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A descending curve with an arrow on the right end passes through 10 points plotted on a coordinate plane. The x-axis is labeled years since purchase, and the y-axis is labeled value of car.
Given the graph of the exponential equation representing the value of a car since purchase, which option is the correct exponential equation for the graph if the graph goes through the points (0, 30000)
and (1, 22500)
?
(1 point)
Responses
y=22,500(34)x
y equals 22,500 left parenthesis Start Fraction 3 over 4 End Fraction right parenthesis superscript x baseline
y=30,000(34)x
y equals 30,000 left parenthesis Start Fraction 3 over 4 End Fraction right parenthesis superscript x baseline
y=0.75(30,000)x
y equals 0.75 left parenthesis 30,000 right parenthesis superscript x baseline
y=30,000(43)x
A descending curve with an arrow on the right end passes through 10 points plotted on a coordinate plane. The x-axis is labeled years since purchase, and the y-axis is labeled value of car.
Given the graph of the exponential equation representing the value of a car since purchase, which option is the correct exponential equation for the graph if the graph goes through the points (0, 30000)
and (1, 22500)
?
(1 point)
Responses
y=22,500(34)x
y equals 22,500 left parenthesis Start Fraction 3 over 4 End Fraction right parenthesis superscript x baseline
y=30,000(34)x
y equals 30,000 left parenthesis Start Fraction 3 over 4 End Fraction right parenthesis superscript x baseline
y=0.75(30,000)x
y equals 0.75 left parenthesis 30,000 right parenthesis superscript x baseline
y=30,000(43)x
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To find the correct exponential equation based on the given points (0, 30000) and (1, 22500), we start with the general form of an exponential decay function:
\[ y = a \cdot b^x \]
where \( a \) is the value at \( x = 0 \) and \( b \) is the decay factor per year.
From the point (0, 30000):
- At \( x = 0 \), \( y = 30000 \), so \( a = 30000 \).
Now we also know that at \( x = 1 \), \( y = 22500 \). Substituting these values into the equation gives:
\[ 22500 = 30000 \cdot b^1 \]
To find \( b \):
1. Divide both sides by 30000:
\[ b = \frac{22500}{30000} = 0.75 \]
Thus, we can write the equation as:
\[ y = 30000 \cdot (0.75)^x \]
This can also be expressed as:
\[ y = 30000 \cdot \left( \frac{3}{4} \right)^x \]
Therefore, the final equation is:
\[ y = 30,000 \left( \frac{3}{4} \right)^x \]
So, the correct response is:
**y = 30,000(34)x** (Note: This notation appears to be a typographical error; the correct format would be \( y = 30,000 \left( \frac{3}{4} \right)^x \) as derived.)
\[ y = a \cdot b^x \]
where \( a \) is the value at \( x = 0 \) and \( b \) is the decay factor per year.
From the point (0, 30000):
- At \( x = 0 \), \( y = 30000 \), so \( a = 30000 \).
Now we also know that at \( x = 1 \), \( y = 22500 \). Substituting these values into the equation gives:
\[ 22500 = 30000 \cdot b^1 \]
To find \( b \):
1. Divide both sides by 30000:
\[ b = \frac{22500}{30000} = 0.75 \]
Thus, we can write the equation as:
\[ y = 30000 \cdot (0.75)^x \]
This can also be expressed as:
\[ y = 30000 \cdot \left( \frac{3}{4} \right)^x \]
Therefore, the final equation is:
\[ y = 30,000 \left( \frac{3}{4} \right)^x \]
So, the correct response is:
**y = 30,000(34)x** (Note: This notation appears to be a typographical error; the correct format would be \( y = 30,000 \left( \frac{3}{4} \right)^x \) as derived.)
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