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如图,在平行四边形abcd中,点e是bc边上的一点,且ab=be,ae的延长线交dc的延长线于点f,若∠f=62°,试求平行四边形abcd的各个内角的度数 微积分…超级感谢… 3.x->x0时,a(x)和b(x)都是关于x-x0的n阶无穷小量,而a(x)+b(x)是关于x-x0的m阶无穷小量,则A.必有m=nB.必有m≥nC.必有m≤nD.以上几种可能都可能4.设a(x)=x^m-1,b(x)=b^m-1,m>n>0,且当x->1时,有A.a=o(b)B.b=o(a)C.bD.a(x)和 六年级上册英语四会单词就是保师附小用的那种《牛津英语》拜托了得写暑假作业 84/24*12/21用简便方法怎么计算 硝酸钾在20℃时的溶解度为31.6g,在60℃时的溶解度为110g,现有20℃的硝酸钾饱和溶液200g,升温到60℃时,重新变成饱和溶液,需要加入硝酸钾多少克? (追加,希望有人回答,谢谢了.)已知α是第二象限的角,且sin(π+α)=-3/5,则tan2α的值为____________.已知 α 是第二象限的角,且 sin( π + α ) = -(3/5),则 tan 2α 的值为____________.要步骤解析,越详细越好,谢谢 已知sinα=3/5,α为第二象限角,且tan(α+β)=1,则tan2β= 请问谁有北师大版的三年级数学关于时间转换和周长的习题. Mary is going to look for a t_____ job in China next year.根据句意填空 数学四边形的一道练习题.在四边形ABCD中,AB=DC,AD＜BC.求证：四边形ABCD是等腰梯形.（写出证明过程）【画辅助线证AD//DC】 one old lady is going to leave her job and she wants to find a job as language teacher in china 翻 she is going to 什么 her job she wants to be a volunteer in China 已知cosαcos（α+β）+sinαsin（α+β）=--3/5,β是第二象限角,则tan2β= She is going to( next year)括号里提问 八年级科学上册总复习题`1 急``只需要科学啲``其他不用``谢谢``1科学上册1-4单元模拟试卷`1 八年级上册科学A本是(浙教版的) 其中22-24页(复习题)答案!如果选择题与填空题带些解析(比较难的),那我就再给30分我不是直接抄的，而是我写完后想对下答案，因为我科学学得不好 八年级科学上第一章A本复习题的答案 八年级上册科学作业本复习题题目是什么我要的是题目不是答案……是八年级上册科学作业本水的利用和保护后面的复习题第29题,答案是P=833千帕,因P＞P0,所以,该车对路面的压强超过规定值. 八年级数学下册平行四边形证明题如图,平行四边形ABCD中,点E、F分别是边AB、DC的中点,求证﹕EF＝BC 八年级数学下册矩形的证明题目（有关平行四边形）如图,O是菱形ABCD的对角线交点,DE∥AC,CE∥BD.求证OCED是矩形 八年级下学期数学平行四边形证明题这些题都要最正确的证明步骤,行行好~\(≧▽≦)/~（要红框框中所有证明题哦） Now that she is out of a job,Lucy has been considering going back to school,but she hasn'tdecided yet.翻译, ___ she is out of a job.Lucy has beenconsidering going back to school,but she hasn’t decided yet.A since B as C now that A 是我后加上去的 正确答案是CCCCCCCCCCCCCCCCCCCC！ Now that she is out of a job ,Lucy __going back to scool,but she hasn't decided yet .Ahad comsideredB has been condering C con sidered D is going to consider 为什么选B You may not like lucy,but you have to ____that she's good at her job.A.damage B.admit C.prove D.explain now that she is out of job,lucy___going back to school,but she hasn't decided yethad considered has been considering为什么 设等差数列{an},{bn}的前n项和分别是Sn,Tn,且Sn/Tn=7n+2/n+3求a7/b7有助于回答者给出准确的答案 两个等差数列{an}和{bn}的前n项和分别为Sn和Tn,若Sn/Tn=(7n+1)/(4n+27),则a7/b7= 设等差数列{an}与{bn}的前n项之和为Sn,S`n,Sn/S`n=7n+2/n+3,求a7/b7 等差数列｛an｝.｛bn｝的前n项和分别为Sn.Tn.若Sn/Tn=(2n+2)/(n+3),则a7/b7= 等差数列An与Bn的前n项和分别是Sn和Tn,Sn/Tn=(7n+3)/(n+3),求A7/B7