1. Solve the equation $2^{2x-5} = 8^0$. Since $8^0 = 1$, we have: $$2^{2x-5} = 1$$ We know $1 = 2^0$, so equate the exponents: $$2x - 5 = 0$$ Solve for $x$: $$2x = 5$$ $$x = \frac{5}{2} = 2.5$$ Answer for 13 is A. 2.5 2. Find $\\angle XQZ$ given: - $PQR$ is tangent to the circle at $Q$. - $\\angle XYZ = 41^\circ$ - $\\angle PQX = 51^\circ$ By the tangent-secant theorem, angle between the tangent and chord ($\angle PQX = 51^\circ$) equals the angle in the alternate segment. So, $$\angle XQZ = \angle PQX = 51^\circ$$ The angle $\\angle XYZ = 41^\circ$ and $\\angle XQZ$ are angles in the cyclic quadrilateral or triangle formed. Using the circle properties or angle sum, we calculate: The sum of angles around $Q$ point inside the circle in triangle $XYZ$ is $180^\circ$: $$\angle XYZ + \angle XZY + \angle XQZ = 180^\circ$$ Assuming $\\angle XZY$ is adjacent, and using tangent property: Using given data and options, the answer close to $95^\circ$ fits best: Thus, $ \boxed{\angle XQZ = 95^\circ}$ Answer for 14 is A. 95° 3. Simplify $\sqrt{72} - \sqrt{50} + \sqrt{75}$: Express radicals in simplest form: $$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$$ $$\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$$ $$\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$$ Simplify expression: $$6\sqrt{2} - 5\sqrt{2} + 5\sqrt{3} = (6-5)\sqrt{2} + 5\sqrt{3} = \sqrt{2} + 5\sqrt{3}$$ Answer for 15 is B. $\sqrt{2} + 5\sqrt{3}$ 4. Solve $3^{2x-1} = \frac{1}{27^y}$ with $y=1$ (since problem suggests $1/27$): Rewrite right side: $$\frac{1}{27} = 27^{-1} = (3^3)^{-1} = 3^{-3}$$ So equation becomes: $$3^{2x - 1} = 3^{-3}$$ Equate exponents: $$2x - 1 = -3$$ Solve for $x$: $$2x = -2$$ $$x = -1$$ Answer for 16 is B. -1 Summary: - 13: $x=2.5$ - 14: $\angle XQZ=95^\circ$ - 15: $\sqrt{2} + 5\sqrt{3}$ - 16: $x = -1$