Answer:

Step-by-step explanation:
So, our equation is:

In exponetial form, this looks like:

Now lets cube the 5:

Next, we can multiply the denominator on the right side by 5:

We need to now get x alone by adding 4 to both sides:

Finally, we divide by x's coefficent, 2, to get:

Hope this helps! :3
Answer:
square with (-2,4) and (3,-1) vertices
Answer:
common ratio: 1.155
rate of growth: 15.5 %
Step-by-step explanation:
The model for exponential growth of population P looks like: 
where
is the population at time "t",
is the initial (starting) population
is the common ratio,
and
is the rate of growth
Therefore, in our case we can replace specific values in this expression (including population after 12 years, and initial population), and solve for the unknown common ratio and its related rate of growth:
![P(t)=P_i(1+r)^t\\13000=2300*(1+r)^{12}\\\frac{13000}{2300} = (1+r)^12\\\frac{130}{23} = (1+r)^{12}\\1+r=\sqrt[12]{\frac{130}{23} } =1.155273\\](https://tex.z-dn.net/?f=P%28t%29%3DP_i%281%2Br%29%5Et%5C%5C13000%3D2300%2A%281%2Br%29%5E%7B12%7D%5C%5C%5Cfrac%7B13000%7D%7B2300%7D%20%3D%20%281%2Br%29%5E12%5C%5C%5Cfrac%7B130%7D%7B23%7D%20%3D%20%281%2Br%29%5E%7B12%7D%5C%5C1%2Br%3D%5Csqrt%5B12%5D%7B%5Cfrac%7B130%7D%7B23%7D%20%7D%20%3D1.155273%5C%5C)
This (1+r) is the common ratio, that we are asked to round to the nearest thousandth, so we use: 1.155
We are also asked to find the rate of increase (r), and to express it in percent form. Therefore we use the last equation shown above to solve for "r" and express tin percent form:

So, this number in percent form (and rounded to the nearest tenth as requested) is: 15.5 %
The answer is 27 but I am not sure I hope my calculations ain't wrong
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: