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What essentially distinguishes semantics and pragmatics is whether in the study of me

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更多“What essentially distinguishes…”相关的问题
第1题
Insurance contract is essentially a contract between the insurer and the insured. ()A.正确

Insurance contract is essentially a contract between the insurer and the insured. ()

A.正确

B.错误

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第2题
Insurance is essentially a contract between two parties, namely the insurer and insured,like the con
signor and consignee.( )
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第3题
Essentially,imports or exports and the vessels carrying the goods are subject to
arrival formalities,departure formalities and customs clearance.()

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第4题
在HTML文件中使用<font>元素来定义文字字体,使用dis属性定义使用何种字体,使用size属性定义字体大小,使用color属性改变文字颜色()
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第5题
汉译英:“卸货港;到达日期”,正确的翻译为:()A.Port Of loading;Date Of departureB.Port Of dis

汉译英:“卸货港;到达日期”,正确的翻译为:()

A.Port Of loading;Date Of departure

B.Port Of discharge;Date Of departure

C.Port Of loading;Date Of arrival

D.Port Of discharge;Date Of arrival

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第6题
汉译英:“途经口岸;处理日期”,正确的翻译为:()A.Port of call; Date of departureB.Port of dis

汉译英:“途经口岸;处理日期”,正确的翻译为:()

A.Port of call; Date of departure

B.Port of discharge; Date of departure

C.Port of call; date of treatment

D.Port of discharge; Date of treatment

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第7题
汉译英:“装货港;到达日期”,正确的翻泽为:()A.Port of loading;Date of departureB.Port of dis

汉译英:“装货港;到达日期”,正确的翻泽为:()

A.Port of loading;Date of departure

B.Port of discharge;Date of departure

C.Port of loading;Date of arrival

D.Port of discharge;Date of arrival

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第8题
●SDLC was invented by IBM to replace the older Bisynchronous protocol for wide area connec
tions between IBM equipment .A varietion of the (71) protocol called NRM is essentially the same as SDLC.SDLC is (72) protocol like HDLC , Frame. Relay or X.25.An SDLC network is made up of (73) that controls all communications , and (74) .It is capable of (75) operation, but almost all practical applications are strictly half duplex.

(71) A.HDLC

B.Frame. Relay

C.X.25

D.X.21

(72) A.a peer to peer

B.a point-to-point

C.not a peer to peer

D.a peer to peers

(73) A.a primary station

B.one or more primary stations

C.a secondary station

D.one or more secondary stations

(74) A.a primary station

B.one or more primary stations

C.a secondary station

D.one or more secondary stations

(75) A.half duplex

B.full duplex

C.half or full duplex

D.half and full duplex

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第9题
阅读以下说明和C++程序,将应填入(n)处的字句写在对应栏内。【说明】 字符串在程序设计中扮演着重要

阅读以下说明和C++程序,将应填入(n)处的字句写在对应栏内。

【说明】

字符串在程序设计中扮演着重要角色。现需要设计字符串基类string,包含设置字 符串、返回字符串长度及内容等功能。另有一个具有编辑功能的串类edlt_string,派生于string,在其中设置一个光标,使其能支持在光标处的插入、删除操作。

【程序】

include <iostream.h>

include <stdio.h>

include <string.h>

class string

{

int length;

char *data;

public:

int get_length() {return length;}

char *get_data() {return data;}

~string() {delete data;}

int set data(int in_length, char *in_data);

int set_data(char *data);

void print() {cout<<data<<endl;}

};

class edit_string: public string

{

int cursor;

public:

int get_cursor() {return cursor;}

void move_cursor(int dis) {cursor=dis;}

int add_data(string *new_data);

void delete_data(int num);

};

int string::set_data(int in_length,char *in_data)

{

length=in_length;

if(!data)

delete data;

(1)

strcpy(data,in_data);

return length;

}

int string::set data(char *in_data)

{

(2)

if(!data)

delete data;

(1)

strcpy(data,in_data);

return length;

}

int edit_string::add_data(string *new_data)

{

int n,k,m;

char *cp,*pt;

n=new_data->get_length();

pt=new_data->get_data();

cp=this->get_data();

m=this->get_length();

char *news=new char[n+m+1];

for(int i=0; i<cursor; i++)

news[i]=cp[i];

k=i;

for(int j=0; j<n; i++,j++)

news[i]=pt[j];

cursor=i;

for(j=k; j<m; j++,i++)

(3)

news[i]='\0';

(4)

delete news;

return cursor;

}

void edit string::delete_data(int num)

{

int m;

char *cp;

cp=this->get_data();

m=this->get_length();

for(int i=cursor; i<m; i++)

(5)

cp[i]='\0';

}

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第10题
What is project management, and what are the steps managers use in planning projects?
What is project management, and what are the steps managers use in planning projects?

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第11题
Can you give me _____ was left?

A.all that

B.that

C.what that

D.it that

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