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重庆市巴蜀中学2020-2021学年高二(下)期末数学试题

重庆市巴蜀中学2020-2021学年高二(下)期末数学试题
重庆市巴蜀中学2020-2021学年高二(下)期末数学试题

高二英语下学期期末综合复习(含答案)

Ⅰ. Complete the following sentences according to the Chinese given. 1. In any game, (合作) is more important than individual skill. 2. The bus came to an (突然的) stop, making some passengers fall off their seats. 3. As most of us know, an object will (加速) at a certain rate while falling. 4. If there were no (规则), some sports would be very dangerous. 5. We should (倡导) an economical lifestyle to reduce the growing waste pollution. 6. Some problems will only get worse if we leave them (忽视). 7. The economy is (繁荣) in China, creating many job opportunities. 8. The volunteers do (值得) our special respect and admiration. 9. It is wise to (寻求) a second opinion before making an important decision. 10. Good health care should not be a (特权), but a right enjoyed by everyone. 11. The great (缺点) to living near an airport is the

重庆巴蜀中学

重庆巴蜀中学 关于开展校本课程总结表彰及等级评定的通知各位校长、各位老师: 为了调动学校广大教职工积极参与新课程改革,开发校本课程,促进学校办出特色,促进教师专业发展;也为了整理、总结学校开展校本课程建设的成果,不断提高教师、学校实施校本课程建设的能力,经学校研究,决定自2011年起,每年进行一次校本课程建设总结表彰。现将《巴蜀中学校本课程评价方案》(见附一)公布于此,并对校本课程等级申报、评定工作,做如下部署,请各位老师参照执行。 1、凡是参与过高2011级高一(下)选修课程开设(具体课程名称及指导教师姓名,见附三)的教师及课程;凡是参与过高2013级高一(下)选修课程开设(具体课程名称及指导教师姓名,见附四)的教师及课程,因相关资料在学校已有存档,故相关课程教师,可以不再提交资料及填写申报表,学校组织专家组根据学校已有资料进行评审。 2、学校除高2011级、高2013级外,在其他年级开设过选修课的老师,或附 3、附4有遗漏、错误的教师,请将开课的相关材料进行整理,于7月29日以前将纸质材料,课程等级申报表(见附二)交到本部行政楼二楼课改处。或发往bashu_zhang@https://www.wendangku.net/doc/4911695687.html,。逾期不予受理申报申请。 重庆巴蜀中学 2011年7月21日

附一: 巴蜀中学校本课程评价方案 (试行) 一、指导思想及评价原则 落实学校教育理念。校本课程开设要符合学校“教育以人为本,校长以教师为本,教师以学生为本”的教育理念,发现和发展学生的潜能,促进学生全面发展和个性成长,引领教师多元发展。 贯穿学校德育主线。校本课程必须体现学校的“善为根、雅为骨、志为魂”的育人理念,校本课程评价必须依托“公正诚朴”的校训,提升学生对“善雅志”的感悟,培养紧跟时代主旋律的合格中学生。 彰显区域发展特色。校本课程在“131”校本课程体系内,教师可以选择自己认为合适的任何内容来设计课程,但课程的选择要体现学校的办学特色及学校所在区域的政治、经济、文化、社会等各方面实际情况,体现区域发展特色。 把握时代发展脉搏。校本课程的提出和教学内容设计必须符合时代发展的特征,扎根经济、政治、文化和社会的丰厚土壤,尽量体现经济发展的方向、政治民主法治建设进程、文化发展的趋势和社会发展的热点。总之,校本课程要引领时代发展潮流。 坚持科学发展思路。校本课程是在国家课程基础上的拓展,学生能否实现知识、能力、视野的拓展,能否实现情感、态度、价值观的感悟是评价校本课程实施成效的重要内容。因此,校本课程的开发必须与国家课程相一致,与学生身心特点相适应,与学生的兴趣爱好相一致,帮助学生认识科学规律、接受人文熏陶。 二、评价策略

重庆市巴蜀中学高二上学期期末考试数学(理)试题

重庆市巴蜀中学高二上期末考试 数学(理科)试题 一、选择题:本大题共12个小题,每小题5分,共60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. 已知函数在处取得极值,则() A. B. C. D. 2. 某空间几何体的三视图如图所示,则该几何体的体积是() A. B. C. D. 3. 命题“,均有”的否定形式是() A. ,均有 B. ,使得 C. ,均有 D. ,使得 4. “”是“”的() A. 充分不必要条件 B. 必要不充分条件 C. 充要条件 D. 既不充分也不必要条件 5. 我国南宋时期的数学家秦九韶是普州(现四川省安岳县)人,秦九韶在其所著的《数书九章》中提出的多项式求值的秦九韶算法,至今仍是比较先进的算法.如图所示的程序框图给出了利用秦九韶算法求某多项式值的一例,则输出的的值为()

A. B. C. D. 6. 函数的导函数的图像如图所示,则的图像可能是() A. B. C. D. 7. 设、是两条不同的直线,、是两个不同的平面,下列命题中错误的() A. 若,,,则 B. 若,,,则 C. 若,,则 D. 若,,,则 8. 已知函数在区间上单调递增,则的取值范围是() A. B. C. D. 9. 如图所示程序框图输出的结果是,则判断狂内应填的条件是()

A. B. C. D. 10. 已知点为椭圆上第一象限上的任意一点,点,分别为椭圆的右顶点和上顶点,直线与交于点,直线与轴交于点,则的值为() A.2 B. C. 3 D. 11. 已知点在正方体的线段上,则最小值为() A. B. C.0.3 D. 12. 已知中心在原点的椭圆与双曲线有公共焦点,左、右焦点分别为,,且两条曲线在第一象限的交点为,若是以为底边的等腰三角形.椭圆与双曲线的离心率分别为,,则的取值范围是() A. B. C. D. 二、填空题(每题5分,满分20分,将答案填在答题纸上) 13. 若双曲线的离心率为,则__________. 14. 已知抛物线,焦点为,为平面上的一定点,为抛物线上的一动点,则的最小值为__________. 15. 三棱锥中,垂直平面,,,,则该三棱锥外接球的表面积为__________. 16. 已知函数,,若对于任意的,,不等式恒成立,则实数的取值范围为__________.

2020-2021学年高二下学期期末考试英语试题

I 听力测试20分(共20小题;每小题1分,满分20分) 第一节(共5题;每小题1分,满分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,并标在试卷的相应位置。听完每段对话后,你有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。 1. Where are the man's sunglasses probably? A. In his bag. B. On the table. C. In the woman's bag. 2. How should the woman pay? A. In cash. B. By cheque. C. By credit card. 3. What does the woman say about her calculator? A. It is broken. B. It has been lost. C. It has been lent to someone else. 4. Where are the speakers? A. On a bus. B. On a plane. C. On a train. 5. What does the man suggest the woman do? A. Enjoy the view. B. Bring her family here. C. Share the photos with her family.

第二节(共15题;每小题1分,满分15分) 听下面5段对话或独白。每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。听每段对话或独白前,你将有时间阅读每个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。每段对话或独白读两遍。 听第6段材料,回答6至8题 6. What does the man do now? A. A cleaner. B. A repairman. C. An electrician. 7. How does the man feel about his job? A. Exhausted. B. Satisfied. C. Frustrated. 8. Who will the man pay medical benefits for? A. His wife. B. His son. C. Himself. 听第7段材料,回答8至11题 9. What is the probable relationship between the speakers? A. Doctor and patient. B. Nurse and patient. C. Doctor and nurse. 10. Why does the man change his appointment?

2020届 重庆巴蜀中学高三适应性月考 卷(二)数学(理)试题(解析版)

2020届重庆巴蜀中学高三适应性月考卷(二)数学(理)试 题 一、单选题 1.已知α是第二象限角,且sin 4 5 α=,则cos α=( ) A . 45 B .45 - C .35 D .35 - 【答案】D 【解析】通过同角三角函数的平方关系,结合α是第二象限角,cos α为负值,直接代入解得答案. 【详解】 ∵α是第二象限角,且sin 45 α= , 可得3cos 5α==-, 故选:D . 【点睛】 本题考查同角三角函数关系,注意象限角的符号即可,属于基础题. 2.集合A ={x |(x ﹣1)(x ﹣7)≤0},集合B ={x |x =2k +1,k ∈N },则A ∩B =( ) A .{1,7} B .{3,5,7} C .{1,3,5,7} D .{1,2,3,4,5,6,7} 【答案】C 【解析】先求出集合A 与B ,求出两集合的交集即可. 【详解】 ∵集合()(){} {}|=17017|A x x x x x ≤≤≤=﹣﹣, 集合B ={x |x =2k +1,k ∈Z }, ∴A ∩B ={1,3,5,7}, 故选:C . 【点睛】 本题考查集合的运算,此类题目一般比较简单,只需将两集合解出,再进行交并补运算即可求解.

3.向量a =r (1,2),b =r (2,λ),c =r (3,﹣1),且(a b +r r )∥c r ,则实数λ= ( ) A .3 B .﹣3 C .7 D .﹣7 【答案】B 【解析】向量a r ,b r ,计算可得a b +r r ,再由c r 和(a b +r r )∥c r ,代入向量平行的性质 公式计算,即可求解. 【详解】 根据题意, 向量=a r (1,2),=b r (2,λ), 则()=32+a b λ+,r r , c =r (3,﹣1),且(a b +r r )∥c r , 则有()()3132+0λ?--=, 解可得=3λ-, 故选:B . 【点睛】 本题考查平面向量的坐标运算和平行的性质,属于平面向量常考题型. 4.已知随机变量X 服从正态分布N (3,σ2),且P (x ≤1)=0.1,则P (3<X ≤5)=( ) A .0.1 B .0.2 C .0.3 D .0.4 【答案】D 【解析】根据已知随机变量X 服从正态分布N (3,σ2),得到正态分布曲线关于=3x 对称,又根据题目P (x ≤1)=0.1,由对称性可得()50.1P x ≥=,因此得到P (1≤X ≤5)的值,再乘1 2 即为所求. 【详解】 ∵随机变量X 服从正态分布N (3,σ2), ∴正态分布曲线关于=3x 对称, 又P (x ≤1)=0.1, ∴()50.1P x ≥=, ∴()() 510.1235= =0.42 2 P X P X ≤≤-?≤1<=,

2018-2019高二下学期期末考试英语试卷含答案

2019学年高二英语期末试卷 (考试时间:100分钟试卷满分:120分) 第一部分阅读理解(共两节,满分40分) 第一节(共15小题;每小题2分,满分30分) 阅读下列短文,从每题所给的A、B、C和D四个选项中,选出最佳选项。 A Female Scientists Who Have Changed the World There are still hundreds of relatively-unknown women who have changed the world with their research throughout history. Tiera Guinn This 21-year-old scientist hasn’t yet graduated from college, but Tiera Guinn’s already doing rocket science. The MIT(麻省理工学院) senior is helping build a rocket for the United States that could be one of the biggest and most powerful ever made. She’s an aerospac e(航天航空)major with high credit points who also works as a Rocket Structural Design and Analysis Engineer for the Space Launch System that aerospace company Boeing is building for the US. Elizabeth Blackwell Elizabeth Blackwell, who was born in Bristol, England in 1821, was the first woman to graduate from medical school in the United States, became an activist for poor American women’s health found a medical school for women in England. Jane Goodall The most famous primate(灵长目) scientist in history, Jane Goodall was known for her work with chimps (黑猩猩)and as a champion of animal rights. And Goodall wasn’t just working in a lab; she climbe trees and imitated the behavior of chimps in Tanzania to gain their trust and study them in their natural habitat. Rachel Carson voice rose above the rest to become central to American In the 1960s, one environmental scientist’s politics, culture, and foreign policy: Rachel Carson’s. Her article, "Silent Spring,"warned of the dangers of pesticides(农药) and chemicals to humans, plants, and animals, and was a landmark in the (周围的)mental history. nation’s environ

重庆市巴蜀中学2020学年高二数学下学期半期考试试题 理(含解析)

重庆市巴蜀中学2020学年高二数学下学期半期考试试题理(含解析) 第Ⅰ卷(共60分) 一、选择题:本大题共12个小题,每小题5分,共60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.命题,的否定是() A. B. C. D. 【答案】B 【解析】 【分析】 按存在性命题的否定的规则写出即可. 【详解】因命题为“,”,它是存在性命题, 故其否定为:,选B. 【点睛】全称命题的一般形式是:,,其否定为.存在性命题的一般形式是,,其否定为. 2.抛物线上的点到其焦点的距离为() A. 3 B. 4 C. 5 D. 6 【答案】C 【解析】 【分析】 利用焦半径公式可得长度. 【详解】,故选C. 【点睛】如果抛物线的方程为,则抛物线上的点到焦点的距离为. 3.圆形铜钱中间有一个边长为4毫米的正方形小孔,已知铜钱的直径为16毫米,现向该铜钱

上随机地投入一粒米(米的大小忽略不计),那么该粒米落入小孔内的概率为() A. B. C. D. 【答案】A 【解析】 【分析】 算出正方形小孔的面积和铜钱的面积,利用几何概型的概率公式可得所求的概率. 【详解】设为“该粒米落入小孔内”,因为正方形小孔的面积为平方毫米,铜钱的面积为平方毫米,故,故选A. 【点睛】几何概型的概率计算关键在于测度的选取,测度通常是线段的长度、平面区域的面积、几何体的体积等. 4.设,是两条不同的直线,,是两个不同的平面,则下列命题正确的是() A. 若,,则 B. 若,,则 C. 若,,,,则 D. 若,,,则 【答案】D 【解析】 【分析】 对于A,B选项均有可能为线在面内,故错误;对于C选项,根据面面平行判定定理可知其错误;直接由线面平行性质定理可得D正确. 【详解】若,,则有可能在面内,故A错误; 若,,有可能面内,故B错误; 若一平面内两相交直线分别与另一平面平行,则两平面平行,故C错误. 若,,,则由直线与平面平行的性质知,故D正确. 故选D. 【点睛】本题考查的知识点是,判断命题真假,比较综合的考查了空间中直线与平面的位置关系,属于中档题.

人教版2020学年高二英语下学期期末考试试题(新版)新人教版

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