Question

as fórmulas c= \frac{5}{9} (f-32) e f= \frac{9}{5}c+32 são usadas para transformar graus Fahrenheit em graus Celsius,e vice-versa.Resolva os problemas utilizando calculadora.
a)Transforme 170°F em graus Celsius.
b)Qual temperatura é mais baixa:0° ou 10°F?
c)Em determinada cidade,no dia primeiro,a temperatura média foi 45°F.No dia 2 a temperatura média foi 43°F.No dia 3,a temperatura média foi 41°F.Se esse padrão continuar,qual será a temperatura média em graus Celsius no dia 4?

Answer 1

Resolução da questão, veja:

Letra ''A'':

$\dfrac{\°C}{5}} = \dfrac{\°F-32}{9}}\\\\\\\\ \dfrac{\°C}{5}} = \dfrac{170-32}{9}}\\\\\\\\ 9\°C = 5~\cdot~138\\\\\\\\\ 9\°C = 690\\\\\\\\ \°C = \dfrac{690}{9}}\\\\\\\\\\\ \large\boxe{\boxed{\boxed{\boxed{\boxed{\°C \approx  76,66}}}}}}}}}}}$

Ou seja, 170 °F equivale à aproximadamente 76,66 °C.

Letra ''B'':

$\dfrac{\°C}{5}} = \dfrac{\°F-32}{9}}\\\\\\\\\\\ \dfrac{0}{5} = \dfrac{\°F-32}{9}}\\\\\\\\ 0 = 5\°F-160\\\\\\\\ 5\°F = 160\\\\\\\\ \°F = \dfrac{160}{5}}\\\\\\\ \°F = 32.$

$\dfrac{\°C}{5}} = \dfrac{\°F-32}{9}}\\\\\\\\ \dfrac{\°C}{5}} = \dfrac{10-32}{9}}\\\\\\\\ 9\°C = 5~\cdot~(-22)\\\\\\\\\ 9\°C = -110\\\\\\\\ \°C = \dfrac{-110}{9}}\\\\\\\ \°C \approx -12,22.$

Ou seja, 10 °F é mais frio do que 0 °C.

Letra ''C'':

$\Delta{\Theta}} = 41 - 2\\\\\\\ \Delta{\Theta}} = 39\°F\\\\\\\\\\\\\ \dfrac{\°C}{5} = \dfrac{\°F-32}{9}}\\\\\\\\\\\ \dfrac{\°C}{5}} = \dfrac{39-32}{9}}\\\\\\\\\ 9\°C = 9~\cdot~7\\\\\\\\\\ 9\°C = 63\\\\\\\\\ \°C=\dfrac{63}{9}}\\\\\\\\ \large{\boxed{\boxed{\boxed{\°C = 7.}}}}}}}}}}$

Ou seja, a variação de temperatura no Quarto dia equivale à 39 °F ou 7 °C.

Espero que te ajude :).