Answer :
To determine which choice can be defined using an exponential function and to find the corresponding function for each choice, let's analyze the given choices.
### Choice A:
The prize amount starts at \[tex]$0.10 on January 1st. Each subsequent day, the amount doubles.
- On January 1: \$[/tex]0.10
- On January 2: \[tex]$0.20 (which is \$[/tex]0.10 2)
- On January 3: \[tex]$0.40 (which is \$[/tex]0.20 2 or \[tex]$0.10 * 2^2)
- On January 4: \$[/tex]0.80 (which is \[tex]$0.40 2 or \$[/tex]0.10 2^3)
This pattern shows exponential growth, where the amount is multiplied by 2 each day.
The general form of an exponential function is:
[tex]\[ A(t) = A_0 \times (r)^{t-1} \][/tex]
where:
- [tex]\( A(t) \)[/tex] is the amount on day t,
- [tex]\( A_0 \)[/tex] is the initial amount,
- [tex]\( r \)[/tex] is the common ratio (in this case, 2),
- [tex]\( t \)[/tex] is the day number.
For Choice A:
- The initial amount, [tex]\( A_0 \)[/tex], is \[tex]$0.10.
- The common ratio, \( r \), is 2.
Thus, the function for Choice A is:
\[ A_t = 0.10 \times (2)^{t-1} \]
### Choice B:
The prize amount starts at \$[/tex]5.00 on the first day. Each subsequent day, the amount increases by \[tex]$5.00.
- On January 1: \$[/tex]5.00
- On January 2: \[tex]$10.00 (which is \$[/tex]5.00 + \[tex]$5.00)
- On January 3: \$[/tex]15.00 (which is \[tex]$10.00 + \$[/tex]5.00 or \[tex]$5.00 + 2 \times \$[/tex]5.00)
- On January 4: \[tex]$20.00 (which is \$[/tex]15.00 + \[tex]$5.00 or \$[/tex]5.00 + 3 \times \[tex]$5.00)
This pattern shows a linear growth, where the amount increases by a fixed sum each day.
The general form of a linear function is:
\[ A(t) = A_1 + d \times (t-1) \]
where:
- \( A(t) \) is the amount on day t,
- \( A_1 \) is the initial amount,
- \( d \) is the daily increase.
For Choice B:
- The initial amount, \( A_1 \), is \$[/tex]5.00.
- The daily increase, [tex]\( d \)[/tex], is \$5.00.
Thus, the function for Choice B is:
[tex]\[ A_t = 5.00 + 5.00(t-1) \][/tex]
### Summary:
- Choice A can be defined using an exponential function: [tex]\( A_t = 0.10 \times (2)^{t-1} \)[/tex]
- Choice B can be defined using a linear function: [tex]\( A_t = 5.00 + 5.00(t-1) \)[/tex]
### Choice A:
The prize amount starts at \[tex]$0.10 on January 1st. Each subsequent day, the amount doubles.
- On January 1: \$[/tex]0.10
- On January 2: \[tex]$0.20 (which is \$[/tex]0.10 2)
- On January 3: \[tex]$0.40 (which is \$[/tex]0.20 2 or \[tex]$0.10 * 2^2)
- On January 4: \$[/tex]0.80 (which is \[tex]$0.40 2 or \$[/tex]0.10 2^3)
This pattern shows exponential growth, where the amount is multiplied by 2 each day.
The general form of an exponential function is:
[tex]\[ A(t) = A_0 \times (r)^{t-1} \][/tex]
where:
- [tex]\( A(t) \)[/tex] is the amount on day t,
- [tex]\( A_0 \)[/tex] is the initial amount,
- [tex]\( r \)[/tex] is the common ratio (in this case, 2),
- [tex]\( t \)[/tex] is the day number.
For Choice A:
- The initial amount, [tex]\( A_0 \)[/tex], is \[tex]$0.10.
- The common ratio, \( r \), is 2.
Thus, the function for Choice A is:
\[ A_t = 0.10 \times (2)^{t-1} \]
### Choice B:
The prize amount starts at \$[/tex]5.00 on the first day. Each subsequent day, the amount increases by \[tex]$5.00.
- On January 1: \$[/tex]5.00
- On January 2: \[tex]$10.00 (which is \$[/tex]5.00 + \[tex]$5.00)
- On January 3: \$[/tex]15.00 (which is \[tex]$10.00 + \$[/tex]5.00 or \[tex]$5.00 + 2 \times \$[/tex]5.00)
- On January 4: \[tex]$20.00 (which is \$[/tex]15.00 + \[tex]$5.00 or \$[/tex]5.00 + 3 \times \[tex]$5.00)
This pattern shows a linear growth, where the amount increases by a fixed sum each day.
The general form of a linear function is:
\[ A(t) = A_1 + d \times (t-1) \]
where:
- \( A(t) \) is the amount on day t,
- \( A_1 \) is the initial amount,
- \( d \) is the daily increase.
For Choice B:
- The initial amount, \( A_1 \), is \$[/tex]5.00.
- The daily increase, [tex]\( d \)[/tex], is \$5.00.
Thus, the function for Choice B is:
[tex]\[ A_t = 5.00 + 5.00(t-1) \][/tex]
### Summary:
- Choice A can be defined using an exponential function: [tex]\( A_t = 0.10 \times (2)^{t-1} \)[/tex]
- Choice B can be defined using a linear function: [tex]\( A_t = 5.00 + 5.00(t-1) \)[/tex]
