已知(L[e^{-at}]=dfrac{1}{s+a})，则(L[te^{-at}])为（）

(y''+4y'+3y=e^{-t})，(y(0)=y'(0)=1)的解为(y(t)=frac{1}{4}[(7+2t)e^{-t}-3e^{-3t}])（）

(2e^{-4t}-e^{-2t})的拉普拉斯变换为(frac{1}{(s+2)(s+4)})

23、抛物线的离心率为e=1.

Part1Reading(Readthefollowingthreepassages,andchoosethebestanswerforeachblank.)Text1.1-5Text1Inayearhelearned1andLatinwellenoughtoreadsimpletexts;oftenhiseyeswereredand2fromstrainandlackofsleep.Sometimeshethoughtofhimselfashehadbeenafewyearsbeforeandasastonishedbythememoryofthatstrangefigure,brownandpassiveastheearth3ithademerged.Hethoughtofhisparents,andtheywerenearlyasstrangeasthechildtheyhadborne;he4amixedpityforthemandadistantlove.NearthemiddleofthehisfourthyearattheUniversity,ArcherSloanestoppedhimonedayafterclassandaskedhimto5hisofficeforachat.（1-5）

PartIReading:TextCompletion(Readthefollowingthreepassagesandchoosethebestanswertoeachoftheblank.)Text1Paula,Thankyouforvolunteeringforthecharitycampaignthisyear.Iknowitisalotofextraworkforyou,butIthinkitiswonderfulthatwecanhelpthepoorchildreninourcity.1_____.Firstofall,ourown2_____regularlymakedonations.Manyofourworkersregularlygivemoneyorvolunteertheirtimetoworkonthischaritycampaign.Secondly,wecollectmoneyfromourcustomers.Wehavecollectionboxesplacedatthecashregisters,socustomerscanputintheirspare3_____.Thethirdsourcecomesfromcorporatedonations,especiallythosethatwedobusinesswith,suchasour4_____,whowebuyfrom.IfthereisanythingIcandotohelp,pleasecometalktome.Let’smakethisyear’scampaignabig5____.1.

弹性模量（E）、切变模量（G）、泊松比（μ）三者关系为G=E/[2(1+μ)]。

1．下面链接外部JavaScript正确的是（）。A、<scriptsrc="animation.js"></script>B、<linksrc="animation.js">C、<scripthref="animation.js"></script>D、<stylesrc="animation.js"></style>

1．为实例对象动态添加一个成员时，该成员将保存在（）。A、该对象中B、该对象的原型对象中C、该对象的构造函数中D、原型对象中不存在时保存到原型对象，否则保存到当前对象

1．以下选项不能用作函数名开头的是（）。A、字母B、数字C、下划线（_）D、$符号

1．下列选项中表示十进制数字9的是（）。A、0001B、1001C、1101D、0101

1．阅读以下代码，输出结果为（）。vari=24;for(leti=0;i<10;++i){}console.log(i);A、24B、9C、10D、undefined

1．达式“['2','1',2,1].includes('1',3)”的输出结果为（）。A、trueB、falseC、1D、0

以下程序的输出结果是ls1=[1,2,3,4,5]ls2=[3,4,5,6,7,8]cha1=[]foriinls2:ifinotinls1:cha1.append(i)print(cha1)

以下代码的输出结果是foriinrange(1,6):ifi%4==0:breakelse:print(i,end=’,’)