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October 6th, 2016, 02:13 PM  #1 
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10  Use the formula F = 9/5 C + 32 to find the corresponding temperature range
Why did I get these wrong? is it just the order? as 0 is warmer than 2 or is it something else?

October 6th, 2016, 02:22 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587  $F = \dfrac{5}{9}C + 32 \implies C = \dfrac{9}{5}(F32)$ for 32F to 50F ... $C = \dfrac{9}{5}(3232) = \dfrac{9}{5}(0) = 0$ $C = \dfrac{9}{5} (5032) = \dfrac{9}{5} (18 ) = 32.4$ now try the other one you missed ... 
October 6th, 2016, 02:40 PM  #3  
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10  Quote:
so for the ones I got right I divided the answer by the reciprocal to get c alone, why do we not do that here? like 9 goes into 18, twice then 2(5) =10? Last edited by GIjoefan1976; October 6th, 2016 at 02:57 PM. Reason: ah well it worked, I think I just missed my signs.  
October 6th, 2016, 03:40 PM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 
I typed the initial equation for F incorrectly (my discalcula). Should be ... $F=\dfrac{9}{5}C + 32 \implies C= \dfrac{5}{9}(F32)$ So, $F=50 \implies C=10$ Apologies. 
October 6th, 2016, 05:29 PM  #5 
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10  

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9 or 5, f95c, find, formula, range, temperature 
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