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Due in 10 hours, 13 minutes Due Mon 10/22/2018 11.59 pm Acar was valued at $39,000 in the year 1990.The value depreciated to 512,000 by theyear 2004.What was the an...

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Due in 10 hours, 13 minutes Due Mon 10/22/2018 11.59 pm Acar was valued at $39,000 in the year 1990.The value depreciated to 512,000 by theyear 2004.What was the annual rate of change between 1990 and 2004? Round the rate of decrease [0 decimal places:B) What is the correct answer t0 part written in percentage form?C) Assume that the car value continues t0 drop by the same percentage. What will the value be in the year 2009 valuc Round t0 the nearest 50 dollars;Get help: VideoPoints poisible 10

Due in 10 hours, 13 minutes Due Mon 10/22/2018 11.59 pm Acar was valued at $39,000 in the year 1990.The value depreciated to 512,000 by theyear 2004. What was the annual rate of change between 1990 and 2004? Round the rate of decrease [0 decimal places: B) What is the correct answer t0 part written in percentage form? C) Assume that the car value continues t0 drop by the same percentage. What will the value be in the year 2009 valuc Round t0 the nearest 50 dollars; Get help: Video Points poisible 10 nmiled alempi Submit 888



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The value of a 2003 Toyota Corolla is given by the function $$v(t)=14,000(0.93)^{t}$$ where $t$ is the number of years since its purchase and $v(t)$ is its value in dollars. (Source: Kelley Blue Book) (a) What was the Corolla's initial purchase price? (b) What percent of its value does the Toyota Corolla lose each year? (c) How long will it take for the value of the Toyota Corolla to reach $\$ 12,000 $

Okay, so this question has two parts were given that a Toyota Camry costs $21,075 at sale after three years. It's worth 11,000 and $981 were asked what the annual depreciation rate is. Basically, this is asking what the rate of decay is, so we can use this equation. P is equal to p sub zero times You raised to the R Times T first to find t We know we know some values that we can already plug in, such as we know that the final rate was 11,981 is equal to, and we know it's that it was initially bought at 21,000 in 75. This is multiplied by E. We're trying to find our and we also know that it has been three years now. If we solve for R, we would get that R is equal to the natural log of 11,981 divided by 21,000 75. We would also get that it's divided by three now. Now that we have are we can plug it back in for part B so this was the answer to part a. R is equal to the natural log of 11,900 anyone over 21,075 all over three. Now for Part B. They're asking what the price of the car would be after five years so we can use the same equation again. This time we would have we would be only finding P, and we know that it started at 21,000 and $75 which is multiplied by E raised to the l N of 11,000 981 divided by 21,075 multiplied by five. Because that's how many years were looking at. If you put this into a calculator, you would get that P is equal to 8200 21 point 99 That is how much the car will be worth after five years

So in this problem were given the formula 100 times one minus the square root of these, divided by C. And it represents the annual rate of depreciation, where C. Is the cost and V. Is the value after two years. Well, for this particular problem, they told us that the car costs are the $8000. So that will be our to see value. And then we're told the value at the end of two years is 24,000, so V. Is equal to 24,000. So what we want to do is find the rate that the car is depreciating it. So all we're gonna do is substitute these values into our formula. So we'll have 100 times one minus the square root of V, which is 24,000 divided by C, which is 38,000. Okay, so now all we need to do is simplify. So we'll start by reducing our fraction. So we have 24,000 divided by 38,000. And if we leave that as a fraction, that will be 12/19. So we have 100 times one minus the square root, A 12/19. Now we're going to take the square root of this. Well the square root of 12/19 is and I'm going to go around 25 places after the decimal. So we'll have one minus 0.79472 Alright, well now we need to subtract well one minus 0.0, sorry one minus 0.79472 is equal to let's put 100 here, 0.205 to eight. Then we'll multiply this by 100. Which is going to give us 20.5 to 8. So now we found the greatest appreciating it. It's depreciating at approximately 20.5 to 8% every two years or after the first two years

We're doing a depreciation problem, which is an exponential problem as it depreciates by the same rate. I'm sorry. Each year. So we're going to use our exponential A. B to the X. Formula. And are a. Is our starting point, which is 38 $1000. So we have F. Of X equals 38 1000 times B to the X. And were given that the car depreciated to 11,000 In by the year. 2009. So we have 20090.10 11,000. So I'm going to plug those in. So I have 11,000 Equals 38,000 Times B to the 10th power. The divide both sides by 38,000. So now I have B to the 10th equals and I'm going to just cancel out the three zeroes in the numerator with the three zeroes in the denominator. So I'm just going to have 11/38 and now I'm going to multiply the exponents by 1/10. So my exponents on that side cancel. So I'm going to take this side And raise it to the 1/10 power. So now I have the equals 11/38 raised to the 1/10 power. Now you have to raise the one, put the 1/10 in parentheses. Otherwise the calculator will raise it to the first power and then divide that answer by 10. Which would be wrong. So here we get .8133 is RB. So plugging that into my formula, I now have F. Of X Equals 38,000 Times .8133 to the X. Power. And they want us to figure out by the year 2013. So that's when x equals 10. So I'm going to put 10 in. So my f. 10 Equals 38,000 Times .8133 raised to the tent. And when I do that I get f. of 10 technique war $4,811.62 sense

Hello. So here we have that are is going to be equal to 1 -1 the quantity here? S oversee to the one over Ed. So given our um, given information, this is going to be our is going to be equal to one minus 11,500 divided by 9 22,090 Raised to the 1/4. So that's going to be approximately equal to 1 -0.841, which is going to be equal to 0.159, which is 15.9%. All right. Take care.


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