To determine which of the given relationships are both linear and proportional, we need to identify which equations can be expressed in the form [tex]\( y = kx \)[/tex], where [tex]\( k \)[/tex] is a constant.
Let's analyze each equation one by one:
1. [tex]\( y = x + 1 \)[/tex]:
- This equation is of the form [tex]\( y = mx + b \)[/tex] where [tex]\( m = 1 \)[/tex] and [tex]\( b = 1 \)[/tex].
- For a relationship to be proportional, [tex]\( b \)[/tex] must be 0. Since in this case [tex]\( b \)[/tex] is not 0, this is not a proportional relationship.
2. [tex]\( y = \frac{x}{12} \)[/tex]:
- This equation is of the form [tex]\( y = kx \)[/tex] where [tex]\( k = \frac{1}{12} \)[/tex].
- There is no constant term added to [tex]\( kx \)[/tex], which means [tex]\( b = 0 \)[/tex] here.
- Therefore, this is a proportional relationship.
3. [tex]\( y = 0.7x \)[/tex]:
- This equation is also of the form [tex]\( y = kx \)[/tex] where [tex]\( k = 0.7 \)[/tex].
- Again, there is no constant term added to [tex]\( kx \)[/tex], which means [tex]\( b = 0 \)[/tex] here.
- Therefore, this is a proportional relationship.
4. [tex]\( y = 6x^2 \)[/tex]:
- This equation involves [tex]\( x^2 \)[/tex], making it a quadratic relationship, not linear.
- Since it is not linear, it cannot be a proportional relationship. Therefore, it is not proportional.
5. [tex]\( y = 1200x \)[/tex]:
- This equation is of the form [tex]\( y = kx \)[/tex] where [tex]\( k = 1200 \)[/tex].
- There is no constant term added to [tex]\( kx \)[/tex], which means [tex]\( b = 0 \)[/tex] here.
- Therefore, this is a proportional relationship.
Summarizing the analysis, the linear relationships that are also proportional are:
- [tex]\( y = \frac{x}{12} \)[/tex]
- [tex]\( y = 0.7x \)[/tex]
- [tex]\( y = 1200x \)[/tex]
Thus, the corresponding indices of these relationships are:
[tex]\[ [2, 3, 5] \][/tex]
