4╱5*(float)3＋（float）3*4/5=?

4╱5*(float)3＋（float）3*4/5=? float (6>5>4)+(FLOAT)(3/2)为什么等于1.0 以下函数值的类型是（ ）.fun ( float x ) { float y; y= 3*x-4; return y; } 求x^3-15x^2+16x-80一个实根,下面程序怎么算不出来#include#includefloat f(float x){return ((x-5)*x-16)*x-80;}float xpoint(float x1,float x2){return (x1*f(x2)-x2*f(x1))/(f(x2)-f(x1));}float root(float x1,float x2){float x,y,y1;y1=f(x1);do{ float (*p)(float)是什么意思 以下数组定义中,合法的是（ ）.A.int float[] B.int 3abc[] C.int apple[] D.float *abc[5] fun(float x) { float y; y=3*x-4; return y; }以上函数值的类型为什么是int?怎么不是float呢? integral(float(*f)(float a,float 计算下列表达式:a%3+b*b/(float)c/5 float是什么意思 float market 以下函数值的类型是 fun ( float x ) { float y; y= 3*x-4; return y; } 以下函数值的类型是fun ( float x ){ float y;y= 3*x-4;return y;} 请斧正c中的错误-1#includefloat search(float (*p)[4],float n);main(){float score[3][4]={{11,22,33,44},{22,22,33,44},{33,44,55,66}};search(score,2);getch();}float search(float (*p)[4],float n){int i;printf(the input num:%d,n);for(i=0;i c语言的问题,求x^3-5x^2+16x-80的根,我用的弦截法,但是出错#include#includefloat f(float x){float y;y=x*(x*x-5*x+16)-80;return y;}float xpoint(float x1,float x2){float root;root=(f(x2)*x1-x2*f(x1))/(f(x2)-f(x1));return root;}void main() sizeof（float）是（）类型表达式? 用弦切法求f(x) =x^3 - 5x^2 + 16x - 80 的根,求证一下那里出问题了,输入-1 和8,结果是6.75.#include #include float t(float x){return (x*x*x - 5.0*x*x + 16*x - 80); }float t(float x1,float x2);{return ((x1 * f(x2) - x2 * f(x1)) / (f(