To find the integer that, when added to -9, gives a sum of 3, we can set up the equation: ( x + (-9) = 3 ). Solving for ( x ), we add 9 to both sides: ( x = 3 + 9 ). Thus, ( x = 12 ). Therefore, the integer is 12.
x + 3x + 3 = 4x + 3
The sum of a negative integer and a positive integer is zero only when the two integers have the same absolute value. For example, if you have -3 (negative integer) and +3 (positive integer), their sum is zero. However, in general, this situation occurs infrequently since it requires specific values. Most of the time, the sum will be a negative integer if the absolute value of the negative integer is greater than that of the positive integer, and a positive integer if the opposite is true.
No, -3 + 8 = 5.
The sum of three consecutive integers starting with ( n ) can be expressed as ( n + (n + 1) + (n + 2) ). Simplifying this expression gives ( 3n + 3 ). Therefore, the sum of the three consecutive integers is ( 3n + 3 ) or ( 3(n + 1) ).
If the sum of each digit of a number is divisible by three ((1 + 1 + 1 + 0)/3 = 1) and it gives an integer, and the number is divisible by 2 (1110/2 = 555) and also gives an integer, then the number is divisible by 6.So, yes.
x + 3x + 3 = 4x + 3
The sum of a negative integer and a positive integer is zero only when the two integers have the same absolute value. For example, if you have -3 (negative integer) and +3 (positive integer), their sum is zero. However, in general, this situation occurs infrequently since it requires specific values. Most of the time, the sum will be a negative integer if the absolute value of the negative integer is greater than that of the positive integer, and a positive integer if the opposite is true.
No, -3 + 8 = 5.
-3
8
It's not always. 7 + (-3) = 4
6 6+(3*62) = 114
The sum is 3*(x-1) or 3x - 3
The sum of three consecutive integers starting with ( n ) can be expressed as ( n + (n + 1) + (n + 2) ). Simplifying this expression gives ( 3n + 3 ). Therefore, the sum of the three consecutive integers is ( 3n + 3 ) or ( 3(n + 1) ).
If the sum of each digit of a number is divisible by three ((1 + 1 + 1 + 0)/3 = 1) and it gives an integer, and the number is divisible by 2 (1110/2 = 555) and also gives an integer, then the number is divisible by 6.So, yes.
The first integer is 17.
By the uniqueness of integers, the only integer for 39 is 39 itself. And that cannot have a sum of 3.