Create a sequence that describes the worth of Aria’s car each year for 10 years. (Show all your work). (2 pts)

Answer 1

To create a sequence that describes the worth of Aria's car each year for 10 years, we will assume that the car depreciates in value over time. A common way to model this depreciation is to use a fixed percentage decrease each year.

Step 1: Determine the Initial Value and Depreciation Rate

Let's assume:

• The initial value of the car (Year 0) is $20,000.
• The car depreciates in value by 15% each year.

Step 2: Calculate the Depreciation

The depreciation can be expressed using a formula. The value of the car after \( n \) years can be calculated as:

\[ V(n) = V_0 \times (1 - r)^n \]

where:

• \( V(n) \) is the value of the car after \( n \) years.
• \( V_0 \) is the initial value of the car ($20,000 in this case).
• \( r \) is the rate of depreciation (0.15 for 15%).
• \( n \) is the number of years.

Step 3: Create the Sequence

Now, we can calculate the worth of the car for each year from \( n = 0 \) to \( n = 10 \):

1. Year 0: \[ V(0) = 20000 \times (1 - 0.15)^0 = 20000 \times 1 = 20000 \]

2. Year 1: \[ V(1) = 20000 \times (1 - 0.15)^1 = 20000 \times 0.85 = 17000 \]

3. Year 2: \[ V(2) = 20000 \times (1 - 0.15)^2 = 20000 \times 0.85^2 = 20000 \times 0.7225 = 14450 \]

4. Year 3: \[ V(3) = 20000 \times (1 - 0.15)^3 = 20000 \times 0.85^3 = 20000 \times 0.614125 = 12282.50 \]

5. Year 4: \[ V(4) = 20000 \times (1 - 0.15)^4 = 20000 \times 0.85^4 = 20000 \times 0.52773125 = 10554.63 \]

6. Year 5: \[ V(5) = 20000 \times (1 - 0.15)^5 = 20000 \times 0.85^5 = 20000 \times 0.449576 = 8991.52 \]

7. Year 6: \[ V(6) = 20000 \times (1 - 0.15)^6 = 20000 \times 0.85^6 = 20000 \times 0.38263752 = 7652.75 \]

8. Year 7: \[ V(7) = 20000 \times (1 - 0.15)^7 = 20000 \times 0.85^7 = 20000 \times 0.32439198 = 6487.84 \]

9. Year 8: \[ V(8) = 20000 \times (1 - 0.15)^8 = 20000 \times 0.85^8 = 20000 \times 0.2764801 = 5529.60 \]

10. Year 9: \[ V(9) = 20000 \times (1 - 0.15)^9 = 20000 \times 0.85^9 = 20000 \times 0.2357945 = 4715.89 \]

11. Year 10: \[ V(10) = 20000 \times (1 - 0.15)^{10} = 20000 \times 0.85^{10} = 20000 \times 0.2008747 = 4017.49 \]

Summary of the Sequence

The sequence representing the worth of Aria’s car for each year for 10 years is:

• Year 0: $20,000
• Year 1: $17,000
• Year 2: $14,450
• Year 3: $12,282.50
• Year 4: $10,554.63
• Year 5: $8,991.52
• Year 6: $7,652.75
• Year 7: $6,487.84
• Year 8: $5,529.60
• Year 9: $4,715.89
• Year 10: $4,017.49

This sequence illustrates how the value of the car decreases over the 10-year period.

Answer 2

simple and smaller

Answer 3

Certainly! Here's a simpler version that summarizes the worth of Aria's car each year for 10 years, assuming an initial value of $20,000 and a depreciation rate of 15% per year.

Initial Parameters

• Initial Value (\( V_0 \)): $20,000
• Depreciation Rate: 15% (or 0.15)

Depreciation Formula

The worth of the car after \( n \) years can be calculated as: \[ V(n) = V_0 \times (1 - 0.15)^n \]

Calculation for Each Year

1. Year 0: \( V(0) = 20000 \)
2. Year 1: \( V(1) = 20000 \times 0.85 = 17000 \)
3. Year 2: \( V(2) = 20000 \times 0.85^2 \approx 14450 \)
4. Year 3: \( V(3) = 20000 \times 0.85^3 \approx 12282.50 \)
5. Year 4: \( V(4) = 20000 \times 0.85^4 \approx 10554.63 \)
6. Year 5: \( V(5) = 20000 \times 0.85^5 \approx 8991.52 \)
7. Year 6: \( V(6) = 20000 \times 0.85^6 \approx 7652.75 \)
8. Year 7: \( V(7) = 20000 \times 0.85^7 \approx 6487.84 \)
9. Year 8: \( V(8) = 20000 \times 0.85^8 \approx 5529.60 \)
10. Year 9: \( V(9) = 20000 \times 0.85^9 \approx 4715.89 \)
11. Year 10: \( V(10) = 20000 \times 0.85^{10} \approx 4017.49 \)

Summary of Worth Each Year

• Year 0: $20,000
• Year 1: $17,000
• Year 2: $14,450
• Year 3: $12,282.50
• Year 4: $10,554.63
• Year 5: $8,991.52
• Year 6: $7,652.75
• Year 7: $6,487.84
• Year 8: $5,529.60
• Year 9: $4,715.89
• Year 10: $4,017.49

This concise summary provides the depreciation of the car over the 10 years.

Answer 4

Create a formula for the worth of Aria’s car over time. (Show all your work) (2 pt

Answer 5

To create a formula for the worth of Aria’s car over time, we’ll use the principles of exponential decay due to depreciation. Here's a step-by-step breakdown of how to formulate the equation.

Step 1: Identify the Initial Parameters

1. Initial value of the car (\( V_0 \)): This is the purchase price of the car. Let's assume \( V_0 = 20000 \) (in dollars).

2. Depreciation rate (\( r \)): This is the percentage by which the car value decreases each year. In this case, the depreciation rate is 15%, or \( r = 0.15 \).

Step 2: Make a Depreciation Formula

The worth of an asset that depreciates over time can typically be modeled with the formula:

\[ V(t) = V_0 \times (1 - r)^t \]

Where:

• \( V(t) \) is the value of the car at time \( t \) (in years).
• \( V_0 \) is the initial value of the car.
• \( r \) is the rate of depreciation (converted to decimal form).
• \( t \) is the time in years.

Step 3: Plug in Known Values

Substituting the known values into the formula:

• \( V_0 = 20000 \)
• \( r = 0.15 \)

The formula becomes:

\[ V(t) = 20000 \times (1 - 0.15)^t \]

This simplifies to:

\[ V(t) = 20000 \times (0.85)^t \]

Final Formula

Thus, the formula for the worth of Aria’s car over time \( t \) (in years) is:

\[ V(t) = 20000 \times (0.85)^t \]

Summary

• Initial Value (\( V_0 \)): $20,000
• Depreciation Rate (\( r \)): 15% (or 0.15)
• Formula: \[ V(t) = 20000 \times (0.85)^t \]

This formula allows you to calculate the car's worth at any given year \( t \). For example, to find the value after 5 years, you would substitute \( t = 5 \) into the formula:

\[ V(5) = 20000 \times (0.85)^5 \]

This approach provides a clear understanding of the depreciation model applied to Aria's car.

Answer 6

Explain how your work shown in Question #2 and #3 demonstrate that the statement “No matter what the growth rate of a linear function is, that of an exponential function will always exceed it” is either true or false. (Use specific data to support and justify your conclusions.) (3 pts)

Answer 7

To assess the statement “No matter what the growth rate of a linear function is, that of an exponential function will always exceed it,” we can compare the growth of a linear function with the formula of an exponential decay function, as demonstrated in the previous responses about Aria's car worth over time.

Step 1: Define the Linear Function

A linear function can be expressed in the following form:

\[ L(t) = mt + b \]

Where:

• \( m \) is the slope (growth rate),
• \( b \) is the initial value,
• \( t \) is time (in years).

For the sake of this example, let’s define Aria’s linear function based on her car value as if it depreciated linearly rather than exponentially. Assuming an initial value (\( V_0 = 20000 \)) and that it loses $1,500 each year (a linear depreciation), the linear function would be:

\[ L(t) = 20000 - 1500t \]

Step 2: Define the Exponential Function

From our previous calculations for Aria's car value, we established that:

\[ V(t) = 20000 \times (0.85)^t \]

Step 3: Compare Values Over Time

Data Points for Linear and Exponential Functions

Let’s calculate the worth of the car over a 10-year period for both functions:

• Linear Function \( L(t) \) Values:

  • Year 0: \( L(0) = 20000 \)
  • Year 1: \( L(1) = 18500 \)
  • Year 2: \( L(2) = 17000 \)
  • Year 3: \( L(3) = 15500 \)
  • Year 4: \( L(4) = 14000 \)
  • Year 5: \( L(5) = 12500 \)
  • Year 6: \( L(6) = 11000 \)
  • Year 7: \( L(7) = 9500 \)
  • Year 8: \( L(8) = 8000 \)
  • Year 9: \( L(9) = 6500 \)
  • Year 10: \( L(10) = 5000 \)

• Exponential Function \( V(t) \) Values:

  • Year 0: \( V(0) \approx 20000 \)
  • Year 1: \( V(1) \approx 17000 \)
  • Year 2: \( V(2) \approx 14450 \)
  • Year 3: \( V(3) \approx 12282.50 \)
  • Year 4: \( V(4) \approx 10554.63 \)
  • Year 5: \( V(5) \approx 8991.52 \)
  • Year 6: \( V(6) \approx 7652.75 \)
  • Year 7: \( V(7) \approx 6487.84 \)
  • Year 8: \( V(8) \approx 5529.60 \)
  • Year 9: \( V(9) \approx 4715.89 \)
  • Year 10: \( V(10) \approx 4017.49 \)

Step 4: Analysis and Conclusions

Observations:

• At Year 0, both the linear and exponential functions start at $20,000.
• For Year 1, the values are:
  • \( L(1) = 18500 \)
  • \( V(1) \approx 17000 \)

The linear function is higher.

• However, as we progress, the exponential function decreases at a faster rate than the linear function. By Year 5, the values are:
  • \( L(5) = 12500 \)
  • \( V(5) \approx 8991.52 \)

At Year 5, the linear function is still greater. However, the exponential decay continues to accelerate.

• By Year 10, the values are:
  • \( L(10) = 5000 \)
  • \( V(10) \approx 4017.49 \)

At Year 10, the linear function is again greater.

Conclusion

The statement “No matter what the growth rate of a linear function is, that of an exponential function will always exceed it” is false in this specific context.

While it’s evident that in general terms, exponential growth will always surpass linear growth when the exponential function increases (e.g., population growth, interest compounded), in the case of depreciation (as was modeled here), a linear depreciation can yield higher values for a considerable time. In our example, the linear depreciation of the car continued to exceed the exponential decay for the duration we calculated.

This illustrates that the statement must be contextually understood: it holds true when considering exponential growth but does not apply equivalently to exponential decay compared with linear depreciation rates.

Answer 8

Create your own Arithmetic Sequence, where your common difference is greater than one and not divisible by 2 or 3, then answer or solve the following:
List at least 5 terms of your sequence. Explain what makes your sequence arithmetic. (2 pts)

Answer 9

Arithmetic Sequence Creation

Let's create an arithmetic sequence with the following characteristics:

• Common difference: greater than 1 and not divisible by 2 or 3.
• I will choose a common difference of 5 (since it's greater than 1 and not divisible by 2 or 3).

Step 1: Define the Sequence

Let’s start the sequence with an initial term (\(a_1\)) of 3.

Step 2: List the Terms

The arithmetic sequence can be defined as follows:

\[ a_n = a_1 + (n - 1) \cdot d \]

where:

• \(a_1 = 3\) (initial term),
• \(d = 5\) (common difference),
• \(n\) is the term number.

Now, we can list at least 5 terms:

1. Term 1 (\(n=1\)): \[ a_1 = 3 + (1 - 1) \cdot 5 = 3 + 0 = 3 \]

2. Term 2 (\(n=2\)): \[ a_2 = 3 + (2 - 1) \cdot 5 = 3 + 5 = 8 \]

3. Term 3 (\(n=3\)): \[ a_3 = 3 + (3 - 1) \cdot 5 = 3 + 10 = 13 \]

4. Term 4 (\(n=4\)): \[ a_4 = 3 + (4 - 1) \cdot 5 = 3 + 15 = 18 \]

5. Term 5 (\(n=5\)): \[ a_5 = 3 + (5 - 1) \cdot 5 = 3 + 20 = 23 \]

Step 3: List of the First 5 Terms

So the first 5 terms of the sequence are:

• \(3, 8, 13, 18, 23\)

Explanation of What Makes this Sequence Arithmetic

An arithmetic sequence is defined as a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is referred to as the common difference.

In this sequence:

• The common difference (\(d\)) between each consecutive term is always 5:
  • \(8 - 3 = 5\)
  • \(13 - 8 = 5\)
  • \(18 - 13 = 5\)
  • \(23 - 18 = 5\)

This consistent addition of 5 to each term indicates that this sequence is indeed an arithmetic sequence.

Characteristics of the Sequence:

1. The common difference is constant: 5.
2. The sequence starts with an initial term of 3.
3. The common difference is greater than 1.
4. The common difference (5) is not divisible by 2 or 3.

Thus, we clearly establish that the constructed sequence is arithmetic based on these properties.