So we have a situation in which a computer at time zero was worth $575 and then two years later it's worth $275. So 5 75 to 2. 75 in two years and we want to start out. We want to write one model as linear, and we want to write one model as exponential and exponential with the base E. And they show that is X could also do. It is T. So I have a couple of things to do here. First of all, let's do the linear model, and we know that that's the Y intercept. We also know that we can figure out what the rate of change is. So let's find the slope. The slope is the difference in the Wise, which is to 75 minus 5 75 over the difference in the exes, which is two years minus zero years. And so this is dropping $300 in two years. So it's dropping $150 per year is what a linear model would predict and, like, get this citizen move up. So let's write our model our model is negative 1 50 times the input variable of time and years plus $575. So there is linear. Now let's find our exponential model. Well, we know little A is 5 75 and then we want to plug in a time. So we know what this value is. That is going to be that 5 75 automatically here, because is our initial amount. And the only thing we have to find is R. K. So we'll plug in a point. We know that it will be worth 275 once we have a time of two years. So we divide both sides by the 5. 75 and we have e toothy two k is equal to this to 75/5 75 and then we can take the natural aga both sides a couple different ways. You can saw that exponential, but I'm gonna move that equal sign there and right, Ellen right there. And then we could do our log of a power. In fact, I'm not even well that well equal sign even more. We'll move away over there. I'm gonna do log of a power and I will get to K times, Elena V. And now that's moved down. But Elena B is one. So we deceptive divide both sides by two. And we'll have what that little K isn't that quick. Hit that on my calculator. So I have Ln of 2 75 divided by 5 75 calls my parentheses and then divide that by two. And I find that that K value is negative. 0.3688 and it's negative because we're decaying. So our model for exponential is that 5 75 times e to the negative 0.3688 times X or T whichever variable you want to use. Then we're asked to compare and grab those on our calculator. So I would set up my window, you know, your decaying. So let's look at a time frame. Really? We're looking at, like, the first two years, so I'm gonna have my windows set up to go from zero to, but I'm gonna go up to five years. What the heck? So 0 to 5 scaling by ones, and then I'm gonna have my why minimum be at zero and my highest value out of school up to 600 and I'll scare by 100. So again, this is going to go up to 600 we can put both of our models in. We've got our negative 1 50 x plus 5 75 and then we have our 5 75 times e to the power of and we have negative 750.3688 x, and we can let the calculator graph those. And when we dio, we naturally end up getting after all this right, We end up getting that linear model. Does that and the exponential model put in red. When I look at it on my screen, it dips below a little bit. Not a lot. But then at two, that's a common intersection point. Because we know two and 2 75 is a common point for both those as is this zero and 5 75 and then this graph levels off. So I believe part C or part part V as you look at the graph, and we can see that during that first year, our first two years that it appears as though the, uh, the exponential model seems to decay faster and drops down faster. Now, if we look at a little table, we can see what happens in those first years. So here's our linear model and here is our exponential model and at time, zero. We know they're both 5 75. We know that, and so I could go to table set and I can start my table at zero. Now I'll just go up by one. So let's just look at a few years. And now we set up table and we have after one year, the linear is worth 24 25. But the exponential is birth 397.65 And then, after two years to 75 we know that their goal is gonna be the same and then notice what happens. Linear keeps dropping until we get thio time. Three were at 1 25 time, four were negative. So apparently you have to pay somebody to take the computer and versus here for part, 390.1 831.52 so we can see that although the exponential does decay faster, then later on, it decays more slowly and linear drops out between three and four years to have absolutely no value with the computer, which is probably the case anyway, so hopefully that's helpful.