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(2)先证明四边形ECFD为平行四边形,再由EC=ED得出四边形ECFD为菱形. 解:(1)证明:∵AD=BC,AE=BF,CE=DF, 在△ACE和△BDF中, , ∴△ACE≌△BDF(SSS), ∴∠A=∠B, ∴AE∥BF; (2)四边形ECFD为菱形, ∵△ACE≌△BDF, ∴EC=DF,∠ACE=∠BDF, ...
eCFD, where "e" stands for e-mail, offers a new concept to circumvent this situation. eCFD facilitates a remote and tailored use of CFD software and computing capacity, using e-mail as basic means for controlling the flow simulations. The concept of eCFD is described in this paper. Its...
如图,在Rt△ABC中,∠ACB=90°,AC=4,BC=3,动点D从点A出发,沿线段AC以每秒1个单位的速度向终点C运动,动点E同时从点B出发,以每秒2个单位的速度沿射线BC方向运动,当点D停止时,点E也随之停止,连结DE,当C、D、E三点不在同一直线上时,以ED、EC我邻边作?ECFD,设点D运动的时间为t(秒). ...
人教物理8上-第05章第1节、电荷_ECFD(中) 647 播放互联网密码 互联网分享 收藏 下载 分享 手机看 登录后可发评论 评论沙发是我的～选集(69) 自动播放 [1] 人教物理8上-第01章第1节、声音... 4522播放 15:37 [2] 人教物理8上-第01章第1节、声音... 1136播放 15:42 [3] 人教物理8...
如图,已知点P是△ABC的重心,过P作AC的平行线DE,分别交AB于点D、交BC于点E;作DF∥BC,交AC于点F,若S△ABC=18,则S四边形ECFD=___.ADF
c调是2 6弦 f调是6 3弦 d调是1 5弦 e调是3 7弦 调不同指法不同
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