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A car is originally worth \(34,450. It takes 13 years for bthis car to totally depreciate. a. Write the straight line depreciation equation for this situation. b. How long will it take for the car to be worth half its value? c. How long will it take for the car to be worth \)10,000? Round your answer to the nearest tenth of a year.

Short Answer

Expert verified
a. The straight line depreciation equation is \( V = 34450 - 2650t \). b. It will take 6.5 years to depreciate to half its value. c. It will take 9.2 years to depreciate to $10,000.

Step by step solution

01

Writing the straight line depreciation equation

The value of the car decreases by $34,450 over 13 years. Thus, the rate of depreciation is \( \frac{34450}{13} = $2650 \) per year. The depreciation equation will be \( V = 34450 - 2650t \), where V is the value of the car at any given time t in years.
02

Calculating the time for the car to be worth half its value

To find the time it takes for the car to depreciate to half its value, set V equal to \( \frac{34450}{2} = $17225 \). Then, solve the equation \( 17225 = 34450 - 2650t \) for t. Solving this, we get \( t = \frac{34450 - 17225}{2650} = 6.5 \) years.
03

Calculating the time for the car to be worth $10,000

To find the time it takes for the car to depreciate to $10,000, set V equal to $10,000. Then, solve the equation \( 10000 = 34450 - 2650t \) for t. Solving this, we get \( t = \frac{34450 - 10000}{2650} = 9.2 \) years.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Depreciation Equation
Understanding the depreciation equation is a fundamental aspect of financial algebra, specifically when it comes to assets such as vehicles. In the case of straight line depreciation, which is the most straightforward depreciation method, the depreciation equation simplifies to a linear relationship between the asset value and time.

Consider a car worth \(34,450; as it ages, its value decreases at a uniform rate until it becomes zero after a certain number of years, in this case, 13 years. The annual depreciation can be expressed as \(\frac{Initial\ Value}{Useful\ Life}\), which calculates to \(\frac{34450}{13} = \)2650\) per year. Thus, the straight line depreciation formula for this situation is \( V = 34450 - 2650t \), where \(V\) represents the value of the car at any given time \(t\), measured in years.

It's this linear equation that allows us to calculate the value of the car at any point during its useful life and is a fundamental concept within financial algebra.
Value Depreciation
The concept of value depreciation is directly tied to the useful life of an asset. Over time, almost every asset loses value, which can be accounted for using various methods of depreciation, with straight line depreciation being one of the simplest. This method assumes a constant rate of depreciation each year.

In our example, the car decreases in value by a fixed annual depreciation of \(2650. This regular reduction in value reflects that the car is moving towards the end of its useful life, which is assumed to be 13 years for the car initially worth \)34,450. To find out when the car will be worth half its original value, we simply plug in half of the initial value into our depreciation equation and solve for \(t\). The same method applies if we want to calculate when its value will drop to $10,000. We see that value depreciation is not just a financial concept; it’s a predictable numeric trend that can be calculated and planned for, making it a crucial consideration in areas such as accounting and asset management.
Financial Algebra
Financial algebra involves applying mathematical techniques to solve problems related to financial scenarios, such as calculating interest rates, loan repayment schedules, and depreciation of assets. In our car depreciation problem, financial algebra is used to establish the relationship between the car’s value, its initial cost, the annual depreciation rate, and time.

The depreciation equation derived from the straight line method is a primary example of using algebra to address a real-world financial problem. The equation itself becomes a tool for making informed economic decisions regarding the asset. For instance, knowing how to calculate the point in time when a car will be worth a certain amount can influence decisions about when to sell, trade, or replace the vehicle. Financial algebra encompasses these practical applications and empowers individuals to strategize around the fiscal aspects of their assets.

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