suraj took a slice of pizza from the freezer and put it in the oven. the pizza was heated at a constant…

suraj took a slice of pizza from the freezer and put it in the oven. the pizza was heated at a constant rate.\nthe table compares the pizzas temperature (in degrees celsius) and the time since suraj started heating it (in minutes).\ntime (minutes) temperature (degrees celsius)\n1 2.5\n7 47.5\n13 92.5\nhow fast was the pizza heated?\ndegrees celsius per minute

suraj took a slice of pizza from the freezer and put it in the oven. the pizza was heated at a constant rate.\nthe table compares the pizzas temperature (in degrees celsius) and the time since suraj started heating it (in minutes).\ntime (minutes) temperature (degrees celsius)\n1 2.5\n7 47.5\n13 92.5\nhow fast was the pizza heated?\ndegrees celsius per minute

Answer

Explanation:

Step1: Select two data - points

Let's take the first (1, 2.5) and the second (7, 47.5) data - points.

Step2: Calculate the rate of change

The rate of change formula is $\frac{\text{Change in temperature}}{\text{Change in time}}$. Change in temperature $=47.5 - 2.5=45$ degrees Celsius, change in time $=7 - 1 = 6$ minutes. So the rate is $\frac{47.5 - 2.5}{7 - 1}$.

Step3: Simplify the expression

$\frac{47.5 - 2.5}{7 - 1}=\frac{45}{6}=7.5$ degrees Celsius per minute.

Answer:

7.5