At noon, the temperature of a container of ether was –109⁰C. At 1 p.m., the temperature of the container of ether was –34⁰C.

What was the change in temperature from noon to 1 p.m.?

⁰C

Question

At noon, the temperature of a container of ether was –109⁰C. At 1 p.m., the temperature of the container of ether was –34⁰C.
 
What was the change in temperature from noon to 1 p.m.?
 
   
 ⁰C

kva1992 · Answer

wouldnt you just subtract –109⁰C by –34⁰C?

anonymous · Answer

@alanabrisa

anonymous · Answer

kva1992 is right, you would subtract to get the difference.

anonymous · Answer

oh, ok

anonymous · Answer

-75? @alanabrisa

anonymous · Answer

Actually hold on a second

anonymous · Answer

Ok so I thinkk, instead of subtracting -109 - (-34), we want to subtract -34 - (-109). Because you want to find the change in temperature from -109 to -34. And since the temperature is rising, it wouldn't make sense for it to be -75. because that means the temperature went down. But if we subtract -34 - (-109) = 75. And it makes more sense if the temperature rose 75 degrees and is now at -34.

anonymous · Answer

Does that kind of make sense or did I ramble too much? lol

kva1992 · Answer

err that what i essentially meant lol

kva1992 · Answer

essentially you could do it the other way and all you would have to do is multiply a negative to make it positive but you would need to know what you are doing.

anonymous · Answer

ok. so now the answer is -34?

anonymous · Answer

Sorry openstudys giving me a hard time lol So just solve the equation -34 - (-109) = ? Sorry if I confused you

kva1992 · Answer

The answer is 75