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2 years ago

14

A patient is instructed to take three 50-mcg tablets of pergolide mesylate (Permax) daily. How many milligrams of the drug would

the patient receive weekly?

1 answer:

Black_prince [1.1K]2 years ago

6 0

Answer:

The patient would receive 1.05mg of the drug weekly.

Step-by-step explanation:

First step: How many mcg of the drug would the patient receive daily?

The problem states that he takes three doses of 50-mcg a day. So

1 dose - 50mcg

3 doses - x mcg

x = 50*3

x = 150 mcg.

He takes 150mcg of the drug a day.

Second step: How many mcg of the drug would the patient receive weekly?

A week has 7 days. He takes 150mcg of the drug a day. So:

1 day - 150mcg

7 days - x mcg

x = 150*7

x = 1050mcg

He takes 1050mcg of the drug a week.

Final step: Conversion of 1050 mcg to mg

Each mg has 1000 mcg. How many mg are there in 1050 mcg? So

1mg - 1000 mcg

xmg - 1050mcg

1000x = 1050

$x = \frac{1050}{1000}$

x = 1.05mg

The patient would receive 1.05mg of the drug weekly.

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andy picked 5 tulips.brianna picked 3 more tulips than andy , carley picked 16 tulips. how many times as many tulips did carley

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The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

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The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

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