httP//dⅩ.10086..cn.gXmh02

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httP//dⅩ.10086..cn.gXmh02(共10篇)

httP//dⅩ.10086..cn.gXmh02（一）：

设矩阵P=-1 -4(第一行）1 1（第二行）.D=-1 0（第一行）0 2（第二行）.A由P^-1AP=D确定,试求A^5

因为 P^-1AP=D
所以 A = PDP^-1
所以 A^5 = (PDP^-1)^5
= PD^5P^-1
= P diag((-1)^5,2^5)P^-1
=
43 44
-11 -12

httP//dⅩ.10086..cn.gXmh02（二）：

Hi.This is the qmail-send program at yahoo.com.
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Hi.This is the qmail-send program at yahoo.com.
I"m afraid I wasn"t able to deliver your message to the following addresses.
This is a permanent error; I"ve given up.Sorry it didn"t work out.
HI 这是雅虎的Qmail发送程序.
我们无法向下列地址发送你的邮件.
这是一个不可消除的错误.我们无法解决它.我们暂时放弃.请谅解.
下面的哪些就不用看了,都是对这个错误用数字进行描述的,跟英文没什么关系.【httP//dⅩ.10086..cn.gXmh02】

httP//dⅩ.10086..cn.gXmh02（三）：

计算（-2）^2002+（-2）^2003=（）
A.-2^4005 B.-2 C.-2^2002 D.2^2002
下列各式中,正确的式子有（）
①x^4•x^2=x^8②x^3•x^3=2x^6③a^4•a^3=a^7④a^5+a^7=a^12⑤(-a)^2•(-a^2)=a^4
A.1个 B.2个 C.3个 D.4个
计算（8•2^n+1）•（8•2^n-1）的结果是（）
A.8•2^2n B.16•2^2n C.8•4^2n D.2^2n+6
计算2x^2•(-3x^3)的结果是（）
A.-6x^5 B.6x^5 C.-2x^6 D.2x^6

c
a
d
a【httP//dⅩ.10086..cn.gXmh02】

httP//dⅩ.10086..cn.gXmh02（四）：

解方程组
1.
(D - 1)x + 2y = 0
3x + (D - 2)y = 0
已知当独立变量t=0时,x=1,y=0
2.
Dx + 4y = 2e^t
(2D - 3)x + (D^2 - D)y = 0
3.
x + (D - 1)y + (D + 2)z = 0
Dy + Dz - u = 0
(D + 1)x - y = 0
4.
(D^2 - 1)x + 2(D + 1)y + (D + 1)z = 0
[(D - 1)^2]x + 4Dy + (2D - 1)z = 0
(D -1)x + 2y - (D -1)z = 0
D全都代表微分代换符号,我觉得应该先化成矩阵做的,但是还是不太会.

当年的数学知识全忘啦...

httP//dⅩ.10086..cn.gXmh02（五）：

已知：1、A、B、C、D、X为5个整数且不为0 2、3A+2=5B+4=9C+8=D 3、D=11X
问：1、X是否有解 2、是唯一解还是N解 3、该解是否有规律

由 3A+2=5B+4 得 3(A-B)=2(B+1) ,
因此 B+1 是 3 的倍数,A-B 是 2 的倍数,
因此 B=3k-1 ,A=5k-1 ；-----------（1）
由 5B+4=9C+8 得 5(B-C)=4(C+1) ,
同理 C=5m-1 ,B=9m-1 ；----------（2）
由于 D=11x=9C+8=45m-1=44m+(m-1) ,因此 m-1=11n ,m=11n+1 ,---------（3）
由（1）（2）（3）可得
A=5k-1=15m-1=15(11n+1)-1=165n+14 ,
B=3k-1=9m-1=9(11n+1)-1=99n+8 ,
C=5m-1=5(11n+1)-1=55n+4 ,
D=45m-1=45(11n+1)-1=495n+44 ,
X=45n+4 .
回答：
1、X 有解 .
2、X 有无穷多解 .
3、该解有规律.公式是
A=5k-1=15m-1=15(11n+1)-1=165n+14 ,
B=3k-1=9m-1=9(11n+1)-1=99n+8 ,
C=5m-1=5(11n+1)-1=55n+4 ,
D=45m-1=45(11n+1)-1=495n+44 ,
X=45n+4 .
其中 n 为任意整数 .

httP//dⅩ.10086..cn.gXmh02（六）：

AAAA+BBB+CC+D=2002求A.B.C.D是多少?

A=1
B=8
C=0
D=3
希望我的回答对你有所帮助
另附：
#include
main()
{
int a,b,c,d;
for(a=0;a

httP//dⅩ.10086..cn.gXmh02（七）：

如图，点0₂是⊙0₁上一点，⊙0₁与⊙0₂相交于A、D两点，BC⊥AD，垂足为D，分别交⊙0₁与⊙O₂于B、C两点，延长D0₂交⊙0₂于E，交BA延长线于F，B0₂交AD于G，连接AC．
（1）求证：∠BGD=∠C；
（2）如果∠D0₂C=45°，求证：AD=AF；
（3）如果BF=6CD，且线段BD、BF的长是关于x的方程x²-（4m+2）x+4m²+8=0的两个实数根，求cos∠ABD的值．

（1）证明：∵BC⊥AD于D，
∴∠BDA=∠CDA=90°，
∴AB、AC分别为⊙O₁、⊙O₂的直径，
∵∠2=∠3，∠BGD+∠2=90°，∠C+∠3=90°，
∴∠BGD=∠C；

（2）证明：∵∠DO₂C=45°，
∴∠ABD=45°，
∵O₂D=O₂C，
∴∠C=∠O₂DC=12（180-∠DO₂C）=67.5°，
∴∠4=22.5°，
∵∠O₂DC=∠ABD+∠F，
∴∠F=∠4=22.5°，
∴AD=AF；

（3）∵BF=6CD，
∴设CD=k，则BF=6k，
连接AE，则AE⊥AD，
∴AE∥BC，
∴

AE

BD

＝

AF

BF

，
∴AE•BF=BD•AF，
又∵在△AO₂E和△DO₂C中，AO=DO₂，∠AOE=∠DOC，O₂E=O₂C，
∴△AO₂E≌△DO₂C，
∴AE=CD=k，
∴6k²=BD•AF=（BC-CD）（BF-AB），
∵∠BO₂A=90°，O₂A=O₂C，
∴BC=AB，
∴6k²=（BC-k）（6k-BC），
∴BC²-7kBC+12k²=0，
解得：BC=3k，或BC=4k，
当BC=3k时，BD=2k，
∵BD、BF的长是关于x的方程x²-（4m+2）x+4m²+8=0的两个实数根，
∴由根与系数的关系知：BD+BF=2k+6k=8k=4m+2，BD•BF=12k²=4m²+8，
整理，得：4m²-12m+29=0，
∵△=（-12）²-4×4×29=-320＜0，此方程无实数根，
∴BC=3k舍去，
当BC=4k时，BD=3k，
∴3k+6k=4m+218k²=4m²+8，
整理，得：m²-8m+16=0，解得：m₁=m₂=4，
∴原方程可化为x²-18x+72=0，
解得：x₁=6，x₂=12，
∴BD=6，BF=12．
∴CD=2，∵AF=AD，
∴设AF=AD=x，
∴BF-x=AB，
∴AB²=AD²+BD²，
∴（12-x）²=x²+36，
解得：x=4.5，
∴AB=12-4.5=7.5，
cos∠ABD=

BD

AB

=

6

7.5

=

4

5

．

httP//dⅩ.10086..cn.gXmh02（八）：

（-0.2）的2002次方×5的2002次方+（-1）2002次方+（-1）2001次方的值是多少?
A、3 B、-2。C、-1。D、1

（-0.2）的2002次方×5的2002次方+（-1）2002次方+（-1）2001次方
=（-0.2×5）的2002次方+1-1
=(-1)的2002次方+0
=1+0
=1

httP//dⅩ.10086..cn.gXmh02（九）：

Hi! My name is Robert and I live in Chicago. My school is called Parkside Elementary.  It is Jake and Elizabeth’s school too.  Michael and Trevor’s school is called Lincoln Middle School.  In the USA many students go to elementary school between the ages of 5-10, then middle school between the ages of 11-13, and then a four year high school, and last but not least, many people go to college or vocational school. Our elementary school starts at 8:30 am and ends at 3:20 pm. We go to school on Monday through Fridays and have the weekends off. Some of the favorite sports here in America are basketball, baseball, football, soccer, hockey, and more. But, my favorite sport is football. My favorite player in the NFL (National Football League) is Brian Urlacher of the Chicago Bears ( the Bears are my favorite team too). Americans wear jeans, a shirt, socks, shoes, and sometimes a hat. I usually wear a baseball hat, my favorite pants, and my favorite shirt. My pants have 4 pockets on it, and my shirt has red and black stripes on it. There is no common agreement in the west about the best method of education. Many views can be found among parents, teachers, and students. http://library.thinkquest.org/CR02www.qzjyzx.com2/italy.html http://www.cfl.cqu.edu.cn/jpkc/kj/xb/xb2/2-1/culturalnotes.htm 小题1:Students in elementary school stay at school for about ________days. A．four B．five C．six D．seven 小题2: The writer’s favorite sport is __________. A．basketball B．baseball C．hockey D．football 小题3:The third passage mainly tells us __________. A．the writer’s favorite sports B．the writer’s favorite food C．the writer’s favorite clothes D．the writer’s favorite color 小题4:“views”in the last passage means _________ in Chinese. A．视野 B．风景 C．见解 D．检查

小题1:B 小题1:D 小题1:C 小题1:C

这是一阅读理解题，目的是考查学生的阅读理解能力。做这种题时大体浏览一下问题，带着问题去读，这样会有的放矢。通读短文，遇到陌生词汇，在不影响理解的情况下可不去管它，如果影响了理解，可根据语言环境、构词法等去猜它的含义。在选择答案时，可应用排除法，这样会提高正确率。1.我个人认为答案不准确，正确的答案应该是D.可根据Our elementary school starts at 8:30 am and ends at 3:20 pm. 来判断。2题可根据my favorite sport is football确定选项。3.这一段主要谈了作者的服饰。4.可应用排除法，ABD不可能.

httP//dⅩ.10086..cn.gXmh02（十）：

------When did you _____this school?----__________
A.come into;Since two years.B.come into ;For two years
C.enter into; In 2002 D.enter;two years ago
分析考察的知识点,

D.
enter是及物动词,直接带宾语,不加介词,所以C排除.而come有其固定搭配：come to,故不选A,D.若从后面的空看,问句以When提问,要用点时间回答,而不用段时间.这样也能排除A,D.

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