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gayaneshka [121]
2 years ago
13

Write the decimal as a fraction 0.272727

Mathematics
1 answer:
Oksana_A [137]2 years ago
7 0
.272727 written as a fraction is 272727/1000000
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An item is on sale for 50% off the regular price of $3.18. A customer buys one and gives the cashier $5.00. The cashier gives th
monitta
The regular price is 3.18.
 50% off half the price, so you can either divide the regular price by 2 or multiply it by 0.50

3.18 / 2 = 1.59   ( 3.18 *0.50 = 1.59)

so the sale price is 1.59

now subtract the sale price from $5 to find the amount of change they got back:

5.00 - 1.59 = $3.14 change back
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3 years ago
Consider the function f(x) = (x + 2)2 + 1. Which of the following functions shifts the graph of f(x) to the right three units?
lakkis [162]

Answer:

g(x) = (x – 1)2 + 1

Step-by-step explanation:

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2 years ago
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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
It's geometry please help me
Gekata [30.6K]

Answer:

x=13

Step-by-step explanation:

C is the midpoint of line BD, BC=CD. 6x-16=4x+10, x=13

6 0
3 years ago
SLOPE AND PARALLELISM
Lostsunrise [7]
The answer is B maybe try it?
6 0
3 years ago
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