Pick a and b as the last two digits (and we may assume a is greater than b). Make the remaining first seven digits the same as each other (any way will be fine). The last two digits will be ab for the first number and ba for the second number. The value of ab is 10a+b, the value of ba is 10b+a. The difference is (10a+b)-(10b+a)= 9a-9b = 9(a-b). To minimise this we just make a and b consecutive integers - there are eight ways of doing this. There are 7! ways of arranging the first seven digits. So 8! ways of choosing the numbers.
The largest possible values for the integers are 47, 49, and 51.
71.(142/2).
At least the following families: all integers; all positive integers; all odd integers; and all "square integers", that is, integers that are squares of other integers.
If they are integers, then the possible answers are {1, 6, 8} and {2, 4, 9}.If not, there are infinitely many possible solutions.If they are integers, then the possible answers are {1, 6, 8} and {2, 4, 9}.If not, there are infinitely many possible solutions.If they are integers, then the possible answers are {1, 6, 8} and {2, 4, 9}.If not, there are infinitely many possible solutions.If they are integers, then the possible answers are {1, 6, 8} and {2, 4, 9}.If not, there are infinitely many possible solutions.
To find the smallest possible value of 20P + 10Q + R when P, Q, and R are different positive integers, we should start by assigning the smallest possible values to P, Q, and R. Since they are different positive integers, we can assign P = 1, Q = 2, and R = 3. Substituting these values into the expression, we get 20(1) + 10(2) + 3 = 20 + 20 + 3 = 43. Therefore, the smallest possible value of 20P + 10Q + R is 43.
490.
1 and -26 is one possible answer.
44 and55
The largest possible values for the integers are 47, 49, and 51.
The set of all real numbers is one possible answer.
71.(142/2).
At least the following families: all integers; all positive integers; all odd integers; and all "square integers", that is, integers that are squares of other integers.
If they are integers, then the possible answers are {1, 6, 8} and {2, 4, 9}.If not, there are infinitely many possible solutions.If they are integers, then the possible answers are {1, 6, 8} and {2, 4, 9}.If not, there are infinitely many possible solutions.If they are integers, then the possible answers are {1, 6, 8} and {2, 4, 9}.If not, there are infinitely many possible solutions.If they are integers, then the possible answers are {1, 6, 8} and {2, 4, 9}.If not, there are infinitely many possible solutions.
To find the smallest possible value of 20P + 10Q + R when P, Q, and R are different positive integers, we should start by assigning the smallest possible values to P, Q, and R. Since they are different positive integers, we can assign P = 1, Q = 2, and R = 3. Substituting these values into the expression, we get 20(1) + 10(2) + 3 = 20 + 20 + 3 = 43. Therefore, the smallest possible value of 20P + 10Q + R is 43.
It is not possible to tell if the difference give an organism an advantage because you have not given the difference.
yes and it is possible to do
well, the square root of 24999999 is 4999.999, and the answer is 4999 • 5001. Hope this helps!