. Water freezes at 0◦ Celsius and 32◦ Fahrenheit and it boils at 100◦C and 212◦F.
(a) Find a linear function F that expresses temperature in the Fahrenheit scale in terms of
degrees Celsius. Use this function to convert 20◦C into Fahrenheit.
(b) Find a linear function C that expresses temperature in the Celsius scale in terms of
degrees Fahrenheit. Use this function to convert 110◦F into Celsius.
(c) Is there a temperature n such that F(n) = C(n)?
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(a) To find the linear function F that expresses temperature in Fahrenheit in terms of degrees Celsius, we can use the formula for converting between Celsius and Fahrenheit:
F = (9/5)C + 32
where F represents the temperature in Fahrenheit and C represents the temperature in Celsius.
To convert 20°C into Fahrenheit, we can substitute C = 20 into the formula:
F = (9/5)(20) + 32
F = 36 + 32
F = 68
Therefore, 20°C is equivalent to 68°F.
(b) To find the linear function C that expresses temperature in Celsius in terms of degrees Fahrenheit, we rearrange the formula from part (a) to solve for C:
F = (9/5)C + 32
F - 32 = (9/5)C
(5/9)(F - 32) = C
Substituting F = 110 into the formula:
C = (5/9)(110 - 32)
C = (5/9)(78)
C ≈ 43.33
Therefore, 110°F is approximately equivalent to 43.33°C.
(c) To find a temperature n such that F(n) = C(n), we can set the two linear equations equal to each other and solve for n:
(9/5)n + 32 = (5/9)(n - 32)
(9/5)n + 32 = (5/9)n - (5/9)(32)
(9/5)n - (5/9)n = - (5/9)(32) - 32
(9/5 - 5/9)n = - (5/9)(32) - (9/5)(32)
(81/45 - 25/45)n = - (160/9) - (288/5)
Simplifying and solving for n:
(56/45)n = - (800/45) - (5184/45)
(56/45)n = - (5984/45)
n = - (5984/45) * (45/56)
n = - (5984/56)
n ≈ -106.86
Therefore, there is no temperature n such that F(n) = C(n).
To find the linear function that relates temperature in Fahrenheit to temperature in Celsius, we can use the standard formula for converting from Celsius to Fahrenheit:
°F = (°C × 9/5) + 32
Let's call this function F(°C), where °C is the temperature in Celsius.
To convert 20°C into Fahrenheit using this function, we substitute 20 into °C:
F(20) = (20 × 9/5) + 32
Simplifying the equation:
F(20) = 36 + 32
F(20) = 68°F
Therefore, to convert 20°C to Fahrenheit, we get 68°F.
Now let's find the linear function that relates temperature in Celsius to temperature in Fahrenheit.
Using the same formula, but solving it for °C, we get:
°C = (°F - 32) × 5/9
Let's call this function C(°F), where °F is the temperature in Fahrenheit.
To convert 110°F to Celsius using this function, we substitute 110 into °F:
C(110) = (110 - 32) × 5/9
Simplifying the equation:
C(110) = 78 × 5/9
C(110) = 43.33°C
Therefore, to convert 110°F to Celsius, we get 43.33°C.
Lastly, let's determine if there is a temperature n such that F(n) = C(n).
To find the temperature n for which F(n) = C(n), we can equate the formulas for F(°C) and C(°F):
(°C × 9/5) + 32 = (°F - 32) × 5/9
Expanding and simplifying:
9°C + 160 = 5°F - 160
9°C - 5°F = -320
Converting temperatures to the same unit:
9°C - 5(°C × 9/5 + 32) = -320
9°C - 9°C - 160 = -320
-160 = -320
Since -160 is not equal to -320, there is no temperature n for which F(n) = C(n).
Therefore, there is no temperature at which the Fahrenheit and Celsius scales are equal.
(212-32) = 180
SO, there are 180°F for 100°C
So
each °F is 5/9 as big as a °C
each °C is 9/5 as big as a °F
That should help you get started.