Answer:
0 ≤ t ≤ 5.
Step-by-step explanation:
In the function 
, 
is the independent variable. The domain of 
is the set of all values of that this function can accept.
In this case, is defined in a real-life context. Hence, consider the real-life constraints on the two variables. Both time and volume should be non-negative. In other words,
.
.
The first condition is an inequality about , which is indeed the independent variable.
However, the second condition is about , the dependent variable of this function. It has to be rewritten as a condition about .
![\begin{aligned} f(t) &\ge 0 &&\text{Assumption} \cr 80000 - 16000\, t& \ge 0 && \text{Definition of} ~ f \cr 80000 & \ge 16000\, t && \begin{aligned}&\text{Add $16000\, t$} \\[-0.5em] & \text{to both sides of the inequality}\end{aligned} \cr 5 &\ge t &&\begin{aligned}&\text{Divide both sides of} \\[-0.5em] & \text{the inequality by $16000$}\end{aligned} \cr t &\le 5 && \text{Flip the inequality}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20f%28t%29%20%26%5Cge%200%20%26%26%5Ctext%7BAssumption%7D%20%5Ccr%2080000%20-%2016000%5C%2C%20t%26%20%5Cge%200%20%26%26%20%5Ctext%7BDefinition%20of%7D%20~%20f%20%5Ccr%2080000%20%26%20%5Cge%2016000%5C%2C%20t%20%26%26%20%5Cbegin%7Baligned%7D%26%5Ctext%7BAdd%20%2416000%5C%2C%20t%24%7D%20%5C%5C%5B-0.5em%5D%20%26%20%5Ctext%7Bto%20both%20sides%20of%20the%20inequality%7D%5Cend%7Baligned%7D%20%5Ccr%205%20%26%5Cge%20t%20%26%26%5Cbegin%7Baligned%7D%26%5Ctext%7BDivide%20both%20sides%20of%7D%20%5C%5C%5B-0.5em%5D%20%26%20%5Ctext%7Bthe%20inequality%20by%20%2416000%24%7D%5Cend%7Baligned%7D%20%5Ccr%20t%20%26%5Cle%205%20%26%26%20%5Ctext%7BFlip%20the%20inequality%7D%5Cend%7Baligned%7D)
.
Hence, t ≤ 5.
Combine the two inequalities to obtain the domain:
0 ≤ t ≤ 5.
Answer:
x = -3
Step-by-step explanation:
The graph is symmetrical about the vertical line through its vertex:
x = -3
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:
the range is 18 and the median is 8
Step-by-step explanation:
