1. The problem asks to find the image of the parallelogram GRAM after a translation 11 units right and 2 units up. 2. Translation rule: To translate a point $(x,y)$ by $a$ units right and $b$ units up, the new coordinates are given by: $$ (x', y') = (x + a, y + b) $$ 3. Since the translation is 11 units right and 2 units up, we have $a=11$ and $b=2$. 4. Let the original vertices be $G(x_G,y_G)$, $R(x_R,y_R)$, $A(x_A,y_A)$, and $M(x_M,y_M)$. 5. Applying the translation to each vertex: $$ G' = (x_G + 11, y_G + 2) $$ $$ R' = (x_R + 11, y_R + 2) $$ $$ A' = (x_A + 11, y_A + 2) $$ $$ M' = (x_M + 11, y_M + 2) $$ 6. This moves the entire parallelogram 11 units to the right and 2 units up, preserving its shape and orientation. 7. The image parallelogram $G'R'A'M'$ has vertices shifted accordingly. Final answer: The coordinates of $G'$, $R'$, $A'$, and $M'$ are each the original coordinates plus 11 in the $x$-direction and plus 2 in the $y$-direction.