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I need ( ) cloth, for I'm going to make ( ) clothes.A、many/ manyB、many/ a lot o{C、a lot of/ manyD、much/ much

题目
I need ( ) cloth, for I'm going to make ( ) clothes.

A、many/ many

B、many/ a lot o{

C、a lot of/ many

D、much/ much

相似考题

1.---- Shall we stay at home watching TV tonight?---- No, I’d like __________ and see a movie.A. to go B. go C. went D. going

2.–Do you ever play tennis?–_______.I have played a few times.I'd rather _____ jogging.A.Hardly ever ⋯goingB.Hardly ever ⋯goC.Hardly never ⋯goD.Hardly never ⋯going

3.It was raining outside, I ( ) one umbrella and walked out.A.picked upB.take upC.haveD.make

4.—_________ I take the book out?—I'm afraid not.A. WillB. MayC. MustD. Need

参考答案和解析
参考答案：C
更多“I need ( ) cloth, for I'm going to make ( ) clothes. ”相关问题

  • 第1题：

    I ________ help. I can do it myself.

    A.need to

    B.don’t need

    C.needn’t

    D.need


    参考答案：B

  • 第2题：

    在Need矩阵中，need（i,j）=a表示（）。

    A．进程i最多需要j类资源a个

    B．进程i最多还需要j类资源a个

    C．进程j最多还需要i类资源a个

    D．进程j最多需要i类资源a个


    进程i最多还需要j类资源a个

  • 第3题：

    下面（）仅输出m的大于1的最小因子。

    A．for (i =2; i＜=m-1; i++) if (m % i == 0) { printf("%d is 最小因子n", i); break; }

    B．for (i =2; i＜=m-1; i++) if (m % i == 0) { printf("%d is 最小因子n", i); continue; }

    C．for (i =2; i＜=m-1; i++) if (m % i == 0) { printf("%d is 最小因子n", i); }

    D．i=2; while (m % i != 0) i++; printf("%d is 最小因子n", i);


    A

  • 第4题：

    I ____ you will write me back soon.

    A、wish

    B、hope

    C、want

    D、need


    正确答案:B

  • 第5题：

    2、在Need矩阵中，need（i,j）=a表示（）。

    A．进程i最多需要j类资源a个

    B．进程i最多还需要j类资源a个

    C．进程j最多还需要i类资源a个

    D．进程j最多需要i类资源a个


    P1;P3

  • 第6题：

    下面（）是正确的判断素数程序（m＞1)。

    A．j=0; for (i =2; i＜=m-1; i++) if (m % i != 0) j++; if(j==m-2) printf(“%d是素数n", m);

    B．j=0; for (i =2; i＜=m-1; i++) if (m % i == 0) j++; if(j==0) printf(“%d是素数n", m);

    C．flag=0; for (i =2; i＜=m-1; i++) if (m % i == 0) flag=1; if(flag==0) printf(“%d是素数n", m);

    D．for (i =2; i＜=m-1; i++) if (m % i == 0) i=m+2; if(i==m+3) printf(“%d是素数n", m);


    D

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