羊的齿式为 A: 2(0 0 3 0/ 4 0 3 0)=20 B: 2(0 0 3 3/ 4 0 3 3)=32 C: 2(4 0 3 0/ 4 0 3 0)=28 D: 2(4 0 3 3/ 4 0 3 3)=40
羊的齿式为 A: 2(0 0 3 0/ 4 0 3 0)=20 B: 2(0 0 3 3/ 4 0 3 3)=32 C: 2(4 0 3 0/ 4 0 3 0)=28 D: 2(4 0 3 3/ 4 0 3 3)=40
如果x+2y-8z=02x-3y+5z=0,其中xyz≠0,那么x:y:z=( ) A: 1:2:3 B: 2:3:4 C: 2:3:1 D: 3:2:1
如果x+2y-8z=02x-3y+5z=0,其中xyz≠0,那么x:y:z=( ) A: 1:2:3 B: 2:3:4 C: 2:3:1 D: 3:2:1
【单选题】如图示代码,下面哪个是正确的输出结果 A. 0 1 2 3 4 5 B. 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 C. 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 D. 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5
【单选题】如图示代码,下面哪个是正确的输出结果 A. 0 1 2 3 4 5 B. 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 C. 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 D. 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5
如果,其中xyz≠0,那么x:y:z= A: 1:2:3 B: 2:3:4 C: 2:3:1 D: 3:2:1
如果,其中xyz≠0,那么x:y:z= A: 1:2:3 B: 2:3:4 C: 2:3:1 D: 3:2:1
int x=3,y,z;<br/>y=-x++;<br/>z=y+8/++x;<br/>Console.WriteLine{{0},{1},{2}",x,y,z);<br/>此程序的输出结果是____。 A: 5,-3,-2 B: 4,-3,-1 C: 4,-4,-2 D: 5,-4,-2
int x=3,y,z;<br/>y=-x++;<br/>z=y+8/++x;<br/>Console.WriteLine{{0},{1},{2}",x,y,z);<br/>此程序的输出结果是____。 A: 5,-3,-2 B: 4,-3,-1 C: 4,-4,-2 D: 5,-4,-2
用符号“∈”或“∉”填空 (1)0 N; (2) 0.6 Z; (3)π R; (4)1/3 Q; (5)0 ∅
用符号“∈”或“∉”填空 (1)0 N; (2) 0.6 Z; (3)π R; (4)1/3 Q; (5)0 ∅
选择符号填空(A:"∈", B:"∉")。(1)0 N;(2) 0.6 Z;(3)π R;(4)1/3 Q;(5)0 ∅
选择符号填空(A:"∈", B:"∉")。(1)0 N;(2) 0.6 Z;(3)π R;(4)1/3 Q;(5)0 ∅
9. 已知函数$z=z(x,y)$由${{z}^{3}}-3xyz={{a}^{3}}$确定,则$\frac{{{\partial }^{2}}z}{\partial x\partial y}=$( ) A: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ B: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-xy)}{{{({{z}^{2}}-xy)}^{2}}}$ C: $\frac{z({{z}^{3}}-2xyz-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ D: $\frac{z({{z}^{3}}-2xy{{z}^{2}}-{{x}^{2}}y)}{{{({{z}^{2}}-xy)}^{3}}}$
9. 已知函数$z=z(x,y)$由${{z}^{3}}-3xyz={{a}^{3}}$确定,则$\frac{{{\partial }^{2}}z}{\partial x\partial y}=$( ) A: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ B: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-xy)}{{{({{z}^{2}}-xy)}^{2}}}$ C: $\frac{z({{z}^{3}}-2xyz-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ D: $\frac{z({{z}^{3}}-2xy{{z}^{2}}-{{x}^{2}}y)}{{{({{z}^{2}}-xy)}^{3}}}$
已知直线的一般方程\( \left\{ {\matrix{ {x - 2y - z + 4 = 0} \cr {5x + y - 2z + 8 = 0} \cr } } \right. \), 则其点向式方程为( ) A: \( { { x - 2} \over 2} = {y \over { - 3}} = { { z - 4} \over {11}} \) B: \( {x \over 5} = {y \over { - 3}} = { { z - 4} \over {11}} \) C: \( { { x - 2} \over 5} = { { y + 1} \over { - 3}} = { { z - 4} \over {11}} \) D: \( { { x - 2} \over 2} = { { y + 1} \over { - 3}} = { { z - 4} \over {11}} \)
已知直线的一般方程\( \left\{ {\matrix{ {x - 2y - z + 4 = 0} \cr {5x + y - 2z + 8 = 0} \cr } } \right. \), 则其点向式方程为( ) A: \( { { x - 2} \over 2} = {y \over { - 3}} = { { z - 4} \over {11}} \) B: \( {x \over 5} = {y \over { - 3}} = { { z - 4} \over {11}} \) C: \( { { x - 2} \over 5} = { { y + 1} \over { - 3}} = { { z - 4} \over {11}} \) D: \( { { x - 2} \over 2} = { { y + 1} \over { - 3}} = { { z - 4} \over {11}} \)
int x=2, y=3, z=4;则表达式x&&y+z的值为() A: 2 B: 3 C: 0 D: 1
int x=2, y=3, z=4;则表达式x&&y+z的值为() A: 2 B: 3 C: 0 D: 1