Using this and the knowledge that 32° F = 0°C and 212°F = 100°C, find an equation that computes Celsius temperature in terms of Fahrenheit temperature; i.e., an equation of the form C = “an expression involving only the variable F.”
c=(f-32)/1.8

Likewise, find an equation that computes Fahrenheit temperature in terms of Celsius temperature; i.e. an equation of the form F = “an expression involving only the variable C.”
f=1.8c+32

(a) How cold was it in Oslo in °F?

(b) What temperature is it the same in °F and °C?

The equations are: [tex]F = 1.8C + 32[/tex] and [tex]C = \frac{(F - 32)}{1.8}[/tex].
The temperature in Oslo is [tex]60.8^oF[/tex].
[tex]-40^oF[/tex] is equivalent to [tex]-40^oC[/tex]

Given that:

[tex](F_1,C_1) = (32,0)[/tex]

[tex](F_2,C_2) = (212,100)[/tex]

First, calculate the slope (m)

[tex]m = \frac{C_2 - C_1}{F_2 - F_1}[/tex]

So we have:

[tex]m = \frac{100 - 0}{212 - 32}[/tex]

[tex]m = \frac{100}{180}[/tex]

[tex]m = \frac{5}{9}[/tex]

So, the equation is calculated using:

[tex]C= m(F - F_1) + C_1[/tex]

So, we have:

[tex]C = \frac 59(F - 32) + 0[/tex]

[tex]C = \frac 59(F - 32)[/tex]

Rewrite as:

[tex]C = \frac{(F - 32)}{9/5}[/tex]

[tex]C = \frac{(F - 32)}{1.8}[/tex]

To solve for F, we multiply both sides by 1.8

[tex]1.8C =F - 32[/tex]

Add 32 to both sides

[tex]F = 1.8C + 32[/tex]

(a) The temperature of Oslo in F

We have:

[tex]C = 16[/tex]

So, we make use of:

[tex]F = 1.8C + 32[/tex]

[tex]F = 1.8 \times 16 + 32[/tex]

[tex]F = 60.8[/tex]

The temperature in Oslo is [tex]60.8^oF[/tex]

(b) When F = C

We have:

[tex]F = 1.8C + 32[/tex]

Substitute F for C

[tex]F = 1.8F +32[/tex]

Collect like terms

[tex]F - 1.8F =32[/tex]

[tex]-0.8F =32[/tex]

Make F the subject

[tex]F = \frac{32}{-0.8}[/tex]

[tex]F = -40[/tex]

Hence,

[tex]-40^oF[/tex] is equivalent to [tex]-40^oC[/tex]