–Would you like to take a look?–()!A、With pleasureB、No, I wouldn’tC、No, pleaseD、Here you are

题目
–Would you like to take a look?–()!

A、With pleasure

B、No, I wouldn’t

C、No, please

D、Here you are

相似考题

1.It is easy to ____ a machine ____ but difficult to put it together.A、tear…apartB、take …awayC、take …apartD、bring…away

2.”You ____ have a wrong number,” she said. “There’s no one of that name here.”A. needB. canC. mustD. would

3.7． If I _______ you，l_______ take an umbrella with me．A． was ; shouldB． was ; wouldC． were ; willD． were ; would

4.A) 连词成句（共5小题，每小题1分，计5分）将所给单词连成完整、正确的句子。（单词不得重复使用，标点已给出）86. do, like, you, English__________________________________________________________________________?

更多“–Would you like to take a look?–()! ”相关问题

  • 第1题：

    If you don’t like to swim, you _____as well stay at home.

    A)should            B)may            C)can             D)would


    选B,因为might/may as well 或者 just as well 都是表达"不妨,做好...的意思."

  • 第2题：

    You can ______the seat belt ____ as soon as the light overhead goes off.

    A.turn…on

    B.take…off

    C.take…out

    D.pick…out


    正确答案：C

  • 第3题：

    “You ____ have a wrong number,” she said. “There’s no one of that name here.”

    A. need

    B. can

    C. must

    D. would


    正确答案：C
    解析：must have意思是肯定有什么，依题意可得，答案是C。

  • 第4题：

    "You _________ finish your homework before the dinner." said mother.

    A.should

    B.have to

    C.must

    D.would


    参考答案：C

  • 第5题：

    设有语句 int n,m; char a[5],b; cin＞＞n＞＞a＞＞m＞＞b; 为了分别将10和20分别赋给n和m，look赋给a，t赋给b，则正确的输入为

    A．10 "look" 20 't'＜CR＞

    B．10 "look" 10

    C．10 look 20 t

    D．10＜CR＞look20＜CR＞t＜CR＞


    正确答案：C
    解析：本题考查C++中流操作符的操作规则，即分割符。首先根据字符不需要加引号可以排除选项A和B，而D用回车作分割符，但却在look和20之间缺少一个回车符，故据此可选出正确答案为C。

  • 第6题：

    1、利用艾拉托斯特尼筛法无穷素数列表primes （参见讲义第8章) 获得介于100000和200000之间素数的方法是

    A．take 200000 (take 100000 primes)

    B．take 100000 (take 200000 primes)

    C．take 100000 (drop 100000 primes)

    D．takeWhile (\x -> x <= 200000) (dropWhile (\x -> x <= 100000) primes)

    E．takeWhile (\x -> x <= 200000) (take 200000 primes)

    F．[x | x <- primes, x >100000, x < 200000]

    G．[x | x <- take 200000 primes, x >100000, x < 200000]


    Sieve(n，P)． 输入：正整数n． 输出：小于等于n的所有素数P． ①if n=1 then P← ，计算结束； ②P←{2}； ③a←2； ④if n=a then计算结束； ⑤b←min{a 2 ，n}； ⑥Q←{x|a＜x≤b}； ⑦for P中的每一个x ⑧for Q中的每一个y ⑨ then从Q中删去y； ⑩P←P∪Q； (11)a←b； (12)转④．

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