Let's analyze this problem.

Given Information:

1. Toothbrush A is described by the equation: $y = 78x$ where $y$ represents the number of rotations after $x$ seconds. This implies that Toothbrush A has a constant rotation rate of 78 rotations per second.

2. Toothbrush B is represented by a graph showing the number of rotations over time.

Analyzing Toothbrush B’s Rate:

Looking at the graph for Toothbrush B:

• At $x = 1$ second, the number of rotations $y$ is about 50.
• At $x = 2$ seconds, $y$ is around 100 rotations.

From these two points, we can calculate the rate of Toothbrush B as: $\text{Rate} = \frac{\text{Change in rotations}}{\text{Change in time}} = \frac{100 - 50}{2 - 1} = 50 \text{ rotations per second}$

Comparing Rates:

• Toothbrush A rotates at 78 rotations per second.
• Toothbrush B rotates at 50 rotations per second.

Since 78 > 50, Toothbrush A has a higher rate.

Correct Answer:

The best answer is:

• Option A: "Toothbrush A, because its unit rate is greater by 3 rotations per second."