The word "banana" has 3 consonants (b, n, n) and 3 vowels (a, a, a).

if (a > b && a > c) printf("%d\n", a);

else if (b > c) printf("%d\n", b);

else printf("%d\n", c);

Uncle B <3 x

a is 3m

b is 3n

m and n are relatively prime - they do not have any common factors.

a2 = 9m2

Since m had no factors in common with n, m2 also has no factors in common with n. We know that 3 is a common factor of a and b. The only new possible factor between a2 and b is the second 3 when a is squared. We do not know whether b had more than one factor of 3. So, the greatest common factor of a2 and b, when the greatest common factor of a and b is 3, is either 3 or 9.

It isnC0*A^n*b^0 + nC1*A^(n-1)*b^1 + ... + nCr*A^(n-r)*b^r + ... + nCn*A^0*b^n

where nCr = n!/[r!*(n-r)!]

// transpose for the sparse matrix void main() { clrscr(); int a[10][10],b[10][10]; int m,n,p,q,t,col; int i,j; printf("enter the no of row and columns :\n"); scanf("%d %d",&m,&n); // assigning the value of matrix for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { printf("a[%d][%d]= ",i,j); scanf("%d",&a[i][j]); } } printf("\n\n"); //displaying the matrix printf("\n\nThe matrix is :\n\n"); for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { printf("%d\t",a[i][j]); } printf("\n"); } t=0; printf("\n\nthe non zero value matrix are :\n\n"); for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { // accepting only non zero value if(a[i][j]!=0) { t=t+1; b[t][1]=i; b[t][2]=j; b[t][3]=a[i][j]; } } } printf("a[0 %d %d %d\n",m,n,t); for(i=1;i<=t;i++) { printf("a[%d %d %d %d\n",i,b[i][1],b[i][2],b[i][3]); } a[0][1]=n; a[0][2]=m; a[0][3]=t; int s[10],u[10]; if(t>0) { for(i=1;i<=n;i++) { s[i]=0; } for(i=1;i<=t;i++) { s[b[i][2]]=s[b[i][2]]+1; } u[1]=1; for(i=2;i<=n;i++) { u[i]=u[i-1]+s[i-1]; } for(i=1;i<=t;i++) { j=u[b[i][2]]; a[j][1]=b[i][2]; a[j][2]=b[i][1]; a[j][3]=b[i][3]; u[b[i][2]]=j+1; } } printf("\n\n the fast transpose matrix \n\n"); printf("a[0 %d %d %d\n",n,m,t); for(i=1;i<=t;i++) { printf("a[%d %d %d %d\n",i,a[i][1],a[i][2],a[i][3]); } getch(); }

Its spelt B-A-N-A-N-A-S its by Gwen Stephani

There are 3 letters: B, A, and N. The A is repeated twice more and the N once more.

The power of a quotient is the quotient of the power!

(a/b)^n = (a^n) / (b^n)

where a/b is the quotient and n is the power.

its worth nothing n stuff (b^o^)b its worth nothing n stuff (b^o^)b its worth nothing n stuff (b^o^)b its worth nothing n stuff (b^o^)b its worth nothing n stuff (b^o^)b its worth nothing n stuff (b^o^)b

2 squared divided by 3 squared ie 4/9 in general, (a/b)^n = a^n/b^n . In this case (2/3)^2=2^2/3^2=4/9 Check: 2/3 = .667 squared = 0.444889; 4/9 =0.44444. Near enough!

#include<stdio.h>

#include<conio.h>

void main()

{

int a[3][3],b[3][3],c[3][3],r,c;

for(r=0;r<=3;r++)

{

for(c=0;c<3;c++)

{

printf("\n enter the value=");

scanf("%d%d",&a[r][c],&b[r][c]);

}

}

printf("\n first matrix=\n");

for(r=0;r<=3;r++)

{

for(c=0;c<3;c++)

{

printf("%d\t",a[r][c]);

}

printf("\n");

}

printf("\n scond matrix=\n");

for(r=0;r<=3;r++)

{

for(c=0;c<3;c++)

{printf("%d\t",b[r][c]);

}

printf("\n");

}

printf("\n sum of given matrix=\n");

for(r=0;r<=3;r++)

{

for(c=0;c<3;c++)

{

c[r][c]=a[r][c]+b[r][c];

printf("%d\t",c[r][c]);

}

printf("\n");

}

getch();

}

In algebra this is a method of solving equations for a variable. It stands to reason that if A=B then A+1=B+1. In fact it is true that if N is a number then

A+N=B+N

A-N=B-N

A x N = B x N

A/N = B/N

We can do anything we want as long as we do the same thing to both sides.

//Matrix multiplication import java.util.Scanner; public class Matrix

{

public static void main(String args[])

{

Scanner s= new Scanner(System.in);

int i,j,k; System.out.println("enter the value of n"); int n=s.nextInt(); int a[][]=new int[n][n]; int b[][]=new int[n][n]; int c[][]=new int[n][n]; System.out.println("enter the array elements of a:"); for(i=0;i<n;i++)

{

for(j=0;j<n;j++)

{

a[i][j]=s.nextInt();

}

}//end of a matrix System.out.println("enter the array elements of b:"); for(i=0;i<n;i++)

{

for(j=0;j<n;j++)

{

b[i][j]=s.nextInt();

}

}//end of b matrix System.out.println("the result matrix is:");

for(i=0;i<n;i++)

{

for(j=0;j<n;j++)

{

for(k=0;k<n;k++)

{ c[i][j]+=a[i][k]*b[k][j]; }

}

} for(i=0;i<n;i++)

{

for(j=0;j<n;j++)

{

System.out.print(+c[i][j]);

}System.out.println();

}

}//end of main

}//end of class

There are 20 Rotary club members in the town. This can be calculated by subtracting the Lions club members who are also Rotary club members (10) from the total Lions club members (30), leaving 20 Lions club members who are not part of the Rotary club.

Thing is, 3 doesn't equal 0.03...

But... using false logic,

it could be proved:

Do we agree:

0/3 = 0/0.03 = 0

We know if:

n/a = n/b

then, a =b.

Therefore, 3 = 0.03

However, you can't do that with a 0, so

3 doesn't equal 0.03

n= 2B2-B

B= y-1

Substitute.

n=2(y-1)2-(y-1)

Foil (y-1)2.

n=2(y2-2y+1)-(y-1)

Distribute the 2.

n=2y2-4y+2-(y-1)

Distribute the negative.

n=2y2-4y+2-y+1

Combine like terms.

n=2y2-5y+3

Final answer:

n=2y2-5y+3

b is equal to 5

main()

{

int i, n;

printf("Enter a positive integer\n");

scanf("%d", &n);

for(i=1;i<=n;i++) printf("%d-",i);

printf("\b\n");

}

The number of links are:

L=b-(n-1)=b-n+1

Where b=Number of branches

n=number of nodes

nick n miley b-up Jo n Taylor swift b-up

nick n miley b-up Jo n Taylor swift b-up

nick n miley b-up Jo n Taylor swift b-up

According to SOWPODS (the combination of Scrabble dictionaries used around the world) there are 3 words with the pattern B--N---N. That is, eight letter words with 1st letter B and 4th letter N and 8th letter N. In alphabetical order, they are:

brinkman

brinkmen

bulnbuln

Any number to the power '0' equals '1'.

Proof ;

Let a^(n) = b

Then dividing

a^(n) / a^(n) = b/b

a^(n-n) = b/b

a^(0) = 1

I think the question is wrong - it should be 3 B M S H T R The answer is 3 blind mice see how they run.

#include<stdio.h>

void main()

{

int a[3],i,b;

printf("Enter 3 numbers\n");

for(i=0;i<=3;i++)

scanf("%d",&a[i]);

b=a[0];

for(i=0;i<=3;i++)

if(a[i]>b)

b=a[i];

printf("The biggest number is %d",b);

}

you use loops to draw out information from an array or a vector. Also they can be used to count up or down from any giving number.

#include <iostream>

using namespace std;

int main()

{

short a = 1;

short b = 1;

short c = 2;

//do

while (a && b && c != 34);

{

a = c + b;

cout << a << "\n";

b = a + c;

cout << b << "\n";

c = b + a;

cout << c << "\n";

}//while (a && b && c != 34);

return 0;

}

the output of this program would be

3

5

8

13

21

34

if i was to rewrite this program with out the loop, i would have to write the body of the loop

a = c + b;

cout << a << "\n";

b = a + c;

cout << b << "\n";

c = b + a;

cout << c << "\n";

over and over again until i reached the number 34. its adding 1 + 1 = 2

2+1=3, 3+2=5 and so on, that's how the program is reaching 34.

..use do{} while{}..

for example..

#include
main(){
int choose;
double a,b,c;
printf("Enter 7 if you want to use the calculator. Otherwise,type any character.\n");
scanf("%d",&choose);
do{
if (choose==7){
printf("Press 1 for addition.\n");
printf("Press 2 for subtraction.\n");
printf("Press 3 for multiplication.\n");
printf("Press 4 for division.\n");
scanf("%d",&choose);
if (choose==1){
printf("Enter the addends\n");
scanf("%lf %lf", &a,&b);
c=a+b,
printf("The sum of %lf and %lf is %lf.\n",a,b,c);
}else if(choose==2){
printf("Enter the minuend and subtrahend\n");
scanf("%lf %lf", &a,&b);
c=a-b,
printf("The difference of %lf and %lf is %lf.\n",a,b,c);
}else if(choose==3){
printf("Enter the multipliers\n");
scanf("%lf %lf", &a,&b);
c=a*b,
printf("The product of %lf and %lf is %lf.\n",a,b,c);
}else if(choose==4){
printf("Enter the dividend and divisor\n");
scanf("%lf %lf", &a,&b);
c=a/b,
printf("The quotient of %lf and %lf is %lf.\n",a,b,c);
}else{
printf("You have entered an invalid digit.\n");}
printf("If you want to continue,press 7 and choose again from 1 to 4.\n");
printf("Do you want to exit? Enter any key.\n");
scanf("%d",&choose);
}
}while(choose==7);
}

B(n)=n(n+1)/2

So B(9)=9x10/2=45

[ Fibonacci series___: ]

#include<stdio.h>

int main(void)

{

int n,i,c,a=0,b=1;

printf("Enter Fibonacci series of nth term : ");

scanf("%d",&n);

printf("%d %d ",a,b);

for(i=0;i<=(n-3);i++)

{

c=a+b;

a=b;

b=c;

printf("%d ",c);

}

}

B. N. Bessonov has written:

'\\'

B. N. Sarma was born in 1887.

B. N. Sircar died in 1980.

B. N. Sircar was born in 1901.

N. B. Hardeman was born in 1874.

N. B. Hardeman died in 1965.

B-J- and the Bear - 1978 S-T-U-N-T- 3-12 was released on:

USA: 31 March 1981

int maxprod (int n)

{

return n/2;

}

int main (void)

{

int n, a, b;

n= 7; /* for example */

a= maxprod (n);

b= n-a;

printf ("%d+%d=%d, %d*%d=%d\n", a, b, a+b, a, b, a*b);

}

Start

dimension A[N], B[N]

For c=1 to N

Input A[N]

Next

For c=1 to N

A[N] = A[N] *10

next

For c+1 to N

B[N] = A[N]

Next

for c=1 to N

print B[N]

Next

End

#include<stdio.h>

int main(){

int a[3][3],b[3][3],c[3][3],i,j;

printf("Enter the First matrix->");

for(i=0;i<3;i++)

for(j=0;j<3;j++)

scanf("%d",&a[i][j]);

printf("\nEnter the Second matrix->");

for(i=0;i<3;i++)

for(j=0;j<3;j++)

scanf("%d",&b[i][j]);

printf("\nThe First matrix is\n");

for(i=0;i<3;i++){

printf("\n");

for(j=0;j<3;j++)

printf("%d\t",a[i][j]);

}

printf("\nThe Second matrix is\n");

for(i=0;i<3;i++){

printf("\n");

for(j=0;j<3;j++)

printf("%d\t",b[i][j]);

}

for(i=0;i<3;i++)

for(j=0;j<3;j++)

c[i][j]=a[i][j]+b[i][j];

printf("\nThe Addition of two matrix is\n");

for(i=0;i<3;i++){

printf("\n");

for(j=0;j<3;j++)

printf("%d\t",c[i][j]);

}

return 0;

}

It is not simple. The only systematic way is to find the prime factorisation of the number and write it in exponential form.

So suppose n = (p1^r1)*(p2^r2)*...*(pk^rk) where p1, p2, ... pk are prime numbers and rk are the indices (or powers).

Then the factors of n are (p1^s1)*(p2^s2)*...*(pk^sk) where 0 ≤ sk ≤ rk. And remember that anything raised to the power 0 is 1.

Example:

n = 72 = 2*2*2*3*3 = (2^3)*(3^2)

so, the factors of n are (2^a)*(3^b) where a = 0, 1, 2 or 3 and b = 0, 1 or 2.

When

(a, b) = (0, 0) the factor is 1.

(a, b) = (1, 0) the factor is 2.

(a, b) = (2, 0) the factor is 4.

(a, b) = (3, 0) the factor is 8.

(a, b) = (0, 1) the factor is 3.

(a, b) = (1, 1) the factor is 6.

(a, b) = (2, 1) the factor is 12.

(a, b) = (3, 1) the factor is 24.

(a, b) = (0, 2) the factor is 9.

(a, b) = (1, 2) the factor is 18.

(a, b) = (2, 2) the factor is 36.

(a, b) = (3, 2) the factor is 72.

Yes, here's how you can show it, for the equivalent proportions a/b & (n*a)/(n*b):

a na

-=--

b nb

Cross multiply: a*n*b ? b*n*a. Because multiplication is commutative, the question mark is an equal sign.

# include<stdio.h>

void main()

{

menu();

}

menu()

{

int i, a =0,b=1,c;

printf("Enter the operation need to be performed" \n);

printf("Enter 1 to perform And operation" \n);

printf("Enter 2 to perform OR operation" \n" \n);

printf("Enter 3 to perform NOT operation" \n);

printf("Enter 4 to exit" \n);

scanf("%d", i);

if (i==1)

c=and(a,b);

else if(i==2)

c=OR(a,b);

else if(i==3)

c=not(a,b);

else if(i==4)

exit(0);

printf ("The result is %d", c);

menu();

}

and(a,b)

{

return (a&&b);

}

or()

{

return (ab);

}

not(a)

{

return (!a);

}

}

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();

}

b-n+1

Ska 'n' B was created in 1980-04.

n my b-day s then december 27

The rate law for this reaction is rate = k[A]^m[B]^n. From the given information, substituting the values for rate, [A], [B], and the exponents m and n, you can solve for the rate constant k. In this case, k = rate / ([A]^m[B]^n), so k = 2 / (10^2 * 3^1).