Question

Please tell me answer​

Answer 1

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• Is there a temperature which is numerically equal in both Fahrenheit and Celcius? If so, find them.

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Given that,

$\sf F=\frac{9}{5}C+32$

Now, Temperature in both scale are numerically equal. So,

$\sf F=C</p><p>[tex]\sf \implies F=\frac{9}{5}F+32$

$\sf \implies F-\frac{9}{5}F =32$

$\sf \implies \frac{5F-9F}{5}=32$

$\sf \implies \frac{-4}{5}F=32$

$\sf \implies F=32\times \frac{-5}{4}$

$\sf \implies F=-40$

Therefore, $-40\degree C$ is the temperature which is numerically equal in both Celcius and Fahrenheit scale, i.e.,

$-40\degree C=-40\degree F$

Answer 2

Now, Temperature in both scale are numerically equal. So,

F+32

32⟹F− 9/5

F=32

⟹ 5×5F−9F =32

F=32⟹ 5−4

F=32

⟹F=32×4−5

⟹F=−40

Therefore, -40\degree C−40°C is the temperature which is numerically equal in both Celcius and Fahrenheit scale, i.e.,

F−40°C=−40°F