3(b5)
To be divisible by 36 the number must be divisible by 9 (and 4). To be divisible by 9, the sum of the digits must be divisible by 9. 5 + 5 = 10 → need an extra 8 → A + B = 8.
5(a+b)-7
Given:a + b = 2a - b = 8Then:a = b + 8∴ (b + 8) + b = 2∴ 2b = -6∴ b = -3∴ a = 5
Assuming that a and b are two non-negative numbers, then their sum is a + b and the difference is |a - b|.
The algebraic expression is 3(b+5).
a + b = 61a - b = 5∴ a = 5 + b∴ (5 + b) + b = 61∴ 2b = 56∴ b = 28∴ a = 33
import javax.swing.JOptionPane; public class Addition { public static void main( String args[] ) { String firstNumber = JOptionPane.showInputDialog( "Enter first integer" ); String secondNumber = JOptionPane.showInputDialog( "Enter second integer" ); int number1 = Integer.parseInt( firstNumber ); int number2 = Integer.parseInt( secondNumber ); int sum = number1 + number2; JOptionPane.showMessageDialog( null, "The sum is " + sum, "Sum of Two Integers", JOptionPane.PLAIN_MESSAGE ); } }
3(b5)
To be divisible by 36 the number must be divisible by 9 (and 4). To be divisible by 9, the sum of the digits must be divisible by 9. 5 + 5 = 10 → need an extra 8 → A + B = 8.
start input a,b,c,d,e sum=a+b+c+d+e avg=sum/5 print"average is"; avg
It is: 5*(b+2) which is the same as 5b+10
5(a+b)-7
1.Start 2. Input a,b,c 3. Sum = a+b+c 4. Average = sum/3 5. Output - Sum,Average 6. Stop
Um, x2+3x-5=0? This is ax2+bx+c where a=1, b=3, and c=-5. The sum of the roots is -b/a so that means the sum of the roots is -3. Also, product of the roots is c/a. That means the product of the roots is -5. -3+(-5)= -8. There you have it.
41
You can write that as 5 / (a + b), or as a fraction, with the 5 on top, and a + b at the bottom. The actual value of this expression will depend on the values of "a" and "b".