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The table shows the amount of pet food in cups remaining in an automatic feeder as a function of the number of meals the feeder...Question
the table shows the amount of pet food in cups remaining in an automatic feeder as a function of the number of meals the feeder has dispensed.
Number of meals dispensed, x: 1, 3, 6, 7
Amount of Pet food remaining, y: 21, 15, 6, 3
write a function rule that models this situation:
Number of meals dispensed, x: 1, 3, 6, 7
Amount of Pet food remaining, y: 21, 15, 6, 3
write a function rule that models this situation:
Answers
Answered by
GPT-4o mini
To model the relationship between the number of meals dispensed \( x \) and the amount of pet food remaining \( y \), we can analyze the provided data points:
- Meals dispensed (x): 1, 3, 6, 7
- Pet food remaining (y): 21, 15, 6, 3
First, let's look for a pattern between \( x \) and \( y \):
1. Between \( x = 1 \) and \( x = 3 \):
\[
y \text{ changes from } 21 \text{ to } 15 \text{ (decrease of 6)}
\]
Here, 2 meals are dispensed, so the average decrease per meal is \( \frac{6}{2} = 3 \).
2. Between \( x = 3 \) and \( x = 6 \):
\[
y \text{ changes from } 15 \text{ to } 6 \text{ (decrease of 9)}
\]
Here, 3 meals are dispensed, so the average decrease per meal is \( \frac{9}{3} = 3 \).
3. Between \( x = 6 \) and \( x = 7 \):
\[
y \text{ changes from } 6 \text{ to } 3 \text{ (decrease of 3)}
\]
Here, one meal is dispensed, so the decrease per meal is 3.
From these observations, we can see that each meal dispensed typically decreases the remaining amount of pet food by a consistent rate of 3 cups per meal dispensed.
We can construct a linear function rule based on this information. The general form of a linear equation is:
\[
y = mx + b
\]
Where:
- \( m \) is the slope (the rate of change, which we found to be -3 in this case since \( y \) decreases as \( x \) increases).
- \( b \) is the y-intercept (the amount of food when no meals have been dispensed).
From our data, we can determine \( b \). When \( x = 0 \) (no meals dispensed), we can substitute into the equation. The slope indicates that if 1 meal decreases \( y \) by 3, then the starting amount of food can be extrapolated. However, based on the provided data, we can see that when 0 meals are dispensed, you would have:
Using the point \( (1, 21) \):
\[
21 = -3(1) + b \implies b = 21 + 3 = 24
\]
So our function rule that models this situation is:
\[
y = -3x + 24
\]
Thus, for this automatic feeder, the amount of pet food remaining \( y \) in cups after \( x \) meals have been dispensed can be expressed by the function:
\[
y = -3x + 24
\]
- Meals dispensed (x): 1, 3, 6, 7
- Pet food remaining (y): 21, 15, 6, 3
First, let's look for a pattern between \( x \) and \( y \):
1. Between \( x = 1 \) and \( x = 3 \):
\[
y \text{ changes from } 21 \text{ to } 15 \text{ (decrease of 6)}
\]
Here, 2 meals are dispensed, so the average decrease per meal is \( \frac{6}{2} = 3 \).
2. Between \( x = 3 \) and \( x = 6 \):
\[
y \text{ changes from } 15 \text{ to } 6 \text{ (decrease of 9)}
\]
Here, 3 meals are dispensed, so the average decrease per meal is \( \frac{9}{3} = 3 \).
3. Between \( x = 6 \) and \( x = 7 \):
\[
y \text{ changes from } 6 \text{ to } 3 \text{ (decrease of 3)}
\]
Here, one meal is dispensed, so the decrease per meal is 3.
From these observations, we can see that each meal dispensed typically decreases the remaining amount of pet food by a consistent rate of 3 cups per meal dispensed.
We can construct a linear function rule based on this information. The general form of a linear equation is:
\[
y = mx + b
\]
Where:
- \( m \) is the slope (the rate of change, which we found to be -3 in this case since \( y \) decreases as \( x \) increases).
- \( b \) is the y-intercept (the amount of food when no meals have been dispensed).
From our data, we can determine \( b \). When \( x = 0 \) (no meals dispensed), we can substitute into the equation. The slope indicates that if 1 meal decreases \( y \) by 3, then the starting amount of food can be extrapolated. However, based on the provided data, we can see that when 0 meals are dispensed, you would have:
Using the point \( (1, 21) \):
\[
21 = -3(1) + b \implies b = 21 + 3 = 24
\]
So our function rule that models this situation is:
\[
y = -3x + 24
\]
Thus, for this automatic feeder, the amount of pet food remaining \( y \) in cups after \( x \) meals have been dispensed can be expressed by the function:
\[
y = -3x + 24
\]
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