Answer :
We are given the relationship between the Fahrenheit and Celsius temperatures:
[tex]$$
F = \frac{9}{5}C + 32.
$$[/tex]
Let’s work through each part step-by-step.
–––––––
Step (a): Convert 50°C to Fahrenheit
Substitute [tex]$C = 50$[/tex] into the equation:
[tex]$$
F = \frac{9}{5}(50) + 32.
$$[/tex]
First, compute the multiplication:
[tex]$$
\frac{9}{5} \times 50 = 90.
$$[/tex]
Then add 32:
[tex]$$
F = 90 + 32 = 122.
$$[/tex]
Thus, a temperature of 50°C is equivalent to [tex]$122^\circ \text{F}$[/tex].
–––––––
Step (b): Convert –60°F to Celsius
We first rewrite the equation to solve for [tex]$C$[/tex]. Starting from
[tex]$$
F = \frac{9}{5}C + 32,
$$[/tex]
subtract 32 from both sides:
[tex]$$
F - 32 = \frac{9}{5}C.
$$[/tex]
Now, multiply both sides by [tex]$\frac{5}{9}$[/tex] to isolate [tex]$C$[/tex]:
[tex]$$
C = \frac{5}{9}(F - 32).
$$[/tex]
Now substitute [tex]$F = -60$[/tex]:
[tex]$$
C = \frac{5}{9}(-60 - 32) = \frac{5}{9}(-92).
$$[/tex]
Multiplying the fraction:
[tex]$$
C = -\frac{460}{9}.
$$[/tex]
This result, expressed as a decimal, is approximately
[tex]$$
C \approx -51.11.
$$[/tex]
Thus, a temperature of [tex]$-60^\circ \text{F}$[/tex] is about [tex]$-\frac{460}{9}^\circ \text{C}$[/tex] (or approximately [tex]$-51.11^\circ \text{C}$[/tex]).
–––––––
Step (c): Find the Temperature Where [tex]$F$[/tex] and [tex]$C$[/tex] Are Equal
We need to find the temperature where
[tex]$$
F = C.
$$[/tex]
Substitute [tex]$F = C$[/tex] into the conversion formula:
[tex]$$
C = \frac{9}{5}C + 32.
$$[/tex]
Now, isolate the terms involving [tex]$C$[/tex]. Subtract [tex]$\frac{9}{5}C$[/tex] from both sides:
[tex]$$
C - \frac{9}{5}C = 32.
$$[/tex]
Factor [tex]$C$[/tex] out of the left side:
[tex]$$
\left(1 - \frac{9}{5}\right) C = 32.
$$[/tex]
Compute the factor:
[tex]$$
1 - \frac{9}{5} = \frac{5}{5} - \frac{9}{5} = -\frac{4}{5}.
$$[/tex]
Thus, we have:
[tex]$$
-\frac{4}{5}C = 32.
$$[/tex]
Solve for [tex]$C$[/tex] by dividing both sides by [tex]$-\frac{4}{5}$[/tex]:
[tex]$$
C = \frac{32}{-\frac{4}{5}}.
$$[/tex]
Dividing by a fraction is equivalent to multiplying by its reciprocal:
[tex]$$
C = 32 \times \left(-\frac{5}{4}\right) = -40.
$$[/tex]
Thus, the temperature at which the Celsius and Fahrenheit scales have the same value is [tex]$-40^\circ$[/tex].
–––––––
Final Answers:
a. [tex]$50^\circ\text{C}$[/tex] is equivalent to [tex]$122^\circ\text{F}$[/tex].
b. [tex]$-60^\circ\text{F}$[/tex] is equivalent to [tex]$-\frac{460}{9}^\circ\text{C}$[/tex] (approximately [tex]$-51.11^\circ\text{C}$[/tex]).
c. The temperature where the Fahrenheit and Celsius scales are equal is [tex]$-40^\circ$[/tex].
[tex]$$
F = \frac{9}{5}C + 32.
$$[/tex]
Let’s work through each part step-by-step.
–––––––
Step (a): Convert 50°C to Fahrenheit
Substitute [tex]$C = 50$[/tex] into the equation:
[tex]$$
F = \frac{9}{5}(50) + 32.
$$[/tex]
First, compute the multiplication:
[tex]$$
\frac{9}{5} \times 50 = 90.
$$[/tex]
Then add 32:
[tex]$$
F = 90 + 32 = 122.
$$[/tex]
Thus, a temperature of 50°C is equivalent to [tex]$122^\circ \text{F}$[/tex].
–––––––
Step (b): Convert –60°F to Celsius
We first rewrite the equation to solve for [tex]$C$[/tex]. Starting from
[tex]$$
F = \frac{9}{5}C + 32,
$$[/tex]
subtract 32 from both sides:
[tex]$$
F - 32 = \frac{9}{5}C.
$$[/tex]
Now, multiply both sides by [tex]$\frac{5}{9}$[/tex] to isolate [tex]$C$[/tex]:
[tex]$$
C = \frac{5}{9}(F - 32).
$$[/tex]
Now substitute [tex]$F = -60$[/tex]:
[tex]$$
C = \frac{5}{9}(-60 - 32) = \frac{5}{9}(-92).
$$[/tex]
Multiplying the fraction:
[tex]$$
C = -\frac{460}{9}.
$$[/tex]
This result, expressed as a decimal, is approximately
[tex]$$
C \approx -51.11.
$$[/tex]
Thus, a temperature of [tex]$-60^\circ \text{F}$[/tex] is about [tex]$-\frac{460}{9}^\circ \text{C}$[/tex] (or approximately [tex]$-51.11^\circ \text{C}$[/tex]).
–––––––
Step (c): Find the Temperature Where [tex]$F$[/tex] and [tex]$C$[/tex] Are Equal
We need to find the temperature where
[tex]$$
F = C.
$$[/tex]
Substitute [tex]$F = C$[/tex] into the conversion formula:
[tex]$$
C = \frac{9}{5}C + 32.
$$[/tex]
Now, isolate the terms involving [tex]$C$[/tex]. Subtract [tex]$\frac{9}{5}C$[/tex] from both sides:
[tex]$$
C - \frac{9}{5}C = 32.
$$[/tex]
Factor [tex]$C$[/tex] out of the left side:
[tex]$$
\left(1 - \frac{9}{5}\right) C = 32.
$$[/tex]
Compute the factor:
[tex]$$
1 - \frac{9}{5} = \frac{5}{5} - \frac{9}{5} = -\frac{4}{5}.
$$[/tex]
Thus, we have:
[tex]$$
-\frac{4}{5}C = 32.
$$[/tex]
Solve for [tex]$C$[/tex] by dividing both sides by [tex]$-\frac{4}{5}$[/tex]:
[tex]$$
C = \frac{32}{-\frac{4}{5}}.
$$[/tex]
Dividing by a fraction is equivalent to multiplying by its reciprocal:
[tex]$$
C = 32 \times \left(-\frac{5}{4}\right) = -40.
$$[/tex]
Thus, the temperature at which the Celsius and Fahrenheit scales have the same value is [tex]$-40^\circ$[/tex].
–––––––
Final Answers:
a. [tex]$50^\circ\text{C}$[/tex] is equivalent to [tex]$122^\circ\text{F}$[/tex].
b. [tex]$-60^\circ\text{F}$[/tex] is equivalent to [tex]$-\frac{460}{9}^\circ\text{C}$[/tex] (approximately [tex]$-51.11^\circ\text{C}$[/tex]).
c. The temperature where the Fahrenheit and Celsius scales are equal is [tex]$-40^\circ$[/tex].
