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

Answer 1

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.