1. **Find OM in terms of p and r.** Since the problem does not provide a diagram or further context, we assume OM is a segment related to p and r by a geometric or algebraic relation. Without additional information, this cannot be solved explicitly. 2. **Evaluate $$27^{3/2} \times \left(\frac{81}{16}\right)^{-1/4}$$ without a calculator or tables.** Step 1: Express bases as powers of primes. $$27 = 3^3, \quad 81 = 3^4, \quad 16 = 2^4$$ Step 2: Rewrite the expression: $$27^{3/2} = (3^3)^{3/2} = 3^{3 \times \frac{3}{2}} = 3^{\frac{9}{2}}$$ $$\left(\frac{81}{16}\right)^{-1/4} = \left(\frac{3^4}{2^4}\right)^{-1/4} = \left(3^4 \times 2^{-4}\right)^{-1/4} = 3^{-1} \times 2^{1} = \frac{2}{3}$$ Step 3: Multiply the two results: $$3^{\frac{9}{2}} \times \frac{2}{3} = 3^{\frac{9}{2} - 1} \times 2 = 3^{\frac{7}{2}} \times 2$$ Step 4: Simplify: $$3^{\frac{7}{2}} = 3^{3 + \frac{1}{2}} = 3^3 \times 3^{\frac{1}{2}} = 27 \times \sqrt{3}$$ Therefore, $$27 \times \sqrt{3} \times 2 = 54 \sqrt{3}$$ **Final answer:** $$54 \sqrt{3}$$ 3. **Fill in dots on empty faces of the cube net given opposite faces sum to 7 dots.** Given: - Face 1: 3 dots - Face 2: 5 dots - Face 4: 1 dot - Faces 3, 5, 6: empty Opposite faces sum to 7, so: - Opposite Face 1 (3 dots) must have 4 dots (7 - 3 = 4) → Face 6 = 4 dots - Opposite Face 2 (5 dots) must have 2 dots (7 - 5 = 2) → Face 5 = 2 dots - Opposite Face 4 (1 dot) must have 6 dots (7 - 1 = 6) → Face 3 = 6 dots 4. **Construct rhombus QRST with angle TQR = 60° and QS = 10 cm using ruler and compass.** Step 1: Draw segment QS = 10 cm. Step 2: At point Q, construct an angle of 60°. Step 3: Since QRST is a rhombus, all sides are equal. Use compass to mark length QS = 10 cm on the ray forming 60° at Q to locate R. Step 4: Complete the rhombus by drawing sides RT and TS equal to 10 cm. 5. **Calculate amount obtained from sales on Saturday.** Step 1: Oranges bought Thursday = 1948 Step 2: Oranges sold Thursday = 750 Step 3: Oranges sold Friday = 750 + 240 = 990 Step 4: Oranges bought Saturday = 560 Step 5: Oranges remaining after Friday = 1948 - 750 - 990 = 208 (Check: 1948 - 750 = 1198; 1198 - 990 = 208) Step 6: Total oranges on Saturday = 208 + 560 = 768 Step 7: Price per orange = 8 Step 8: Amount obtained = 768 × 8 = 6144 6. **Simplify $$\frac{x^2 + x - 4xy - 4y}{(x + 1)(4y^2 - xy)}$$** Step 1: Factor numerator: Group terms: $$x^2 + x - 4xy - 4y = (x^2 + x) - (4xy + 4y) = x(x + 1) - 4y(x + 1) = (x + 1)(x - 4y)$$ Step 2: Factor denominator: $$4y^2 - xy = y(4y - x)$$ So denominator is: $$(x + 1) y (4y - x)$$ Step 3: Cancel common factor $(x + 1)$: $$\frac{(x + 1)(x - 4y)}{(x + 1) y (4y - x)} = \frac{x - 4y}{y(4y - x)}$$ Step 4: Note that $4y - x = -(x - 4y)$, so: $$\frac{x - 4y}{y(4y - x)} = \frac{x - 4y}{y \times -(x - 4y)} = -\frac{1}{y}$$ **Final simplified expression:** $$-\frac{1}{y}$$ 7. **Given $3\theta$ is acute and $\sin 3\theta = \cos 2\theta$, find $\theta$.** Step 1: Use identity $\cos \alpha = \sin (90^\circ - \alpha)$ (degrees assumed). So, $$\sin 3\theta = \cos 2\theta = \sin (90^\circ - 2\theta)$$ Step 2: Since $\sin A = \sin B$, then either $$3\theta = 90^\circ - 2\theta$$ or $$3\theta = 180^\circ - (90^\circ - 2\theta) = 90^\circ + 2\theta$$ Step 3: Solve first equation: $$3\theta + 2\theta = 90^\circ \Rightarrow 5\theta = 90^\circ \Rightarrow \theta = 18^\circ$$ Step 4: Solve second equation: $$3\theta = 90^\circ + 2\theta \Rightarrow 3\theta - 2\theta = 90^\circ \Rightarrow \theta = 90^\circ$$ But $3\theta$ must be acute, so $3 \times 90^\circ = 270^\circ$ is not acute. Step 5: Therefore, $\theta = 18^\circ$. 8. **A cylinder with radius = height has surface area 154 cm². Find diameter to 2 decimal places. ($\pi = 3.142$)** Step 1: Let radius = height = $r$. Step 2: Surface area of cylinder: $$SA = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r^2 = 4\pi r^2$$ Step 3: Given $SA = 154$: $$4 \pi r^2 = 154 \Rightarrow r^2 = \frac{154}{4 \pi} = \frac{154}{4 \times 3.142} = \frac{154}{12.568} \approx 12.25$$ Step 4: Calculate $r$: $$r = \sqrt{12.25} = 3.5$$ Step 5: Diameter $d = 2r = 7.00$ cm 9. **Two sectors with arcs CD and EF of concentric circles, center O. Angle COD = $\frac{2}{3}$ radians, CE = DF = 5 cm.** No question is posed, so no solution is provided.