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October 6th, 2016, 02:13 PM   #1
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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?
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October 6th, 2016, 02:22 PM   #2
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$F = \dfrac{5}{9}C + 32 \implies C = \dfrac{9}{5}(F-32)$

for 32F to 50F ...

$C = \dfrac{9}{5}(32-32) = \dfrac{9}{5}(0) = 0$

$C = \dfrac{9}{5} (50-32) = \dfrac{9}{5} (18 ) = 32.4$

now try the other one you missed ...
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October 6th, 2016, 02:40 PM   #3
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Quote:
Originally Posted by skeeter View Post


$F = \dfrac{5}{9}C + 32 \implies C = \dfrac{9}{5}(F-32)$

for 32F to 50F ...

$C = \dfrac{9}{5}(32-32) = \dfrac{9}{5}(0) = 0$

$C = \dfrac{9}{5} (50-32) = \dfrac{9}{5} (18 ) = 32.4$

now try the other one you missed ...

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.
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October 6th, 2016, 03:40 PM   #4
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I typed the initial equation for F incorrectly (my discalcula). Should be ...

$F=\dfrac{9}{5}C + 32 \implies C= \dfrac{5}{9}(F-32)$

So, $F=50 \implies C=10$

Apologies.
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October 6th, 2016, 05:29 PM   #5
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Quote:
Originally Posted by skeeter View Post
I typed the initial equation for F incorrectly (my discalcula). Should be ...

$F=\dfrac{9}{5}C + 32 \implies C= \dfrac{5}{9}(F-32)$

So, $F=50 \implies C=10$

Apologies.
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