1. **State the problem:** We have an isosceles trapezoid GHIJ with $m\angle H = 6z + 28^\circ$ and $m\angle I = 130^\circ$. We need to find the value of $z$. 2. **Recall properties:** In an isosceles trapezoid, the base angles are congruent. Since $GH \parallel IJ$, angles $H$ and $I$ are consecutive angles between the parallel sides and are supplementary. 3. **Write the supplementary angle equation:** $$m\angle H + m\angle I = 180^\circ$$ Substitute the given values: $$6z + 28 + 130 = 180$$ 4. **Simplify the equation:** $$6z + 158 = 180$$ 5. **Isolate $z$:** $$6z = 180 - 158$$ $$6z = 22$$ 6. **Solve for $z$:** $$z = \frac{22}{6}$$ Show cancellation: $$z = \frac{\cancel{22}}{\cancel{6}} = \frac{11}{3} \approx 3.67$$ **Final answer:** $$z = \frac{11}{3} \text{ or approximately } 3.67$$