Answer :
To find the temperature at noon yesterday, we need to start with the known temperature at midnight and work backward to account for the temperature drop.
1. Identify the given information:
- Temperature at midnight: [tex]\( -12.8 \)[/tex]°F
- Temperature drop from noon to midnight: [tex]\( 28.2 \)[/tex]°F
2. Understand the problem:
- The temperature at noon was higher than the temperature at midnight because it dropped by [tex]\( 28.2 \)[/tex]°F over the course of the day.
3. Set up the problem:
- Let [tex]\( T \)[/tex] be the temperature at noon.
- We know that after the drop of [tex]\( 28.2 \)[/tex]°F, the temperature became [tex]\( -12.8 \)[/tex]°F, i.e.,
[tex]\[ T - 28.2 = -12.8 \][/tex]
4. Solve for [tex]\( T \)[/tex]:
- Add [tex]\( 28.2 \)[/tex] to both sides of the equation to isolate [tex]\( T \)[/tex]:
[tex]\[ T = -12.8 + 28.2 \][/tex]
5. Calculate the result:
- Perform the addition:
[tex]\[ T = 15.399999999999999 \][/tex]
The resulting temperature at noon, when rounded to an appropriate number of significant figures, is approximately [tex]\( 15.4 \)[/tex]°F.
Therefore, the temperature at noon yesterday was approximately [tex]\( 15.4 \)[/tex]°F.
1. Identify the given information:
- Temperature at midnight: [tex]\( -12.8 \)[/tex]°F
- Temperature drop from noon to midnight: [tex]\( 28.2 \)[/tex]°F
2. Understand the problem:
- The temperature at noon was higher than the temperature at midnight because it dropped by [tex]\( 28.2 \)[/tex]°F over the course of the day.
3. Set up the problem:
- Let [tex]\( T \)[/tex] be the temperature at noon.
- We know that after the drop of [tex]\( 28.2 \)[/tex]°F, the temperature became [tex]\( -12.8 \)[/tex]°F, i.e.,
[tex]\[ T - 28.2 = -12.8 \][/tex]
4. Solve for [tex]\( T \)[/tex]:
- Add [tex]\( 28.2 \)[/tex] to both sides of the equation to isolate [tex]\( T \)[/tex]:
[tex]\[ T = -12.8 + 28.2 \][/tex]
5. Calculate the result:
- Perform the addition:
[tex]\[ T = 15.399999999999999 \][/tex]
The resulting temperature at noon, when rounded to an appropriate number of significant figures, is approximately [tex]\( 15.4 \)[/tex]°F.
Therefore, the temperature at noon yesterday was approximately [tex]\( 15.4 \)[/tex]°F.