The quantities \( x \) and \( y \) are proportional. \( \begin{array}{ll}x & y \\ 11 & 1 \frac{2}{9}\end{array} \) \( 21 \quad 2 \frac{1}{3} \) \( 45 \quad 5 \) Find the constant of proportionality \( (r) \) in the equation \( y^{k}-r \) : \( r= \) \( \square \)

To find the constant of proportionality \( r \), we first express the relationship as \( y = kx \), where \( k \) is the constant of proportionality. We have the pairs of \( (x, y) \): 1. \( (11, 1 \frac{2}{9}) = (11, \frac{11}{9}) \) 2. \( (21, 2 \frac{1}{3}) = (21, \frac{7}{3}) \) 3. \( (45, 5) \) Now, we'll calculate \( k \) using the first pair: The proportionality constant \( k \) can be computed by rearranging the equation: \[ k = \frac{y}{x} \] For the first pair: \[ k = \frac{\frac{11}{9}}{11} = \frac{11}{9} \cdot \frac{1}{11} = \frac{1}{9} \] Next, we check \( k \) with the second pair: \[ k = \frac{2 \frac{1}{3}}{21} = \frac{\frac{7}{3}}{21} = \frac{7}{3 \cdot 21} = \frac{7}{63} = \frac{1}{9} \] And for the third pair: \[ k = \frac{5}{45} = \frac{1}{9} \] Since \( k \) remains the same across all pairs, the constant of proportionality \( r \) in the equation \( y^{k} - r \) simply equals \( 0 \) when \( k = \frac{1}{9} \) because \( r \) represents an added or subtracted constant in a proportional relationship. So, \( r = 0 \).