solve+their+problem
@祖瑞5480：matlab中solve函数的用法.悬赏20分
郦卖19383072823…… solve Symbolic solution of algebraic equations. solve('eqn1','eqn2',...,'eqnN') solve('eqn1','eqn2',...,'eqnN','var1,var2,...,varN') solve('eqn1','eqn2',...,'eqnN','var1','var2',...'varN') The eqns are symbolic expressions or strings specifying equations. The ...
@祖瑞5480：Matlab中,求解如下一元六次方程的根:3*x^6+12*x^5+4*x^4+7*x^3+8*x+1=0 -
郦卖19383072823…… solve('3*x^6+12*x^5+4*x^4+7*x^3+8*x+1')
@祖瑞5480：在Matlab中用什么函数求解非线性方程的根?写出求解如下非线性方程根的Matlab代码. x^2+exp(x) - 3=0 -
郦卖19383072823…… 利用solve函数,其格式为:a=solve('等式')代码如下:a=solve(' x^2+exp(x)-3=0')结果a = 0.83448...
@祖瑞5480：matlab 中solve循环比如: i=1:10 s=solve('x^3+x^2+x=1=i','x') -
郦卖19383072823…… n = 10; i = 1:n; s = cell(1,n); for i = 1:n ep = ['solve(''x^3+x^2+x =' num2str(i) ''')']; s{i} = eval(ep); end; vpa(s{1},4) ans = 0.5437 - 0.7718 + 1.115*i - 0.7718 - 1.115*i vpa(s{2},4) ans = 0.8105 - 0.9053 + 1.284*i - 0.9053 - 1.284*i vpa(s{3},4) ans = 1.0 - 1.0 + 1.414*i - 1.0 - 1.414*i
@祖瑞5480：MATLAB用solve解方程时,比如x=solve('ax+b=c),假如在前面赋值N=ax+b
郦卖19383072823…… 没看明白,你是想求解【N=c,其中N=ax+b】这样以x为未知数的方程,还是说想要求出来x=solve('ax+b=c')之后,用N=ax+b代入,得到关于N的表达式?如果是前者,使用符号表达式(而不是字符串)调用solve即可:123 syms a x b N=a*x+b; x=solve(N-c,x)
@祖瑞5480：only+状语提前+倒装 -
郦卖19383072823…… 部分倒装1. only +状语或状语从句置于句首,句子用部分倒装. 例1:Only in this way can you solve this problem. 只有用这种方法,你才可以解决这个问题. 例2:Only after he had spoken out the word did he realize he had made a big mistake.只有当他已经说出那个字后才意识到自己犯了个大错误.
@祖瑞5480：有关matlab中solve,如下:solve('a*x+b=0','b*y+a=0','x','y');其中a=3,b=4,要求求得的x,y为数值
郦卖19383072823…… > a=3;b=4; >> [x,y]=solve('a*x+b=0','b*y+a=0','x','y'); >> eval(x),eval(y) ans = -1.333333333333333 ans = -0.750000000000000
@祖瑞5480：MATLAB中solve如何去掉中间变量 -
郦卖19383072823…… 给你一个例子吧.solve('a*x=1','a') solve('a*x=1','x') 即后面的是要求的表达式.你的式子比较复杂,我就不知道能不能得到答案.方程组可以参考:[xx,yy]=solve('a*x+b*y-c=0','e*x+f*y-g=0','x','y') 不过你的问题有点怪怪的.x是变量,四个方程应当只能消掉三个中间变量.
@祖瑞5480：有关matlab中solve,如下:solve('a*x+b=0','b*y+a=0','x','y');其中a=3,b=4,要求求得的x,y为数值 -
郦卖19383072823…… >> a=3;b=4;>> [x,y]=solve('a*x+b=0','b*y+a=0','x','y');>> eval(x),eval(y) ans = -1.333333333333333 ans = -0.750000000000000
@祖瑞5480：英语翻译solve their problems - 作业帮
郦卖19383072823…… [答案] 玛丽总是尽力帮助朋友解决问题 翻译 Marry___always____ ______ _______ _______help her friends. 答案:is trying her best to
郦卖19383072823…… solve Symbolic solution of algebraic equations. solve('eqn1','eqn2',...,'eqnN') solve('eqn1','eqn2',...,'eqnN','var1,var2,...,varN') solve('eqn1','eqn2',...,'eqnN','var1','var2',...'varN') The eqns are symbolic expressions or strings specifying equations. The ...
@祖瑞5480：Matlab中,求解如下一元六次方程的根:3*x^6+12*x^5+4*x^4+7*x^3+8*x+1=0 -
郦卖19383072823…… solve('3*x^6+12*x^5+4*x^4+7*x^3+8*x+1')
@祖瑞5480：在Matlab中用什么函数求解非线性方程的根?写出求解如下非线性方程根的Matlab代码. x^2+exp(x) - 3=0 -
郦卖19383072823…… 利用solve函数,其格式为:a=solve('等式')代码如下:a=solve(' x^2+exp(x)-3=0')结果a = 0.83448...
@祖瑞5480：matlab 中solve循环比如: i=1:10 s=solve('x^3+x^2+x=1=i','x') -
郦卖19383072823…… n = 10; i = 1:n; s = cell(1,n); for i = 1:n ep = ['solve(''x^3+x^2+x =' num2str(i) ''')']; s{i} = eval(ep); end; vpa(s{1},4) ans = 0.5437 - 0.7718 + 1.115*i - 0.7718 - 1.115*i vpa(s{2},4) ans = 0.8105 - 0.9053 + 1.284*i - 0.9053 - 1.284*i vpa(s{3},4) ans = 1.0 - 1.0 + 1.414*i - 1.0 - 1.414*i
@祖瑞5480：MATLAB用solve解方程时,比如x=solve('ax+b=c),假如在前面赋值N=ax+b
郦卖19383072823…… 没看明白,你是想求解【N=c,其中N=ax+b】这样以x为未知数的方程,还是说想要求出来x=solve('ax+b=c')之后,用N=ax+b代入,得到关于N的表达式?如果是前者,使用符号表达式(而不是字符串)调用solve即可:123 syms a x b N=a*x+b; x=solve(N-c,x)
@祖瑞5480：only+状语提前+倒装 -
郦卖19383072823…… 部分倒装1. only +状语或状语从句置于句首,句子用部分倒装. 例1:Only in this way can you solve this problem. 只有用这种方法,你才可以解决这个问题. 例2:Only after he had spoken out the word did he realize he had made a big mistake.只有当他已经说出那个字后才意识到自己犯了个大错误.
@祖瑞5480：有关matlab中solve,如下:solve('a*x+b=0','b*y+a=0','x','y');其中a=3,b=4,要求求得的x,y为数值
郦卖19383072823…… > a=3;b=4; >> [x,y]=solve('a*x+b=0','b*y+a=0','x','y'); >> eval(x),eval(y) ans = -1.333333333333333 ans = -0.750000000000000
@祖瑞5480：MATLAB中solve如何去掉中间变量 -
郦卖19383072823…… 给你一个例子吧.solve('a*x=1','a') solve('a*x=1','x') 即后面的是要求的表达式.你的式子比较复杂,我就不知道能不能得到答案.方程组可以参考:[xx,yy]=solve('a*x+b*y-c=0','e*x+f*y-g=0','x','y') 不过你的问题有点怪怪的.x是变量,四个方程应当只能消掉三个中间变量.
@祖瑞5480：有关matlab中solve,如下:solve('a*x+b=0','b*y+a=0','x','y');其中a=3,b=4,要求求得的x,y为数值 -
郦卖19383072823…… >> a=3;b=4;>> [x,y]=solve('a*x+b=0','b*y+a=0','x','y');>> eval(x),eval(y) ans = -1.333333333333333 ans = -0.750000000000000
@祖瑞5480：英语翻译solve their problems - 作业帮
郦卖19383072823…… [答案] 玛丽总是尽力帮助朋友解决问题 翻译 Marry___always____ ______ _______ _______help her friends. 答案:is trying her best to
