Answer :
Sure, let's go through the process step-by-step:
### Part (a):
To find a linear function [tex]\( F \)[/tex] that expresses temperature in Fahrenheit in terms of degrees Celsius, we use the two given points: [tex]\((0^\circ \text{C}, 32^\circ \text{F})\)[/tex] and [tex]\((100^\circ \text{C}, 212^\circ \text{F})\)[/tex].
1. Calculate the slope ([tex]\(m\)[/tex]):
[tex]\[
m = \frac{212 - 32}{100 - 0} = \frac{180}{100} = \frac{9}{5}
\][/tex]
2. Use the point-slope form of a linear equation, [tex]\( F(C) = mC + b \)[/tex]. We know the slope [tex]\( m = \frac{9}{5} \)[/tex] and one point [tex]\((0, 32)\)[/tex]:
[tex]\[
F(C) = \frac{9}{5}C + 32
\][/tex]
Now, to convert [tex]\(20^\circ \text{C}\)[/tex] into Fahrenheit, substitute [tex]\(C = 20\)[/tex] into the formula:
[tex]\[
F(20) = \frac{9}{5} \times 20 + 32 = 36 + 32 = 68^\circ \text{F}
\][/tex]
### Part (b):
To find a linear function [tex]\( C \)[/tex] that expresses temperature in Celsius in terms of degrees Fahrenheit, we use the same two points and reverse the variables.
1. Use the inverse slope:
The inverse of the slope [tex]\( \frac{9}{5} \)[/tex] is [tex]\( \frac{5}{9} \)[/tex].
2. Use the formula [tex]\( C(F) = \frac{5}{9}(F - 32) \)[/tex].
Now, to convert [tex]\(110^\circ \text{F}\)[/tex] into Celsius, substitute [tex]\(F = 110\)[/tex] into the formula:
[tex]\[
C(110) = \frac{5}{9}(110 - 32) = \frac{5}{9} \times 78 = 43.33^\circ \text{C} \, (\text{rounded to two decimal places})
\][/tex]
### Part (c):
To find if there is a temperature [tex]\( n \)[/tex] such that [tex]\( F(n) = C(n) \)[/tex], we set the two expressions equal:
[tex]\[
\frac{9}{5}n + 32 = \frac{5}{9}(n - 32)
\][/tex]
Solving for [tex]\( n \)[/tex]:
1. Simplify both sides:
[tex]\[
\frac{9}{5}n + 32 = \frac{5}{9}n - \frac{160}{9}
\][/tex]
2. Multiply through to eliminate fractions. The least common multiple of 5 and 9 is 45:
[tex]\[
45 \left(\frac{9}{5}n + 32 \right) = 45 \left(\frac{5}{9}n - \frac{160}{9} \right)
\][/tex]
Which simplifies to:
[tex]\[
81n + 1440 = 25n - 800
\][/tex]
3. Solve for [tex]\( n \)[/tex]:
[tex]\[
81n - 25n = -800 - 1440
\][/tex]
[tex]\[
56n = -2240
\][/tex]
[tex]\[
n = -\frac{2240}{56} = -40
\][/tex]
So, there is a temperature [tex]\( n = -40 \)[/tex], where the temperature is the same in both Celsius and Fahrenheit scales.
### Part (a):
To find a linear function [tex]\( F \)[/tex] that expresses temperature in Fahrenheit in terms of degrees Celsius, we use the two given points: [tex]\((0^\circ \text{C}, 32^\circ \text{F})\)[/tex] and [tex]\((100^\circ \text{C}, 212^\circ \text{F})\)[/tex].
1. Calculate the slope ([tex]\(m\)[/tex]):
[tex]\[
m = \frac{212 - 32}{100 - 0} = \frac{180}{100} = \frac{9}{5}
\][/tex]
2. Use the point-slope form of a linear equation, [tex]\( F(C) = mC + b \)[/tex]. We know the slope [tex]\( m = \frac{9}{5} \)[/tex] and one point [tex]\((0, 32)\)[/tex]:
[tex]\[
F(C) = \frac{9}{5}C + 32
\][/tex]
Now, to convert [tex]\(20^\circ \text{C}\)[/tex] into Fahrenheit, substitute [tex]\(C = 20\)[/tex] into the formula:
[tex]\[
F(20) = \frac{9}{5} \times 20 + 32 = 36 + 32 = 68^\circ \text{F}
\][/tex]
### Part (b):
To find a linear function [tex]\( C \)[/tex] that expresses temperature in Celsius in terms of degrees Fahrenheit, we use the same two points and reverse the variables.
1. Use the inverse slope:
The inverse of the slope [tex]\( \frac{9}{5} \)[/tex] is [tex]\( \frac{5}{9} \)[/tex].
2. Use the formula [tex]\( C(F) = \frac{5}{9}(F - 32) \)[/tex].
Now, to convert [tex]\(110^\circ \text{F}\)[/tex] into Celsius, substitute [tex]\(F = 110\)[/tex] into the formula:
[tex]\[
C(110) = \frac{5}{9}(110 - 32) = \frac{5}{9} \times 78 = 43.33^\circ \text{C} \, (\text{rounded to two decimal places})
\][/tex]
### Part (c):
To find if there is a temperature [tex]\( n \)[/tex] such that [tex]\( F(n) = C(n) \)[/tex], we set the two expressions equal:
[tex]\[
\frac{9}{5}n + 32 = \frac{5}{9}(n - 32)
\][/tex]
Solving for [tex]\( n \)[/tex]:
1. Simplify both sides:
[tex]\[
\frac{9}{5}n + 32 = \frac{5}{9}n - \frac{160}{9}
\][/tex]
2. Multiply through to eliminate fractions. The least common multiple of 5 and 9 is 45:
[tex]\[
45 \left(\frac{9}{5}n + 32 \right) = 45 \left(\frac{5}{9}n - \frac{160}{9} \right)
\][/tex]
Which simplifies to:
[tex]\[
81n + 1440 = 25n - 800
\][/tex]
3. Solve for [tex]\( n \)[/tex]:
[tex]\[
81n - 25n = -800 - 1440
\][/tex]
[tex]\[
56n = -2240
\][/tex]
[tex]\[
n = -\frac{2240}{56} = -40
\][/tex]
So, there is a temperature [tex]\( n = -40 \)[/tex], where the temperature is the same in both Celsius and Fahrenheit scales.
