• 2022-07-27 问题

    已知动点的运动方程为x=t,y=2l2,则其轨迹方程为: A: x=t2-t B: y=2t C: y-2x2=2 D: y+2t2=0

    已知动点的运动方程为x=t,y=2l2,则其轨迹方程为: A: x=t2-t B: y=2t C: y-2x2=2 D: y+2t2=0

  • 2022-05-29 问题

    下列语句语法正确的是( ) A: if x<2*y and x>y then y=x**2 B: if x<2*y : x>y then y=x^2 C: if x<2*y and x>y then y=x2 D: if x<2*y and x>y then y=x^2

    下列语句语法正确的是( ) A: if x<2*y and x>y then y=x**2 B: if x<2*y : x>y then y=x^2 C: if x<2*y and x>y then y=x2 D: if x<2*y and x>y then y=x^2

  • 2021-04-14 问题

    【单选题】对任意实数x 1 , y 1 , x 2 , y 2 , x 1 < x 2 , y 1 < y 2 , 分布函数P{x 1 <X≤x 2 , y 1 <Y≤y 2 }=? A. F(x 2 , y 2 )+ F(x 1 , y 1 )+ F(x 1 , y 2 )+ F(x 2 , y 1 ) B. F(x 2 , y 2 )- F(x 1 , y 1 )+ F(x 1 , y 2 )- F(x 2 , y 1 ) C. F(x 2 , y 2 )+ F(x 1 , y 1 )- F(x 1 , y 2 )- F(x 2 , y 1 ) D. F(x 2 , y 2 )- F(x 1 , y 1 )- F(x 1 , y 2 )+ F(x 2 , y 1 )

    【单选题】对任意实数x 1 , y 1 , x 2 , y 2 , x 1 < x 2 , y 1 < y 2 , 分布函数P{x 1 <X≤x 2 , y 1 <Y≤y 2 }=? A. F(x 2 , y 2 )+ F(x 1 , y 1 )+ F(x 1 , y 2 )+ F(x 2 , y 1 ) B. F(x 2 , y 2 )- F(x 1 , y 1 )+ F(x 1 , y 2 )- F(x 2 , y 1 ) C. F(x 2 , y 2 )+ F(x 1 , y 1 )- F(x 1 , y 2 )- F(x 2 , y 1 ) D. F(x 2 , y 2 )- F(x 1 , y 1 )- F(x 1 , y 2 )+ F(x 2 , y 1 )

  • 2021-04-14 问题

    F(x,y)=(y-x2)(y-2x2)在原点取得极值

    F(x,y)=(y-x2)(y-2x2)在原点取得极值

  • 2022-06-03 问题

    4.已知二元函数$z(x,y)$满足方程$\frac{{{\partial }^{2}}z}{\partial x\partial y}=x+y$,并且$z(x,0)=x,z(0,y)={{y}^{2}}$,则$z(x,y)=$( ) A: $\frac{1}{2}({{x}^{2}}y-x{{y}^{2}})+{{y}^{2}}+x$ B: $\frac{1}{2}({{x}^{2}}{{y}^{2}}+xy)+{{y}^{2}}+x$ C: ${{x}^{2}}{{y}^{2}}+{{y}^{2}}+x$ D: $\frac{1}{2}({{x}^{2}}y+x{{y}^{2}})+{{y}^{2}}+x$

    4.已知二元函数$z(x,y)$满足方程$\frac{{{\partial }^{2}}z}{\partial x\partial y}=x+y$,并且$z(x,0)=x,z(0,y)={{y}^{2}}$,则$z(x,y)=$( ) A: $\frac{1}{2}({{x}^{2}}y-x{{y}^{2}})+{{y}^{2}}+x$ B: $\frac{1}{2}({{x}^{2}}{{y}^{2}}+xy)+{{y}^{2}}+x$ C: ${{x}^{2}}{{y}^{2}}+{{y}^{2}}+x$ D: $\frac{1}{2}({{x}^{2}}y+x{{y}^{2}})+{{y}^{2}}+x$

  • 2021-04-14 问题

    分解因式()x()3()y()-()2()x()2()y()2()+()xy()3()正确的是A.()xy()(()x()+()y())()2()B.()xy()(()x()2()﹣()2()xy()+()y()2())()C.()xy()(()x()2()+2()xy()﹣()y()2())()D.()xy()(()x()﹣()y())()2

    分解因式()x()3()y()-()2()x()2()y()2()+()xy()3()正确的是A.()xy()(()x()+()y())()2()B.()xy()(()x()2()﹣()2()xy()+()y()2())()C.()xy()(()x()2()+2()xy()﹣()y()2())()D.()xy()(()x()﹣()y())()2

  • 2022-05-27 问题

    ‏求解方程组[img=218x63]1803072f0e0e849.png[/img]接近 (2,2) 的解‌ A: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] B: NSolve[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] C: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,y},{2,2}] D: FindRoots[{x^2+y^2=5Sqrt[x^2+y^2]-4x,y=x^2},{x,2},{y,2}]

    ‏求解方程组[img=218x63]1803072f0e0e849.png[/img]接近 (2,2) 的解‌ A: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] B: NSolve[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] C: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,y},{2,2}] D: FindRoots[{x^2+y^2=5Sqrt[x^2+y^2]-4x,y=x^2},{x,2},{y,2}]

  • 2022-05-27 问题

    求解方程组[img=218x63]1803072e5daced1.png[/img]接近 (2,2) 的解 A: NSolve[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] B: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] C: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,y},{2,2}] D: FindRoots[{x^2+y^2=5Sqrt[x^2+y^2]-4x,y=x^2},{x,2},{y,2}]

    求解方程组[img=218x63]1803072e5daced1.png[/img]接近 (2,2) 的解 A: NSolve[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] B: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] C: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,y},{2,2}] D: FindRoots[{x^2+y^2=5Sqrt[x^2+y^2]-4x,y=x^2},{x,2},{y,2}]

  • 2022-06-26 问题

    设二维随机变量 (X , Y )服从二维正态分布,则随机变量X + Y与X – Y不相关的充要条件为( ) A: E (X ) = E (Y ) B: E (X 2) – [E (X )]2 = E (Y 2 ) – [E (Y )]2 C: E (X 2 ) = E (Y 2) D: E (X 2) + [E (X )]2 = E (Y 2 ) + [E (Y )]2

    设二维随机变量 (X , Y )服从二维正态分布,则随机变量X + Y与X – Y不相关的充要条件为( ) A: E (X ) = E (Y ) B: E (X 2) – [E (X )]2 = E (Y 2 ) – [E (Y )]2 C: E (X 2 ) = E (Y 2) D: E (X 2) + [E (X )]2 = E (Y 2 ) + [E (Y )]2

  • 2022-06-19 问题

    方程$(x^2+1)(y^2-1) + xy y' = 0$的通解为 A: $y^2 = C \frac{e^{-x^2}}{x^2}$ B: $y = C \frac{e^{-x^2}}{x^2}$ C: $y^2 = C \frac{e^{-x^2}}{x^2}+1$ D: $y=C \frac{e^{-x^2}}{x^2}+1$

    方程$(x^2+1)(y^2-1) + xy y' = 0$的通解为 A: $y^2 = C \frac{e^{-x^2}}{x^2}$ B: $y = C \frac{e^{-x^2}}{x^2}$ C: $y^2 = C \frac{e^{-x^2}}{x^2}+1$ D: $y=C \frac{e^{-x^2}}{x^2}+1$

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