Question

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

Explanation:

The radicand of the square root function \( f(t)=\sqrt{3t - 9} \) is \( 3t-9 \). Since the square root of a negative number is not a real number, the radicand must be greater than or equal to zero. So the inequality representing this is \( 3t - 9\geq0 \).

Answer:

A. \( 3t - 9\geq0 \)