To solve the problem, let's break it down step by step and understand the changes in profits for both locations over time.
### First Location Analysis:
1. Initial profit = [tex]$3,000.
2. Decrease in profit = 1.5% per week.
We can express a 1.5% decrease as a multiplication factor. A 1.5% decrease corresponds to multiplying by (100% - 1.5%) = 98.5%, or 0.985.
So, the profit \( y \) after \( x \) weeks can be represented by the equation:
\[ y_1 = 3000 \times 0.985^x \]
### Second Location Analysis:
1. Initial profit = $[/tex]1,500.
2. Increase in profit = 1.2% per week.
We can express a 1.2% increase as a multiplication factor. A 1.2% increase corresponds to multiplying by (100% + 1.2%) = 101.2%, or 1.012.
So, the profit [tex]\( y \)[/tex] after [tex]\( x \)[/tex] weeks can be represented by the equation:
[tex]\[ y_2 = 1500 \times 1.012^x \][/tex]
### System of Equations:
To determine when the profits from the two locations are the same, we need a system of equations that sets the two profit equations equal to each other. From the above analysis, we have:
[tex]\[ y = 3000 \times 0.985^x \][/tex]
[tex]\[ y = 1500 \times 1.012^x \][/tex]
Now, let's look at the provided answer choices:
A. [tex]\( y = 3000 \times (1.015)^x \)[/tex]
[tex]\( y = 1500 \times (1.012)^{\infty} \)[/tex]
B. [tex]\( y = -3000 \times (1.015)^a \)[/tex]
[tex]\( y = 1500 \times (1.012)^2 \)[/tex]
C. [tex]\( y = -3000 \times (0.985)^x \)[/tex]
[tex]\( y = 1500 \times (1.012)^2 \)[/tex]
D. [tex]\( y = 3000 \times (0.985)^x \)[/tex]
[tex]\( y = 1500 \times (1.012)^x \)[/tex]
The correct system of equations for the profit [tex]\( y \)[/tex] to be equal for both locations is:
[tex]\[ y = 3000 \times 0.985^x \][/tex]
[tex]\[ y = 1500 \times 1.012^x \][/tex]
Thus, the correct answer is:
D. [tex]\( y = 3000 \times (0.985)^x \)[/tex]
[tex]\[ y = 1500 \times (1.012)^x \][/tex]
