Answer:
a. dQ/dt = -kQ
b.
c. k = 0.178
d. Q = 1.063 mg
Step-by-step explanation:
a) Write a differential equation for the quantity Q of hydrocodone bitartrate in the body at time t, in hours, since the drug was fully absorbed.
Let Q be the quantity of drug left in the body.
Since the rate of decrease of the quantity of drug -dQ/dt is directly proportional to the quantity of drug left, Q then
-dQ/dt ∝ Q
-dQ/dt = kQ
dQ/dt = -kQ
This is the required differential equation.
b) Solve your differential equation, assuming that at the patient has just absorbed the full 9 mg dose of the drug.
with t = 0, Q(0) = 9 mg
dQ/dt = -kQ
separating the variables, we have
dQ/Q = -kdt
Integrating we have
∫dQ/Q = ∫-kdt
㏑Q = -kt + c
when t = 0, Q = 9
So,
c) Use the half-life to find the constant of proportionality k.
At half-life, Q = 9/2 = 4.5 mg and t = 3.9 hours
So,
taking natural logarithm of both sides, we have
d) How much of the 9 mg dose is still in the body after 12 hours?
Since k = 0.178,
when t = 12 hours,
Answer:
x = 87
Step-by-step explanation:
37 + 56 = 93
triangles always equal to 180
180 - 93 = 87
Step-by-step explanation:
2x + 3 y = 6 by 2
4x - 6y = 3
____________
4x+ 6 y = 12
4x - 6 y = 3
_____________
12 y = 9
y= 12/ 9 = 4 / 3
2X+ 3( 4 / 3 ) = 6
X = I
Answer:
A. 16 3/4
Step-by-step explanation:
it just is
Answer:
-2.1y + 12.9
Step-by-step explanation: