F(x) = 9/5 (x -273.15) + 32; The function F gives the temperature, in degrees Fahrenheit, that.....

Bluebook Digital SAT Test 2 Module 2 (Hard) Question 14:

F(x) = 9/5 (x -273.15) + 32
The function F gives the temperature, in degrees Fahrenheit, that corresponds to a temperature of x kelvins. If a temperature increased by 2.10 kelvins, by how much did the temperature increase, in degrees Fahrenheit?

KEEP IN TOUCH:

[clau]

I’m confused so I’m just gonna write it out 😭.

So x is equal to the temp in kelvin, but they give us the amount that the temperature increased by.

So in order to calculate how much the temperature increased in Fahrenheit, we’d have to subtract the temp in Fahrenheit before and after an increase in 2.10 kelvin.

By making the kelvin equal to 273.15, were conveniently making the Fahrenheit equal to 32.
After making an increase in 2.10 to the kelvin, 275.25, it will give us the temperature in Fahrenheit after an increase in 2.10 kelvin.

If we subtract those 2 values, it gives us what we’re looking for; how much the temperature increased in Fahrenheit.

Realistically you could have plugged in any value for x as long as you added 2.10 the second time to x. But by cancelling out the first part of the equation altogether, you simplified it significantly.

I get it now 😅 thank you for this!! The explanation given by college board is a nightmare lol.

[ZaidAli-lint]

how did you know where to put the 2.10 kelvins? Why'd you also replace the 273.15 with the 2.10 kelvins? Thanks!

[Cleaps]

I still don't understand why you don't add the 32 to 3.78.

[celinemneimne]

how do you know when you can plug in a random value for x or any other variable (in any question)