Ask your own question, for FREE!
Mathematics 16 Online
OpenStudy (anonymous):

Anyone good with temperature problems? involving writing an algebraic inequality? I got an answer but I dont understand how it was done. Please help.

OpenStudy (anonymous):

I will attempt to help you out post the problem

OpenStudy (anonymous):

what you got?

OpenStudy (anonymous):

Solve. Show an algebaric inequality. In order for a chemical reaction to remain stable, its celsius temperature must be no more than 111.4 deg. C. Find the Fahrenheit temp. at which this reaction will remain stable. F = 9/5C + 32.

OpenStudy (anonymous):

If you can please show me how you did the problem so I can know how to do these types later on as well. Thank you! :D

OpenStudy (anonymous):

this is easy

OpenStudy (anonymous):

you ready for this?

OpenStudy (anonymous):

yes :D

OpenStudy (anonymous):

here it comes

OpenStudy (anonymous):

F=9/5c+32 F=9/5*111.4+32 F=1,002.6/5+32 F=200.52+32 F=232.52 DEGREES ANS

OpenStudy (anonymous):

you know the drill make it easy for meeeee!!

OpenStudy (anonymous):

satisfied??

OpenStudy (anonymous):

ok im sorry but you lost me when you skipped a line. you just put 111.4 / 1?

OpenStudy (anonymous):

and multiplied 9/5 * 111.4/1?

OpenStudy (anonymous):

exactly

OpenStudy (anonymous):

so for the ineqauilty would it be F < or F >?

OpenStudy (anonymous):

F<??

OpenStudy (anonymous):

F<232.52 right?

OpenStudy (anonymous):

It would be : Temperature>232.52F

OpenStudy (anonymous):

ahhhhh, i seee i seee. THANK YOU A TON!!!!! :)

OpenStudy (anonymous):

Fahrenheit temperature must be no more than 232.52 degrees Fahrenheit

OpenStudy (anonymous):

satisfied?

OpenStudy (anonymous):

yes thank you!! :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!