How can I get legal advice for free?()A. You can get nothing for free.B. To be frank, I don't know for sure.C. You can go to a law firm and ask about that.

题目

How can I get legal advice for free?()

A. You can get nothing for free.

B. To be frank, I don't know for sure.

C. You can go to a law firm and ask about that.

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2.---Good morning. I've got an appointment with Miss Smith in the PersonnelDepartment.--Ah, good morning. You ____be Mrs. Peters.A. mightB. mustC. wouldD. can

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更多“How can I get legal advice for free?()A. You can get nothing for free.B. To be frank, I ”相关问题

  • 第1题：

    7．I_______ dance． It's too difficult for me．

    A． can

    B． am

    C． can’t

    D． am not


    正确答案：C

  • 第2题：

    —_______ I return the book to the library this week?— No, you _______.You can keep it until the end of this month.

    A.Must; needn’t

    B.Can; can’t

    C.May; mustn’t

    D.Can; mustn’t


    答案：A

  • 第3题：

    If he had really been there, I ___________ have seen him

    A.will
    B.shall
    C.should
    D.can

    答案：C
    解析：
    考查应拟语气。根据if条件从句中的had really been there可知，这是与过去事实相反的虚拟语气．因此从句谓语动词用had+过去分词的形式，主句谓语动词用would／should／could／might+have+过去分词的形式。故C项正确。

  • 第4题：

    I() packing in wooden cases.

    A、think

    B、prefer

    C、get


    参考答案：B

  • 第5题：

    If any beasts comes at you, I()stay with you and help you.

    A、am going to

    B、will

    C、can


    正确答案:B

  • 第6题：

    下面的代码用于输出字符数组ch中每个字符出现的次数，应该填入的代码是（）public static void main（String[] args) { char[] ch = { 'a', 'c', 'a', 'b', 'c', 'b' }; HashMap map = new HashMap（); for （int i = 0; i ＜ ch.length; i++) { ＜ 填入代码 ＞ } System.out.println（map); }

    A．if (map.contains(ch[i])) { map.put(ch[i], map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }

    B．if (map.contains(ch[i])) { map.put(ch[i], (Integer) map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }

    C．if (map.containsKey(ch[i])) { map.put(ch[i], (int) map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }

    D．if (map.containsKey(ch[i])) { map.put(ch[i], (Integer) map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }


    if (map.containsKey(ch[i])) { map.put(ch[i], (Integer)map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }

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