To find the point that would have an open circle on the graph of the piecewise function \( f(x) \), we need to look at where the function changes from one piece to another.
According to the piecewise function:
\[ f(x) = \begin{cases} 2.5x & \text{if } x < 10 \ 2.25x & \text{if } x \geq 10 \end{cases} \]
At \( x = 10 \), the function transitions from the first piece \( 2.5x \) (for \( x < 10 \)) to \( 2.25x \) (for \( x \geq 10 \)).
To determine the value of the function at \( x = 10 \):
- From the first part \( f(10) \) with \( x = 10 \) gives \( f(10) = 2.5 \times 10 = 25 \).
- From the second part \( f(10) \) gives \( f(10) = 2.25 \times 10 = 22.5 \).
Since the function changes at \( x = 10 \), the point \( (10, 25) \) would not be included in the graph (hence, the open circle), while \( (10, 22.5) \) would be included (solid circle) since \( f(10) = 22.5 \).
Thus, the point that has an open circle is \( (10, 25) \).