1. The problem asks to find the inequality that determines the domain of the function $f(x) = \sqrt{\frac{1}{2}x - 10} + 3$. 2. The domain of a function with a square root requires the radicand (expression inside the root) to be greater than or equal to zero because the square root of a negative number is not a real number. 3. The radicand here is $\frac{1}{2}x - 10$. 4. Therefore, the inequality to find the domain is: $$\frac{1}{2}x - 10 \geq 0$$ 5. This inequality ensures the expression inside the square root is non-negative, so the function is defined for all $x$ satisfying this. 6. To solve for $x$, add 10 to both sides: $$\frac{1}{2}x - 10 + 10 \geq 0 + 10$$ $$\frac{1}{2}x \geq 10$$ 7. Multiply both sides by 2 to isolate $x$ (since 2 is positive, inequality direction stays the same): $$\cancel{2} \times \frac{1}{2}x \geq 10 \times \cancel{2}$$ $$x \geq 20$$ 8. So the domain of $f(x)$ is all real numbers $x$ such that $x \geq 20$. 9. Among the given options, the correct inequality to find the domain is: $$\frac{1}{2}x - 10 \geq 0$$