巧用公共弦方程解题

众所周知 ,若相交两圆的方程分别为x2 y2 D1x E1y F1=0 ,x2 y2 D2 x E2 y F2 =0 ,则它们的公共弦所在直线的方程为( D1- D2 ) x ( E1- E2 ) y ( F1- F2 ) =0 .这个方程应用很广 ,它不仅使解有关两圆相交问题简捷方便 ,而且还有利于解有关圆锥曲线的弦的方程问题 .例 1　在椭圆 x21 6 y24 =1内有一定点A( 1 ,1 ) ,过点 A作一直线与椭圆相交于 B,C两点 ,且使得点 A恰好是弦 BC的中点 ,求此直线的方程 .解　设 B,C两点的坐标分别为 B( x,y) ,C( x1,y1) ,则由中点坐标公式得x1=2 - x,　 y1=2 - y,因为 B,C两点…     

Possible Spectrums of 3×3 Upper Triangular Operator Matrices

Let H1, H2 and H3 be infinite dimensional separable complex Hilbert spaces. We denote by M(D,E,F) a 3×3 upper triangular operator matrix acting on H1⊕H2⊕H3 of the form M(D,E,F)=(A D E 0 B F 0 0 C). For given A ∈ B(H1), B ∈ B(H2) and C ∈ B(H3), the sets UD,E,F σp(M(D,E,F)), ∪D,E,F σr(M(D,E,F)), ∪D,E,F σc(M(D,E,F)) and ∪D,E,F σ(M(D,E,F)) are characterized, where D ∈ B(H2,H1), E ∈ B(H3, H1), F ∈ B(H3, H2) and σ(·), σp(·), σr(·),σc(·) denote the spectrum, the point spectrum, the residual spectrum and the continuous spectrum, respectively.     

2002年数学中考模拟测试题(二)

一、选择题:(本大题满分36分,每小题3分) 1、计算(-2)2+(-1)101所得的结果是( ). A.-3 B.-2 C.3 D.4 2、下列运算①(x-6)(x+b)=x2-b2,②(am)n=amn,③(a/b)n=an/bn,④a2=a,正确的个数是( ). A.1 B.2 C.3 D.4. 3、方程组{x2+y2=6 y=5x的解的个数是( ). A.1 B.2 C.3 D.4 4、不等式组{x>-3 x>4的解集是( ). A.x>-3 B.x>4 C.x>1 D.x>-1 5、反比例函数y=-1/x的图象不经过第( )象A.一、三 B.二、四     

第38届美国中学数学竞赛试题(AHSME)

1.(1+x~2)(1-x~8)等于(A)1-x~5;(B)1-x~6;(C)1+x~2 -x~3;(D)1+x~2-x~3-x~5;(E)1 +x~2-x~3-x~6。 2.如图所示,从边长为3的正三角形ABC中切去一角,得三角形BDE,其边长DB=EB=1,则剩余的四边形ADEC的周长是(A)6:(B)61/2;(C)7;(D)71/2;(E)8. 3.小于100且个位数是7的质数的个数是(A)4;(B)5;(C)6;(D)7;(E) 8.(A)6;(B)8:(C)31/2;(D)24;(E)512, 5.一个学生将一堆数据的数值分布的精确的百     

初二年级数学科第一学期期末自测题

一、选择题1 .多项式 3a2 b-6ab2 c+3abd的公因式是 (　 ) .A .ab　　　　　　　　B .3abC .3abc　D .3abd2 .下列图形中 ,既是轴对称图形 ,又是中心对称图形的是 (　 ) .A .( 1 ) ( 2 ) ( 3 ) ( 4 ) ( 5 )　B .( 2 ) ( 3 ) ( 4 ) ( 5 )C .( 2 ) ( 3 ) ( 5 )　　D .( 3 ) ( 5 )3 .下列不等式 :①x -2 >0 ;② 2 -x >0 ;③x -2≥ 0 ;④ 2 -x≥ 0 .解集可用右图表示的是(　 ) .A .①　　　B .②　C .③　　　D .④4.在矩形ABCD中 ,点E ,G在AC上 ;点H ,F在BD上 ;且AE =BF =CG =DH .顺次连结E ,F ,G ,H ,则四边形EFGH是 (　 ) .…     

九年级数学科第二学期自测题

A组一、填空题1 .抛物线y=-2x2 -x+1的顶点在第象限 .2 .把函数y =-12 x2 的图像向右平移 1个单位 ,再向下平移 2个单位 ,所得图像的解析式为.3 .对于反比例函数y =-2x 与二次函数y =-x2+3 .请说出它们的两个相同点① ,②;再说出它们的两个不同点① ,② .4.函数y =x2 -2x -1 ,当x =时 ,y有最小值.5 .如图 ,△ABC中 ,BC =a .若D1,E1分别是AB ,AC的中点 ,则D1E1=;若D2 ,E2 分别是D1B ,E1C的中点 ,则D2 E2 =;若D3 ,E3 分别是D2 B ,E2 C的中点 ,则D3 E3 =;……依此类推 .若Dn,En 分别是Dn -1B ,En-1C的中点 ,则DnEn= ≠= .(n为…     

九年级数学(华东师大版)第一学期期中自测题

A组一、选择题1.下列各式中,正确的是().A.(12)-3=8B.a3÷a2=a5C.a2+a3=a5D.(-2a3)-3=6a-92.若分式x2+x-6x2-4的值等于零,则x的值是().A.2或-3B.3或-2C.2D.-33.若圆的一条弦把圆分成度数的比为1∶3的两条弧,则劣弧所对的圆周角等于().A.45°B.90°C.135°D.270°4.将方程2-x2-4x+1=3x+1去分母并化简后,得到的方程是().A.x2-5=0B.x2-3=0C.x2-2x-5=0D.x2-2x-3=05.已知AB是⊙O的直径,弦CD⊥AB,垂足是E.如果AB=10,CD=6,那么AE的长为().A.1B.3C.9D.1或96.若x2-9=0,则x2-5x+6x-3的值为().A.1B.-5C.0D.1或-57.如图,四边形ABCD内…     

2007——2008学年度第一学期七年级数学科期终检测题(配华东师大版)

(考试时间:100分钟满分:110分)一1选择题(每小题2分,共20分)1.计算-2 5的结果是()1A17B1-7C13D1-321计算(-1)3的结果是()1A1-1B11C1-3D1331当x=-1时,代数式x2 2x的值是()1A.-2B.-1C.0D144.一个整式减去x2-y2后所得的结果是x2 y2,则这个整式是()1A12x2B1-2y2C1-2x2D12y25.如图1所示,AB∥CD,∠ABE=110°,则∠ECD=()1A1140B1110C170D12061如图2,已知AB⊥CD,垂足为O,图中∠1与∠2的关系是()1A1∠1 ∠2=90°B1∠1 ∠2=180°C1∠1=∠2D1无法确定71一条公路两次转弯后又回到原来的方向(即AB∥CD,如图3)1如果第一次转弯时的∠B=14…     

考点15 向量的概念及运算

1.(北京卷,3)若a=1,b=2,c=a+b,且c⊥a则向量a与b的夹角为().(A)30°(B)60°(C)120°(D)150°2.(山东卷,7)已知向量a、b,且AB=a+2b,BC=-5a+6b,CD=7a-2b,则一定共线的三点是().(A)A、B、D(B)A、B、C(C)B、C、D(D)A、C、D3.(全国卷,8)已知点A(3,1),B(0,0),C(3,0).设∠BAC的平分线AE与BC相交于E,那么有BC=λCE,其中λ等于().(A)2(B)21(C)-3(D)-314.(辽宁卷,9)若直线2x-y+c=0按向量a=(1,-1)平移后与圆x2+y2=5相切,则c的值为().(A)8或-2(B)6或-4(C)4或-6(D)2或-85.(全国卷,10)点P在平面上作匀速直线运动,速度向量v=(4,-3)(即…     

名校基础知识自测

初一年级一、选择题1.若 a2 + 1的相反数是 3 ,则 -3a的相反数为(　　 ) .(A) 9　 (B) 4　 (C) -9　 (D) -2 42 .在数轴上 -2与 -5之间的有理数有(　　 ) .(A)无数个 (B) 3个 (C) 2个 (D) 4个3 .计算 (-2 ) 99+ (-2 ) 1 0 0 的结果为 (　　 ) .(A) -2　 (B) 2 99　 (C) -2 99　 (D) 24.平方结果是 4的有理数的立方为 (　　 ) .(A) 8　　　　 (B) -8(C) -8或 8(D) 16或 -165 .欲使x2 yn2 +2 和 -x2 yn - 1 是同类项 ,则n应取 (　　 ) .(A) 32 　 (B) 6　 (C) 23 　 (D) 26.已知a <0 ,-1相似文献   

Estimation of the number of components of finite mixtures of multivariate distributions

An estimator of the number of components of a finite mixture ofk-dimensional distributions is given on the basis of a one-dimensional independent random sample obtained by a transformation of ak-dimensional independent random sample. A consistency of the estimator is shown. Some simulation results are given in a case of finite mixtures of two-dimensional normal distributions.     

Eigenvector p-Bases of Rings of Constants of Derivations

Let A be a UFD of characteristic p > 0, let 𝒵 be a set of some eigenvectors of a derivation of A. We prove, under some additional assumptions, a necessary and sufficient condition for 𝒵 to be a p-basis of the minimal ring of constants containing 𝒵. The main preparatory result is the unique decomposition theorem with respect to a factor from a given subalgebra containing A^p.     

Galois Structure of K-Groups of Rings of Integers

N/Kbe a Galois extension of number fields with finite Galois group G.We describe a new approach for constructing invariants of the G-module structure of the K groups of the ring of integers of N in the Grothendieck group of finitely generated projective Z[G]modules. In various cases we can relate these classes, and their function field counterparts, to the root number class of Fröhlich and Cassou-Noguès.     

有资格限制的指派问题的求解方法

在实际的指派工作中，常会遇到某个人有没有资格去承担某项工作的问题，因此，本建立了有资格限制的指派问题的数学模型。在此数学模型中，将效益矩阵转化为判定矩阵，由此给出了判定此种指派问题是否有解的方法；在有解的情况下，进一步将效益矩阵转化为求解矩阵，从而将有资格限制的指派问题化为传统的指派问题来求解。最后给出了一个数值例子来说明这样的处理方法是有效的。     

On Sums of Coefficients of Products of Polynomials

We study the nilpotency of the sums of all coefficients of some sorts of products of polynomials over reversible, IFP, and NI rings, and introduce an SCN ring as a generalization. We characterize SCN rings in relation with related ring properties, and also provide several useful properties and ring extensions of SCN rings.     

Convergence of sequences of sets of associated primes

It is a well-known result of M. Brodmann that if is an ideal of a commutative Noetherian ring , then the set of associated primes of the -th power of is constant for all large . This paper is concerned with the following question: given a prime ideal of which is known to be in for all large integers , can one identify a term of the sequence beyond which will subsequently be an ever-present? This paper presents some results about convergence of sequences of sets of associated primes of graded components of finitely generated graded modules over a standard positively graded commutative Noetherian ring; those results are then applied to the above question.

     

Rates of Convergence of Means of Euclidean Functionals

Let L be the Euclidean functional with p-th power-weighted edges. Examples include the sum of the p-th power-weighted lengths of the edges in minimal spanning trees, traveling salesman tours, and minimal matchings. Motivated by the works of Steele, Redmond and Yukich (Ann. Appl. Probab. 4, 1057–1073, 1994, Stoch. Process. Appl. 61, 289–304, 1996) have shown that for n i.i.d. sample points {X ₁,…,X _n } from [0,1] ^d , L({X ₁,…,X _n })/n ^(d−p)/d converges a.s. to a finite constant. Here we bound the rate of convergence of EL({X ₁,…,X _n })/n ^(d−p)/d . Y. Koo supported by the BK21 project of the Department of Mathematics, Sungkyunkwan University. S. Lee supported by the BK21 project of the Department of Mathematics, Yonsei University.     

Number of connected components of groups of real points of adjoint groups

We present a unified approach to compute the number of connected components in the group of real points of adjoint almost simple real algebraic groups.     

Enumeration of perfect matchings of a type of Cartesian products of graphs

Let G be a graph and let Pm(G) denote the number of perfect matchings of G.We denote the path with m vertices by P_m and the Cartesian product of graphs G and H by G×H. In this paper, as the continuance of our paper [W. Yan, F. Zhang, Enumeration of perfect matchings of graphs with reflective symmetry by Pfaffians, Adv. Appl. Math. 32 (2004) 175-188], we enumerate perfect matchings in a type of Cartesian products of graphs by the Pfaffian method, which was discovered by Kasteleyn. Here are some of our results:1. Let T be a tree and let C_n denote the cycle with n vertices. Then Pm(C₄×T)=∏(2+α²), where the product ranges over all eigenvalues α of T. Moreover, we prove that Pm(C₄×T) is always a square or double a square.2. Let T be a tree. Then Pm(P₄×T)=∏(1+3α²+α⁴), where the product ranges over all non-negative eigenvalues α of T.3. Let T be a tree with a perfect matching. Then Pm(P₃×T)=∏(2+α²), where the product ranges over all positive eigenvalues α of T. Moreover, we prove that Pm(C₄×T)=[Pm(P₃×T)]².