Question

consider the function $f(t) = \sqrt{3t - 9}$. you cannot take the square root of a negative number, so $3t - 9$ must be greater than or equal to zero. complete set up an inequality showing that the radicand cannot be negative. $3t - 9 \geq 0$ (checked), $3t + 9 \geq 0$, $3t \geq 0$, $t \geq 0$ complete what is the domain of the function? domain: $t > 3$; domain: $t \geq -3$; domain: $t \geq 0$; domain: $t \geq 3$ done

Explanation:

Step1: Solve the inequality \(3t - 9 \geq 0\)

Add 9 to both sides of the inequality: \(3t - 9 + 9 \geq 0 + 9\), which simplifies to \(3t \geq 9\).

Step2: Divide both sides by 3

Divide each side of the inequality \(3t \geq 9\) by 3: \(\frac{3t}{3} \geq \frac{9}{3}\), which gives \(t \geq 3\).

Answer:

Domain: \(t \geq 3\) (corresponding to the option "Domain: \(t \geq 3\)")