Question

Quy đổi 25⁰C,40⁰C,68⁰C,75⁰C sang ⁰F

Quy đổi 68⁰F,109 ⁰F; 136,4^oF,226,4^oF sang ^oC

Answer 1

Quy tắc : ^oC = \(\dfrac{^oF-32}{1,8}\)

^oF = ( ^oC x 1,8 ) + 32

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Answer 2

a) \(25^0C=77^0F\)

\(40^0C=104^0F\)

\(68^0C=154,4^0F\)

\(75^0C=167^0F\)

b) \(68^0F=20^0C\)

\(109^0F=47,78^0C\)

\(136,4^0F=58^0C\)

\(226,4^0F=108^0C\)

Answer 3

25\(^0\)C= 0\(^0\)C+25\(^0\)C. Vậy:

25\(^0\)C= 32\(^0\)F+ ( 25 x 1,8\(^0\)F)=77\(^0\)F

40\(^0\)C= 0\(^0\)C+40\(^0\)C. Vậy:

40\(^0\)C= 32\(^0\)F+ (40 x 1,8\(^0\)F)=104\(^0\)F

68\(^0\)C= 0\(^0\)C+68\(^0\)C. Vậy:

68\(^0\)C= 32\(^0\)F+ (68 x 1,8\(^0\)F)= 154,4\(^0\)F

75\(^{^{ }0}\)C= 0\(^{^{ }0}\)C + 75\(^0\)C. Vậy:

75\(^0\)C= 32\(^0\)F+ (75 x 1,8\(^0\)F)=167\(^0\)F

68\(^0\)F=\(\dfrac{68-32}{1,8}\)=\(\dfrac{36}{1,8}\)=20\(^0\)C

109\(^0\)F= \(\dfrac{109-32}{1,8}\)=\(\dfrac{77}{1,8}\)= 42,777778\(^0\)C

136,4\(^0\)F=\(\dfrac{136,4-32}{1,8}\)=\(\dfrac{104,4}{1,8}\)= 58\(^0\)C

226,4\(^0\)F=\(\dfrac{226,4-32}{1,8}\)=\(\dfrac{194,4}{1,8}\)=108\(^0\)C

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