2
if abc is 0 then at least one of the factors must be zero. since a and b are both nonzero, c must be zero.
The negative integers greater than -10 are -9, -8, -7, -6, -5, -4, -3, -2, and -1. These integers are all closer to zero than -10, making them greater on the number line.
The integers greater than 10 but less than 100 start from 11 and go up to 99. To find the count of these integers, subtract 10 from 99, which gives 89, and then subtract 1 (to exclude 10 itself), resulting in 89 - 10 = 89. Thus, there are 89 integers greater than 10 but less than 100.
The integers are -14, -12 and -10.
1
if abc is 0 then at least one of the factors must be zero. since a and b are both nonzero, c must be zero.
The negative integers greater than -10 are -9, -8, -7, -6, -5, -4, -3, -2, and -1. These integers are all closer to zero than -10, making them greater on the number line.
The integers greater than 10 but less than 100 start from 11 and go up to 99. To find the count of these integers, subtract 10 from 99, which gives 89, and then subtract 1 (to exclude 10 itself), resulting in 89 - 10 = 89. Thus, there are 89 integers greater than 10 but less than 100.
No, there are an infinite number of integers. So, there would be an infinite (infinity/2-1) number of positive integers. And, there would be an infinite (infinity-10) number of integers greater than ten.
There are no "two consecutive integers" that can do that.But there are two consecutive even integers that can: 8 and 10 .
10
The integers are -14, -12 and -10.
integers are x and x + 2, so x + 2 = 2x - 10 ie x = 12. Integers are 12 and 14
It is not possible.
The numbers are 9 and 10.
No. To be divisible by a number it must be greater than, or equal to, that number. 10 is less than 66, so 10 cannot be divisible by 66.
1