向列表框list的0行添加字符串变量text内容的代码为list.insert(0,text)

According to the text, list some business activities of entertaining clients. Can you list some other creative activities of entertaining clients?

对数螺线$r={{\text{e}}^{\theta }}$在$\theta =\frac{\text{ }\!\!\pi\!\!\text{ }}{2}$对应点处的切线的直角坐标方程为( )。 A: $y+x={{\text{e}}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{2}}}$ B: $y-x={{\text{e}}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{2}}}$ C: $y={{\text{e}}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{2}}}(x+1)$ D: $y={{\text{e}}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{2}}}(x-1)$

When you have a list of numbers or of text, you can present them in a table.

代码如下 #函数定义 def chanageList(list): list.append(" end") print("list",list) #调用 strs =['1','2'] chanageList() print("strs",strs) 下面对 strs 和 list 的值输出正确的是 A: strs ['1','2'] list ['1','2'] B: strs ['1','2','end'] list ['1','2','end'] C: list ['1','2','end'] strs ['1','2','end'] D: list ['1','2','end'] strs ['1','2']

函数$y={{\ln }^{3}}{{x}^{2}}$的微分为（ ）。 A: $\text{d}y=6x{{\ln }^{2}}{{x}^{2}}\ \text{d}x$ B: $\text{d}y=\frac{6}{x}{{\ln }^{2}}{{x}^{2}}\ \text{d}x$ C: $\text{d}y=3{{\ln }^{2}}{{x}^{2}}\ \text{d}x$ D: $\text{d}y=2x{{\ln }^{3}}{{x}^{2}}\ \text{d}x$

从原点向曲线$$y=1-\ln x$$作切线，则由切线、曲线和$$x$$轴围成图形的面积为()． A: $$\frac{1}{2}{{\text{e}}^{2}}+\text{e}$$ B: $$\frac{1}{2}{{\text{e}}^{2}}-\text{e}$$ C: $${{\text{e}}^{2}}+\text{e}$$ D: $${{\text{e}}^{2}}-\text{e}$$

$\int_{0}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{4}}{[\cos (2t)\mathbf{i}+\sin (2t)\mathbf{j}+t\sin t\mathbf{k}]}\operatorname{dt}=$( ) A: $(\frac{1}{2},\frac{1}{2},\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ B: $(1,\frac{1}{2},\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ C: $(\frac{1}{2},1,\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ D: $(1,1,\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$

1802cdfa0c72530.png图中第一个列表的创建语句是（ ）。 A: L = list(range(2, 100, 2)) B: L = list(range(2, 101, 2)) C: L = list(range(1, 100, 2)) D: L = list(range(1, 101, 2))

Is it right or false?when we cite the reference,we should number the list of references and match with the citation in text