请问各位大侠,下面的程序那里有问题?
#include<stdio.h>
int main(void)
{
    float weight;
    float value;
    printf("Are you worth your weight in rhodium?\n");
    printf("Let's check it out.\n");
    printf("please enter your weight in pounds:");
    scanf("%f",&weight);
    value=770*weight*14.5833;
    printf("Your weight in rhodium is $%.2f.\n",value);
    printf("You are easily worch that! If rhodium prices drop.\n");
    printf("eat more to maintain your value.\n);
    getchar();
    getchar();
}
这些事错误提示
E:\C\C-3\C-3-1\C3-1.c(10) : warning C4244: '=' : conversion from 'double ' to 'float ', possible loss of data
E:\C\C-3\C-3-1\C3-1.c(13) : error C2001: newline in constant
E:\C\C-3\C-3-1\C3-1.c(14) : error C2064: term does not evaluate to a function
E:\C\C-3\C-3-1\C3-1.c(14) : warning C4047: 'function' : 'const char *' differs in levels of indirection from 'int '
E:\C\C-3\C-3-1\C3-1.c(14) : warning C4024: 'printf' : different types for formal and actual parameter 1
E:\C\C-3\C-3-1\C3-1.c(14) : error C2143: syntax error : missing ')' before ';'
执行 cl.exe 时出错.
  膜拜了

#include<stdio.h>
int main(void)
{
    float weight;//用户体重
    float value;//相等重量的铑的价值
    printf("Are you worth your weight in rhodium?\n");
    printf("Let's check it out.\n");
    printf("please enter your weight in pounds:");
    scanf("%f",&weight);//从用户处获取输入，为这个示例的关键处
    value=770*weight*14.5833;
    printf("Your weight in rhodium is $%.2f.\n",value);
    printf("You are easily worch that! If rhodium prices drop.\n");
    printf("eat more to maintain your value.\n);
    getchar();
    getchar();
}

多谢了啊