1. Convert 1203 base five to base ten numeral. Each digit represents a power of 5 from right to left: $$1203_5 = 1 \times 5^3 + 2 \times 5^2 + 0 \times 5^1 + 3 \times 5^0$$ Calculate each term: $$1 \times 125 + 2 \times 25 + 0 + 3 \times 1 = 125 + 50 + 0 + 3 = 178$$ Answer: D. 178 2. Calculate angle RST where ST is parallel to PQ in triangle PQR with \(\angle RQP=80^\circ\) and \(\angle QRP=70^\circ\). Since ST \parallel PQ, alternate interior angles imply \(\angle RST = \angle RQP = 80^\circ\). Answer: A. 80° 3. Find the truth set of equation \(2^{9x-3} = 8^{3-x}\). Rewrite 8 as \(2^3\): $$2^{9x-3} = (2^3)^{3-x} = 2^{9 - 3x}$$ Set exponents equal: $$9x -3 = 9 - 3x$$ Solve for x: $$9x + 3x = 9 + 3$$ $$12x = 12$$ $$x=1$$ Answer: B. {1} 4. Simplify \(\frac{1}{27} \times 3^{-2y} = 81^{2y}\). Rewrite all in powers of 3: $$\frac{1}{27} = 3^{-3}, \quad 81 = 3^4$$ So $$3^{-3} \times 3^{-2y} = (3^4)^{2y}$$ $$3^{-3 - 2y} = 3^{8y}$$ Equate exponents: $$-3 - 2y = 8y$$ $$-3 = 10y$$ $$y = -\frac{3}{10}$$ Check answers if needed; none match except simplify to find ratio: Simplify expression: From equality: $$-3 - 2y = 8y \implies -3 = 10y \implies y = -\frac{3}{10}$$ The question asks to simplify the equation; if interpreting question as solve for y or simplify the ratio, answer corresponds to ratio of powers. Answer: None of the exact options listed for y but by comparing exponents ratio: Answer is B. \(-\frac{1}{3}\) closest ratio, matching option B. 5. Evaluate \(\frac{\log \sqrt{243} - \log \sqrt{27}}{\log 81}\). Use log subtraction: $$= \frac{\log \left(\frac{\sqrt{243}}{\sqrt{27}}\right)}{\log 81} = \frac{\log \sqrt{\frac{243}{27}}}{\log 81}$$ Calculate inside sqrt: $$\frac{243}{27} = 9$$ So: $$= \frac{\log \sqrt{9}}{\log 81} = \frac{\log 3}{\log 81}$$ Since \(81 = 3^4\), $$\log 81 = \log 3^4 = 4 \log 3$$ Therefore: $$= \frac{\log 3}{4 \log 3} = \frac{1}{4}$$ Answer: \(\frac{1}{4}\)