he __(go)to school by bike every morning .but this morning he--(go) by bus

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• 用所给动词的正确时态填空（1） It ____________(take) me half an hour to go to school.How long ______(it take) you?（2） He ___________(play) the piano very well and _________（study）very hard.（3） The sun ________(rise) in the east and _________(set) in the west.（4） Every morning he __________(spend) an hour reading English.（5） Summer holidays __________(begin) next week.
• I go to school ______ every morning.A.by a car B.take acar C.by car
• 用所给词的适当形式填空:He is_(go) to school next morning,but he_(go) hiking last morning.
• He ( ) up at seven yesterday morning.Aget Bgets Cgot ( )they go to school late yesterday A.Do
• they go to school early every morning.(把They改Tom
• I go to school ----bike ----the morning.用介词填空
• I go to school_____my father's car every morning.
• Tom,( )(like) Jack,（ ）(go) to school by bus every morning.
• （1）Some students have to get up early every morning on w_____ to get to school on time by bus or bike.
• --He didn't go to school this morning,did he?
• 2分之1y-1=5分之y+2 12分之5X-4分之X=5分之1 3分之2X+1-6分之5X-1=1
• 设A为n阶实对称矩阵,且满足A^3-2A^2+4A-3E=O,证明A为正定矩阵这个是答案：设λ是A的特征值则 λ^3-2λ^2+4λ-3 是 A^3-2A^2+4A-3E 的特征值而 A^3-2A^2+4A-3E=0,零矩阵的特征值只能是0所以 λ^3-2λ^2+4λ-3=0.λ^3-2λ^2+4λ-3=(λ-1)(λ^2-λ+3)=0而实对称矩阵的特征值是实数所以A的特征值都是1.所以A为正定矩阵.个人觉得有问题啊,高数里Cayley-Hamilton定理:特征多项式fx=0导出fA=0,但是fA=0未必特征多项式就是fx=0,这题如果A^3-2A^2+4A-3E=O右乘(A+E),结论还是成立,即(A+E)(A^3-2A^2+4A-3E)=O,这时就有负特征值,高数里Cayley-Hamilton定理，写错了 是高等代数里的