Answer:
Step-by-step explanation:
im not sure if it is right,but here is my thought since its gonna be half first you do the 4x2x2=16 divide 16 by 2 witch will give you have of 16 witch will also give you 8
Answer:
C
Step-by-step explanation:
a = 4
b = 5
c = 2
C = arccos((a² + b² - c²) / 2ab)
C = arccos((16 + 25 - 4) / 2(4)(5))
C = arccos(37 / 40)
C = 22.33°
2abcosC
2(4)(5)cos(22.33)
40(0.925)
37
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:
DO IT ON YOUR OWN
Step-by-step explanation:
KIDDING UM CAN WE BE FRIENDS
AND TRUE
Answer:
y=x/2+1/2
Step-by-step explanation: