carbon atom, monomer, macromolecule, and polymer.
This element is carbon.
c. hydrogen
B: two atoms of oxygen is the correct answer
2 more neutrons.
largest of a, b, c :a > b ? a > c ? a : c : b > c ? b : c
// return the larger of any 2 numbers int larger (int a, int b) { return a>b?a:b; } // return the largest of any 3 numbers int largest (int a, int b, int c) { return larger (larger (a, b), c)); } // return the middle value of any 3 numbers int middle (int a, int b, int c) { if (a>b) a^=b^=a^=b; // swap a and b (b is now the larger of the two) if (b>c) b^=c^=b^=c; // swap b and c (c is now the largest of all three) return larger (a, b); // return the larger of a and b }
As a general rule atomic radius decreases as you read across a period, so from largest to smallest the order would be Li, B, C, F.
void main() { int a,b,c; clrscr(); printf("Enter the value of a:"); scanf("%d",&a); printf("\nEnter the value of b:"); scanf("%d",&b); printf("\nEnter the value of c:"); scanf("%d",&c); if(a>b) { if(a>c) { if(b>c) { printf("c is smallest\n"); printf("b is middle\n"); printf("a is largest\n"); } else { printf("b is smallest\n"); printf("c is middle\n"); printf("a is largest\n"); } } else { printf("b is smallest\n"); printf("a is middle\n"); printf("c is largest\n"); } } else if(b>c) { if(a>c) { printf("c is smallest\n"); printf("a is middle\n"); printf("b is largest\n"); } else { printf("a is smallest\n"); printf("c is middle\n"); printf("b is largest\n"); } } else { printf("a is smallest\n"); printf("b is middle\n"); printf("c is largest\n"); } getch(); }
Let the numbers be A,B and C and assume that A ≠ B ≠ C 1) Subtract B from A. If the result is a positive number then A > B otherwise B > A. 2) Subtract the greater of A and B from C. If the result is a positive number then C is the largest number otherwise the subtrahend (either A or B) is the largest number.
Find the largest of two, then find the largest of that value and the third value. int* max (int* a, int* b) { return (a*) > (b*) ? a : b; } int* max_of_three (int* a, int* b, int* c) { return max (max (a, b), c); }
for the largest number: #include<stdio.h> void main() { int a,b,c,number,largestnumber; a=99; b=9; c=77; if(a>b) { number=a; } else if(b>c) { number=b; } else { number=c; } largestnumber=number; printf("%d",largestnumber); }
#define max2(a,b) (b>a?b:a) #define max3(a,b,c) (max2(a,max(b,c)))
Atom economy or atom utilization is simply the Mr or molecular weight of desired product formed divided by the Mr of all the reactants. Say we have a reaction A + B -----> C C + D ------> E AU = (Mr of E/(Mr of A + B + C)) x 100
#define max (a, b) ((a) >= (b)) ? (a) : (b)
#include <iostream.h> #include <conio.h> void main() { clrscr(); int largest(int,int,int); cout<<"Enter 3 Integer Numbers\n"; int a,b,c; cin>>a>>b>>c; int result; result=largest(a,b,c); cout<<"\n\nLargest Value of Inputed is "<<result; getch(); } inline largest(int a,int b,int c) { int z; z=(a>b)?((a>c)?a:c):((b>c)?b:c); return(z); }
A plus b plus c equals d. A is the largest answer b is the smallest answer and d is less than 6?''