A + B is also a multiple of C. ------------------------------------------- let k, m and n be integers. Then: A = nC as A is a multiple of C B = mC as B is a multiple of C → A + B = nC + mC = (n + m)C = kC where k = n + m kC is a multiple of C. Thus A + B is a multiple of C.
#include<iostream.h> void main() { cout<<' '<<' '<<"a"<<'\n'; cout<<' '<<"a"<<"b"<<"a"<<'\n'; cout<<'a'<<'b'<<'c'<<'b'<<'a'<<"\n"; }
#include<stdio.h> int main() { int a,b,c,d; for(a=1; a<5; a++) { for(b=1; b<5; b++) { for(c=1; c<5; c++) { for(d=1; d<5; d++) { if(!(a==b a==c a==d b==c b==d c==d)) printf("dd\n",a,b,c,d); } } } } return 0; }
To factor a trinomial (three-term expression) of the form n2+bn+c, find the two numbers h and k that are factors of c and add up to b. Then, write those numbers in this template: (n+h)(n+k) For the trinomial n2+7n-44, c = -44 and b = 7. The two numbers that are factors of -44 and add up to 7 are 11 and -4. So, the factored form would be (n+11)(n-4).
Use Euclid's formula. Take any two integers m and n that are coprime and let m > n (if not, just swap them around). Then a = m2 - n2 b = 2mn and c = m2 + n2 form a primitive Pythogorean triple. And since (a, b, c) is a Pythagorean triple then so is (k*a, k*b, k*c) for any integer k, except that the latter is not primitive.
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The full form of n C r is "n choose r" which represents the number of ways to select r items from a set of n items without regard to the order of selection.
If the question is what is the full form of n b. then it is nota bene........It is latin for note well or mark well
The full form for NCERT is National Council of Educational Research and Training. This is a schooling.
Electronic Numerical Integrator and Computer
C - Car H - Hanging I - In N - North A - America
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 number be X, then B% = B/100 → B% of X = C → B/100 x X = C → X = C ÷ (B/100) = C x 100/B = 100C ÷ B So to find the number, divide C by B percent.
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If A and B are multiples of C, then A + B is also a multiple of C: If A is a multiple of C then A = mC for some integer m If B is a multiple of C, then B = nC for some integer n → A + B = mC + nC = (m + n)C = kC where k = m + n and is an integer → A + B is a multiple of C
A + B is also a multiple of C. ------------------------------------------- let k, m and n be integers. Then: A = nC as A is a multiple of C B = mC as B is a multiple of C → A + B = nC + mC = (n + m)C = kC where k = n + m kC is a multiple of C. Thus A + B is a multiple of C.