编程计算下列函数的值X>0时,Y=1

importjava.util.Scanner;publicclassDemo{publicstaticvoidmain(String[]args){Scannersc=newScanner(Syst

voidmain(void){doublex,y;printf("pleaseinputX:");scanf("%lf",&x);if(x>10)y=x/10.0;elseif(x>0)y=x-2;e

#include<stdio.h>#include<math.h>void main(){float x,y;scanf("%f",&a

X的区间你划分不对,x<0没有最后一个表达式倒是-3还是=3再问：y=-7x³+3(x

这个是你要的程序,vc6调试通过了#includevoidmain(){doublex,y;printf("x=");scanf("%lf",&x);if(x

#include"stdio.h"doubleY(doublex){returnx-1?1.0/(1+x):0;}intmain(){\x05doublex;\x05scanf("%lf",&x);p

#include#includefloatf(floatx){doubley;if(x

PrivateSubCommand1_Click()DimnAsInteger,iAsIntegerDimsumAsDoublen=Val(InputBox("PleaseEnterN:"))sum=

用matlab吧别用C自虐了………………

你这个题目可能没有写全,根据你提供的条件,Y在X5时=1,如果X在(1,5)之间,Y无解这样就可以做了,在窗体上画一个Command控件,编写如下代码PrivateSubCommand1_Click(

窗体上添加2个text,1个commandPrivateSubCommand1_Click()x=Text1.TextIfx>=0Theny=1+x^2Elsey=1-2*xEndIfText2.Te

基于c#的classProgram{staticvoidMain(string[]args){Console.WriteLine("Pleaseenteranumber");inta=int.Pars

clc;clear;x=linspace(0,5,10);ep(1,:)=ones(size(x));fori=1:5ep(i+1,:)=x.^(i)/factorial(i);endy=sum(ep

double是双精度的浮点型数据,输出使用　%lf格式声明.另外如果为了减少程序执行多分支流程,建议swichcase进行分支处理,这样可以只走一个分支,而不是你现在的走三个if分支.switch()

#include#include#includevoidmain(){doublex,y,temp;cin>>x;if(x==0){exit(0);cout

定义变量x,y;输入x;if(xy=fabs(2*x+5);}elseif(x=0){y=sqrt(x+1);}else{y=3*x+1;}输出y;return1;大体就是这样,因为我不是学c语言的,

#include#includedoublefun(doublex,intn){\x09inti;\x09intj;\x09doublefaci=1.0;\x09doubles=1.0;\x09for

ifx>=0theny=xelsey=-xendifprinty

#includedoublepow(doublex,intn){if(1==n){returnx;}else{doubletemp=pow(x,n-1);return(x*temp);}}voidma

1:symsxy;y=sin(x)./x;ezplot(y,[-10,10]);%-10,10是画图像x的取值范围,可以改.subs(y,x,2);%计算x=2时的函数值.即sin(2)/2.2.sy