Answer:
Cardiac output:
Step-by-step explanation:
Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.
To Find : Find the cardiac output.
Solution:
Formula of cardiac output:
---1
A = 3 mg
Do, integration by parts
![[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0](https://tex.z-dn.net/?f=%5B%5Cint%7B20te%5E%7B-0.6t%7D%7D%20%5C%2C%20dt%5D%5E%7B10%7D_0%3D%5B20t%5Cint%7Be%5E%7B-0.6t%7D%20%5C%2Cdt%7D-%5Cint%5B%5Cfrac%7Bd%5B20t%5D%7D%7Bdt%7D%5Cint%20%7Be%5E%7B-0.6t%7D%20%5C%2C%20dt%5Ddt%5D%5E%7B10%7D_0)
![[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0](https://tex.z-dn.net/?f=%5B%5Cint%7B20te%5E%7B-0.6t%7D%7D%20%5C%2C%20dt%5D%5E%7B10%7D_0%3D%5B%5Cfrac%7B-20te%5E%7B-0.6t%7D%7D%7B0.6%7D%2B%5Cfrac%7B20%7D%7B0.6%7D%5Cint%20%7Be%5E%7B-0.6t%7D%20%5C%2Cdt%5D%5E%7B10%7D_0)
![[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}](https://tex.z-dn.net/?f=%5B%5Cint%7B20te%5E%7B-0.6t%7D%7D%20%5C%2C%20dt%5D%5E%7B10%7D_0%3D%5B%5Cfrac%7B-20te%5E%7B-0.6t%7D%7D%7B0.6%7D%2B%5Cfrac%7B20e%5E%7B-0.6t%7D%7D%7B%280.6%29%5E2%7D%5D%5E%7B10%7D_%7B0%7D)
![[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}](https://tex.z-dn.net/?f=%5B%5Cint%7B20te%5E%7B-0.6t%7D%7D%20%5C%2C%20dt%5D%5E%7B10%7D_0%3D%5B%5Cfrac%7B-200e%5E%7B-6%7D%7D%7B0.6%7D%2B%5Cfrac%7B20e%5E%7B-6%7D%7D%7B%280.6%29%5E2%7D%5D%2B%5Cfrac%7B20%7D%7B%280.60%5E2%7D)
![[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}](https://tex.z-dn.net/?f=%5B%5Cint%7B20te%5E%7B-0.6t%7D%7D%20%5C%2C%20dt%5D%5E%7B10%7D_0%3D%5Cfrac%7B20%281-e%5E%7B-6%7D%7D%7B%280.6%29%5E2%7D-%5Cfrac%7B200e%5E%7B-6%7D%7D%7B0.6%7D)
![[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}](https://tex.z-dn.net/?f=%5B%5Cint%7B20te%5E%7B-0.6t%7D%7D%20%5C%2C%20dt%5D%5E%7B10%7D_0%5Csim%20%7B54.49%7D)
Substitute the value in 1
Cardiac output:
Cardiac output:
Hence Cardiac output:
Answer:
option 1
Step-by-step explanation:
(-2) (25/7)
-50/7 = -7 1/7
Part A.
The trip starts at 8am which corresponds to 0 hrs, point (0hr, 0mi)
2hrs later it's 10am. .point (2hr, 140mi)
The average speed is the slope between 0 and 2 hrs. Remember the slope formula m = Δy/Δx
m = (140 - 0) / (2 - 0)
m = 70 mph
Part B. Average speed from 11am - 2pm
11am is point (3hr, 140mi)
2pm is point (6hr, 300mi)
As you can see from the graph, the speed or slope changes at 1pm (5,260). You Can just use the start and end points.
m = (300-140) / (6-3)
m = 160/3
53.3 mph
* It comes out the same solution as if you average the two different slopes. 2hrs at 60mph + 1 hr at 40mph = (120 + 40)/3 = 160/3
Part C. Total average speed = total distance / total time driving
He went 70 mph for 2 hrs
stopped for an hour (slope is zero, no speed)
60 mph for 2hrs
40mph for 1 hr
300mi /5hr = 60mph
Part D. No Question....
