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,
Step-by-step explanation:
-2, -8/3, -10/3, -4, -14/3
Write as multiples of 1/3.
-6/3, -8/3, -10/3, -12/3, -14/3
This is an arithmetic sequence where the first term is -6/3 and the common difference is -2/3.
Therefore, the recursive formula is:
aᵢ₊₁ = aᵢ − 2/3, a₁ = -2
Answer:
<h3>(1,1) and (2.5)=3.6</h3>
Step-by-step explanation:
i think it is 3.6
Answer:
y-intercept is (0, 2) and the slope is 1