Answer :
We are given two choices:
1. In Choice A, the prize amount starts at \[tex]$0.10 on January 1 and doubles each day. This type of growth is exponential.
2. In Choice B, the prize increases by a constant \$[/tex]5.00 each day, which is a linear pattern.
Since exponential functions have the form
[tex]$$
A_t = a \cdot r^{\,t-1},
$$[/tex]
where:
- [tex]$a$[/tex] is the initial amount,
- [tex]$r$[/tex] is the common ratio (multiplier) for each day,
- [tex]$t$[/tex] is the day number,
for Choice A we have:
- Initial amount [tex]$a = 0.10$[/tex],
- Doubling each day means [tex]$r = 2$[/tex].
Thus, the exponential function for Choice A is:
[tex]$$
A_t = 0.10 \cdot 2^{\,t-1}.
$$[/tex]
To verify:
- On day 1 ([tex]$t=1$[/tex]):
[tex]$$ A_1 = 0.10 \cdot 2^{\,1-1} = 0.10 \cdot 2^0 = 0.10, $$[/tex]
- On day 2 ([tex]$t=2$[/tex]):
[tex]$$ A_2 = 0.10 \cdot 2^{\,2-1} = 0.10 \cdot 2^1 = 0.20, $$[/tex]
- On day 3 ([tex]$t=3$[/tex]):
[tex]$$ A_3 = 0.10 \cdot 2^{\,3-1} = 0.10 \cdot 2^2 = 0.40, $$[/tex]
and so on.
Therefore, Choice A is the one that can be defined by an exponential function, and the function is:
[tex]$$
A_t = 0.10 \cdot 2^{\,t-1}.
$$[/tex]
1. In Choice A, the prize amount starts at \[tex]$0.10 on January 1 and doubles each day. This type of growth is exponential.
2. In Choice B, the prize increases by a constant \$[/tex]5.00 each day, which is a linear pattern.
Since exponential functions have the form
[tex]$$
A_t = a \cdot r^{\,t-1},
$$[/tex]
where:
- [tex]$a$[/tex] is the initial amount,
- [tex]$r$[/tex] is the common ratio (multiplier) for each day,
- [tex]$t$[/tex] is the day number,
for Choice A we have:
- Initial amount [tex]$a = 0.10$[/tex],
- Doubling each day means [tex]$r = 2$[/tex].
Thus, the exponential function for Choice A is:
[tex]$$
A_t = 0.10 \cdot 2^{\,t-1}.
$$[/tex]
To verify:
- On day 1 ([tex]$t=1$[/tex]):
[tex]$$ A_1 = 0.10 \cdot 2^{\,1-1} = 0.10 \cdot 2^0 = 0.10, $$[/tex]
- On day 2 ([tex]$t=2$[/tex]):
[tex]$$ A_2 = 0.10 \cdot 2^{\,2-1} = 0.10 \cdot 2^1 = 0.20, $$[/tex]
- On day 3 ([tex]$t=3$[/tex]):
[tex]$$ A_3 = 0.10 \cdot 2^{\,3-1} = 0.10 \cdot 2^2 = 0.40, $$[/tex]
and so on.
Therefore, Choice A is the one that can be defined by an exponential function, and the function is:
[tex]$$
A_t = 0.10 \cdot 2^{\,t-1}.
$$[/tex]
