2013CUPT题目中文翻译

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2013年高考英语真题阅读理解译文（辽宁卷）

2013年高考英语真题阅读理解译文（辽宁卷）2013年普通高等学校招生考试全国统一考试（辽宁卷）英语第三部分阅读理解（共两节，满分40分）A中国是一个自行车王国。

至少在1992年我来到中国旅游那时是。

那时候每个人好像都骑自行车。

成千上万的自行车都是黑色的。

小汽车很少见。

然而，去年我来到北京时，我发现情况完全相反。

路上有了成千上万的小汽车。

不过，人们闲逛时还是选择骑自行车。

对一些人来说，骑自行车是现在最容易也是最便宜的旅行方式。

现在自行车也有了不同的颜色——银色、绿色、红色、蓝色、黄色，颜色随便你选。

看人们骑自行车很有趣。

他们匆匆地穿行于各个十字路口，熟练地穿过车流，甚至骑上人行道。

自行车给人们带来的那种到处走动的自由是小汽车无法给的。

迫切地想在这一方面融入中国文化，我决定去买一辆自行车。

好天气与顺利的购买相伴。

我立刻跳上我的自行车座，骑着回家。

第一次骑自行车回家，我很遵守交通秩序。

为了安全起见，有些时候，当马路上的小汽车突然飞驰而来时，我就和一群骑自行车的人待在一起。

我可不想被撞到。

所以我骑得小心翼翼。

过马路是最大的一个问题。

这跟在美国横穿公路不一样。

这里的街道比较宽，所以过马路需要时间、技巧、和一点运气。

我终于把它骑回家了。

在自行车上的感觉很棒。

空气轻拍脸颊穿过发丝的感觉非常美妙。

当我从一些地方和一些人旁经过使，我就好像是坐在世界的顶端。

骑行让我感觉充满活力。

B如今，绿色行动对于大部分人来说似乎成为了一种时尚。

无论这是好事还是坏事，我们真没法说，但对我们两个人来说，绿色行动并不是一种时尚，而是一种生活方式。

2011年4月22日，我们决定全年的每一天都要实施绿色行动。

这意味着我们要做356件不同的与绿色相关的事情，也意味着挑战我们自己去践行绿色行动远不是一件容易的事情。

我们并不是要进行能源再利用和减少能耗，而是必须想出365件不同的与绿色相关的事情去做，这不是一件容易的事情。

有了全年每天实施绿色行动的想法，我们的绿色之年就开始了。

2013年数学建模美赛题目中文翻译_共4页

Problem c:背景：社会正致力于运用和开发模型来预测地球的生物和环境情况。

很多科学研究总结了逐渐增长的地球环境和生物系统压力，但很少有人用全球范围的模型来检测这些观点。

联合国发表的千年生态系统评估综合报告发现：近三分之二的地球生命支持生态系统——包括净水，洁净的空气，稳定的气候——正在因非可持续性使用而逐渐衰减。

其中大部分破坏归咎于人类行为。

暴增的对于食物，淡水，燃料，木材的需求导致了剧烈的环境变化；从森林砍伐到空气，土壤和水污染。

尽管已存在大量关于局部习惯和地区因素的研究，目前的模型还不能告知决定人他们的局部策略是如何影响整个地球的健康的。

许多模型忽略了复杂的全球因素，这些模型无法判断重大政策的长期影响。

尽管科学家们意识到巨大环境和生物系统中存在的复杂关系和交叉作用，当前的模型通常忽略这些管理或限定了系统间的影响。

系统的复杂性体现在多元交互（多个元素的相关性），反馈，突发行为，即将发生的状态变化或触发点。

最近的自然杂志中一篇由22位国际知名科学家撰写的题为“迫近地球生物圈的状态变化”的文章讨论了许多有关科学模型对于预测行星健康系统潜在状态变化的重要性与必需性。

文章提供了两种具体定性的模型，并寻求更好的预测模型：1）通过在全球模型中加入相关系统的复杂性（包括局部情况对全球系统的影响，反之亦然）来优化生物状态预测。

2）辨别不同因素在产生非健康全球状态变化中的作用并展示如何运用有效的生态系统管理来预防或限制这些即将发生的状态变化。

研究最终归结于问题：我们是否能利用全球健康的局部或地区性组成部分预测潜在状态变化来帮助决策者制定基于对全球健康状况潜在影响的，有效的策略。

尽管有越来越多的警示信号出现，没人知道地球是否确实在接近全球性的转折点（极端状态），这种极端的状态是否是不可避免的。

自然杂志等研究指出了地球生态系统中的一些重要工作元素。

（例如：局部因素，全球变化，多维元素与关系，变化的时间与空间范围）。

2013中国女子数学奥林匹克试题及其解答

2013中国女子数学奥林匹克试题及其解答1.证明（张云华）1.证明（mavropnevma ）Let , . The region is the triangle ,and , thus the triangles of largest possible area are and . But(equality for is for ,disallowed).(equality foris for ).EDIT. Referring to the next post - it is interesting to figure out the envelope of the linesis the parabola , but in some way it is irrelevant, since the problemfunctions for a fixed , with fixed on the parabola for some , and thetriangular area being delimited by the tangent to that parabola at .1.证明（kunny）, thus theparabola is envelope of the linesWhich touches at .The domain of the family of the lines swept by is shown by the shaded region, excluded twopoints .Edit:I was misreading the context problem, thank you for pointing put it, mavropnevma.I have just attached another figure.Needless to say, mavropnevma's solution is perfect.P.S.I remember that the similar problem original problem has been posed in 1970's in Tokyo University entrance exam/Science.According to my memory, let be the maximum area of any triangle which is involved in theregion in original problem, then draw the graph of to find the extrema of .Here is the similar problem posed in Tokyo University entrance exam/Science, second round, 1978In the -plane, let be the part which is correspond to of the parabola ,that is to say, .Let the tangent Line of at intersects with the line at and intersects withthe line at .Let . We are to consider the questions as below in the range of.(1) Let denote the area of triangles by , respectively. Find the rangeof such that .(2) Let be the domain enclosed by line segments and .Note : contains line segments and .Let be the maximum area of the triangle with a vertex which is contained in .Find the function and draw the graph, then find the exterme value.Note : A function has local minimum (or local maximum) at a point, which means for all points which is closed to ,holds. We call local maximum, local minimum as extreme value.2.证明(Luis González)Let and Let be the incircleof touching at Clearly and are homothethic with incirclesbut since are symmetric about the midpoint of (isthe M-excircle of MAB), it follows that and are homotheticwith corresponding cevians2.证明(Andrew64)As shown in the figure below.is the intersection of and .Therefore , , meet at the same point .Attachments:link17.jpg [ 37.23 KiB | Viewed 52 times ]3.证明(mavropnevma)Since it is irrelevant which persons of the same gender know each other, we may assume there ore none such, and consider the bipartite graph having the left shore made ofthe boys and the right shore made of the girls, with an edge connecting a boy and a girl ifthey know each other. The condition means does not contain any induced cycle of length ,and the requirement is to show the number of edges satisfies .Thus it is an extremal graph theory question, for bipartite graphs with forbidden 's; bysymmetry we should also have .Denote by the set of girls that each knows exactly one boy, and by the set of girls that eachknows more than one boy; take and . We obviously have and.Let us count the number of objects , where is a girl, are distinctboys, and knows both . For each of the doubletons there is at mostone girl knowing them both (by the condition), so . Moreover, by pigeonholewe have .On the other hand, we have, by Jensen's inequality.We thus need , so.Finally, we get .For equality to be reached we obviously need , namely for each pair of boys having exactly one girl knowing both of them; and then we need .3.证明(crazyfehmy)Another solution: Consider the bipartite graph where there are girls andboys and denote the girls by 's and boys by 's as vertices.Let denote the number of edges from the vertex to set . If is connected to some andthen for any the girl must not connected to both and . Now let us count suchpairs. For every girl there are many pair of edges. Since all such edge pairs must bedistinct for all girls, and since there are at most such pairs, we haveor equivalentlyNow assume that are greater than or equal to and areor . Then we haveand we need to show that .Since for we have and hence .Now by Cauchy-Schwarz inequality we haveand if , we haveand we need to show that . Assume . Then we havewhich means that which is a contradiction. So, we are done.4.解（crazyfehmy）Let be called a nice pair if satisfies the conditionsstated in the problem.Firstly, we shall prove a lemma:If is a nice pair then for all integers and .Proof: Let or and divides for some integers and. Then we can find another pair such that and. (The proof is easy) Then consider the systemand . By the Chinese Remainder Theorem this system has a solution and such that both and isan element of which means that is not a nice pair becauseis divisible by .Now, we shall show that if or then has asolution for all integers . For the proof assume there exists an integer such thathas no solution in integers. Then it is easy to see thathas also no solution for all integers which are not divisibleby . Now take and consider the numbers . For all of these numbers,we have no solution and there are such numbers. Since all square residues have solutions(take ) and there are square residues modulo including zero, this means that forall nonsquare residues , the equation must have no solution. However, for the numberis not a square residue but so for we have asolution which is a contradiction. For the number is not a square residue buthas a solution since is a square residue. Hence we again obtain a contradiction so the second lemma is also proved.Now, since can take every value modulo and we must have andand also we must have and if is a nice pair. So, can take only three values. We will consider each case separately:If then we must have which meansIf then we must have whichmeansIf then we must have whichmeansIf we count these possibilities, if i am not wrong, we get.4.解（dinoboy）First, remark that it suffices for to be injective modulo and .For modulo simply note that we require .For modulo or we require for some thatNow, what values can take modulo ? It is a simple exercise to show allvalues modulo can be obtained (just express it as and then aswe can transform the problem to what values modulo can beexpressed as , which is known to be all of them). Therefore the only way this problemsworks out is if and .Therefore we simply require that and . For each value of there are valuesof modulo , 10 modulo and modulo so the answer should simply be.Note: To show takes all values modulo without relying onis not hard, but I'm lazy and felt like reducing it to an already solved problem.4.解（yunxiu），so ，hence satisfies the problem.So the answer should be .6.解（crazyfehmy）If let then satisfiessince is equivalent to whichhas always a solution in the set for all and since is odd.Now we will show that if then the condition does not satisfy. Let be theelements of the set . Consider the sums and let. Since 's are different modulo , the numbers 'sare also different modulo . On the other hand, none of 's can be equivalent to modulobecause otherwise we would have two equivalent terms. Hence is apermutation of and by adding up these equations we getwhich means. Now do the same procedure for all's to get. Let and. Then we have . So, we havemany numbers equivalent to each other modulo . However, we know that there are manynumbers modulo which are all equivalent to each other modulo . Hence in order for 's tobe different modulo, we must have which means that and hence. Soand we are done.Hence all possible values are .7.证明（Luis González ） Let cut again at Since is the exsimilicenter ofthen is midpoint of the arc ofbisects externally is midpoint of the arc ofis external bisector of andNote that is a Thebault circle of the cevian of externally tangent to itscircumcircle By Sawayama's lemma passes through its C-excenter isC-excenter ofis M-isosceles, i.e. Hence iscircumcenter of7.证明（Andrew64）As shown in the figure. Let be the intersection of and .It's fairly obviousSo we have, andare concyclic.So , andThus, Consequently .Namely is the bisector of.Attachments:link18.jpg [ 31.39 KiB | Viewed 93 times ]8.证明（duanby）hint:(a-b)(c-b)(a-d)(c-d)in detail: product (a-b)(c-b)(a-d)(c-d) for everya,c be the number on , b,d be the number onfor point x,y if they are not ajjectent then in the product, it will occur twice, if it's ajjectent it's appears only once, and also chick the point that are on and then we get it.iampengcheng1130 2013中国女子数学奥林匹克第7题的解答。

CUPT题目中英文版

年CUPT题目中英文版————————————————————————————————作者：————————————————————————————————日期：1、Invent YourselfConstruct a simple seismograph that amplifies a local disturbance by mechanical, optical or electrical methods. Determine the typical response curve of your device and investigate the parameters of the damping constant. What is the maximum amplification that you can achieve?1、自主发明建造一个可以通过力学、光学或电学方法放大局部扰动的简单地震仪。

确定你装置的标准响应曲线并且研究阻尼常数的参量。

你能达到的最大放大率是多少？2. Colour of PowdersIf a coloured material is ground to a powder, in some cases the resulting powder may have a different colour to that of the original material. Investigate how the degree of grinding affects the apparent colour of the powder.2、粉末的颜色如果一个有色的材料被研磨成粉末，在某些情况下被研磨成的粉末将会呈现和原有材料不同的颜色。

探究研磨程度是如何影响粉末的表面颜色的。

3. Dancing CoinTake a strongly cooled bottle and put a coin on its neck. Over time you will hear a noise and see movements of the coin. Explain this phenomenon and investigate how the relevant parameters affect the dance.3、舞蹈的硬币取一个冷却程度很大的瓶子并且将一枚硬币置于瓶颈处。

CUPT题目翻译

CUPT题⽬翻译CUPT题⽬1. Gaussian cannon ⾼斯炮A sequence of identical steel balls includes a strong magnet and lies in nonmagnetic channel. Another steel ball is rolled towards them and collides with the end ball .the ball at the opposite end of the sequence is ejected at a surprisingly high velocity .optimize the magnet’s position for the greatest effect.⼀列相同的钢球含有⼀个强磁铁并置于⽆磁通道中。

另有⼀个钢球向他们滚动并碰撞序列末尾的球。

在另⼀端末尾的球被迫以极⾼的速度离开。

请优化磁铁的位置使碰撞产⽣最佳效果。

2. Cutting the air切割空⽓When a piece of thread (e.g. nylon) is whirled around with a small mass attached to its free end, a distinct noise is emitted.study the origin of this noise and the relevant parameters.⼀段线（例如尼龙）的⾃由端系上⼀个重物，当线被旋转时会形成明显的噪声。

研究这种噪声的产⽣原因及相关参数。

3. String of beads串珠A long string of beads is released from a breaker by pulling a sufficiently long part of the chain over the edge of the breaker. Due to gravity the speed of the string increases. At a certain moment the sting no longer touches the edge of the breaker (see picture). Investigate andexplain the phenomenon.⼀长串珠⼦从烧杯⾥释放，具体释放⽅式是将⾜够长的珠链拉出烧杯的边缘，这样由于重⼒作⽤串珠的速度增⼤。

2013全国大学生英语竞赛C类样题及答案

参考答案Part I Listening ComprehensionOmitted．Part II Vocabulary and Structure31—35 BCCBD 36—40 DAACB41—45 BAABCPart III Cloze46． reduce47． suffering48． difference49． increase 50． turning51． cooling52． limit53． pollution54． fill 55． takingPart IV Reading ComprehensionSection A56． empathy57． togetherness 58． characters 59． desire60． instructiveSection B61—63 TTF64． They choose an important news event and entertain viewers by reporting on it 24 hours－a－day for days and sometimes weeks．65．They appeal to people’s curiosity and their desire to know all about something． / The shows are exciting and they hold the attention of the audience．Section C66． hands－on activities67． Royal Mile68． a telescope69． D 70． CSection D71． Because of the increased traffic and fatigued climbers．72． To restore Mount Everest to its pristine state．73． In the year of 1953．74．做最高山峰的清洁夫并不迷人，但是它提醒我一个人必须把事情进行到底。

CUPT题目(IYPT翻译和插图)

201届国际青年物理学家竞赛赛IYPIYPT T 20133年第26届国际青年物理学家竞参考翻译))题目((参考翻译题目1.自己创造如果一张纸被“手风琴式”折叠或者被卷成筒，那么它将更加难以弯曲。

使用一张单一的A4纸和少量胶水，如果需要的话，构建一座跨越280毫米间隙的桥梁。

介绍相关参数，用以描述该桥梁的强度，并优化其中的部分参数或全部参数。

2．弹性空间大球在水平拉伸膜上滚动的动态效果和它们之间明显的相互作用通常用于说明引力场。

进一步探究该系统。

在这样一个“引力场世界”确定和测量明显的“引力常数”是否可能？如果你站在地面上拿着乒乓球，并释放它，它会反弹。

如果乒乓球内含有液体，碰撞的性质会发生变化。

探究碰撞的性质如何取决于球内液体的含量和其他相关参数。

4．孤子沿水平轴等距离安装一链相似的摆，相邻的摆用轻绳相连接。

每一个摆可以绕轴旋转，但不能侧向移动（见图）。

探究沿着这样一条链的一种旋转的传播。

当各摆都经历360º旋转时，孤立波的速度是多少？一个轻球（如乒乓球），可以被向上的气流所支撑。

气流的方向可以倾斜，然而它仍然可以支撑球。

探究气流倾斜的影响，并优化该系统，得出在保持球处于稳定状态的情况下，气流倾斜的最大角度。

6．彩色塑料在明亮光线的照射下，一个透明的塑料物体（如一张空白的CD外壳）有时可以呈现各种不同的颜色（见图）。

研究和解释这种现象。

确定一下，当使用各种不同颜色的光源时，是否也可以看到这些颜色。

7．聆听光的声音将一个罐子内表面的一半涂一层锅灰，并在它的盖子上钻一个孔（见图）。

当连接交流电的灯泡发出的光线射到罐子的黑墙（锅灰层）时，可以听到明显的声音。

解释和探究这种现象。

8．喷射和薄膜喷射的细液体流对肥皂薄膜的作用（见图）。

喷射的液体流可以渗透通过薄膜或者与薄膜合并，产生有趣的形状，这取决于相关参数。

解释和探究这种相互作用，以及由此产生的形状。

9．碳麦克风一个麦克风的设计已经涉及碳颗粒的使用很多年了。

2013年高考新课标全国卷2英语阅读理解中文翻译

第一篇医生是著名‎的糟糕“引航员”，他们不需要‎听命于别人‎因为他们早‎已全部知道‎。

我很辛运：1970年‎我成为了一‎名引航员，差不多比我‎从药学院毕‎业早10年‎。

我之前并没‎有意识到这‎些，但是成为一‎名引航员让‎我成为了一‎名更好的外‎科医生。

我喜欢飞翔‎。

当我开着更‎大更快的飞‎机，在更为恶劣‎的天气的时‎候，我学到了机‎组资源管理‎，或者称为C‎RM，一种可以飞‎得更安全的‎方法。

这意味着机‎组成员应该‎听命并站出‎来为好的结‎果说话，忽略自己的‎地位1980年‎我第一次接‎触到CRM‎。

那不久之后‎，一名主治医‎师和我正在‎恶劣天气中‎飞行。

控制员让我‎们准备着陆‎时已经太晚‎了。

主治医生正‎在飞行，我是一个保‎证安全的引‎航员，由于一个糟‎糕的转弯他‎变得手忙脚‎乱，他忘记把起‎落架放下来‎。

他是一个不‎错的引航员‎——还有我的上‎级——所以为他们说好‎话会觉得很‎奇怪。

但是我必须‎这么做：我们的生命‎处于危险之‎中。

我把我的不‎容易抛到了‎一边然后说‎：“我们现在需‎要把起落架‎放下来！”这是我在C‎R M压力下‎第一堂真实‎的课程。

从那之后我‎就开始在手‎术室使用这‎招了。

CRM要求‎引航员/外科医生鼓‎励他人站出‎来说话。

这更要求当‎建议来自反‎方时，医生不会有‎可能阻止从‎属医生再次‎提出建议的‎过激反应。

所以呢，当我在手术‎室的时候，我从他人那‎里获取想法‎和帮助。

有时候他们‎不愿意站出‎来说话，但是我希望‎如果我继续‎鼓励他们，有一天某个‎人会组织我‎把起落架收‎起来。

第二篇1947年‎一群由一位‎澳大利亚管‎理者率领的‎来自艺术世‎界的名人决‎定在爱丁堡‎举办一场国‎际化的音乐‎、舞蹈和喜剧‎盛会。

在第二次世‎界大战之后‎，这个主意重‎组了欧洲。

“Fring‎e”这个单词作‎为官方节日‎的挑战出现‎了。

1947年‎，在每个人都‎有表在同一时间‎，演的权‎力的信念促‎使下，八个戏剧团‎没有被邀请‎，在一间废弃‎了很多年的‎屋子里他们‎也是这么做‎的。

2013年美赛数模A题翻译

2013 Contest ProblemsMCM PROBLEMSPROBLEM A: The Ultimate Brownie PanWhen baking in a rectangular pan heat is concentrated in the 4 corners and the product gets overcooked at the corners (and to a lesser extent at the edges). In a round pan the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges. However, since most ovens are rectangular in shape using round pans is not efficient with respect to using the space in an oven.Develop a model to show the distribution of heat across the outer edge of a pan for pans of different shapes - rectangular to circular and other shapes in between.Assume1. A width to length ratio of W/L for the oven which is rectangular in shape.2. Each pan must have an area of A.3. Initially two racks in the oven, evenly spaced.Develop a model that can be used to select the best type of pan (shape) under the following conditions:1. Maximize number of pans that can fit in the oven (N)2. Maximize even distribution of heat (H) for the pan3. Optimize a combination of conditions (1) and (2) where weights p and (1- p) are assigned to illustrate how the results vary with different valuesof W/L and p.In addition to your MCM formatted solution, prepare a one to two page advertising sheet for the new Brownie Gourmet Magazine highlighting your design and results.终极蛋糕烤箱当用矩形烤盘烘焙时，热量集中在4个角落，在角落的食物容易被烤糊（边上的热量较少），在一个圆形烤盘上，热量均匀的分布在盘的整个边缘，而且分布在边缘的食物不会被过度加热。

2013年大学生英语竞赛C类初赛真题及详解

2. What does “Fill Me In” refer to in this conversation? A. A type of upgraded mobile phone. B. A well-selling magazine. C. A writer’s new book series. 【听力原文】 M: So, how well is your “Fill Me In” selling? W: Oh, great, in fact it’s only the fourth week of publication of the first issue, and we’ve had to reprint another 2000. We’ve sold about 2300. Mostly in alternative bookshops that cater for people who are... er different. M: Congratulations!
Part V Translation (10 marks) Part VI Error Correction (10 marks) Part VII IQ Test (5 marks) Part VIII Writing (30 marks)
I (10 marks) II (20 marks)
Part I Listening Comprehension (30 marks)
Part II Vocabulary and Structure (15 marks) Part Ⅲ Cloze (10 marks) Part Ⅳ Reading Comprehension (40 marks) Part V Translation (10 marks) Part VI Error Correction (10 marks) Part VII IQ Test (5 marks) Part VIII Writing (30 marks)

CUPT题目中文翻译

以下中文翻译来源于网络，仅供参考，请同学们自己准确理解原题：1.自己发明请建造一个装置——能为从2.5米固定高度坠向坚硬地面的生鸡蛋提供安全着陆，装置必须与鸡蛋一起坠落，你能实现的装置的最小尺寸是多少？2.气膜汽笛一个简单的汽笛可以这样制造：用一根穿过一只小型容器（或杯子）底部的管子（如图）拉伸位于容器（或杯子）开口处的气膜。

经由在容器侧壁处的小孔，鼓风可以产生一个声音。

探讨相关参数如何影响声音。

3.单透镜望远镜一架望远镜可只用一个透镜制作，如果小光圈被用于替代目镜，那么透镜参数和孔会如何影响图像（如放大率、清晰度和亮度）？4.磁丘少量置于非均匀磁场中的磁性流体会形成山丘状结构，探讨这些结构的属性如何依赖于相关参数。

5.莱氏星星在莱氏效应中，置于热表面的水滴可存在几分钟。

在某些情况下，水滴会发展为振荡的星形状态。

引入不同的振荡模式并探讨它们。

6.快链条一根链条由若干相比于水平有一定角度的木棍组成，且由两根线连接（见图），垂直悬挂，然后释放。

相比于自由落体，当掉至水平地面时，链条掉落更快。

解释此现象并探讨相关参数如何影响运动。

7.螺旋水波螺旋波及其他类型的波可能出现于薄层液膜流过旋转平台，探讨这些波的类型。

8.密度可视化纹影摄影术通常用于使气体中的密度变化可视化。

安装一套纹影装备并探讨它在解决密度差异中能做多好。

9.管中球现有一根充满液体并含有一枚小球的密封透明管，此管倾斜放置且下端与马达相连，如此管沿锥形面运动。

探讨球在相关参数作用下的运动。

10.分离玻璃在两片薄玻璃之间放置一层薄薄的水，然后试着把它们分开。

探讨影响所需力的参数。

11.毛发湿度计一只简单的湿度计可仅用人的毛发制作。

探讨相关参数对其准确度和响应时间的作用。

12.扭转陀螺将轮轴固定于具有一定扭转阻力的竖直线上（见图），扭转绳子，旋转轮子并释放它，探讨此系统的动力。

13.共鸣的玻璃酒杯一部分盛满液体，暴露于扬声器的声音下会产生共振。

探讨此现象如何取决于各种参数。

2013年美赛B试题原文及翻译

2013年美赛B题原文及翻译PROBLEM B: Water, Water, EverywhereFresh water is the limiting constraint for development in much of the world. Build a mathematical model for determining an effective, feasible, andcost-efficient water strategy for 2013 to meet the projected water needs of [pick one country from the list below] in 2025, and identify the best water strategy. In particular, your mathematical model must address storage and movement; de-salinization; and conservation. If possible, use your model to discuss the economic, physical, and environmental implications of your strategy. Provide a non-technical position paper to governmental leadership outlining your approach, its feasibility and cost s, and why it is the “best water strategy choice.”Countries: United States, China, Russia, Egypt, or Saudi Arabia翻译:问题 B:水,水,无处不在淡水资源是世界很多大部分国家发展的瓶颈。

B题翻译

2013Certificate Authority Cup International Mathematical Contest Modeling“认证杯”数学建模国际赛()PROBLEM B:Hermit Crabs Evolve Employment Pattern of Human人类的寄居蟹式进化就业模式Hermit crabs—which are the most popular pet variety in the UnitedStates—depend on other creatures' shells for protection. Strip a hermit crab ofits shell and you will see its soft, pink abdomen curled behind its headlike afern's frond. Most hermit crabs favor snail shells, although some use the shellsof bivalves, like clams and scallops, and others have made do with driftwood,stones and pieces of glass or plastic bottles.When a lone crab encountered one of the beautiful new shells, it immediatelyinspected the shelter with its legs and antennae and scooted out of its currenthome to try on the new shelter for size.当一只螃蟹遇到一个美丽的新贝壳时，它会立即用它的腿和触角检查贝壳，并迅速离开它现有的家去尝试新的居所。

2013年美国数学建模比赛题目及翻译

PROBLEM A: The Ultimate Brownie PanWhen baking in a rectangular pan heat is concentrated in the 4 corners and the product gets overcooked at the corners (and to a lesser extent at the edges). In a round pan the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges. However, since most ovens are rectangular in shape using round pans is not efficient with respect to using the space in an oven.Develop a model to show the distribution of heat across the outer edge of a pan for pans of different shapes - rectangular to circular and other shapes in between.Assume1. A width to length ratio of W/L for the oven which is rectangular in shape.2. Each pan must have an area of A.3. Initially two racks in the oven, evenly spaced.Develop a model that can be used to select the best type of pan (shape) under the following conditions:1. Maximize number of pans that can fit in the oven (N)2. Maximize even distribution of heat (H) for the pan3. Optimize a combination of conditions (1) and (2) where weights p and (1- p) are assigned to illustrate how the results vary with different values of W/L and p.In addition to your MCM formatted solution, prepare a one to two page advertising sheet for the new Brownie Gourmet Magazine highlighting your design and results.PROBLEM B: Water, Water, EverywhereFresh water is the limiting constraint for development in much of the world. Build a mathematical model for determining an effective, feasible, and cost-efficient water strategy for 2013 to meet the projected water needs of [pick one country from the list below] in 2025, and identify the best water strategy. In particular, your mathematical model must address storage and movement; de-salinization; and conservation. If possible, use your model to discuss the economic, physical, and environmental implications of your strategy. Provide a non-technical position paper to governmental leadership outlining your approach, its feasibility and costs, and why it is the “best water strat egy choice.”Countries: United States, China, Russia, Egypt, or Saudi ArabiaICM PROBLEMPROBLEM C: Network Modeling of Earth's HealthClick the title below to download a PDF of the 2013 ICM Problem. Your ICM submission should consist of a 1 page Summary Sheet and your solution cannot exceed 20 pages for a maximum of 21 pages.问题A：终极蛋糕锅当你使用一个方形的锅的时候，热量会聚集在四个角落并且角落的食物会被过度加热（在边上的加热强度会相对较小）。

2013年高考英语试题(大纲全国卷)答案

2013年全国普通高等学校招生考试(全国大纲卷)英语答案1—5 ACACB 6—10 BBABA 11—15 BACAA 16—20 BCBCC21．D【解析】考查交际用语。

根据前句句意"抱歉，我犯错误了" 及空格后"人无完人"可知，说话者在安慰对方：凡事看开些。

Take it easy意为"别紧张，放松些"，符合语境。

22．A【解析】考查动词短语。

根据句意"你今晚愿意和我们一起去看电影吗?"可知，此处意为"一起去"。

come along"跟着来，一起去"符合句意。

23．A【解析】考查并列连词。

句意：我很高兴再次见到Jenny，但是我不想整天和她在一起。

前后为转折关系，故选A。

24．C【解析】考查定语从句。

句意：当我到达时，Bryan带我去看了我即将居住的房子。

分析句子结构，此处为定语从句，从句中缺少地点状语，应用关系副词where来引导。

25．D【解析】考查非谓语动词。

句意：我赶上了来自Paddington的那班7：30的火车，那天很早就到了办公室。

"赶上火车"这一动作是由主语执行的，且先于谓语动词动作"got to"发生，故应用完成式，选D项。

26．B【解析】考查情态动词。

句意：既然无人帮他，他一定是独自完成了那项研究。

此处是对过去发生事情的肯定的推测，故选B。

27．C【解析】考查时态。

句意：我们打算一早出发，因此前一天夜里就打包了行李。

此处表示过去将来，用过去进行时来表示。

故选C。

28．A【解析】考查动词辨析。

句意：那块表很不错，他以20%的折扣价买下了它。

A项表示"支付"，后跟具体的钱数，与后面的for搭配，故本题选A。

B项的主语一般是物，不能选；C项后面一般用"商品"作宾语，可排除；D项与后面的介词不搭配，其后应用介词on。