a. 250 °C + 273.15 = 523.15 K
b. 90 °C + 273.15 = 363.15 K
c. (10 °C × 9/5) + 32 = 50 °F
d. (33 °C × 9/5) + 32 = 91.4 °F
e. (166 °F − 32) × 5/9 + 273.15 = 347.594 K
f. (0 °F − 32) × 5/9 + 273.15 = 255.372 K
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Verified answer
a. 250 °C + 273.15 = 523.15 K
b. 90 °C + 273.15 = 363.15 K
c. (10 °C × 9/5) + 32 = 50 °F
d. (33 °C × 9/5) + 32 = 91.4 °F
e. (166 °F − 32) × 5/9 + 273.15 = 347.594 K
f. (0 °F − 32) × 5/9 + 273.15 = 255.372 K