试计算[tex=2.357x1.143]qLEI0A0oW5AAtpN27x+kixPwUjjkZeDHSl9xO9wXjgQ=[/tex]键的伸缩振动频率(它的力常数[tex=0.571x1.0]rFc/sfAAuCOtzhevhoREeA==[/tex]为 [tex=4.571x1.214]3kicDfQ9LYk3ZCOaCzWChS5ADN89diyp/O7/tse9xxhSyJollnV7TNj5iePa2ycS[/tex] )为多少
举一反三
- set1 = {x for x in range(10) if x%2!=0} print(set1) 以上代码的运行结果为？ A: {1, 3, 5, 7, 9} B: {1, 3, 5, 7} C: {3, 5, 7, 9} D: {3, 5, 7}
- set1 = {x for x in range(10) if x%2!=0} set1.remove(1) print(set1) 以上代码的运行结果为？ A: {1, 3, 5, 7, 9} B: {1, 3, 5, 7} C: {3, 5, 7, 9} D: {3, 5, 7}
- 表示x是5的倍数或是9的倍数的逻辑表达式为 ____。 A: x \ 5 = 0 Or x \ 9 = 0 B: x \ 5 = 0 And x \ 9 = 0 C: x Mod 5 = 0 Or x Mod 9 = 0 D: x Mod 5 = 0 And x Mod 9 = 0
- 计算下列矩阵的 [tex=0.571x1.0]rFc/sfAAuCOtzhevhoREeA==[/tex] 次幂, 其中 [tex=0.571x1.0]rFc/sfAAuCOtzhevhoREeA==[/tex] 为正整数:[tex=7.929x4.786]SG13E7iu2HdaLVWfWJMdasNcssnOsnpcSXP9pfv8ZVudX8uBxPyIW+BW1iuKqBWPQy19xF0hvC5K+ZJXm49WVAb1VZdwsjQNiE6Ohf5lij4=[/tex]
- [tex=0.571x1.0]FGGpnaR8m8C48rN8O0c7aw==[/tex]是不等于 2 和 5 的质数，[tex=0.571x1.0]rFc/sfAAuCOtzhevhoREeA==[/tex] 是自然数，证明:[tex=5.143x2.643]9ZG142R87UZpWk+DznzbSQal4BSvAr69lHZKETg6dhBcGaNQr9t99v+4iW7u7hVUeO+W/nUuchAm9kbqae1S7A==[/tex]