Answer :
For this case, the first thing we must do is define variables.
We have then:
c: temperature in degrees celsius
f: temperature in Fahrenheit
The generic equation of the line is:
[tex]f-fo = m (c-co)
[/tex]
Where, the slope is given by:
[tex]m = \frac{f2-f1}{c2-c1} [/tex]
Substituting values:
[tex]m = \frac{212-122}{100-50} [/tex]
Rewriting:
[tex]m = \frac{90}{50} [/tex]
[tex]m = \frac{9}{5} [/tex]
We choose an ordered pair:
[tex] (co, fo) = (100, 212)
[/tex] Substituting values in the generic equation:
[tex]f-212 = \frac{9}{5} (c-100)
[/tex]
Rewriting:
[tex]f-212 = \frac{9}{5}c - 180 [/tex]
[tex]f = \frac{9}{5}c - 180 + 212
[/tex]
[tex]f = \frac{9}{5}c + 32
[/tex]
Answer:
An equation of the line in slope-intercept form is:
[tex]f = \frac{9}{5}c + 32 [/tex]
See attached image.
We have then:
c: temperature in degrees celsius
f: temperature in Fahrenheit
The generic equation of the line is:
[tex]f-fo = m (c-co)
[/tex]
Where, the slope is given by:
[tex]m = \frac{f2-f1}{c2-c1} [/tex]
Substituting values:
[tex]m = \frac{212-122}{100-50} [/tex]
Rewriting:
[tex]m = \frac{90}{50} [/tex]
[tex]m = \frac{9}{5} [/tex]
We choose an ordered pair:
[tex] (co, fo) = (100, 212)
[/tex] Substituting values in the generic equation:
[tex]f-212 = \frac{9}{5} (c-100)
[/tex]
Rewriting:
[tex]f-212 = \frac{9}{5}c - 180 [/tex]
[tex]f = \frac{9}{5}c - 180 + 212
[/tex]
[tex]f = \frac{9}{5}c + 32
[/tex]
Answer:
An equation of the line in slope-intercept form is:
[tex]f = \frac{9}{5}c + 32 [/tex]
See attached image.
The graph that models the relationship between the Fahrenheit and Celsius scales is attached below.
An equation of the line in slope-intercept form
is F= 9/5 C+ 32.
I am hoping
that this answer has satisfied your query and it will be able to help you in
your endeavor, and if you would like, feel free to ask another question.