马歇尔—勒纳条件假定供给弹性(Se)为( )。 A: 0≤Se≤1 B: Se=∞ C: Se=0 D: Se=1
马歇尔—勒纳条件假定供给弹性(Se)为( )。 A: 0≤Se≤1 B: Se=∞ C: Se=0 D: Se=1
$\int_{0}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{4}}{[\cos (2t)\mathbf{i}+\sin (2t)\mathbf{j}+t\sin t\mathbf{k}]}\operatorname{dt}=$( ) 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: $(1,1,\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$
Text 1
Text 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: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}+1$
Skim the text and answer the following questions. 1) What type of writing is the text?
Skim the text and answer the following questions. 1) What type of writing is the text?
Part 1 : The text begins with an _________.
Part 1 : The text begins with an _________.
Fast reading text 1
Fast reading text 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: $${{\text{e}}^{2}}-\text{e}$$
函数TEXT("2018/1/1","mm")返回值为1。
函数TEXT("2018/1/1","mm")返回值为1。
函数TEXT('2018/1/1','mm')返回值为1。
函数TEXT('2018/1/1','mm')返回值为1。