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Answer:

[tex]x\geq -6[/tex]

Step-by-step explanation:

The domain of a function is all of the values that [tex]x[/tex] can be under that specific function. In this case, we're asking what values of [tex]x[/tex] allow [tex]\sqrt{\frac{1}{3} x+2}[/tex] to exist.

In order for square roots to exist, the quantity under the square root must be greater than or equal to [tex]0[/tex], because you can't take the square root of a negative number. Therefore, we can write the following inequality to solve for [tex]x[/tex]:

[tex]\frac{1}{3}x+2\geq 0[/tex]

Solving this inequality, we get:

[tex]\frac{1}{3}x+2\geq 0[/tex]

[tex]\frac{1}{3}x\geq -2[/tex] (Subtract [tex]2[/tex] from both sides of the inequality to isolate [tex]x[/tex])

[tex]x\geq -6[/tex] (Multiply both sides of the inequality by [tex]3[/tex] to get rid of [tex]x[/tex]'s coefficient)

Hope this helps!

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