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Which of the following gives all of the sets that contain √9?

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

Respuesta :

I dont think its b because one of the square roots 9 (-3) is not a natural number

I think its choice c
JeanaShupp

Answer: c.the set of all integers, the set of all rational numbers, and the set of all real numbers

Step-by-step explanation:

The given number is [tex]\sqrt{9}[/tex].

We know that [tex](3)^2=9[/tex] and  [tex](-3)^2=9[/tex]

Therefore, [tex]\sqrt{9}=\pm3[/tex]

Therefore, the value of [tex]\sqrt{9}[/tex] does not belongs to the set of natural numbers since -3 does not belongs to it.

Since 3,-3 belongs to the set of integers.

Thus, the value of [tex]\sqrt{9}[/tex] belongs to the set of integers  the set of all rational numbers, and the set of all real numbers because the set of integers is contained in the set of rationals and the set of rational is contained in the set of real numbers.

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