在环形区域[img=136x26]18030733be53638.png[/img]上, 绘制函数图形[img=129x27]18030733c6c9cd6.png[/img]
A: Plot3D[x^2+y^2,{x,-2,2},{y,-2,2},Exclusions→Function[{x,y},0.5B: Plot3D[x^2+y^2,{x,-2,2},{y,-2,2},RegionFunction→Function[{x,y},0.5C: Plot3D[x^2+y^2,{x,-2,2},{y,-2,2},RegionFunction→Function[{x,y},2>x^2+y^2>0.5]]
D: Plot3D[x^2+y^2,{y,-2,2},{x,-2,2},Exclusions→Function[{x,y},0.5
A: Plot3D[x^2+y^2,{x,-2,2},{y,-2,2},Exclusions→Function[{x,y},0.5
D: Plot3D[x^2+y^2,{y,-2,2},{x,-2,2},Exclusions→Function[{x,y},0.5
B,C
举一反三
- 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]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}]
- 设置曲面边界样式为红色粗线条。 A: Plot3D[x^2+y^2,{x,-2,2},{y,-2,2},BoundaryStyle→Directive[Red,Thick]] B: Plot3D[x^2+y^2, BoundaryStyle→Directive[Red,Thick],{x,-2,2},{y,-2,2}] C: Plot3D[x^2+y^2, BoundaryStyle→Directive[Red,Thick],{y,-2,2},{x,-2,2}] D: Plot3D[x^2+y^2,{y,-2,2},{x,-2,2},BoundaryStyle→Directive[Red,Thick]]
- 设置曲面边界样式为红色粗线条。 A: Plot3D[x^2+y^2,{x,-2,2},{y,-2,2},BoundaryStyle→Directive[Red,Thick]] B: Plot3D[x^2+y^2, BoundaryStyle→Directive[Red,Thick],{x,-2,2},{y,-2,2}] C: Plot3D[x^2+y^2, BoundaryStyle→Directive[Red,Thick],{y,-2,2},{x,-2,2}] D: Plot3D[x^2+y^2,{y,-2,2},{x,-2,2},BoundaryStyle→Directive[Red,Thick]]
- 计算二重积分[img=159x48]18030731271aaff.png[/img], D 是单位圆盘[img=89x26]180307312f6708b.png[/img],应使用的语句是 A: Integrate[Sqrt[x^2+y^2 ], {x^2+y^2≤1}] B: Integrate[Sqrt[x^2+y^2 ]Boole[x^2+y^2≤1],{x,-1,1},{y,-1,1}] C: NIntegrate[Sqrt[x^2+y^2 ]Boole[x^2+y^2≤1],{x,-1,1},{y,-1,1}] D: Integrate[Sqrt[x^2+y^2 ],{x^2+y^2≤1,{x,-1,1},{y,-1,1}}]
内容
- 0
5、(4分)下列哪个曲线表示球面( )。 A、y=x^2+z^2 B、y^2=x^2+z^2 C、y=x^2-z^2 D、x^2+y^2+z^2=6
- 1
求解偏微分方程[img=178x28]18030731a73d552.png[/img], 应用的语句是 A: DSolve[(x^2+y^2)D[u,x]+x yD[u,y]==0,u,{x,y}] B: DSolve[(x^2+y^2)Dt[u[x,y],x]+xyDt[u[x,y],y]==0,u[x,y],{x,y}] C: DSolve[(x^2+y^2)D[u[x,y],x]+xyD[u[x,y],y]==0,u[x,y]] D: DSolve[(x^2+y^2)D[u[x,y],x]+xyD[u[x,y],y]==0,u[x,y],{x,y}]
- 2
应用Matlab软件计算行列式[img=110x88]17da5d7b00219d6.png[/img]为( ). A: x^2 - 6*x^2*y^2 + 8*x*y^3 - 3*y^4 B: x^3 - 6*x^2*y^2 + 8*x*y^3 - 3*y^4 C: x^4 - 6*x^2*y^2 + 8*x*y^3 - 3*y^4 D: x^5- 6*x^2*y^2 + 8*x*y^3 - 3*y^4
- 3
设有int x=2,y;,执行语句x=(y=3,2+y,2*y);后,则变量x的值是______。 A: 2 B: 6 C: 10 D: 3
- 4
\({\lim_{x\to0}}\)\({\lim_{y\to0}}\)\(\frac{tan(x^2+y^2)}{x^2+y^2}\)= <br/>______