The graph below shows your speed at different times while riding a bicycle.

A coordinate graph is shown. Horizontal axis is labeled Time from the Start in minutes. Vertical axis is labeled Speed in kilometers per hour.
The first line segment starts at 0 comma 5 and ends at 4 comma 9.
The second line segment starts at 4 comma 9 and ends at 7 comma 7.
The third line segment starts at 7 comma 7 and ends at 10 comma 7.

For how many minutes did your speed remain constant?
A. 3
B. 4
C. 7
D. 10

Your speed remained constant for 4 minutes between 4 and 7 on the horizontal axis. Therefore, the answer is B. 4.

Well, I'm not the fastest bicycle rider out there, but I can still figure this one out! Looking at the graph, it seems like your speed remained constant for a total of 7 minutes. So, I would have to go with option C, 7 minutes. Keep pedaling!

To find the answer, we need to identify the segments where the speed remains constant.

The first line segment starts at (0, 5) and ends at (4, 9). In this segment, the speed increases from 5 km/h to 9 km/h.

The second line segment starts at (4, 9) and ends at (7, 7). In this segment, the speed decreases from 9 km/h to 7 km/h.

The third line segment starts at (7, 7) and ends at (10, 7). In this segment, the speed remains constant at 7 km/h.

Therefore, the speed remains constant for 3 minutes during the third line segment.

So, the correct answer is A. 3.

To determine for how many minutes your speed remained constant, we need to identify the line segments on the graph where the vertical values (speed) are the same.

Looking at the given graph, we can see that there are two line segments where the speed remains constant: the second line segment (from 4,9 to 7,7) and the third line segment (from 7,7 to 10,7).

The second line segment has a speed of 9 km/hr for a duration of 7 - 4 = 3 minutes.
The third line segment has a speed of 7 km/hr for a duration of 10 - 7 = 3 minutes.

Therefore, your speed remained constant for a total of 3 + 3 = 6 minutes.

Answer: The correct option is B. 4