Reconstruction of Centrally Symmetrical Convex Bodies by Projection Curvature Radii

China &Foreign Medical Treatment 中外医疗DOI:10.16662/ki.1674-0742.2020.07.005微血管减压术治疗三叉神经痛疗效的影响因素分析成振林,周勤仁,康东,蔺俭,王斌甘肃省河西学院附属张掖人民医院神经外科,甘肃张掖734000[摘要]目的探讨影响微血管减压术治疗三叉神经痛手术疗效的相关因素。

方法回顾性分析2009年1月—2016年1月河西学院附属张掖人民医院120例原发性三叉神经痛患者的病例资料,用χ2检验行单因素分析评价患者术后疗效的影响因素。

结果患者男性67例,术后有效65例(97.0%),女性53例,术后有效42例(97.6%)(χ2=0.175,P =0.807),年龄>60岁者12例,有效10例(83.3%),年龄≤60岁者108例,有效102例(94.4%),(χ2=0.187,P =0.665),左侧面部疼痛者35例,有效31例(88.5%),右侧面部疼痛者85例,有效81例(95.2)(χ2=0.029,P =0.864),疼痛持续时间>1年者10例,有效7例(70.0%),疼痛持续时间≤1年110例,有效107例(97.2%),(χ2=0.069,P =0.793),三叉神经受压部位位于脑干91例,有效88例(96.7%),三叉神经受压部位位于脑池段29例,有效26例(89.6%)(χ2=0.267,P =0.605),这些一般因素对手术疗效无明显影响(P>0.05)。

典型三叉神经痛97例,有效97例(100.0%),非典型三叉神经痛23例,有效21例(91.3%)(χ2=4.034,P =0.02),术前MRTA 检查有血管压迫者102例,有效100例(98.0%),术前MRTA 检查无血管压迫者18例,有效17例(94.4%)(χ2=5.029,P =0.03),术中有责任血管者98例,有效98例(100.0%),术中无责任血管22例,有效20例(90.9%)(χ2=4.061,P =0.00),以上是影响患者术后临床疗效的主要因素(P<0.05)。

语言学试题集英语语言学往年试题集锦语言学的资料很少，看完书后想找些习题或往年试题做做，可以起到练兵，巩固所看书本知识的作用。

我收集了一些高校语言学的往年试题供大家参考，讨论，交流一下做题的感受，也希望大家可以把自己所考学校的往年试题发表在上面，给大家一起讨论。

1 One of the main features of our human language is arbitrariness .Can you briefly expla in what is this feature refers to ? Give examples if necessary(10 points). <北师大2003年试题）2 In english we can describe a story as "a successful story" or "a success story ".Do yo u think they mean the same ? Please explain and give your reasons(10 points) ,<同上》3 Expain the following terms ,giving examples where necessary.(50 points) <中山2003》 de sign feature macrolinguistics vowel minimal pair folk etymology aspect anopho r error ana lysisr metaphor4 Language can change through blending ,metanalysis ,back-formation, analogical creation and borrowing.Give two english words for each of them (5 points) 清华2000年试题5 Answer the following question briefly.clearly,grammatically and correctly.(10 Points )湖南师大2003年What is it wrong to assume that the meaning of a sentence is the sum of the meaning of t he words which compose it ?7 Define the following terms.(10 points) 中国海洋大学1999Phoneme ,consonant,morpheme,lexicon,syntax,endocentric construction,semantics,hyponymy , language ,design feature8 Define the following terms .(20 points) 苏州大学1997 allophone morpheme assimilation i nternal authority interlanguage phatic communionclosed-class word government semantica triangic lingua francaWhat is the main grammatical difference between a sentence and a clause ? 同上6 Translate into chinese and exemplify each of the following.(10 points )Example : dialectal synonymsAnswer , 方言同义词， Fall and autumn are dialectal synonyms .homography homophony gradable opposites endocentric constuctionexocentric construction9 大连外国语学院1992年语言学全部试题 100 POINTSList the six important characteristics of human language .What are the types of morphemes ?Illustrate the deep and surface structures .What do u know about the semantic features ?How does language change ?10 Words in our mental lexicon are known to be related to one another .Discuss the relat ionships between words ,using examples from the english language .(15 points ) 北外2003年试题11 What do you think are the similarities and dissimilarities between learning a first a nd a second language? ( 30 points) 同上04年10月自考现代语言学试题Ⅰ.Directions: Read each of the following statements carefully. Decide which one of the four choices best completes the statement and put the letter A,B,C or D in the brackets.(2%×10= 20%)1.Chomsky uses the ter m ( ) to refer to the actual realization of a language user’s knowled ge of the rules of his language in linguistic communication.A. langueB. competenceC. paroleD. performance2.In terms of the place of articulation, the following sounds [t][d][s][z][n] share the feat ure of ( ).A. palatalB. alveolarC. bilabialD. dental3.Transformational Generative Grammar was introduced by ( ) in 1957.A. L. BloomfieldB. F. SaussureC. N. ChomskyD.M. A. K. Halliday4.Natural languages are viewed to vary according to ( ) set on UG principles to particular v alues.A. Adjacent ConditionB. parametersC. Case ConditionD. Case requirement5. Synonyms are classified into several kinds. The kind to which“girl”and“lass” belong i s called ( ) synonyms.A. stylisticB. dialectalC. emotiveD. collocational6. The illocutionary point of ( ) is to express the psychological state specified in the utt erance.A. representativesB. commissivesC. expressivesD. declaratives7. Modern English words man, woman, child, eat, fight, ect. originate from ( ).A. Middle EnglishB. Old EnglishC. FrenchD. Norman French8. In a diglossic country, the two diglossic forms of a language are generally two varieties of the same language, but there are situations in which the H-variety may have no ( ) relat ionship with the L-variety.A. geneticB. socialC. directD. close9.Many aphasics do not show total language loss. Rather, different aspects of language are i mpaired. Aphasics in ( ) area reveal word-finding difficulties and problems with syntax.A. Werniker’sB. visualC. motorD. Broca’s10. ( ) motivation occurs when the learner desires to learn a second language in order to co mmunicate with native speakers of the target language.A. InstrumentalB. FunctionalC. IntegrativeD. SocialⅡ. Directions: Fill in the blank in each of the following statements with one word, the fir st letter of which is already given as a clue. Note that you are to fill in ONE word only, a nd you are not allowed to change the letter given.(1%×10=10%)11. If a linguistic study describes and analyzes the language people actually use, it is said to bed .12. Stops, fricatives, affricates, liquids, and glides all have some degree of o and are the refore consonants.13. M is the smallest meaningful unit of language.14. A is the movement of an auxiliary verb to the sentence-initial position, such as “be”, “have”, “do” etc.15. R is what a linguistic form refers to in the real world; it is a matter of the relations hip between form and the reality.16. In Austin’s early speech act theory, c were statements that either state or describe, a nd were thus verifiable.17. In the process of first language acquisition, children usually construct their personal grammars, and their language develops in stages until it a the grammatical rules of the adul t language.18. A s community is one group, all of whose members share the same language or at least a s ingle language variety.19. People may communicate their feelings or thoughts via n signals such as facial expressio ns, gestures, postures, or proxemic space.20. Although the development of a communicative system is not unique to human beings, the natural acquisition of l as a system of highly abstract rules and regulations for creative com munication is what distinguishes humans from all other animal species.Ⅲ.Directions: Judge whether each of the following statements is true or false. Put a T for true or F for false in the brackets in front of each statement. If you think a statement is false, yo u must explain why you think so and give the correct version. (2%×10=20%)21. ( ) The writing system of a language is always a later invention used to record speech; thus there are still many languages in today’s world that can only be spoken, but not wri tt en.22. ( ) In such sound combinations as /bi:p/, /geip/ and /su:p/, the voiceless stop /p/, occ urring in the final position, is unaspirated, i.e. pronounced with the strong puff of air wi thheld to some extent.23. ( ) The part of speech of the compound is always determined by the part of speech of the second element, without exception.24. ( ) The relationship between the embedded clause and its matrix clause is one of a part to the whole.25. ( ) The contextualist view of meaning holds that meaning should be studied in terms of t he situational context and linguistic context.26. ( ) Searle’s classification of illocutionary acts is based on the classification of per formative verbs.27. ( ) One kind of language change results in an increase of the number of exceptional or i rregular morphemes. This kind of change has been called internal borrowing—that is, we “bo rrow” from one part of the grammar and apply the rule generally.28. ( ) There are differences in the way people of various age categories speak. The differe nces most easily noted by the layman are likely to be grammatical in nature.29. ( ) The left hemisphere of the brain is superior to the right hemisphere because the lef t hemisphere is language-dominant.30 ( ) A child born to a Chinese or English speaking family takes about the same number of y ears to acquire their native tongue, regardless of their general intelligence.Ⅳ.Directions: Explain the following terms, using one or two examples for illustration.(3%×10=30%)31. assimilation rule32. root33. bound morphemes34. surface structure35. grammaticality36. elaboration37. bilingualism38. creole39. the Sapir-Whorf hypothesis40. fossilizationⅤ. Answer the following questions.(10%×2=20%)41. The phonological features that occur above the level of individual sounds are called sup rasegmental features. Discuss the main suprasegmental features, illustrating with examples h ow they function in the distinction of meaning.42. Explain and give examples to show in what way componential analysis is similar to the an alysis of phonemes into distinctive features.2002年10月份全国高等教育自学考试现代语言学试题第一部分选择题Ⅰ.Directions: Read each of the following statements carefully. Decide which one of the fou r choices best completes the statement and put the letter A,B,C or D in the brackets.(2%×1 0=20%)1.The fact that different languages have different words for the same object is good proof t hat human language is ______.A. arbitraryB. non-arbitraryC. logicalD. non-productive2.All the back vowels in English are pronounced with rounded-lips,i.e. rounded, EXCEPT _____ _.3.The level of syntactic representation that exists before movement takes place is commonly termed the ______.A. phrase structureB. surface structureC. syntactic structureD. deep structure4.The theory of ______ accounts for the fact that noun phrases appear only in subject and ob ject positions.A. Case ConditionB. Adjacent ConditionC. parameterD. Adjacent parameters5.The phenomenon that words having different meanings have the same form is called ______.A. polysemyB. hyponymyC. antonymyD. homonymy6.The utterance "We're already working 25 hours a day, eight days a week." obviously violate s the maxim of ______.A. qualityB. quantityC. relationD. manner7.In first language acquisition children usually ______ grammatical rules from the linguistic information they hear.A. useB. acceptC. generalizeD. reconstruct8.Standardization known as ______ is necessary in order to facilitate communications.A. language interpretationB. language identificationC. language choiceD. language planning9.Which of the following choices is not the key biological basis for human language acquisit ion?______.A. Cerebral cortexB. NeuronsC. EyesD. Angular gyrus10.Basically all the following categories except ______ are always missing in the children's telegraphic speech stage.A. the copula verb "be"B. inflectional morphemesC. function wordsD. content words第二部分非选择题Ⅱ.Directions: Fill in the blank in each of the following statements with one word, the fir st letter of which is already given as a clue. Note that you are to fill in ONE word only, and you are not allowed to change the letter given.(1%×10=10%)nguage exists in time and changes through time. The description of a language at some p oint of time is called a _s_______ study of language.12.An essential difference between consonants and vowels is whether the air coming up from t he lungs meets with any _o________ when a sound is produced.13.The morphemes that cannot be used by themselves, but must be combined with other morpheme s to form words are called _b________ morphemes.14.XP may contain more than just X.For example, the NP "the boy who likes his puppy" consist s of Det,N and S,with Det being the _s________,N the head and S the complement.15.According to Searle's classification of illocutionary acts,"to suggest that someone should see the doctor" should fall into the category of _d________.16.Hyponymy is the relationship which obtains between specific and general lexical items.The word that is more general in meaning is called _s________.17.Vowels can be nasalized.The vowel nasalization rule is an _a________ rule,which,for the m ost part, is caused by articulatory or physiological process in which successive sounds are made identical, or more similar, to one another.18.One mark of an informal style is the frequent occurrence of _s________ words and expressi ons, which make sense only to the people of particular social groups and serve as a mark of membership and solidarity within a given social group.19.The brain is divided into two roughly symmetrical halves, called _h________, one on the r ight and one on the left.20.Linguists often use the term native language or mother tongue instead of first language, and _t________ language instead of second language in second language acquisition literatur e.Ⅲ.Directions: Judge whether each of the following statements is true or false. Put a T for true or F for false in the brackets in front of each statement. If you think a statement i s false, you must explain why you think so a nd give the correct version.(2%×10=20%)21.( )Human capacity for language has a genetic basis,i.e. we are all born with the ability to acquire language and the details of a language system are genetically transmitted.22.( )A general difference between phonetics and phonology is that phonetics is focused on t he production of speech sounds while phonology is more concerned with how speech sounds dist inguish meaning.23.( )Only words of the same parts of speech can be combined to form compounds.24.( )Sentences are not formed by randomly combining lexical items, but by following a set of syntactic rules that arrange linguistic elements in a particular order.25.( )The same semantic feature occurs in one part of speech only. For example, "female" occ urs only in nouns such as "mother", "woman" "girl" "tigress" and so on but not in other part s of speech.26.( )According to Searle's classification of illocutionary acts, inviting, ordering, advisi ng, promising and apologizing all fall into the category of directives.27.( )New words may be formed from existing words by subtracting an affix thought to be part of the old word; that is, ignorance sometimes can be creative. Thus "peddle" was derived fr om "peddler" on the mistaken assumption that the "-er" was the agentive suffix.28.( )Women in Western countries at least appear to be more status-conscious and sensitive t o the social significance of certain linguistic variables.29.( )The case of Genie confirms that the language faculty of an average human degenerates a fter the critical period and consequently, most linguistic skills cannot develop.30.( )Conscious knowledge of linguistic rules does ensure acquisition of the rules and there fore an immediate guidance for actual performance.Ⅳ.Directions: Explain the followin g terms, using one or two examples for illustration. (3%×10=30%)31.narrow transcription32.stem33.derivational affixes34.grammatical relation35.predication36.semantic narrowing37.nonstandard languages38.linguistic taboo39.angular gyrus40.interlanguageⅤ.Answer the following questions.(10%×2=20%)41.Explain with examples the three notions of phone, phoneme and allophone, and also how the y are related.42.Explain what is sense and what is reference with examples.语言学试题第一部分选择题一、单项选择题(本大题共10小题，每小题2分，共20分)在每小题列出的四个选项中只有一个选项是符合题目要求的，请将正确选项前的字母填在题后的括号内。

木仓医学考研复试SCI长难句循环内科第一章—二尖瓣返流Ischemic mitral regurgitation(IMR)is a frequent complication of left ventricular(LV)global or regional pathological remodeling due to chronic coronary artery disease.It is not a valve disease but represents the valvular consequences of increased tethering forces and reduced closing forces.IMR is defined as mitral regurgitation caused by chronic changes of LV structure and function due to ischemic heart disease and it worsens the prognosis.缺血性二尖瓣返流(IMR)是慢性冠脉疾病引起的左心室(LV)整体或局部病理性重塑的常见并发症。

它不是一种瓣膜病，但它表现了系留力增加和关闭力减少所导致的瓣膜症状。

IMR的定义为由缺血性心脏病引起的左室结构和功能的慢性改变所导致的二尖瓣返流，使患者的预后更差。

知识点总结：①ischemic adj.缺血性的，局部缺血的②mitral adj.二尖瓣的③regurgitation n.回流，返流④ventricular n.心室/adj.心室的⑤pathological adj.病理的⑥remodeling n.重塑，重建⑦coronary artery冠状动脉⑧valve/valvular n.瓣膜/adj.瓣膜的⑨tether v.系，拴住⑩prognosis n.预后木仓医学考研复试SCI长难句循环内科第二章—急性心力衰竭A diagnosis of acute heart failure（AHF）is made when patients present acutely with signs and symptoms of heart failure,often with decompensation of pre-existing cardiomyopathy.The most current guidelines classify based on clinical features at initial presentation and are used to both risk stratify and guide the management of haemodynamic compromise.Despite this,AHF remains a diagnosis with a poor prognosis and there is no therapy proven to have long-term mortality benefits.急性心力衰竭(AHF)的诊断是基于患者突然表现出心力衰竭的体征和症状，通常伴有原有心肌病的失代偿。

a rX iv:mat h /51367v1[mat h.CA]18O ct25A Weierstrass-type theorem for homogeneous polynomials David Benko,Andr´a s Kro´o †February 2,2008Abstract By the celebrated Weierstrass Theorem the set of algebraic polyno-mials is dense in the space of continuous functions on a compact set in R d .In this paper we study the following question:does the density hold if we approximate only by homogeneous polynomials?Since the set of homogeneous polynomials is nonlinear this leads to a nontrivial problem.It is easy to see that:1)density may hold only on star-like 0-symmetric surfaces;2)at least 2homogeneous polynomials are needed for approxi-mation.The most interesting special case of a star-like surface is a convex surface.It has been conjectured by the second author that functions con-tinuous on 0-symmetric convex surfaces in R d can be approximated by a pair of homogeneous polynomials.This conjecture is not resolved yet but we make substantial progress towards its positive settlement.In particu-lar,it is shown in the present paper that the above conjecture holds for 1)d =2,2)convex surfaces in R d with C 1+ǫboundary.1Introduction The celebrated theorem of Weierstrass on the density of real algebraic polyno-mials in the space of real continuous functions on an interval [a,b ]is one of the main results in analysis.Its generalization for real multivariate polynomi-als was given by Picard,subsequently the Stone-Weierstrass theorem led to the extension of these results for subalgebras in C (K ).In this paper we shall consider the question of density of homogeneous poly-nomials .Homogeneous polynomials are a standard tool appearing in many areas of analysis,so the question of their density in the space of continuous functions is a natural problem.Clearly,the set of homogeneous polynomials is substantially smaller relative to all algebraic polynomials.More importantly,this set is nonlinear,so its density can not be handled via the Stone-Weierstrass theorem.Furthermore,due to the special structure of homogeneous polynomi-als some restrictions should be made on the sets were we want to approximate (they have to be star-like),and at least2polynomials are always needed for approximation(an even and an odd one).On the5-th International Conference on Functional Analysis and Approxi-mation Theory(Maratea,Italy,2004)the second author proposed the following conjecture.Conjecture1Let K⊂R d be a convex body which is centrally symmetric to the origin.Then for any function f continuous on the boundary Bd(K)of K and anyǫ>0there exist two homogeneous polynomials h and g such that |f−h−g|≤ǫon Bd(K).From now on we agree on the terminology that by“centrally symmetric”we mean“centrally symmetric to the origin”.Subsequently in[4]the authors verified the above Conjecture for crosspoly-topes in R d and arbitrary convex polygons in R2.In this paper we shall verify the Conjecture for those convex bodies in R d whose boundary Bd(K)is C1+ǫfor some0<ǫ≤1(Theorem2).Moreover, the Conjecture will be verified in its full generality for d=2(Theorem3).It should be noted that parallel to our investigations P.Varj´u[13]also proved the Conjecture for d=2.In addition,he gives in[13]an affirmative answer to the Conjecture for arbitrary centrally symmetric polytopes in R d,and for those convex bodies in R d whose boundary is C2and has positive curvature.We also would like to point out that our method of verifying the Conjecture for d=2is based on the potential theory and is different from the approach taken in[13] (which is also based on the potential theory).Likewise our method of treating C1+ǫconvex bodies is different from the approach used in[13]for C2convex bodies with positive curvature.2Main ResultsLet K be a centrally symmetric convex body in R d.We may assume that 2≤d and dim(K)=d.The boundary of K is Bd(K)which is given by the representationBd(K):={u r(u):u∈S d−1}where r is a positive even real-valued function on S d−1.Here S d−1stands for the unit sphere in R d.We shall say that K is C1+ǫ,written K∈C1+ǫ,if the first partial derivatives of r satisfy a Lipǫproperty on the unit sphere,ǫ>0. Furthermore denote byc k x k:c k∈RH d n:=k1+...+k d=nthe space of real homogeneous polynomials of degree n in R d.Ourfirst main result is the following.2Theorem2Let K∈C1+ǫbe a centrally symmetric convex body in R d,where 0<ǫ≤1.Then for every f∈C(Bd(K))there exist h n∈H d n+H d n−1,n∈N such that h n→f uniformly on Bd(K)as n→∞.Thus Theorem2gives an affirmative answer to the Conjecture under the additional condition of C1+ǫsmoothness of the convex surface.For d=2we can verify the Conjecture in its full generality.Thus we shall prove the following. Theorem3Let K be a centrally symmetric convex body in R2.Then for every f∈C(Bd(K))there exist h n∈H2n+H2n−1,n∈N such that h n→f uniformly on Bd(K)as n→∞.We shall see that Theorem3follows fromTheorem4Let1/W(x)be a positive convex function on R such that|x|/W(−1/x)is also positive and convex.Let g(x)be a continuous function which has the same limits at−∞and at+∞.Then we can approximate g(x) uniformly on R by weighted polynomilas W(x)n p n(x),n=0,2,4,...,deg p n≤n.3Proof of Theorem2The proof of Theorem2will be based on several lemmas.The main auxiliary result is the next lemma which provides an estimate for the approximation of unity by even homogeneous polynomials.In what follows||...||D stands for the uniform norm on D.Our main lemma to prove Theorem2is the following.Lemma5Letτ∈(0,1).Under conditions of Theorem2there exist h2n∈H d2n,n∈N,such that||1−h2n||Bd(K)=o(n−τǫ).The next lemma provides a partition of unity which we shall need below. In what follows a cube in R d is called regular if all its edges are parallel to the coordinate axises.We denote the set{0,1,2,...}d by Z d+.Lemma6Given0<h≤1there exist non-negative even functions g k∈C∞(R d)such that their support consists of2d regular cubes with edge h,at most2d of supports of g k’s have nonempty intersection,andg k(x)=1,x∈R d,(1)k∈Z d+|∂m g k(x)/∂x m j|≤c/h m,x∈R d,m∈Z1+,1≤j≤d,(2) where c>0depends only on m∈Z1+and d.3For the centrally symmetric convex body K let|x|K:=inf{a>0:x/a∈K}be its Minkowski functional and setδK:=sup{|x|/|x|K:x∈R d}=max{|x|:x∈Bd(K)}. Moreover for a∈Bd(K)denote by L a a supporting hyperplane at a.Lemma7Let a∈Bd(K),h n∈H d2n be such that for any x∈L a,|x−a|≤4δK we have|h n(x)|≤1.Then whenever x∈L a satisfies|x−a|>4δK and x/t∈K we have|h n(x/t)|≤(2/3)2n.(3) Lemma8Consider the functions g k from Lemma6.Then for at most8d/2h d of them their support has nonempty intersection with S d−1.We shall verifyfirst the technical Lemmas6-8,then the proof of Lemma 5will be given.Finally it will be shown that Theorem2follows easily from Lemma5.Proof of Lemma6.The main step of the proof consists of verifying the lemma for d=1.Let g∈C∞(R)be an odd function on R such that g=1for x<−1/2and monotone decreasing from1to0on(−1/2,0).Further,let g∗(x) be an even function on R such that g∗(x)equals1on[0,1],g(x−3/2)/4+3/4on [1,2],and g(x−5/2)/4+1/4on[2,3].Then it is easy to see that g∗∈C∞(R),it equals1on[−1,1],0for|x|>3and is monotone decreasing on[1,3].Moreoverg∗(x)+g∗(x−4)=1,x∈[−1,5].(4) Set nowg k(x):=g∗(x−4k)+g∗(x+4k),k∈Z1+.Then g k’s are even functions which by(4)satisfy relation∞k=0g k(x)=1,x∈R.In addition,the support of g k equals±[−3+4k,3+4k]and at most2of g k’s can be nonzero at any given x∈R.Finally,for afixed0<h≤1,x∈R d and k=(k1,...,k d)∈Z d+setg k(x):=dj=1g kj(6x j/h).It is easy to see that these functions give the needed partition of unity.Proof of Lemma7.Clearly the conditions of lemma yield that whenever |x−a|>4δK1/|x|K≤δK/|x|≤δK/(|x−a|−|a|)≤δK/(|x−a|−δK)≤4δK/3|x−a|.(5) It is well known that for any univariate polynomial p of degree at most n such that|p|≤1in[−a,a]it holds that|p(x)|≤(2x/a)n whenever|x|>a.Therefore using(5)and the assumption imposed on h n we have|h n(x)|≤(2|x−a|/4δK)2n≤(2|x|K/3)2n.(6) Now it remains to note that by x/t∈K it follows that|x|K≤|t|,and thus we obtain(3)from(6).This completes the proof of the lemma.Proof of Lemma5.Denote by g k,1≤k≤N those functions from Lemma 6whose support A k has a nonempty intersection with S d−1.Then by Lemma8N≤8d/2h d.(9) Moreover,by(1)Ng k=1on S d−1.(10)k=1SetB k:=A k∩S d−1,C k:={u r(u):u∈B k}⊂Bd(K),1≤k≤N. For each1≤k≤N choose a point u k∈B k and set x k:=u k r(u k)∈Bd(K). Furthermore let L k be the supporting plane to Bd(K)at the point x k and set for1≤k≤N,L∗k:=L k∪(−L k)D k:={x∈L∗k:x=t u for some u∈B k,t>0};5f k(x):=g k(u),x∈Bd(K),x=u r(u),u∈S d−1q k(x):=g k(u),x∈L∗k,x=t u,u∈S d−1,t>0. Clearly,q k∈C∞(L∗k)is an even positive function which by property(2)canbe extended to a regular centrally symmetric cube I⊃K so that we have on I |∂m q k/∂x m j|≤c/h m,1≤j≤d,1≤k≤N.(11) Here and in what follows we denote by c(possibly distinct)positive constants depending only on d,m and K.We can assume that I is sufficiently large so thatI⊃G k:={x∈L k:|x−x k|≤4δK},1≤k≤N.Then by the multivariate Jackson Theorem(see e.g.[10])applied to the even functions q k satisfying(11)for arbitrary m∈N(to be specified below),there exist even multivariate polynomials p k of total degree at most2n such that||q k−p k||G∗k≤c/(hn)m≤1,1≤k≤N,(12) where G∗k:=G k∪(−G k),h:=n−γ(0<γ<1is specified below),and n is sufficiently large.We claim now that without loss of generality it may be assumed that each p k is in H d2n.Indeed,since G∗k⊂L∗k it follows that the homogeneous polynomial h2:=<x,w>2∈H d2is identically equal to1on G∗k(here w is a properly normalized normal vector to L k),so multiplying the even degree monomials of p k by even powers of h2we can replace p k by a homogeneous polynomial from H d2n so that(12)holds.Thus we may assume that p k∈H d2n and relations(12) hold.In particular,(12)also yields that||p k||G∗k≤2,1≤k≤N.(13) Now consider an arbitrary x∈Bd(K)\C k.Then with some t>1we have t x∈L∗k and q k(t x)=0.Hence if t x∈G∗k then by(12)it follows that|p k(x)|≤|p k(t x)|≤c/(hn)m.On the other hand if t x/∈G∗k then by(13)and Lemma7we obtain|p k(x)|≤2(2/3)2n.The last two estimates yield that for every x∈Bd(K)\C k we have|p k(x)|≤c((2/3)2n+(hn)−m),1≤k≤N.(14) Now let us assume that x∈C k.Clearly,the C1+ǫproperty of Bd(K)yields that whenever x∈Bd(K),t x∈L∗k,t>1we have for every1≤k≤N(t−1)|x|=|x−t x|≤c min{|x−x k|,|x+x k|}1+ǫ.(15)6Obviously,for every u∈B kmin{|u−u k|,|u+u k|}≤√1+m+ǫ+d >τǫand letγ:=1+mProof of Theorem2.First we use the classical Weierstrass Theorem to approximate f∈C(Bd(K))by a polynomialp m=mj=0h∗j,h∗j∈H d j,0≤j≤mof degree at most m so that||f−p m||Bd(K)≤δ7with any givenδ>0.Letτ∈(0,1)be arbitrary.According to Lemma5 there exist h n,j∈H d2n−2[j/2]such that||1−h n,j||Bd(K)=O(n−τǫ),0≤j≤m.Clearly,h∗:=mj=0h n,j h∗j∈H d2n+H d2n+1and||f−h∗||Bd(K)≤δ+O(n−τǫ).W(t)is positive and convex on R(19)|t|t )is positive and convex on R.(20)Remark10Equivalently,instead of(20)we may assume that(21)below holds and lim t→+∞t(tQ′(t)−1)≤lim t→−∞t(tQ′(t)−1).We also remark that(19) implies that(20)is satisfied on(−∞,0)and on(0,+∞).We mention the function W(t)=(1+|t|m)−1/m,1≤m,as an example which satisfies(19)and(20).We say that a property is satisfied inside R if it is satisfied on all compact subsets of R.Some consequences of(19)and(20)are as follows.8limt→±∞|t|W(t)=ρ∈(0,+∞)exists.(21) Since exp(Q(t))is convex,it is Lipschitz continuous inside R.So exp(Q(t)) is absolutely continuous inside R which implies that both W(t)and Q(t)are absolutely continuous inside R.Q′(t)is bounded inside R a.e.because by(19)exp(Q(t))Q′(t)is increasing a.e.We collected below some frequently used definitions and notations in the paper.Definitions11Let L⊂R and let f:L→R∪{−∞}∪{+∞}.f is H¨o lder continuous with H¨o lder index0<τ≤1if with some K constant |f(x)−f(y)|≤K|x−y|τ,x,y∈L.In this case we write f∈Hτ(L).The L p norm of f is denoted by||f||p.When p=∞we will also use the ||f||L notation.We say that an integral or limit exists if it exists as a real number.Let x∈R.If f is integrable on L\(x−ǫ,x+ǫ)for all0<ǫthen the Cauchy principal value integral is defined asP V L f(t)dt:=limǫ→0+ L\(x−ǫ,x+ǫ)f(t)dt,if the limit exists.It is known that P V L g(t)/(t−x)dt exists for almost every x∈R if g: L→R is integrable.For0<ιand a∈R we definea+ι:=max(a,ι)and a−ι:=max(−a,ι).For a>b the interval[a,b]is an empty set.We say that a property is satisfied inside L if it is satisfied on all compact subsets of L.o(1)will denote a number which is approaching to zero.For example,we may write10x=100+o(1)as x→2.Sometimes we also specify the domain(which may change withǫ)where the equation should be considered.For example, sin(x)=o(1)for x∈[π,π+ǫ]whenǫ→0+.The equilibrium measure and its support S w is defined on the next page.Let [aλ,bλ]denote the support S Wλ(see Lemma(15)).For x∈(aλ,bλ)let Vλ(x):=0,and for a.e.x∈(aλ,bλ)letVλ(x):=P V bλaλλ√t−x dt(x−aλ)(bλ−x)+1(x−aλ)(bλ−x).(22)Let x∈[−1,1].Depending on the value of c∈[−1,1]the following integrals may or may not be principal value integrals.v c(x):=−P V c−1λ√π2√h c(x):=P V 1cλ√π2√1−t2e−Q(t)dt,x∈[−1,1].1−x2(t−x)P n(x)and p n(x)denote polynomials of degree at most n.Functions with smooth integrals was introduced by Totik in[11].Definitions12We say that f has smooth integral on R⊂L,if f is non-negative a.e.on R andf=(1+o(1)) J f(23)Iwhere I,J⊂R are any two adjacent intervals,both of which has length0< epsilon,andǫ→0.The o(1)term depends onǫand not on I and J.We say that a family of functions F has uniformly smooth integral on R, if any f∈F is non-negative a.e.on R and(23)holds,where the o(1)term depends onǫonly,and not on the choice of f,I or J.Cleary,if f is continuous and it has a positive lower bound on R then f has smooth integral on R.Also,non-negative linear combinations offinitely many functions with smooth integrals on R has also smooth integral on R.From the Fubini Theorem it follows that ifνis afinite positive Borel measure on T⊂R and{v t(x):t∈T}is a family of functions with uniformly smooth integral on R for which t→v t(x)is measureable for a.e.x∈[a,b],thenv(x):= T v t(x)dν(t)has also smooth integral on R.Finally,if f n→f uniformly a.e.on R,f n has smooth integral on R and f has positive lower bound a.e.on R then f has smooth integral on R. Remark13Since exp(−Q)is absolutely continuous inside R and(exp(−Q))′=−exp(−Q)Q′is bounded a.e.on[−1,1],by the fundamental theorem of calcu-√1+t∈H0.5([−1,1]), lus we see that exp(−Q(t))∈H1([−1,1]).And√1+t exp(−Q(t))∈H0.5([−1,1])so√soLet w(x)≡0be a non-negative continuous function on¯R such thatlimx→∞|x|w(x)=α∈[0,+∞)exists.(24) Whenα=0,then w belongs to the class of so called“admissible”weights.We write w(x)=exp(−q(x))and call q(x)externalfield.Ifµis a positive Borel unit measure on¯R-in short a“probability measure”,then its weighted energy is defined byI w(µ):= log1|t−x|dµ(t).This definition makes sense for a signed measureν,too,if log|t−x| d|ν|(t) exists.LetS w:=supp(µw)denote the support ofµw.Whenα=0,then S w is a compact subset of R.In this case with some F w constant we haveUµw+Q(x)=F w,x∈S w.Let Bd(K)be the boundary of a two dimensional convex region K⊂R2 which is centrally symmetric to the origin(0,0).For t∈R let(x(t),y(t))be any of the two points on Bd(K)for whichy(t)Lemma14W(t)satisfies properties(19),(20).And S W=¯R.Proof.W is positive on R.We may assume that x(t)>0,t∈R.Let t1,t3∈R and t2:=αt1+(1−α)t3,where0<α<1.Let(x2,y2)be the intersection of the line segments(0,0)(x(t2),y(t2)).Note that1/x(t2)≤1/x2and by elementary calculations:1x(t1)+(1−α)1Lemma15Let1<λ.Then S Wλis afinite interval[aλ,bλ],andµWλis absolutely continuous with respect to the Lebesgue measure and its density is dµWλ(x)=Vλ(x)dx.Proof.Let1<p.Note that exp(λQ(x))is a convex function because it is the composition of two continuous convex functions.So by[2],Theorem5, S Wλis an interval[aλ,bλ],which isfinite since lim x→±∞|x|Wλ(x)=0.The density function(dµWλ(x))/dx exists,since(Wλ)′=−exp(−λQ)λQ′∈L p(R),see Theorem IV.2.2of[8].The integral at(22)is the Hilbert transform on R of the function defined as λF Wλ,x∈[aλ,bλ],we get Uµ(x)=UµWλ(x),x∈(aλ,bλ).But(27)shows that Uµ+(x)and Uµ−(x)arefinite for all x∈[aλ,bλ].So Uµ+(x)=UµWλ+µ−(x), x∈(aλ,bλ).Hereµ+andµWλ+µ−are positive measures which have the same mass.µWλ,µ−(andµ+)all havefinite logarithmic energy(see(27)),henceµWλ+µ−has it,too.Applying Theorem II.3.2.of[8]we get Uµ+(z)=UµWλ+µ−(z)for all z∈C.By the unicity theorem([8],Theorem II.2.1.)µ+=µWλ+µ−.Henceµ=µWλand our lemma is proved.||µW [−N,N]||.We remark thatµW({∞})=0which implies that||µW [−N,N]||→1as N→+∞.(29) By([9],pp.3)there exists K∈R such that1K≤log|z−t|Wλ1(z)Wλ1(t)dνN(t)dνN(z)isfinite.(31) By(30)the double integral at(31)is bounded from below.It equals to: log1Here the first double integral is finite because V W is finite ([9],Theorem1.2).And thesecondintegral is bounded from above since νN has compact support.So (31)is established.Choose 0<τsuch that ||τW (x )||∞≤1.Now,I W (µ)−log(τ2)=lim M →+∞ min M,log 1|z −t |(τW (z ))(τW (t ))dµW λn (t )dµW λn (z )≤lim n →+∞ log 1|z −t |(τW (z ))λn (τW (t ))λndνN (t )dνN (z )= log 1|z −t |W (z )W (t )−K dνN (t )dνN (z )≤ log 1||µW[−N,N ]||dµW (z )||µW [−N,N ]||2 +1λ2ωS W λ S W λ2,14whereωSWλis the classical equilibrium measure of the set S Wλ(with no externalfield present).(We remark that S Wλ⊃SWλ2.)It follows that ifλis so close to1that SWλ2⊃I holds,then[a,b]⊂(aλ,bλ)and Vλ(x)has positive lower bound a.e.on[a,b].c−x,as x→c−.Here o(1)depends on c−x only.Lemma18Let−1<a<b<1and0<ιbefixed.Let0<ǫ<1/10and δ:=√v c(x2)+ι,v c(x1)−ιh c(x2)+ι,h c(x1)−ιt−x2≤1+x2−x1t−x1=(1+o(1))11−t2exp(−Q(t))/π2and integrating on [c,1]we gainh c(x2)1−x22/v c(x1)+ι=1+o(1).(34)Returning to the case of x1,x2≤c−δ,from v c(x)=h c(x)+B(x),from (33)and from B(x2)=B(x1)+o(1)we get|v c(x2)−v c(x1)|=|o(1)|(1+|v c(x1)−B(x1)|)≤|o(1)|(|v c(x1)|+1+||B||[a,b]).(35)15Assuming|v c(x1)|≤1,we have|v c(x2)+ι−v c(x1)+ι|≤|v c(x2)−v c(x1)|≤|o(1)|,so(34)holds again.Finally,if|v c(x1)|≥1,then from(35)v c(x2)|v c(x1)|=|o(1)|,from which(34)again easily follows.The proof of the rest of our lemma is similar.J v c(t)+ιdt=1+o(1),asǫ→0+,where o(1)is independent of I,J and c.Letδ:=√ǫ,c+√1−t2exp(−Q(t))/π2.Applying Lemma17(with A:=(a−1)/2,B:=(b+1)/2)we have√ǫ,c)asǫ→0+,which easily leads toh c(x)=(f(c)1−c2+o(1))(−log|c−x|)for x∈[c−√√ǫ,c)asǫ→0+.(36)Clearly,(36)also holds for x∈(c,c+√J v c(t)+ιdt=(f(c)1−c2+o(1)) I log1(f(c)1−c2+o(1)) J log1where we used that log(1/|x|)has smooth integral on[−2,2]([1],Proposition20).Following the proof of Lemma24of[1]we will prove the following lemma. But we remark that the absolutely continuous hypothesis of Lemma24is un-necessary at[1].Lemma21Let N(x)be a bounded,increasing,right-continuous function on [−1,1]and let f(x)∈L1([−1,1])be non-negative.ThenP V 1−1f(t)N(t)t−xdt, a.e.x∈[−1,1].Proof.Let us denote the left hand side of(38)by F(x).Since f(x)and f(x)N(x)are in L1[−1,1]and N(x)is increasing,there is a set of full mea-sure in(−1,1)where f1(x),F(x)and N′(x)all exist.Let x be chosen from this set.It follows that f c(x)exist for all c∈[−1,1]\{x}.Also,F(x)=limǫ→0+ x−ǫ−1f(t)N(t)t−xdt .(39) 17t→f t(x)is a continuous increasing function on[−1,x)and it is a continuous decreasing function on(x,1]so at(39)we can use integration by parts to get x−ǫ−1+ 1x+ǫ=−f x−ǫ(x)N(x−ǫ)+f−1(x)N(−1)+ (−1,x−ǫ]f t(x)dN(t)+f x+ǫ(x)N(x+ǫ)−f1(x)N(1)+ (x+ǫ,1]f t(x)dN(t)But above f−1(x)=0andf x+ǫ(x)N(x+ǫ)−f x−ǫ(x)N(x−ǫ)=[f x+ǫ(x)−f x−ǫ(x)]N(x+ǫ)+f x−ǫ(x)[N(x+ǫ)−N(x−ǫ)].(40) Note thatf x+ǫ(x)−f x−ǫ(x)=−P V x+ǫx−ǫf(t)Lemma22Let[a,b]be arbitrary and let1<λbe chosen to satisfy the conclu-sion of Lemma16.Then Vλ(x)has smooth integral on[a,b].Proof.To keep the notations simple we will assume that−1<a<b<1, and aλ=−1,bλ=1,that is,the support ofµWλis[−1,1].This can be done without loss of generality.Definev(t):=λ√π2√where v(t)also depends on the choice of x.Note that M(t),t∈[−1,1],is a bounded,increasing,right-continuous function which agrees with exp(Q(t))Q′(t) almost everywhere.Applying Lemma21for f(t):=v(t)and N(t):=M(t),let usfix an x∈[a,b] value for which both(38)and dµWλ(x)=Vλ(x)dx are satisfied.(These are satisfied almost everywhere.)From(22)and Lemma21we haveVλ(x)=11−x2+P V 1−1λ√π2√π√t−x dt=L(x)+(−1,1]v t(x)dM(t),where L(x):=1/(π√Vλ(x)(ι)has also positive lower bound a.e.on[a,b],assumingιis small enough. In addition,v t(x)≥0when t∈[0,x],whereas v t(x)≥B(x)≥−||B||[a,b]whent∈(x,1],so Vλ(x)(−)(ι)is bounded a.e.on[a,b].It follows that Vλ(x)(−)(ι)≤(1−η)Vλ(x)(+)(ι)a.e.x∈[a,b]for someη∈(0,1).Applying Lemma20we conclude that Vλ(x)(ι)has smooth integral on[a,b] (ifιis small enough).Therefore Vλ(x)has smooth integral by(44)and(45).We now restate Theorem4and prove it.Theorem24For a weight satisfying(19)and(20)we have Z¯R(W)=∅.That is,any continuous function g:¯R→R can be uniformly approximated by weighted polynomials W n p n(n=0,2,4,...)on¯R.Proof.Let x0∈¯R.We show that x0∈Z¯R(W).20First let us assume that x0isfinite.Choose J:=[a,b]such that a<x0<b holds.Let f(x)be a continuous function which is zero outside J and f(x0)=0. Let1<λ=u/v(u,v∈N+)be a rational number for which the conclusion of Lemma16holds.Now we use a powerful theorem of Totik.Since Vλhas a positive lower bound a.e.on J and it has smooth integral on J(see Lemma22), by[11],Theorem1.2,(a,b)∩Z(Wλ)=∅.So we canfind P n(n=0,1,2,...) such that(Wλ)n P n→f uniformly on¯R.So for n:=Nv,we haveW Nu p Nu→f,N=0,1,2,...,uniformly on¯R,(46) where p Nu:=P Nv and deg(p Nu)≤Nv≤Nu.For allfixed s∈{0,...,u−1}if we approximate f/W s instead of f at(46),it easily follows that there exist p k (k=0,1,2,...)such thatW k p k→f,k=0,1,2,...,uniformly on¯R.(47) Using only k=0,2,4,...,we get x0∈Z¯R(W)by Lemma23.Now let x0=∞.DefineW0(x):=1x).Note that1/W0(x)(=|x|/W(−1/x))and|x|/W0(−1/x)(=1/W(x))are posi-tive and convex functions because W satisfies(20)and(19).Let g be a continuous function on¯R.Define−1/∞to be0and−1/0to be∞.(So g(x)is continuous on¯R if and only if g(−1/x)is continuous on¯R.) Observe that for some p n we haveW n(x)p n(x)→g(x)(n=0,2,4,...)uniformly on¯R,iffW n(−1/x)p n(−1/x)→g(−1/x)(n=0,2,4,...)uniformly on¯R,iffW0n(x)q n(x),→g(−1/x)(n=0,2,4,...)uniformly on¯R,where q n(x):=x n p n(−1/x)are polynomials,deg q n≤n.Now let f(x)be a continuous function on¯R which is zero in a neighborhood of0but f(∞)=0.By what we have already proved,q n polynomials exist such that W0n(x)q n(x)(n=0,2,4,...)tends to f(−1/x)uniformly.Therefore we can approximate f(x)uniformly by W n(x)p n(x)(n=0,2,4,...),where p n(x):=x n q n(−1/x).Proof.Recall the definition:y(t)/x(t)=t,t∈¯R,where(x(t),y(t))∈Bd(K) and W(t):=|x(t)|.Definef(t):=f(x(t),y(t))=f(−x(t),−y(t)),t∈¯R.Note that if n is an even number(and a(n)kare unknowns)thennk=0a(n)kx n−k(t)y k(t)=x n(t)nk=0a(n)k y(t)Proof of Theorem3.Define f(x,y):=1,(x,y)∈Bd(K).By Lemma25there exist h2n∈H22n, n∈N,such that||1−h2n||Bd(K)→0.From here Theorem3follows the same way Theorem2follows from Lemma5.[7]E.B.Saff,Incomplete and orthogonal polynomials.In C.K.Chui,L.L Schu-maker,and J.D.Ward,editors,Approximation Theory IV,pp219-255.Academic Press,New York,1983[8]E.B.Saff,V.Totik,Logarithmic Potentials with External Fields,Springer-Verlag,Berlin,1997[9]P.Simeonov,A minimal weighted energy problem for a class of admissibleweights,Houston Journal of Mathematics,to appear[10]A.F.Timan,Theory of Functions of a Real Variable,(Moscow,1960)(inRussian).[11]V.Totik,Weighted polynomial approximation for convex externalfields,Constr.Approx.16(2)(2000)261-281.[12]V.Totik,Weighted approximation with varying weight.Lecture Notes inMathematics,1569.Springer-Verlag,Berlin,1994.[13]P.Varj´u,Approximation by homogeneous polynomials,submitted to Con-str.Approx.D.BenkoDepartment of MathematicsWestern Kentucky UniversityBowling Green,KY42101USAE-mail:dbenko2005@A.Kro´oAlfr´e d R´e nyi Institute of MathematicsHungarian Academy of SciencesH-1053Budapest,Re´a ltanoda u.13-15HungaryE-mail:kroo@renyi.hu23。

小学上册英语第5单元期末试卷[含答案]英语试题一、综合题(本题有50小题，每小题1分，共100分.每小题不选、错误，均不给分)1 What do we call a shape with four equal sides?A. RectangleB. SquareC. TriangleD. Pentagon2 The _____ (ocean/lake) is calm.3 What do we call the part of the plant that absorbs water and nutrients from the soil?A. StemB. LeafC. RootD. Flower答案：C4 A small ___ (小虾) swims in the ocean.5 The chemical symbol for silver is ______.6 What do you call a vehicle with two wheels?A. CarB. TruckC. BicycleD. Bus7 What is the name of the famous landmark in Egypt?A. ColosseumB. Great WallC. PyramidsD. Stonehenge答案: C8 My ______ loves to dance.9 We visited the ________ (动物园) last week.10 They like to ________ outside.11 A __________ (溶液) is a homogeneous mixture of two or more substances.12 My dad is a skilled __________ (机械师).13 A baby cat is called a ______.14 A _____ (草甸) is a grassy area with wildflowers.15 What is the primary color that mixes with blue to create green?A. RedB. YellowC. BlackD. White答案:B16 A __________ is a change that produces one or more new substances.17 A _______ (小鼹鼠) digs tunnels in the soil.18 What is the process of producing electricity using water?A. HydropowerB. Solar powerC. Wind powerD. Nuclear power答案:A19 What do you call the first meal of the day?A. LunchB. DinnerC. BreakfastD. Snack答案:C20 The color of cabbage juice changes with pH; it can be red or ______.21 The __________ (历史的精神) lives on in our traditions.22 My ________ (玩具名称) is a colorful rainbow.23 The _____ (海豹) is often seen on rocky shores.24 The teacher is very ________.25 The process of ______ can lead to the formation of new soil.26 A lizard can change color to adapt to its ______ (环境).27 What do we call the act of making a choice?A. Decision-makingB. PlanningC. OrganizingD. Arranging答案:A28 How many sides does a hexagon have?A. FourB. FiveC. Six29 The __________ (历史的实证) supports arguments.30 She is a great ___. (singer)31 The __________ is a historical site in Turkey. (特洛伊)32 The _______ (The Great Depression) led to widespread economic turmoil.33 My dad is known for his __________ (智慧).34 Goats like to climb _______ (山).35 What is the primary purpose of a map?A. To tell timeB. To guide travelersC. To measure distancesD. To show weather答案: B. To guide travelers36 What is the term for the time between sunset and night?A. DawnB. DuskC. NoonD. Midnight答案:B37 During summer, we often have ______ (野餐) in the park. We bring sandwiches and ______ (饮料) to enjoy.38 We visit the _____ (水族馆) often.39 What is the capital of Japan?A. BeijingB. SeoulD. Bangkok40 I call my grandmother __________. (奶奶/外婆)41 In spring, flowers start to __________ as the weather gets warmer. (盛开)42 What do we use to cut paper?A. GlueB. ScissorsC. TapeD. Pencil答案:B43 I planted ______ (花) in my garden. I hope they bloom ______ (快).44 The __________ can help us understand the dynamics of natural systems.45 What is the capital of Venezuela?A. CaracasB. QuitoC. BogotáD. Lima答案: A46 What is the opposite of 'hot'?A. WarmB. ColdC. CoolD. Mild答案:B47 The __________ helps some animals to see in the dark.48 The _____ (歌声) is beautiful.49 A __________ is a region known for its cultural heritage.50 My _____ (小兔) is very fluffy.51 What is the main ingredient in chocolate?A. CocoaB. SugarC. MilkD. Flour答案:A52 The ______ is a great motivator.53 My _______ (狗) barks at everyone who passes by.54 The __________ is a famous natural landmark in Canada. (尼亚加拉瀑布)55 The study of rocks can reveal information about the Earth's ______.56 The bird is ________ in the sky.57 We have a _____ (梦) of traveling.58 Certain plants have unique characteristics that help them ______ in their native habitats. (某些植物有独特的特征，帮助它们在原生栖息地生存。

中西医结合治疗慢性胰腺炎的疗效观察李明发布时间：2021-09-16T05:54:14.146Z 来源：《健康世界》2021年9期作者：李明胡枫[导读] 探讨联合中西医治疗慢性胰腺炎疾病的效果。

方法：研究对象以2018年2月至2020年10月来我院接受诊治的98例慢性胰腺炎患者进行研究，根据不同的治疗方法分为常规治疗组（予以常规西医治疗）与联合治疗组（予以中西医联合治疗），对两组的临床疗效及症状改善情况进行比较。

射洪市人民医院四川射洪 629200摘要：目的：探讨联合中西医治疗慢性胰腺炎疾病的效果。

方法：研究对象以2018年2月至2020年10月来我院接受诊治的98例慢性胰腺炎患者进行研究，根据不同的治疗方法分为常规治疗组（予以常规西医治疗）与联合治疗组（予以中西医联合治疗），对两组的临床疗效及症状改善情况进行比较。

结果：联合治疗组临床治疗总有效率（95.9%）显著高于常规治疗组（81.6%），联合治疗组腹痛、脂肪泻、腹胀及消化不良评分显著低于常规治疗组腹痛评分，组间比较有明显差异性（P＜0.05）。

结论：对慢性胰腺炎患者采用中西医结合治疗，能有效改善患者症状，提高临床疗效，值得临床推广。

关键词：慢性胰腺炎；中西医；联合治疗；症状积分；疗效Observation on Curative Effect of Integrated Traditional Chinese and Western Medicine on Chronic PancreatitisLi Ming Hu Feng（Shehong People's Hospital，Shehong，Sichuan 629200）【Abstract】Objective：To explore the effect of combined Chinese and Western medicine in the treatment of chronic pancreatitis.Methods：The study subjects were 98 patients with chronic pancreatitis who came to our hospital for diagnosis and treatment from February 2018 to October 2020.According to different treatment methods，they were divided into conventional treatment group（conventional western medicine treatment）and combined treatment group（Combined treatment of Chinese and Western medicine），the clinical efficacy and symptom improvement of the two groups were compared.Results：The total effective rate of clinical treatment in the combined treatment group（95.9%）was significantly higher than that in the conventional treatment group（81.6%）.The scores of abdominal pain，steatorrhea，bloating and dyspepsia in the combined treatment group were significantly lower than those of the conventional treatment parison between groups There are significant differences（P<0.05）.Conclusion：The use of integrated traditional Chinese and western medicine for patients with chronic pancreatitis can effectively improve the symptoms of patients and improve the clinical efficacy.It is worthy of clinical promotion.[Keywords] chronic pancreatitis；Chinese and Western medicine；combined therapy；symptom score；curative effect慢性胰腺炎是临床常见且多发的慢性炎症疾病，具有病情凶险、发病迅速、易反复发作等特点，最终易造成患者全部或部分丧失胰腺内外分泌功能[1-2]。

生物英语测试题及答案一、选择题（每题2分，共20分）1. Which of the following is not a characteristic of living organisms?A. ReproductionB. ResponsivenessC. GrowthD. Inertia2. What is the basic unit of life?A. CellB. TissueC. OrganD. Organ system3. What is the process by which organisms convert sunlight into chemical energy?A. PhotosynthesisB. RespirationC. FermentationD. Digestion4. What is the term for the study of the structure of organisms?A. PhysiologyB. AnatomyC. PathologyD. Embryology5. Which of the following is not a kingdom in the classification of life?A. PlantaeB. AnimaliaC. FungiD. Crystallia6. What is the function of chlorophyll in plants?A. To store waterB. To store foodC. To absorb lightD. To release oxygen7. What is the process by which organisms break down food into energy?A. PhotosynthesisB. RespirationC. FermentationD. Digestion8. What is the term for the genetic material found in the nucleus of a cell?A. DNAB. RNAC. ProteinD. Enzyme9. What is the term for the study of the classification of organisms?A. TaxonomyB. AnatomyC. PhysiologyD. Ecology10. Which of the following is not a type of tissue in plants?A. Meristematic tissueB. Vascular tissueC. Connective tissueD. Protective tissue二、填空题（每空1分，共10分）1. The process by which organisms use oxygen to break down glucose is called ________.2. The largest organ in the human body is the ________.3. The study of the interactions between organisms and their environment is known as ________.4. The basic unit of heredity in all organisms is the________.5. The process by which plants produce food is called________.三、简答题（每题10分，共20分）1. Explain the difference between mitosis and meiosis in terms of their roles in cell division.2. Describe the role of mitochondria in a cell and why they are often referred to as the "powerhouses" of the cell.四、论述题（每题15分，共30分）1. Discuss the importance of biodiversity and how human activities can affect it.2. Explain the concept of natural selection and provideexamples of how it has shaped species over time.答案：一、选择题1. D2. A3. A4. B5. D6. C7. B8. A9. A10. C二、填空题1. Respiration2. Skin3. Ecology4. Gene5. Photosynthesis三、简答题1. Mitosis is the process of cell division that results in two daughter cells each having the same number of chromosomes as the parent cell. It is essential for growth, repair, and asexual reproduction. Meiosis, on the other hand, is a type of cell division that reduces the chromosome number by half, resulting in four non-identical daughter cells. This process is crucial for sexual reproduction, as it allows for the combination of genetic material from two parents, resultingin genetic variation in offspring.2. Mitochondria are organelles found in the cells of eukaryotic organisms and are often referred to as the "powerhouses" of the cell because they generate most of the cell's supply of adenosine triphosphate (ATP), which is used as a source of chemical energy. Mitochondria are unique among organelles because they contain their own DNA and can replicate independently of the cell.四、论述题1. Biodiversity is crucial for maintaining the balance of ecosystems and for the health of the planet. It provides a variety of services, such as pollination, water purification, and climate regulation. Human activities, such as deforestation, pollution, and climate change, cansignificantly reduce biodiversity. These activities can lead to the loss of species, disruption of ecosystems, and a decrease in the services that nature provides to humans.2. Natural selection is the process by which organisms that are better adapted to their environment tend to survive and produce more offspring. Over time, this leads to the evolution of species. For example, the peppered moth in England underwent a change in coloration due to industrial pollution, where darker moths became more common because they were less visible to predators against the soot-darkened trees. This is a classic example of how natural selection can lead to changes in species over time.。

小学上册英语第6单元期中试卷(有答案)英语试题一、综合题(本题有100小题，每小题1分，共100分.每小题不选、错误，均不给分)1.What is the capital of France?A. ParisB. LondonC. TokyoD. Berlin答案:A Paris2.My ________ (玩具名称) is a great companion.3.The ________ was a major event in the history of technological evolution.4.Which insect makes honey?A. AntB. FlyC. BeeD. Mosquito答案: C. Bee5.The butterfly's wings are delicate and ______ (美丽).6.She is ___ (laughing/sobbing) at the movie.7.The playground is ______ (full) of children.8.What do you call a collection of songs performed by a musician?A. AlbumB. PlaylistC. MixtapeD. Compilation答案: A9.The chemical formula for arachidic acid is ______.10.My pet _____ loves to cuddle and play.11.The dog is ________ outside.12.My dad encourages me to be __________ (开放的) to opportunities.13.What is the name of the famous landmark in Paris?A. ColosseumB. Eiffel TowerC. Statue of LibertyD. Big Ben答案: B14.In a chemical reaction, the products are formed from the _____.15.What is the term for plants that grow in water?A. TerrestrialB. AquaticC. XerophyticD. Epiphytic答案:B16. A __________ is a landform that rises above the surrounding area.17.The Cold War was a period of tension between _______ and the West.18.I enjoy ______ (旅行) during the summer.19.She has a _____ (colorful) backpack.20.The Earth's surface features are shaped by ______ forces.21. A chemical reaction can be identified by the release of ______.22.The ________ was a major conflict that defined a generation.23.The chemical symbol for zinc is ______.24. A __________ is a substance made of two or more elements.25.连词成句。