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Engineering, 26.02.2020 04:17 samehajamal1234567

Using the celsius_to_kelvin function as a guide, create a new function, changing the name to kelvin_to_celsius, and modifying the function accordingly 1 def celsius_to_kelvin(value_celsius): value_kelvin -. 4 value_kelvin- value_celsius 273.15 return value_kelvin Your solution goes here 9 value_c 0.e 10 valuek=0.0 - 12 value_c 10.8 13 15 16 print (value-c, 'Cis', celsius-to-kelvin(value-c), "K.) valuek# 283.15 print (value-k, .İS', kelvin-to-celsíus (value-k), 'C') -

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Using the celsius_to_kelvin function as a guide, create a new function, changing the name to kelvin_...

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