To determine when the costs of the two landscaping companies are equal, we can set up equations for each company's charges based on the information provided.
For Company A:
- Cost \( A(t) = 50t + 150 \)
For Company B:
- Cost \( B(t) = 25t + 300 \)
To find the number of hours \( t \) where the costs are equal, we set the two equations equal to each other:
\[ 50t + 150 = 25t + 300 \]
Now, let's solve for \( t \):
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Subtract \( 25t \) from both sides: \[ 50t - 25t + 150 = 300 \] \[ 25t + 150 = 300 \]
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Subtract 150 from both sides: \[ 25t = 150 \]
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Divide by 25: \[ t = 6 \]
So, the two companies charge the same amount when \( t = 6 \) hours.
Graphing the Functions
To graph the functions:
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Create a table of values:
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For Company A \( A(t) = 50t + 150 \):
- If \( t = 0 \), \( A(0) = 150 \)
- If \( t = 6 \), \( A(6) = 50 \cdot 6 + 150 = 300 \)
- If \( t = 12 \), \( A(12) = 50 \cdot 12 + 150 = 600 \)
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For Company B \( B(t) = 25t + 300 \):
- If \( t = 0 \), \( B(0) = 300 \)
- If \( t = 6 \), \( B(6) = 25 \cdot 6 + 300 = 450 \)
- If \( t = 12 \), \( B(12) = 25 \cdot 12 + 300 = 600 \)
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Plot the points:
- For Company A: (0, 150), (6, 300), (12, 600)
- For Company B: (0, 300), (6, 450), (12, 600)
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Graph the lines:
- The linear function for Company A will have a steeper slope than Company B due to the higher hourly rate.
- The intersection point of \( A(t) \) and \( B(t) \) will be at \( (6, 300) \) representing 6 hours at a cost of $300.
Writing it on a Graph
To display on the graph:
- Draw the y-axis (cost) and x-axis (time in hours).
- Label each axis appropriately.
- Mark the y-values for costs and plot the points for both companies.
- Draw the lines for both functions.
- Clearly mark the intersection point at \( (6, 300) \) and label it accordingly.
This point indicates that both companies charge the same amount ($300) for 6 hours of landscaping.