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

because 1.2 isnt half of 6 duh

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3