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could someone please show me how to work this problem out .... C = 5/9 (F - 32) solve for F?
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you cant?
oh wait you just want F on its own?
it's 5 over 9 not 5 divided by 9
ya i just don't fully get the concept
\[C = \frac{5}{9}(F - 32)\]
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celcius to farenheight?
ya but i wan't to find out what F is
" farenheight"
its the same as this \[C = \frac{5(F-32)}{9}\]
\[9C = 5(F-32)\]
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see what i did there? multiply by 9
then divide by 5 \[\frac{9C}{5} = F -32\]
then add 32 to both sides \[\frac{9C}{5} + 32 = F\]
get what i did?
sorta
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\[F= 32+\frac{ 9C }{ 5 }\]so you got
yah
COOL THANKS FOR EXPLAINING IT
np
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