Respuesta :
From the given information, there are some missing tables and the correct format is outlined in the table below.
When the data in the table were graphed: the slope (tangent line) and the time in minutes are represented as follows:
t(mins) 36 38 40 42 44
heartbeats 2524 2661 2880 2934 3067
Let us assume that P refers to the point (42, 2948) which is located on the graph of heartbeats. Then, the objective is to find the slope of the secant line PQ.
where;
- Q is that point located on the graph for t = 36, 38, 40, 44
∴
For the first slope, where P = (42, 2948) & Q = (36,2524)
The slope;
[tex]\mathbf{\Delta S = \dfrac{2524 - 2948}{36 - 42}}[/tex]
ΔS = 70.667
For the 2nd slope, where P = (42, 2948) & Q = (38,2661)
The slope;
[tex]\mathbf{\Delta S = \dfrac{2661-2948}{38 - 42}}[/tex]
ΔS = 71.75
For the 3rd slope, where P = (42, 2948) & Q = (40,2880)
The slope;
[tex]\mathbf{\Delta S = \dfrac{2880-2948}{40 - 42}}[/tex]
ΔS = 34.00
For the 4th slope, where P = (42, 2948) & Q = (44, 3067)
The slope;
[tex]\mathbf{\Delta S = \dfrac{3067-2948}{44 - 42}}[/tex]
ΔS = 59.50
Thus, the points and the slope of the secant lines can be computed as:
Points (36,2524) (38,2661) (40,2880) (44, 3067)
Slope of secant lines 70.7 71.8 34.0 59.5
Therefore, we can conclude that the cardiac monitor that is being used to measure the heart rate of the patient after surgery starts to decrease and reduce from 71.8 to 59.5 after being stable for a while. This shows that the heart rate is dropping after 42 minutes.
Learn more about cardiac monitor here:
https://brainly.com/question/4179914?referrer=searchResults