Question

as a certain engines rotation speed increases, its temperature increases at a constant rate.
the table compares the engines rotation speed (in cycles per second) and its temperature (in degrees celsius).
rotation speed (cycles per second) temperature (degrees celsius)
9 22.2
12 24.6
15 27.0
what is the engines temperature at rest?
degrees celsius
show calculator

Explanation:

Step1: Find the slope of the linear - relationship

The slope $m$ of a line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $m=\frac{y_2 - y_1}{x_2 - x_1}$. Let $(x_1,y_1)=(9,22.2)$ and $(x_2,y_2)=(12,24.6)$. Then $m=\frac{24.6 - 22.2}{12 - 9}=\frac{2.4}{3}=0.8$.

Step2: Use the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$. Using the point $(9,22.2)$ and $m = 0.8$, we have $y-22.2=0.8(x - 9)$.

Step3: Simplify the equation

Expand the right - hand side: $y-22.2=0.8x-7.2$. Then $y=0.8x + 15$.

Step4: Find the temperature at rest

When the engine is at rest, $x = 0$. Substitute $x = 0$ into the equation $y=0.8x + 15$. We get $y=15$.

Answer:

15