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: $\text{d}y=2x{{\ln }^{3}}{{x}^{2}}\ \text{d}x$
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: $\text{d}y=2x{{\ln }^{3}}{{x}^{2}}\ \text{d}x$
如何一次性创建 text/1/2/3/4
如何一次性创建 text/1/2/3/4
Text Analysis(2)— Unit 3 Please match these examples with each suggestion.
Text Analysis(2)— Unit 3 Please match these examples with each suggestion.
According to the text, the tertiary industry produces approximately 2/3 of the national wealth.
According to the text, the tertiary industry produces approximately 2/3 of the national wealth.
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: $y={{\text{e}}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{2}}}(x-1)$
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: $y={{\text{e}}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{2}}}(x-1)$
已知齐次方程$(x-1){{y}^{''}}-x{{y}^{'}}+y=0$的通解为$Y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}$,则方程$(x-1){{y}^{''}}-x{{y}^{'}}+y={{(x-1)}^{2}}$的通解是( ) A: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-({{x}^{2}}+1)$ B: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-({{x}^{3}}+1)$ C: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}$ D: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}+1$
已知齐次方程$(x-1){{y}^{''}}-x{{y}^{'}}+y=0$的通解为$Y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}$,则方程$(x-1){{y}^{''}}-x{{y}^{'}}+y={{(x-1)}^{2}}$的通解是( ) A: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-({{x}^{2}}+1)$ B: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-({{x}^{3}}+1)$ C: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}$ D: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}+1$
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: $${{\text{e}}^{2}}-\text{e}$$
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: $${{\text{e}}^{2}}-\text{e}$$
How many pieces of advice are there in the text? A: 5 B: 4 C: 2 D: 3
How many pieces of advice are there in the text? A: 5 B: 4 C: 2 D: 3
Lesson 3 Text A Words Matching Lesson 3 Text A Words Matching
Lesson 3 Text A Words Matching Lesson 3 Text A Words Matching
$\int_{0}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{4}}{[\cos (2t)\mathbf{i}+\sin (2t)\mathbf{j}+t\sin t\mathbf{k}]}\operatorname{dt}=$( ) 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: $(1,1,\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$
$\int_{0}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{4}}{[\cos (2t)\mathbf{i}+\sin (2t)\mathbf{j}+t\sin t\mathbf{k}]}\operatorname{dt}=$( ) 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: $(1,1,\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$