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Common Core: High School - Functions : Growth and Decay by Contant Percent Rate: CCSS.Math.Content.HSF-LE.A.1c

Study concepts, example questions & explanations for Common Core: High School - Functions

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All Common Core: High School - Functions Resources

6 Diagnostic Tests 82 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #31 : Linear, Quadratic, & Exponential Models*

An particular medicine, has a half-life of about $\displaystyle 4$ hours. If a patient was administered $600 mg $\displaystyle $600 mg$ of the drug at $\displaystyle $9:00 am$, how much is left at $\displaystyle $2:00 pm$?

Note: The half life formula is

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

Possible Answers:

$A=250.24\textup{ mg}$ $\displaystyle A=250.24\textup{ mg}$

$A=252.44\textup{ mg}$ $\displaystyle A=252.44\textup{ mg}$

$A=252.24\textup{ mg}$ $\displaystyle A=252.24\textup{ mg}$

$A=254.24\textup{ mg}$ $\displaystyle A=254.24\textup{ mg}$

$A=251.24\textup{ mg}$ $\displaystyle A=251.24\textup{ mg}$

Correct answer:

$A=252.24\textup{ mg}$ $\displaystyle A=252.24\textup{ mg}$

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.

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

$\\A_0=\textup{Intial Amount}=600\textup{ mg} \\h=\textup{half life}=4\textup{ hours} \\t_f=2:00\textup{ pm} \\t_i=9:00\textup{ am} \\A=?$ $\displaystyle \\A_0=\textup{Intial Amount}=600\textup{ mg} \\h=\textup{half life}=4\textup{ hours} \\t_f=2:00\textup{ pm} \\t_i=9:00\textup{ am} \\A=?$

Step 2: Calculate $\displaystyle t$.

$2:00-9:00\rightarrow 5 \textup{ hours }$ $\displaystyle 2:00-9:00\rightarrow 5 \textup{ hours }$

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

$A=600\textup{ mg}\cdot \left(\frac{1}{2}\right)^{\frac{5}{4}}$ $\displaystyle A=600\textup{ mg}\cdot \left(\frac{1}{2}\right)^{\frac{5}{4}}$

$A=600\textup{ mg}\cdot \left(\frac{1}{2}\right)^{1.25}$ $\displaystyle A=600\textup{ mg}\cdot \left(\frac{1}{2}\right)^{1.25}$

$A=600\textup{ mg}\cdot 0.4204$ $\displaystyle A=600\textup{ mg}\cdot 0.4204$

$A=252.24\textup{ mg}$ $\displaystyle A=252.24\textup{ mg}$

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Example Question #32 : Linear, Quadratic, & Exponential Models*

An particular medicine, has a half-life of about 5.8 $\displaystyle 5.8$ hours. If a patient was administered $400 mg $\displaystyle $400 mg$ of the drug at $\displaystyle $9:00 am$, how much is left at $\displaystyle $2:00 pm$?

Note: The half life formula is

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

Possible Answers:

$A=220.08\textup{ mg}$ $\displaystyle A=220.08\textup{ mg}$

$A=222.08\textup{ mg}$ $\displaystyle A=222.08\textup{ mg}$

$A=210.08\textup{ mg}$ $\displaystyle A=210.08\textup{ mg}$

$A=228.02\textup{ mg}$ $\displaystyle A=228.02\textup{ mg}$

$A=218.18\textup{ mg}$ $\displaystyle A=218.18\textup{ mg}$

Correct answer:

$A=220.08\textup{ mg}$ $\displaystyle A=220.08\textup{ mg}$

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.

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

$\\A_0=\textup{Intial Amount}=400\textup{ mg} \\h=\textup{half life}=5.8\textup{ hours} \\t_f=2:00\textup{ pm} \\t_i=9:00\textup{ am} \\A=?$ $\displaystyle \\A_0=\textup{Intial Amount}=400\textup{ mg} \\h=\textup{half life}=5.8\textup{ hours} \\t_f=2:00\textup{ pm} \\t_i=9:00\textup{ am} \\A=?$

Step 2: Calculate $\displaystyle t$.

$2:00-9:00\rightarrow 5 \textup{ hours }$ $\displaystyle 2:00-9:00\rightarrow 5 \textup{ hours }$

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

$A=400\textup{ mg}\cdot \left(\frac{1}{2}\right)^{\frac{5}{5.8}}$ $\displaystyle A=400\textup{ mg}\cdot \left(\frac{1}{2}\right)^{\frac{5}{5.8}}$

$A=400\textup{ mg}\cdot \left(\frac{1}{2}\right)^{0.8621}$ $\displaystyle A=400\textup{ mg}\cdot \left(\frac{1}{2}\right)^{0.8621}$

$A=400\textup{ mg}\cdot 0.5502$ $\displaystyle A=400\textup{ mg}\cdot 0.5502$

$A=220.08\textup{ mg}$ $\displaystyle A=220.08\textup{ mg}$

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All Common Core: High School - Functions Resources

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Common Core: High School - Functions : Growth and Decay by Contant Percent Rate: CCSS.Math.Content.HSF-LE.A.1c

Study concepts, example questions & explanations for Common Core: High School - Functions

All Common Core: High School - Functions Resources

Example Questions

Example Question #31 : Linear, Quadratic, & Exponential Models*

Example Question #32 : Linear, Quadratic, & Exponential Models*

All Common Core: High School - Functions Resources

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