Which of the following gives all of the sets that contain sqrt9? A. the set of all irrational numbers
B. the set of all natural numbers, the set of all whole numbers, and the set of all integers
C. the set of all integers, the set of all rational numbers, and the set of all real numbers
D. the set of all natural numbers, the set of all whole numbers, the set of all integers, the set of all rational numbers, and the set of all real numbers

C the set of all integers, the set of all rational numbers, and the set of all real numbers

Step-by-step explanation: The value of square root of 9 is ±3. This means that there two answers to square root of 9: 3 and -3.

As far as 3 is concerned, it is one of the elements of the natural number set. It is already known that natural numbers are also whole numbers, also integers, also rational numbers, and also real numbers.

As far as -3 is concerned, it is not a natural number since the set of natural numbers does not contain negative numbers. -3 is an integer. Which means that -3 is also a rational number and also a real number.

After identifying the common sets and taking the intersection of both the above classifications, it yields the set of all integers, the set of all rational numbers, and the set of all real numbers

Hopes this helps bro

step-by-step explanation:

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).