考题
过点(4,-1,3)且平行于直线L:(x-3)/2=y=(z-1)/5的直线方程为( ).A.(x-4)/2=(y+1)/0=(z-3)/5
B.(x-4)/2=(y+1)/1=(z-3)/5
C.(x+4)/2=(y-1)/0=(z+3)/5
D.(x+4)/2=(y-1)/1=(z+3)/5答案:B解析:
考题
填空题设f(x,y,z)=exyz2,其中z=z(x,y)是由x+y+z+xyz=0确定的隐函数,则fx′(0,1,-1)=____。正确答案:1解析:构造函数F(x,y,z)=x+y+z+xyz,则有∂z/∂x=-Fx′/Fz′=-(1+yz)/(1+xy),(∂z/∂x)|(0,1,-1)=0,又由f(x,y,z)=exyz2 ,得fx′=exyz2+exy·2z·zx′,代入(0,1,-1),得fx′(0,1,-1)=e0×1×(-1)2+e0×1×2×(-1)×0=1。
考题
单选题过点(4,-1,3)且平行于直线L:(x-3)/2=y=(z-1)/5的直线方程为().A
(x-4)/2=(y+1)/0=(z-3)/5B
(x-4)/2=(y+1)/1=(z-3)/5C
(x+4)/2=(y-1)/0=(z+3)/5D
(x+4)/2=(y-1)/1=(z+3)/5正确答案:A解析:暂无解析
考题
单选题Which of the following choices could be equal to set Z ifX = {2, 5, 6, 7, 9} and Y = {2, 5, 7}X ∪ Y ∪ Z = {1, 2, 3, 4, 5, 6, 7, 8, 9}X ∩ Z = {2, 6}Y ∩ Z = {2}A
Z= {1, 4, 8}B
Z= {1, 3, 8}C
Z= {1, 3, 4, 8}D
Z= {1, 2, 3, 4, 6, 8}E
Z= {1, 2, 3, 5, 6, 8}正确答案:B解析:We know that: X = {2, 5, 6, 7, 9} and Y = {2, 5, 7}, X ∪ Y ∪ Z = {1, 2, 3, 4, 5, 6, 7, 8, 9}. This tells us that Z must contain {1, 3, 4, 8} because these elements are in the union of X, Y, and Z but not in set X and not in set Y.We also know that: X ∩ Z = {2, 6}, Y ∩ Z = {2}. This tells us that Z must also contain {2, 6}. Only choice D contains all these elements. Z = {1, 2, 3, 4, 6, 8}
考题
填空题设f(x,y,z)=exyz2,其中z=z(x,y)是由x+y+z+xyz=0确定的隐函数,则fx′(0,1,-1)=____。正确答案:1解析:构造函数F(x,y,z)=x+y+z+xyz,则有∂z/∂x=-Fx′/Fz′=-(1+yz)/(1+xy),(∂z/∂x)|(0,1,-1)=0,又由f(x,y,z)=exyz2 ,得fx′=exyz2+exy·2z·zx′,代入(0,1,-1),得fx′(0,1,-1)=e0×1×(-1)2+e0×1×2×(-1)×0=1。
考题
填空题设f(x,y,z)=exyz2,其中z=z(x,y)是由x+y+z+xyz=0确定的隐函数,则fx′(0,1,-1)=____。正确答案:1解析:构造函数F(x,y,z)=x+y+z+xyz,则有∂z/∂x=-Fx′/Fz′=-(1+yz)/(1+xy),(∂z/∂x)|(0,1,-1)=0,又由f(x,y,z)=exyz2 ,得fx′=exyz2+exy·2z·zx′,代入(0,1,-1),得fx′(0,1,-1)=e0×1×(-1)2+e0×1×2×(-1)×0=1。
考题
填空题若有int x=3,y=4,z=5; ,则表达式 !(x+y)+z-1 && y+z/2的值是()正确答案:1解析:暂无解析
考题
单选题Which of the following choices could be equal to set Z ifX = {2, 5, 6, 7, 9} and Y = {2, 5, 7}X ∪ Y ∪ Z = {1, 2, 3, 4, 5, 6, 7, 8, 9}X ∩ Z = {2, 6}Y ∩ Z = {2}A
Z= {1, 4, 8}B
Z= {1, 3, 8}C
Z= {1, 3, 4, 8}D
Z= {1, 2, 3, 4, 6, 8}E
Z= {1, 2, 3, 5, 6, 8}正确答案:E解析:We know that: X = {2, 5, 6, 7, 9} and Y = {2, 5, 7}, X ∪ Y ∪ Z = {1, 2, 3, 4, 5, 6, 7, 8, 9}. This tells us that Z must contain {1, 3, 4, 8} because these elements are in the union of X, Y, and Z but not in set X and not in set Y.We also know that: X ∩ Z = {2, 6}, Y ∩ Z = {2}. This tells us that Z must also contain {2, 6}. Only choice D contains all these elements. Z = {1, 2, 3, 4, 6, 8}
考题
单选题Which of the following choices could be equal to set Z ifX = {2, 5, 6, 7, 9} and Y = {2, 5, 7}X ∪ Y ∪ Z = {1, 2, 3, 4, 5, 6, 7, 8, 9}X ∩ Z = {2, 6}Y ∩ Z = {2}A
Z= {1, 4, 8}B
Z= {1, 3, 8}C
Z= {1, 3, 4, 8}D
Z= {1, 2, 3, 4, 6, 8}E
Z= {1, 2, 3, 5, 6, 8}正确答案:B解析:We know that: X = {2, 5, 6, 7, 9} and Y = {2, 5, 7}, X ∪ Y ∪ Z = {1, 2, 3, 4, 5, 6, 7, 8, 9}. This tells us that Z must contain {1, 3, 4, 8} because these elements are in the union of X, Y, and Z but not in set X and not in set Y.We also know that: X ∩ Z = {2, 6}, Y ∩ Z = {2}. This tells us that Z must also contain {2, 6}. Only choice D contains all these elements. Z = {1, 2, 3, 4, 6, 8}
考题
填空题设f(x,y,z)=exyz2,其中z=z(x,y)是由x+y+z+xyz=0确定的隐函数,则fx′(0,1,-1)=____。正确答案:1解析:构造函数F(x,y,z)=x+y+z+xyz,则有∂z/∂x=-Fx′/Fz′=-(1+yz)/(1+xy),(∂z/∂x)|(0,1,-1)=0,又由f(x,y,z)=exyz2 ,得fx′=exyz2+exy·2z·zx′,代入(0,1,-1),得fx′(0,1,-1)=e0×1×(-1)2+e0×1×2×(-1)×0=1。