Thank you for visiting In countries like the USA and Canada temperature is measured in Fahrenheit whereas in countries like India it is measured in Celsius Here is a. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
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].
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