Question

suraj took a slice of pizza from the freezer and put it in the oven. the pizza was heated at a constant rate.
the table compares the pizzas temperature (in degrees celsius) and the time since suraj started heating it (in minutes).
time (minutes) temperature (degrees celsius)
1 2.5
7 47.5
13 92.5
how fast was the pizza heated?
degrees celsius per minute

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