As a certain engine's rotation speed increases, its temperature increases at a constant rate.

The table compares the engine's rotation speed (in cycles per second) and its temperature (in degrees Celsius).

Rotation (cycles per second): 11, 13, and 15.

Temperature (degrees Celsius): 23.8, 25.4, and 27.

What is the increase in temperature that comes with an increase of 1 cycle per second in the rotation?

Moreover, the increase in temperature that comes with an increase of 1 cycle per second in the rotation corresponds to the rate of change of the relationship.

The table of values shows that for each increase of in Rotation, Temperature increases by 1.6 degrees Celsius. The rate of change is the ratio of those corresponding differences.

ΔRotation =1.6

ΔTemperature =2

1.6/2=0.8

In conclusion, the increase in temperature that comes with an increase of 1 cycle per second in the rotation is 0.8 degrees Celsius.

boffeemadrid boffeemadrid · Answer 2 · 2021-10-13T10:10:14+02:00

The change in temperature is required when there is an increase of 1 cycle of rotation per second.

The required rate is is 0.8 degrees Celsius per increase of 1 cycle per second in the rotation.

Let the rotation be represented by the x axis

and the temperature be represented by the y axis.

The temperature increases at a constant rate with increase engine's certain rotation speed so it is a linear relationship.

As the increase in temperature is required when there is an increase of 1 cycle of rotation per second we need to find the slope.

The means the ratio of change in y divided by the change in x.

Change in y = [tex]\Delta y=25.4-23.8=1.6[/tex]

Change in x = [tex]\Delta x=13-11=2[/tex]

The slope is

[tex]m=\dfrac{\Delta y}{\Delta x}\\\Rightarrow m=\dfrac{1.6}{2}\\\Rightarrow m=0.8[/tex]

So, the required rate is is 0.8 degrees Celsius per increase of 1 cycle per second in the rotation.

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