Answer :
Sure! Let's find the stopping distance for the given air temperatures using the function provided.
The function given is [tex]\( D(F) = 2F + 115 \)[/tex], where [tex]\( F \)[/tex] is the temperature in degrees Fahrenheit and [tex]\( D(F) \)[/tex] is the stopping distance.
1. Find [tex]\( D(0) \)[/tex]:
- Plug in [tex]\( F = 0 \)[/tex] into the function.
- [tex]\( D(0) = 2 \times 0 + 115 \)[/tex]
- [tex]\( D(0) = 0 + 115 \)[/tex]
- [tex]\( D(0) = 115 \)[/tex]
So, the stopping distance at 0 degrees Fahrenheit is 115.
2. Find [tex]\( D(-30) \)[/tex]:
- Plug in [tex]\( F = -30 \)[/tex] into the function.
- [tex]\( D(-30) = 2 \times (-30) + 115 \)[/tex]
- [tex]\( D(-30) = -60 + 115 \)[/tex]
- [tex]\( D(-30) = 55 \)[/tex]
So, the stopping distance at -30 degrees Fahrenheit is 55.
Therefore, the answers are:
- [tex]\( D(0) = 115 \)[/tex]
- [tex]\( D(-30) = 55 \)[/tex]
The function given is [tex]\( D(F) = 2F + 115 \)[/tex], where [tex]\( F \)[/tex] is the temperature in degrees Fahrenheit and [tex]\( D(F) \)[/tex] is the stopping distance.
1. Find [tex]\( D(0) \)[/tex]:
- Plug in [tex]\( F = 0 \)[/tex] into the function.
- [tex]\( D(0) = 2 \times 0 + 115 \)[/tex]
- [tex]\( D(0) = 0 + 115 \)[/tex]
- [tex]\( D(0) = 115 \)[/tex]
So, the stopping distance at 0 degrees Fahrenheit is 115.
2. Find [tex]\( D(-30) \)[/tex]:
- Plug in [tex]\( F = -30 \)[/tex] into the function.
- [tex]\( D(-30) = 2 \times (-30) + 115 \)[/tex]
- [tex]\( D(-30) = -60 + 115 \)[/tex]
- [tex]\( D(-30) = 55 \)[/tex]
So, the stopping distance at -30 degrees Fahrenheit is 55.
Therefore, the answers are:
- [tex]\( D(0) = 115 \)[/tex]
- [tex]\( D(-30) = 55 \)[/tex]