If the reading in Fahrenheit scale is twice t | vedclass

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(D) The relationship between the Fahrenheit scale $(F)$ and the Celsius scale $(C)$ is given by the formula: $\frac{F-32}{9} = \frac{C}{5}$.Given that

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$320$

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(D) The relationship between the Fahrenheit scale $(F)$ and the Celsius scale $(C)$ is given by the formula: $\frac{F-32}{9} = \frac{C}{5}$.Given that the reading in Fahrenheit is twice the reading in Celsius, we have $F = 2C$, which implies $C = \frac{F}{2}$.Substituting $C = \frac{F}{2}$ into the temperature conversion formula:$\frac{F-32}{9} = \frac{F/2}{5}$$\frac{F-32}{9} = \frac{F}{10}$Cross-multiplying gives: $10(F - 32) = 9F$.$10F - 320 = 9F$.$10F - 9F = 320$.$F = 320^{\circ} F$.

If the reading in Fahrenheit scale is twice the reading in Celsius scale, then the reading in Fahrenheit scale is (in $^{\circ} F$)

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