1. The problem states that a polygon with vertices at points $D(-10,-8)$, $G(-6,8)$, $F(8,8)$, and $E(5,-4)$ is dilated by a factor of $\frac{3}{4}$ centered at the origin. 2. The formula for dilation centered at the origin with scale factor $k$ is: $$ (x', y') = (kx, ky) $$ This means each coordinate of the original point is multiplied by the scale factor. 3. Apply the dilation factor $\frac{3}{4}$ to each vertex: - For $D(-10,-8)$: $$ D' = \left(\frac{3}{4} \times -10, \frac{3}{4} \times -8\right) = \left(-\frac{30}{4}, -\frac{24}{4}\right) = (-7.5, -6) $$ - For $G(-6,8)$: $$ G' = \left(\frac{3}{4} \times -6, \frac{3}{4} \times 8\right) = \left(-\frac{18}{4}, \frac{24}{4}\right) = (-4.5, 6) $$ - For $F(8,8)$: $$ F' = \left(\frac{3}{4} \times 8, \frac{3}{4} \times 8\right) = \left(\frac{24}{4}, \frac{24}{4}\right) = (6, 6) $$ - For $E(5,-4)$: $$ E' = \left(\frac{3}{4} \times 5, \frac{3}{4} \times -4\right) = \left(\frac{15}{4}, -3\right) = (3.75, -3) $$ 4. The resulting polygon after dilation has vertices at $D'(-7.5,-6)$, $G'(-4.5,6)$, $F'(6,6)$, and $E'(3.75,-3)$. 5. This completes the dilation process by scaling all points towards the origin by $\frac{3}{4}$. Final answer: The dilated polygon vertices are $D'(-7.5,-6)$, $G'(-4.5,6)$, $F'(6,6)$, and $E'(3.75,-3)$.