ex5-2
挖掘机维修资料大全 ppt课件
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）SH200-3零件手册SH200-3维护手册(电气线路篇） SH200-3维护手册（电气线路篇）SH200-3维护手册 （电气线路篇）SH200-3维护手册（概述篇）SH200-3 维护手册（维护篇）SH
ppt课件
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200-3维护手册（液压油路篇）SH200-3维护手册（主 体篇）SH200GT-3维修手册（完全篇）SH450LHD， SH300-2维修手册中俊住友-3电气系统原理讲座视频 住友200，300EC电机的
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零件目录6 现代挖掘机电气系统（245-278）7 电气系 统主要功能综述（279-288）-7系列机服务指南9 主 要机型电气回路图（311-344）R210LC-7维修手册 R220-5故障代码诊断现代
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装修手册-5型主控阀管路
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w j 0 l 2 s 二 手 挖 掘 机 w w w . e r s h o u w a j u e j i . o r g
2PC270-7PC300-
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6PC300,400-5PC300，350，360-7PC400-6，4506PC600-7PC650-5SEPC750,800-7PC1250-7内部故障案 例小松发动机小松内部培训资料
EX系列
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35EX30UR-2EX60EX100-2,120-2EX120-3EX1205EX125WD-5EX200-2EX200-3EX200-5EX2005HHEEX200LCEX220-2EX22
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挖掘机液压系统三一液压挖掘机培训教材（刘军士） 挖掘机培训教材-液压挖掘机结构及使用液压泵及其 控制系统液压泵进油压力对工程机械考能的影响液 压挖掘机的检查、调整
第五、六章作业解答
4.0
x(m)
0.10 0.05
P
B
t (s)
P
4.0
O
o0
B
AB
O
AO
o0
B
t (s)
1
由比例关系求振动的周期T
(2) 画出P点对应的旋转矢量AP，可知P点相位： P
0
t BO BO 5 6 5 T 2 2 12
T
PO P O 0 ( 3) 3 OP两点的相位差：
2
(1) 由振动曲线可知振动的振幅： 解： A 0.10m
BO B O 2 ( 3) 5 6
画出t =0时刻的旋转矢量AO，可知振动的初相：
OB两点的时间间隔：t BO 4.0s
x(m)
AO
x(m)
0.10 0.05
O 5 3 或 3
Ex5-20
3 2k 1 2k 0.75
x2+x3振幅最小的条件是： 32 3 2 ( 2k 1)
两个同频率简谐运动1和2的振动曲线如图 所示，求：(1)两简谐运动的运动方程；(2) 在同一图中画出两简谐运动的旋转矢量； (3) 若两简谐运动叠加，求合振动方程。
o0
t (s)
2 rad s 1 T
2
画出两个简谐运动在初始时刻的旋转矢量A1和A2。 由矢量图可知两个简谐运动的初相分别为： 1 2 因此两简谐运动方程分别为：
2
两简谐运动的相位差： 21 2 1 合振动振幅： A
2 1 2 2
新视野大学英语第三版读写教程-Book2-Unit1-SectionA课后练习答案
Pre-reading activities P2
• Script: I'm Michal. I live in Poland, and Polish is my native language. I started learning English by listening to the Beatles everyday. At the beginning, I didn’t know what they were singing, but soon I started to understand individual words, and later, more. Each time I hear a Beatles song, I am delighted and feel proud because now I can understand a lot better. I listen to many rock bands, all in English, and I really enjoy them. Besides listening to English songs, I also started reading science fiction books in English. I still remember how excited I was when I successfully finished reading the first English book, a great breakthrough in my life! When I read in English, I not only have the pleasure of reading but also get the feeling of self-accomplishment. English has given me so much satisfaction and pleasure. I feel I’m free.
5-2 最大似然估计
i =1
又设 x1, x2 , , xn 为样本 X1, X 2 , , X n 的一组
观察值.
n
L(x;θ ) = L(x1,
,
xn
;θ
)
=
∏
i =1
p(
xi
;θ
),
θ ∈ Θ,
则 L(x;θ )称为样本似然函数.
8
例 X~P (λ), 即 P( X = m) = λm e−λ , m = 0,1,
的解.
∂θi
18
定义法
当似然函数L(θ )有不连续点时，似然方程没有
意义，须从定义出发求最大似然估计. 性质
如果 θˆ 是θ 的极大似然估计，则对任一函数 g(θ )，其极大似然估计为 g(θˆ) .
该性质称为极大似然估计的不变性.
19
例 设 X = ( X1, , X n ) 是来自两点分布 b(1, p) 的
浙江财经学院本科教学课程 ----经济数学(三)
概率统计
最大似然估计
§5.2 最大似然估计
思想方法：一次试验就出现的事件有较大的概率
例如: 有两外形相同的箱子,各装100个球 一箱 99个白球 1 个红球 99 1 一箱 1 个白球 99个红球 1 99
现从两箱中任取一箱, 并从箱中任取一球, 结果所取得的球是白球.
但不知道是黑球多还是红球多，则从中抽出一
球为黑球的概率θ为 ¼ 或 ¾ .
现从罐子里有放回地抽出 n 个球，试根据样本
数据，估计 θ 的值为 ¼ ，还是 ¾？
解：令Xi 表示第i 次抽球的结果，即
⎧1, 黑球 ,
Xi
=
⎨ ⎩
0
,
否则 .
EN45545-2材料及元件的防火要求
EN45545-2材料及元件得防火要求EN45 5 4 5-2最新标准为2013版，其主要关注于材料得燃烧性能、热释放量、烟密度、毒性等性能。
EN45 5 45 — 2标准根据材料得最终用途，划分为 R1—R 2 6共26个不同得类别。
根据车辆类型及操作环境,EN 4 5 5 45-2—2013 标准将火灾得危险等级划分为HL1、HL2、HL3,共3个火灾风险等级。
根据车辆运行环境得风险程度,EN45 5 4 5-2标准将风险划分为EN 4 5545- 2 II L 1 ,EN 4 554 5—2 HL2 , EN 4 5545—2 HL 3 ,这三类风险等级。
其中EX4 5545—2 HL 3要求最高，最为严格.为了减少车辆运行时得风灾危险程度，要求铁路车辆内装饰材料最好都达到EN45545 HL 3类。
EN4 5 5 4 5-2 HL1-IIL3风险等级根据车辆运行环境及车辆类别来划分等级 EN4 5 5 45-2 HL1:标准车辆，没有急救培训得员工得自动列车，双屮板车辆在0 P era t i o n category 1得环境下运行。
EN 4 5 545— 2 HL 2 :卧铺车辆在0pe r ation categor y 1得环境下运行；标准车辆，没有急救培训得员工得自动列车，双屮板车辆以及卧铺车辆在Ope r at 1 on C a t egor y 2得环境下运行；标准车辆，没有电救培训得员工得自动列车，双屮板车辆在 Op e ration c a teg ory 3 得环境下运行•EN45545 —2 HL 3 :卧铺车辆在 Op e rat i on c a tego r y 3 得环境下运行；标准车辆，没有急救培训得员工得自动列车，双屮板车俩以及卧铺车辆在Oper a tion cate g ory 4得环境下运行。
不同风险等级得EX4554 5 -2测试要求就是完全不同得。
C语言程序设计教程 清华大学出版社 陈明主编 第5章答案
第5章习题解答一、选择题1．以下叙述中错误的是。
A）对于double类型数组，不可以直接用数组名对数组进行整体输入或输出B）数组名代表的是数组所占存储区的首地址，其值不可改变C）在程序执行中，当数组元素的下标超出所定义的下标范围时，系统将给出“下标越界”的出错信息D）可以通过赋初值的方式确定数组元素的个数解析：对于double类型数组，不能直接用数组名对数组进行整体输入或输出。
所以选项A的说法是正确的。
数组名代表数组所占内存区的首地址，是一个固定指针，其值不可改变，因此选项B也是正确的。
在C语言一维数组定义时可以不指定数组长度，由赋初值的初值个数决定数组的长度。
因此选项D也是正确的。
C语言程序在执行时，不检查数组元素的下标是否越界，如果出现越界，可能会引起系统死机，因此选项C不正确。
答案：C2．有以下程序：/* 文件路径名:ex5_1_2\main.c */#include <stdio.h> /* 标准输入/出头文件*/int main(void) /* 主函数main() */{char s[] = "abcde"; /* 定义字符数组*/s += 2; /* s自加2 */printf("%d\n", s[0]); /* 输出s[0] */return 0; /* 返回值0, 返回操作系统*/}执行后的结果是。
A）输出字符a的ASCII码B）输出字符c的ASCII码C）输出字符c D）程序出错解析：数组s[]的首地址值s是不可改变的，故该程序编译时会出错。
答案：D3．有以下程序：/* 文件路径名:ex5_1_3\main.c */#include <stdio.h> /* 标准输入/出头文件*/int main(void) /* 主函数main() */{int a[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, *p = &a[3], *q = p + 2; /* 定义数组及变量*/printf("%d\n", *p + *q); /* 输出*p + *q之值*/return 0; /* 返回值0, 返回操作系统*/}程序运行后的输出结果是。
电磁场电磁波第2版答案
电磁场电磁波第2版答案【篇一：电磁场与电磁波答案(第四版)谢处方】给定三个矢量a、b和c如下： a?ex?ey2?ez3b??ey4?ezc?ex5?ez2求：（1）aa；（2）a?b；（3）a?b；（4）?ab；（5）a在b上的分量；（6）a?c；（7）a?（8）(a?b)?c和a?(b?c)。
(b?c)和(a?b)?c；解（1）aa?e?e2?e3a??ex?ey?eza（2）a?b?(ex?ey2?ez3)?(?ey4?ez)?ex?ey6?ez4?（3）a?b?(ex?ey2?ez3)?(?ey4?ez)?－11a?b??1 5，得 ??cos??()?135.abab8a?b （5）a在b上的分量 ab?aco? ?sab?bexeyez（4）由 co?sab?（6）a?c?12?3??ex4?ey13?ez10 0?2ex5exeyez1?ex8?ey5?ez20 ez5（7）由于b?c?0?40?2eya?b?12?3??ex10?ey1?ez40?41所以 a?(b?c)?(ex?ey2?ez3)?(ex8?ey5?ez20)??42(a?b)?c?(?ex10?ey1?ez4)?(ex5?ez2)??42ex5exa?(b?c)?1eyez（8）(a?b)?c??10?1?4?ex2?ey40?ez50?2ey5ez202?3?ex55?ey44?ez1181.2三角形的三个顶点为p(0,1,?2)、p(4,1,?3)和p(6,2,5)。
123（1）判断?ppp是否为一直角三角形；123（2）求三角形的面积。
解（1）三个顶点p(0,1,?2)、p(4,1,?3)和p(6,2,5)的位置矢量分别为123r1?ey?ez2，r2?ex4?ey?ez3，r3?ex6?ey2?ez5 则r12?r2?r1?ex4?ez， r23?r3， ?r?2ex2?ey?ez8r31?r1?r3??ex6?ey?ez7由此可见r12?r23?(ex4?ez)?(ex2?ey?ez8)?0故?pp为一直角三角形。
C语言程序设计-第5章全解
函数返回的值，程序可以使用它， 也可以不使用它
while(…) { getchar(); c=getchar(); … } /* 返回值不被使用 */ /* 返回值被使用 */
2018/10/21 华中科技大学计算机学院C语言课程组 18
5.2.1
函数的定义
函数定义的一般形式为： 类型名 函数名（参数列表） { 声明部分 语句部分 }
2018/10/21 华中科技大学计算机学院C语言课程组 19
类型名说明函数返回值（即出口参数）的数据类型 （简称为函数的类型或函数值的类型），可以是除 数组以外的任何类型。当返回值类型为void，函数 将不返回任何值。 参数列表说明函数入口参数的名称、类型和个数， 它是一个用逗号分隔的变量名及其类型列表，它描 述了在调用函数时函数所接收的参数。一个函数可 能没有参数，在没有参数的情况下，参数列表说明 为void，否则必须明确地列出每一个参数的类型。
16
图5.1 C语言程序的基本结构
2018/10/21
华中科技大学计算机学院C语言课程组
17
5.2
函数的定义与函数的声明
程序中若要使用自定义函数实现所需的功能， 需要做三件事： ① 按语法规则编写完成指定任务的函数，即 定义函数； ② 有些情况下在调用函数之前要进行函数声 明； ③ 在需要使用函数时调用函数
2018/10/21
华中科技大学计算机学院C语言课程组
9
在函数的顶端用“/*……*/”格式包含的部分是函 数头部注释，包括函数名称、函数功能、函数参数、 函数返回值等内容，如有必要还可增加作者、创建 日期、修改记录（备注）等相关项目。虽然函数头 部注释在语法上不是必需的，但可以提高程序的质 量和可维护性，在程序设计时要遵从这一编程规范。 GetNum是函数名，其后的void说明函数调用时不 接收任何参数，即没有入口参数，函数执行完应该 返回所产生的随机数，即该随机数是函数的出口参 数，函数名前的int说明出口参数的类型为整型。 函数体内的rand是接口stdlib.h中的一个函数，它 返回一个非负并且不大于常量RAND_MAX的随机整 数，RAND_MAX的值取决于计算机系统。 MAX_NUMBER是用#define定义的符号常量，其 值为1000。当执行return语句时，其后表达式的值 被带回到调用函数中。
2024届云南省云南师范大学附属中学高三第二次教学质量监测(数学试题文)试题
2024届云南省云南师范大学附属中学高三第二次教学质量监测（数学试题文）试题注意事项1．考试结束后，请将本试卷和答题卡一并交回．2．答题前，请务必将自己的姓名、准考证号用0．5毫米黑色墨水的签字笔填写在试卷及答题卡的规定位置． 3．请认真核对监考员在答题卡上所粘贴的条形码上的姓名、准考证号与本人是否相符．4．作答选择题，必须用2B 铅笔将答题卡上对应选项的方框涂满、涂黑；如需改动，请用橡皮擦干净后，再选涂其他答案．作答非选择题，必须用05毫米黑色墨水的签字笔在答题卡上的指定位置作答，在其他位置作答一律无效． 5．如需作图，须用2B 铅笔绘、写清楚，线条、符号等须加黑、加粗．一、选择题：本题共12小题，每小题5分，共60分。
在每小题给出的四个选项中，只有一项是符合题目要求的。
1．设变量,x y 满足约束条件22390x y x y x +≤⎧⎪-≤⎨⎪≥⎩，则目标函数2z x y =+的最大值是（ ）A ．7B ．5C ．3D ．22．已知实数x ，y 满足约束条件2211x y y x y kx +≥⎧⎪-≤⎨⎪+≥⎩，若2z x y =-的最大值为2，则实数k 的值为（ ）A ．1B ．53C ．2D ．733．我国数学家陈景润在哥德巴赫猜想的研究中取得了世界领先的成果．哥德巴赫猜想是“每个大于2的偶数可以表示为两个素数（即质数）的和”，如16511=+，30723=+．在不超过20的素数中，随机选取两个不同的数，其和等于20的概率是（ ） A ．114B ．112C ．328D ．以上都不对4．已知直线1:240l ax y ++=，2:(1)20l x a y +-+=，则“1a =-”是“12l l ”的A ．充分不必要条件B ．必要不充分条件C ．充分必要条件D ．既不充分也不必要条件5．已知椭圆C ：()222210x y a b a b +=>>的左，右焦点分别为1F ，2F ，过1F 的直线交椭圆C 于A ，B 两点，若290ABF ∠=︒，且2ABF 的三边长2BF ，AB ，2AF 成等差数列，则C 的离心率为（ ）A ．12BCD6．已知,a b ∈R ，3(21)ai b a i +=--，则|3|a bi +=（ ）7．圆心为()2,1且和x 轴相切的圆的方程是（ ） A ．()()22211x y -+-= B ．()()22211x y +++= C ．()()22215x y -+-=D ．()()22215x y +++=8．如图是来自古希腊数学家希波克拉底所研究的几何图形，此图由三个半圆构成，三个半圆的直径分别为直角三角形ABC 的斜边BC ，直角边,AB AC .已知以直角边,AC AB 为直径的半圆的面积之比为14，记ABC α∠=，则sin 2α=（ ）A ．925B ．1225C ．35D ．459．若集合{|2020}A x N x =∈=，22a =，则下列结论正确的是（ ）A ．{}a A ⊆B ．a A ⊆C ．{}a A ∈D ．a A ∉10．若复数z 满足2(13)(1)i z i +=+，则||z =（ ）A ．54B ．55C ．102D ．10511．已知函数e 1()e 1x x f x -=+，()0.32a f =，()0.30.2b f =，()0.3log 2c f =，则a ，b ，c 的大小关系为（ ）A ．b a c <<B ．c b a <<C ．b c a <<D ．c a b <<12．若函数()y f x =的定义域为M ＝{x|－2≤x≤2}，值域为N ＝{y|0≤y≤2},则函数()y f x =的图像可能是（ ）A ．B ．C ．D ．13．已知三棱锥P ABC -的四个顶点都在球O 的球面上，PA PB PC ==，2AB =，5BC =，3AC =，E ，F分别为AC ，PB 的中点，32EF =，则球O 的体积为______. 14．已知函数f(x)＝322{102x x x x ≥，，（－），＜＜，若关于x 的方程f(x)＝kx 有两个不同的实根，则实数k 的取值范围是________．15．如图，是一个四棱锥的平面展开图，其中间是边长为2的正方形，上面三角形是等边三角形，左、右三角形是等腰直角三角形，则此四棱锥的体积为_____．16．已知实数x 、y 满足121y y x x y m ≥⎧⎪≤-⎨⎪+≤⎩，且可行域表示的区域为三角形，则实数m 的取值范围为______，若目标函数z x y =-的最小值为-1，则实数m 等于______.三、解答题：共70分。
北京航空航天大学C语言c第三讲(第五章) 选择结构程序设计
if (money>=500) cost=0.8; else if (money>=300) cost=0.85; else if (money>=100) cost=0.9 else if (money>=50) cost=0.95; else cost=1.0; price=money*cost;
高 低
3,优先级 <,<=,>,>= 为同一级别,== 和!= 为同一级别. 在同一级别中,采取由左至右的结合方向.
算术运算符 -> 关系运算符 -> 赋值运算符
4,关系表达式
用关系运算符将两个表达式连接起来叫关 系表达式。 例：关系运算的优先级举例
c>a+b a>b!=c a==b<c a=b>c
if (表达式) 语句1 else 语句2
说明： (1)表达式是指关系表达式或逻辑表达式，条
件满足指的是表达式的值为真，即表达式的 值是一个非０值。 例如： if (3) printf(“O.K”); 输出O.K(永真） if („a‟) printf(“%d”,„a‟); 输出97 (永真) if (0) printf(“false”); 不执行printf语句(永假)
能被4整除，但不能被100整除； 能被400整除。
用year表示年份，判断闰年的逻辑表达式为： (year%4==0 && year%100 != 0)||(year % 400 ==0)
如果给year输入一个的值，使上面的表达式的值为１
时，则year所代表的年份就是闰年。
• 课堂练习: 写出逻辑表达式。 10 1. 判断一个点是否 10 -10 落入圆形中 2. 求落入阴影部 -10 分的任意一点 x*x+y*y<=100 (小园的半径为1.)
大学体验英语4课后习题参考答案上
Unit1Passage AEx.5 1. obligation 2. applauded 3. fulfilled 4. mirror 5. flexibility6. devalue7. striving8. entailed9. supposedly 10. ConsequentlyEx6. 1. refer to as 2. at best 3. by the same token 4. at large 5. helpup …asEx.8 Translate the following sentences into English.1. With his promotion, he has taken on greater responsibilities.2. He felt he did not have to make such a commitment to John any more.3. Mary likes to go shopping in her spare time, as opposed to Lucy, who prefers to stay at home reading.4. At best he’s ambitious, at worst a power-seeker without conscience or qualifications.5. We have striven to the full to convince him, but we have made no headway.Ex14. 1. equality 2. sustained 3. clue 4. expansiveness 5. was sacked6. defined7. sphere8. sour9. depressing 10. divisionEx15. 1. in the way 2. falling apart 3. have ruled out 4. at this rate 5. took up Passage B Ex. 17 T ranslate the following sentences into English.1. The popularity of Internet provides people with quick and convenient access to information.2. Since he doesn’t agree to this plan, there is not much point in thinking about it any more.3. I haven't seen it myself, but it is supposed to be a very good film.4. If you don't work hard, you'll end up nowhere.5. It's up to the United States to take the initiative in banning nuclear weapons.Unit2Ex5. 1.awaited, 2.have overtaken, 3.pose, 4.cues, 5.has undermined,6.soptted,7.judged,8.distracted,9.offend,10.maskEx6. 1.brought out, 2.with the exception of, 3.thanks to, 4.taken in, 5.capable ofEx8. 1. He is no more fit to be a headmaster than any schoolboy would be.2. As for her father, she is not sure whether he will accept her and her baby.3. Staying up late will undermine one’s health while going to bed early and getting up early will benefit it.4. The ambassador personally conveyed the president’s message to the premier.5. The girl decided to open up and tell the police what she had seen.Ex14. 1.embarrassed, 2.tutor, 3.potential, 4.hence, 5.yield,6.affection,7.appropriate,8.presentation,9.romantic,10.recommendationEx15. 1.in terms of, 2.participating in, 3.get down to, 4.base...on, 5.acts onEx17. 1. Excessive exercises does more harm than good to one’s health. Therefore we must control the amount of exercise we do.2. He made a few attempts to join in their conversation, but he gave up out of timidness.3. Trains cannot compete with planes in terms of convenience and speed.4. Making a comparison between Chinese and American cultures will help you to better understand and adapt to your study environment in the USA.5. It is a traditional Chinese virtue for the young on buses to yield their seats to the old, the weak, the sick and the disabled.Unit3Key to Exercises5. 1. violated 2. derive 3. blocked 4. default 5. rationalized6. quoted7. exclusive8. contacted9. anonymous 10. charge6 1. is up to 2.resulted in 3.was charged with 4.was unlikely 5. handing out 8. 1. It is simply unbelievable for him to change 500 dollars for a haircut.2. People grant you the privilege, so you should serve the people wholeheartedly.3. The weather forecast is very important so that we can decide when to go to sea.4. The sound of guns violated the usual calmness of Sunday morning, and people hada strong feeling that the war was coming.5. Even though you read three pages per minute, you will by no means finish the book by the end of this weekend.Key to Exercises14. 1.have witnessed 2. trample 3. abusing 4.manual 5. duplicated6. knowingly7. tactic8. patrolling9. budget 10. enforce15. 1. end up 2. give...short shrift 3. step up ...to 4. infringed on5. doesn’t fit into16. 1 a fine 2. rights/law 3. to violate law/rights 4. money 5.deadlines6.a presentation7. rehearsals8. a copy/money9. permission 10.an interview17. 1. He can beat his opponent easily without giving a second thought.2. With little or no knowledge of driving, he drove the car onto the road.3. It is illegal to make an unauthorized change in the construction of the bridge.4. I could accept this fact calmly because I knew if I were not able to finish the mission, chances were that the team would be defeated.5. Children in remote mountain areas are lacking in scientific and cultural knowledge.Unit4Key to Ex.5. 1. sensitive 2. classification 3. sufficiently 4. steer 5. familiarity6. consequently7. reward8. qualifications9. fundamentals 10. evaluate6. 1. As for 2. focus…on 3. benefit…from 4. failed to 5. relevan t to8. 1. My grandfather’s interests rang from playing chess to climbing mountains although he is 70 years.2. Surprisingly, Tom’s grandmother plays with dolls at the age of 80.3. His selflessness and talent qualify him for the job most students dream to get.4. What interpretation would you put on his odd/strange dream?5. The brand-new car our boss has just bought id capable of making 200 kilometers an hour.Key to EX.14. 1. beyond 2. champions 3. ignored 4. obliged 5. sheer 6. lagged 7. prospects 8. proficiency 9. prejudice 10. claim15. 1. in a way 2. drawing up 3. calls for 4. signed up for 5. set…back17. 1. He suggested at the board meeting that the easy task should be left to Mary whose health is poor.2. Mr Brown, who is in Hong Kong now, has a lot of teaching experience and will come to teach our students in the spring.3. This is the best hotel in our city: the food is good and the same is true of the service.4. In order to attract more passengers, the caption assured all the passengers of the safety of traveling by ship.5. His new car, for which he had paid ￥150 000, was seriously damaged in an accident and was beyond repair.。
机械连接无梭管阀门55-MY#M系列II2G Ex h IIC T6 T5 Gb +5°C ≤ Ta
Instruction ManualMechanically Jointed Rodless CylinderII 2G Ex h IIC T6/T5 GbMarking Description:II2provided by compressed air into a force which causes mechanical linearsubject to “Special Conditions of Use”, please see Section 2.3.1 Safety InstructionsThese safety instructions are intended to prevent hazardous situationsand/or equipment damage. These instructions indicate the level ofpotential hazard with the labels of “Caution,” “Warning” or “Danger.”They are all important notes for safety and must be followed in additionto International Standards (ISO/IEC) *1), and other safety regulations.*1) ISO 4414: Pneumatic fluid power - General rules relating to systems.ISO 4413: Hydraulic fluid power - General rules relating to systems.IEC 60204-1: Safety of machinery - Electrical equipment of machines.(Part 1: General requirements)ISO 10218-1: Robots and robotic devices - Safety requirements forindustrial robots - Part 1: Robots.•Refer to product catalogue, Operation Manual and HandlingPrecautions for SMC Products for additional information.• Keep this manual in a safe place for future reference.Warning•Always ensure compliance with relevant safety laws andstandards.•All work must be carried out in a safe manner by a qualified person incompliance with applicable national regulations.2 SpecificationsThis product is suitable for use in Zones 1 and 2 only.2.1 Product Specifications:Refer to the operation manual for this product;2.2 Production Batch Code:The batch code printed on the label indicates the month and the year ofproduction as per the following table;2.3 Special Conditions of Use:•Products are suitable for sub-divisions IIC.•Products are suitable for Zones 1 & 2 only.2.3.1 Temperature Marking:2.3.1.1 Standard Product:•In the normal ambient temperature range (+5°C to +40°C) the productis rated to temperature class T6.•In the special ambient temperature range (+40°C to +60°C) the productis rated to temperature class T5.2.3.2 Static:•Danger of electrostatic discharge.3 Installation3.1 InstallationWarning•Do not install the product unless the safety instructions have been readand understood.•Do not twist or bend the cylinder, or mount the product when subjectto tension.•Do not use in an application where the product is stopped mid-stroke,via an external stop.•Do not use where cylinders are being synchronised to move a singleload.•In order to install the product, use one of the brackets available foundin the standard product catalogue;SketchSee the product catalogue for the exact code to order which relates to thebore size of your product.•When replacing the side supports, use the hexagonal socket head capscrews of the following sizes, and the hexagon wrenches shown below.3.2 EnvironmentWarning•Do not use in an environment where corrosive gases, chemicals, saltwater or steam are present.•Do not use in an explosive atmosphere except within the specifiedrating.•Do not expose to direct sunlight. Use a suitable protective cover.•Do not install in a location subject to vibration or impact in excess ofthe product’s specifications.•Do not mount in a location exposed to radiant heat that would result intemperatures in excess of the product’s specifications.•Do not use in a place subject to heavy vibration and/or shock.•Do not use in wet environments, where water can remove the presenceof the lubrication.•Do not use in case of heavy dusty environments where dust canpenetrate into the cylinder and dry the grease.•Do not allow dust layers to build up on the cylinder surface and insulatethe product.3.3 PipingCaution•Before connecting piping make sure to clean up chips, cutting oil, dustetc.•When installing piping or fittings, ensure sealant material does notenter inside the port. When using seal tape, leave 1 thread exposedon the end of the pipe/fitting.•3.4 LubricationCaution•SMC products have been lubricated for life at manufacture, and do notrequire lubrication in service.•If a lubricant is used in the system, refer to catalogue for details.Basic Circuituse, and could relate to an increase in maximum surface temperatureabove what the product specification declares.3.6 Electrical Connection•The product should be grounded by the piston rod and the body inorder to create an electrically conductive path to thesystem/application.•Ground the product in accordance with applicable regulations.•Do not pass an electrical current through the product.4 Settings4.1 Air Cushion adjustment•For air-cushion adjustment, tighten or loosen the cushion valve usinga hexagon socket wrench or a flat head screwdriver (excluding ø10).Warning•Do not operate the cushion valve in the fully closed or fully openedstate.Using it in the fully closed state will cause the cushion seal to bedamaged. Using it in the fully opened state will cause the piston assemblyor the cover to be damaged.•Be certain to activate the air cushion at the stroke end.When the cylinder is used with the cushion valve in a fully open position,a suitable external device should be installed to absorb all of the kineticenergy of the mechanism, of which the actuator is part, before reachingeach end of stroke. If this is not done, the piston assembly will bedamaged.5 How to OrderRefer to product catalogue for ‘How to Order’.6 Outline DimensionsRefer to the standard product catalogue for general dimensions.ORIGINAL INSTRUCTIONS7.1General maintenanceCaution•Not following proper maintenance procedures could cause the product to malfunction and lead to equipment damage.•If handled improperly, compressed air can be dangerous.• Maintenance of pneumatic systems should be performed only by qualified personnel.• Before performing maintenance, turn off the power supply and be sure to cut off the supply pressure. Confirm that the air is released to atmosphere.• After installation and maintenance, apply operating pressure and power to the equipment and perform appropriate functional and leakage tests to make sure the equipment is installed correctly.• If any electrical connections are disturbed during maintenance, ensure they are reconnected correctly and safety checks are carried out as required to ensure continued compliance with applicable national regulations.• Do not make any modification to the product.• Do not disassemble the product, unless required by installation or maintenance instructions.• Do not use a product which looks or contains damage, this will invalidate the certification. If damage is seen, please replace the product immediately.• Periodically check the product for any damage or rust appearing. This could result in an increase in friction and lead to dangerous conditions. Replace the whole actuator if any of these conditions appear.• Replace the product, when air leakage is above the allowable value 7.2 Replacement of Dust Seal BandWarning• The dust seal band is the only customer replaceable part.• If other parts are deemed to have failed please replace the product.Use only original SMC dust seal bands, given in the table below.7.3 Disassembly procedure• Disassemble the cylinder, remove the old grease and place all the parts on a clean cloth in a clean environment. The following flat head screwdriver or hexagon socket wrench shall be used to loosen the set• The following screwdriver or hexagon socket wrenches shall be used• Remove the old dust seal band, scrapers and if necessary, bearings and side scrapers.1 Dust seal band2 Scraper3 Bearing S4 Bearing R 7.4 Lubrication procedure▪ Dust seal band inner and outer surface ▪ scrapers ▪ bearings▪ side scrapers ▪ tube top surface7.5 Assembly procedure• The cylinder is assembled in the following order: bearings, dust seal band, side scraper, parallel key, scraper, stopper, spacer and end cover. Tighten the cross recessed binding head screws or the hexagon socket button bolt according to the torque values given.• Finally tighten the two set screws at each side with tightening torque of 0.1 Nm.• Check for cylinder smooth movement and for air leakage.8 Limitations of Use8.1 Limited warranty and disclaimer/compliance requirements Refer to Handling Precautions for SMC Products.Caution8.2 Obligations of the end-user• Ensure the product is used within the specification outlined.• Ensure that the maintenance periods are suitable for the application. • Ensure any cleaning processes to remove dust layers are made with the atmosphere in mind (e.g. using a damp cloth to avoid static build up).• Ensure that the application does not introduce additional hazards by mounting, loading, impacts or other methods.• Ensure that there is sufficient ventilation and air circulation around the product.• If the product is subject to direct heat sources in the application, they should be shielded so that the actuator temperature stays within the stated operating range.Danger• Do not exceed any of the specifications listed in Section 2 of this document as this will be deemed improper use.• Air equipment has an air leakage during operation within certain limits. Do not use this equipment when the air itself introduces additional hazards and could lead to an explosion.• Use only Ex certified auto switches. These should be ordered separately.• Do not use this product in the presence of strong magnetic fields that could generate a surface temperature higher than the product specification.• Avoid applications where the piston rod end and the adjoining part in the application can create a possible ignition source.• Do not install or use these actuators where there is the possibility for the piston rod to impact foreign objects.• In the event of damage or failure of any parts located in the vicinity where this product has been installed, it is the responsibility of the user to determine whether or not this has compromised the safety and condition of this product and/or the application.• External impact on the cylinder body could result in a spark and/or cylinder damage. Avoid any application where foreign objects can hit or impact the cylinder. In such situations the application should install a suitable guard to prevent this occurrence.• Do not use this equipment where vibration could lead to failure.9 Product DisposalThis product shall not be disposed of as municipal waste. Check your local regulations and guidelines to dispose this product correctly, in order to reduce the impact on human health and the environment.Refer to or www.smc.eu for your local distributor/importer.URL : https:// (Global) https:// www.smc.eu (Europe) SMC Corporation, 4-14-1, Sotokanda, Chiyoda-ku, Tokyo 101-0021, JapanSpecifications are subject to change without prior notice from the manufacturer. © 2022 SMC Corporation All Rights Reserved. Template DKP50047-F-085M。
题44：编程输入10个正整数,然后自动按从大到小的顺序输出
题44：编程输入10个正整数，然后自动按从大到小的顺序输出。
题44：编程输入10个正整数，然后自动按从大到小的顺序输出。
Input输入只有一行，包括10整数.Output输出只有一行,包括10个整数。
Sample Input1 2 3 4 5 7 6 8 9 10Sample Output10 9 8 7 6 5 4 3 2 1提示：数组与排序，注意输出最后一个空格要考虑。
处理方法最后一个单独输出。
program ex_44;var a:array[1..10] of integer;i,j,t:integer;beginfor i:=1 to 10 do read(a[i]);for i:=1 to 9 dofor j:=i+1 to 10 doif a[i]<a[j] then begin t:=a[i];a[i]:=a[j];a[j]:=t; end;for i:=1 to 9 do write(a[i],' ');write(a[10]);end.题46：统计字母出现次数输入一串小写字母（以"."为结束标志），统计出每个字母在该字符串中出现的次数(若某字母不出现，则不要输出)。
Input输入只有一行，包括若干个字符。
Output输出只有两行，第一行为出现的小写字母，第二行为字母的出现次数。
Sample Inputabdceeef.Sample Outputabcdef111131提示：注意利用字符做下标，可以让程序简单。
program ex_45;var a:array['a'..'z'] of integer;i,ch:char;beginread(ch);for i:='a' to 'z' do a[i]:=0;while ch<>'.' dobegina[ch]:=a[ch]+1;read(ch);end;for i:='a' to 'z' doif a[i]<>0 then write(i);writeln;for i:='a' to 'z' doif a[i]<>0 then write(a[i]);end.题48：求序列中所有递增或递减子序列的个数并输出子序列题48：求序列中所有递增或递减子序列的个数并输出子序列。
MATLAB数学实验课后答案
数学实验MATLAB参考答案（重要部分）P20,ex1(5) 等于[exp(1),exp(2);exp(3),exp(4)](7) 3=1*3, 8=2*4(8) a为各列最小值，b为最小值所在的行号(10) 1>=4,false, 2>=3,false, 3>=2, ture, 4>=1,ture(11) 答案表明：编址第2元素满足不等式(30>=20)和编址第4元素满足不等式(40>=10)(12) 答案表明：编址第2行第1列元素满足不等式(30>=20)和编址第2行第2列元素满足不等式(40>=10)P20, ex2(1)a, b, c的值尽管都是1，但数据类型分别为数值，字符，逻辑，注意a 与c相等，但他们不等于b(2)double(fun)输出的分别是字符a,b,s,(,x,)的ASCII码P20,ex3>> r=2;p=0.5;n=12;>> T=log(r)/n/log(1+0.01*p)T =11.5813P20,ex4>> x=-2:0.05:2;f=x.^4-2.^x;>> [fmin,min_index]=min(f)fmin =-1.3907 %最小值min_index =54 %最小值点编址>> x(min_index)ans =0.6500 %最小值点>> [f1,x1_index]=min(abs(f)) %求近似根--绝对值最小的点f1 =0.0328x1_index =24>> x(x1_index)ans =-0.8500>> x(x1_index)=[];f=x.^4-2.^x; %删去绝对值最小的点以求函数绝对值次小的点>> [f2,x2_index]=min(abs(f)) %求另一近似根--函数绝对值次小的点f2 =0.0630x2_index =65>> x(x2_index)ans =1.2500P20,ex5>> z=magic(10)z =92 99 1 8 15 67 74 51 58 4098 80 7 14 16 73 55 57 64 414 81 88 20 22 54 56 63 70 4785 87 19 21 3 60 62 69 71 2886 93 25 2 9 61 68 75 52 3417 24 76 83 90 42 49 26 33 6579 6 13 95 97 29 31 38 45 7210 12 94 96 78 35 37 44 46 5311 18 100 77 84 36 43 50 27 59>> sum(z)ans =505 505 505 505 505 505 505 505 505 505 >> sum(diag(z))ans =505>> z(:,2)/sqrt(3)ans =57.157746.188046.765450.229553.693613.85642.88683.46416.928210.3923>> z(8,:)=z(8,:)+z(3,:)z =92 99 1 8 15 67 74 51 58 4098 80 7 14 16 73 55 57 64 414 81 88 20 22 54 56 63 70 4785 87 19 21 3 60 62 69 71 2886 93 25 2 9 61 68 75 52 3423 5 82 89 91 48 30 32 39 6683 87 101 115 119 83 87 101 115 11910 12 94 96 78 35 37 44 46 5311 18 100 77 84 36 43 50 27 59P 40 ex1先在编辑器窗口写下列M函数，保存为eg2_1.m function [xbar,s]=ex2_1(x)n=length(x);xbar=sum(x)/n;s=sqrt((sum(x.^2)-n*xbar^2)/(n-1));例如>>x=[81 70 65 51 76 66 90 87 61 77];>>[xbar,s]=ex2_1(x)xbar =72.4000s =12.1124P 40 ex2s=log(1);n=0;while s<=100n=n+1;s=s+log(1+n);endm=n计算结果m=37clear;F(1)=1;F(2)=1;k=2;x=0;e=1e-8; a=(1+sqrt(5))/2;while abs(x-a)>ek=k+1;F(k)=F(k-1)+F(k-2); x=F(k)/F(k-1); enda,x,k计算至k=21可满足精度P 40 ex4clear;tic;s=0;for i=1:1000000s=s+sqrt(3)/2^i;ends,toctic;s=0;i=1;while i<=1000000s=s+sqrt(3)/2^i;i=i+1;ends,toctic;s=0;i=1:1000000;s=sqrt(3)*sum(1./2.^i);s,tocP 40 ex5c=[15 14 14 14 14 15 16 18 20 22 23 25 28 ...31 32 31 29 27 25 24 22 20 18 17 16];plot(t,c)P 40 ex6(1)clear;fplot('x^2*sin(x^2-x-2)',[-2,2])x=-2:0.1:2;y=x.^2.*sin(x.^2-x-2);plot(x,y)y=inline('x^2*sin(x^2-x-2)');fplot(y,[-2 2])(2)参数方法t=linspace(0,2*pi,100);x=2*cos(t);y=3*sin(t); plot(x,y)(3)x=-3:0.1:3;y=x;[x,y]=meshgrid(x,y);z=x.^2+y.^2;surf(x,y,z)(4)x=-3:0.1:3;y=-3:0.1:13;[x,y]=meshgrid(x,y);z=x.^4+3*x.^2+y.^2-2*x-2*y-2*x.^2.*y+6;surf(x,y,z)(5)t=0:0.01:2*pi;x=sin(t);y=cos(t);z=cos(2*t);plot3(x,y,z)(6)theta=linspace(0,2*pi,50);fai=linspace(0,pi/2,20);[theta,fai]=meshgrid(theta,fai); x=2*sin(fai).*cos(theta);y=2*sin(fai).*sin(theta);z=2*cos(fai);surf(x,y,z)(7)x=linspace(0,pi,100);y1=sin(x);y2=sin(x).*sin(10*x);y3=-sin(x);plot(x,y1,x,y2,x,y3)page41, ex7x=-1.5:0.05:1.5;y=1.1*(x>1.1)+x.*(x<=1.1).*(x>=-1.1)-1.1*(x<-1.1);plot(x,y)page41,ex8分别使用which trapz, type trapz, dir C:\MATLAB7\toolbox\matlab\datafun\ page41,ex9clear;close;x=-2:0.1:2;y=x;[x,y]=meshgrid(x,y);a=0.5457;b=0.7575;p=a*exp(-0.75*y.^2-3.75*x.^2-1.5*x).*(x+y>1);p=p+b*exp(-y.^2-6*x.^2).*(x+y>-1).*(x+y<=1);p=p+a*exp(-0.75*y.^2-3.75*x.^2+1.5*x).*(x+y<=-1);mesh(x,y,p)page41, ex10lookfor lyapunovhelp lyap>> A=[1 2 3;4 5 6;7 8 0];C=[2 -5 -22;-5 -24 -56;-22 -56 -16];>> X=lyap(A,C)X =1.0000 -1.0000 -0.0000 -1.00002.0000 1.0000 -0.0000 1.0000 7.0000Chapter 3%Exercise 1>> a=[1,2,3];b=[2,4,3];a./b,a.\b,a/b,a\bans =0.5000 0.5000 1.0000ans =2 2 1ans =0.6552 %一元方程组x[2,4,3]=[1,2,3]的近似解ans =0 0 00 0 00.6667 1.3333 1.0000%矩阵方程[1,2,3][x11,x12,x13;x21,x22,x23;x31,x32,x33]=[2,4,3]的特解Exercise 2(1)>> A=[4 1 -1;3 2 -6;1 -5 3];b=[9;-2;1];>> rank(A), rank([A,b]) %[A,b]为增广矩阵ans =3ans =3 %可见方程组唯一解>> x=A\bx =2.38301.48942.0213Exercise 2(2)>> A=[4 -3 3;3 2 -6;1 -5 3];b=[-1;-2;1];>> rank(A), rank([A,b]) ans =3ans =3 %可见方程组唯一解>> x=A\bx =-0.4706-0.2941Exercise 2(3)>> A=[4 1;3 2;1 -5];b=[1;1;1];>> rank(A), rank([A,b])ans =2ans =3 %可见方程组无解>> x=A\bx =0.3311-0.1219 %最小二乘近似解Exercise 2(4)>> a=[2,1,-1,1;1,2,1,-1;1,1,2,1];b=[1 2 3]';%注意b的写法>> rank(a),rank([a,b])ans =3ans =3 %rank(a)==rank([a,b])<4说明有无穷多解>> a\bans =110 %一个特解Exercise 3>> a=[2,1,-1,1;1,2,1,-1;1,1,2,1];b=[1,2,3]';>> x=null(a),x0=a\bx =-0.62550.6255-0.20850.4170x0 =11%通解kx+x0 Exercise 4>> x0=[0.2 0.8]';a=[0.99 0.05;0.01 0.95];>> x1=a*x, x2=a^2*x, x10=a^10*x>> x=x0;for i=1:1000,x=a*x;end,xx =0.83330.1667>> x0=[0.8 0.2]';>> x=x0;for i=1:1000,x=a*x;end,xx =0.83330.1667>> [v,e]=eig(a)v =0.9806 -0.70710.1961 0.7071e =1.0000 00 0.9400>> v(:,1)./xans =1.17671.1767 %成比例，说明x是最大特征值对应的特征向量Exercise 5%用到公式(3.11)(3.12)>> B=[6,2,1;2.25,1,0.2;3,0.2,1.8];x=[25 5 20]'; >> C=B/diag(x)C =0.2400 0.4000 0.05000.0900 0.2000 0.0100 0.1200 0.0400 0.0900 >> A=eye(3,3)-CA =0.7600 -0.4000 -0.0500 -0.0900 0.8000 -0.0100 -0.1200 -0.0400 0.9100 >> D=[17 17 17]';x=A\D x =37.569625.786224.7690%Exercise 6(1)>> a=[4 1 -1;3 2 -6;1 -5 3];det(a),inv(a),[v,d]=eig(a) ans =-94ans =0.2553 -0.0213 0.04260.1596 -0.1383 -0.22340.1809 -0.2234 -0.0532v =0.0185 -0.9009 -0.3066-0.7693 -0.1240 -0.7248-0.6386 -0.4158 0.6170d =-3.0527 0 00 3.6760 00 0 8.3766%Exercise 6(2)>> a=[1 1 -1;0 2 -1;-1 2 0];det(a),inv(a),[v,d]=eig(a) ans =1ans =2.0000 -2.0000 1.00001.0000 -1.0000 1.00002.0000 -3.0000 2.0000v =-0.5773 0.5774 + 0.0000i 0.5774 - 0.0000i-0.5773 0.5774 0.5774-0.5774 0.5773 - 0.0000i 0.5773 + 0.0000id =1.0000 0 00 1.0000 + 0.0000i 00 0 1.0000 - 0.0000i%Exercise 6(3)>> A=[5 7 6 5;7 10 8 7;6 8 10 9;5 7 9 10]A =5 76 57 10 8 76 8 10 95 7 9 10>> det(A),inv(A), [v,d]=eig(A)ans =1ans =68.0000 -41.0000 -17.0000 10.0000-41.0000 25.0000 10.0000 -6.0000-17.0000 10.0000 5.0000 -3.000010.0000 -6.0000 -3.0000 2.0000v =0.8304 0.0933 0.3963 0.3803-0.5016 -0.3017 0.6149 0.5286-0.2086 0.7603 -0.2716 0.55200.1237 -0.5676 -0.6254 0.5209d =0.0102 0 0 00 0.8431 0 00 0 3.8581 00 0 0 30.2887%Exercise 6(4)、(以n=5为例)%关键是矩阵的定义%方法一（三个for）n=5;for i=1:n, a(i,i)=5;endfor i=1:(n-1),a(i,i+1)=6;endfor i=1:(n-1),a(i+1,i)=1;enda%方法二（一个for）n=5;a=zeros(n,n);a(1,1:2)=[5 6];for i=2:(n-1),a(i,[i-1,i,i+1])=[1 5 6];enda(n,[n-1 n])=[1 5];a%方法三（不用for）n=5;a=diag(5*ones(n,1));b=diag(6*ones(n-1,1));c=diag(ones(n-1,1));a=a+[zeros(n-1,1),b;zeros(1,n)]+[zeros(1,n);c,zeros(n-1,1)] %下列计算>> det(a)ans =665>> inv(a)ans =0.3173 -0.5865 1.0286 -1.6241 1.9489-0.0977 0.4887 -0.8571 1.3534 -1.62410.0286 -0.1429 0.5429 -0.8571 1.0286-0.0075 0.0376 -0.1429 0.4887 -0.5865 0.0015 -0.0075 0.0286 -0.0977 0.3173 >> [v,d]=eig(a)v =-0.7843 -0.7843 -0.9237 0.9860 -0.9237 0.5546 -0.5546 -0.3771 -0.0000 0.3771-0.2614 -0.2614 0.0000 -0.1643 0.0000 0.0924 -0.0924 0.0628 -0.0000 -0.0628-0.0218 -0.0218 0.0257 0.0274 0.0257d =0.7574 0 0 0 00 9.2426 0 0 00 0 7.4495 0 00 0 0 5.0000 00 0 0 0 2.5505%Exercise 7(1)>> a=[4 1 -1;3 2 -6;1 -5 3];[v,d]=eig(a) v =0.0185 -0.9009 -0.3066-0.7693 -0.1240 -0.7248-0.6386 -0.4158 0.6170d =-3.0527 0 00 3.6760 00 0 8.3766>> det(v)ans =-0.9255 %v行列式正常, 特征向量线性相关，可对角化>> inv(v)*a*v %验算ans =-3.0527 0.0000 -0.00000.0000 3.6760 -0.0000-0.0000 -0.0000 8.3766>> [v2,d2]=jordan(a) %也可用jordanv2 =0.0798 0.0076 0.91270.1886 -0.3141 0.1256-0.1605 -0.2607 0.4213 %特征向量不同d2 =8.3766 0 00 -3.0527 - 0.0000i 00 0 3.6760 + 0.0000i>> v2\a*v2ans =8.3766 0 0.00000.0000 -3.0527 0.00000.0000 0.0000 3.6760>> v(:,1)./v2(:,2) %对应相同特征值的特征向量成比例ans =2.44912.44912.4491%Exercise 7(2)>> a=[1 1 -1;0 2 -1;-1 2 0];[v,d]=eig(a)v =-0.5773 0.5774 + 0.0000i 0.5774 - 0.0000i-0.5773 0.5774 0.5774-0.5774 0.5773 - 0.0000i 0.5773 + 0.0000id =1.0000 0 00 1.0000 + 0.0000i 00 0 1.0000 - 0.0000i>> det(v)ans =-5.0566e-028 -5.1918e-017i %v的行列式接近0, 特征向量线性相关，不可对角化>> [v,d]=jordan(a)v =1 0 11 -1 0d =1 1 00 1 10 0 1 %jordan标准形不是对角的，所以不可对角化%Exercise 7(3)>> A=[5 7 6 5;7 10 8 7;6 8 10 9;5 7 9 10]A =5 76 57 10 8 76 8 10 95 7 9 10>> [v,d]=eig(A)0.8304 0.0933 0.3963 0.3803-0.5016 -0.3017 0.6149 0.5286-0.2086 0.7603 -0.2716 0.55200.1237 -0.5676 -0.6254 0.5209d =0.0102 0 0 00 0.8431 0 00 0 3.8581 00 0 0 30.2887>> inv(v)*A*vans =0.0102 0.0000 -0.0000 0.00000.0000 0.8431 -0.0000 -0.0000-0.0000 0.0000 3.8581 -0.0000-0.0000 -0.0000 0 30.2887%本题用jordan不行, 原因未知%Exercise 7(4)参考6(4)和7(1), 略%Exercise 8 只有(3)对称, 且特征值全部大于零, 所以是正定矩阵. %Exercise 9(1)>> a=[4 -3 1 3;2 -1 3 5;1 -1 -1 -1;3 -2 3 4;7 -6 -7 0]>> rank(a)ans =3>> rank(a(1:3,:))ans =2>> rank(a([1 2 4],:)) %1,2,4行为最大无关组3>> b=a([1 2 4],:)';c=a([3 5],:)'; >> b\c %线性表示的系数ans =0.5000 5.0000-0.5000 1.00000 -5.0000%Exercise 10>> a=[1 -2 2;-2 -2 4;2 4 -2]>> [v,d]=eig(a)0.3333 0.9339 -0.12930.6667 -0.3304 -0.6681-0.6667 0.1365 -0.7327d =-7.0000 0 00 2.0000 00 0 2.0000>> v'*vans =1.0000 0.0000 0.00000.0000 1.0000 00.0000 0 1.0000 %v确实是正交矩阵%Exercise 11%设经过6个电阻的电流分别为i1, ..., i6. 列方程组如下%20-2i1=a; 5-3i2=c; a-3i3=c; a-4i4=b; c-5i5=b; b-3i6=0; %i1=i3+i4;i5=i2+i3;i6=i4+i5;%计算如下>> A=[1 0 0 2 0 0 0 0 0;0 0 1 0 3 0 0 0 0;1 0 -1 0 0 -3 0 0 0;1 -1 0 0 0 0 -4 0 0;0 -1 1 0 0 0 0 -5 0;0 1 0 0 0 0 0 0 -3;0 0 0 1 0 -1 -1 0 0;0 0 0 0 -1 -1 0 1 0;0 0 0 0 0 0 -1 -1 1];>>b=[20 5 0 0 0 0 0 0 0]'; A\bans =13.34536.44018.54203.3274-1.18071.60111.72630.42042.1467%Exercise 12>> A=[1 2 3;4 5 6;7 8 0];>> left=sum(eig(A)), right=sum(trace(A))left =6.0000right =6>> left=prod(eig(A)), right=det(A) %原题有错, (-1)^n应删去left =27.0000right =27>> fA=(A-p(1)*eye(3,3))*(A-p(2)*eye(3,3))*(A-p(3)*eye(3,3))fA =1.0e-012 *0.0853 0.1421 0.02840.1421 0.1421 0-0.0568 -0.1137 0.1705>> norm(fA) %f(A)范数接近0ans =2.9536e-013%Exercise 1(1)roots([1 1 1])%Exercise 1(2)roots([3 0 -4 0 2 -1])%Exercise 1(3)p=zeros(1,24);p([1 17 18 22])=[5 -6 8 -5];roots(p)%Exercise 1(4)p1=[2 3];p2=conv(p1, p1);p3=conv(p1, p2);p3(end)=p3(end)-4; %原p3最后一个分量-4roots(p3)%Exercise 2fun=inline('x*log(sqrt(x^2-1)+x)-sqrt(x^2-1)-0.5*x'); fzero(fun,2)】%Exercise 3fun=inline('x^4-2^x');fplot(fun,[-2 2]);grid on;fzero(fun,-1),fzero(fun,1),fminbnd(fun,0.5,1.5)%Exercise 4fun=inline('x*sin(1/x)','x');fplot(fun, [-0.1 0.1]);x=zeros(1,10);for i=1:10, x(i)=fzero(fun,(i-0.5)*0.01);end;x=[x,-x]%Exercise 5fun=inline('[9*x(1)^2+36*x(2)^2+4*x(3)^2-36;x(1)^2-2*x(2)^2-20*x(3);16*x(1)-x(1)^3-2*x(2)^2-16*x(3)^2]','x');[a,b,c]=fsolve(fun,[0 0 0])%Exercise 6fun=@(x)[x(1)-0.7*sin(x(1))-0.2*cos(x(2)),x(2)-0.7*cos(x(1))+0.2*sin(x(2))];[a,b,c]=fsolve(fun,[0.5 0.5])%Exercise 7clear; close; t=0:pi/100:2*pi; x1=2+sqrt(5)*cos(t); y1=3-2*x1+sqrt(5)*sin(t);x2=3+sqrt(2)*cos(t); y2=6*sin(t);plot(x1,y1,x2,y2); grid on; %作图发现4个解的大致位置，然后分别求解y1=fsolve('[(x(1)-2)^2+(x(2)-3+2*x(1))^2-5,2*(x(1)-3)^2+(x(2)/3)^2-4]',[1.5,2])y2=fsolve('[(x(1)-2)^2+(x(2)-3+2*x(1))^2-5,2*(x(1)-3)^2+(x(2)/3)^2-4]',[1.8,-2])y3=fsolve('[(x(1)-2)^2+(x(2)-3+2*x(1))^2-5,2*(x(1)-3)^2+(x(2)/3)^2-4]',[3.5,-5])y4=fsolve('[(x(1)-2)^2+(x(2)-3+2*x(1))^2-5,2*(x(1)-3)^2+(x(2)/3)^2-4]',[4,-4])%Exercise 8(1)clear;fun=inline('x.^2.*sin(x.^2-x-2)');fplot(fun,[-2 2]);grid on; %作图观察x(1)=-2;x(3)=fminbnd(fun,-1,-0.5);x(5)=fminbnd(fun,1,2);fun2=inline('-x.^2.*sin(x.^2-x-2)');x(2)=fminbnd(fun2,-2,-1);x(4)=fminbnd(fun2,-0.5,0.5);x(6)=2feval(fun,x)%答案: 以上x(1)(3)(5)是局部极小，x(2)(4)(6)是局部极大，从最后一句知道x(1)全局最小， x(2)最大。
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ENC-510EX/520EX ENC-710EX/720EX ENC-1010EX/1020EX ENC-1652EX ENC-2200EX
ENC-2900EX
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1. 部件名称 ............................................................................................................4 2. 安装 ...................................................................................................................6
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ENC-710EX，720EX
外气吸入口
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5YyX5Lqs5pyN6KOF5a2m6ZmiMjAxMeW5tOaZrumAmuaWh+WPsuOAgeeQhuW3peex_b7NMkb[1]
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北京服装学院2011年普通文史、理工类专业招生报考指南学校简介北京服装学院是全国唯一以服装命名，艺工为主，艺、工、经、管等多学科协调发展，具有鲜明办学特色的全日制普通高等学校。
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学校现有各类在校生7000余人，教职员工800余人，设有5个二级学院（服装艺术与工程学院、艺术设计学院、材料科学与工程学院、商学院、信息工程学院）、2个系（外语系、造型艺术系）、2个教学部（思想政治理论课教学部、基础教学部）、1个教学中心（计算机信息中心）及国际学院、继续教育学院和研究生部。
有21个本科专业，7个一级学科硕士点、20余个二级学科硕士点，1个第二学士学位点、4个双学位点。
有1个北京市重点建设一级学科（纺织科学与工程）, 3个北京市重点建设二级学科和4个市级科研机构（北京市服装材料研究开发与评价重点实验室、首都服饰文化与服装产业研究基地、服装北京市技术转移中心、数字与交互媒体实验室）。
有全国“十佳”特色博物馆之一的民族服饰博物馆。
学校坚持以本科教育为主，积极发展研究生、留学生和继续教育，形成了以本科为主的多层次办学格局。
尤其自2007年教育部实施高等教育“质量工程”以来，学校以“质量工程”为契机，坚持科学发展观，不断深入教育教学改革，通过科学规划、精心组织，扎实推进，各项工程均取得了标志性成果，使我校的综合办学实力和社会影响力得到进一步的提升：目前拥有国家级特色专业建设点4个——艺术设计、服装设计与工程、高分子材料与工程、轻化工程；北京市级特色建设专业点6个——艺术设计、服装设计与工程、高分子材料与工程、工业设计、表演、轻化工程；北京市级精品课程5门——《服装艺术设计》、《中国服装史》、《数码设计基础》、《成衣纸样与工艺》、《服装画技法》；北京高等教育精品教材6部——《服装艺术设计》和《服装纸样设计原理与技术(男装／女装)》等；国家级优秀教学团队1个——“纺织材料与纺织品设计艺工结合教学团队”；北京市级优秀教学团队4个——“纺织材料与纺织品设计艺工结合教学团队”、“服装设计与营销管理教学团队”、“计算机公共课教学团队”和“国际经济与贸易教学团队”；北京市教学名师3人；北京市级实验教学示范中心2个——“服装材料与工程实验教学中心”和“服装服饰实验教学中心”；国家级人才培养模式创新实验区1个——“艺工融合应用型现代服装高级人才培养模式创新实验区”；北京高等学校市级校外人才培养基地2个——“北服•爱慕服装专业校外人才培养基地”和“北服•新百丽校外人才培养基地”；学校还承担了教育部“新世纪高等教育教学改革工程”项目、教育部英语教学改革项目、北京市级教改立项多项，多次获得国家级、省部级以及北京市级优秀教学成果奖等奖项。
03日立EX系列挖掘机性能参数
EX210LC-5
EX220-5
EX230LC-5
EX300-5
EX350LC-5
A
轴距
mm
3370
3660
3460
10140
3730
4050
B
行走机构总长度
mm
4170
4460
4260
3190
4640
4940
C
配重离地间隙
mm
1030
1030
1090
3100
1140
1160
D
尾端回转半径
6000
6400
6400
斗杆长度(mm)
2910
2910
2960
3200
3200
铲斗挖掘ቤተ መጻሕፍቲ ባይዱ(kN)
134
134
158
158
218
218
斗杆挖掘力(kN)
99
99
110
110
157
157
工作重量(kg)
18800
19900
22500
23800
31000
32600
接地压力(kpa)
42
41
49
47
63
61
SuperEX系列履带式挖掘机
（合肥日立）
（1）主要技术性能表
型号
项目
EX200-5
EX210LC-5
EX220-5
EX230LC-5
EX300-5
EX350LC-5
规格（标准配置）
铲斗型式
反铲
反铲
反铲
反铲
反铲
反铲
防爆等级Exib
防爆等级ExibⅡBT是什么意思？Ex是防爆电气产品的标志，ib是可以应用于一区二区的本质安全型电气产品，ⅡB是可以应用于ⅡB类气体与空气形成的爆炸性混合物环境中，防爆等级划分，主要分为I类（矿用）、II类（厂用）。
其中II类又分为：IIA、IIB、IIC（安全级别:A〈B〈C)T是温度组别,T4是指设备表面温度不超过135度,T6安全级别最高,也就是设备表面温度越低越好.综合来说就是本防爆产品为可用于ⅡB类气体的一区二区，设备表面温度不超过135度的本质安全型防爆电气产品。
淀粉粉尘被引燃的最大温度：410℃～430℃名词解释：1、T：爆炸性环境用电气设备按其最高表面温度划分为T1—T6组别交换机，光端机，等弱电电子设备的标准一般设为T6；2 ia 等级：在正常工作、一个故障和二个故障时均不能点燃爆炸性气体混合物的电气设备。
正常工作时，安全系数为 2.0；一个故障时，安全系数为1。
5;二个故障时，安全系数为1。
0 。
注:有火花的触点须加隔爆外壳、气密外壳或加倍提高安全系数.ib 等级：在正常工作和一个故障时不能点燃爆炸性气体混合物的电气设备。
正常工作时,安全系数为2。
0 ；一个故障时，安全系数为1。
5 。
正常工作时，有火花的触点须加隔爆外壳或气密外壳保护,并且有故障自显示的措施，一个故障时安全系数为1。
0 .EExd：是指将爆炸包起来的意思弱电设备自我感觉防爆等级是ia?3、Ⅰ类是矿类,Ⅱ类是厂用类.Ⅱ类又可分为ⅡA、ⅡB、ⅡC类,标志ⅡB的设备可适用于ⅡA设备的使用条件；ⅡC可适用于ⅡA、ⅡB的使用条件。
4、1、国际电工委员会/欧洲电工委员会划分的危险区域的等级分类0区(Zone 0）：易爆气体始终或长时间存在；连续地存在危险性大于1000小时/每年的区域；1区（Zone 1）：易燃气体在仪表的正当工作过程中有可能发生或存在；断续地存在危险性10～1000小时/每年的区域；2区（Zone 2）：一般情形下,不存在易燃气体且即使偶尔发生，其存在时间亦很短；事故状态下存在的危险性0.1~10小时/每年的区域；5、ⅡA、ⅡB、ⅡC类的区别：气体组别险性气体标准化委员会EN50014EC NEC500条款CLASS1表气GB-3836-1 能量（微焦)乙炔ⅡC A ⅡC 20氢气ⅡC A ⅡC 20乙烯ⅡB C ⅡB 60丙烷ⅡA D ⅡA 180注:中国GB3836标准规定ⅡC级最小点燃能量为19微焦耳，ⅡA级最小点燃能量为200微焦耳。
5-2定积分计算
2 1
1 ex
1
d (e x
1)
ln
ex
1
2 1
ln(e 1)
24
定积分的分部积分法
练习 求下列定积分
(1) e x2 ln xdx 1
解 u ln x, v x2 ,
u 1 ,
x3 v
x
3
e x2 ln xdx
x3
e
ln x
e 1 x3 dx
2
21
2) 1 xe2xdx 1 1 xde2x
0
20
1 ( xe2 x 2
1 0
1 e2 xdx)
0
1 (e2 1 1 e2xd 2x)
2
20
1
1 (e2 1 e2x )
2
20
1 (e2 1) 4
22
定积分的换元法
练习 求下列定积分
(1)
ln 2
14
例2 求定积分：
(1)
2
1
(
x
1 x
2
)
dx
(2)
1 ex
1 1 ex
dx
(3)
2
| x 1 | dx 1
解
(1)
2
1
(
x
1
2
)
dx
x
2 (x2
1
2
1 )dx x2
( x3 2x 1) 2 4 5
3
x1 6
(2)
1 ex dx