A car costs $30,000 when purchased as new and depreciates at a rate of 18% per year. 1. Write an exponential decay model to represent the situation. Be sure to define the variables.
2. Find the value of the car after 10 years.

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v(t) = 30000×0.82^t . . . value in dollars after t years
v(10) ≈ 4123 . . . . dollars
Step-by-step explanation:
The general form of the decay model is ...
f(t) = f(0)×(1 -decay rate)^time
where the decay rate is a fraction between 0 and 1, and the time is expressed in the same time units as the decay rate. The multiplier f(0) is the initial value that is decaying.
In this problem, the initial value is $30,000 and the decay rate is 18% per year. So, we can write a function v(t), where v is the car's value in dollars, and t is the number of years after it starts depreciating.
v(t) = 30,000×(1 -0.18)^t
v(t) = 30,000×0.82^t
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The car's value in 10 years is ...
v(t) = 30,000×0.82^10 ≈ 4123.44
The value of the car after 10 years is about $4,123.

I’m going to infer you said parallel.solve for yy= -1/2x+7/4lines that are parallel have the same slope, different y intercept.so we know m is -1/2then from there solve for y intercept.