To calculate the total cost, \( C \), of purchasing \( p \) pounds of rice and a bottle of seasoning, we can use the cost information provided:
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The cost of rice is $6.50 for 5 pounds. Therefore, the cost per pound of rice, \( r \), is: \[ r = \frac{6.50}{5} = 1.30 \text{ dollars per pound} \]
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The cost of one bottle of seasoning remains constant at $1.00.
The total cost \( C \) as a function of pounds of rice \( p \) can be expressed as: \[ C(p) = 1.30p + 1.00 \]
Graphing the Equation
- X-axis (Rice in pounds, \( p \)): This ranges from 1 to 10 pounds.
- Y-axis (Total cost, \( C \)): We need to calculate the total cost for different values of \( p \).
Calculations for Different Values of \( p \):
- For \( p = 1 \): \[ C(1) = 1.30(1) + 1.00 = 2.30 \]
- For \( p = 2 \): \[ C(2) = 1.30(2) + 1.00 = 3.60 \]
- For \( p = 3 \): \[ C(3) = 1.30(3) + 1.00 = 4.90 \]
- For \( p = 4 \): \[ C(4) = 1.30(4) + 1.00 = 6.20 \]
- For \( p = 5 \): \[ C(5) = 1.30(5) + 1.00 = 7.50 \]
- For \( p = 6 \): \[ C(6) = 1.30(6) + 1.00 = 8.80 \]
- For \( p = 7 \): \[ C(7) = 1.30(7) + 1.00 = 10.10 \]
- For \( p = 8 \): \[ C(8) = 1.30(8) + 1.00 = 11.40 \]
- For \( p = 9 \): \[ C(9) = 1.30(9) + 1.00 = 12.70 \]
- For \( p = 10 \): \[ C(10) = 1.30(10) + 1.00 = 14.00 \]
Summary of Points to Plot:
- (1, 2.30)
- (2, 3.60)
- (3, 4.90)
- (4, 6.20)
- (5, 7.50)
- (6, 8.80)
- (7, 10.10)
- (8, 11.40)
- (9, 12.70)
- (10, 14.00)
Plotting the Graph
- Plot these points on your graph, where the x-axis (pounds of rice) ranges between 1 and 10 and the y-axis (total cost) includes values from about 2.20 to 14.00.
- Connect the points with a straight line, since the relationship is linear.
Conclusion
The graph \( C \) vs. \( p \) will be a straight line starting from (1, 2.30) and increasing linearly to (10, 14.00). The slope of the line represents the cost per additional pound of rice, which is $1.30, while the y-intercept (the cost when no rice is purchased) is $1.00, which is the cost of the seasoning.
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