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Mathematics 9 Online
OpenStudy (anonymous):

could someone please show me how to work this problem out .... C = 5/9 (F - 32) solve for F?

OpenStudy (anonymous):

you cant?

OpenStudy (anonymous):

oh wait you just want F on its own?

OpenStudy (anonymous):

it's 5 over 9 not 5 divided by 9

OpenStudy (anonymous):

ya i just don't fully get the concept

OpenStudy (anonymous):

\[C = \frac{5}{9}(F - 32)\]

OpenStudy (anonymous):

celcius to farenheight?

OpenStudy (anonymous):

ya but i wan't to find out what F is

OpenStudy (anonymous):

" farenheight"

OpenStudy (anonymous):

its the same as this \[C = \frac{5(F-32)}{9}\]

OpenStudy (anonymous):

\[9C = 5(F-32)\]

OpenStudy (anonymous):

see what i did there? multiply by 9

OpenStudy (anonymous):

then divide by 5 \[\frac{9C}{5} = F -32\]

OpenStudy (anonymous):

then add 32 to both sides \[\frac{9C}{5} + 32 = F\]

OpenStudy (anonymous):

get what i did?

OpenStudy (anonymous):

sorta

OpenStudy (anonymous):

\[F= 32+\frac{ 9C }{ 5 }\]so you got

OpenStudy (anonymous):

yah

OpenStudy (anonymous):

COOL THANKS FOR EXPLAINING IT

OpenStudy (anonymous):

np

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