The population of Gatham City is 750. The population can be represented by the equation p(t)=750(1.05)t

Question

is the amount of years from today. If I check the population in 5 years, what would be an appropriate domain to represent this?

Answer 1

To determine an appropriate domain for the population function \( p(t) = 750(1.05)^t \), where \( t \) is the number of years from today, we need to consider the context of the problem.

Since you're looking at the population in 5 years, the values of \( t \) should start from today (when \( t = 0 \)) and go up to 5 years in the future. Therefore, the appropriate domain for \( t \) in this context would be:

\[ t \in [0, 5] \]

This means that \( t \) can take any real number value starting from 0 (today) and going up to 5 (five years from today).

Answer 2

Ruby bought a new car for $28,000.00. If the car depreciates in value by 5%

each year, the situation can be modeled by the equation f(x)=28,000(0.95)x

. Complete the table to find the value of the car, to the nearest cent, after three years.

0 28,000.00
1 26,600.00
2 25,270.00

Answer 3

To find the value of the car after three years using the depreciation model given by the equation \( f(x) = 28,000(0.95)^x \), we can calculate \( f(3) \) where \( x = 3 \):

\[ f(3) = 28,000(0.95)^3 \]

First, calculate \( (0.95)^3 \):

\[ (0.95)^3 = 0.857375 \]

Now, substitute this value back into the function:

\[ f(3) = 28,000 \times 0.857375 \]

Now compute the final value:

\[ f(3) \approx 28,000 \times 0.857375 \approx 24,000.50 \]

Therefore, the value of the car after three years, to the nearest cent, is approximately $24,000.50.

Here's the completed table:

| Year (x) | Car Value (f(x)) | |----------|-------------------------| | 0 | $28,000.00 | | 1 | $26,600.00 | | 2 | $25,270.00 | | 3 | $24,000.50 |