eliza+janes+bad+day绘本故事
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濮昨13028957738：
<<野天鹅>>的英文故事梗概 -
66159石乳
：The Wild Swans(野天鹅) Long ago and far away there lived a King. He was very proud of his eleven sons and one daughter. All of his children were good, kind a...
濮昨13028957738：
培训学校的英语对话稿
66159石乳
： ELIZA:I heard you were teaching English over there.Tell me about it. Did you like it?SUE:Oh,yes,it was very interesting.ELIZA:What were the schools like?SUE:Oh,I didn't ...
濮昨13028957738：
设A,B都是n阶可逆矩阵,且ABA=B - 1.证明:秩(E - AB)+秩(E+AB)=n. -
66159石乳
：[答案] 证明:因为ABA=B-1, 所以(E-AB)(E+AB)=E+AB-AB-ABAB=E-E=0 所以r(E-AB)+r(E+AB)≤n 又因为n=r(2E)=r[(E-AB)+(E+AB)]≤r(E-AB)+r(E+AB) 所以r(E-AB)+r(E+AB)=n 即秩(E-AB)+秩(E+AB)=n.
濮昨13028957738：
电子基础的逻辑函数化剪Y=ABC+A+B+C,ABC上有一横Y=ABC+ABC,A,B,C,上各有一横Y=A+(B)上一横+BD+CDE+D(A)上一横 -
66159石乳
：[答案] Y=A'B'C'+A+B+C =(A+B+C)'+A+B+C =1 Y=ABC+A'B'C' 已经最简 Y=A+B'+BD+CDE+DA' =(A+A'D)+(B'+BD)+CDE =A+D+B'+D+CDE =A+B'+D(1+1+CE) =A+B'+D
濮昨13028957738：
美少女啦啦队2中蒂娜的扮演者是谁? -
66159石乳
：主演】 Kirsten Dunst .... Torrance Shipman Eliza Dushku .... Missy Pantone Jesse Bradford .... Cliff Pantone Gabrielle Union .... Isis Clare Kramer...
濮昨13028957738：
英语翻译是电影的影评:贫穷的卖花女Eliza,聪明且漂亮.她的出现引起了语言学家Higgins的注意,Higgins的朋友Pickering和他打赌,经过教授的训练,卖... -
66159石乳
：[答案] Poor flower girl Eliza, intelligent and beautiful. Her appearance attracted the attention of linguists Higgins, HigginsFriends of Pickering and his bet, after the professor's train, flower-girls can be a lady manners are very elegant. He readily accepted the ...
濮昨13028957738：
查一部动画片,很早以前的. -
66159石乳
： 名称:【丽莎和她的朋友们】百度百科:http://baike.baidu.com/view/1215535.htm介绍:故事的主角丽莎(Eliza Thornberry),12岁,拥有普通得不能再普通的外表,两条高高翘起的小辫,脸上长...
濮昨13028957738：
cbc+bcb=abba然后c+b=a,b+c=b,c+b=ab.abc各等于多少? -
66159石乳
：[答案] cbc+bcb=abba 那么,b+c=xa,b0+c0=xa0,b00+c00=xa00 则,xa+xa0+xa00=abba 所以,x=a,b=2a 就有,b+c=aa=11a,2a=b 所以,c=9a,则a=1,b=2,c=9
濮昨13028957738：
高中不等式中,有一常用不等式是ab≤[(a+b)/2]²或ab≤(a²+b²)/2.高中不等式,有一常用不等式是ab≤[(a+b)/2]²或ab≤(a²+b²)/2.当中的[(a+b)/2]² 与... -
66159石乳
：[答案] 两个不等式都对,在不同的情况下使用,只不过确定的ab的范围不同,比如说ab=3,可能[(a+b)/2]²的值为4,(a²+b²)/2的值为5,都是成立的 PS:ab≤[(a+b)/2]²和ab≤(a²+b²)/2都是由(a-b)²≥0推导出来的,但[(a+b)/2]²≤(a²+b²)/2
濮昨13028957738：
数字电路证明(AB+A'B')C+B'+C'=A+B'+C' -
66159石乳
：[答案] B'=B'+A'B' = B'+A'B'+A'B'C' = B'+A'B'C'C'=C'+AC' = C'+AC'+ABC' = C'+ABC'原式=ABC+A'B'C+B'+C'=ABC+(C'+ABC')+A'B'C+(B'+A'B'C')=C'+AB+(B'+A'B') (B'+A'B'=B')=C'+AB+(B'+AB') (B'=B'+AB')=A+B'+C
