (迫害),with the result that by the 1920s half of all indigenous (土著的) languages in Australia, the US, South Africa and Argentina were extinct. The climate crisis is now considered the “final nail in the coffin” for many indigenous languages and the knowledge they represent.“Languages are already endangered,” says Anastasia Richl, director of the Strathy language unit at Queen’s University in Kingston, Ontario. Huge factors are globalization and migration, as communities move to regions where their language is not spoken or valued, according to Richl. “It seems particularly cruel, ” she says, “ that most of the world’s languages are in parts of the world that are growing unpleasant to people. ” Vanuatu, a South Pacific island nation measuring 12, 189 km2, has 110 languages, the highest density (密度) of languages on the planet. It is also one of the countries most at risk of sea level rise. “Many small language communities are on islands and coastlines easily subject to hurricanes and sea level rise,” she says. Others live on lands where rising temperature threatens traditional farming and fishing practices, leading to migration.In response to the crisis, the UN launched the International Decade of Indigenous Languages in 2022. Promoting and conserving languages of indigenous communities is “ not only important for them, but for all humanity, ” said Csaba Korosi, the UN general assembly president, urging countries to allow access to education in indigenous languages.12．What does the underlined phrase “final nail in the coffin” in paragraph 2 mean?A．The last straw.B．The last challenge.C．The last possibility.D．The last opportunity.13．Why is Vanuatu mentioned in paragraph 3?A．To explain the main reason for language density.B．To show the common features of endangered languages.C．To stress the impact of geographical position on migration.D．To illustrate the situation of minority language communities.14．What does the International Decade of Indigenous Languages aim to do?A．Support migrants to access local education.B．Discourage people from massive migration.C．Improve the living conditions of minority groups.D．Preserve the languages of indigenous communities.15．Which of the following is the best title for the text?A．Faced with Disasters: Communities Have to LeaveB．Lost for Words: Climate Crisis Brings Threat of CatastropheC．Upset at Extinction: UN Urges International CooperationD．Involved in Action: Experts Seek Solutions to Climate Crisis二、七选五Feeling upset is a normal part of life. Something stressful may occur everywhere, andare some useful ways to keep our cool when we feel frustrated.17Conflicts with others can be difficult to deal with. Sometimes we need to leave from the unpleasant condition. It is OK to say, “I’m going to take a break” and go for a walk or find a quiet place to sit and cool off. The goal is to give yourself some space to calm down and consider what to do next.Do breathing exercises to relaxWhen we’re upset, our nervous system will speed up our heart rate and muscles tense. It prepares the body as if to meet an attack. 18 If possible, find a safe place, sit comfortably and focus on your breath, breathing deeply and evenly. Mindful breathing helps us calm our feelings.Lift your mood with laughterLaughter relieves the tension we feel when depressed. We can think of a joke or watch a video that always makes us roar with laughter. Afterward, you may be able to lift the fog of frustration. Make sure what you use is not mean-spirited, though. 19Change your languageOnce things appear disappointing, many of us might punish ourselves with unkind words. In fact, it doesn’t have to be this way. 20 For example, instead of saying “I always mess things up,” we could say, “I made a mistake, but I’ll do it better next time.”A．Keep yourself away from the noiseB．Remove yourself from the situationC．We should turn negative thoughts into positive statements.D．Those kinds of humor can actually make us feel more down.E．Thus, we can be physically and mentally entertained from that.F．You can soften such physical reaction by using practical techniques.G．However, we can decide how we behave and react to the troubles we face.三、完形填空Devi had never been to a swimming lesson in her life. But now that her mother had28．A．imagined B．concluded C．noticed D．wondered 29．A．content B．grateful C．curious D．doubtful 30．A．hide B．overcome C．support D．convey 31．A．Still B．Therefore C．Otherwise D．Moreover 32．A．fantastic B．ridiculous C．tolerant D．accessible 33．A．parent B．friend C．partner D．instructor 34．A．at ease B．in trouble C．at risk D．in need 35．A．competing B．playing C．trying D．winning四、用单词的适当形式完成短文阅读下面短文，在空白处填入一个单词或者用所给单词的正确形式填空。

广东省大湾区2025届高三上学期9月统一调研考试数学试题(解析版)

★启用前注意保密大湾区2025届普高毕业级统一调研考试数学2024.9本试卷共4页，满分150分.考试时间120分钟.★祝考试顺利★注意事项：1．答卷前，考生务必将自己的学校、班级、姓名、考场号、座位号和准考证号填写在答题卡上，将条形码横贴在答题卡“条形码粘贴处”.2．作答选择题时，选出每小题答案后，用2B 铅笔在答题卡上将对应题目选项的答案信息点涂黑；如需改动，用橡皮擦干净后，再选涂其他答案.答案不能答在试卷上.3．非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内相应位置上；如需改动，先划掉原来的答案，然后再写上新答案；不准使用铅笔和涂改液.不按以上要求作答无效.4．考生必须保证答题卡的整洁.考试结束后，将试卷和答题卡一并交回.一、选择题：本题共8小题，每小题5分，共40分.在每小题列出的四个选项中，只有一项是符合题目要求的.1. 若24log log 2m n +=，则2m n =（）A. 3B. 4C. 9D. 16【答案】D 【解析】【分析】利用对数的运算性质化简给定式子求解即可.【详解】因为24log log 2m n +=，所以221log log 22m n +=，故得12222log log log 4m n +=，化简得1222log log 4mn =，所以124mn =，故216m n =，故D 正确. 故选：D.2. 设复数z 满足|1|2z −=，z 在复平面内对应的点为(),x y ，则（）A. 22(1)2x y −+=B. 22(1)2x y +−=C. 22(1)4x y −+=D. 22(1)4x y +−=【答案】C 【解析】【分析】i z x y =+2=，两边平方得到答案.【详解】i z x y =+，则()|1|2|1i |2z x y −=⇒−+=，2=，故22(1)4x y −+=. 故选：C3. 若2{1,3,4,}m m ∈，则m 可能取值的集合为（） A. {0,1,4} B. {0,3,4}C. {1,0,3,4}−D. {0,1,3,4}【答案】B 【解析】【分析】根据给定条件，利用元素与集合的关系列式计算并验证即得. 【详解】由2{1,3,4,}m ，得21m ≠，则1m ≠，由2{1,3,4,}m m ∈，得3m =，此时29m =，符合题意；或4m =，此时216m =，符合题意；或2m m =，则0m =，此时20m =，符合题意，所以m 可能取值的集合为{0,3,4}. 故选：B4. 已知随机变量~(,)X B n p ，若(2)2()D X E X =，则p =（）A.116B.18C.14D.12【答案】D 【解析】【分析】根据二项分布的期望、方差公式列方程，从而求得p . 【详解】依题意X 满足二项分布，且(2)2()D X E X =，即()()()()42,2D X E X D X E X ==，即()21np p np −=，解得12p =，（0p =舍去）.故选：D5. 甲、乙等6人围成一圈，且甲、乙两人相邻，则不同排法共有（） A. 6种 B. 12种C. 24种D. 48种【答案】D 【解析】【分析】将甲、乙两人看成一个人，根据n 个不同元素围成的环状共有()1!n − 种排法求解.【详解】因为由于环状排列没有首尾之分，将n 个不同元素围成的环状排列剪开看成n 个元素排成一排，即共有!n 种排法，由于n 个不同元素共有n 种不同的剪法，则环状排列共有()!1!n n n=− 种排法.甲、乙两人相邻而坐，可将此2人当作1人看，即5人围一圆桌，有()51!−种坐法，又因为甲、乙2人可换位，有2!种坐法，故所求坐法为()51!2!48−×=种. 故选：D6. 已知函数()f x 的定义域为R ，且(1)(5)f f =，函数(1)f ax −的图象关于直线2x =对称，则a =（） A. 1 B. 2C. 3D. 4【答案】B 【解析】【分析】函数(1)f ax −的图象关于直线2x =对称，可得到()()(1)41f ax f a x −=−−，再根据(1)(5)f f =列出方程式可求解【详解】根据题意知，函数(1)f ax −的图象关于直线2x =对称，则得到()()(1)41f ax f a x −=−−，又因(1)(5)f f =，则令11415ax ax a −= −+−=或15411ax ax a −=−+−= 解之可得2a =.故选：B7. 已知正(3)n n ≥棱锥的侧棱长为3，则其体积可能为（） A. 10 B. 11 C. 12 D. 13【答案】A 【解析】的【分析】设正棱锥的底面正多边形的外接圆的半径为(03)R R <<，利用棱锥的体积公式，可得正棱锥的体积21π3V R <2(0,9)x R =∈，设()239f x x x =−，利用导数求得函数的单调性与最大值，结合选项，即可求解.【详解】设正棱锥的底面正多边形的外接圆的半径为R ，可得外接圆的面积为2πS R = 因为正棱锥的侧棱长为3，所以底面正多边形的外接圆的半径03R <<，又由正棱锥的高为h=设正棱锥的底面多边形的面积为1S ，所以正棱锥的体积21111π333V S h R =⋅<，其中03R <<，令2(0,9)x R =∈，可得11133V S h =⋅< 设()239,(0,9)f x x x x =−∈，可得()()218336f x x x x x =−=′−，当(0,6)x ∈时，ff ′(xx )>0，函数()f x 单调递增；当(6,9)x ∈时，ff ′(xx )<0，函数()f x 单调递减，所以，当6x =时，函数()f x 取得最大值，最大值为()6108f =，所以1113V <<，结合选项，只有A 选项符合题意. 故选：A.8. 记n S 为数列{}n a 的前n 项和，且10a =，2n k a n k −=−1(02)n k −<≤，则31S =（） A. 26− B. 31−C. 36−D. 40−【答案】B 【解析】【分析】根据2n k a n k −=−写出各项值，直接求和. 【详解】10a =，1220101a a −==−=， 2321211a a −==−=， 2420202a a −==−=，故12344a a a a +++=；的3523330a a −==−=， 3622321a a −==−=， 3721312a a −==−=， 3820303a a −==−=，故56786a a a a +++=； 4927473a a −==−=−， 41026462a a −==−=−， 41125451a a −==−=−，⋅⋅⋅ ， 41620404a a −==−=，故9101634842a a a −+++⋅⋅⋅+=×=； 51721551510a a −==−=−， 5182145149a a −==−=−，⋅⋅⋅ ， 53121514a a −==−=，故17183110415452a a a −+++⋅⋅⋅+=×=−；故31S =()()()()12345678916173131a a a a a a a a a a a a +++++++++⋅⋅⋅+++⋅⋅⋅+=−. 故选：B二、选择题：本题共3小题，每小题6分，共18分.在每小题列出的选项中，有多项符合题目要求.全部选对的得6分，部分选对的得部分分，有选错的得0分.9. 已知样本数据7,3,5,3,10,8，则这组数据的（） A. 众数为3 B. 平均数为6.5 C. 上四分位数为8 D. 方差为203【答案】ACD 【解析】【分析】利用众数，平均数，方差，上四分位数公式逐个选项分析求解即可. 【详解】首先，我们把数据从小到大排列，得到3,3,5,7,8,10，对于A ：观察得数据3出现的次数最多，所以众数为3，故A 正确；对于B ：平均数为335781036666+++++==，故B 错误；对于C ：因为一共有6个数据，且675 4.75×%=，所以上四分位数为第5个数，故上四分位数为8，故C 正确；对于D ：方差为2222221(36)(36)(56)(76)(86)(106)6−+−+−+−+−+− ， []1120991141640663=+++++=×=，故D 正确. 故选：ACD.10. 若某等腰直角三角形的其中两个顶点恰为椭圆C 的两个焦点，另一个顶点在C 上，则C 的离心率可能为（） A.12B.C. 1−D.【答案】BC 【解析】【分析】首先不影响答案情况下可固定直角和椭圆的焦点所处的坐标轴，再设AB AC k ==，最后分,B C 为焦点和,A C 为焦点并结合椭圆定义和离心率公式讨论即可.【详解】在等腰直角ABC 中，在不影响离心率的情况下不妨设π2A ∠=，AB AC k ==，0k >，且椭圆焦点位于x 轴上，当椭圆以B ，C 为焦点时，根据椭圆和等腰直角三角形对称性知点A 为椭圆上顶点，则22,a k a k ==；2,c ck =，离心率ce a==当椭圆以,A C或,A B为焦点时，2(1,a k k a=+=+，2,2kc k c==，离心率1cea==−，1.故选：BC.11. 记函数()sin cos2sin3f x x x x=在区间π0,2的极值点分别为1α，2α()12αα<，函数()()()2143g x x x x=−−的极值点分别为1β，2β()12ββ<，则（）A.1256ββ+= B. ()()i if gαβ=()1,2i=C. ()()()21f f x f αα≤≤D. 2114αβ<【答案】ABD 【解析】【分析】选项A ：根据导数可得1β，2β为方程2242030x x −+=的两个根，进而可得；选项B ：()6428sin 10sin 3sin f x x x x =−+，根据换元设2sin t x =得()328103h t t t t =−+，与()328103g x x x x =−+解析式相同，进而可判断；选项C ：由()1π12f f α=>可判断；选项D ：根据先求出1105β<=<，2π5034α<<<根据不等式的性质进而可得.【详解】选项A ：()()()322143=8103g x x x x x x x =−−−+，()2=24203g x x x ′−+，故由题意可知1β，2β为方程2242030x x −+=的两个根，故1256ββ+=，A 正确；选项B ：()()()23642sin cos 2sin 3sin 12sin 3sin 4sin 8sin 10sin 3sin f x x x x x x x x x x x ==−−=−+，设2sin t x =，因π0,2x∈，则()0,1t ∈，此时函数yy =ff (xx )可化为()28103h t t t t =−+，由题意此函数的极值点分别为1β，2β()12ββ<，当π0,2x∈时，函数2sin t x =单调递增，故112sin βα=，222sin βα=，故 ()()11fg αβ=，()()22f g αβ=，故B 正确；选项C ：由2242030x x −+=解得1β=2β=()()111f g g αβ==<，由题意函数()f x 在()10,α上单调递增，在()12,αα上单调递减，在2π,2α上单调递增，而π12f =，故()()0201π,,2x f x f αα∃∈>，故C 错误；选项D ：由A可知，1105β<=<，2223sin 4βα==<，因2π0,2α∈，故20sin α<<2π5034α<<<，故2114αβ<，故D 正确，故选：ABD三、填空题：本题共3小题，每小题5分，共15分.12. 已知等比数列{}n a 的前6项和为63，其中偶数项和是奇数项和的两倍，则1a =______．【答案】1 【解析】【分析】设出公比，根据()2461352a a a a a a ++=++，求出公比2q ，故13521a a a ++=，得到11a =. 【详解】设公比为q ，则12345663a a a a a a +++++=，其中()2461352a a a a a a ++=++，又()246135a a a q a a a ++=++，故2q ，()135363a a a ++=故13521a a a ++=，即2411111141621a a q a q a a a ++=++=，解得11a =. 故答案：113. 已知球O 是某圆锥内可放入的最大的球，其半径为该圆锥底面半径的一半，则该圆锥的体积与球O 的体积之比为______．【答案】83##223【解析】【分析】根据题意作出相应的截面图形，设AE x =，利用勾股定理，用r 表示AE ，结合圆锥体积和球的体积公式即可求解.【详解】球O 是某圆锥内可放入的最大的球，则该球为圆锥的内切球，截面如图所示：设球O 的半径为r ，则圆锥底面半径为2r ，为可得在ABC 中，,AD BC OF AC ⊥⊥，2CD CF r ==，设AE x =，由勾股定理得AF ===222AD CD AC +=，即()())222222x r r r +++，化简得223440x rx r +−=，即()()3220x r x r −+=， 0x ，则23x r =，即23AE r =，则圆锥体积为()321232ππ22339r rr r +=，球O 的体积为34π3r ，所以圆锥的体积与球O 的体积之比为3332π894π33r r =.故答案为：83.14. 设A ，B ，C 三点在棱长为2的正方体的表面上，则AB AC ⋅的最小值为______．【答案】2− 【解析】【分析】法一：可初步确定A 点所在的平面，作B ，C 在这个面的射影1B ，1C ，利用AB AC⋅()()1111AB B B AC C C =+⋅+把空间向量问题转化为平面向量问题，结合向量数量积的性质和基本不等式求最小值.法二：建立空间直角坐标系，不妨假设A 在平面xOy 中，设()12,,0A a a ，()123,,B b b b ，()123,,C c c c ，()112,,0B b b 和()112,,0C c c 分别是点B ，C 在平面xOy 上的投影，利用向量不等式可得：()211113311114AB AC AB AC b c AB AC AB AC +⋅+≥⋅≥−⋅≥−，即可求解.【详解】法一：如图：不防设点A 在正方体的下底面内，B ，C 在正方体的表面的任何位置，它们在下底面的射影分别为1B ，1C .则11AB C C ⊥，11AC B B ⊥.所以110AB C C ⋅= ，110AC B B ⋅= ，110B B C C ⋅≥. 所以AB AC ⋅()()1111AB B B AC C C =+⋅+11111111AB AC AB C C AC B B B B C C =⋅+⋅+⋅+⋅ 1111AB AC B B C C =⋅+⋅11AB AC ≥⋅11AB AC ≥−⋅ （当1AB 与1AC 方向相反时取“=”）.又()211114AB AC AB AC +⋅≤（当且仅当1AB = 1AC时取“=”）.分析两个“=”成立的条件，可知A 为11B C 中点时，AB AC ⋅有最小值.此时1111AB AC B C +=≤（当11B C 为下底面的面对角线时取“=”）.所以112AB AC ⋅≤=，AB AC ⋅ 11AB AC ≥−⋅ ⇒2AB AC ⋅≥− （当A 位于下底面中心，B ，C 在下底面的射影是下底面的面对角线端点时取“=”）.法二：将正方体置于空间直角坐标系O xyz −中，且A 在平面xOy 中，点O 和点()2,2,2的连线是一条体对角线．设()12,,0A a a ，()123,,B b b b ，()123,,C c c c ，()112,,0B b b 和()112,,0C c c 分别是点B ，C 在平面xOy 上的投影．可得()130,0,B B b = ，()130,0,C C c = ，110AB C C ⋅= ，110AC B B ⋅=则()()111111111111AB AC AB B B AC C C AB AC AB C C AC B B B B C C ⋅=+⋅+=⋅+⋅+⋅+⋅1133AB AC b c =⋅+，因为()211113311114AB AC AB AC b c AB AC AB AC +⋅+≥⋅≥−⋅≥−，当且仅当点C 为11B C 的中点时，等号成立，可得()2211111244AB AC B C +−=−≥− ，所以2AB AC ⋅≥−，当()1,1,0A ，11222b c b c −=−=，且330b c =时等号成立．故答案为：2−.【点睛】关键点点睛：本题的关键是利用AB AC ⋅()()1111AB B B AC C C =+⋅+ ，把空间向量的数量积转化成平面向量的数量积，“降维”是解决该题的关键思想.四、解答题：本题共5小题，共77分.解答应写出文字说明、证明过程或演算步骤.15. 记ABC 中角A ，B ，C 所对的边分别为a ，b ，ccos 1A A −=．（1）求A ；（2）记ABC 的外接圆半径为R ，内切圆半径为r ，若3a =，求rR的取值范围．【答案】（1）π3A = （2）1(0,]2【解析】【分析】（1）利用辅助角公式整理得到π1sin 62A−=结合角A 范围即可求解；（2）根据正弦定理确定ABC 的外接圆半径为R ，根据等面积确定内切圆半径为r ，从而可得rR的不等式，进而可求其取值范围．【小问1详解】cos 1A A −=，11cos 22A A ∴−=，则π1sin 62A−= ， ()0,πA ∈ ， ππ66A ∴−=，解得π3A =，π3A ∴=；【小问2详解】根据正弦定理得：2sin aRA==，设ABC 的内心为O ，易知2π3BOC ∠=，由11sin 22BOC S ar OB OC BOC ==⋅⋅∠，则r OC ⋅，由余弦定理得：2222cos a OB OC OB OC BOC +−⋅⋅∠，即2293OB OC OB OC OB OC ++⋅≥⋅，当且仅当OB OC =时取等号，3OB OC ∴⋅≤，0r ∴<≤∴12r R=≤， ∴1(0,]2Rr ∈． 16. 已知函数21()exx x f x +−=．的（1）求()f x 的极值；（2）讨论()f x 在区间[,m m +上的最大值．【答案】（1）极小值为e −，极大值为25e；（2）答案见解析. 【解析】【分析】（1）求出函数()f x 的导数，探讨导数值正负求出极值.（2）借助（1）求出的函数()f x 的单调性，再对m 进行分类讨论，结合单调性得到最大值. 【小问1详解】函数21()e x x x f x +−=定义域为R ，求导得221(1)(1)(2)()e e x xx x x x x f x +−+−+−′==−，当1x <−或2x >时，()0f x ′<，当12x −<<时，()0f x ′>，因此函数()f x 在1x =−处取得极小值(1)e f −=−，在2x =处取得极大值25(2)ef =，所以函数()f x 的极小值为e −，极大值为25e . 【小问2详解】由（1）知，函数()f x 在(,1),(2,)−∞−+∞上单调递减，在(1,2)−上单调递增，①当1m +≤−，即1m ≤−时，()f x 在[,m m 上单调递减，max ()()f x f m =；②当11m <−<−时，()f x 在[,1)m −上单调递减，在(1,m −+上单调递增，由()0f x =，得12x x ，21x x −，当1m −<≤时，1m −<≤12()()()(f m f x f x f m ≥≥+，max ()()f x f m =；当1m <<−1m <+<，12()()()(f m f x f x f m <≤，max()(f x f m =；③当12m −≤≤时，()f x 在[,m m 上单调递增，max()(f x f m =+；④当22m −<<时，()f x 在[,2)m上单调递增，在(2,m +上单调递减，max 25()(2)e f x f ==； ⑤当2m ≥时，()f x在[,m m 上单调递减，max ()()f x f m =，所以当m ≤2m ≥时，函数()f x 的最大值为()21e mm m f m +−=；当2m <≤−时，函数()f x的最大值为(f m+；当22m <<时，函数()f x 的最大值为25(2)ef =.17. 如图，在四面体ABCD 中，ABC 是正三角形，ACD 是直角三角形，ABD CBD ∠=∠，AB BD =．（1）证明：平面ACD ⊥平面ABC ；（2）若二面角D AE C −−的正切值为ACDE 与四面体ABCD 的体积之比．【答案】（1）证明见解析（2）45【解析】【分析】（1）取AC 的中点O ，连接DO ，BO ，得到DO AC ⊥，再由ABC 是正三角形，得到BO AC ⊥，利用面面垂直的判定证明；（2）以O 为坐标原点，OA 为x 轴，OB 为y 轴，OD 为z 轴，建立的空间直角坐标系，分别求得平面ADC 和平面ACE 的法向量，结合向量的夹角公式列出方程，即可求解. 【小问1详解】由题设得，ABD CBD ≅ ，从而AD DC =. 又ACD 是直角三角形，所以=90ADC ∠°.取AC 的中点O ，连接DO 、BO ，则DO ⊥AC 且OD OA =，又ABC 是正三角形，故BO AC ⊥.则Rt AOB 中，22222BO AO AB BO DO +==+，又AB BD =，所以222BO DO BD +=，故OD OB ⊥.而AC OB O ∩=且都在面ABC ，故OD ⊥面ABC ，而OD ⊂面ACD ，所以平面ACD ⊥平面ABC .【小问2详解】设2AB =， DE mDB =，结合（1）结论，以O 为坐标原点，OA 为x 轴，OB 为y 轴，OD 为z 轴，建立如图所示的空间直角坐标系，则(0,0,1),(0,0,0),(1,0,0),(1,0,0)D O B C A −，易知平面ADC 的法向量为1(0,1,0)n =，设(,,)E x y z ，由DE mDB =，可得,1)E m −，得,1),(1,0,0)OE m OA =−= ，设面ACE 的法向量为2(,,)n x y z =，则()22010n OA x n OEm z ⋅== ⋅=+−=，取1y m =−，得0,x z ==，所以2(0,1)n m =−，因为二面角D AE C −−的正切值为，则121cos ,7n n =，又01m ≤≤，解得45m =，所以45DE DB = ，所以E 到底面ACD 距离与B 到底面ACD 的距离之比为45，所以四面体ACDE 与四面体ABCD 的体积之比45.18. 在平面直角坐标系xOy 中，等轴双曲线1C 和2C 的中心均为O ，焦点分别在x 轴和y 轴上，焦距之比为2，1C 的右焦点F 到1C 的渐近线的距离为2．（1）求1C ，2C 的方程；的（2）过F 的直线交1C 于A ，B 两点，交2C 于D ，E 两点，AB 与DE的方向相同．（ⅰ）证明：||||AD BE =；（ⅱ）求AOD △面积的最小值．【答案】（1）22224;1x y y x −=−=（2）（ⅰ）证明见解析；（ⅱ【解析】【分析】（1）根据双曲线特征设22:i i C x y t −=，结合已知列方程求解；（2）（ⅰ）先设直线再联立方程应用两根的和结合中点M ，即可证明；（ⅱ）先把面积转化为122S S S −=再设函数()[)0,9f x x =∈借助导函数正负得出函数的单调性进而求出最小值. 【小问1详解】由题设可设 22:i i C x y t −=,这里120,0t t ><. 易知i C 渐近线为y x =±，焦距为i C的右焦点)F，由题设可知22=× , 解得124,1t t ==−. 所以1C 的方程为224x y −=,2C 的方程为221x y −=−. 【小问2详解】（ⅰ）设直线 ()()()()11223344:2,,,,,,,AB x my A x y B x y C x y D x y =+，, 联立直线 AAAA 和 i C 的方程22i x my x y t =+−=,得()22180i m y t −++−=. 为使直线 AAAA 和 i C 均有2个交点，必须有210m −≠，()()22324180i i m m t ∆=−−−> , 解得29m <且21m ≠．由韦达定理可得1234121234228811y y y y t t y y y y m m +=+=−− == −−注意到 1234y y y y +=+，因此线段 AAAA 和线段 BE 具有相同的中点．记上述中点为 M ，注意到,AD DM AM BE EM BM =−=−,所以AD BE = . （ⅱ）由（ i ）可知AOD 和BOE 的面积相等．记AOD 的面积为S ,AOB 的面积为1S ,DOE 的面积为2S .由 AB 与 DE 的方向相同可知122S S S −= . 因为11212S OF y y =××−=，同理2S =所以122S S S −==−, 设()[)0,9f x x =∈, 则()f x ′当[)0,7x ∈时，()()0,f x f x ′>单调递增，当()7,9x ∈时，()()0,f x f x ′>单调递减，因此S ≥当且仅当27m =时，等号成立，因此，AOD【点睛】关键点点睛：解题的关键点时把面积转化为122S S S −=，设函数()[)0,9f x x =+∈借助导函数正负得出函数的单调性进而求出最小值. 19. 设离散型随机变量X ，Y 的取值分别为12{,,,}p x x x ，12{,,,}q y y y (),N p q ∗∈．定义X 关于事件“j Y y =”(1)j q ≤≤的条件数学期望为：1(|)(|)pj i i i i E X Y y x P X x Y y =====∑．已知条件数学期望满足全期望公式：1()(|)()qi i i E X E X Y y P Yy ====∑．解决如下问题：为了研究某药物对于微生物A 生存状况的影响，某实验室计划进行生物实验．在第1天上午，实验人员向培养皿中加入10个A 的个体．从第1天开始，实验人员在每天下午向培养皿中加入该种药物．当加入药物时，A 的每个个体立即以相等的概率随机产生1次如下的生理反应（设A 的每个个体在当天的其他时刻均不发生变化，不同个体的生理反应相互独立）： ①直接死亡；②分裂为2个个体．设第n 天上午培养皿中A 的个体数量为n X ．规定1()10E X =，1()0D X =．（1）求65(|6)E X X =；（2）求()n E X ；（3）已知21(|)(1)n n E X X k k k −==+(N )k ∗∈，证明：()n D X 随着n 的增大而增大．【答案】（1）6 （2）()10n E X = （3）证明见解析【解析】【分析】（1）如果在第五天下午加入药物后，有K 个个体分裂，可得16,2K B∼，可求65(|6)E X X =；（2）随机变量Z 表示第1n −天下午加入药物之后分裂的个体数目，则1,2Z B k∼且2n X Z =，可得1(|)n n E X X k k −==设1n X −的取值集合为{}12,,,r x x x ，则由全期望公式可求得结论；（3）由（2）可知2()n E X ()2110n E X −+，可求得()()2100101n E X n =+−，进而可得()n D X .【小问1详解】在事件56X =发生的条件下，如果在第五天下午加入药物后，有K 个个体分裂，则16,2K B∼ ，()1632E K =×=，所以62X K =，()()6562236E X X E K ===×=．【小问2详解】由（1）可类似得到：在事件1n X k −=发生的条件下，如果在第1n −天下午加入药物之后，有m 个个体分裂，则n X 的取值为()2k m k m m +−−=．在事件1n X k −=发生的条件下，令随机变量Z 表示第1n −天下午加入药物之后分裂的个体数目，则1,2Z B k∼且2n X Z =．因此11001(|)2(2|)2()2()22r rn n n n m m E X X k m P X m X k m P Z m E Z k k −−==⋅⋅××∑∑．设1n X −的取值集合为{}12,,,r x x x ，则由全期望公式可知111100()(|)()()()r rn n n i n i i n i n t t E X E X X x P X x x P X x E X −−−−=====⋅==∑∑．这表明(){}n E X 是常数列，所以()()110n E X E X ==．【小问3详解】由（2）可知22111()(|)()rnnn i n i i E X E X X x P X x −−====∑ ()()()()2221111110ri i n i n n n i x x P X x E X X E X −−−−==+==+=+∑，这表明(){}2nE X 是公差为10的等差数列．第21页/共21页又因为()()()22111100E X D X E X =+= ，所以()()2100101n E X n =+−，从而()()()()22101n n n D X E X E X n =−=− ．可以看出，()n D X 随着n 的增大而增大．【点睛】关键点点睛：本题的关键之一是理解期望与方差的计算公式以及题意，尤其是二项分布的期望公式.。

广东省八校2024-2025学年高三上学期9月联合检测英语试题

广东省八校2024-2025学年高三上学期9月联合检测英语试题一、阅读理解China’s Best New Outdoor AttractionsWhen it comes to tourist attractions, here are a few of our favorites.The Grand Canal, the world’s longest human-made riverStretching 1, 782 kilometers, China’s Grand Canal is the world’s longest human-made river. Inscribed as a UNESCO World Heritage Site in 2014, the Grand Canal-also known as Jing-Hang Canal-runs from Beijing to Hangzhou and connects s to multiple waterways, including the Yellow River and the Yangtze River.Jiangsu Garden Expo ParkNanjing’s Garden Expo Park Scenic Spot showcases the region’s famous Jiangsu-style gardens. It is home to beautiful, classic Jiangsu-style gardens and other attractions, including a water botanical garden, a theater and a pedestrian shopping street. Among the highlights is Suzhou Garden, which was inspired by the Canglang Pavilion—a UNESCO World Heritage Site in nearby Suzhou.Universal Beijing ResortOpened in September 2021, the 169-hectare Universal Beijing Resort is the world’s fifth and largest Universal Studios theme park. There are seven themed lands offering dozens of shows and rides, many of them outdoors: The Wizarding g World of Harry Potter；Transformers Metrobase; Hollywood; WaterWorld; Minion Land；Jurassic World Isla Nublar; and Kung Fu Panda Land of Awesomeness.Glamping (豪华露营) and stargazing at Ningxia’s Desert Star Hotel Ningxia Hui Autonomous Region experiences little rainfall and enjoys around 300 days of clear skies yearly,making it one of the best places to enjoy China’s desert landscapes and engage in some serious stargazing. One of the top ways to experience Ningxia is by checking into the Desert Star Hotel in Ningxia’s Zhongwei City. The complex offers indoor and outdoor dining options, a swimming pool, an astronomy-themed theater and family-friendly sand activities. 1．Which attraction is listed as a UNESCO World Heritage Site?A．Jing-Hang Canal.B．Jiangsu Garden Expo Park.C．Universal Beijing Resort.D．Ningxia’s Desert Star Hotel.2．What does Universal Beijing Resort offer?A．Several dozen themed lands.B．Astronomy-themed movies.C．Outdoor shows and rides.D．A pedestrian shopping street.3．What can visitors do in Ningxia’s Desert Star Hotel?A．Attend a concert.B．Play with sand.C．Swim in the ocean.D．Enjoy classic gardens.Rowan Atkinson was born in a middle-class family. He had a speaking disability right from his childhood. This gave him a tough time in his childhood and at the start of his career. He was constantly bullied and laughed at because of his look and his speaking disability. This made him very shy and quiet. He didn’t have a lot of friends and the several rejections left him feel lonely. Rowan was very much interested in science. And he decided to pursue a career in this field.After he completed his Master’s degree, he realized acting was something he really wanted to pursue. He auditioned for several TV shows but faced back-to-back rejections, because he didn’t have a good face and grand body in addition to his stammering (口吃的) problem as well. Rowan kept his passion alive and worked hard towards reaching his dreams. He aimed at making people laugh and with persistent efforts over the years he has successfully established this.After the several rejections he started creating original comedy sketches. Interestingly he realized that whenever he played some characters, he spoke fluently. He used it as an inspiration for his acting. This is exactly where Rowan’s life changed.Rowan continued pursuing his dreams, despite all the hardships and rejections he had faced because of his looks and disorder. He had a major breakthrough when he started his own show Mr. Bean. Mr. Bean was “strange, surreal and non-speaking character”. He proved that even without a Hollywood face or a heroic body, you can become one of the most loved and respected actors in the world.Life always rewards those who are willing to keep moving forward. He teaches us that for success all you really need is hard work. Never let your fears and disabilities stop your life. Walk past these and make yourself better. Your hard work and efforts are sure to pay off one day. 4．What can we know about Rowan when he was young?A．He liked bullying his classmates.B．He had a few friends and felt lonely.C．He showed little interest in science.D．He recovered from his speaking disability.5．Why did Rowan want to act?A．He wanted to become popular.B．He enjoyed staying in the stage.C．He intended to make people happy.D．He was eager to overcome his drawback. 6．What made Rowan’s life change?A．That he was respected by the audience.B．That he could speak fluently in some characters.C．That he got a Hollywood face and a herbic body.D．That he received a. Master’s degree in science.7．What contributed to Rowan’s success according to the text?A．His wisdom B．His independence.C．His honesty.D．His willpower.Human beings are not alone in having invented vaccination, while honeybees got there first and they can run what look like vaccination programmes, which has been confirmed by Gyan Harwood of the University of Illinois, Urbana-Champaign.Queen bees vaccinate their eggs before they are laid. But the question is how the queen receives her antigen supply, for she lives purely on royal jelly, a substance secreted by nurse bees when they are in the life stage of feeding the young. Dr Harwood wondered if the nurses combined the royal jelly they produced with pieces from pathogens they had consumed while eating something brought in from outside.To test this idea, they collected about 150 nurse bees and divided them among six queenless mini hives equipped with the young to look after. They fed the nurses on sugar-water, and for three of the hives they added Paenibacillus larvae, a bacterium causing a disease, to sugar-water.Dr Harwood and Dr Salmela labeled the bacteria with a certain dye, to make them easy to track. And, sure enough, microscope confirmed that Paenibacillus larvae were getting into royal jelly secreted by those bees which had been fed with the sugar-water. Moreover, examination of this royal jelly revealed higher levels of defensive substance, compared with royal jelly from bees that had not been mixed with Paenibacillus larvae. This substance is thought to help bee immune systems fight against bacterial infections.All told, these findings suggest that nurse bees are indeed, via their royal jelly, passing antigens on to the queen, then into eggs. They also mean, because the young receive royal jelly forthe first few days after they hatch, the nurses are giving the young the second antigens. Each young bee is therefore being vaccinated twice.8．What puzzled Dr Harwood from paragraph 2?A．What the royal jelly consists of.B．Where nurse bees receive pathogens.C．How the antigen come into the queen bees' bodies.D．Whether honeybees run vaccination earlier than man.9．How did Dr Harwood develop his experiment?A．By dividing bees into different roles.B．By keeping track of the special bacterium.C．By changing the components of royal jelly.D．By observing nurse bees' different behaviors.10．What can we infer according to the results of the experiment?A．Nurse bees are the key to vaccination for bee group.B．The nurse bees pass the antigen only to the queen.C．Bacteria-used royal jelly has fewer defense substances.D．Two vaccinations are given to young bees by caregivers directly.11．Where is the text probably taken from?A．A pet guide.B．A social website.C．An official document.D．A medical magazine.The British has obeyed the “keep to the left” rule for long. Have you ever wondered why? There is a historical reason for this: it’s all to do with keeping your sword hand free! In the Middle Ages you never knew who you were going to meet when travelling on horseback. Most people are right-handed, so if a stranger passed by on the right of you, your right hand would be free to use your sword if required.Indeed the “keep to the left” rule goes back even further in time; archaeologists have discovered evidence suggesting that the Romans drove carts and wagons on the left, and it is known that Roman soldiers always marched on the left. This “rule of the road” was officially sanctioned in 1300 AD when Pope (教皇) Boniface Ⅷ declared that all pilgrims (朝圣者)travelling to Rome should keep to the left.This continued until the late 1700s when large wagons became popular for transporting goods. These wagons were drawn by several pairs of horses and had no driver’s seat. Instead, in order to control the horses, the driver sat on the horse at the back left, thus keeping his whip hand free. Sitting on the left however made it difficult to judge the traffic coming the other way, as anyone who has driven a left-hand drive car along the winding lanes of Britain will agree!In Britain there wasn’t much call tor these massive wagons and the smaller British vehicles had seats for the driver to sit on behind the horses. As most people are right-handed, the driver would sit to the right of the seat so his whip hand was free. Traffic congestion in 18th century London led to a law being passed to make all traffic on London Bridge keep to the left in order to reduce collisions. This rule was incorporated (并入) into the Highway Act of 1835 and was adopted throughout the British Empire.12．What probably happened in the Middle Ages?A．Passers-by walked on the right.B．Strangers fought every time they met.C．People would be attacked by strangers.D．Tourists travelled everywhere without limit. 13．What does the underlined word “sanctioned” in Paragraph 2 mean?A．Recognized.B．Designed.C．Explained.D．Postponed. 14．What was the attitude of drivers in the late 18th century to driving on the left?A．Unclear.B．Indifferent.C．Negative.D．Doubtful. 15．Which of the following can be the best title for the text?A．What Is the “Keep to the Left” Rule?B．Why Do the British Drive on the Left?C．Where Was “Drive on the Left” Carried Out?D．When Is the “Keep to the Left” Rule Popular?There are so many definitions of wisdom. Simply put, wisdom is a mix of insight, common sense, experience, and results to sound judgement. It is not merely a desirable quality but an absolute necessity in a world filled with chaos, controversy, and conflicts. 16 It shines our way through darkness.Growing up with, fairly takes and fantasy movies, I always associated wisdom with Gandalf, Yoda, and Albus Dumbledore. Although these characters have given every one of us tonsof inspiration, wisdom actually lies within ourselves. 17 And how?18 You will never grow from being in your comfort zone, which is why you need to get out of it. Do change your routine and experience as much as what life takes you to! In fact, there are always several sides to a life story. To identify the truth, you need to be receptive to different views. Never base your perspective on the most popular opinions. It is not emotion-based either. Instead, train your mind to be a judgment-free space for ideas. Open-mindedness opens new pathways to profound insights.What if you are desperate for personal growth and improvement? 19 The more time you spend with them, the more transfer of knowledge there will be. Your tutors can be whoever is wiser than you. Yet prior to that, engage in dialogue with them to find out why.Rome was not built in a day. 20 It is important to acknowledge that it takes a span of your life and continuous effort to mature in wisdom as you navigate (找到正确的方法) life’s challenges.A．Individuals need to bring it out.B．It isn’t something you are born with.C．Knowledge speaks, but wisdom listens.D．Remember developing wisdom is beyond a destination.E．A shortcut to do that is to have wise people as your tutors.F．In order to navigate in it, wisdom becomes our guiding light.G．Starting up new experiences is an avenue for acquiring wisdom.二、完形填空V olunteering at Greenpeace is more than a simple individual action. It is a collective force 21 a vision of a greener and fairer world.When I took my first step as a volunteer with Greenpeace Brazil, I had no idea how this experience would 22 my life and my commitment to our planet. Through Greenpeace. I 23 other people, understood how the community functioned, and exchanged 24 with fellow members. Today, looking back, I see how 25 impactful this journey has been, not only for me but also for the communities in the local groups that, we embraced(拥抱) with26 .What makes this experience even more 27 is how our campaigns come to life through the 28 of everyone and the passion each of us nurtures for the environment. Whether in the fight against climate change, the defense of forests, or the protection of oceans, each of us plays a 29 role in our own way.Over time, I have seen our local community 30 awareness and taking action on local environmental issues, especially regarding the mining 31 we face in our state. I witnessed the building of strong connections with other organizations, 32 our movement for a sustainable planet.My initial motivation for volunteering at Greenpeace was to do my part in 33 the environment. Today, I know that I am doing much more. I am part of a global movement that is changing the world. In this 34 , I have found not only an incredible community of people but also a purpose that 35 me every day.21．A．improving B．increasing C．discovering D．shaping 22．A．introduce B．transform C．control D．create 23．A．connected with B．competed against C．went through D．got across 24．A．impression B．money C．experience D．work 25．A．incredibly B．strangely C．unfairly D．fortunately 26．A．creativity B．friendship C．success D．passion 27．A．expensive B．welcoming C．meaningful D．inviting 28．A．kindness B．support C．agreement D．consideration 29．A．general B．familiar C．normal D．decisive 30．A．raising B．encouraging C．developing D．building 31．A．benefits B．businesses C．challenges D．operations 32．A．starting B．strengthening C．designing D．reducing 33．A．decorating B．constructing C．changing D．protecting 34．A．journey B．career C．environment D．campaign 35．A．orders B．surprises C．inspires D．calls三、语法填空阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。

2024-2025学年外研版(2024)第二册地理上册月考试卷727

2024-2025学年外研版(2024)第二册地理上册月考试卷727考试试卷考试范围：全部知识点；考试时间：120分钟学校：______ 姓名：______ 班级：______ 考号：______总分栏题号一二总分得分评卷人得分一、选择题(共8题，共16分)1、图中显示的地理含义是（）A. 聚落人口规模越大，其数量就会越多B. 聚落人口规模越小，其数量就会越少C. 城市人口在16万人时，城市数量最多D. 城市人口在32万人时，城市数量最多2、有关中国人口、民族的叙述，正确的是（）A. 我国是统一的多民族国家，共56个少数民族B. 人口分布西部密集，东部稀疏C. 民族分布“大散居，小聚居，交错杂居”D. 人口总数世界第二位3、下列能正确反映一个地区文化特点的是（）①南方的居民夏天睡竹床，东北农村房屋使用双层玻璃②溜冰是北方人冬季爱好的运动项目，游泳是南方人喜好和擅长的运动项目③彝族的那达慕大会、傣族的泼水节、蒙古族的火把节④新疆人到国内的其他地方后说普通话A. ①②B. ②③C. ③④D. ①④4、为解决城市发展过程中的人口集聚、土地紧缺等问题，也为了控制城市的无序蔓延，围绕轨道交通站点进行高强度高密度开发，利用轨道交通线路引导城市的合理发展。

下图为东京都市圈围绕交通站、点开发的三种模式。

下列关于三种开发模式的叙述正确的是（）①模式A——居住用地集聚于站点②模式B——人口向轨道沿线集聚③模式C——工业用地集聚于站点④模式C——郊区形成新的增长极A. ①②B. ①③C. ②③D. ②④5、关于日本和英国农业的叙述，正确的是（）A. 英国气候温和多雨，农业以种植业为主B. 日本人少地多，农业侧重于精耕细作C. 两国均有丰富的渔业资源D. 英国畜牧业发达，种植业以水稻、小麦为主6、2010年1月1日，中国---东盟自由贸易区正式启动，是世界人口最多第三大自由贸易区。

有关中国---东盟贸易叙述正确的是：①中国---东盟粮食、水果类互补性强；②东盟多以资源型工业为主，劳动密集型工业处于劣势；③中国机械、电子等产品对东盟具有优势，东盟林矿产品出口对中国具有优势；④中国劳动密集型产品对东盟具有优势。

2024-2025学年(外研版)九年级英语上学期第一次月考试卷(广东卷)(含解析)

2024-2025学年九年级上学期第一次月考（广东卷）英语说明:1. 全卷满分为90分,考试用时为70分钟。

2. 答卷前,考生务必用黑色字迹的签字笔或钢笔在答题卡填写自己的准考证号、姓名、考场号、座位号。

用2B铅笔把对应该号码的标号涂黑。

3. 选择题每小题选出答案后,用2B铅笔把答题卡上对应题目选项的答案信息点涂黑,如需改动,用橡皮擦干净后,再选涂其他答案,答案不能答在试题上。

4. 非选择题必须用黑色字迹钢笔或签字笔作答,答案必须写在答题卡各题目指定区域内相应位置上;如需改动,先划掉原来的答案,然后再写上新的答案；不准使用铅笔和涂改液。

不按以上要求作答的答案无效。

一、语法选择（本题有10小题，每小题1分，共10分）通读下面短文，掌握其大意，然后按照句子结构的语法和上下文连贯的要求，从每题所给的三个选项中选出一个最佳答案，并将答题卡上对应题目所选的选项涂黑。

Peggy used to say very little with others. But that all changed when her family moved to a new house.When they got there, she found 1 old book outside their house. To 2 surprise, the book shone(发光) in the dark. “How strange! What is the book 3 ?” she asked her parents, but they had no idea.That night, she started reading the book with her pet dog beside her. But she could hardly understand it. Just then, her dog started speaking, “ 4 interesting book you’ve found!” She couldn’t believe it, 5 the dog continued telling her all kinds of interesting things. After some time, she asked, “So how can you be talking?” “I don’t know.” said the dog.Peggy 6 to find out the truth. So she showed the book to some other animals. All of them started talking with her 7 a friendly way, telling her some pretty interesting stories.During the next several days, Peggy was having fun with her new friends. However, one day, the book disappeared and so did the “voices of the animals”. She 8 looked everywhere, but she couldn’t find 9 the book was.When she returned to school again, Peggy was surprised 10 most of her schoolmates special. Little by little, she started talking more to them. And now she has got more friends than anyone else in school. 1．A．a B．an C．the2．A．she B．her C．hers3．A．make B．make of C．made of4．A．What B．What a C．What an5．A．or B．but C．if6．A．decides B．will decide C．decided7．A．in B．at C．for8．A．careful B．carefully C．care9．A．where B．that C．what10．A．finding B．find C．to find二、完形填空（本大题有10小题，每小题1分，共10分）通读下面短文，掌握其大意，然后在每小题所给的四个选项中，选出一个最佳答案，并将答题卡对应题目所选的选项涂黑。

广东四校联考2023-2024学年高三上学期9月月考物理试题及参考答案

2023-2024A 15B81km/h第 1 题图B C1.2mA0.75C0.6B 0.8 D 0.5第 3 题图A 2m/sB 6m/sC 11m/sD 20m/s第 4 题图ABCDAD-tA BC DMBmb第7 题图1 s1491351 s3 s 1231AmgCmgD(mg)2 + f2t =v − t.6m/s240 m55 mD第10 题图B B == m/s350gFA 0 = c m.A..c m.= 10m / s8m/s60m/s第13 题图θ= 37= 10m / s2sin 37。

= cos37。

=21.60.2 m/s2第14 题图151= 30°OA2第15 题图高三四校联考（一）物理参考答案12345678910DDACDBCADBDBD11．（8分，每空2分）【答案】（1）交流（2）AB （3）0.810 1.97解：（1）交流电源。

（2）A ．为减小实验误差，应调节定滑轮的高度使细线与长木板平行，A 正确；B ．释放小车前，应让小车靠近打点计时器，B 正确；C ．先接通电源，再释放小车，C 错误；D ．用刻度尺测量纸带上两点间距离时，使刻度尺的零刻线对准纸带上的第一个计数点，依次记录下纸带上两点间距离，不应移动刻度尺分别测量每段长度，这样会产生较大的误差，D 错误。

故选AB 。

（3）相邻两计数点间还有四个计时点未标出，打点计时器所接电源的频率为50Hz ，则有纸带上相邻两点的时间间隔是T =0.02×5s=0.1s由∆x =aT 2，可得小车运动的加速度2222215.0112.9810.969.107.10 5.0010m /s 1.97m /s 90.1x a T -∆++---==⨯=⨯B 点时的速度27.109.100.162010m s s 0.810m s 220.10.2B x v T -∆+==⨯==⨯12．（10分，每空2分）【答案】（1）7.00（2）在弹性限度内，弹簧的弹力大小与弹簧的形变量成正比（3）A 、10（4）27.00【详解】（1）图中刻度尺的分度值为1mm ，要估读到分度值的下一位，已知刻度尺的零刻度线与弹簧上端平齐，弹簧长度为07.00cm=L （2）由图可知，弹簧的弹力大小与弹簧的伸长量成正比。

专题04 选词填空(含解析)-(2025年)外研版(三起)英语四年级上册期中专项练习

期中专项复习04 选词填空学校:___________姓名：___________班级：___________考号：___________一、填空题1．(Excuse / Help) me! Where’s the cinema?2．I like the elephant best. It is (between / beside) the panda and the monkey. 3．Look! Sam, (here / here’s) a dog.4．(Where’s / Where) No. 2 Park Street?5．I live (at / on) No.2 Park Street.6．The bus is (at / on) the station.7．The car is __________ (up / down) the hill.8．Let's (get on / turn on) the bus.9．She (likes / like) running so much.10．Mum is (swim / swimming) in the pool.11．English is __________ (I / my) favourite subject.12．He's reading a book (about / to) China.13．She is talking to (her / she) friend.14．— (What / Where) are they doing now?—They’re playing under the tree.15．Those (child / children) are flying kites.16．The children (are / is) doing homework.17．My sister is talking her friends. (on / to)18．The people (in / on) the lake. They’re rowing a boat. 19．We can see lots (of / for) interesting things.20．Look! The boy is between the big (tree / trees). 21．Look at (this / these) girls.22．The elephant is listening (to / at) music.23．He’s drawing (an / a) egg.24．Juice (and / with) ice is also very nice.25．Do you want (any / some) milk?26．Here are noodles (and / with) tomato and egg. 27．What do you want to (buy / by)?28．(What / What’s) the grandma doing?29．Can (me / I) help you?30．How (many / much) is the flower?31．—How much (is / are) the flowers?—Twelve yuan.32．— (How much / How many) is it?—Five yuan.33．Juice with ice is (too / also) very nice.34．Look at the kites up (in / at) the sky.35．Can you jump (fast / far)?36．I’m the (win / winner).37．Jack is very tall and (strong / lost). He can play basketball. 38．Can you (swim / swimming)?39．She is (we / our) star.40．—Can I have some soup?—No, I’m (afraid / happy) you can’t.41．I can (make / making) noodles. Can you?42．Can Lingling (play / plays) football? 43．—Let’s (get / gets) on the bus, Lingling.—OK.二、读一读，选一选。

2024-2025学年外研版八年级科学上册阶段测试试卷918

2024-2025学年外研版八年级科学上册阶段测试试卷918考试试卷考试范围：全部知识点；考试时间：120分钟学校：______ 姓名：______ 班级：______ 考号：______总分栏一、单选题(共5题，共10分)1、要使马铃薯和甘薯高产应当注意使用哪种无机盐（）A. 含氮无机盐B. 含磷无机盐C. 含钾无机盐D. 含硼无机盐2、2014年08月03日16时30分在云南省昭通市鲁甸县（北纬27.1度，东经103.3度）发生6.5级地震，震源深度12千米．医疗救助人员打算给当地居民提供一些关于饮用水的建议，以下建议不合理的是（）A. 用明矾使悬浮颗粒沉降下来B. 用漂白粉进行消毒杀菌C. 只要无色透明，就可以放心饮用D. 饮用前加热煮沸3、教室里有4盏电灯，去掉其中任意一盏灯泡，另外三盏电灯都能继续发光，则这4盏灯( )A. 一定是串联B. 一定是并联C. 可能是串联，也可能是并联D. 无法判断4、在光下检测植物的呼吸作用不明显的原因是（）A. 光合作用强，吸收了二氧化碳B. 光下不进行呼吸作用C. 光合作用抑制了呼吸作用D. 光下呼吸作用减弱下列说法正确的是（）A. 反应后原子的数目减少B. a=0.6C. B，D的相对分子质量之比为9：1D. 该反应一定是复分解反应评卷人得分二、填空题(共7题，共14分)6、下列物质中：①空气②酱油③食醋④汞⑤水⑥矿泉水⑦氧气⑧氢气⑨二氧化碳⑩铁。

纯净物的是____，混合物的是____。

7、根吸收水分的主要部位是根尖的____，其上有大量的根毛，这大大增加了根吸水的____，提高了根的吸水效率。

8、呼吸系统由____和____组成，前者由____共同组成，主要作用是将吸入的空气变得____、____、____减少对肺的刺激．9、物体在月球上受到的重力只有地球上的六分之一，1.2吨质量的嫦娥三号探测器在月球上受到的重力是____N。

10、发光二极管只允许电流从二极管的正极流入，负极流出．如图所示，使磁铁在线圈中左右移动，此时会看见两只发光二极管轮流发光。

广东省深圳市罗湖区2024-2025学年高三上学期9月质量检测试题地理(含答案)

绝密★启用前试卷类型：A2024-2025 学年第一学期学业质量检测高三地理说明：2024.91.答题前，请将姓名、班级和学校用黑色字迹的钢笔或签字笔填写在答题卡指定的位置上，并正确粘贴条形码。

2.全卷共 8 页，考试时间 75 分钟，满分 100 分。

3.作答单项选择题时，选出每题答案后，用 2B 铅笔把答题卡上对应题目答案标号的信息点框涂黑。

如需改动，用橡皮擦干净后，再选涂其它答案。

作答非选择题时，用黑色字迹的钢笔或签字笔把答案写在答题卡指定区域内，写在本试卷或草稿纸上，其答案一律无效。

4.考试结束后，请将答题卡交回。

第 Ⅰ 卷一、选择题（本大题共 16 小题，总分 48 分）咖啡树喜静风、高温，抗寒性差，畏强光。

四川省攀枝花市是目前世界纬度最高的咖啡生产区，当地咖农将咖啡种植区主要集中于河谷两侧地势较高的坡地上，且常将咖啡树与芒果树套种（芒果树一般较咖啡树高大）。

图 1 示意攀枝花咖啡种植位置，据此完成 1～2 题。

图 11.影响攀枝花成为世界上纬度最高咖啡生产区的主导因素是A . 降水B . 地形C . 纬度D . 水源2. 芒果树给咖啡树生长提供的有利环境是①削弱光照②减少虫害③降低风速④提高气温A . ①②④B . ②③④C . ①③④D .①②③青藏高原面积达250 万km2，大部分地区海拔在4000～6000 m，在夏季是个巨大的热源，具有一定的增温效应，高原内部的气温要高于同纬度相同海拔的气温，这种增温效应对高原及其周围的地理生态格局有深远的影响。

图2 是青藏高原及其周边地区不同地区的高程及7 月份均温数据，据此完成3～4 题。

图23.在海拔5200 m 的高度上，7 月大气温度最高的是A. 甲B. 乙C. 丙D. 丁4. 与四川盆地相比，青藏高原地区①林线更低②林线更高③雪线更高④雪线更低A. ①④B. ①③C. ②④D. ②③长海县位于辽东半岛东侧黄海北部海域，全县由195 个海岛组成，是东北地区唯一海岛边境县，该县借址普兰店市皮口镇荒芜海岸建设渔业加工园，发展渔产品加工业，图3 为两地区域简图。

专题08 匹配题(含解析)-(2025年)外研版(三起)英语五年级上册期中专项练习

期中专项复习08 匹配题学校:___________姓名：___________班级：___________考号：___________一、根据句意，选择对应的图片。

A．B．C．D． E.( )1．There are enough rulers for five children.( )2．There are not enough cars for fifteen people.( )3．There is so much rice for a little boy.( )4．There are so many clothes on the line.( )5．There is only one pen in the box.二、从方框中选出正确的答语。

( )7．How many dogs can you see?( )8．How did you go to the Great Wall?( )9．Does Sam like reading books?( )10．How much juice did you buy?( )11．What’s the matter with Daming?( )12．There are only nine caps.三、根据句意，选择对应的图片。

A．B．C．D． E.( )13．He lost his shoes.( )14．We visited the London Eye at the weekend.( )15．Daming took a photo of his father.( )16．Sam’s shorts are clean.( )17．We need two bottles of juice.四、从方框中选出正确的答语。

( )18．How much water did you drink? ( )19．Did you buy bananas? ( )20．How many cats can you see? ( )21．Is this your coat?( )22．How many pupils are there in your class?五、根据句意，选择对应图片，将序号填在括号内。

2024届广东金太阳高三上学期9月联考日语试卷及答案

高三日语试卷参考答案第一部分：听力(2分×15＝30分)1．Ａ2．Ｂ3．Ｃ4．Ａ5．Ｂ6．Ｂ7．Ｃ8．Ａ9．Ａ10．Ｃ11．Ｂ12．Ａ13．Ｃ14．Ｂ15．Ｃ第二部分：日语知识运用(1分×40＝40分)16．Ｂ17．Ａ18．Ｃ19．Ｄ20．Ａ21．Ｂ22．Ｃ23．Ｄ24．Ａ25．Ａ26．Ｂ27．Ａ28．Ｄ29．Ｃ30．Ａ31．Ｄ32．Ｂ33．Ａ34．Ｃ35．Ｂ36．Ｄ37．Ｃ38．Ｂ39．Ａ40．Ｃ41．Ｄ42．Ｃ43．Ａ44．Ｂ45．Ｄ46．Ｄ47．Ａ48．Ｂ49．Ｄ50．Ｃ51．Ｂ52．Ｂ53．Ｄ54．Ａ55．Ｃ第三部分：阅读理解(2.5分×20＝50分)（一）56．Ｃ57．Ａ58．Ｄ59．Ｂ60．Ｄ（二）61．Ｃ62．Ｄ63．Ｂ64．Ｃ65．Ｄ（三）66．Ｂ67．Ｂ68．Ｃ69．Ａ70．Ｄ（四）71．Ｃ72．Ｂ73．Ｄ74．Ｄ75．Ｃ第四部分：写作(满分30分)参考范文自分に拍手喝采を自分に拍手喝采を送らない人は、成功しにくいでしょう。

この世で一番自分を理解しているのは自分で、最大の敵も自分自身です。

自分に勝って、自分に拍手喝采をするのは成功への始まりだと思います。

だから、自分で自分を応援しましょう。

きっと、夢や目標に向かって頑張りたい気持ちがますます強くなると思います。

私は真面目に勉強している自分に喝采を送りたいです。

この2年間、一生懸命頑張って成績が伸びました。

まだ他人と差があるが、2年前の私と比べたらだいぶ進歩したと思います。

また、私は親孝行をして、礼儀正しく人に接する自分に喝采を送りたいです。

家族や友達を大切にして、人と仲良くして過ごした日々は幸せで、自分も成長してきたと感じました。

これから他人を応援するだけでなく、自分も応援して人生を盛り上げていきたいです。

（348字）作文评分标准1.评分方法：先根据短文的内容和语言的表达能力初步确定其所属档次，然后按照该档次的标准并结合评分说明确定或调整档次，最后给分。

2023-2024学年广东省广州市高三上学期9月月考物理质量检测试题(含答案)

2023-2024学年广东省广州市高三上学期9月月考物理质量检测试题一、单选题1.2020年11月10日8时12分，中国自主研发的万米载人潜水器“奋斗者”号，在马里亚纳海沟下沉至接近海底时，向水底发射出持续时间为1.00s的某脉冲声波信号，最终在深度10909m处成功坐底。

在该深度，“奋斗者”号每平方厘米要承受41.0710N的海水压力，并停留6h进行了一系列的深海探测科考活动。

下列说法正确的是（）A.m、N、s是国际单位制中的基本单位B.在下沉过程中，潜水器的位移一定是10909mC.“8时12分”指的是时间间隔D.采集海底矿物时，不能将潜水器视为质点2.如图所示，在水平路面上匀速直线行驶的汽车内，一个手机通过吸盘式支架固定在倾斜的前挡风玻璃上。

若吸盘和玻璃间的空气已被排空，则手机和支架（）A.共受3个力作用B.共受4个力作用C.受前挡风玻璃的作用力方向一定竖直向上D.受前挡风玻璃的摩擦力方向一定竖直向上3.2019年10月15日，第十届环太湖国际公路自行车赛在无锡结束。

比赛中甲、乙两赛车(可视为质点)在一条直线上运动，其速度－时间图象如图所示，下列对甲、乙运动描述中正确的是A.t0时刻甲乙相遇B.0～t0时间内乙的加速度逐渐增大C.0～t0时间内甲乙间的距离先增大后减小D.0～t0时间内的某时刻甲、乙加速度大小相等4.“长征七号”A运载火箭于2023年1月9日在中国文昌航天发射场点火升空，托举“实践二十三号”卫星直冲云霄，随后卫星进入预定轨道，发射取得圆满成功。

已知地球表面的重力加速度大小为g，地球的半径为R，“实践二十三号”卫星距地面的高度为h（h小于同步卫星距地面的高度），入轨后绕地球做匀速圆周运动，则（）A.该卫星的线速度大小大于7.9km/sB.该卫星的动能大于同步卫星的动能C.该卫星的加速度大小等于gD.该卫星的角速度大小大于同步卫星的角速度5.滑板运动是年轻人喜爱的一种新兴极限运动，如图，某同学腾空向右飞越障碍物，若不计空气阻力，并将该同学及滑板看着是质点，则该同学及板在空中运动的过程中（）A.做匀变速运动B.先超重后失重C.在最高点时速度为零D.在向上和向下运动通过空中同一高度时速度相同6.近年来，机器人与智能制造行业发展迅速，我国自主研发的机器人在北京冬奥会和疫情防控中均发挥了重要作用。

2024高三九省联考英语试卷及答案解析

2024高三九省联考英语试卷及答案解析2024高三九省联考英语试卷及答案文字版注意事项：1. 答卷前，考生务必将自己的考生号、姓名、考点学校、考场号及座位号填写在答题卡上。

2. 回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。

如需要改动，用橡皮擦干净后，再选涂其他答案标号。

回答非选择题时，将答案写在答题卡上。

写在本试卷上无效。

3. 考试结束后，将本试卷和答题卡一并交回。

第一部分听力(共两节，满分30分)做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节(共5小题;每小题1. 5分，满分 7. 5分)听下面5 段对话。

每段对话后有一个小题，从题中所给的 A、B、C三个选项中选出最佳选项。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

例: How much is the shirt?A. ?19. 15.B. ?9. 18.C. ?9. 15.答案是C。

1. What will Chris do next?A. Drink some coffee.B. Watch the World Cup.C. Go to sleep.2. What is the probable relationship between the speakers?A. Strangers.B. Classmates.C. Relatives.3. What is the woman’s attitude to the man’s suggestion?A. Favorable.B. Tolerant.C. Negative.4. What can we learn about Tom?A. He’s smart for his age.B. He’s unwilling to study.C. He’s difficult to get along with.5. What did Kevin do yesterday?A. He went swimming.B. He cleaned up his house.C. He talked with his grandparents.第二节(共15小题;每小题1. 5分，满分22. 5 分)听下面5段对话或独白。

广东省深圳2024-2025学年高三上学期9月月考试题物理含答案

深圳2024-2025学年度高三第一学期第二次月考物理试题（答案在最后）卷面满分100分，考试用时75分钟。

注意事项：1.答卷前，考生务必将自己的姓名、准考证号码填写在答题卡上。

2.作答时，务必将答案写在答题卡上。

写在本试卷及草稿纸上无效。

3.考试结束后，将答题卡交回。

一、单项选择题：本题共7小题，每小题4分，共24分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

1.深中通道是国家重大工程和世界级跨海集群工程，全长约24千米，北距虎门大桥约30千米，小明计划周末从家坐车去深圳宝安机场，他事先用导航软件搜索路线，若选择走A路线，经过虎门大桥，路程约103千米，耗时约1小时50分钟；若选择走B路线，经过深中通道，路程约53千米，耗时约1小时10分钟，假设汽车实际行驶路程和时间与导航软件预计的相同，则下列说法正确的是（）A.汽车通过两条路线的平均速度相同B.汽车通过A路线的平均速度更大C.汽车通过A路线的位移更大D.汽车通过A路线的平均速率更大2.如图，潜艇从海水的高密度区驶入低密度区过程称为“掉深”。

图a，某潜艇在高密度区水平向右匀速航行；t=0时，该捞艇开始“掉深”，图b为其竖直方向的速度时间图像，水平速度v保持不变。

若以水平向右为x轴，竖直向下为y轴，则带艇“掉深”后的0~30s内。

能大致表示其运动轨迹的图形是（）A. B.C. D.3.2023年的春晚舞蹈《锦绣》，艺术地再现了古代戍边将士与西域各民族化干戈为玉帛并建立深厚友谊的动人故事。

图（a ）是一个优美且难度极大的后仰动作，人后仰平衡时，可粗略认为头受到重力G 、肌肉拉力2F 和颈椎支持力1F 。

如图（b ），若弯曲后的头颈与水平方向成60︒角，2F 与水平方向成45︒角，则可估算出1F 的大小为（）A.)1G +B.)1G -C.)2GD.)2G -4.图甲是北京冬奥会单板滑雪大跳台比赛项目中运动员在空中姿态的合成图。

比赛场地分为助滑区、起跳台、着陆坡和终点区域四个部分。

外研版英语高考试卷及答案指导

外研版英语高考仿真试卷及答案指导一、听力第一节（本大题有5小题，每小题1.5分，共7.5分）1、Listen to the conversation and answer the question.A. The man is planning to go to the movies with his friends.B. The woman is suggesting that the man should take a break.C. The man is tired of working all day and wants to relax.Question: What are they talking about?Answer: BExplanation: In the conversation, the woman says, “You’ve been working all day. How about we go out for a coffee break?” This indicates that she is suggesting the man should take a break, which is why answer B is correct.2、Listen to the dialogue and fill in the blanks with the correct words.Tom: Hi, Jane. How was your vacation last week?Jane: Oh, it was amazing! ___________1_. We visited the Great Wall and the Forbidden City.Tom: That sounds great! _________2_. I’ve always wanted to go there. Jane: Yeah, it’s definitely worth a visit. _________3___. The food there is also very delicious.Answer: 1. We had a fantastic time, 2. I wish I could go, 3. The culture and history are fascinating.Explanation:1.The woman is describing her vacation as “amazing,” so “We had a fantastic time” is the appropriate response.2.The man expresses his wish to visit the places the woman mentioned, so “I wish I could go” fits the context.3.The woman is emphasizing the positive aspects of her trip, including the culture and history, and the food, so “The culture and history are fascinating” is the correct completion.3.You hear a conversation between two students discussing their weekend plans.Student A: Hey, do you have any plans for this weekend?Student B: Yeah, I’m planning to go hiking. How about you?Student A: I was thinking of visiting my grandparents. It’s been a while since I last saw them.Student B: That soun ds nice. They’re always so happy to see me.What does Student A plan to do this weekend?A) Go hiking.B) Visit their grandparents.C) Stay home and relax.D) Go on a trip.Answer: BExplanation: Student A mentions that they are planning to visit their grandparents, which indicates that this is their weekend plan. Therefore, thecorrect answer is B) Visit their grandparents.4.You hear a teacher introducing a new project to the class.Teacher: Good morning, class. I’m pleased to announce that we will besta rting a new project this week. It’s called the “Community Service Project.” Each of you will be expected to complete a minimum of 10 hours of community service by the end of the month. This can be anything from volunteering at a local shelter to helping out at a park cleanup. Remember, this is not only a great way to give back to our community but also an opportunity to learn new skills and make new friends. Who would like to share some ideas on how we can make this project successful?What is the main purp ose of the “Community Service Project”?A) To learn new skills.B) To make new friends.C) To give back to the community.D) To complete school assignments.Answer: CExplanation: The teacher explicitly states that the purpose of the project is “to give back to our community.” This makes C) To give back to the community the correct answer. While the other options may be outcomes of the project, they are not the primary goal as stated by the teacher.5.How old is the woman when she graduates from university?A)22 years old.B)23 years old.C)24 years old.D)25 years old.Answer: B) 23 years old.Explanation: In the conversation, the woman mentions that she will be 22 when she starts her first job, and she graduates from university one year before that, making her 23 years old at the time of graduation.二、听力第二节（本大题有15小题，每小题1.5分，共22.5分）1、Section B: Listening ComprehensionDirections: In this section, you will hear 8 short conversations and 2 long conversations. At the end of each conversation, one or more questions will be asked about what was said. Both the conversation and the questions will be spoken only once. After each question, there will be a pause. During the pause, you must read the four choices marked A), B), C), and D), and decide which is the best answer. Then mark the corresponding letter on Answer Sheet 2 with a single line through the centre.2、Question: What is the main topic of the conversation?A)The importance of exercise.B)The benefits of a healthy diet.C)The relationship between stress and health.D)The effects of sleep on productivity.Answer: C)Explanation: The conversation discuss es how stress can affect one’s health and the importance of finding ways to manage stress. The speakers mention different methods of stress relief and its impact on overall well-being, making option C) the best answer. Options A), B), and D) are not the central themes of the conversation.3.You are going to listen to a conversation between a student and a teacher about a school project. Listen carefully and answer the question.Question: What is the main purpose of the school project?A. To study the local history.B. To learn about different cultures.C. To improve environmental awareness.D. To practice team building skills.Answer: BExplanation: The main purpose of the school project is discussed in the conversation, and the teacher mentions that the project is meant to help the students learn about different cultures. Therefore, the correct answer is B.4.You will hear a short interview with a famous author. Listen carefully and answer the question.Question: What inspired the author to write their latest book?A. A personal experience.B. A historical event.C. A dream they had.D. A conversation with a reader.Answer: AExplanation: During the interview, the author discusses their inspiration for the new book, and they mention that it was a personal experience that sparked their interest in writing the story. Therefore, the correct answer is A.5、What does the man suggest the woman do?A) Go to the library.B) Take a break from studying.C) Study together with him.D) Go for a walk.Answer: C)Explanation: In the conversation, the man suggests that the woman should study together with him rather than continue studying alone. He says, “Why don’t we go over the material together? It might help us both.”6、How does the woman respond to the man’s suggestion?A) She declines because she is too tired.B) She agrees but suggests they meet tomorrow.C) She accepts immediately.D) She hesitates but then agrees.Answer: D)Explanation: The woman initially seems uncertain about changing her plansbut eventually agrees when she says, “I was just going to keep reading on my own, but maybe you’re right. Let’s do it.”7.W: Have you seen the new movie “Invisible” that just came out? It’s supposed to be very exciting.M: I heard about it, but I haven’t had the time to go see it yet. How was it?Q: What can we infer about the man from the conversation?A: The man has not seen the movie yet.解析：男士回答中提到他没有时间去看电影，说明他还没有看过这部电影。

2024-2025学年外研版(2019)九年级物理上册阶段测试试卷240

2024-2025学年外研版(2019)九年级物理上册阶段测试试卷240考试试卷考试范围：全部知识点；考试时间：120分钟学校：______ 姓名：______ 班级：______ 考号：______总分栏一、选择题(共6题，共12分)1、下列数据符合实际的是（）A. 家用普通冰箱冷冻室的温度约-50℃B. 正常情况下，人的脉搏跳动一次的时间约为10sC. 中学生的平均身高约为160dmD. 滨海地区夏天最高气温约37℃2、岸边的树，在地上会出现树影，在水中会出现倒影，下列关于这两个“影”的说法正确的是（）A. 地上的树影是光的直线传播形成的，水中的倒影是光的反射形成的B. 地上的树影是光的反射形成的，水中的倒影是光的直线传播形成的C. 地上的树影是光的反射形成的，水中的倒影是光的折射形成的D. 地上的树影是光的折射形成的，水中的倒影是光的反射形成的3、甲、乙两个灯分别标有“[2.5V0.3A<]”和“[3.8V0.3A<]”的字样，将它们串联接人同一电路中，若甲灯正常发光，则下列说法中正确的是A. 乙灯也一定能正常发光，甲灯比乙灯亮B. 乙灯也一定能正常发光，乙灯比甲灯亮C. 乙灯不能正常发光，甲灯比乙灯亮D. 乙灯不能正常发光，乙灯比甲灯亮4、下列说法中，正确的是[( <][) <]A. 摩擦起电就是创造了电荷B. 一个带正电的物体能够吸引另一个物体，另一个物体一定带负电C. 规定正电荷定向移动的方向为电流的方向D. 与任何物体摩擦后的玻璃一定带正电5、在如图所示的电路中，电源电压保持不变，闭合开关S，当滑动变阻器的滑片P向右移动时，下列分析正确的是（）A. 电流表A示数与电流表A2示数的差变大B. 电流表A1示数与电流表A2示数的和变大C. 电压表V示数与电流表A2示数的比值变大D. 电压表V示数与电流表A1示数的乘积变大6、如图所示的四种现象中，属于光的直线传播的是（）A.筷子好像在水面处“折断”B.放大镜把文字放大C.桥在水中形成“倒影”D.日全食现象评卷人得分二、多选题(共9题，共18分)7、关于四冲程汽油机的压缩冲程，下列说法正确的是（）A. 汽缸里面的气体质量不变B. 汽缸里面的气体压强增大C. 汽缸里面气体的内能转化为活塞的机械能D. 汽缸里面的气体温度升高8、在如图所示中能反映电动机的工作原理的是（）A.B.C.D.9、对于功的理解，下列说法错误的是（）A. 作用在物体上的力越大，做的功就越大B. 物体移动的距离越长，做的功就越多C. 有力作用在运动的物体上，该力必然对物体做功D. 有力作用在物体上，物体在力的方向上移动距离才算力对物体做了功10、如图所示，电源电压不变，只闭合S1与只闭合S3电流表的示数之比为3：2，下列说法正确的是（）A. R1、R2之比为1：2B. 只闭合S1与只闭合S2电流表的示数之比为2：1C. 只闭合S2与只闭合S3电阻R2消耗的功率之比为9：1D. 闭合S1、S2与只闭合S3整个电路的功率之比为9：211、如图所示为小金同学在百联商务宾馆里看到的旅行牙膏，根据你的观察和生活经验判断，下列说法中正确的是（）A. 牙膏盖中间的尖锐物可增大压强B. 这支新牙膏受到的重力是3NC. 挤牙膏反映力能使物体发生形变D. 牙膏盖上的条纹是为了增大摩擦12、小玲和小亮面对面站在水平地面上，小玲穿着旱冰鞋，她所受的重力为G，小玲对地面的压力为N，地面对小玲的支持力为N′．小玲推了小亮一下，如图所示，小玲自己却向后运动，而小亮依然在原地静止，则下列说法中正确的是（）A. G与N′是一对平衡力B. N与N′是一对相互作用力C. 小玲推小亮时，小玲所受的合力为零D. 小玲推小亮时，小亮受地面的摩擦力大于小玲给他的推力13、下列做法中不符合安全用电规范的是（）A. 控制电灯的开关接在零线上B. 家庭电路中用铜丝代替熔丝C. 使用测电笔时，手要接触笔尾的金属电极D. 家用电器金属外壳不要接地14、如图所示，“奔马”模型的后蹄能稳稳地站立在手指上，下列分析错误的是（）A. “奔马”受到的重力与手指对它的支持力是一对平衡力B. “奔马”受到的重力与它对手指的压力是一对平衡力C. “奔马”对手指的压力与手指对它的支持力是一对平衡力D. “奔马”受到的重力与它对手指的压力是一对相互作用力15、如图所示的电路中，电源电压为4.5V，小灯泡上标有“4.5V2.25W”的字样（设灯丝电阻不变），滑动变阻器的最大阻值为30Ω，电流表量程为“0～0.6A”，两块电压表的量程为“0～3V”．以下说法正确的是（）A. 滑片向左移动时，电流表示数增大，小灯泡越亮B. 滑片向右移动时，电压表V1示数的减小值等于电压表V2示数的增加值C. 滑动变阻器允许接入电路的阻值范围是4.5～18ΩD. 滑动变阻器接入电路的阻值最小时，电路消耗的总功率为1.5W评卷人得分三、填空题(共5题，共10分)16、（2015•晋江市二模）如图所示，是A、B两种燃料完全燃烧释放的热量Q与其质量m 的关系图，从图中可看出A燃料的热值 B燃料的热值（选填：“大于”、“小于”或“等于”）．在标准大气压下，若一定质量的B燃料完全燃烧释放的热量为4.9×106J，其中有60%的热量被质量为10kg、初始温度为38℃的水吸收（c水=4.2×l03J/（kg•℃），则水的温度将升高℃．17、沸腾时观察到的现象是：水中形成，上升时，到水面，里面的水蒸气散发到空气中．18、下列常见物理现象各属于哪种物态变化：A．严冬玻璃窗上结有冰花；B．铸造车间里铁水铸成零件．19、如图所示，容器A、B内盛有液面在同一水平面的清水，用带有阀门K的斜管将两容器相连，当将阀门K 打开时水将流动．（填“会”或“不会”）20、滑动变阻器是靠改变连入电路中的电阻线的______来改变______，从而改变电路中的______。

2025届广东罗湖区高三9月质检化学答案

2024-2025学年第一学期学业质量检测高三化学评分参考一、选择题：本题共16小题，共44分。

第1～10题，每小题2分；第11～16题，每小题4分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

题号 1 2 3 4 5 6 7 8 9 10答案 D C D A D B C A C B题号11 12 13 14 15 16答案 D B A D C B二、非选择题：本大题共4小题，共56分。

17. （14分）（1）①AD （2分，各1分。

错一个对一个给0分，错一个对两个给1分，全对给2分）②3.55×10-2 或0.0355（2分）（2）HClO、ClO- （2分，各1分。

错一个对一个给0分，错一个对两个给1分，全对给2分）（3）① b （1分）光照②2HClO===2HCl+O2↑（2分。

化学式正确给1分，条件符合正确给0.5分，配平正确给0.5分）（4）①有固体未溶解（1分）②NaOH（1分，答案合理即可）（5）① 0.3 （1分）②溶液pH越大，CO2转化为CO32-的比例越大（或“利于HCO3-转化为CO32-”），则容易生成CaCO3（（2分，仅部分合理给1分。

）18. （14分）（1）PbSO4 （1分）；2Fe2+ + H2O2 + 2H+ = 2Fe3+ + 2H2O（2分。

化学式正确给1分，条件符合正确给0.5分，配平正确给0.5分）（2）[2.8~3.5）或 2.8≤pH＜3.5（2分，答案合理即可。

对一半给1分）（3）盐酸（2分，浓或稀都给分）（4）H2A2或萃取剂（2分）（5）粉状物粒度更小，比表面积．．置换生成海绵状的铟．（2分，加点部分表．．．．更大，更有利于述合理各1分。

答“表面积”给分。

其他正确，不回答“海绵状”给1.5分。

表述合理、但不准确给1分。

）（6）①4（1分）②190×4aa3·NN AA× 1021（2分，列式正确未化简给分，计算出最终结果且正确给分）19. （14分）（1）Na（s）＋12Cl2（g）=NaCl（s）ΔH＝－410.8 kJ/mol或者 2Na（s）＋Cl2（g）=2NaCl（s）ΔH＝－821.6 kJ/mol （2分。

广东八校2025届高三上学期9月联合检测数学试卷(解析版)

数学注意事项：1. 本试卷满分150分，考试时间120分钟.2. 答题前，考生务必将自己的姓名、准考证号等填写在答题卡的相应位置.3. 全部答案在答题卡上完成，答在本试题卷上无效.4. 回答选择题时，选出每小题答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑.如需改动，用橡皮擦干净后，再选涂其他答案标号.5. 考试结束后，将本试题卷和答题卡一并交回.一、选择题：本题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的.1. 样本数据45，50，51，53，53，57，60的下四分位数为（） A. 50 B. 53 C. 57 D. 45【答案】A 【解析】【分析】根据百分位数的概念即可求解.【详解】由这组数据共7个，则7725%4i =×=，所以这组数据的下四分位数为第250. 故选：A ．2. 已知集合3{|278}A x x =−<<，{||2|3,Z}B x x x =−≤∈，则A B = （） A. {}1,0− B. {}0,1C. {}1,0,1−D. {}0,1,2【答案】C 【解析】【分析】解不等式化简集合,A B ，则再利用交集的定义求解即得.【详解】由3278x −<<，解得32x −<<，即{|32}Ax x =−<<，由|2|3x −≤，得323x −≤−≤，解得15x −≤≤，则{|15,Z}{1,0,1,2,3,4,5}B x x x =−≤≤∈=−，所以{1,0,1}A B =− . 故选：C3. 已知向量()1,2a =−，()3,b λ= ，若2a b + 与2a b −平行，则实数λ的值为（）A. 23−B.23C. 6D. 6−【答案】D 【解析】【分析】先求2a b + 与2a b −的坐标，然后由向量平行的坐标表示可得.【详解】因为()1,2a =−，()3,b λ=，所以2(5,22)a b λ++，2(5,4)a b λ−=−−又2a b + 与2a b −平行，所以5(4)5(22)λλ−=−+，解得6λ=−. 故选：D4. 已知函数()()()2sin 2xf x x x x a =++为偶函数，则a =（）A. －2B. －1C. 0D. 2【答案】A 【解析】【分析】利用偶函数的性质进行求解即可. 【详解】由2200x x x +≠⇒≠且2x ≠−，由0x a x a +≠⇒≠−，因为该函数是偶函数，所以定义域关于原点对称，因此有22a a −=⇒=−，即()()()sin 22xf x x x x =+−，定义域为()()()(),22,00,22,−∞−∪−∪∪+∞，因为()()()()()()()()()sin sin sin 222222x x xf x f x x x x x x x x x x −−−====−−+−−−−+−+，所以该函数是偶函数，符合题意，故选：A 5. 已知曲线()ln x y e ax x −在点()1,ae 处的切线方程为y kx =，则k =（）A. 1−B. 0C. 1D. e【答案】D 【解析】【分析】利用导数的几何意义，求得在点()1,ae 处的切线方程，再根据切线过原点，即可求得参数k . 【详解】令()()ln x yf x e ax x =−，则()()1ln x xf x e ax x e a x′−+−（），所以()12f ea e =′−，因为曲线()ln x ye ax x −在点()1,ae 处的切线方程为y kx =，所以该切线过原点，所以()12f ea e ae =′−=，解得1a =，即k e =. 故选：D.【点睛】本题考查导数的几何意义，只需准确写出切线方程即可，属基础题.6. 已知函数()()1cos 02f x x x ωωω+>在区间π0,2 内有最大值，但无最小值，则ω的取值范围是（） A. 25,36B. 28,33C. 15,66D. 18,63【答案】B 【解析】【分析】利用辅助角公式可得()πsin 6f x x ω=+，根据自变量取值范围以及正弦函数图象性质可得πππ3π2262ω<+≤，解不等式可得28,33ω ∈.【详解】易知()1πcos sin 26f x x x x ωωω+=+，当π0,2x∈时，ππππ,6626x ωω+∈+，又因为()f x 在区间π0,2内有最大值，但无最小值，所以πππ3π2262ω<+≤，解得2833ω<≤. 故选：B7. 已知抛物线2:8C y x =，圆22(2):4F x y −+=，直线:(2)(0)l y k x k =−≠自上而下顺次与上述两曲线交于1234,,,M M M M 四点，则下列各式结果为定值的是 A. 1324M M M M ⋅ B. 14FM FM ⋅ C. 1234M M M M ⋅ D. 112FM M M ⋅【答案】C 【解析】【详解】【分析】由()228y k x y x=−= 消去y 整理得2222(48)40(0)k x k x k k −++=≠，设111422(,),(,)M x y M x y ，则21212248,4k x x x x k++==．过点14,M M 分别作直线:2l x ′=−的垂线，垂足分别为,A B ，则11422,2M F x M F x =+=+．对于A ，13241412(2)(2)(4)(4)M M M M M F M F x x ⋅=++=++12124()16x x x x =+++，不为定值，故A 不正确．对于B ，14121212(2)(2)2()4FM FM x x x x x x ⋅=++=+++，不为定值，故B 不正确．对于C ，12341412(2)(2)4M M M M M F M F x x ⋅=−−==，为定值，故C 正确．对于D ，1121111(2)(2)FM M M M F M F x x ⋅=⋅−=+，不为定值，故D 不正确．选C ．点睛：抛物线定义的两种应用：（1）当已知曲线是抛物线时，抛物线上的点M 满足定义，它到准线的距离为d ，则|MF |＝d ，有关距离、最值、弦长等是考查的重点；（2）利用动点满足的几何条件符合抛物线的定义，从而得到动点的轨迹是抛物线．8. 已知函数()f x 是定义域为R 的函数，()()20f x f x ++−=，对任意1x ，[)21,x ∈+∞()12x x <，均有()()210f x f x −>，已知a ，b ()a b ≠为关于x 的方程22230x x t −+−=的两个解，则关于t 的不等式()()()0f a f b f t ++>的解集为（） A. ()2,2− B. ()2,0−C. ()0,1D. ()1,2【答案】D 【解析】【分析】由题可得函数()f x 关于点()1,0对称，函数()f x 在R 上单调递增，进而可得()()01f t f >=，利用函数的单调性即得.【详解】由()()20f x f x ++−=，得()10f =且函数()f x 关于点()1,0对称．由对任意1x ，[)21,x ∈+∞()12x x <，均有()()210f x f x −>，可知函数()f x 在[)1,+∞上单调递增．又因为函数()f x 的定义域为R ，所以函数()f x 在R 上单调递增．因为a ，b ()a b ≠为关于x 的方程22230x x t −+−=的两个解，所以()2Δ4430t =−−>，解得22t −<<，且2a b +=，即2b a =−．又()()20f x f x ++−=，令x a =−，则()()0f a f b +=，则由()()()0f a f b f t ++>，得()()01f t f >=，所以1t >．综上，t 的取值范围是()1,2. 故选：D ．二、选择题：本题共3小题，每小题6分，共18分.在每小题给出的四个选项中，有多项符合要求.全部选对的得6分，部分选对的得部分分，有选错的得0分.9. 已知复数()i z a a =−∈R ，且6iz的虚部为3，则（） A. 1a =B.3z= C. ()()213i z +⋅−为纯虚数 D.2i2z ++在复平面内对应的点在第二象限【答案】AC 【解析】【分析】利用向量的除法运算和虚部为3，即可求出1a =，再利用复数乘除运算和模的运算以及复平面内对应点的表示，就能作出选项判断.【详解】由()()()226i i 6i 6i 66i i i i 11a a z a a a a a +===−+−−+++的虚部为3，则2631aa =+，解得1a =，所以选项A 正确.()()()31i 33331i,i 1i 1i 1i 22z z +=−===+−−+，所以3z，所以选项B 错误. 由()()()()213i 3i 13i 10i z +⋅−=−⋅−=−为纯虚数，所以选项C 正确. 由()()()()2i 3i 2i 2i11i 23i3i 3i 22z ++++===++−−+，所以复数2i 2z ++在复平面内对应的点为11,22，位于第一象限，所以选项D 错误，故选：AC.10. 欧拉函数()()*n n ϕ∈N的函数值等于所有不超过正整数n ，且与n 互质的正整数的个数（公约数只有1的两个正整数称为互质整数），例如：()32ϕ=，()42ϕ=，则（） A.()()()468ϕϕϕ⋅=B. 当n 为奇数时，()1n n ϕ=−C. 数列(){}2nϕ为等比数列D. 数列()()23nn ϕϕ的前n 项和小于32 【答案】ACD 【解析】【分析】根据函数新定义即可知()42ϕ=，()62ϕ=，()84ϕ=，可得A 正确，由()968ϕ=≠可得B 错误；易知小于2n 的所有正奇数与2n 均互质,共有12n −个，可得C 正确；同理可得()1323nn ϕ−=×，利用等比数列前n 项和公式即可求得D 正确.【详解】对于A ，因为()42ϕ=，()62ϕ=，()84ϕ=，所以()()()468ϕϕϕ⋅=，故A 正确；对于B ，由于()968ϕ=≠，故B 错误；对于C ，因为小于2n 2n 均互质，且小于2n 的所有正奇数有12n −个，所以()122nn ϕ−=，因此数列(){}2nϕ为等比数列，故C 正确；对于D ，同理小于3n 的所有3的倍数与3n 均不互质，共有13n −个，因此小于3n 的所有与3n 互质的数共有123n −×个，即()1323nn ϕ−=×，所以()()111221223233n n n n nϕϕ−−−==⋅ × ，令11223n n a −=⋅，则122113322213n n n S a a a−=++⋅⋅⋅+=⋅< −，故D 正确，故选：ACD.11. 已知双曲线22:13y E x −=的左､右焦点分别为1F 、2F，过点C 的直线l 与双曲线E 的左､右两支分别交于P 、Q 两点，下列命题正确的有（） A. 当点C 为线段PQ 的中点时，直线lB. 若(1,0)A −，则222QF A QAF ∠=∠ C. 212PF PF PO ⋅> D. 若直线lB ，则11PF QF PB QB +=+ 【答案】BCD 【解析】【分析】对于A 选项，设122(,),(,)P x y Q x y ，代入双曲线，用点差法即可判断；对于B 选项，设00(,)Q x y ，表示出020tan 2y QF A x ∠=−−和020tan 22y QAF x ∠=−−，得出22tan tan 2QF A QAF ∠=∠，再结合22,(0,π)QF A QAF ∠∠∈即可得出结论；对于C 选项，设(,)P P P x y ，其中1P x ≤−，由双曲线方程，得出223(1)P P y x =−，利用两点之间距离公式，分别表示出12PF PF ⋅和2||PO ，通过做差即可得出结论；对于D 选项，根据双曲线的定义，得出1212=F P Q PF F F Q ++，再证出点B 与点2F 关于直线l【详解】选项A ：设122(,),(,)P x y Q x y ，代入双曲线得，221122221313y x y x −= −= ，两式相减得， 121212121()()()()03x x x x y y y y −+−−+=，∵点C 为线段PQ 的中点， ∴1212x x +=，122y y +=，即122x x +=，12y y +∴12121()2()03x x y y −⋅−−=，1212y y kx x −∴==−A 错误；选项B ：设00(,)Q x y ，020tan 2y QF A x ∠=−− ，020tan 1y QAF x ∠=+，002200021tan 221()1y x yQAF y x x ⋅+∴∠==−−−+，22tan tan 2QF A QAF ∴∠=∠，又22,(0,π)QF A QAF ∠∠∈ ，222QF A QAF ∴∠=∠，故B 正确；选项C ：设(,)P P P x y ，其中1P x ≤−，则2213P Py x −=，即223(1)P P y x =−，121P PF x ==+，221P PF x ==−，212(21)(21)41P P P PF PF x x x ∴⋅=+⋅−=−，2222223(1)43P P P P P PO x y x x x =+=+−=− ， 2221243(41)20P P PO PF PF x x ∴−⋅=−−−=−<，212PO PF PF ∴<⋅，故C 正确；选项D ：212PF PF a −= ，122QF QF a −=，122PF PF a ∴=−，122QF a QF =+， 22212122=PF QF PF a a QF PF QF ∴−++++=，∵直线lPQ k =，且过点C ，∴直线l方程为：1202x −=，又∵B ，2(2,0)F ，2BF k ∴2(1PQ BF k k ∴⋅=−，即2BF PQ ⊥，又∵点B 到直线l的距离：1d ，点2F 到直线l的距离：2d ，即12d d =，∴点B 与点2F 关于直线l 对称，22PB QB PF QF ∴+=+，11PF QF PB QB +=+∴，故D 正确；故选：BCD ．【点睛】结论点睛：本题涉及到双曲线中的有关结论：（1）若点00(,)M x y 是双曲线22221x y a b−=上一条弦PQ 的中点，则直线PQ 的斜率2020PQx b k a y =⋅；的（2）若双曲线上有两点P 、Q ，且位于不同两支，则1122PF QF PF QF +=+．三、填空题：本题共3小题，每小题5分，共15分.12. 52)a x−的展开式中的常数项是10，则a =____________. 【答案】2− 【解析】【分析】根据给定条件，求出二项式展开式的通项公式，再确定常数项即得.【详解】二项式52)a x −展开式的通项为55521552C ()()C ,N,5rr r r r rr a T a x r r x−−+=−=−∈≤，由5502r−=，得1r =，于是125C 510T a a =−=−=，所以2a =−. 故答案为：2−13. 中国传世数学著作《九章算术》卷五“商功”主要讲述了以立体问题为主的各种形体体积的计算公式.例如在推导正四棱台（古人称方台）体积公式时，将正四棱台切割成九部分进行求解.下图（1）为俯视图，图（2）为立体切面图.E 对应的是正四棱台中间位置的长方体，,,,B D H F 对应四个三棱柱，A C I G ,,,对应四个四棱锥.若这四个三棱柱的体积之和为12，四个四棱锥的体积之和为4，则该正四棱台的体积为__________.【答案】28 【解析】【分析】令四棱锥的底面边长为a ，高为h ，三棱柱的高为b ，由四个三棱柱的体积之和与四个四棱锥的体积之和，可得23a h =和6abh =，则有2b a =，求出中间长方体的体积，即可得该正四棱台的体积. 【详解】如图，令四棱锥的底面边长为a ，高为h ，三棱柱的高为b ，依题意，四棱锥的体积为2113a h =，即23a h =，三棱柱的体积为132ahb =，即6abh =，因此2,b a =于是长方体的体积22412V b h a h===，所以该正四棱台的体积为1241228++=.故答案为：2814. 随机数表是人们根据需要编制出来的，由0，1，2，3，4，5，6，7，8，9这10个数字组成，表中每一个数都是用随机方法产生的，随机数的产生方法主要有抽签法、抛掷骰子法和计算机生成法．现有甲、乙、丙三位同学合作在一个正二十面体（如图）的各面写上0~9这10个数字（相对的两个面上的数字相同），这样就得到一个产生0~9的随机数的骰子．依次投掷这个骰子，并逐个记下朝上一面的数字，就能按顺序排成一个随机数表，若甲、乙、丙依次投掷一次，按顺序记下三个数，三个数恰好构成等差数列的概率为______．【答案】120##0.05【解析】【分析】甲投1次，记下数字有10种可能，乙投1次也有10种可能；丙投1次也有10种可能，所以甲、乙、丙依次投掷1次，由分步乘法原理可得所有记下数字的总情况数，再列举出等差数列的公差为0，1，2，3，4的所有情况，将公差为1，2，3，4的等差数列中的第1项和第3项的数字交换，分别构成公差为1−，2−，3−，4−的等差数列，可得出构成等差数列的可能情况数，根据古典概率公式计算可得选项. 【详解】甲投1次，记下数字有10种可能，乙投1次也有10种可能；丙投1次也有10种可能，所以甲、乙、丙依次投掷1次，记下数字有1010101000××=种情况，0~9这10个数字中选3个，能构成等差数列的情况如下：公差为0的等差数列有：0，0，0；1，1，1；2，2，2；；9，9，9共10种情况；公差为1的等差数列有：0，1，2；1，2，3；2，3，4；3，4，5；4，5，6；5，6，7；6，7，8；7，8，9共8种情况；公差为2的等差数列有：0，2，4；1，3，5；2，4，6；3，5，7；4，6，8；5，7，9共6种情况；公差为3的等差数列有：0，3，6；1，4，7；2，5，8；3，6，9共4种情况；公差为4的等差数列有：0，4，8；1，5，9共2种情况；公差为1，2，3，4的等差数列中的第1项和第3项的数字交换，分别构成公差为1−，2−，3−，4−的等差数列，所以构成等差数列的可能情况有()102864250+×+++=种，所以若甲、乙、丙依次投掷一次，按顺序记下三个数，三个数恰好构成等差数列的概率为501100020=. 故答案为：120【点睛】关键点点睛：本题解决的关键是要细致地分类讨论，做到不重不漏列举出所有构成等差数列的情况，从而得解.四、解答题：本题共5小题，共77分.解答应写出必要的文字说明、证明过程及演算步骤.15. 记ABC 的内角A ，B ，C 的对边分别为a ，b ，c ，已知2125b ac =，π3B =.（1）求sin sin A C +的值；（2）若ABC ABC 的周长.【答案】（1（2）5 【解析】【分析】（1）利用余弦定理可得32a cb +=，再由正弦定理计算可得结果；（2）由三角形面积公式可得2b =，可得332a cb +==，所以ABC 的周长为5. 【小问1详解】根据题意由余弦定理可得2222cos b a c ac B =+−，又π3B =可得π1cos cos 32B ==，即可得222a c b ac +=+，所以()222225933124a c b ac b b b +++×，可得32a cb +=，由正弦定理可得33sin sin sin 22A CB +==；【小问2详解】易知2115sin 2212ABC S ac B b ==× 解得24b =，即2b =；由（1）中32a cb +=可得3ac +=，所以ABC 的周长为5a b c ++=.16. 已知函数()()2exf x xax a =−−，a ∈R .（1）当2a >−时，研究()f x 的单调性；（2）若0a ≥，当1x x =时，函数()f x 有极大值m ；当2x x =时，()f x 有极小值n ，求m n −的取值范围.【答案】（1）()f x ()2,a −上单调递减，在()(),2,,a ∞∞−−+上单调递增；（2）)24e ,− +∞【解析】【分析】（1）对函数求导并结合2a >−即可判断出()f x 的单调性；（2）根据（1）中结论可得()24e e a m n a a −−=++，构造函数()g a 并求导得出其单调性即可求得m n−的取值范围.小问1详解】易知函数()f x 的定义域为x ∈R ，则()()()e2xf x x x a ′=+−，又因为2a >−，所以当()2,x a ∈−时，()0f x ′<，当(),2x ∞∈−−或(),x a ∈+∞时，()0f x ′>；因此可得()f x 在()2,a −上单调递减，在()(),2,,a ∞∞−−+上单调递增；【小问2详解】若0a ≥，由（1）可知()f x 在2x =−处取得极大值，在x a =处取得极小值，所以()()()2e ,e 24aa m f n f a a −=+=−=−=，即()24e e a m n a a −−=++；在【设函数()()2,04e eag a a a a −=+≥+，则()()21e e 0a g a a −′+=+>，所以()g a 在[)0,+∞上单调递增，所以()()204e g a g −≥=，即m n −的取值范围为)24e ,− +∞. 17. 如图，四棱锥P ABCD −中，PC ⊥底面ABCD ，底面ABCD 为菱形，60BAD ∠= ，2AB PC ==，,M N 分别为,PD PB 的中点.（1）证明：MN ⊥平面PAC ；（2）求二面角C PB D −−的正弦值. 【答案】（1）答案见解析；（2 【解析】【分析】（1）利用线面垂直的性质可得PC BD ⊥，再利用线面垂直的判定定理以及性质定理即可得出结论；（2）建立空间直角坐标系，利用空间向量求得两平面的法向量，利用法向量即可得出二面角C PB D −−的正弦值. 【小问1详解】连接,AC BD ，如下图所示：由PC ⊥底面ABCD ，BD ⊂底面ABCD ，可得PC BD ⊥，又因为底面ABCD 为菱形，所以AC BD ⊥，显然,,PC AC C PC AC ∩=⊂平面PAC ，因此BD ⊥平面PAC ，又,M N 分别为,PD PB 的中点，所以MN BD ⁄⁄；即MN ⊥平面PAC ；【小问2详解】记,AC BD 的交点为O ，取AP 的中点为E ，连接OE ，易知OE PC ⁄⁄，由（1）可得OE ⊥平面ABCD ；又,AC BD ⊂平面ABCD ，所以,OE AC OE BD ⊥⊥，因为AC BD ⊥，所以,,AC BD OE 两两垂直；以O 为坐标原点，以,,AC BD OE 所在直线分别为x 轴，y 轴，z 轴建立如图所示空间直角坐标系；又60BAD ∠= ，2AB PC ==，所以2AC BD =；则()()()(),2,0,1,0,0,1,0C P B D −，所以())()0,0,2,2,0,2,0CP PB BD =−=−；设平面CPB 的一个法向量为mm��⃗=(xx 1,yy 1,zz 1)，可得11112020m CP z m PBy z ⋅== ⋅=+−=，解得10z =，令11x =，则1y =；即()1,m=为平面CPB 的一个法向量，设平面PBD 的一个法向量为nn�⃗=(xx ,yy ,zz )，的可得22222020n BD y n PBy z ⋅=−= ⋅=+−=，解得20y =，令22x =，则2z =即(n =为平面PBD 的一个法向量，可得cos ,m nmn m n ⋅==，设二面角C PB D −−为θ，可得sin θ 所以二面角C PB D −−. 18. 已知椭圆2222:1(0)x y C a b a b+=>>的右焦点为F ，右顶点为A ，上顶点为B ，点O 为坐标原点，线段OA 的中点恰好为F ，点F 到直线AB．（1）求C 的方程；（2）设点E 在直线4x =上，过F 作EF 的垂线交椭圆C 于,M N 两点．记MOE △与NOE △面积分别为12,S S ，求12S S 的值．【答案】（1）22143x y +=（2）1 【解析】【分析】（1）根据题意可得2a c =，又点F 到直线AB列式计算求得,a b ；（2）设线段MN 中点()00,G x y ，利用点差法可得OG OE k k =，,,O G E 三点共线，即直线OE 过线段MN 的中点G ，得解. 【小问1详解】设FF (cc ,0)，则c=，由线段OA 的中点恰好为F ，得2ac =，所以2224a ab −=，整理得2234b a =，的由()(),0,0,A a B b 得直线AB 方程为0bx ay ab +−=，所以点F 到直线AB所以2b a ，椭圆C 的方程为22143x y +=．【小问2详解】设()()()11224,,,,,E t M x y N x y ，线段MN 的中点()00,G x y ，则121200,22x x y y x y ++==．由（1）知FF (1,0)，直线EF 的斜率13tk =，当0t ≠时，直线MN 的斜率122123y y k t x x −=−=−．因为点,M N 在椭圆C 上，所以22112222143143x y x y += += ，两式相减，整理得()()()()12121212043x x x x y y y y +−+−+=，又12120,22x x y y x y ++==，所以004y t x =，直线OG 的斜率为4OG t k =，因为直线OE 的斜率为4OE tk =，所以,,O G E 三点共线，即直线OE 过线段MN 的中点G ，当0t =时，直线OE 也过线段MN 的中点G ，所以,M N 到直线OE 的距离相等，即OME 与ONE 等底等高．所以1122,1S S S S ==．【点睛】思路点睛：设()()()11224,,,,,E t M x y N x y ，线段MN 的中点()00,G x y ，利用点差法可得OG OE k k =，,,O G E 三点共线，即线段MN 的中点G 在直线OE 上，得解.19. 某制药公司研制了一款针对某种病毒的新疫苗.该病毒一般通过病鼠与白鼠之间的接触传染，现有n 只白鼠，每只白鼠在接触病鼠后被感染的概率为12，被感染的白鼠数用随机变量X 表示，假设每只白鼠是否被感染之间相互独立（1）若()()595P X P X===，求数学期望()E X ；（2）接种疫苗后的白鼠被病鼠感染的概率为p ，现有两个不同的研究团队理论研究发现概率p 与参数()01θθ<<的取值有关.团队A 提出函数模型为()22ln 13p θθ=+−，团队B 提出函数模型为()11e 2p θ−=−.现将100只接种疫苗后的白鼠分成10组，每组10只，进行实验，随机变量()1,2,,10i X i =⋅⋅⋅表示第i 组被感染的白鼠数，将随机变量()1,2,,10i X i =⋅⋅⋅的实验结果()1,2,,10i x i =⋅⋅⋅绘制成频数分布图，如图所示.（i ）试写出事件“11221010,,,X x X x X x ==⋅⋅⋅=”发生的概率表达式（用p 表示，组合数不必计算）；（ⅱ）在统计学中，若参数0θθ=时使得概率()11221010,,,P X x X x X x ==⋅⋅⋅=最大，称0θ是θ的最大似然估计.根据这一原理和团队A ，B 提出的函数模型，判断哪个团队的函数模型可以求出θ的最大似然估计，并求出最大似然估计.参考数据：3ln0.40552≈.【答案】（1）50 （2）（i ）()()()()()33227512342510101010C C C C 1p p −；（ⅱ）团队B 可以求出θ的最大似然估计，0ln 2θ= 【解析】【分析】（1）由题意可得1~,2X B n，再根据()()595P XP X ===求解即可；（2）（i ）设11221010,,","A X x X x X x ==== ，依题意得()P A ()()()()()332987649223344661010101010C 1C 1C 1C 1C 1p p p p p p p p p p =−−−−−，化简即可；（ⅱ）记1323324210101010()ln()()()()25ln 75ln(1)g p C C C C p p =++−，求导分析单调性可得最大值，分别在团体A ，B 中提出函数模型即可得答案．【小问1详解】由题知，随机变量X 服从二项分布，1~,2X B n，由()()595P X P X===，即555595955595111111C 1C 1C 1222222n n n n n n n −−−− −=−=−，得100n =，所以()50E X np ==；【小问2详解】（i ）A =“11221010,,,X x X x X x ==⋅⋅⋅=”， ()P A()()()()()332987649223344661010101010C 1C 1C 1C 1C 1p p p p p p p p p p =−−−−−，所以()()()()()()33227512342510101010C C C C 1P A p p −；（ii ）记()()()()()()33221023410101010ln C C C C 25ln 75ln 1g p p p =++−，则()()25752510011pg p p p p p −′=−=−−，当104p <<时，()0g p ′>，()g p 单调递增；第21页/共21页当114p <<时，()0g p ′<，()g p 单调递减；当14p =时，()g p 取得最大值，即P 取得最大值，在团队A 提出的函数模型()22ln 13p θθ=+−，()01θ<<中，记函数()()212ln 13f x x x =+−，()01x <<，()()21144431331x x f x x x x −−+′−++，当102x <<时，()10f x ′>，()1f x 单调递增；当112x <<时，()10f x ′<，()1f x 单调递减，当12x =时，()1f x 取得最大值3113ln ln 0.40552642 −<≈，则θ不可以估计，在团体B 提出的函数模型()11e 2p θ−=−中，记函数()()211e 2x x f −=−，()2f x 单调递增，令()214f x =，解得ln 2x =，则团队B 可以求出θ的最大似然估计，且0ln 2θ=是θ的最大似然估计.（1）根据题中条件确定随机变量的可能取值；（2）求出随机变量所有可能取值对应的概率，即可得出分布列；（3）根据期望的概念，结合分布列，即可得出期望（在计算时，要注意随机变量是否服从特殊的分布，如超几何分布或二项分布等，可结合其对应的概率计算公式及期望计算公式，简化计算）．。

广东省八校2024-2025学年高三上学期9月联合检测地理试题

7.为了能将陆地上制造的沉管管节在海底一次性安装到位建设方需要利用的地理信息技术是（）
A.地理信息系统B.北斗卫星导航系统
C.遥感技术D.现实虚拟技术
8.西人丁岛隧道出入口上方设有多条横梁组成的结构，其目的是（）
A.辅助提升行车安全B.降低海水灌入风险
C.节省工程物料成本D.巩固隧道支撑结构
海雾是指海上或沿海地区上空低层大气.中出现大量微小水滴而形成的一种天气田2019年12月22—24日渤海沿海出现一次大雾天气。下图为22日20时的海平面等压线分钟图。据此完成下面小题。
15.随深度增加，该采样点不同土层的变化特征是（）
A.颜色变浅B.质地增粗C.有机质增加D.空隙变大
16.野外考察发现，该采样点表层有机质含量低，其主要影响因素是（）
A.地形B.生物C.成土母质D.气候
二、非选择题：本题共3小题，共52分。
17.阅读图文材料，完成下列要求。
近年来粤港澳大湾区积极推动农业产业聚集发展。“万田农谷”现代农业产业园位于广州增城区新塘镇万田水库旁。该园区借助湾区“总部经济”资源，积极承接珠三角农业资源，全力打造集新能源建设、种植培育、农产品加工、都市休闲、农旅于一体的现代化农业产业园和大湾区都市农业科创基地。园区以光伏生态农业项目为试点，通过在大棚及各建筑表面铺设光伏电板利用光罷发电，充分满足园区生产及周边地区用电需求。园区的建设和投产，促进城乡融合发展。下图为万田农谷园区的区域图。
5.1975——2020年黄河“几字湾”沙漠化空间变化幅度最大的是（）
A.轻度沙漠化B.中度沙漠化
C.重度沙漠化D.极重度ቤተ መጻሕፍቲ ባይዱ漠化
6.推测1975——2000年,黄河“几字湾”河段（）
A.洪涝灾害增多B.河道横向变宽

1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。

(下转第2版)■主管、主办山西省外语教师协会 ■编辑出版学英语报社 ■国内统一刊号 CN14-0702/(F) ■第9期完形填空作者由养一只鹦鹉的经历告诉我们：如果我们的内心不能释然，那么生命就失去了存在的意义。

1. B 。

根据下文fly 提示可知，它竟然不会飞。

2. D 。

根据上文which made him very upset 可知，今天天气很好，于是我就把它放到了一棵树的树枝上，想让它开心。

3. C 。

根据上文它茫然不知所措可知，它来回走动，就是不扇动翅膀飞翔。

4. A 。

有一天，在树枝上它使劲地拍打着翅膀。

5. D 。

根据上文语境可知D 项正确。

6. A 。

由下文freedom 可知，它一直在等待着时机成熟，以便能挣脱牢笼。

7. B 。

直到第二天晚上它才回来，但是一直呆在远处，不让我抓住。

8. C 。

我用食物引诱它，但是没有成功。

于是我就把鸟笼子藏起来，以免让它产生回来就意味着被锁进笼子的感觉。

9. B 。

参见第8题解析。

10. A 。

下文就是我对它许诺的事情。

11. C 。

根据第一段weather 及逻辑推理可知，从此以后，只要天气晴朗，我就会放它出去。

12. D 。

根据本段最后一句died 可知，它从附近的鸽子身上传染了一种疾病。

13. C 。

根据句中的虚拟语气及逻辑含义可知，如果当初我一直把它关在笼子里不让它外出飞行，那么它现在也不至于死掉。

14. A 。

从作者对人生的态度及下一句What sense is there in being a bird if you can’t fly 可知，如果你是一只鸟却不能飞，那有什么意义呢。

据此可推知，作者认为一个人的生命质量比生命长短更重要。

15. B 。

根据文章大意可知，如果我们无法使自己轻松自在，那么生命就失去了存在的意义。

语法填空一位父亲拍摄影片记录女儿的成长。

16. a 。

此处需要填不定冠词a 表泛指。

17. his 。

根据句子结构可知，此处需要填形容词性物主代词his 。

18. h as changed/has been changing 。

根据句子语法结构可知，此处需要填谓语动词。

由于动作发生在过去并且持续到现在，故填has changed 或has been changing 。

19. to keep 。

根据句子结构可知，此处需要填动词不定式表目的。

20. w hose 。

whose 引导定语从句。

先行词是film ，先行词与关系词之间有从属关系，故填whose 。

21. as 。

此处应填as 。

feature sb as 为固定搭配，意为“以……为特色”。

22. showing 。

根据句子结构可知，此处需要填现在分词。

23. but 。

此处需要填but 表转折。

24. surprised 。

由于主语是人，故此处填surprised ，表示“感到惊讶的”。

25. I n 。

此处需要填介词In 与后面的meantime 搭配。

阅读理解A为了防止孩子逃学，巴西的一个城市让孩子穿上一种植入了芯片的T 恤。

26. C 。

推理判断题。

文章第一段说明，这种定位芯片可以防止孩子逃学，故C 项正确。

27. A 。

细节理解题。

从文章倒数第三段可知，如果孩子旷课三次，父母应该向学校说明孩子缺课的原因。

28. D 。

细节理解题。

从文章最后一段The T-shirts can be washed and ironedwithout damaging the chips 可知，这种T 恤可以像其他衣服那样正常清洗。

29. D 。

推理判断题。

从文章倒数第二段可知，放在芯片上面的这段话是在说明教育的重要性。

30. B 。

推理判断题。

综合分析全文内容可知，本文属科技新闻报道类文章。

B数码时代，那些胶卷相机看起来已经过时，但如今还有人依旧对这些旧物恋恋不舍。

31. D 。

词义理解题。

根据第四段可知，已经过时的柯达胶卷依旧能够拍摄出美丽生动的照片，而这些照片是一些智能工具无法比拟的。

据此可知，smart tools 指的是数码相机。

32. B 。

细节理解题。

从第五段a technical dinosaur 可知，对大部分美国人来说，胶卷已经是过时的产品。

33. B 。

综合理解题。

从倒数第三段可知，Robert 必须驱车前往550公里以外的冲洗中心去冲洗照片。

34. D 。

推理判断题。

从最后两段可知，那些真实生动的照片对Robert 而言，有非常重要的意义。

35. A 。

主旨大意题。

Robert 对已经过时的胶片相机情有独钟，为了拍摄美丽的风景依旧乐此不疲。

C最近，研究人员成功地利用电脑技术使四肢瘫痪的人实现了他们“活动肢体”的愿望。

36. B 。

细节理解题。

从第一段第二句及文章主要内容可知，此处是指四肢瘫痪的人。

37. B 。

综合理解题。

从第二段第二句可知，BrainGate2目的是帮助瘫痪者重获活动肢体的能力。

38. A 。

综合理解题。

从第四段可知，传感器检测到大脑的活动之后，把信息传输到一台电脑里，电脑里的软件把这信息转化成指令，引导机器人手臂活动。

39. D 。

推理判断题。

从倒数第二段可知，Leigh Hochberg 对这项技术比较满意。

40. A 。

推理判断题。

从最后一句可知，此项技术还需要进一步的完善。

总共进行了约200次的试验，成功率最高为66%(132次)，故D 项错误。

D文章主要介绍了是否可以把老虎等一些凶猛动物当宠物来养。

41. C 。

推理判断题。

从最后一段But Felicia isn’t concerned 可知，她不介意。

据此推知，她将继续养Will 。

42. D 。

推理判断题。

根据文章But that can have the opposite effect…until one day, Baby Luv bit her so deeply that she had to go to the hospital 可知，养这种动物具有危险性。

43. A 。

综合理解题。

结合倒数第三段As a result, many owners end up neglecting or even abandoning their animals 可知，当人们不要这些奇异的宠物后，就会抛弃它们。

44. B 。

细节理解题。

根据文章For some, these pets are status symbols — a way to make a parade of wealth, etc 可知，养这种异国宠物是他们社会身份的象征，一种炫耀方式。

45. C 。

标题归纳题。

本文主要讨论是否可以把老虎等一些凶猛动物当宠物来养，以及为什么美国人热衷于饲养这种异国宠物的原因。

信息匹配46. A 。

题中trembles 与A 项the shakes…stop movement 相匹配。

47. C 。

题中skin…pregnancy…breastfeeding 与C 项skin…I’m feeding my baby 相匹配。

48. B 。

题中a pair of gloves and avoid being blown by cold wind 与B 项Rheumatoid Arthritis (风湿性关节炎)… red hand joints 相匹配。

49. F 。

题中asthma patients 与F 项asthma 相匹配。

50. D 。

题中sunlight factor…If it reduces in sunlight 与D 项in the sun it reduces slightly 相匹配。

基础写作One possible version:Mr. Johnson,According to the arrangements by our school, I will be responsible for your activities during your stay in Shenzhen. On the morning of December 8, the students from the two sides will have a gathering so that they can get to know about each other’s school life, culture, customs and so on. At noon, the students from the two sides will have their lunch together and your students will learn to make Chinese dumplings etc. After that, the students are to pay a visit to the city of Shenzhen together for the purpose of learning about the great achievements made in the past thirty years. If you have any opinions about the schedule, please don’t hesitate to let me know.Yours,Li Hua读写任务One possible version:Faced with a life-or-death struggle, two frogs took different attitudes to it: the fat one failed to change his bad luck because he gave up halfway while the skinny one succeeded because of his persistence.As far as I’m concerned, I had similar experience last year. I used to be the top student in my class and was very proud of myself. I was especially good at math and regarded as the “Little Math Master” in my class. However, one day, I failed one important math exam. You can imagine how depressed I was then at the bad news. So I hated to talk about it with anyone, including my parents and my friends. As a result, I fell ill because of bad mood and lack of appetite for meals. Soon this was discovered and corrected by my math teacher, Mr. Wu. He talked with me for hours, encouraging me to study harder to get out of trouble. Now I am still good at math thanks to his help.I learned from this experience that one should face troubles bravely and try to believe in himself. Since then it has been my philosophy of life thatwell instead of only giving out orders or punishing us.In my opinion, school rules should be made to benefit most students. Under the rules, they can study and play well enough. In a word, rules are made to benefit them, not to harm them.第12期完形填空在前进的道路上，如果遇到了挫折或障碍，不要灰心丧气，要相信自己一定能行。

外研高三广东9-12期 解答案析

广东省深圳市富源学校2023-2024学年高三上学期9月调研考试英语试题

广东省大湾区2025届高三上学期9月统一调研考试数学试题(解析版)

广东省八校2024-2025学年高三上学期9月联合检测英语试题

2024-2025学年外研版(2024)第二册地理上册月考试卷727

2024-2025学年(外研版)九年级英语上学期第一次月考试卷(广东卷)(含解析)

广东四校联考2023-2024学年高三上学期9月月考物理试题及参考答案

专题04 选词填空(含解析)-(2025年)外研版(三起)英语四年级上册期中专项练习

2024-2025学年外研版八年级科学上册阶段测试试卷918

广东省深圳市罗湖区2024-2025学年高三上学期9月质量检测试题 地理(含答案)

专题08 匹配题(含解析)-(2025年)外研版(三起)英语五年级上册期中专项练习

2024届广东金太阳高三上学期9月联考日语试卷及答案

2023-2024学年广东省广州市高三上学期9月月考物理质量检测试题(含答案)

2024高三九省联考英语试卷及答案解析

广东省深圳2024-2025学年高三上学期9月月考试题 物理含答案

外研版英语高考试卷及答案指导

2024-2025学年外研版(2019)九年级物理上册阶段测试试卷240

2025届广东罗湖区高三9月质检化学答案

广东八校2025届高三上学期9月联合检测数学试卷(解析版)

广东省八校2024-2025学年高三上学期9月联合检测地理试题

外研高三广东9-12期解答案析

广东省深圳市罗湖区2024-2025学年高三上学期9月质量检测试题地理(含答案)

广东省深圳2024-2025学年高三上学期9月月考试题物理含答案