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At the coffee shop, the cost for using the internet is proportional to the time it's used. It costs $2 per hour to use the inte...
At the Cop Shop the cost for using the internet is proportional to the time it is used it cost $2 per hour to use the internet so let the graph drawn to show this relationship where X represents the number of hours and why represents the cost graph one why goes from 0 to 3 x goes from 0 to 6 with points at one two and three diagonal Craft 2 why goes from 0 to 4:00 it goes from 0 to 4 with 0.602 and 4 graph 3 why goes from -3 to 0 x goes from 0 to 6 with Point set negative one and two negative 2 and 4 breath for graph goes from -4 to 0 and Y and 0 to 4 on x with points at 0 negative 2 and 1 -4 and 2
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Answered by
GPT-4o mini
To visualize the relationships you've described, I will help you map out the equations and plot the graphs accordingly.
### Graph 1
For the first graph, where the cost \( y \) is proportional to the number of hours \( x \), if the cost is $2 per hour, we have:
\[
y = 2x
\]
#### Points to Plot:
- \( (0, 0) \) at \( x = 0 \)
- \( (1, 2) \) at \( x = 1 \)
- \( (2, 4) \) at \( x = 2 \) (this point is outside of the given range but helps to plot the line)
- \( (3, 6) \) at \( x = 3 \) (this point is also outside of the given range)
**Graph Range:** \( x = 0 \) to \( x = 6 \) and \( y = 0 \) to \( y = 4 \).
### Graph 2
For the second graph, you want points at (0, 2) and (4, 0):
Let's assume a linear relationship connecting these two points.
To find the slope \( m \):
- Points: \( (0, 2) \) and \( (4, 0) \)
- Slope \( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0-2}{4-0} = -\frac{1}{2} \)
Thus, the equation would be:
\[
y = -\frac{1}{2}x + 2
\]
#### Points to Plot:
- \( (0, 2) \)
- \( (4, 0) \)
### Graph 3
For the third graph, you have a range where \( y \) goes from -3 to 0 and \( x \) goes from 0 to 6 with the points of \((-1, 2)\) and \((-2, 4)\):
This might represent a different context but assuming you meant points within the range of positive coordinates, a linear equation isn't well-defined. Let's assume a reflective notion to give a straight line through \( (0, -3) \):
You might want:
\[
y = 0 \quad \text{at} \quad x = 0
\]
But let's use the points. It might be a mistake in pairing but a simple line downward.
### Graph 4
In graph four, you mention you have points \( (0, -2) \) and \( (1, -4) \) and an endpoint at \( y = 0 \).
Assuming the line connects directly between these two points, it would be:
Calculate the slope:
Points: \( (0, -2) \) and \( (1, -4) \)
\[
m = \frac{-4 - (-2)}{1 - 0} = -2
\]
Equation of the line is:
\[
y = -2x - 2
\]
This describes a line starting from \( x = 0 \).
### Summary
To summarize, let’s organize:
1. **Graph 1:** \( y = 2x \) with valid points (0,0), (1,2), (2,4), but should limit \( y \) at 4
2. **Graph 2:** \( y = -\frac{1}{2}x + 2 \) with points (0,2) and (4,0)
3. **Graph 3:** Needs better clarification, can assume linear down to the limit you need.
4. **Graph 4:** \( y = -2x - 2 \) will depict a line through the negative.
## Notes:
- Graphs should be plotted based on the linear equations derived. Make sure to properly use a graphing tool or graph paper to visualize them collectively.
- The points should be within the right quadrant based on what you’re trying to observe.
### Graph 1
For the first graph, where the cost \( y \) is proportional to the number of hours \( x \), if the cost is $2 per hour, we have:
\[
y = 2x
\]
#### Points to Plot:
- \( (0, 0) \) at \( x = 0 \)
- \( (1, 2) \) at \( x = 1 \)
- \( (2, 4) \) at \( x = 2 \) (this point is outside of the given range but helps to plot the line)
- \( (3, 6) \) at \( x = 3 \) (this point is also outside of the given range)
**Graph Range:** \( x = 0 \) to \( x = 6 \) and \( y = 0 \) to \( y = 4 \).
### Graph 2
For the second graph, you want points at (0, 2) and (4, 0):
Let's assume a linear relationship connecting these two points.
To find the slope \( m \):
- Points: \( (0, 2) \) and \( (4, 0) \)
- Slope \( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0-2}{4-0} = -\frac{1}{2} \)
Thus, the equation would be:
\[
y = -\frac{1}{2}x + 2
\]
#### Points to Plot:
- \( (0, 2) \)
- \( (4, 0) \)
### Graph 3
For the third graph, you have a range where \( y \) goes from -3 to 0 and \( x \) goes from 0 to 6 with the points of \((-1, 2)\) and \((-2, 4)\):
This might represent a different context but assuming you meant points within the range of positive coordinates, a linear equation isn't well-defined. Let's assume a reflective notion to give a straight line through \( (0, -3) \):
You might want:
\[
y = 0 \quad \text{at} \quad x = 0
\]
But let's use the points. It might be a mistake in pairing but a simple line downward.
### Graph 4
In graph four, you mention you have points \( (0, -2) \) and \( (1, -4) \) and an endpoint at \( y = 0 \).
Assuming the line connects directly between these two points, it would be:
Calculate the slope:
Points: \( (0, -2) \) and \( (1, -4) \)
\[
m = \frac{-4 - (-2)}{1 - 0} = -2
\]
Equation of the line is:
\[
y = -2x - 2
\]
This describes a line starting from \( x = 0 \).
### Summary
To summarize, let’s organize:
1. **Graph 1:** \( y = 2x \) with valid points (0,0), (1,2), (2,4), but should limit \( y \) at 4
2. **Graph 2:** \( y = -\frac{1}{2}x + 2 \) with points (0,2) and (4,0)
3. **Graph 3:** Needs better clarification, can assume linear down to the limit you need.
4. **Graph 4:** \( y = -2x - 2 \) will depict a line through the negative.
## Notes:
- Graphs should be plotted based on the linear equations derived. Make sure to properly use a graphing tool or graph paper to visualize them collectively.
- The points should be within the right quadrant based on what you’re trying to observe.
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