A: 1.apparent 2.affix 3.appearances
B: 1.arrogant 2.affix 3.aesthetics
C: 1.apparent 2.affect 3.aesthetics
D: 1.arrogant 2.affect 3.appearances
举一反三
- They don’t affix much value to their looks, or think of them in terms of aesthetics
- If __________ thought passes through their minds at all, they like to think of themselves as average-looking.
- 下面代码的输出结果是（ ）。 t=[1,2,3] s=tuple(t) print(t,s) A: [1, 2, 3] [1, 2, 3] B: (1, 2, 3) (1, 2, 4) C: [1, 2, 3] (1, 2, 3) D: (1, 2, 6)[1, 2, 3]
- In the author’s opinion, most men ( ) A: do not pay any attention to their looks. B: are arrogant about their looks. C: think of themselves as average-looking. D: think of themselves as bad-looking.
- 设有如下关系表： R S T A B C A B C A B C 1 1 2 3 1 3 1 1 2 2 2 3 2 2 3 3 1 3
内容
- 0
有以下程序： main() int i, j; for(i=1; i＜4; i++) for(j=i; j＜4; j++)printf("%d* %d=%d", i, j, i*j); printf("\n"); 程序运行后的输出结果是 A: A) 1*1=1 1*2=2 1*3=3 B: 2*1=2 2*2=4 C: 3*1=3 D: B) 1*1=1 1*2=2 1*3=3 E: 2*2=4 2*3=6 F: 3*3=9 G: C) 1*1=1 H: 1*2=2 2*2=4 I: 1*3=3 2*3=6 3*3=9 J: D) 1*1=1 K: 2*1=2 2*2=4 L: 3*1=3 3*2=6 3*3=9
- 1
设α1=(1，4，3，-1)T，α2=(2，t，-1，-1)T，α3=(-2，3，1，t+1)T，则 A: 对任意的t，α1，α2，α3必线性无关． B: 仅当t=-3时，α1，α2，α3线性无关． C: 若t=0，则α1，α2，α3线性相关． D: 仅t≠0且t≠-3，α1，α2，α3线性无关．
- 2
设集合A={1, 2, 3}, 不是A上的等价关系。 A: {(1, 1), (2, 2), (3, 3)} B: {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)} C: {(1, 1), (2, 2), (3, 3), (2, 3)} D: {(1, 1), (2, 2), (2, 3), (3, 2), (2, 1), (1, 2), (3, 3), (1, 3), (3, 1)}
- 3
设集合A ={1, 2, 3}, 下列关系R中哪些是等价关系？A、{<1, 1>, <2, 2>, <3, 3>}B、{<1, 1>, <2, 2>, <3, 3>, <3, 2>, <2, 3>}C、{<1, 1>, <2, 2>, <3, 3>, <1, 3>}D、{<1, 1>, <2, 2>, <1, 2>, <2, 1>, <1, 3>, <3, 1>, <3, 3>, <2, 3>, <3, 2>}
- 4
下面程序的功能是输出以下9阶方阵。请填空。 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 1 1 2 3 3 3 3 3 2 1 1 2 3 4 4 4 3 2 1 1 2 3 4 5 4 3 2 1 1 2 3 4 4 4 3 2 1 1 2 3 3 3 3 3 2 1 1 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 # include int main( ) { int a[10][10]，n，i，j，m； scanf("%d"，&n)； if(n%2= =0) m=n/2； else( )； for(i=0；i m=n/2+1 n–i–1 n–i–1