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Salim purchased a new car for $21,000, and it depreciates in value each year. The function y=21,000(1β0.15)x π¦ = 21 , 000 ( 1 β...Asked by b
Salim purchased a new car for $21,000, and it depreciates in value each year. The function y=21,000(1β0.15)x
y
=
21
,
000
(
1
β
0.15
)
x
represents the depreciation. He also purchases a motorcycle, which depreciates each year as well. Its depreciation is represented by the function y=14,000(1β0.09)x
y
=
14
,
000
(
1
β
0.09
)
x
. Graph these two functions using GeoGebra. Which statement best describes when the motorcycle will be worth $7,000?
Statement #1: The motorcycle will have a value of $7,000 after roughly 7.3 years.
Statement #2: The motorcycle will have a value of $7,000 after roughly 6.7 years.
(1 point)
Statement #
y
=
21
,
000
(
1
β
0.15
)
x
represents the depreciation. He also purchases a motorcycle, which depreciates each year as well. Its depreciation is represented by the function y=14,000(1β0.09)x
y
=
14
,
000
(
1
β
0.09
)
x
. Graph these two functions using GeoGebra. Which statement best describes when the motorcycle will be worth $7,000?
Statement #1: The motorcycle will have a value of $7,000 after roughly 7.3 years.
Statement #2: The motorcycle will have a value of $7,000 after roughly 6.7 years.
(1 point)
Statement #
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To determine when the motorcycle will be worth $7,000, we can analyze the depreciation function for the motorcycle:
The value of the motorcycle after \( x \) years is given by:
\[ y = 14,000(1 - 0.09)^x \]
We set \( y \) equal to $7,000 to find \( x \):
\[ 7,000 = 14,000(1 - 0.09)^x \]
Dividing both sides by 14,000 gives:
\[ \frac{7,000}{14,000} = (1 - 0.09)^x \]
This simplifies to:
\[ 0.5 = (0.91)^x \]
Next, we can solve for \( x \) by taking the logarithm of both sides:
\[ \log(0.5) = x \cdot \log(0.91) \]
Now, we can find \( x \):
\[ x = \frac{\log(0.5)}{\log(0.91)} \]
Calculating \( x \):
1. \(\log(0.5) \approx -0.3010\)
2. \(\log(0.91) \approx -0.0434\)
Thus:
\[ x \approx \frac{-0.3010}{-0.0434} \approx 6.93 \]
Rounding to one decimal place gives:
\[ x \approx 6.9 \]
This indicates that the motorcycle will depreciate to $7,000 in approximately 6.9 years.
Now, reviewing the original statements:
- Statement #1: The motorcycle will have a value of $7,000 after roughly 7.3 years.
- Statement #2: The motorcycle will have a value of $7,000 after roughly 6.7 years.
Since the calculated value is close to 6.9 years, which is nearer to Statement #2, we can conclude that:
**Statement #2** is the better choice.
The value of the motorcycle after \( x \) years is given by:
\[ y = 14,000(1 - 0.09)^x \]
We set \( y \) equal to $7,000 to find \( x \):
\[ 7,000 = 14,000(1 - 0.09)^x \]
Dividing both sides by 14,000 gives:
\[ \frac{7,000}{14,000} = (1 - 0.09)^x \]
This simplifies to:
\[ 0.5 = (0.91)^x \]
Next, we can solve for \( x \) by taking the logarithm of both sides:
\[ \log(0.5) = x \cdot \log(0.91) \]
Now, we can find \( x \):
\[ x = \frac{\log(0.5)}{\log(0.91)} \]
Calculating \( x \):
1. \(\log(0.5) \approx -0.3010\)
2. \(\log(0.91) \approx -0.0434\)
Thus:
\[ x \approx \frac{-0.3010}{-0.0434} \approx 6.93 \]
Rounding to one decimal place gives:
\[ x \approx 6.9 \]
This indicates that the motorcycle will depreciate to $7,000 in approximately 6.9 years.
Now, reviewing the original statements:
- Statement #1: The motorcycle will have a value of $7,000 after roughly 7.3 years.
- Statement #2: The motorcycle will have a value of $7,000 after roughly 6.7 years.
Since the calculated value is close to 6.9 years, which is nearer to Statement #2, we can conclude that:
**Statement #2** is the better choice.
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