Question

25ºR= ºF
25ºF= K
333K= ºC
333ºC= ºR
333ºR= ºF
yg baik tolong

Answer 1

25ºR= 9/4 x 25 = 56,25 + 32 = 88,25ºF
25ºF= (25-32)x 5/9 = 3,89 + 273 = 276,89 °K
333K= 333-273 = 60 ºC
333ºC= 4/5 x 333 = 266,4 ºR
333ºR= 9/4 x 333 = 749,25 + 32 = 781,25 ºF

Answer 2

25°R=(25×9/4)+32
=(225/4)+32
=56.25+32
=88.25°F
25°F=((25-32)×5/9)+273
=(-7×5/9)+273
=(3/9)+273
=3.89+273
=276.89°K
333°K=333-273
=60°C
333°C=333×4/5
=1332/5
=266.4°R
333°R=(333×9/4)+32
=(2997/4)+32
=749.25+32
=781.25°F