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高中英语 双基限时练3 新人教版必修1

高中英语 双基限时练3 新人教版必修1
高中英语 双基限时练3 新人教版必修1

双基限时练(三) Unit 1 Part Ⅲ

Ⅰ.单词拼写

1.________(少年) in some schools are driven to work hard at their lessons.

答案Teenagers

2.That's ________ (确切地) what I wanted.

答案exactly

3.We are ________ (感激的) to your help.

答案grateful

4.In the past, we used to ________ (交换) letters.

答案swap

5.I heard a(n) ________ (条) of news over the radio.

答案item

Ⅱ.选词填空

get along/on well with, join sb. in, get/be tired of, fall in love with

答案getting along/on well with

2.I'll live in Shanghai, because I went to Shanghai last year and ________ it.

答案fell in love with

3.Would you like to ________ playing football?

答案join us in

4.The st udents ________ being asked so many questions at a time.

答案get/are tired of

Ⅲ.完成句子

1.I soon ________ ________ ________ ________ ________ the problem.

不久,我就发现难以解决这个问题。

答案found it difficult to settle

2.I am glad to hear that you ________ ________ ________ an operation.

很高兴,听说你手术已康复。

答案have recovered from

3.We ________ ________ ________ her when we were doi ng shopping in the mall.

我们在商场购物时碰巧遇见了她。

答案happened to meet

4.You must understand that it is no pleasure ________ ________ ________ ________.

你必须理解,遭受这样的痛苦是件不愉快的事情。

答案sufferi ng from such pain

5.It was in New Zealand ________ ________ ________ ________ for the first time.

我是在新西兰第一次遇见他的。

答案that I met him

Ⅳ.阅读理解

Our lifestyles today are very busy. We have family, school, sports, entertainment and social activities to fit into a time that seems never enough. We need to be healthy to meet the demands of daily life. But what does it mean to have a healthy lifestyle?

To have a healthy lifestyle, we need to:

◆ eat different kinds of healthy food most of the time

◆ do exercise often

◆ have time to relax

◆ get enough sl eep to give our bodies time to grow healthy and strong

Read about a normal day in the lives of two children.

Abbey gets up at 7:00 am, feeds the dog, and has cereal and a glass of juice for breakfast. She walks to school with her friend, Julia. She has a bag of chips for a morning snack, drinks water, and has a chicken and lettuce sandwich with a banana for lunch. She likes to play soccer with her friend s at lunch and morning break t ime. She walks home with Julia, has some crackers with cheese and juice for afternoon snack and plays with the dog for a while. She plays computer games for an hour or two before dinner, then has a shower and does her homework. She watches her favorite television show for an hour, then usually goes to bed at about 9:30 pm.

Carl gets up at 8:30 am and has two pieces of bread with jam and a glass of milk for breakfast. His mother drives him to school on her way to work. He eats cookies with juice for morning snack and a pot pie ordered from the school lunch room with juice for lunch. He likes to play card games with his friends at lunchtime and climb the fixed equipment at morning break time. He catches the bus home, has a cereal bar and a can of soda for afternoon snack, then watches some television. He has a shower before dinner, then plays the computer for an hour or two. He goes to bed at about 10:30 pm

1.About a healthy lifestyle, the writer doesn't talk about ________.

人教新课标版语文高一人教版必修三双基限时练 12动物游戏之谜

双基限时练(十二)动物游戏之谜 一、基础测试 1.下列词语中,加点字的注音完全正确的一项是() A.缅.甸(miàn)聒.噪(ɡuō) 潜.游(qián) 嬉.闹(xī) B.脊.背(jǐ) 挨.挤(ái) 倒.立(dào) 消耗.(hào) C.汲.取(jí) 安抚.(fǔ) 尾鳍.(qí) 格.斗(ɡé) D.反馈.(kuì) 模.式(mó) 捉.摸(zuó) 嚼.烂(jiáo) 解析A.缅miǎn;B.挨āi;D.捉zhuō。 答案 C 2.下列词语中,没有错别字的一项是() A.伙伴兴高彩烈平衡史无前例 B.默契得意扬扬厮打销声匿迹 C.跳越猩猩相惜消融铁打金刚 D.元气自娱自乐调剂杂草老藤 解析A.兴高彩.烈—采;B.得意扬扬 ..—洋洋;C.跳越.—跃,猩猩 ..相惜—惺惺。 答案 D 3.下列加点的成语使用不正确的一项是() A.在动物身上,无论从形态结构、生理过程,还是行为方面去分析,尽可能节省能量 的例子几乎俯拾皆是 ....。 B.单独游戏时,动物常常兴高采烈 ....地独自奔跑、跳跃,在原地打圈子。 C.每当刮起大风时,成群的露脊鲸把尾鳍高高举出水面,正对着大风,以便像船帆似 的,让大风推着它们,得意洋洋 ....地“驶”向海岸。 D.研究者们各执己见,众说纷纭 ....。 解析A项中应是“比比皆是”。“比比皆是”和“俯拾皆是”都表示相同的事物很多,到处都是。前者侧重表示多得很,到处都是;后者侧重表示容易得到。 答案 A 4.下列各句没有语病的一句是() A.动物的游戏行为,是动物行为研究中被认为最复杂、最难以捉摸、引起争论最多的

行为。 B.在动物身上,无论从形态结构、生理过程和行为方面去分析,尽可能节省能量的例子比比皆是。 C.地震发生后,当地政府及解放军部队全力救助,目前灾区群众已住进了临时帐篷,防止余震再次发生。 D.有人认为科学家终日埋头科研,不问家事,有点不近人情,然而事实却是对这种偏见的最好说明。 解析B项中应将“和”改为“还是”;C项不合逻辑,应为“预防余震发生”;D项中“事实是说明”正好说反了,依句意应是“事实正是对上述说法的反驳”。 答案 A 5.下列句子中标点符号的使用,不正确的一项是() A.马驹常常欢快地连续扬起前蹄,轻盈地蹦跳;猴类喜欢在地上翻滚,拉着树枝荡秋千……单独游戏时动物显得自由自在。 B.这些科学家认为,动物游戏是为了“自我娱乐”,而自我娱乐是动物天性的表现,正像捕食、逃避敌害,繁殖行为等是动物的天性一样。 C.这几种假说,哪一种更有道理?动物的游戏,究竟是为了“演习”,为了“自娱”,为了“学习”,还是为了“锻炼”?研究者们各执己见,众说纷纭。 D.电视剧《水浒传》标明作者为施耐庵、罗贯中,引起许多观众的疑惑:《水浒传》的作者不是施耐庵吗?怎么又多出个罗贯中? 解析B项中,第二个“自我娱乐”也应加引号,“逃避敌害”后的逗号应改为顿号,因为三者处在同一等级的并列上。 答案 B 6.下列语句排列正确的一项是() ①否则,我就只能相信,这些曲调是真正的音乐。 ②那些歌也许是有关航行,或有关浮游节肢动物的来源,或有关领地界限的简单而实打实的叙述和声明。 ③当然还有其他方法来解释鲸鱼之歌。 ④除非有一天有人证明,这些长长的、缭绕的、执著的曲调,被不同的歌唱者重复着,又加上了它们各自的修饰,这不过是为了向海面下数百英里之外传递像“鲸鱼在这儿”之类寻常的信息。 ⑤但迄今证据还没有得到。 A.③①②⑤④B.②③①④⑤ C.⑤②③①④D.③②⑤④① 解析注意“否则”“除非”“但”等词。

双基限时练26

双基限时练(二十六) 1.已知下列四个等式: ①sin(α+β)=sin αcos β+cos αsin β; ②cos(α+β)=cos αcos β-sin αsin β; ③cos ? ?? ??π2+α=-sin α; ④tan(α-β)=tan α-tan β1+tan αtan β . 其中恒成立的等式有( ) A .2个 B .3个 C .4个 D .5个 解析 ①,②,③对任意角α,β恒成立,④中的α,β还要使正 切函数有意义. 答案 B 2.1-tan15°1+tan15° 的值为( ) A. 3 B.33 C .1 D .- 3 解析 原式=tan45°-tan15°1+tan45°tan15° =tan(45°-15°)=tan30°=33. 答案 B 3.设tan(α+β)=25,tan ? ????β-π4=14,则tan ? ?? ??α+π4等于( ) A.1328 B.1322 C.322 D.16 3.已知α,β为锐角,cos α=45,tan(α-β)=-13,则tan β的值为 ( )

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高二数学 双基限时练3

双基限时练(三) 1.设f ′(x 0)=0,则曲线y =f (x )在点(x 0,f (x 0))处的切线( ) A .不存在 B .与x 轴垂直 C .与x 轴平行 D .与x 轴平行或重合 答案 D 2.一木块沿某一斜面自由下滑,测得下滑的水平距离s 与时间t 之间的函数关系为s =18t 2,则当t =2时,此木块在水平方向的瞬时速 度为( ) A. 2 B. 1 C.12 D .14 解析 s ′=lim Δt →0 Δs Δt =lim Δt →0 18(t +Δt )2-18t 2Δt =lim Δt →0 14tΔt +18(Δt )2Δt =lim Δt →0 (14t +18Δt )=14t . ∴当t =2时,s ′=12. 答案 C 3.若曲线y =h (x )在点P (a ,h (a ))处切线方程为2x +y +1=0,则 ( ) A .h ′(a )<0 B .h ′(a )>0 C .h ′(a )=0 D .h ′(a )的符号不定 解析 由2x +y +1=0,得h ′(a )=-2<0. ∴h ′(a )<0. 答案 A

4.曲线y =9x 在点(3,3)处的切线方程的倾斜角α等于( ) A .45° B .60° C .135° D .120° 解析 k =y ′=lim Δx →0 Δy Δx =lim Δx →0 9x +Δx -9x Δx =lim Δx →0 -9x (x +Δx )=-9x 2. ∴当x =3时,tan α=-1.∴α=135°. 答案 C 5.在曲线y =x 2 上切线倾斜角为π4的点是( ) A .(0,0) B .(2,4) C .(14,116) D .(12,14) 解析 y ′=lim Δx →0 Δy Δx =lim Δx →0 (x +Δx )2-x 2Δx =lim Δx →0 2xΔx +(Δx )2Δx =lim Δx →0 (2x +Δx )=2x . 令2x =tan π4=1,∴x =12,y =14. 故所求的点是(12,14). 答案 D 6.已知曲线y =2x 2上一点A (2,8),则过点A 的切线的斜率为________. 解析 k =f ′(2)=lim Δx →0 2(2+Δx )2-2×22Δx

2019-2020年高中数学 双基限时练12 新人教A版必修3

2019-2020年高中数学双基限时练12 新人教A版必修3 1.问题:①有1000个乒乓球分别装在3个箱子内,其中红色箱子内有500个,蓝色箱子内有200个,黄色箱子内有300个,现从中抽取一个容量为100的样本;②从20名学生中选出3名参加座谈会. 方法:Ⅰ.简单随机抽样法Ⅱ.系统抽样法Ⅲ.分层抽样法.其中问题与方法能配对的是( ) A.①Ⅰ,②ⅡB.①Ⅲ,②Ⅰ C.①Ⅱ,②ⅢD.①Ⅲ,②Ⅱ 解析读题知①用分层抽样法,②用简单随机抽样法. 答案 B 2.一个单位有职工160人,其中有业务员104人,管理人员32人,后勤服务人员24人,要从中抽取一个容量为20的样本,用分层抽样方法抽出样本,则在20人的样本中管理人员人数为( ) A.3 B.4 C.12 D.7 解析由题意可得20 160 ×32=4. 答案 B 3.某地区为了解居民家庭生活状况,先把居民按所在行业分为几类,然后每个行业抽1 100的居民家庭进行调查,这种抽样是( ) A.简单随机抽样B.系统抽样 C.分层抽样D.分类抽样 答案 C 4.一个总体分为A,B两层,其个体数之比为,用分层抽样方法从总体中抽取一个容量为10的样本,则A层中抽取的样本个数为( ) A.8 B.6 C.4 D.2 答案 A 5.某大学数学系共有本科生5000人,其中一、二、三、四年级的学生数之比为4:3:2:1.要用分层抽样的方法从所有本科生中抽取一个容量为200的样本,则应抽三年级的学生( ) A.80人B.40人 C.60人D.20人

解析 分层抽样应按比例抽取,所以应抽取三年级的学生人数为200×2 10=40. 答案 B 6.一个单位共有职工200人,其中不超过45岁的有120人,超过45岁的有80人.为了调查职工的健康状况,用分层抽样的方法从全体职工中抽取一个容量为25的样本,应抽取超过45岁的职工________人. 解析 依题意得,抽取超过45岁的职工人数为25 200×80=10. 答案 10 7.某工厂生产A ,B ,C 三种不同型号的产品,产品数量之比依次为 现用分层 抽样方法抽出一个容量为n 的样本,样本中A 种型号产品有16件,那么此样本的容量n =________. 解析 由题意得n =16×10 2=80. 答案 80 8.课题组进行城市空气质量调查,按地域把24个城市分成甲、乙、丙三组,对应城市数分别为4,12,8.若用分层抽样抽取6个城市,则丙组中应抽取的城市数为________. 答案 2 9.某企业有三个车间,第一车间有x 人,第二车间有300人,第三车间有y 人,采用分层抽样的方法抽取一个容量为45人的样本,第一车间被抽取20人,第三车间被抽取10人,问:这个企业第一车间、第三车间各有多少人? 解 x =20×30045-20-10=400(人),y =10×300 45-20-10 =200(人). 10.某单位有工程师6 人,技术员12 人,技工18 人,要从这些人中抽取一个容量为 n 的样本.如果采用系统抽样和分层抽样方法抽取,都不用剔除个体;如果样本容量增加1 个,则在采用系统抽样时,需要在总体中先剔除1个个体,求样本容量n . 解 解法1:总体容量为6+12+18=36(人).当样本容量是n 时,由题意知,系统抽样的间隔为36n ,分层抽样的比例是n 36,抽取工程师人数为n 36×6=n 6人,技术人员人数为n 36×12 =n 3人,技工人数为n 36×18=n 2 人,所以n 应是6的倍数,36的约数,即n =6,12,18. 当样本容量为(n +1)时,总体容量是35 人,系统抽样的间隔为35n +1,因为35 n +1 必须是整数,所以n 只能取6,即样本容量n =6. 解法2:总体容量为6+12+18=36(人). 当抽取n 个个体时,不论是系统抽样还是分层抽样,都不用剔除个体,所以n 应为

2016高中数学人教B版必修3双基限时练12

双基限时练(十二) 用样本的频率分布估计总体的分布 基础强化 1、下列关于频率分布直方图的说法,正确的就是() A、直方图的高表示取某数的频率 B、直方图的高表示该组上的个体在样本中出现频数与组距的比值 C、直方图的高表示该组上的个体在样本中出现的频率 D、直方图的高表示该组上的个体在样本中出现频率与组距的比值 解析频率分布直方图的纵坐标表示错误!,故选D、 答案D 2、从一群学生中抽取一个一定容量的样本对她们的学习成绩进 行分析,前三组就是不超过80分的人,其频数之与为20人,其频率之与(又称累积频率)为0、4,则所抽取的样本的容量就是() A、100 B、80 C、40 D、50 解析样本容量为亠 = 50、故选D、 0、4 答案D 3、下列说法不正确的就是() A、频率分布直方图中每个小矩形的高就就是该组的频率 B、频率分布直方图中各个小矩形的面积之与等于1

C、频率分布直方图中各个小矩形的宽一样大 D、频率分布折线图就是依次连接频率分布直方图的每个小矩形 上端中点得到的 解析在频率分布直方图中各个小矩形的高就就是该组的错误!、 答案A 4、某市教育行政部门为了对2012届高中毕业生的学生水平进行评价,从该市高中毕业生中抽取1000名学生的数学成绩作为样木进行统计,其频率分布直方图如图所示、则这1000名学生的数学平均成绩的最大可能值为() 频率 O 40 50 60 70 80 90 100分数(分) A、67、50 B、72、05 C、76、50 D、77、50 解析由题意得平均成绩的最大可能值为0、05X50 + 0、1X60 + 0、25X70 + 0、35X80 + 0、15X90 + 0、1X100 = 77、50、 答案D 5、一个社会调查机构就某地居民的月收入调查了10000人,并根据所得数据画了样本的频率分布直方图(如下图)、为了分析居民的收入与年龄、学历、职业等方而的关系,要从这10000人中再用分?层抽样方法抽出100人作进一步调查,则在[2500, 3000)(元)月收入段应

【精品】人教版高中英语必修四:双基限时练【1】(含答案解析)

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高中数学双基限时练21

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