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All Common Core: High School - Functions Resources

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Flashcards: Growth and Decay by Contant Percent Rate: CCSS.Math.Content.HSF-LE.A.1c

An anti-diabetic drug, has a half-life of about hours. If a patient was administered $200 mg of the drug at $8:00 am , how much is left at $1:00 pm ?

Note: The half life formula is

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

$A=82\textup{ mg}$

$A=90\textup{ mg}$

$A=84\textup{ mg}$

$A=94\textup{ mg}$

$A=184\textup{ mg}$

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Flashcards: Growth and Decay by Contant Percent Rate: CCSS.Math.Content.HSF-LE.A.1c

An anti-diabetic drug, has a half-life of about hours. If a patient was administered $200 mg of the drug at $8:00 am , how much is left at $1:00 pm ?

Note: The half life formula is

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

Your answer:

$A=82\textup{ mg}$

Your answer:

$A=90\textup{ mg}$

Your answer:

$A=94\textup{ mg}$

Your answer:

$A=184\textup{ mg}$

Correct answer:

$A=84\textup{ mg}$

Explanation:

This question is testing one's ability to recognize real life situations that have a exponential growth or decay over a certain interval and how to deal with them in function form.

For the purpose of Common Core Standards, recognize situations that have a exponential growth or decay over a certain interval, falls within the Cluster A of construct and compare linear, quadratic, and exponential model and solve problems concept (CCSS.Math.content.HSF.LE.A).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify the known values given in the question.

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

$\\A_0=\textup{Intial Amount}=200\textup{ mg} \\h=\textup{half life}=4\textup{ hours} \\t_f=1:00\textup{ pm} \\t_i=8:00\textup{ am} \\A=?$

Step 2: Calculate .

$1:00-8:00\rightarrow 5 \textup{ hours }$

Step 3: Substitute in known values into the half life formula to solve for .

$A=200\textup{ mg}\cdot \left(\frac{1}{2}\right)^{\frac{5}{4}}$

$A=200\textup{ mg}\cdot \left(\frac{1}{2}\right)^{1.25}$

$A=200\textup{ mg}\cdot 0.420$

$A=84\textup{ mg}$

Flashcards: Growth and Decay by Contant Percent Rate: CCSS.Math.Content.HSF-LE.A.1c

An anti-diabetic drug, has a half-life of about 6.2 hours. If a patient was administered $500 mg of the drug at $8:21 am , how much is left at $4:34 pm ?

Note: The half life formula is

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

$A=197.57\textup{ mg}$

$A=128.57\textup{ mg}$

$A=159.57\textup{ mg}$

$A=199.57\textup{ mg}$

$A=199.99\textup{ mg}$

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Flashcards: Growth and Decay by Contant Percent Rate: CCSS.Math.Content.HSF-LE.A.1c

An anti-diabetic drug, has a half-life of about 6.2 hours. If a patient was administered $500 mg of the drug at $8:21 am , how much is left at $4:34 pm ?

Note: The half life formula is

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

Your answer:

$A=197.57\textup{ mg}$

Your answer:

$A=128.57\textup{ mg}$

Your answer:

$A=159.57\textup{ mg}$

Your answer:

$A=199.99\textup{ mg}$

Correct answer:

$A=199.57\textup{ mg}$

Explanation:

This question is testing one's ability to recognize real life situations that have a exponential growth or decay over a certain interval and how to deal with them in function form.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify the known values given in the question.

$A=A_0\cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}$

$\\A_0=\textup{Intial Amount}=500\textup{ mg} \\h=\textup{half life}=6.2\textup{ hours} \\t_f=4:34\textup{pm} \\t_i=8:21\textup{ am} \\A=?$

Step 2: Calculate .

$4:34-8:21\rightarrow 8 \textup{ hours }13\textup{ minutes}$

Recall that there are sixty minutes in an hour therefore,

$8\textup{ hours }13\textup{ minutes}=8+\frac{13}{60}=8.216\overline{6}$

Step 3: Substitute in known values into the half life formula to solve for .

$A=500\textup{ mg}\cdot \left(\frac{1}{2}\right)^{\frac{8.216\overline{6}}{6.2}}$

$A=500\textup{ mg}\cdot \left(\frac{1}{2}\right)^{1.325}$

$A=500\textup{ mg}\cdot 0.399$

$A=199.57\textup{ mg}$

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All Common Core: High School - Functions Resources

Flashcards: Growth and Decay by Contant Percent Rate: CCSS.Math.Content.HSF-LE.A.1c

Flashcards: Growth and Decay by Contant Percent Rate: CCSS.Math.Content.HSF-LE.A.1c

Flashcards: Growth and Decay by Contant Percent Rate: CCSS.Math.Content.HSF-LE.A.1c

Flashcards: Growth and Decay by Contant Percent Rate: CCSS.Math.Content.HSF-LE.A.1c

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