精校Word版答案全---江苏省扬州市江都区大桥高级中学2019_2020学年高二英语上学期期中
江苏省扬州市江都区大桥高级中学2019-2020学年高二4月学情调研地理试题 Word版含答案
大桥高级中学2019—2020学年度第二学期高二学情调研(1)地理试题命题:校审: 2020.4第I卷(选择题)一、单选题区域·是客观存在的,具有一定的范围和界线,有的界线明确,有的具有一定的过渡性或模糊性,如图所示。
据此完成下面小题。
1.下列有关区域含义的说法,错误的是A.区域是指一定的地域空间B.区域之间具有一定的差异性C.区域都具有明确的界线D.区域之间是相互联系的2.与图示区域A的边界类型相同的是①行政区②热量带③干湿区④三江平原⑤吉林省和吉林市A.②③B.①⑤C.③④D.④⑤人均GDP和人均GDP增长率分别是衡量区域经济发展水平和发展速度的重要指标。
下面为近年来五省市人均GDP和人均GDP增长率与全国平均值之比的统计图,图中X轴表示人均GDP增长率与全国平均值之比,Y轴表示人均GDP与全国平均值之比。
各省市括号中的数值为其万元产值能耗,全国平均值为0.74(单位:吨标准煤/万元)。
据此回答下列各题。
3.关于五省市经济发展状况的叙述,正确的是()A.山西经济发展水平高于湖北B.广西经济发展速度低于全国C.湖北经济发展水平高于江苏D.上海经济发展速度低于江苏4.从万元产值能耗看()A.山西最高,应优化工业结构以降低能耗B.广西最低,应承接东部地区高耗能工业C.上海和江苏较低,应大力发展重型工业D.湖北较高,应发展资源密集型工业以降低能耗读我国甲、乙两区域略图,完成下列各题。
5.关于甲、乙两区域河流特征的描述,不正确的是( )A.甲区域以冰雪融水补给为主,乙区域以雨水补给为主B.甲区域以内流河为主,乙区域以外流河为主C.甲区域以春汛为主,乙区域以夏汛为主D.甲区域水系呈向心状,乙区域水系呈放射状6.甲、乙两区域分别盛产棉花和天然橡胶,其共同的区位优势是( )①夏季热量充足②劳动力价格较低③农业科技发达④农业机械化程度高A.①②B.①③C.②③D.③④读“某城市近30年来产业结构变化示意图”完成下面小题。
江苏省扬州市江都区大桥高级中学2019-2020学年高二下学期期中考试英语试题 Word版含答案
大桥高级中学2019-2020学年度第二学期英语期中试卷(考试时间:120分钟总分:150分)一听力(共20小题,每小题1.5分,满分30分)1.What drink did the woman order?A.A Pepsi.B.A Coke.C.A beer.2.How does the man probably feel toward the woman?A.Angry.B.Disappointed.C.Grateful.3.Why does the man have to pack carefully?A.So the woman can clean easily.B.So nothing breaks.C.So they don't forget anything.4.What did the speakers see last night?A.A little but noisy bird.B.A dark-colored bird.C.A bird that flew very quietly.5.Why does the woman need directions?A.She can't read her map.B.She doesn't have a map anymore.C.The museum is not in the map.听第6段材料,回答第6、,7题。
6.What's special about the people riding bikes?A.They are all men.B.They have no clothes on.C.They wear the same clothes.7.Why does the woman want to go somewhere else?A.She's too cold.B.She hates what she saw.C.She wants to get something to eat.听第7段材料,回答第8至10题。
江苏省扬州市江都区大桥高级中学2019-2020学年高一下学期学情检测(二)英语试题 Word版含答案
大桥高级中学2019-2020学年度第二学期高一学情检测(二)英语试卷命题人:审核人: 2020,5考试时间120分钟,试卷满分150分。
第一部分:听力(共两节,满分30分)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。
1.Where was the boy hiding?A. Behind the door.B. Under the bed.C. In the closet.2.What will the man do during winter break?A. Go skiing.B. Stay home with his dog.C. Visit some overseas friends.3.Where did the man get the vegetables?A. From his school.B. From the community garden.C. From the grocery store.4.What kind of food does the woman want the man to buy?A. Junk food.B. Cheap food.C. Good quality food.5.What size does the woman need?A. Large.B. Medium.C. Small.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。
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。
听完每段对话或独白前,你将有时间阅读各个小题;每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
每段对话或独白读二遍。
江苏省扬州市江都区大桥高级中学2019-2020学年高二4月学情调研化学试题 Word版含解析
大桥高级中学2019—2020学年度第二学期高二学情调研化学试题1、本试卷分第I卷(选择题)和第II卷(非选择题)两部分,全卷满分100分,考试时间90分钟。
2、本试卷可能用到的相对原子质量 C:12 H :1 O:16 N:14 Mn-55第I卷(选择题共40分)单项选择题(本题包括10小题,每小题2分,共20分。
每小题只有一个....选项符合题意)1.正丁烷与异丁烷互为同分异构体的依据是A. 具有相似的物理性质B. 具有相似的化学性质C. 具有不同的分子结构D. 分子式相同,但分子内碳原子的连接方式不同【答案】D【解析】【分析】同分异构体是分子式相同,但结构不同的化合物.【详解】A、因二者的结构不同,为不同的物质,则物理性质不同,故A错误;B、因结构不同,则性质不同,但都属于烷烃,化学性质相似,但不能作为互为同分异构体的依据,故B错误;C、二者的结构不同,都属于烷烃,C原子连接的原子,具有相似的空间结构,但不能作为互为同分异构体的依据,故C错误;D、正丁烷与异丁烷的为分子式相同的烷烃,分子内碳原子的连接方式不同,则结构不同,二者互为同分异构体,故D正确;答案选D。
2.下列有关化学用语表示正确的是A. CH2F2的电子式:B. 羟基的电子式:C. 对硝基甲苯的结构简式:D. 异丙醇的结构简式:CH3CH2CH2OH 【答案】B【解析】【详解】A.CH2F2的电子式中,氟原子最外层达到8电子稳定结构,其正确的电子式为:,故A错误;B.羟基是电中性基团,氧原子与氢原子以1对共用电子对连接,电子式为,故B正确;C.对硝基甲苯的结构简式为:,故C错误;D.异丙醇的结构简式:CH3CH(OH)CH3,故D错误;故答案B。
3.下列有关工业生产叙述正确的是A. 工业上通常使用电解法制备金属钠、镁、铝等B. 合成氨工业中,将NH3及时液化分离有利于加快反应速率C. 硫酸工业中,采用常压条件的原因是此条件下催化剂活性最高D. 电解精炼铜时,将粗铜与电源的负极相连【答案】A【解析】【详解】A、钠、镁、铝是活泼的金属,用电解法制备,选A;B、合成氨工业中,将氨气及时液化分离,浓度减小,速率减小,是为了提高反应物的转化率,不选B;C、制取硫酸的工业中,常采用常压条件的原因是常压条件想二氧化硫的转化率已经很大,加压转化率增大程度不大,所以不选C;D、电解精炼铜时,将粗铜与电源的正极相连,做电解池的阳极,失去电子发生氧化反应,不选D。
江苏省扬州市江都区大桥高级中学2019-2020学年高一下学期学情检测(二)语文试题答案
大桥高级中学2019—2020学年度第二学期高一学情调查(二)语文审校:2020.5.13一、现代文阅读(36分)(一)论述类文本阅读(本题共3小题,9分)阅读下面的文字,完成1~3题。
①继唱歌选秀、户外真人秀等类型之后,喜剧正在成为当前电视综艺节目的重要组成部分。
近年来,《笑傲江湖》《欢乐喜剧人》等数十档喜剧类电视综艺节目,轮番上演,成为荧屏新风景。
②喜剧以夸张的手法、巧妙的结构、诙谐的台词及对人物性格的刻画,引起人们对丑和滑稽的嘲笑,对正常的人生和美好的理想予以肯定。
基于描写对象和手法的不同,喜剧可分为讽刺喜剧、抒情喜剧、荒诞喜剧和闹剧等样式。
在中国艺术舞台上,从王实甫的《西厢记》、关汉卿的《救风尘》到当代的《抓壮丁暗恋桃花源》等,一直都蕴藏着喜剧的基因,散落着喜剧的种子。
随着艺术表现形式的多样化,除了传统的舞台喜剧,还出现了喜剧电影、情景喜刷、喜剧型广播剧、喜剧类综艺节目等。
③现代人生活、工作压力前所未有,喜剧让人或捧腹或开怀,为生存于压力之下的人们提供了宣泄情绪、释放压力的出口,尤其在当下社会转型期,喜剧更为社会转型期的中国观众提供了心灵按摩和温情安慰。
在话剧市场不景气的情况下,“开心麻花”系列却一票难求;在电影市场激烈的竞争中,《泰囧》等喜剧片票房大卖;在电视剧产量供过于求的情况下,都市轻喜剧也总能取得不错的收视成绩,这些都反映出观众对喜剧的强烈需求。
④虽然电视荧屏上出现了大量喜剧类节目,可在喜剧研究专家蔡体良看来,近些年,中国喜剧的发展却并不乐观,“真正意义上的喜剧精品,屈指可数”。
蔡体良的话指出了当前中国喜剧的最大问题:量多质不高。
另外,随着收视竞争的加剧,一些喜剧节目“喜”得有些变了味。
比如,有的节目拿残疾人的生理缺陷开涮,有的节目用低俗的段子吸引眼球,还有的节目用社会阴暗面刺激观众的神经。
所有这些,也能让观众笑,但观众获得了欢笑,却被玷污了心灵。
⑤观众需要笑,但不是肤浅的笑。
喜剧的本质不能被低俗小品所取代,也不是插科打诨所能表达的。
江苏省扬州市江都区大桥高级中学2019-2020学年高二下学期期中考试英语试题+Word版含答案
大桥高级中学2019-2020学年度第二学期英语期中试卷(考试时间:120分钟总分:150分)一听力(共20小题,每小题1.5分,满分30分)1.What drink did the woman order?A.A Pepsi.B.A Coke.C.A beer.2.How does the man probably feel toward the woman?A.Angry.B.Disappointed.C.Grateful.3.Why does the man have to pack carefully?A.So the woman can clean easily.B.So nothing breaks.C.So they don't forget anything.4.What did the speakers see last night?A.A little but noisy bird.B.A dark-colored bird.C.A bird that flew very quietly.5.Why does the woman need directions?A.She can't read her map.B.She doesn't have a map anymore.C.The museum is not in the map.听第6段材料,回答第6、,7题。
6.What's special about the people riding bikes?A.They are all men.B.They have no clothes on.C.They wear the same clothes.7.Why does the woman want to go somewhere else?A.She's too cold.B.She hates what she saw.C.She wants to get something to eat.听第7段材料,回答第8至10题。
江苏省扬州市江都区大桥高级中学2019-2020学年高一下学期学情调研(一)数学答案
17. 【答案】(1)
;(2)
【详解】(1) 边上的高过
,因为 边上的高所在的直线与 所在的直线
互
相垂直,故其斜率为 3,方程为:
(2) 由题 点坐标为
,
的中点
是
的一条中位线,所以
,
,
其斜率为:
,所以 的斜率为
所以直线 的方程为:
化简可得:
.
18. 证明:(1) 因为点 E 、 F 分别是棱 BC 、 BD 的中点,所以 EF 是 BCD 的中位线,
1 2
,
2 2
,则
sin
A
2 sin B
2 2
,1
,即
sin
A
2 2
,1
由于在三角形中, A 0, ,由正弦函数的图像可得: A [ , 3 ] ;故答案选 D
44
12. 【详解】由
,及余弦定理
得
,因为 ,所以
,
利用正弦定理可得
,得
得 因为 所以
,
为三角形内角,所以
,所以
,
,
因为
1. C
2.【详解】对于答案 A、B、D 分别是公理 1、3、2;答案 C 不是公理,故选 C 【点睛】本题考查了点、线、面的公理,熟悉公理是解题关键,属于基础题.
3. 由题将直线化成斜截式,可得答案.
【详解】由题将直线的化简可得
,所以斜率为 故选 D
【点睛】本题考查了直线的方程,一般式化为斜截式,属于基础题.
4. 【详解】
边 最短.由正弦定理得
.故选 A.
5. C 6. A 7. C
8. A
9. 【详解】依题意可得
,
在三角形 中,由余弦定理可得:
江苏省扬州市江都区大桥高级中学2019-2020学年高一上学期期中考试语文试题+Word版含答案
大桥高级中学2019—2020学年度第一学期高一期中考试语文 (教师版)审核:刘剑峰、舒少清、童阜兰 2019.11.10一、现代文阅读(32分)(一)论述类文本阅读(本题共3小题,9分)【命题人:吕蒙蒙】阅读下面的文字,完成1-3题①谈田园诗离不开陶渊明。
陶的田园诗对农村事物和恬静的环境给予了由衷的赞颂,朴实自然毫无喧饰,充分表达了诗人的真实感受,同时将农村那宁静平和的生活表现得如仙境般优美,令人神往。
诗人对田园生活的赞美未直接说出,而是通过草屋茅舍、榆柳桃李、远村炊烟、鸡鸣狗吠、山气飞鸟的白描,流露出对田园生活的由衷喜爱和深切依恋,将诗人那纯净的心地和平静的心境,与简朴恬静的田园风光交融为了一体。
②在诗歌创作中,情、景、理三者交融至关重要,而情又是最重要的,离开情的景就没有了生气,离开情的理更是“淡乎寡味”的空理。
而陶诗总能透过人人可见之物,普普通通之事,表达高于世人之情,写出人所未必悟出之理,又呈现新的意境,给人以美感。
他善于寓情于理,把自己对人生、对现实的深刻认识形象化,把诗情与哲理、与景物紧密结合起来,因而给人以清新自然、毫不枯燥的感觉,可谓发乎事,源于景,缘乎情,而以理为统摄。
③陶渊明的田园诗不是对现实生活作单层次的、平面的再现,而是在情、事、景、理的统中构成了多层次的艺术整体,给人们以丰富而深刻的审美享受。
如《饮酒·其五》:结庐在人境,而无车马喧。
问君何能尔?心远地自偏。
采菊东篱下,悠然见南山。
山气日タ佳,飞鸟相与还。
此中有真意,欲辨已忘言。
④在这首诗里,“悠然”采菊,南山和飞鸟,还有对“真意”和“心远地自偏”的感叉,概括起来就是外在的事与景和内在的情与理的统一,构成深远浑厚的意境。
⑤陶渊明的田园诗能够千古流传,与它语言的质朴关系很大。
陶渊明独特的生活经历,朴素的农村生活和平淡的田园景色,要求尽可能采用近似“田家语”的补素的语言和白描手法,从而形成田园诗平淡自然的风格,达到“一语天然万古新,豪华落尽见真淳”。
江苏省扬州市江都区大桥高级中学2019-2020学年高二上学期期中考试(11.3)化学(必修)试题 Word版含答案
大桥高级中学2019—2020学年度第一学期高二期中考试高二化学(必修)本试卷分第一卷(选择题)和第二卷(非选择题)两部分,满分100分 时间 70分钟可能用到的相对原子质量:H -1,C -12, O -16,Cl -35.5,N -14,Na -23,Al -27, Fe -56, Cu -64,Ag -108 Ca -40 Zn -65一、单项选择题:每题只有1个选项是符合要求的(本部分26题,每题3分,共78分) 1.江苏省将大力实施“清水蓝天”工程。
下列不利于...“清水蓝天”工程实施的是 A .迅速提升我省的家用汽车保有量B .加强城市生活污水脱氮除磷处理,遏制水体富营养化C .积极推广太阳能、风能、地热能及水能等的使用,减少化石燃料的使用D .大力实施矿物燃料“脱硫、脱硝技术”,减少硫的氧化物和氮的氧化物污染 2. 下列变化属于化学变化的是A .碘的升华B .氮气液化C .钢铁锈蚀D .汽油挥发3. 18 8O 作为“标记原子”被广泛应用于化学、医药学等领域,下列关于18 8O 说法正确的是A. 质量数为18B. 核电荷数为10C. 中子数为8D. 核外电子数为104.公安部发布法令:从2011年5月1日起,醉酒驾驶机动车将一律入刑。
下列关于酒精的叙述错误..的是 A .化学名称为乙醇 B .易挥发 C .常用作有机溶剂 D .不能燃烧 5. 下列气体可用右图所示装置收集的是( )A. CO 2B. CH 4C. NH 3D. H 2 6.下列化学用语或模型表示正确的是 A .Cl -离子的结构示意图:2818+ B .CH 4分子的比例模型:C .CCl 4的电子式:ClCl:C Cl:Cl ⋅⋅⋅⋅ D .氮气的结构式:N=N7.设N A 为阿伏加德罗常数的值,下列说法正确的是 A .常温常压下,71 g Cl 2含有2 N A 个Cl 原子B .把40 g NaOH 固体溶于1 L 水中,所得溶液中NaOH 的物质的量浓度为1 mol·L -1C .物质的量浓度为1 mol/L 的K 2SO 4溶液中,含2 N A 个K +D .常温常压下,11.2 L N 2中含有的分子数为0.5 N A 8.下列分散系能产生丁达尔效应的是 ( )A .硫酸铜溶液B .稀硫酸C .溴水D .氢氧化铁胶体9.在pH =1的无色溶液中能大量共存的离子组是A .NH 4+、Mg 2+、SO 42-、Cl -B .Ba 2+、K +、OH -、NO 3-C .Al 3+、Cu 2+、SO 42-、Cl-D .Na +、Ca 2+、Cl -、CO 32-10.K 2FeO 4是一种高效水处理剂。
江苏省扬州市江都区大桥高级中学2019-2020学年高二下学期期中考试数学试题 Word版含解析
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数学试卷
一:选择题(本大题共12小题,每小题5分,共60分)
1.若复数z 满足()341z i +-=,则z
的虚部是( ) A. -2
B. 4
C. -3
D. 3
【答案】B
【解析】
【分析】
由()341z i +-=得到24z i =-+即可 【详解】因为()341z i +-=,所以24z i =-+,所以z 的虚部是4
故选:B
【点睛】本题考查的是复数的计算及其概念,较简单.
2.若回归直线的方程为ˆ2 1.5y
x =-,则变量x 增加一个单位时 ( ) A. y 平均增加1.5个单位
B. y 平均增加2个单位
C. y 平均减少1.5个单位
D. y 平均减少2个单位
【答案】C 【解析】
【分析】 根据回归直线方程 1.52ˆy
x =-+的斜率为负,可得出正确选项. 【详解】由于回归直线方程为 1.52ˆy
x =-+,其斜率为 1.5-,故变量x 增加一个单位时,y 平均减少1.5个单位.故选C.
【点睛】本小题主要考查对回归直线方程系数的理解,考查直线的斜率,属于基础题.
3.若()cos f x x =,则2f π⎛⎫'
⎪⎝⎭的值为( ) A . 0
B. 1
C. 1-
D. 12 【答案】C
【解析】。
江苏省扬州市江都区大桥高级中学2019-2020学年高一化学6月学情检测试题[含答案]
A.电解法
B.铝热反应法
C.氢气或 CO 还原法
D.热分解法
6.下列对化学知识的应用不正确的是
A.鉴别棉纺线与羊毛线可用灼烧法鉴别
B.糖尿病人病情可用氢氧化铜悬浊液检验 C.沾有油脂的器皿可用热的氢氧化钠溶液洗 D.做过银镜反应的试管可用稀硫酸洗 7.关于化学能与热能的叙述,正确的是 A.在化学反应中,化学能只可以转化为热能
(2)强于(1 分)
2HNO3 + Na2CO3 = CO2 ↑+ H2O +2NaNO3(其它合理答案也可)(2 分)
(3)
(2 分) (4)NH3 + H2O + CO2 = NH4HCO3(每空 2 分)
27.(共 6 分) (1)饱和碳酸钠溶液(1 分) 溶解乙醇、中和乙酸、降低乙酸乙酯的溶解度(2 分)
23、下列说法正确的是 A.苯分子中含有三个碳碳双键 C.葡萄糖能够发生银镜反应
B. 将乙酸和乙醇混合加热制取乙酸乙酯 D. 糖类、油脂、蛋白质均能发生水解
二、非选择题(共 31 分) 24.(5 分)下面列出了几组物质,请将物质的合适组号填写在空格上。 A.金刚石与石墨; B.淀粉和纤维素; C.氕 与 氘 D. 甲烷与戊烷;
下图表示产生 H2 的体积(V)与时间(t)的关系正确的是( )
V
V
V
V
a b
ba
a b
a b
t A
t B
t C
A
A
15.下列物质能与溴水发生化学反应并使溴水褪色的是
t D A
A.乙醇
B.乙烯
C.苯
D.甲烷
16.下列物质中,只含共价键的是
A.H2O
B.NaOH
江苏省扬州市江都区大桥高级中学2019-2020学年高二下学期4月学情调研数学试题 Word版含解析
大桥高级中学2019—2020学年度第二学期高二学情调研数学试题 命题: 校审:一、选择题:(5 ′×12) 1.复数32iz i-+=+的共轭复数是( ) A. 2i + B. 2i -C. 1i -+D. 1i --【答案】D 【解析】 试题分析:()()()()3235512225i i i iz i i i i -+--+-+====-+++-,1z i =--,故选D. 考点:复数的运算与复数相关的概念.2.若f (x )3x f ′(-1)的值为( ) A. 0 B. 13-C. 3D.13【答案】D 【解析】 【分析】求函数f (x )的导数,代值计算即可. 【详解】解:f (x )3x∴()2'313f x x -=, ()'113f ∴-=. 故选:D.【点睛】本题考查导数的运算,属于基础题.3.设f (x )=x 2(2-x ),则f (x )的单调增区间是( ) A. 40,3⎛⎫ ⎪⎝⎭B. 4,3⎛⎫+∞⎪⎝⎭C. (-∞,0)D. 4(,0),3⎛⎫-∞⋃+∞ ⎪⎝⎭【答案】A 【解析】 【分析】先求出函数的导函数,令导函数大于0,解不等式求出即可. 【详解】f (x )=x 2(2﹣x ), ∴f ′(x )=x (4﹣3x ), 令f ′(x )>0,解得:0<x <43, 故选:A .【点睛】本题考查了利用导数求函数的单调性,是一道基础题.4.复数(),z x yi x y =+∈R 满足条件42z i z -=+,则24x y +的最小值为( ) A. 2 B. 4C. 42D. 16【答案】C 【解析】 【分析】 将z x yi =+代入42z i z -=+中,可得()42x y i x yi+-=++,则()()222242x y x y +-=++,整理得到23x y +=,进而利用均值定理求解即可【详解】由42z i z -=+,得()42x y i x yi +-=++, 所以()()222242x y x y +-=++,即23x y +=,所以2232422222242x y x y x y ++=+≥==,当且仅当322x y ==时取等号, 故24x y +的最小值为2【点睛】本题考查复数的模的应用,考查利用均值定理求最值5.抛掷2颗骰子,所得点数之和ξ是一个随机变量,则P (ξ≤4)等于 ( ) A.16B.13C.12D.23【答案】A 【解析】 点数之和ξ可能取值有2,3,4,,12,所以P(ξ≤4)1231(2)(3)(4)3636366p p p ξξξ=+=+==++=,选A. 6.已知21nx x ⎛⎫ ⎪⎝⎭+的二项展开式的各项系数和为32,则二项展开式中x 的系数为( ) A. 5 B. 10 C. 20 D. 40【答案】B 【解析】 【分析】首先根据二项展开式的各项系数和012232n n n n n n C C C C +++==,求得5n =,再根据二项展开式的通项为211()()r rn rr n T C x x-+=,求得2r,再求二项展开式中x 的系数.【详解】因为二项展开式的各项系数和012232n n n n n n C C C C +++==,所以5n =,又二项展开式的通项为211()()r rn rr n T C x x-+==3r r n n C x -,351r -=,2r所以二项展开式中x 的系数为2510C =.答案选择B .【点睛】本题考查二项式展开系数、通项等公式,属于基础题.7.设(1+x )3+(1+x )4+(1+x )5+…+(1+x )50=a 0+a 1x +a 2x 2+a 3x 3+…+a 50x 50,则a 3的值是( )A. 450CB. 2350CC. 351CD. 451C【答案】D 【解析】 【分析】由题意可得a 3的值是x 3的系数,而x 3的系数为 C 33+C 43+C 53+…+C 503=C 44+C 43+C 53+…+C 503利用二项式系数的性质求得结果.【详解】解:由题意可得a 3的值是x 3的系数,而x 3的系数为 C 33+C 43+C 53+…+C 503=C 44+C 43+C 53+…+C 503=C 514, 故选:D .【点睛】本题考查二项式系数的性质的应用,求展开式中某项的系数,求出x3的系数为C33+C43+C53+…+C503,是解题的关键.8.已知函数f(x)=x3+ax2+(a+6)x+1有极大值和极小值,则实数a的取值范围是()A. (-1,2)B. (-3,6)C. (-∞,-3)∪(6,+∞)D. (-∞,-3]∪[6,+∞)【答案】C【解析】【分析】由题意可知:导数有两个不等的实根,利用二次方程根的判别式可解决.【详解】解:由于f(x)=x3+ax2+(a+6)x+1,有f′(x)=3x2+2ax+(a+6).若f(x)有极大值和极小值,则△=4a2﹣12(a+6)>0,从而有a>6或a<﹣3,故选:C.【点睛】本题主要考查已知函数存在极值求参,涉及导数与极值的关系,属于基础题.9.已知函数f(x)=x3-ax-1,若f(x)在(-1,1)上单调递减,则a的取值范围为( )A. a≥3B. a>3C. a≤3D. a<3【答案】A【解析】∵f(x)=x3−ax−1,∴f′(x)=3x2−a,要使f(x)在(−1,1)上单调递减,则f′(x)⩽0在x∈(−1,1)上恒成立,则3x2−a⩽0,即a⩾3x2,在x∈(−1,1)上恒成立,在x∈(−1,1)上,3x2<3,即a⩾3,本题选择A 选项.10.函数()f x 在定义域R 内可导,若()()2f x f x =-,且当(),1x ∈-∞时,()()10x f x '-<,设()0a f =,12b f ⎛⎫=⎪⎝⎭,()3c f =,则( ) A. a b c <<B. c b a <<C. c a b <<D.b c a <<【答案】C 【解析】 【分析】先由()()2f x f x =-,确定()()31f f =-,再由(),1x ∈-∞时,()()10x f x '-<得到()f x 在(),1-∞上单调递增,进而可判断出结果.【详解】因为()()2f x f x =-,所以()()31f f =-; 又当(),1x ∈-∞时,()()10x f x '-<,所以()0f x '>, 即函数()f x 在(),1-∞上单调递增; 所以()()()03121f f f f ⎛⎫=<< ⎝-⎪⎭,即c a b <<. 故选:C.【点睛】本题主要考查根据函数单调性与周期性比较大小,涉及导数的方法判断函数单调性,属于常考题型.11.楼道里有12盏灯,为了节约用电,需关掉3盏不相邻的灯,则关灯方案有( ) A. 72种 B. 84种 C. 120种 D. 165种【答案】D 【解析】 【分析】本题看做模型问题,相当于在在9盏亮灯的10个空隙中插入3个不亮的灯,问题得以解决. 【详解】解:当3个中任意2个都不相邻时,把此问题当作一个排队模型在9盏亮灯的10个空隙中插入3个不亮的灯有310C =120种.当有两个相邻时,把2个相邻的捆绑在一起,和上面的做法一样,9盏亮灯的10个空隙中插入2个有21045c =,共有,120+45=165种. 故选:D.【点睛】本题考查了组合中构造模型问题,考查用插空法解决不相邻问题,属于基础题. 12.从1,2,3,4,5中任取2个不同的数,事件A =“取到的2个数之和为偶数”,事件B =“取到两个数均为偶数”,则()|P B A =( ) A.18B.14C.25D.12【答案】B 【解析】 【分析】先求得()P A 和()P AB 的值,然后利用条件概率计算公式,计算出所求的概率.【详解】依题意()22322542105C C P A C +===,()22251=10C P AB C =,故()|P B A =()()1110245P AB P A ==.故选B. 【点睛】本小题主要考查条件概型的计算,考查运算求解能力,属于基础题.二、填空题:(5 ′×4) 13.求i +i 2+…+2019i =_________【答案】1- 【解析】 【分析】利用等比数列的求和公式、复数的周期性即可得出.【详解】解:z =()201911i i i --=()5044311i i i i⎡⎤-⎢⎥⎣⎦-=(1)11i i i +=--. 故答案为:1-.【点睛】本题考查了等比数列的求和公式、复数的周期性,考查了推理能力与计算能力,属于基础题.14.甲乙两名教师和三名学生参加毕业拍照合影,排成一排,甲老师在正中间且甲乙教师相邻的排法共有______种.(用数字作答) 【答案】12 【解析】 【分析】由排列、组合及简单计数问题得:甲乙两名教师和三名学生参加毕业拍照合影,排成一排,甲老师在正中间且甲乙教师相邻的排法共有132312C A =,得解.【详解】解:甲乙两名教师和三名学生参加毕业拍照合影,排成一排,甲老师在正中间且甲乙教师相邻的排法共有132312C A =,故答案为12.【点睛】本题考查了排列、组合及简单计数问题,属中档题. 15.若2019(12)x -=a 0+ a 1x+…+a 2019x 2019(x∈R ),则12a +222a +…+201920192a 的值为_______.【答案】1- 【解析】试题分析:令等式中得;再令,则,所以,故应填.考点:二项式定理与赋值法的综合运用.16.设方程33x x k -=有3个不等的实根,则实数k 的取值范围是__________. 【答案】(2,2)- 【解析】设3()3f x x x =-,对函数求导,2()333(1)(1)f x x x x '=-=+-,令()0f x '>,即3(1)(1)0x x +->,得1x <-或1x >,令()0f x '<,得11x -<<, ∴函数()f x 在(,1)-∞-和(1,)+∞上是增函数,在(1,1)-上是减函数,且(1)2f-=,(1)2f=-,∴可作出()f x的大致图像,如图所示:要使方程33x x k-=有3个不等的实根,则y k=与()f x的图像有3个交点,∴22k-<<,即常数k取值范围是(2,2)-.三、解答题:17.已知复数z满足|z|2,2z的虚部为2,z所对应的点在第一象限,(1)求z;(2)若z,z2,z-z2在复平面上对应的点分别为A,B,C,求cos∠ABC. 【答案】(1) z=1+i.(2)255【解析】分析:(1)设z=x+yi(x,y∈R),根据题意得到x,y的方程组,即得z.(2)先求z,z2,z-z2在复平面上对应的点,再利用向量的夹角公式求cos∠ABC.详解:(1)设z=x+yi(x,y∈R).∵|z|2,=∴x2+y2=2. ①又z2=(x+yi)2=x2-y2+2xyi,∴2xy=2,∴xy=1. ②由①②可1,-1,1-1.x xy y==⎧⎧⎨⎨==⎩⎩解得或∴z=1+i或z=-1-i.又x>0,y>0,∴z=1+i. (2)z 2=(1+i)2=2i, z-z 2=1+i-2i=1-i. 如图所示,∴A(1,1),B(0,2),C(1,-1),()()BA 1,1,BC 1,3,∴=-=-∴cos ∠ABC BA?BC 25|BA||BC|21025====⨯点睛:(1)本题主要考查复数的求法和复数的几何意义,考查向量的夹角,意在考查学生对这些知识的掌握水平. (2) 设a =11(,)x y ,b =22(,)x y ,θ为向量a 与b 的夹角,则121222221122cos x y x yθ=+⋅+.18.已知函数f(x)=e x -ax -1,其中e 是自然对数的底数,实数a 是常数.(1)设a =e ,求函数f(x)的图象在点(1,f(1))处的切线方程; (2)讨论函数f(x)的单调性.【答案】(1)1y =- ; (2)答案见解析. 【解析】 分析】(1)求函数f(x)的导数,可写出对应切线方程 (2) 对函数f(x)的导数值的正负分类,讨论单调性. 【详解】(1)∵a=e ,∴f(x)=e x-ex -1, ∴f′(x)=e x -e ,f(1)=-1,f′(1)=0.∴当a =e 时,函数f(x)的图象在点(1,f(1))处的切线方程为y =-1. (2)∵f(x)=e x -ax -1,∴f′(x)=e x -a.当a≤0时,f′(x)>0,故f(x)在R 上单调递增; 当a >0时,由f′(x)=e x-a =0,得x =ln a ,∴当x <ln a 时,f′(x)<ln a e a - =0,当x >ln a 时,f′(x)>ln a e a - =0, ∴f(x)在(-∞,ln a)上单调递减,在(ln a ,+∞)上单调递增. 综上,当a≤0时,f(x)在R 上单调递增;当a >0时,∴f(x)在(-∞,ln a)上单调递减,在(ln a ,+∞)上单调递增. 【点睛】(1)理解导数的几何意义,可得出切线斜率,求解(2)导数的应用:利用导数值的正负可以判断函数的单调性,能够合理的分类解决问题.19.已知在42nx x 的展开式中,前三项的系数成等差数列. (1)求n 的值;(2)求展开式中的有理项.【答案】(1)8n = (2)00441812T C x x ⎛⎫=⋅⋅= ⎪⎝⎭,4415813528T C x x ⎛⎫=⋅⋅= ⎪⎝⎭,882982112256T C x x -⎛⎫=⋅⋅= ⎪⎝⎭. 【解析】 【分析】(1)由于412nx x ⎫展开式中的前三项系数为:0n C ,112n C ,214n C ,这三数成等差数列⇒2×112n C =0214n n C C +,从而可求得n ;(2)由(1)求得n =8,利用4812x x ⎫⎪⎪⎭展开式的通项公式T r +1=8r C •812rx -⎛⎫ ⎪⎝⎭•12r⎛⎫ ⎪⎝⎭•4r1x -⎛⎫ ⎪⎝⎭=12r ⎛⎫ ⎪⎝⎭•8rC •1634r x -,由1634r -为有理数,求得r ,从而可求得展开式中的有理项.【详解】解:(1)因为12nx x ⎫⎪⎪⎭展开式中的前三项系数为:0nC ,112n C ,214n C ,这三数成等差数列,所以2×112n C =0214n n C C +,即n 2﹣9n +8=0, ∴n =8或n =1(舍去), ∴n =8;(2)812x x ⎫⎪⎪⎭展开式的通项公式T r +1=8r C •812rx -⎛⎫ ⎪⎝⎭•12r ⎛⎫ ⎪⎝⎭•4r1x -⎛⎫ ⎪⎝⎭=12r⎛⎫ ⎪⎝⎭•8r C •1634r x -,∴要使T r +1项为有理项,则16﹣3r =4k , ∴r =0,4,8,∴有理项为:00441812T C x x ⎛⎫=⋅⋅= ⎪⎝⎭,4415813528T C x x ⎛⎫=⋅⋅= ⎪⎝⎭,882982112256T C x x -⎛⎫=⋅⋅= ⎪⎝⎭. 【点睛】本题考查二项式定理的应用与等差数列的性质,关键是掌握好二项展开式的通项公式,属于中档题.20.某大学志愿者协会有6名男同学,4名女同学,在这10名同学中,3名同学来自数学学院,其余7名同学来自物理﹑化学等其他互不相同的七个学院,现从这10名同学中随机选取3名同学,到希望小学进行支教活动(每位同学被选到的可能性相同). (1)求选出的3名同学是来自互不相同学院的概率;(2)设X 为选出的3名同学中女同学的人数,求随机变量X 的分布列. 【答案】(1)4960(2)详见解析 【解析】 【分析】(1)利用排列组合求出所有基本事件个数及选出的3名同学是来自互不相同学院的基本事件个数,代入古典概型概率公式求出即可(2)随机变量X 的所有可能值为0,1,2,3,346310(=k)=(0,1,2,3)k k C C P X k C -=,列出随机变量X 的分布列即可. 【详解】(1)设“选出的3名同学是来自互不相同的学院”为事件A ,则()1203373731049.60C C C C P A C +== (2)随机变量X 的所有可能值为0,1,2,3.()()03124646331010110,1,62C C C C P X P X C C ======()()21304646331010312,3,1030C C C C P X P X C C ======X ∴的分布列为X 0123P16 12 310 130【点睛】本题主要考查了古典概型及其概率公式,互斥事件,离散型随机变量的分布列,属于中档题.21.已知函数()ln f x x x =. (Ⅰ)求()f x 的最小值;(Ⅱ)若对所有1x 都有()1f x ax -,求实数a 的取值范围. 【答案】(Ⅰ)最小值1e-;(Ⅱ)(],1-∞【解析】 【分析】(Ⅰ)由导数的应用,研究函数的单调性,再求其最值, (Ⅱ)构造函数()1ln g x x x=+,由导数的应用求函数的最值即可得解. 【详解】解:(Ⅰ)()f x 的定义域为()0,∞+,()f x 的导数()1ln f x x '=+. 令()0f x '>, 解得1e x >;令()0f x <,解得10e x <<.从而()f x 在10,e ⎛⎫⎪⎝⎭单调递减,在1,e ⎛⎫+∞ ⎪⎝⎭单调递增. 所以,当1ex =时,()f x 取得最小值1e -.(Ⅱ)依题意,得()1f x ax -在[)1,+∞上恒成立,即不等式1ln a x x+对于[)1,x ∈+∞恒成立.令()1ln g x x x=+, 则()211111g x x x x x ⎛⎫'=-=- ⎪⎝⎭. 当1x >时,因为()1110g x x x ⎛⎫'=-> ⎪⎝⎭, 故()g x 是()1,+∞上的增函数,所以()g x 的最小值是()11g =, 从而a 的取值范围是(],1-∞.【点睛】本题考查了利用导数求函数的最值及利用导数研究不等式,属中档题.22.甲、乙两名射手互不影响地进行射击训练,根据以往的数据统计,他们射击成绩的分布列如下表所示. 射手甲 射手乙环数 8 9 10 环数 8 910 概率 131313概率131216(1)若甲射手共有5发子弹,一旦命中10环就停止射击,求他剩余3发子弹的概率; (2)若甲、乙两名射手各射击2次,求4次射击中恰有3次命中10环的概率; (3)若甲、乙两名射手各射击1次,记所得的环数之和为ξ,求ξ的概率分布. 【答案】(1)29 (2)7162(3)见解析. 【解析】 【分析】(1)事件A :射手甲剩下3颗子弹,则第一次不能命中第二次必须命中,按独立事件的概率计算即可得出结果.(2)若甲乙两射手各射击两次,四次射击中恰有三次命中10环分两类:甲命中1次10环,乙命中两次10环和甲命中2次10环,乙命中1次10环,分别求概率再求和;(3)ξ的取值分别为16,17,18,19,20,利用独立事件的概率求法分别求ξ取每个值的概率即可.【详解】解:(1)记事件A :射手甲剩下3颗子弹,∴P (A )=212339⨯= (2)记事件C :甲命中1次10环,乙命中两次10环,事件D :甲命中2次10环,乙命中1次10环,则四次射击中恰有三次命中10环为事件C +D ,∴P (C +D )=2212212222211151336366C C C C +⎛⎫⎛⎫⨯⨯⨯⨯⨯⨯⨯⨯ ⎪ ⎪⎝⎭⎝⎭=7162. (3)ξ的取值分别为16,17,18,19,20,(9分)P (ξ=16)=131139⨯=,P (ξ=17)=11115323318⨯+⨯=, P (ξ=18)=111111636323318⨯+⨯+⨯=,P (ξ=19)=11114363218⨯+⨯=,P (ξ=20)=1113618⨯=∴ξ的分布列为 ξ16 17 18 19 20 P19518618418118【点睛】本题考查独立事件、互斥事件的概率、离散型随机变量的分布列、期望等知识,考查学生的计算能力,属于中档题.。
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江苏省扬州市江都区大桥高级中学2019-2020学年上学期期中试题高二英语(考试时间:120分钟总分:150分)第一部分听力(共两节,每小题1.5分,满分30分)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,并标在试卷的相应位置。
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。
第二部分:阅读理解(共15小题;每小题2分,共30分)AWant to explore new cultures, meet new people and do something worthwhile at the same time? You can do all the three with Global Development Association(GDA).Whatever stage of life you’re at, wherever you go and whatever project you do in GDA, you’ll create positive c hanges in a poor and remote community(社区).We work with volunteers of all ages and backgrounds. Most of our volunteers are aged 17-24. Now we need volunteer managers aged 25-75. They are extremely important in the safe and effective running of our programmes. We have such roles as project managers, mountain leaders, and communication officers.Depending on which role you choose, you could help to increase a community’s access to safe drinking water, or help to protect valuable local cultures. You might also design an adventure challenge to train young volunteers.Not only will you help our young volunteers to develop personally, you’ll also learn new skills and increase your cultural awareness. You may have chances to meet new people who’ll become your lifelong friends.This summer we have both 4-week and 7-week programmes:GDA ensures that volunteers work with community members and local project partners where our help is needed. All our projects aim to promote the development of poor and remote communities.There is no other chance like a GDA programme. Join us as a volunteer manager to develop your own skills while bringing benefits to the communities.Find out more about joining a GDA programme:Website:Email:humanresources@21. What is the main responsibility of volunteer managers?A. To seek local partners.B. To take in young volunteers.C. To raise cultural awareness.D. To carry out programmes.A. EgyptB. South AfricaC. KenyaD. AlgeriaA. explore new culturesB. protect the environmentC. help communities in needD. gain corporate benefitsBFor Western designers, China and its rich culture have long been an inspiration for Western creative.“It’s no secret that China has always been a source(来源)of inspiration for designers,” says Amanda Hill, chief creative officer at A+E Networks, a global media company and home to some of the biggest fashion shows.Earlier this year, the China Through A Looking Glass exhibition in New York exhibited 140 pieces of China-inspired fashionable clothing alongside Chinese works of art, with the aim of exploring the influence of Chinese aesthetics(美学)on Western fashion and how China has fueled the fashionable imagination for centuries. The exhibition had record attendance, showing that there is huge interest in Chinese influences.“China is impossible to overlook,” says Hill. “Chinese models are the faces of beauty andfashion campaigns that sell dreams to women all over the world, which means Chinese women are not just consumers of fashion — they are central to its movement.” Of course, only are today’s top Western designers being influenced by China — some of the best designers of contemporary fashion are themselves Chinese. “V era Wang, Alexander Wang, Jason Wu are taking on Galiano, Albaz, Marc Jacobs-and beating them hands down in design and sales,” adds Hill.For Hill, it is impossible not to talk about China as the leading player when discussing fashion. “The most famous designers are Chinese, so are the models, and so are the consumers," she says. “China is no longer just another market; in many senses it has become the market. If you talk about fashion today, you are talking about China —its influences, its direction, its breathtaking clothes, and how young designers and models are finally acknowledging that in many ways.”24.What can we learn about the exhibition in New York?A. It promoted the sales of artworks.B. It showed ancient Chinese clothes.C. It attracted a large number of visitors.D. It aimed to introduce Chinese models.25.What does Hill say about Chinese women?A. They do business all over the world.B. They start many fashion campaigns.C. They admire super models.D. They are setting the fashion.26.What do the underlined words “taking on” in paragraph 4 mean?A. learning fromB. looking down onC. competing againstD. working with27.What can be a suitable title for the text?A. Young Models Selling Dreams to the WorldB. A Chinese Art Exhibition Held in New YorkC. Differences Between Eastern and Western AestheticsD. Chinese Culture Fueling International Fashion TrendsCShark attacks not only disturb beach activities, but can affect associated tourist industries. Shark nets are a common solution to preventing shark attacks on beaches, but they cause dangers to sea ecosystems.Seeking a cost-effective way to monitor beach safety over large areas, we have developed a system called Shark Spotter. It combines artificial intelligence (AI), computing power, and drone (无人机) technology to identify and warn lifesavers to sharks near swimmers. The project is a cooperation between the University of Technology Sydney and The Ripper Group, which is pioneering the use of drones—called “Westpac Little Ripper Lifesavers”—in the search and rescue movement in Australia.Shark Spotter can detect sharks and other potential threats using real-time aerial imagery. The system analyses video from a camera attached to a drone to monitor beaches for sharks, send warnings, and conduct rescues. Developed with techniques known as “deep learning”, the Shark Spotter system receives imagery from the drone camera and attempts to identify all objects in the scene. Once certain objects are detected, they are put into one of 16 categories: shark, whale, dolphin, rays, different types of boats, surfers, and swimmers.If a shark is detected, Shark Spotter provides both a visual sign on the computer screen and an audible warning to the operator. The operator confirms the warning and sends text messages from the Shark Spotter system to the Surf Life Savers for further action. In an emergency, the drone is equipped with a lifesaving flotation pod (漂浮仓) together with an electronic shark repellent (驱逐装置) that can be dropped into the water in cases where swimmers are in severe trouble, trapped in a rip, or if there are sharks close by.In January 2018, the Westpac Little Ripper Lifesavers was used to rescue two young swimmers caught in a rip at Lennox Head, NSW. The drone flew down the beach some 800 meters from the lifeguard station, and a lifesaving flotation pod was dropped from the drone. The complete rescue operation took 70 seconds.We believe Shark Spotter is a win-win for both marine life and beachgoers. This unique technology combines dynamic video image processing AI and advanced drone technology to creatively deal with the global challenge of ensuring safe beaches, protecting environments, and promoting tourism.28. A Shark Spotter is ________.A. a solution to monitor sharksB. an equipment to identify lifesaversC. a technology to prevent shark attacksD. a project to pioneer the use of drones29. When a shark is spotted near a swimmer, the system will ________.A. take timely actionB. analyze the visual dataC. classify the identified objectsD. turn on “deep learning” mode30. The example in the 5th paragraph shows us that the system is .A. efficient in saving livesB. effective in detecting sharksC. smart in driving sharks awayD. practical over the whole sea area31. What is the author’s attitude towards the future of Shark Spotter?A. Doubtful.B. Optimistic.C. Negative.D. ObjectiveDThe Life-changing Antique Navajo BlanketA California man and his family went from rags to riches after discovering that the blanket given to him by his grandmother was worth a small fortune. Loren Krylzer was living in a small hut and barely getting by on his disability payments. One day, he happened to be watching Antiques Roadshow on TV when he learned that the forgotten old blanket in his closet might be valuable. The Krytzer family heirloom (传家宝) turned out to be an antique (古老的) Navajo weaving from the19th century that fetched US$1.5 million at auction (拍卖会).Krytzer's blanket was prized for much more than its antiquity. Among Native American tribes the Navajo are recognized as the most skillful weavers of blankets and rugs. The weaving style is characterized by vivid, varied patterns and exceptional durability. From shearing the sheep, spinning, preparing, and dyeing the wool to the actual weaving process, it takes around 345 hours to create one blanket.Women traditionally wove Navajo blankets while men built the weaving devices. To this day, Navajo people still make high quality blankets and rugs to sell. They believe that, since only God is perfect, their creations should have some imperfection. Another Navajo belief is that their souls are sewn into every weaving, so they intentionally include a hidden loose thread into each piece. This ensures that it isn't too perfect, and their souls can still escape into the afterlife.The Krylzers also had a loose thread on their road to riches that allowed them to escape the trap of financial ruin. In an unfortunate twist, the family's life-changing windfall (意外之财) came with a huge tax bill from the government and a lot of imploring relatives. After taking a family vacation in Mexico, buying a new sports car and a couple of real estate purchases, Krytzer lost his disabilitypayments and now gets big bills for property taxes and insurance instead. Ironically, the family decided to relocate to a less expensive state to save money.32. How did the Krytzer family go from rags to riches?A. A family member purchased a winning lottery ticket.B. Loren was invited to have an interview on a TV show.C. Loren’s disability payments were unexpectedly increased.D. Loren auctioned off a blanket his grandmother gave him.33. A loose thread woven in a Navajo blanket indicates that ________.A. God’s creation is also imperfectB. people can exit from horrible situationsC. its owner is sure to have an everlasting lifeD. the blanket is really made by hand34. Why did the family finally decide to move to another state?A. To reduce the cost of living.B. To be hidden from their relatives.C. To change for a better environment.D. To apply for disability payments again.35. According to the article, which of the following statements is TRUE?A. Navajo blankets have nice patterns but don’t last long.B. Loren Krytzer made millions of dollars from disability.C. Weaving a blanket involves many processes and skills.D. Loren’s life went from bad to worse because o f the windfall.第三部分:七选五阅读(共5小题,每小题2分,满分10分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。