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Reptile [31]
3 years ago
5

How to solve 0.225555555555 into a fraction mental math strategy​

Mathematics
1 answer:
Naddika [18.5K]3 years ago
6 0

Answer:

Convert to a decimal by dividing the numerator by the denominator.

4.43349753 just easy

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9.423 rounded to the nearest one is 9

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Find sin A <br>a 4/5<br>b 5/4<br>c 3/5<br>d 5/3<br>e 3/4<br>f 4/3​
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Read 2 more answers
Hydrocodone bitartrate is used as a cough suppressant. After the drug is fully absorbed, the quantity of drug in the body decrea
maw [93]

Answer:

a. dQ/dt = -kQ

b. Q = 9e^{-kt}

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

Q = e^{-kt + c}\\Q = e^{-kt}e^{c}\\Q = Ae^{-kt}               (A = e^{c})

when t = 0, Q = 9

Q = Ae^{-kt}               \\9 = Ae^{-k0}\\9 = Ae^{0}\\9 = A\\A = 9

So, Q = 9e^{-kt}

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,

Q = 9e^{-kt}\\4.5 = 9e^{-kX3.9}\\\frac{4.5}{9} = e^{-kX3.9}\\\frac{1}{2} = e^{-3.9k}\\

taking natural logarithm of both sides, we have

ln\frac{1}{2} = ln(e^{-3.9k})\\\\-ln2 = -3.9k\\k = -ln2/-3,9 \\k = -0.693/-3.9\\k = 0.178

d) How much of the 9 mg dose is still in the body after 12 hours?

Since k = 0.178,

Q = 9e^{-0.178t}

when t = 12 hours,

Q = 9e^{-0.178t}\\Q = 9e^{-0.178X12}\\Q = 9e^{-2.136}\\Q = 9 X 0.1181\\Q = 1.063 mg

7 0
3 years ago
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