Answer :
Sure, let's solve the given problem step-by-step.
### Given:
The linear equation to convert Fahrenheit (F) to Celsius (C) is:
[tex]\[ 5F = 9C + 160 \][/tex]
### Part (i):
If the temperature is [tex]\( 30^{\circ} C \)[/tex], what is the temperature in Fahrenheit?
We need to find [tex]\( F \)[/tex] when [tex]\( C = 30 \)[/tex].
1. Substitute [tex]\( C = 30 \)[/tex] into the equation:
[tex]\[ 5F = 9(30) + 160 \][/tex]
[tex]\[ 5F = 270 + 160 \][/tex]
[tex]\[ 5F = 430 \][/tex]
2. Solve for [tex]\( F \)[/tex]:
[tex]\[ F = \frac{430}{5} \][/tex]
[tex]\[ F = 86 \][/tex]
So, the temperature in Fahrenheit is [tex]\( 86^{\circ} F \)[/tex].
### Part (ii):
1. If the temperature is [tex]\( 0^{\circ} C \)[/tex], what is the temperature in Fahrenheit?
We need to find [tex]\( F \)[/tex] when [tex]\( C = 0 \)[/tex].
1. Substitute [tex]\( C = 0 \)[/tex] into the equation:
[tex]\[ 5F = 9(0) + 160 \][/tex]
[tex]\[ 5F = 0 + 160 \][/tex]
[tex]\[ 5F = 160 \][/tex]
2. Solve for [tex]\( F \)[/tex]:
[tex]\[ F = \frac{160}{5} \][/tex]
[tex]\[ F = 32 \][/tex]
So, the temperature in Fahrenheit is [tex]\( 32^{\circ} F \)[/tex].
2. If the temperature is [tex]\( 0^{\circ} F \)[/tex], what is the temperature in Celsius?
We need to find [tex]\( C \)[/tex] when [tex]\( F = 0 \)[/tex].
1. Substitute [tex]\( F = 0 \)[/tex] into the equation:
[tex]\[ 5(0) = 9C + 160 \][/tex]
[tex]\[ 0 = 9C + 160 \][/tex]
2. Solve for [tex]\( C \)[/tex]:
[tex]\[ 9C = -160 \][/tex]
[tex]\[ C = \frac{-160}{9} \][/tex]
[tex]\[ C \approx -17.78 \][/tex]
So, the temperature in Celsius is approximately [tex]\( -17.78^{\circ} C \)[/tex].
### Part (iii):
Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
We need to find a temperature where [tex]\( F = C \)[/tex].
1. Substitute [tex]\( F = C \)[/tex] into the equation:
[tex]\[ 5C = 9C + 160 \][/tex]
2. Rearrange the equation:
[tex]\[ 5C - 9C = 160 \][/tex]
[tex]\[ -4C = 160 \][/tex]
3. Solve for [tex]\( C \)[/tex]:
[tex]\[ C = \frac{160}{-4} \][/tex]
[tex]\[ C = -40 \][/tex]
Therefore, the temperature which is numerically the same in both Fahrenheit and Celsius is [tex]\( -40 \)[/tex].
### Summary of Results:
(i) [tex]\( 30^{\circ} C \)[/tex] is [tex]\( 86^{\circ} F \)[/tex].
(ii)
- [tex]\( 0^{\circ} C \)[/tex] is [tex]\( 32^{\circ} F \)[/tex].
- [tex]\( 0^{\circ} F \)[/tex] is approximately [tex]\( -17.78^{\circ} C \)[/tex].
(iii) The temperature that is numerically the same in both Fahrenheit and Celsius is [tex]\( -40^{\circ} \)[/tex].
### Given:
The linear equation to convert Fahrenheit (F) to Celsius (C) is:
[tex]\[ 5F = 9C + 160 \][/tex]
### Part (i):
If the temperature is [tex]\( 30^{\circ} C \)[/tex], what is the temperature in Fahrenheit?
We need to find [tex]\( F \)[/tex] when [tex]\( C = 30 \)[/tex].
1. Substitute [tex]\( C = 30 \)[/tex] into the equation:
[tex]\[ 5F = 9(30) + 160 \][/tex]
[tex]\[ 5F = 270 + 160 \][/tex]
[tex]\[ 5F = 430 \][/tex]
2. Solve for [tex]\( F \)[/tex]:
[tex]\[ F = \frac{430}{5} \][/tex]
[tex]\[ F = 86 \][/tex]
So, the temperature in Fahrenheit is [tex]\( 86^{\circ} F \)[/tex].
### Part (ii):
1. If the temperature is [tex]\( 0^{\circ} C \)[/tex], what is the temperature in Fahrenheit?
We need to find [tex]\( F \)[/tex] when [tex]\( C = 0 \)[/tex].
1. Substitute [tex]\( C = 0 \)[/tex] into the equation:
[tex]\[ 5F = 9(0) + 160 \][/tex]
[tex]\[ 5F = 0 + 160 \][/tex]
[tex]\[ 5F = 160 \][/tex]
2. Solve for [tex]\( F \)[/tex]:
[tex]\[ F = \frac{160}{5} \][/tex]
[tex]\[ F = 32 \][/tex]
So, the temperature in Fahrenheit is [tex]\( 32^{\circ} F \)[/tex].
2. If the temperature is [tex]\( 0^{\circ} F \)[/tex], what is the temperature in Celsius?
We need to find [tex]\( C \)[/tex] when [tex]\( F = 0 \)[/tex].
1. Substitute [tex]\( F = 0 \)[/tex] into the equation:
[tex]\[ 5(0) = 9C + 160 \][/tex]
[tex]\[ 0 = 9C + 160 \][/tex]
2. Solve for [tex]\( C \)[/tex]:
[tex]\[ 9C = -160 \][/tex]
[tex]\[ C = \frac{-160}{9} \][/tex]
[tex]\[ C \approx -17.78 \][/tex]
So, the temperature in Celsius is approximately [tex]\( -17.78^{\circ} C \)[/tex].
### Part (iii):
Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
We need to find a temperature where [tex]\( F = C \)[/tex].
1. Substitute [tex]\( F = C \)[/tex] into the equation:
[tex]\[ 5C = 9C + 160 \][/tex]
2. Rearrange the equation:
[tex]\[ 5C - 9C = 160 \][/tex]
[tex]\[ -4C = 160 \][/tex]
3. Solve for [tex]\( C \)[/tex]:
[tex]\[ C = \frac{160}{-4} \][/tex]
[tex]\[ C = -40 \][/tex]
Therefore, the temperature which is numerically the same in both Fahrenheit and Celsius is [tex]\( -40 \)[/tex].
### Summary of Results:
(i) [tex]\( 30^{\circ} C \)[/tex] is [tex]\( 86^{\circ} F \)[/tex].
(ii)
- [tex]\( 0^{\circ} C \)[/tex] is [tex]\( 32^{\circ} F \)[/tex].
- [tex]\( 0^{\circ} F \)[/tex] is approximately [tex]\( -17.78^{\circ} C \)[/tex].
(iii) The temperature that is numerically the same in both Fahrenheit and Celsius is [tex]\( -40^{\circ} \)[/tex].