Does this help ? /////////////
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:
The numbers are 20 and - 13
Step-by-step explanation:
Given as ;
The sum of two numbers = 7
Two times a number and three times another number = 13
So ,Let The first number = x
and The second number = y
According to question
x + y = 7 and ...A
2 × x + 3 × y = 1 ....B
Multiply eq A by 2
I.e 2 × ( x + y ) = 2 × 7
Or, 2 × x + 2 × y = 14
Now solve the both equation
( 2 × x + 3 × y ) - ( 2 × x + 2 × y ) = 1 - 14
Or. ( 2 x - 2 x ) + ( 3 y - 2 y ) = - 13
Or, 0 + y = - 13
∴ y = - 13
Put The value of y in eq A
x + y = 7
Or , x = 7 - y
Or, x = 7 - ( - 13 )
so, x = 7 + 13
∴ x =20
Hence The numbers are 20 and - 13 Answer
✓ –28 in simplest radical form is 27!!!
Answer:
|−3| < |−4|
Step-by-step explanation: