Answer :
Let's solve the problem step-by-step, using the information we have about temperatures in Celsius and Fahrenheit.
### Part a: Express [tex]\( F \)[/tex] as a linear function of [tex]\( C \)[/tex]
The problem gives us two points for temperature conversion:
- When Celsius ([tex]\( C \)[/tex]) is 30, Fahrenheit ([tex]\( F \)[/tex]) is 86.
- When Celsius ([tex]\( C \)[/tex]) is 110, Fahrenheit ([tex]\( F \)[/tex]) is 230.
We can represent these points as [tex]\((30, 86)\)[/tex] and [tex]\((110, 230)\)[/tex].
Step 1: Find the slope [tex]\( m \)[/tex] of the line
The slope [tex]\( m \)[/tex] of a line through two points [tex]\((C_1, F_1)\)[/tex] and [tex]\((C_2, F_2)\)[/tex] is given by:
[tex]\[
m = \frac{F_2 - F_1}{C_2 - C_1}
\][/tex]
Substitute the given points:
[tex]\[
m = \frac{230 - 86}{110 - 30} = \frac{144}{80} = 1.8
\][/tex]
Step 2: Find the y-intercept [tex]\( b \)[/tex]
Using the slope-intercept form [tex]\( F = mC + b \)[/tex], substitute one of the points, say [tex]\((30, 86)\)[/tex], to find [tex]\( b \)[/tex]:
[tex]\[
86 = 1.8 \times 30 + b
\][/tex]
[tex]\[
86 = 54 + b
\][/tex]
[tex]\[
b = 86 - 54 = 32
\][/tex]
Thus, the linear function is:
[tex]\[
F = 1.8C + 32
\][/tex]
### Part b: Solve for [tex]\( C \)[/tex] to express [tex]\( C \)[/tex] as a function of [tex]\( F \)[/tex]
We have [tex]\( F = 1.8C + 32 \)[/tex]. To express [tex]\( C \)[/tex] as a function of [tex]\( F \)[/tex], solve for [tex]\( C \)[/tex]:
1. Subtract 32 from both sides:
[tex]\[
F - 32 = 1.8C
\][/tex]
2. Divide both sides by 1.8:
[tex]\[
C = \frac{F - 32}{1.8}
\][/tex]
This can be simplified to:
[tex]\[
C = 0.5556F - 17.7778
\][/tex]
### Part c: Find the temperature where [tex]\( F = C \)[/tex]
Set the two expressions for [tex]\( F \)[/tex] and [tex]\( C \)[/tex] equal to each other:
[tex]\[
1.8C + 32 = C
\][/tex]
Rearrange terms:
[tex]\[
1.8C - C = -32
\][/tex]
[tex]\[
0.8C = -32
\][/tex]
Divide by 0.8:
[tex]\[
C = -40
\][/tex]
Therefore, the temperature at which [tex]\( F = C \)[/tex] is [tex]\(-40\)[/tex] degrees.
### Part a: Express [tex]\( F \)[/tex] as a linear function of [tex]\( C \)[/tex]
The problem gives us two points for temperature conversion:
- When Celsius ([tex]\( C \)[/tex]) is 30, Fahrenheit ([tex]\( F \)[/tex]) is 86.
- When Celsius ([tex]\( C \)[/tex]) is 110, Fahrenheit ([tex]\( F \)[/tex]) is 230.
We can represent these points as [tex]\((30, 86)\)[/tex] and [tex]\((110, 230)\)[/tex].
Step 1: Find the slope [tex]\( m \)[/tex] of the line
The slope [tex]\( m \)[/tex] of a line through two points [tex]\((C_1, F_1)\)[/tex] and [tex]\((C_2, F_2)\)[/tex] is given by:
[tex]\[
m = \frac{F_2 - F_1}{C_2 - C_1}
\][/tex]
Substitute the given points:
[tex]\[
m = \frac{230 - 86}{110 - 30} = \frac{144}{80} = 1.8
\][/tex]
Step 2: Find the y-intercept [tex]\( b \)[/tex]
Using the slope-intercept form [tex]\( F = mC + b \)[/tex], substitute one of the points, say [tex]\((30, 86)\)[/tex], to find [tex]\( b \)[/tex]:
[tex]\[
86 = 1.8 \times 30 + b
\][/tex]
[tex]\[
86 = 54 + b
\][/tex]
[tex]\[
b = 86 - 54 = 32
\][/tex]
Thus, the linear function is:
[tex]\[
F = 1.8C + 32
\][/tex]
### Part b: Solve for [tex]\( C \)[/tex] to express [tex]\( C \)[/tex] as a function of [tex]\( F \)[/tex]
We have [tex]\( F = 1.8C + 32 \)[/tex]. To express [tex]\( C \)[/tex] as a function of [tex]\( F \)[/tex], solve for [tex]\( C \)[/tex]:
1. Subtract 32 from both sides:
[tex]\[
F - 32 = 1.8C
\][/tex]
2. Divide both sides by 1.8:
[tex]\[
C = \frac{F - 32}{1.8}
\][/tex]
This can be simplified to:
[tex]\[
C = 0.5556F - 17.7778
\][/tex]
### Part c: Find the temperature where [tex]\( F = C \)[/tex]
Set the two expressions for [tex]\( F \)[/tex] and [tex]\( C \)[/tex] equal to each other:
[tex]\[
1.8C + 32 = C
\][/tex]
Rearrange terms:
[tex]\[
1.8C - C = -32
\][/tex]
[tex]\[
0.8C = -32
\][/tex]
Divide by 0.8:
[tex]\[
C = -40
\][/tex]
Therefore, the temperature at which [tex]\( F = C \)[/tex] is [tex]\(-40\)[/tex] degrees.
