Step-by-step explanation:
a. We have v as the amount of drug(mg) in the blood at time t (hrs)
b. Flow rate in = k litre/he
For a circulatory system we have the flow rate in and flow rate out to be the same
Therefore,
Flow rate out = k litre/hr
concentration of drug going in = 0mg/litre
Concentration of drug going out = v mg/litre
V(0) = 20mg since 20mg was taken at midnight
V(0) = 20
Half life t1/2 = 36 hours
V(36) = v(0)/2
= 20/2
= 10mg
C. ![ivp = \frac{dv}{dt}](https://tex.z-dn.net/?f=ivp%20%3D%20%5Cfrac%7Bdv%7D%7Bdt%7D)
![= -kv](https://tex.z-dn.net/?f=%3D%20-kv)
![v(0) = 20mg\\v(36) = 10mg](https://tex.z-dn.net/?f=v%280%29%20%3D%2020mg%5C%5Cv%2836%29%20%3D%2010mg)
d. solution
![\frac{dv}{dt} = -kt\\ln(v) = ln(c) - kt](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%20%20%3D%20-kt%5C%5Cln%28v%29%20%3D%20ln%28c%29%20-%20kt)
![\frac{v}{c} = e^{-kt} = v=ce^{-kt}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%7D%7Bc%7D%20%3D%20e%5E%7B-kt%7D%20%20%3D%20v%3Dce%5E%7B-kt%7D)
v(0) = 20
![ce^{-k(0)} =20](https://tex.z-dn.net/?f=ce%5E%7B-k%280%29%7D%20%3D20)
c = 20
so
![v(t) = 20e^{-kt}](https://tex.z-dn.net/?f=v%28t%29%20%3D%2020e%5E%7B-kt%7D)
e.
![t\frac{1}{2}=36hours\\ v(36)= 10](https://tex.z-dn.net/?f=t%5Cfrac%7B1%7D%7B2%7D%3D36hours%5C%5C%20v%2836%29%3D%2010)
![10 = 20e^{-36k}](https://tex.z-dn.net/?f=10%20%3D%2020e%5E%7B-36k%7D)
![\frac{1}{2} =e^{-36k}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%3De%5E%7B-36k%7D)
we take log
![k=\frac{ln(2)}{36}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Bln%282%29%7D%7B36%7D)
please check attachment for answer f
Answer:
Refractive index of a medium depends upon the refractive index of the surroundings
Answer:
He received $6.40 in change.
Explanation:
Since he bought 2.25 pounds, you multiply that by 1.6 to get a sum of 3.6. Since he paid with a ten dollar bill, and he only has to pay 3.60, they will give him back 6.40 in change.
Hope this helped :)
Answer:
Step-by-step explanation:
Given the approximate demand function of night drink expressed as;
p^2+200q^2=177,
p is the price (in dollars) and;
q is the quantity demanded (in thousands).
Given
p = $7
q = 800
Required
dq/dp
Differentiating the function implicitly with respect to p shown;
2p + 400d dq/dp = 0
400q dq/dp = -2p
200qdq/dp = -p
dq/dp = -p/100q
substitute p and q into the resulting equation;
dq/dp = -7/100(800)
dq/dp = -7/80000
dq/dp = -0.0000875
This means that the rate of change of quantity demanded with respect to the price is -0.0000875