2022-2023 英语周报 八年级 外研提升 45答案

第四部分写作第一节Dear Jack,Welcome to our school to study as an exchange studentAs for how to borrow books from the school libraryyou need to have your student ID card and present it to thelibrarian, who can register for you. Then you can go to thebookshelves to look for what you need. You can borrow 2books once and keep them for a week in order that you canhave enough time to finish reading them. If you haventread the book through, you can renew it. You' ll have topay for it if you get the book damaged or lost.hope all this can be of help to you. For morenformation, don,t hesitate to ask meLi Hua

20.(理科)解析】(1)证明:取CD的中点O,连接OP,PC=PDOP⊥CD,∵侧面PCD⊥底面ABCD,平面PCD∩平面ABCDOPC平面PCD,OP⊥平面ABCD,又BCC平面ABCD,OP⊥BC四边形ABCD是正方形,BC⊥CD,又CD∩OP=O,∴BC⊥平面PCD,又PDC平面PCD,BC⊥PD(2)解:以O为原点,以AB的过点O的垂线为x轴,以OC为y轴,以OP为:轴建立空间直角坐标系O-xy,则P(0,0,√3),4(2,-1,0)B(2.0),D(0.-0),C(0,0),M(0.1,y3),AB=(0,2.0p=(-2.13),0B=(2,20),0M=0.3.),设平面PAB的法向量为m=(x1,y1,1),平面BDM的法向量为n=(x2,y2z)物/mAB=0示8则+x5x=03,2y,=0AP-0n. DM=0令x1=√3可得m=(√3,0,2),令y2=-1可得n=(1,-1,√3),mn=33comm目nN7xN51=35平面PAB与平面MBD夹角余弦值为103(文科)【解析)证明:在△FED中EF=√FD+ED-2FD. ED-cos60°=3,FD+EF=ED则∠EFD=90°,即EF⊥ADC在底面上的投影E恰为CD的中点CE⊥平面ABCD,又ADC平面ABCD,CE⊥AD,又EF⊥AD, EmCE=E,EF,CEC平面CEFAD⊥平面CEF;(2)解:平面A4BB∥平面CCDD,A到平面CDD的距离即A到平面CCDD的距离,取DE的中点G,连接AG,由△ADE是边长为2的正三角形,知AC=√3,AG⊥CD.又CE⊥平面ABCD,CE⊥AG,又CECD=E,AG⊥平面CD1D.C,E=、C1C2-CE2=2√3∴SD2CE·C1E=23==·AG·SlI D,CE4,-D1△DCE=元Xv3x2v3=22√3=2

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