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Answer :
Sure! Let's analyze the problem step by step to determine which choice represents an exponential function.
Choice A:
In Choice A, you start with [tex]$0.10 on January 1st, and the amount doubles each successive day. The sequence of payments on each day goes:
- January 1: $[/tex]0.10
- January 2: [tex]$0.20 (which is 0.10 * 2)
- January 3: $[/tex]0.40 (which is 0.20 2)
- January 4: [tex]$0.80 (which is 0.40 2)
- and so on.
The pattern here is that each day's amount is double the previous day's amount. This type of growth is called exponential growth because the amount is multiplied by a constant factor (in this case, 2) each day.
This can be expressed with the formula:
\[ A_t = 0.10 \times (2)^{t-1} \]
where \( t \) is the day number (e.g., \( t = 1 \) for January 1).
Choice B:
In Choice B, you start with $[/tex]5.00 on the first day, and then the amount increases by exactly [tex]$5.00 every subsequent day. The series of payments looks like this:
- January 1: $[/tex]5.00
- January 2: [tex]$10.00
- January 3: $[/tex]15.00
- January 4: [tex]$20.00
- and so on.
This pattern is linear because each term increases by a constant addition, which is $[/tex]5.00 every day. This type of change is expressed by a linear function.
The formula for Choice B is:
[tex]\[ A_t = 5.00 + 5.00(t-1) \][/tex]
where [tex]\( t \)[/tex] is the day number.
Conclusion:
- Choice A is defined by an exponential function: [tex]\( A_t = 0.10 \times (2)^{t-1} \)[/tex].
- Choice B is defined by a linear function: [tex]\( A_t = 5.00 + 5.00(t-1) \)[/tex].
So, the choice that is defined using an exponential function is Choice A, with the function [tex]\( A_t = 0.10 \times (2)^{t-1} \)[/tex].
Choice A:
In Choice A, you start with [tex]$0.10 on January 1st, and the amount doubles each successive day. The sequence of payments on each day goes:
- January 1: $[/tex]0.10
- January 2: [tex]$0.20 (which is 0.10 * 2)
- January 3: $[/tex]0.40 (which is 0.20 2)
- January 4: [tex]$0.80 (which is 0.40 2)
- and so on.
The pattern here is that each day's amount is double the previous day's amount. This type of growth is called exponential growth because the amount is multiplied by a constant factor (in this case, 2) each day.
This can be expressed with the formula:
\[ A_t = 0.10 \times (2)^{t-1} \]
where \( t \) is the day number (e.g., \( t = 1 \) for January 1).
Choice B:
In Choice B, you start with $[/tex]5.00 on the first day, and then the amount increases by exactly [tex]$5.00 every subsequent day. The series of payments looks like this:
- January 1: $[/tex]5.00
- January 2: [tex]$10.00
- January 3: $[/tex]15.00
- January 4: [tex]$20.00
- and so on.
This pattern is linear because each term increases by a constant addition, which is $[/tex]5.00 every day. This type of change is expressed by a linear function.
The formula for Choice B is:
[tex]\[ A_t = 5.00 + 5.00(t-1) \][/tex]
where [tex]\( t \)[/tex] is the day number.
Conclusion:
- Choice A is defined by an exponential function: [tex]\( A_t = 0.10 \times (2)^{t-1} \)[/tex].
- Choice B is defined by a linear function: [tex]\( A_t = 5.00 + 5.00(t-1) \)[/tex].
So, the choice that is defined using an exponential function is Choice A, with the function [tex]\( A_t = 0.10 \times (2)^{t-1} \)[/tex].
Thanks for taking the time to read You win a prize and are offered two choices Choice A tex 0 10 tex on January 1 tex 0 20 tex on January 2. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
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