一、A型选择题 1.维生素A理论上有几种立体异构体 A.4种 B.8种 C.16种 D.32种 E.64种 2.已经得到的维生素A立体异构体有几种 A.2种 B.4种 C.6种 D.8种 E.10种

如果在x的某一去心邻域内f(x)≥0(或f(x)≤0),而且 f(x)→A(x→x),那么A≥0(或A≤0) 证明设在x的某一去心邻域内f(x)≥0. 假设上述论断不成立,即设A<0,那么由函数极限的 局部保号性就有x的某一去心邻域,在该邻域内f(x)<0,这 与f(x)≥0的假定矛盾.所以A≥0

Fred bergsten explains why policymakers need to act now in order to avert the danger of serious damage to the world economy FIVE major risks threaten the world economy. Three centre on the United States: renewed sharp increases in the current-account deficit leading to a crash of the dollar; a budget profile that is out of control; and an outbreak of trade protectionism a fourth relates to China which faces a possible hard landing from its recent overheating The fifth is that oil prices could rise to $60-70 per barrel even without a major political or terrorist disruption, and much higher with one

What is Packaging? 1. A definition of packaging Packaging is best described as a coordinated system of preparing goods for transport, distribution, storage, retailing, and use of the goods 2. The many things a package might be asked to do Packaging is a complex, dynamic, scientific, artistic, and controversial business function Fundamental function of packaging: contain protects/preserves transports informs/sells

一、单项选择 例体液调节的特点是(C) A.迅速B准确C.持久D短暂 1.迷走神经传出纤维的冲动可看作是(C) A.控制系统B.受控系统C.控制信息D.反馈信息 2.下列生理过程中,属于负反馈调节的是(B) A.排尿反射B.减压反射C.排便反射D.血液凝固 3.关于易化扩散的叙述错误的是(A) A通道蛋白质对被转运的物质没有特异性 B.以通道为中介的易化扩散如K+由高浓度一侧向低浓度一侧的扩散 C.以载体为中介的易化扩散,如葡萄糖入膜过程 D.以载体为中介的易化扩散其特点是具有饱和现象

3.1.2 A general uniqueness theorem The key issue for uniqueness of solutions turns out to be the maximal slope of a=a(a) to guarantee uniqueness on time interval T=[to, t,, it is sufficient to require existence of a constant M such that

1、书写标识符时,忽略了大小写字母的区别。 编译程序把a和A认为是两个不同的变量名,而显示出错信息。C认为大写字母和小写字母是两个不同的字符。习惯上,符号常量名用大写,变量名用小写表示,以增加可读性 。 2、忽略了变量的类型,进行了不合法的运算。 printf(\%d\, a%b) 是求余运算,得到a/b的整余数。整型变量a和b可以进行求余运算,而实型变量则不允许进行「求 余」运算 。 3、将字符常量与字符串常量混淆。 在这里就混淆了字符常量与字符串常量,字符常量是由一对单引号括起来的单个字符,字符串常量是一对 双引号括起来的字符序列

第一章: 选择题: 1、A2.a3.c4.a5.b6.A 第二章: 选择题: 1B2.C3.A(不确定)4.B 第三章: 1.c2.b3.d4.c5.b6.b

一、选择题 A型题 1.d2.c3.e4.e5.d6.a7.e8.c9.c10.b11.d12.a 13.b14.a15.d16.e17.e18.e19.c20.c21.b22.d 23.c24.c25.c26.d27.e8.c29.c30.c31.b32.b 33.c34.b35.c36.d37.e38.e39.c40.c41.b42.c

一.(本题20分)设K为数域.给定K4的两个子空间 M={(x1,2,3,4)|21-x2+4x3-3x4=0,x1+x3-x4=0 N={1,x2,x3,4)3x1+x2+x3=0,7x1+7x3-3x4=0} 求子空间MN和M+N的维数和一组基 二(本题10分)在K4内给定 a1=(1,-1,1,1),a2=(2,-2,0,1). 令M=L(a1,a2).试求商空间K4/M的维数和一组基 三.(本题20分)给定数域K上的3阶方阵 1-11 A=24-2 3-35 1.求K上的3阶可逆方阵T,使T-1AT为对角矩阵 2.对于任意正整数m,求Am