Here is a simple C program to find the multiplication of three numbers:
#include <stdio.h>
int main() {
int num1, num2, num3, result;
printf("Enter three numbers: ");
scanf("%d %d %d", &num1, &num2, &num3);
result = num1 * num2 * num3;
printf("Multiplication of the three numbers is: %d\n", result);
return 0;
}
This program prompts the user to enter three numbers, multiplies them together, and then prints out the result.
Sure, darling. Here's a simple C program to find the multiplication of three numbers:
#include <stdio.h>
int main() {
int num1, num2, num3;
printf("Enter three numbers: ");
scanf("%d %d %d", &num1, &num2, &num3);
int result = num1 * num2 * num3;
printf("The multiplication of the three numbers is: %d", result);
return 0;
}
Run along now and give it a try!
Cls input "enter two no.s ",a,b sum=a+b print "sum = ";sum end
(defun max3 (a b c) (cond ((> a b) (cond ((> a c) a) (t c))) ((> b c) b) (t c) ) )
You write it exactly the same as you would write it in any other verions of C++, by taking user input to determine the three sides of your triangle. In other words, input three real numbers. What you do with those three numbers is entirely up to you, but presumably you'd want to calculate the angles of a triangle given the length of its three sides. For that you would need to use the cosine rule which states that for any triangle with angles A, B and C whose opposing sides are a, b and c respectively, cos A = (b2 + c2 - a2)/2bc and cos B = (c2 + a2 - b2)/2ca. Knowing two angles, A and B, you can easily work out that angle C must be 180 - (A + B).
start input A & B if A>B print A is greatest if B>A print B is greatest stop james ola writes.....SOT.
create a program that iterates until it finds a perfect number, then store that perfect number into an array. Continue iterating until you find three more. Then, you have an array of four perfect numbers.
write a shell program for finding out gcd of three given numbers? write a shell program for finding out gcd of three given numbers? write a shell program for finding out gcd of three given numbers? check bellow link http://bashscript.blogspot.com/2009/08/gcd-of-more-than-two-numbers.html
multiplicand_________multiplier / multiplicand x multiplier=product_______product______
The associative power of multiplication states that for any three numbers a, b and c, (a * b) * c = a * (b * c) and so we can write either as a * b * c without ambiguity.
Addition and multiplication: yes
The associative property of multiplication states that for any three numbers a, b and c, (a * b) * c = a * (b * c) and so we can write either as a * b * c without ambiguity. ie, when multiplying three numbers together, you can multiply the first two together and then multiply the result of that by the third, or multiply the second two numbers together and multiply that result by the first, and you will get the same answer.
The associative property of multiplication states that for any three numbers a, b and c, (a * b) * c = a * (b * c) and so we can write either as a * b * c without ambiguity. The associative property of multiplication means that you can change the grouping of the expression and still have the same product.
Using division or multiplication or addition??
largest of a, b, c :a > b ? a > c ? a : c : b > c ? b : c
Multiplication is repeated addition. 3 x 3 = 3 added three times.
Multiplication is repeated addition. 3 x 3 = 3 added three times.
3000
3,000,000,000,000