以下标准溶液可以用直接法配制的是
未知类型:{'options': ['[tex=3.429x1.214]sonuzpBCv7Kt90pHdgmkizd0+pS8gZBK67KucBPnRtU=[/tex]', '[tex=2.786x1.0]fAb4FpUCsaTvk2oI8CN7tg==[/tex]', '[tex=3.786x1.214]4pYEdQ6bu7+lUbnJYqE9etZxlJ9PQepPn3OVcXJGNhIXusiSluC/Ht3ikY3wShr6[/tex]', '[tex=2.714x1.214]QT/fShUhKbOnvkiuSwknFLHUrthFVGT9rLwkA8q3Q/c=[/tex]', '[tex=1.786x1.0]6DbVkdVWKo1p7cRks+Ra5A==[/tex]', '[tex=2.643x1.214]4o3S0LbXYfTXzpVtjYQZ8FWqfTW34Uhrju6qCtejiLs=[/tex]', '[tex=4.5x1.214]Ixqj8Fgi0ShYA5cQcJ+gMOelXofrKVRQgynlr3KLDDic2jGCGKldwqZp+JQ+Aaok[/tex]', '[tex=6.429x1.214]fefqjYOm+vsntSmwwfQligEG2whXsEq6JbMNb9lQgWs7cCyCQzCBoqBzVP7R0Bl7nwj+1R0rm5KWcDTm5x/S85/q0s3DetHVoIA4kjcpgBTdop2GigHkzHp0Mmcw8+ra[/tex]'], 'type': 102}
未知类型:{'options': ['[tex=3.429x1.214]sonuzpBCv7Kt90pHdgmkizd0+pS8gZBK67KucBPnRtU=[/tex]', '[tex=2.786x1.0]fAb4FpUCsaTvk2oI8CN7tg==[/tex]', '[tex=3.786x1.214]4pYEdQ6bu7+lUbnJYqE9etZxlJ9PQepPn3OVcXJGNhIXusiSluC/Ht3ikY3wShr6[/tex]', '[tex=2.714x1.214]QT/fShUhKbOnvkiuSwknFLHUrthFVGT9rLwkA8q3Q/c=[/tex]', '[tex=1.786x1.0]6DbVkdVWKo1p7cRks+Ra5A==[/tex]', '[tex=2.643x1.214]4o3S0LbXYfTXzpVtjYQZ8FWqfTW34Uhrju6qCtejiLs=[/tex]', '[tex=4.5x1.214]Ixqj8Fgi0ShYA5cQcJ+gMOelXofrKVRQgynlr3KLDDic2jGCGKldwqZp+JQ+Aaok[/tex]', '[tex=6.429x1.214]fefqjYOm+vsntSmwwfQligEG2whXsEq6JbMNb9lQgWs7cCyCQzCBoqBzVP7R0Bl7nwj+1R0rm5KWcDTm5x/S85/q0s3DetHVoIA4kjcpgBTdop2GigHkzHp0Mmcw8+ra[/tex]'], 'type': 102}
举一反三
- set1 = {x for x in range(10)} print(set1) 以上代码的运行结果为? A: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} B: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10} C: {1, 2, 3, 4, 5, 6, 7, 8, 9} D: {1, 2, 3, 4, 5, 6, 7, 8, 9,10}
- >>>x= [10, 6, 0, 1, 7, 4, 3, 2, 8, 5, 9]>>>print(x.sort()) 语句运行结果正确的是( )。 A: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] B: [10, 6, 0, 1, 7, 4, 3, 2, 8, 5, 9] C: [10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0] D: ['2', '4', '0', '6', '10', '7', '8', '3', '9', '1', '5']
- 【计算题】5 ×8= 6×4= 7×7= 9×5= 2×3= 9 ×2= 8×9= 7×8= 5×5= 4×3= 5+8= 6 ×6= 3×7= 4×8= 9×3= 1 ×2= 9×9= 6×8= 8×0= 4×7=
- 假设“☆”是一种新的运算,若3☆2=3×4,6☆3=6×7×8,x☆4=840(x>0),那么x等于: A: 2 B: 3 C: 4 D: 5 E: 6 F: 7 G: 8 H: 9
- 以下程序段实现的输出是()。for(i=0;i<;=9;i++)s[i]=i;for(i=9;i>;=0;i--)printf("%2d",s[i]);[/i][/i] A: 9 7 5 3 1 B: 1 3 5 7 9 C: 9 8 7 6 5 4 3 2 1 0 D: 0 1 2 3 4 5 6 7 8 9