At most -80 (negative 80).
Choose any integer. Let's call it "n". Then subtract 8 - n, to get the other integer. (For the two integers to have different signs, one of the integers must be greater than 8, the other will be negative.)
It is a whole number and an integer but, since it is only one number, it cannot be "a integers"!
8
The smaller integer is 6, the larger integer is 32
They're not. Rational numbers are those that can be written as the ratio of two integers -- as a fraction with an integer on top and bottom, in other words. The number 1.5 is not an integer, but it is rational because you can write it as 3/2.
The other integer is -18 because -5 times -18 = 90
The integers are 14 and 7.
Choose any integer. Let's call it "n". Then subtract 8 - n, to get the other integer. (For the two integers to have different signs, one of the integers must be greater than 8, the other will be negative.)
When you add, subtract, or multiply integers, you get integers. When you divide one integer by another one, you may or may not get an integer.
Yes, the integers are 12 and 13.
Suppose one of the integers is x. Then the other is x+36 x+36 = 3x so 2x = 36 so x = 18 and then the other integer is x+36 = 54. Answer: 18 and 54
The two integers are 7 and 14. 7 x 14 = 98.
No. An irrational number is one that is not a rational number. A rational number is once that equals one integer divided by another. So an irrational number cannot be represented by one integer divided by another integer, so it cannot be an integer!
You get another integer that will take the sign from the larger of the two integers that were combined.
The sum of two positive integers is positive
Let's denote the unknown integer as "x". So now we have two integers, "x" and "4x" because one integer is 4 times the other. So the sum of x+4x= 5x 5x = 5 So x=1
The rule in dividing integers is to divide the absolute values. Two positive integers or two negative integers equals positive product. If one integer is positive and the other is negative, the product is negative.