1. The problem asks to graph the image of a triangle after applying a scale factor of $k=\frac{3}{4}$ to its vertices. 2. The formula for scaling a point $(x,y)$ by a factor $k$ about the origin is: $$ (x', y') = (kx, ky) $$ This means each coordinate of the point is multiplied by $k$. 3. Given the original vertices of the triangle are $A'(-5.3, 10.6)$, $B'(5.3, 8)$, and $C'(0, -5.3)$, we apply the scale factor $k=\frac{3}{4}$ to each coordinate: 4. Calculate the scaled coordinates: $$ A'' = \left(\frac{3}{4} \times -5.3, \frac{3}{4} \times 10.6\right) = \left(-\frac{15.9}{4}, \frac{31.8}{4}\right) = (-3.975, 7.95) $$ $$ B'' = \left(\frac{3}{4} \times 5.3, \frac{3}{4} \times 8\right) = \left(\frac{15.9}{4}, 6\right) = (3.975, 6) $$ $$ C'' = \left(\frac{3}{4} \times 0, \frac{3}{4} \times -5.3\right) = (0, -3.975) $$ 5. These new points $A''(-3.975, 7.95)$, $B''(3.975, 6)$, and $C''(0, -3.975)$ represent the vertices of the scaled triangle. 6. The scale factor $k=\frac{3}{4}$ reduces the size of the triangle to 75% of the original, keeping the shape similar and the origin as the center of scaling. Final answer: The image of the triangle after scaling by $\frac{3}{4}$ has vertices at $$ A''(-3.975, 7.95), B''(3.975, 6), C''(0, -3.975) $$