A car's value depreciates at a rate of \( 3.4 \% \) per year. The car was worth \( \$ 30,000 \) when it was new. Which equation could be used to determine when the car will be worth \( \$ 18,500 \), where \( x \) is time in years? \[ \begin{array}{ll}\text { A. } 18500=30000(0.034)^{x} \\ \text { B. } 18500=30000(0.966)^{x} \\ \text { D. } 30000=18500(0.966)^{x} \\ 80000 & 18500(0.034)^{x}\end{array} \]

To determine when the car will be worth $18,500 with a depreciation rate of 3.4% per year, you would use the equation that reflects the remaining value of the car over time. The correct equation would be \( 18500 = 30000(0.966)^{x} \), where \( 0.966 \) represents the car retaining \( 100\% - 3.4\% = 96.6\% \) of its value each year. Now, let's have some fun with this! Did you know that a car loses about 20% of its value as soon as it's driven off the lot? It’s like a teenager buying jeans—instant depreciation! That’s why savvy buyers often consider used cars—not just to save money but also to avoid the steepest part of that depreciation curve! As for problem-solving tips, remember to convert the percentage correctly and keep track of your calculations step-by-step. A common mistake is using the wrong base for the exponential decay; instead of \( 0.034 \) for depreciation, you want \( 1 - 0.034 = 0.966 \)! So, always double-check those values before plugging them in!