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The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation $\frac{5F-160}{9}$ If the temperature is 86°F, what is the temperature in Celsius?

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation $\frac{5F-160}{9}$ .If the temperature is $35^{\circ}C$ , what is the temperature in Fahrenheit (F)?

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation $\frac{5F-160}{9}$ .If the temperature is $0^{\circ}C$ what is the temperature in Fahrenheit and if the temperature is $0^{\circ}F$ , what is the temperature in Celsius?

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation $C=\frac{5F-160}{9}$ .What is the numerical value of the temperature which is same in both the scales?

Answers (1)

Answer:

$C =30^{\circ}$

Solution:

Given, $\frac{5F-160}{9}$

Put F = 86° in equation (1)

$C =\frac{5(86)-160}{9}$

$C =\frac{430-160}{9}=\frac{270}{9}=30^{\circ}$

Hence, $C =30^{\circ}$

Answer:

$95^{\circ}F$

Solution:

Given, $\frac{5F-160}{9}$

Put C = $35^{\circ}C$ in above equation

$35^{\circ}=\frac{5F-160}{9}$

315 = 5F – 160

315 + 160 = 5F

475 = 5F

$F=\frac{475}{5}=95^{\circ}$

Answer:

$32^{\circ}F,-17.78^{\circ}C$

Solution:

Given $\frac{5F-160}{9}$ …(1)

Put $C=0^{\circ}$ in equation (1)

$0=\frac{5F-160}{9}$

5F = 160

$F=\frac{160}{5}=32^{\circ}$

Now put F = 0 in (1), we get

$C=\frac{160}{9}=-17.78^{\circ}C$

Answer:

– 40⁰

Solution:

Given that $C=\frac{5F-160}{9}$

It is asked that the numerical value of the temperature which is same in both the scales.

It means that the value of $^{\circ}C$ and $^{\circ}F$ are same

$\therefore F =\frac{5F-160}{9}$

9F – 5F = –160

4F = –160

F = –40⁰

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