Strange, I don't know where these "divide by 5, multiply by 8" quotes came from. It's actually divide by 5, multiply by *9*. That's why they're giving you the wrong answer. Odd that 2 people made the same mistake.
If you have a calculator simply multiply by 1.8 and add 32...
26 * 1.8 = 46.8
46.8 + 32 = 78.8
If you find yourself needing to work it out without a calculator then you need to come up with a way that works for you. For me, it's...
Take your °C temperature...
26
Double it...
52
Subtract 10% (10% is easily worked out by simply moving the decimal point one place to the left so 10% of 52 is 5.2)...
52 - 5.2 = 46.8
Then add 32...
46.8 + 32 = 78.8
Easy! :)
2006-11-18 00:37:06
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answer #1
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answered by amancalledchuda 4
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Centigrade is converted to Fahrenheit by multiplying the Centigrade figure by 9, dividing by 5, and adding 32
Answer 78.8 Fahrenheit
2006-11-19 20:08:13
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answer #2
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answered by Digital E 3
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9/5 (°Centigrade) + 32 = °Fahrenheit, the same method as the one described above.
alternatively plug you value on this site and an answer will be generated for you:
http://www.albireo.ch/temperatureconverter/
2006-11-19 06:56:11
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answer #3
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answered by Tala 3
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Multilpy by 9, divide by 5. Then add 32
=46.8 + 32
=78.8degs F
2006-11-18 09:16:19
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answer #4
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answered by rosie recipe 7
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If you use the conversion method above (divide by 5, multiply by 8, add 32) then it comes out at 73.6.
However, using a converter it comes out at 78.8 which is correct.
(it should be multiplied by 9 for the traditional method)
2006-11-18 05:25:03
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answer #5
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answered by Bror Jace 2
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26 Celsius (as its now politically correct to call it) is 78.8 Fahrenheit or 299.15 Kelvin
2006-11-19 11:28:34
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answer #6
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answered by WavyD 4
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About 80 degree's
2006-11-18 05:53:45
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answer #7
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answered by Anonymous
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78 .8 says the converter
2006-11-18 05:37:14
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answer #8
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answered by thirtymick 1
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divide by 5, multiply by 8 then add 32.
2006-11-18 05:19:25
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answer #9
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answered by Anonymous
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78 F (I don't know how the others came to 73 - but I just made a comparison on the dial)..........
2006-11-18 05:28:08
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answer #10
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answered by thomasrobinsonantonio 7
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