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Brandon is analyzing the profits from two of his restaurant locations. The first location showed a profit of [tex]\(\$ 3,000\)[/tex] in the first week of the year but has shown a steady decrease in profits of [tex]\(1.5\%\)[/tex] each week since then. His second location showed a profit of [tex]\(\$ 1,500\)[/tex] in the first week of the year and has shown a steady increase in profits of [tex]\(1.2\%\)[/tex] each week since then.

Which system of equations can Brandon use to determine the number of weeks, [tex]\(x\)[/tex], until the profits, [tex]\(y\)[/tex], from the two locations are the same?

A. [tex]\(y = 3,000(1.015)^x\)[/tex]
[tex]\(y = 1,500(1.012)^\infty\)[/tex]

B. [tex]\(y = -3,000(1.015)^a\)[/tex]
[tex]\(y = 1,500(1.012)^2\)[/tex]

C. [tex]\(y = -3,000(0.985)^x\)[/tex]
[tex]\(y = 1,500(1.012)^2\)[/tex]

D. [tex]\(y = 3,000(0.985)^x\)[/tex]
[tex]\(y = 1,500(1.012)^x\)[/tex]



Answer :

GinnyAnswer
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]

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