Answer :
Let's analyze the given problem step-by-step to find which values of [tex]\( b \)[/tex] make sense given the perimeter and the triangle inequality theorem.
1. We know the perimeter of the isosceles triangle is 15.7 inches. So, the equation for the perimeter is:
[tex]\[ 2a + b = 15.7 \][/tex]
2. Let's check each of the given possible values for [tex]\( b \)[/tex] and see if they fit the requirements.
3. The values given to check are:
[tex]\[ \{-2, 0, 0.5, 2, 7.9\} \][/tex]
### Checking Each Value:
- For [tex]\( b = -2 \)[/tex] inches:
[tex]\[ 2a - 2 = 15.7 \implies 2a = 17.7 \implies a = 8.85 \][/tex]
Given [tex]\( a = 8.85 \)[/tex] and [tex]\( b = -2 \)[/tex], [tex]\( b \)[/tex] cannot be negative for a triangle, so this value is not valid.
- For [tex]\( b = 0 \)[/tex] inches:
[tex]\[ 2a + 0 = 15.7 \implies 2a = 15.7 \implies a = 7.85 \][/tex]
Given [tex]\( a = 7.85 \)[/tex] and [tex]\( b = 0 \)[/tex], a side cannot be zero, so this value is not valid.
- For [tex]\( b = 0.5 \)[/tex] inches:
[tex]\[ 2a + 0.5 = 15.7 \implies 2a = 15.2 \implies a = 7.6 \][/tex]
Given [tex]\( a = 7.6 \)[/tex] and [tex]\( b = 0.5 \)[/tex]:
[tex]\[ 2a > b \implies 2 \cdot 7.6 > 0.5 \implies 15.2 > 0.5 \][/tex]
This is true, and since [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are positive, this value is valid.
- For [tex]\( b = 2 \)[/tex] inches:
[tex]\[ 2a + 2 = 15.7 \implies 2a = 13.7 \implies a = 6.85 \][/tex]
Given [tex]\( a = 6.85 \)[/tex] and [tex]\( b = 2 \)[/tex]:
[tex]\[ 2a > b \implies 2 \cdot 6.85 > 2 \implies 13.7 > 2 \][/tex]
This is true, and since [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are positive, this value is valid.
- For [tex]\( b = 7.9 \)[/tex] inches:
[tex]\[ 2a + 7.9 = 15.7 \implies 2a = 7.8 \implies a = 3.9 \][/tex]
Given [tex]\( a = 3.9 \)[/tex] and [tex]\( b = 7.9 \)[/tex]:
[tex]\[ 2a > b \implies 2 \cdot 3.9 > 7.9 \implies 7.8 \ngtr 7.9 \][/tex]
This does not satisfy the triangle inequality theorem, so this value is not valid.
### Conclusion:
The two values for [tex]\( b \)[/tex] that make sense given the perimeter and the rules of triangle formation are:
[tex]\[ \boxed{0.5 \text{ inches}, 2 \text{ inches}} \][/tex]
1. We know the perimeter of the isosceles triangle is 15.7 inches. So, the equation for the perimeter is:
[tex]\[ 2a + b = 15.7 \][/tex]
2. Let's check each of the given possible values for [tex]\( b \)[/tex] and see if they fit the requirements.
3. The values given to check are:
[tex]\[ \{-2, 0, 0.5, 2, 7.9\} \][/tex]
### Checking Each Value:
- For [tex]\( b = -2 \)[/tex] inches:
[tex]\[ 2a - 2 = 15.7 \implies 2a = 17.7 \implies a = 8.85 \][/tex]
Given [tex]\( a = 8.85 \)[/tex] and [tex]\( b = -2 \)[/tex], [tex]\( b \)[/tex] cannot be negative for a triangle, so this value is not valid.
- For [tex]\( b = 0 \)[/tex] inches:
[tex]\[ 2a + 0 = 15.7 \implies 2a = 15.7 \implies a = 7.85 \][/tex]
Given [tex]\( a = 7.85 \)[/tex] and [tex]\( b = 0 \)[/tex], a side cannot be zero, so this value is not valid.
- For [tex]\( b = 0.5 \)[/tex] inches:
[tex]\[ 2a + 0.5 = 15.7 \implies 2a = 15.2 \implies a = 7.6 \][/tex]
Given [tex]\( a = 7.6 \)[/tex] and [tex]\( b = 0.5 \)[/tex]:
[tex]\[ 2a > b \implies 2 \cdot 7.6 > 0.5 \implies 15.2 > 0.5 \][/tex]
This is true, and since [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are positive, this value is valid.
- For [tex]\( b = 2 \)[/tex] inches:
[tex]\[ 2a + 2 = 15.7 \implies 2a = 13.7 \implies a = 6.85 \][/tex]
Given [tex]\( a = 6.85 \)[/tex] and [tex]\( b = 2 \)[/tex]:
[tex]\[ 2a > b \implies 2 \cdot 6.85 > 2 \implies 13.7 > 2 \][/tex]
This is true, and since [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are positive, this value is valid.
- For [tex]\( b = 7.9 \)[/tex] inches:
[tex]\[ 2a + 7.9 = 15.7 \implies 2a = 7.8 \implies a = 3.9 \][/tex]
Given [tex]\( a = 3.9 \)[/tex] and [tex]\( b = 7.9 \)[/tex]:
[tex]\[ 2a > b \implies 2 \cdot 3.9 > 7.9 \implies 7.8 \ngtr 7.9 \][/tex]
This does not satisfy the triangle inequality theorem, so this value is not valid.
### Conclusion:
The two values for [tex]\( b \)[/tex] that make sense given the perimeter and the rules of triangle formation are:
[tex]\[ \boxed{0.5 \text{ inches}, 2 \text{ inches}} \][/tex]