Thinking of thal, my face felt hot. To our amazement, Mrs. Ferguson replied, " You are welcometo plug it in here. When saying that, she pointed to the mirror in her room.That' s really nice ofyou."Alex and I said and walked into her room. When we were fixing our hair Mrs. Ferguson chattedwith us about our tryout for cheerleaders. We had a pleasant chat, during which I found MrsFerguson genuinely cared about her studentsParagraph 2:When we were to leave, Mrs. Ferguson said,"Can I watch your tryout? " Of course, MrsFerguson, we'd be honored, I replied, feeling amazed again. During the tryout, both Alex and Iperformed well and passed the test. Mrs. Ferguson sincerely congratulated us. Meanwhile, sheinded us to study hard when we tried to develop our personality by being cheerleaders. ThrouglFerguson, we'd be honored, " I replied, feeling amazed again. Duringperformed well and passed the test. Mrs. Ferguson sincerely congratulated us. Meanwhile, shereminded us to study hard when we tried to develop our personality by being cheerleaders. Througlthese and many other nice gestures, Mrs. Ferguson gradually gained love and support from all of usHow blessed we were to have another great teacher

18.【命题意图】本题以四面体为载体,考查面面垂直的判定、二面角的佘弦值的求解,体现了直观想象、逻輯推理、嶽学运算等核心素养(1)【证明】∵平面ABD⊥平面BCD,∠CBD=90°,平面ABD∩平面BCD=BD,CBC平面BCD∴CB⊥平面ABD(1分)ADC平面ABD,∵CB⊥AD(2分)又∠BAD=90°,即AD⊥BA,且CB∩AB=B,∴AD⊥平面ABC(3分)∵ADC平面ACD,∴平面ABC⊥平面ACD4分)(2)【解】方法一如图(1),过点A作AO⊥BD,垂足为点O平面ABD⊥平面BCD,平面ABD∩平面BCD=BD,AOC平面ABD,AO⊥平面BCD在Rt△ABD中,AB=2,BD=4,则AD=23,∠ABD=60°BO=1,OD=3,OA=√3(6分)以O为坐标原点,OB所在直线为x轴,过点O且与BC平行的直线为y轴,OA所在直线为z轴,建立空间直角坐标系.(7分)图(1易得A(0,0,3),B(1,0,0),C(1,2√3,0),D(-3,0,0)DA=(3,0,3),DC=(4,23,0).(8分)设平面ACD的法向量为n=(x,y,x)0,阝3x+√3z=00,4x+23y=0.令x=3,则y=-2,x=-3平面ACD的一个法向量为n=(3,-2,-3)(9分)又平面BCD的一个法向量为m=(0,0,1),(10分)∴cos(m,nIn . nIml Inl12=-(1分)由图可知,二面角A-CD-B为锐角,故二面角A-CD-B的余弦值为(12分)方法二如图(2),过点A作AO⊥BD于点O,过点O作OM⊥CD于点M,连接AM∴平面ABD⊥平面BCD,AOC图(2)平面ABD,平面ABD∩平面BCD=BD,∴AO⊥平面BCD(5分)∵CDC平面BCD,AO⊥CD又CD⊥OM,AO∩OM=0,∴CD⊥平面AOM又AMC平面AOM,CD⊥AM(6分)∴∠AMO为二面角A-CD-B的平面角.(7分)在Rt△ABD中,AB=2,BD=4,则AD=√BD-AB2=23,∠ABD=60°,BO=1,OD=3,.0A=vAD'-OD'=3在Rt△BCD中,BC=23,BD=4,CD=√BD+CB2=27(9分)∴sin∠ BDC. OM BCOD CD 73√21OM=3x77OM∴,cos∠AMO=√OA+On4(11分)∴二面角A-CD-B的余弦值为(12分)

以上就是2022七年级下册新目标第28期英语周报答案，更多英语周报答案请关注本网站。

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