Part BDirections:Write an essay with the title “Reading Books in Printed Form. or on Computer” in which you should1) analyze the strong points and weak points of each one, and2) tell your own preference and your reasons.You should write 160—200 words neat

题目

Part B

Directions:

Write an essay with the title “Reading Books in Printed Form. or on Computer” in which you should

1) analyze the strong points and weak points of each one, and

2) tell your own preference and your reasons.

You should write 160—200 words neatly on ANSWER SHEET 2.(20 points)


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    A.CREATE UNIQUE INDEX idx_partno ON parts(part_no)

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    C.CREATE UNIQUE INDEX idx_partno ON parts(part_name, part_no ASC)

    D.CREATE UNIQUE INDEX idx_partno ON parts(part_no, part_name ASC)


    参考答案:A

  • 第2题:

    What publications should a GMDSS Operator consult regarding the proper set-up and operation of vessel equipment ________.

    A.ITU Publications

    B.The manufacturers instruction manuals

    C.Part 90 of the FCC Rules and Regulations

    D.Code of Federal Regulations,Title 47,Part 80,Subpart W


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  • 第3题:

    语素的主要作用是构词,根据其构词作用可分三部分,例如汉语中“老鹰、阿爸”中的“老、啊”和英语happiness、writer中的ness和er就是()、“鹰、爸、happy、write”是(),英语books、doing、walked中的s、ing和ed是()。
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  • 第4题:

    有下列SQL SELECT语句: SELECT * FORM. 成绩表 WHERE 物理 BETWEEN 80 AND 90 下列与该语句等价的( )。

    A.SELECT * FORM. 成绩表 WHERE 物理<=90 AND 物理 >=80

    B.SELECT * FORM. 成绩表 WHERE 物理 <90 AND 物理>80

    C.SELECT * FORM. 成绩表 WHERE 物理>=90 AND 物理<=80

    D.SELECT * FORM. 成绩表 WHERE 物理>90 AND 物理<80


    正确答案:A
    解析:BETWEEN..AND...是SQL查询语句中的一个特殊运算符,意思为“...和...之间”,相当与逻辑与运算中的(A>=AND>=B,其中A>B)。

  • 第5题:

    Which of the following literary forms is regarded as the most common and influential form that English ( ) poetry has taken since 16th century?

    A.Sonnet
    B.Blank Verse
    C.Free Verse
    D.Essay

    答案:B
    解析:
    考查诗歌形式。无韵体诗是用无韵五步抑扬格写成的诗歌,它被称为“可能是自16世纪以来英国诗歌最常见和最具影响力的形式”。

  • 第6题:

    30、如下程序的输出结果是 int main() { char books[][20]={"English","Math","Physical"}; int i,j; for(i=0;i<3;i++) { for(j=0;books[i][j]!=0;j++){ if(books[i][0]<books[i][j]) books[i][0]= books[i][j]; } } printf("%c",books[0][0]); return 0; }

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    B.s

    C.n

    D.h


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