Billion
One billion
Five less than three times a number ( b ) can be expressed mathematically as ( 3b - 5 ). This means you first multiply the number ( b ) by 3 and then subtract 5 from the result.
Let the 2 digit number be 10a +b. Then: a + b = 15 10a + b - 27 = 10b + a => 9a -9b = 27 => a - b = 3 adding the first and last equations gives: 2a = 18 => a = 9 and substituting in the first gives: 9 + b = 15 => b = 6 meaning the original number is 96.
There is nothing to explain much. Here is the code of that kind of application:#include int main() {int firstNum, secondNum, delta = 0;printf("Please enter first number (a): ");scanf("%d", &firstNum);printf("Please enter second number (b): ");scanf("%d", &secondNum);delta = firstNum - secondNum;printf("a - b = %d\n", delta);return 0;}Testing:Please enter first number (a): 46Please enter second number (b): 2a - b = 44Please enter first number (a): 1Please enter second number (b): 45a - b = -44
It is b*b
Class B. The first octet of a Class B license is a number between 128 and 191.
B is not a number.
If We are looking for A, B and C If the first number is A then we know B = A+2 C = A+4 and 3A = B+C or 3A = (A+2) + (A+4) so 3A = 2A +6 3A - 2A = 6 A = 6 B= 8 C= 10
2a + b = 47; a + 3b = 81 ie a = 81 - 3b so - 162 - 6b + b = 47 ie 5b = 115 so b = 23 and a = 12
To add a squared number to a cubed number, first calculate the square of a number (e.g., (a^2)) and the cube of the same or another number (e.g., (b^3)). Then, simply sum the two results: (a^2 + b^3). For example, if (a = 2) and (b = 3), then (2^2 + 3^3 = 4 + 27 = 31).
B + 1 or B - 1
yes b/c 1 is neither prime or composite