Answer :
To determine which choice can be defined using an exponential function, let's analyze both Choice A and Choice B step-by-step:
### Choice A:
- On January 1, you receive [tex]$0.10.
- On January 2, the amount doubles to $[/tex]0.20.
- On January 3, it doubles again to [tex]$0.40.
- On January 4, it doubles to $[/tex]0.80.
Choice A shows a pattern where the amount doubles each day. This type of growth can be defined by an exponential function because the amount increases by a constant factor (in this case, 2) each day.
To define this mathematically, we use an exponential function:
[tex]\[ A_t = 0.10 \times (2)^{t-1} \][/tex]
Here, [tex]\( A_t \)[/tex] is the amount on day [tex]\( t \)[/tex], [tex]\( 0.10 \)[/tex] is the starting amount on January 1, and [tex]\( (2)^{t-1} \)[/tex] represents the exponential growth factor.
### Choice B:
- On January 1, you receive [tex]$5.00.
- On January 2, you receive $[/tex]10.00 (which is [tex]$5.00 more than the previous day).
- On January 3, you receive $[/tex]15.00 (again [tex]$5.00 more than the previous day).
- The increase in amount is linear, as you are receiving an additional $[/tex]5.00 each day.
Choice B represents an arithmetic progression, which is a linear pattern of growth. The function for this growth can be defined as:
[tex]\[ A_t = 5.00 + 5.00(t-1) \][/tex]
Here, [tex]\( A_t \)[/tex] is the amount on day [tex]\( t \)[/tex], with a starting amount of [tex]$5.00 on January 1 and an additional $[/tex]5.00 added for each subsequent day.
### Conclusion:
- Choice A can be defined using an exponential function, is represented by [tex]\( A_t = 0.10 \times (2)^{t-1} \)[/tex].
- Choice B is a linear progression and is represented by a linear function which is [tex]\( A_t = 5.00 + 5.00(t-1) \)[/tex].
Therefore, the correct exponential function for Choice A is [tex]\( A_t = 0.10(2)^{t-1} \)[/tex].
### Choice A:
- On January 1, you receive [tex]$0.10.
- On January 2, the amount doubles to $[/tex]0.20.
- On January 3, it doubles again to [tex]$0.40.
- On January 4, it doubles to $[/tex]0.80.
Choice A shows a pattern where the amount doubles each day. This type of growth can be defined by an exponential function because the amount increases by a constant factor (in this case, 2) each day.
To define this mathematically, we use an exponential function:
[tex]\[ A_t = 0.10 \times (2)^{t-1} \][/tex]
Here, [tex]\( A_t \)[/tex] is the amount on day [tex]\( t \)[/tex], [tex]\( 0.10 \)[/tex] is the starting amount on January 1, and [tex]\( (2)^{t-1} \)[/tex] represents the exponential growth factor.
### Choice B:
- On January 1, you receive [tex]$5.00.
- On January 2, you receive $[/tex]10.00 (which is [tex]$5.00 more than the previous day).
- On January 3, you receive $[/tex]15.00 (again [tex]$5.00 more than the previous day).
- The increase in amount is linear, as you are receiving an additional $[/tex]5.00 each day.
Choice B represents an arithmetic progression, which is a linear pattern of growth. The function for this growth can be defined as:
[tex]\[ A_t = 5.00 + 5.00(t-1) \][/tex]
Here, [tex]\( A_t \)[/tex] is the amount on day [tex]\( t \)[/tex], with a starting amount of [tex]$5.00 on January 1 and an additional $[/tex]5.00 added for each subsequent day.
### Conclusion:
- Choice A can be defined using an exponential function, is represented by [tex]\( A_t = 0.10 \times (2)^{t-1} \)[/tex].
- Choice B is a linear progression and is represented by a linear function which is [tex]\( A_t = 5.00 + 5.00(t-1) \)[/tex].
Therefore, the correct exponential function for Choice A is [tex]\( A_t = 0.10(2)^{t-1} \)[/tex].
