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C = 5/9 (f-32) = 5/9 (85-32) = 5/9 (53) = 29.4c

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Q: If the temperature outside is 85f what is the temperature on the celsius scale?
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Is there a temperature at which the Celsius reading is 20 degrees higher that the Fahrenheit reading?

Yes, there is one. The conversion factor from the English (Fahrenheit) to the System International metric (Celsius) can be expressed as follows: F = 1.8C + 32 (also 9/5C + 32) Here are some examples: at one atmosphere of pressure, water freezes at 0 degrees Celsius. Plugging into the equation, 1.8 x 0 equals 0, plus 32 equals 32 degrees Fahrenheit, also the freezing point of water on that scale. Water begins to boil at 100 C, also a well-known fact. Plugging in again, 1.8 x 100 equals 180, plus 32 equals 212. 212 is the boiling point of water on the Fahrenheit scale. Using algebra, we can find out where the two scales meet. Recalling the first equation: F = 1.8C + 32, but at the point where the scales meet F must be equal to C. Therefore, let F = C and C = F. Substituting... F = 1.8F + 32 F - 1.8F = 32 F(1 - 1.8) = 32 F = 32/(1-1.8) = 32/-0.8 F = -40 -40 Fahrenheit equals -40 Celsius Now let us set F as being equal to C + 20 (or F + 20) more degrees: F = 1.8(F + 20) + 32 F = 1.8F + 36 + 32 F = 1.8F + 68 F - 1.8F = 68 F(1-1.8) = 68 F(-0.8) = 68 F = 68/-0.8 F = -85 ;... At a Fahrenheit temperature of -85F, the Celsius reading is exactly 20 degrees higher at -65C. That's colder than it gets where I'm from!


At what temperature is celsius 20 degrees higher than Fahrenheit?

Ordinay conversion of 'F' to 'C':C = 5/9 (F - 32)But you said that the 'C' number is 20 more than the 'F' number, soF + 20 = 5/9 (F - 32)Multiply each side by 9/5:9/5 (F + 20) = F - 32(9/5)F + 36 = F - 32Subtract F from each side:(4/5)F + 36 = -32Subtract 36 from each side:(4/5)F = -68Multiply each side by (5/4) = 1.25F = (1.25) x (-68)= -85.The temperature is -85F or -65C. (They're the same temperature.)==================================Check:C = 5/9 (F - 32)C = 5/9 (-85 - 32) = 5/9 (-117) = -65-65 - 20 = -85yay !


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