Y1=b*sqrt(-Z(1)+X1.^2/a.2); Error:Missing MATLAB operator.怎么回事呢clcclear allhold onaxis([-25 25 -25 25 -25 25])axis ongrid ona=4;b=4;x=-a-17:a+17;y=-b-17:b+17;[X,Y]=meshgrid(x,y);Z=zeros(size(X))+20;view([0.3848 0.7627 0.5199])for i=1:150Z=Z

matlab画出雪花曲线function y=koch(x1,x2,y1,y2,n)if n>1koch(x1,y1,(2*x1+x2)/3,(2*y1+y2)/3,n-1);koch((2*x1+x2)/3,(2*y1+y2)/3,(x1+x2)/2+(y1-...y2)*sqrt(3)/6,(y1+y2)/2+(x2-x1)*sqrt(3)/6,n-1);koch((x1+x2)/2+(y1-y2)*sqrt(3)/6,(y1+y2)/2+(x2-...x1)*sqr Y1=b*sqrt(-Z(1)+X1.^2/a.2); Error:Missing MATLAB operator.怎么回事呢clcclear allhold onaxis([-25 25 -25 25 -25 25])axis ongrid ona=4;b=4;x=-a-17:a+17;y=-b-17:b+17;[X,Y]=meshgrid(x,y);Z=zeros(size(X))+20;view([0.3848 0.7627 0.5199])for i=1:150Z=Z 帮忙看看哪出错了.lingomin=a+b+c+5*pi;a=@sqrt(x^2+y^2);b=@sqrt((300-x1)^2-(300-y1)^2);c=(x-x1)^2+(y-y1)^2;(x-x1)^2+(y-y1)^2=p^2; Matlab 单叶双曲线 总是画不出图!x1=-10:0.2:10;y1=x1;[x,y]=meshgrid(x1,y1);a=1;b=1;c=1;z1=sqrt(c.^2*(x.^2/(a.^2)+y.^2/(b.^2)-1));z2=-sqrt(c.^2*(x.^2/(a.^2)+y.^2/(b.^2)-1));mesh(x,y,z1)hold on;mesh(x,y,z2)------------------------------------- 杭电2001.为什么是wrong answer?#include#includemain(){float x1,x2,y1,y2;while(scanf(%f %f %f %f,&x1,&x2,&y1,&y2)!=EOF){printf(%.2f,sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)));putchar(' ');}} matlab一直出现Warning:Divide by zero.Warning:Rank deficient,rank = 0,tol = 1.#INFe+000.clear all;clc;L=0.235; %Lp/Lm=0.235x1=0:0.2:2;y1=((0.1*sqrt(2))/pi).*sqrt((1+((0.235/0.1)^2).*((1./x1).^2 ))/((( L+1-L./(x1.^2)).^2)+(0.1^2).*((x1-1./x1).^2))) matlab如何求由点以及拟合成的曲线以及真实曲线的误差之和 x1=0:1:10; y1=3*x1.^2+2*x1+6; z=0:0.001:10; y=spline(x1,y1,z); plot(x1,y1,z,y,'MarkerSize',12);N=length(z);RMES=sqr(sum((y1-y).^2)/(N-1));但是问题是y1 用matlab求曲线斜率大于某值的区域,如何求以下曲线斜率大于某值的区域?最好能给程序出来 x1=1:0.1:10;x2=1:0.1:10;x3=1:0.1:10;y1=100./sqrt(x1);y2=150./sqrt(x2);y3=200./sqrt(x3);x=[x1;x2;x3]';y=[y1;y2;y3]';plot(x,y) e^z=1+sqrt(3)i 求z. 求高手解答用Matlab 如何在matlab编程求这个方程组的解啊?已知a,b,c方程组72.8*(1+cos(a)）=2*(SQRT(x*21.8)+SQRT(25.5*y)+SQRT(25.5*z))50.8*(1+cos(b))=2*SQRT(x*50.8)48*(1+cos(c))=2*(SQRT(x*29)+SQRT(47*y)+SQRT(1.92*z))求x,y,z假设 解析几何椭圆问题：椭圆方程X^2+1/2Y^2=1 直线y=x+b,椭圆上存在两点关于直线对称,求b的取值范围所以x1-y1+b=x2-y2+b或x1-y1+b=-（x2-y2+b）即x1-x2=y1-y2或x1+x2=y1+y2怎么x1-y1+b=-（x2-y2+b）到x1+x2=y1+y22b没有了 解未知数二元一次方程!1、y2-kx2=kx1+2b-y12、k((y2-y1)(x2-x1))用含有k、b、x1、y1的关系式来表示x2,y22、k((y2-y1)(x2-x1))=-1 谁能帮我看下,这段哪错了呀>> clear>> x=1.5;y=-4.4;z=0.15;a=7;b=0.35;c=0.5;d=0.7;m0=-1/7;m1=-40/7;for i=1:10000g(x)=m0*x+0.5*(m1-m0)*(abs(x+1)-abs(x-1));x1=a(y-x-g(x));y1=b*(x-y)+c*z;z1=-d*y;if i>3000plot3(x1,y1,z1);endhold on;x=x1;y=y1;z=z 设空间两个不同的单位向量a=(x1,y1,0),b=(x2,y2,0)与向量c=(1,1,1)的夹角都等于π/4,求x1+y1和x1·y1的值 已知一次函数y=kx+b的图像过点(x1,y1),(x2,y2),x2-x1=1 y2-y1= -2 则k=求教！ 已知一次函数y=(m-1)x+1的图像上两点A（x1,y1),B(x2,y2),当x1>x2时,有y1 若A（x1,y1)、B(x2,y2)在函数y=1/2x的图像上,则当x1,x2满足-------时,y1>y2 哪位大哥用Maple帮忙解一下这个方程组呢Solve[{(x2 - x1)*(x2 - x3) + (y2 - y1)*(y2 - y3) == 0,(x2 - x1)^2 + (y2 - y1)^2 == d1^2,(x2 - x3)^2 + (y2 - y3)^2 == d2^2,k*z*a*(1 - 3*l0/((x3 - a)^2 + (y3 - b)^2 + z^2)^(1/2)) == m*g*x,k*z*b*(1 -