At what temperature is the Fahrenheit scale reading equal to (a) twice (b) half of the Celsius?

We know that,

TC = (TF – 32) x \(frac{5}{9}\) (equ.1)

(a) Twice the Celsius

TF = 2TC

From the equ.1, we can substitute for TF

TC = (2TC – 32) x \(frac{5}{9}\)

9TC = 10TC – 160

TC = 160°C

(b) Half of the Celsius

TF = \(frac{T_{C}}{2}\)

From the equ.1, we can substitute for TF

TC = (\(frac{T_{C}}{2}\) – 32) x \(frac{5}{9}\)

9TC = \(frac{5}{2}T_{C}\) – 160

TC = -24.6°C

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