The bond stretching frequency increases with increasing bond strength. Therefore, the order of increasing bond stretching frequency is: F-H < O-H < N-H < C-H.
ITS EASY...TRY THIS OUT..TRAPEZOIDAL METHOD#include#include#includefloat valcal(float x){return (x*x*x);}int main(){float a,b,h,c,I;int n,i;printf("THE TRAPEZOIDAL RULE:\n");printf("---------------------");printf("\n\n\nEnter the two limits and the no. of divisions:\n");scanf("%f %f %d",&a, &b, &n);h=(b-a)/n;//printf("\nVALUE of h: %f\n", h);c=a;I=valcal(a)+valcal(b);//printf("\nVALUE FOR a: %f\n", valcal(a));//printf("\nVALUE FOR b: %f\n", valcal(b));for(i=1;i=b){printf("\n\nc>b\n\n");break;}//printf("\nVALUE FOR %f: is %f\n",c, valcal(c));I=I+(2*valcal(c));//printf("\nI right now is %f", I);}printf("\n\n\nThe integration of x*x*x is: %f",(h*I)/2);printf("\n\n\n");system("pause");}SIMPSON'S 1/3RD METHOD#include#include#includefloat valcal(float x){return (1/(1+x*x));}int main(){float a,b,h,c,I;int n,i;printf("THE SIMPSON'S ONE-THIRD RULE:\n");printf("------------------------------");printf("\n\n\nEnter the two limits and the no. of divisions:\n");scanf("%f %f %d",&a, &b, &n);h=(b-a)/n;//printf("\nVALUE of h: %f\n", h);c=a;I=valcal(a)+valcal(b);//printf("\nVALUE FOR a: %f\n", valcal(a));//printf("\nVALUE FOR b: %f\n", valcal(b));for(i=1;ib){printf("\n\nc>b\n\n");break;}//printf("\nVALUE FOR %f: is %f\n",c, valcal(c));if(i%2==0)I=I+(2*valcal(c));elseI=I+4*valcal(c);//printf("\nI right now is %f", I);}printf("\n\n\nThe integration of x*x*x is: %f",(h*I)/3);printf("\n\n\n");system("pause");}NEED MORE HELP...MAIL ME YOUR PROB... SEE YA
#include<stdio.h> #include<conio.h> #include<math.h> float f(float x) { float y; if(x!=0) //y=sin(x)-log(x)+exp(x); y=1/(1+x); else printf("answer can not be derived"); return(y); } float g(float x) { float k; k=(-2)/(pow((1+x),3)); return(k); } void main() { float i,n,k,z,s=0,err=0; float h,x,a[100],b,c; clrscr(); printf("\n enter the range of the integration"); scanf("%f%f",&b,&c); printf("\n enter n so to divide the range into n parts"); scanf("%f",&n); h=(c-b)/n; printf("\n\n%f",h); k=(f(b)+f(c)); printf("\nk=%f\n\n",k); for(i=b+h;i<c;i+=h) { z+=f(i); printf("\n%f",z); } s=(h*(k+2*z))/2; printf("\n\n the value of the integration=%f\n\n",s); for(i=b;i<=c;i+=h) err+=g(i); err=(err*(pow(h,3)))/12; printf("\n\n approximate error= %f\n\n",err); getch(); }here...
a b c d e f g h i j k l m n OOOOOOOOOOOOOOO a b c d e f g h i j k l m n OOOOOOOOOOOOOOO
Al-H < C-H < N-H < O-H
occultly
F/U/C/K/I/N REPORT ME B/1/T/C/H/E/S
I'll give you a hint, Finch. Ditch it.
Some words that you can make with H V N N R F Q O are:hornforfrohononorofohonor
Isocyanate is a compound containing O, C, H, and N.
According to SOWPODS (the combination of Scrabble dictionaries used around the world) there are 1 words with the pattern -H--F-NC-. That is, nine letter words with 2nd letter H and 5th letter F and 7th letter N and 8th letter C. In alphabetical order, they are: chaffinch
Given an integer n, an integer f is a fraction of n if f goes into n evenly. That is, n/f is an integer or n = f*x for some integer x.m is a multiple of n if m = n*c for some integer c.