Reamur ke Fahrenheit
[tex] \frac{9}{4} R + 32 \\ \\ \frac{9}{4} (20°) + 32 \\ \\ 45° + 32° \\ \\ 77°F[/tex]
20°R = °F
= (9/4(20) + 32
= (9(20/4) + 32
= (9(5) + 32
= 45 + 32
= 77°F
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Verified answer
Reamur ke Fahrenheit
[tex] \frac{9}{4} R + 32 \\ \\ \frac{9}{4} (20°) + 32 \\ \\ 45° + 32° \\ \\ 77°F[/tex]
20°R = °F
= (9/4(20) + 32
= (9(20/4) + 32
= (9(5) + 32
= 45 + 32
= 77°F