A: △N=9x△x-8y△y
B: △N=3△x+2△y
C: △N=9x△x+8y△y
D: △N=9△x-8△y
举一反三
- N=3x[sup]3[/]－2y[sup]4[/]其绝对误差为（） A: △N=9x<sup>2</sup>△x-8y<sup>3</sup>△y B: △N=3△x+2△y C: △N=9x<sup>2</sup>△x+8y<sup>3</sup>△y D: △N=9△x-8△y
- 设 X ~ N(3, 12)，Y ~ N(2, 4)，且 X，Y 独立，则 X − Y ~ N(1, 8) .
- int x,y,z; x=7; y=8; z=9; if(x>y) x=y; y=z; z=x; printf(“x=%d y=%d z=%d\n”,x,y,z);以上程序段的输出结果是：（） A: x=7 y=8 z=9 B: x=7 y=9 z=7 C: x=8 y=9 z=7 D: x=8 y=9 z=8
- 双曲线x^2/16-y^2/9=1的渐近线方程为() A: y=±16x/9 B: y=±9x/16 C: x/3±y/4=0 D: x/4±y/3=0
- 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
内容
- 0
已知x(n)={1, 2, 3}，y(n)={1, 2, 1}，则x(n)*y(n)=________。（下划线表示n=0） A: {1, 4, 8, 8, 3} B: {1, 4, 8, 8, 3} C: {1, 4, 8, 8, 3} D: {1, 4, 8, 8, 3}
- 1
执行下列程序段的结果是（） int x=3,y=4; printf("x=%d,y=%d",3*x,2*y); A: 3,4 B: x=3,y=4 C: x=9,y=8 D: 9,8
- 2
如下C程序的输出是什么？#include [stdio.h]void Func1 (int x, int y);void Func2 (int *x, int *y); int main() { int x = 3; int y = 4;Func1 (x, y); printf ("x = %d, y = %d\n", x, y);Func2(&x, &y); printf ("x = %d, y = %d\n", x, y);} void Func1 (int x, int y) { x = x + y; y = x - y; x = x - y; printf ("x = %d, y = %d\n", x, y);} void Func2 (int *x, int *y) { *x = *x + *y; *y = *x - *y; *x = *x - *y;;} A: x = 3, y = 4x = 3, y = 4x = 3, y = 4 B: x = 4, y = 3x = 4, y = 3x = 4, y = 3 C: x = 3, y = 4x = 3, y = 4x = 4, y = 3 D: x = 4, y = 3x = 3, y = 4x = 4, y = 3
- 3
Consider the following sequence: x(n)={3,-6,5,-1,0,7,8}, -1≤n≤5.suppose the sequence y(n)=x(-n-2),then y(0)=______ , y(-2)=______ , y(-4)=______ , y(-6)=______ , y(2)=______ .
- 4
下列程序的结果为().change(intx,inty){intt;t=x;x=y;y=t;}main(){intx=2,y=3;change(x,y);printf("x=%d,y=%d\n",x,y);} A: A)x=3,y=2 B: B)x=2,y=3 C: C)x=2,y=2 D: D)x=3,y=3