Just use the coefficients for each variable and put it in a matrix. Is the second question set equal to something? Because then you would put it in place as the question mark. (Did the pic come through?)
So the 3x3 matrix is made up of the coefficients and then to augment the matrix, you add another column at the end with the solutions. Good luck!
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:
Is the question how long would the 185 petri dishes last because if so it would be about 15 days
Hope this helped ;)
Answer:
The new can can hold upto volume of 476.69 cubic centimeters.
Step-by-step explanation:
Diameter of the can = d = 6 cm
Radius of the can = r = 0.5d = 0.5 × 6 cm = 3 cm
Height of the can = h = 12 cm
Company wants to increase the dimensions of their cans by a multiple of 1.12. So, the new dimension will be:
r'= 3 cm × 1.12 = 3.36 cm
h' = 12 cm × 1.12 =13.44 cm
Volume of the = can = V
Volume of a cylinder = 
The new can can hold upto volume of 476.69 cubic centimeters.
