Let's assume we have a certain quantity V that increases or decreases over time at a rate of p% over t units of time (t can be hours, days, months, years, etc). The value V'(t) after t units of time is then given by:
[tex]V^{\prime}(t)=V\cdot(1\pm\frac{p}{100})^t[/tex]Where we use the + sign if we are talking about an increasing and the - sign if we are talking about a decreasing.
We are told that Brianna's car was bought by $35543. We know that it decreases at a rate of 19% per year which means that p=19 and the price of Brianna's car after t years is given by:
[tex]\begin{gathered} V^{\prime}(t)=35543\cdot(1-\frac{19}{100})^t \\ V^{\prime}(t)=35543\cdot0.81^t \end{gathered}[/tex]Then the approximate value of the car in 10 years is V'(10):
[tex]V^{\prime}(10)=35543\cdot0.81^{10}=4321.199[/tex]Then the answer is $4321.