To model the situation with an equation, we can let \( y \) represent the temperature of the oven in degrees Fahrenheit and \( x \) represent the time in minutes. Since the oven heats up at a constant rate, we can express the relationship as a linear equation in the form:
\[ y = mx + b \]
where \( m \) is the rate of change (slope), and \( b \) is the initial temperature of the oven when \( x = 0 \).
Given the points (4, 150) and (12, 350):
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Calculate the slope \( m \): \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{350 - 150}{12 - 4} = \frac{200}{8} = 25 \] This means the temperature increases by 25 degrees Fahrenheit per minute.
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Find the y-intercept \( b \): We can use one of the points to find \( b \). Using point (4, 150): \[ 150 = 25(4) + b \ 150 = 100 + b \ b = 150 - 100 = 50 \]
Now, we can write the equation for the temperature of the oven: \[ y = 25x + 50 \]
Predictions:
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The point (4, 150) means that after 4 minutes, the oven’s temperature was 150 degrees Fahrenheit.
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The point (12, 350) means that after 12 minutes, the oven’s temperature was 350 degrees Fahrenheit.
Using the equation to make a prediction:
To predict the temperature after, say, 10 minutes: \[ y = 25(10) + 50 \ y = 250 + 50 = 300 \] So, after 10 minutes, the oven’s temperature is predicted to be 300 degrees Fahrenheit.