1. **Stating the problem:** We are given a quadratic function $$y=2.5(x-5)^2+0.5$$ and a linear equation $$\frac{5}{22}x + \frac{1}{33}y = 1$$. We also have a right triangle EFB with given segment lengths and perpendicularities, and an inequality $$5x - 4y + 14 > 0$$. We need to analyze these elements step-by-step. 2. **Analyzing the quadratic function:** The function is in vertex form $$y = a(x-h)^2 + k$$ where $$a=2.5$$, $$h=5$$, and $$k=0.5$$. This means the parabola opens upwards with vertex at $$(5, 0.5)$$. 3. **Analyzing the linear equation:** Multiply both sides by 66 (LCM of 22 and 33) to clear denominators: $$66 \times \left(\frac{5}{22}x + \frac{1}{33}y\right) = 66 \times 1$$ $$3 \times 5x + 2y = 66$$ $$15x + 2y = 66$$ Rearranged to slope-intercept form: $$2y = 66 - 15x$$ $$y = \frac{66 - 15x}{2} = 33 - 7.5x$$ 4. **Triangle properties:** Given: - $$mEF = \sqrt{20} = 2\sqrt{5}$$ units - $$mFH = 2$$ units - $$EB \perp HF$$ - $$EF \perp FB$$ Since EF is vertical and FB is horizontal, EF and FB are perpendicular by definition. 5. **Inequality:** $$5x - 4y + 14 > 0$$ can be rewritten as: $$5x - 4y > -14$$ or $$-4y > -5x - 14$$ Dividing by -4 (and reversing inequality): $$y < \frac{5}{4}x + 3.5$$ This inequality describes the region below the line $$y = \frac{5}{4}x + 3.5$$. 6. **Summary:** - The parabola $$y=2.5(x-5)^2+0.5$$ opens upward with vertex at $$(5,0.5)$$. - The line $$y=33 - 7.5x$$ is a decreasing linear function. - The triangle EFB has right angles at F and H with given segment lengths. - The inequality $$5x - 4y + 14 > 0$$ describes the half-plane below $$y=\frac{5}{4}x + 3.5$$. Final answers: - Vertex of parabola: $$(5, 0.5)$$ - Line equation simplified: $$y=33 - 7.5x$$ - Inequality region: $$y < \frac{5}{4}x + 3.5$$