answer:I4.1. Find the area of triangle QRT. 1. Given that the area of the parallelogram PQRS is 80 , text{cm}^2. 2. triangle QRT has the same base and the same height as the parallelogram PQRS. 3. The area of triangle QRT is half of the area of parallelogram PQRS. [ A = frac{1}{2} cdot 80 = 40 , text{cm}^2 ] I4.2. Find B if B = log_2left(frac{8A}{5}right). 1. Substitute A = 40 from the previous step into the equation: [ B = log_2 left( frac{8 cdot 40}{5} right) ] 2. Simplify inside the logarithm: [ B = log_2 left( frac{320}{5} right) ] 3. Further simplify: [ B = log_2 (64) ] 4. Since 64 = 2^6, we use the property of logarithms log_b (b^k) = k: [ B = log_2 (2^6) = 6 ] I4.3. Given x + frac{1}{x} = B, find C = x^3 + frac{1}{x^3}. 1. From the previous step, B = 6: [ x + frac{1}{x} = 6 ] 2. To find x^2 + frac{1}{x^2}, square both sides: [ left( x + frac{1}{x} right)^2 = 6^2 ] 3. Expand the left side: [ x^2 + 2 + frac{1}{x^2} = 36 ] 4. Isolate x^2 + frac{1}{x^2}: [ x^2 + frac{1}{x^2} = 36 - 2 = 34 ] 5. To find x^3 + frac{1}{x^3}, use the formula: [ x^3 + frac{1}{x^3} = left( x + frac{1}{x} right) left( x^2 + frac{1}{x^2} - 1 right) ] 6. Substitute the known values: [ x^3 + frac{1}{x^3} = 6 left( 34 - 1 right) = 6 cdot 33 = 198 ] I4.4. If (C, 2) = 212, find D. 1. Given (C, 2) = qD + p where (C, 2) = 212: [ 2D + C = 212 ] 2. Substitute C = 198: [ 2D + 198 = 212 ] 3. Solve for D: [ 2D = 212 - 198 ] [ 2D = 14 ] [ D = frac{14}{2} = 7 ] Conclusion. (boxed{7})

answer:The area of a square with side length 2 units is (2^2 = 4) square units. The area of a right triangle with both legs of length 2 units is (frac{1}{2} times 2 times 2 = 2) square units. Let's calculate the total area for each proposed polygon: - **Polygon A:** Consists of 2 squares and 1 right triangle. Total area = (2 times 4 + 2 = 10) square units. - **Polygon B:** Consists of 3 squares. Total area = (3 times 4 = 12) square units. - **Polygon C:** Consists of 1 square and 4 right triangles. Total area = (4 + 4 times 2 = 12) square units. - **Polygon D:** Consists of 3 right triangles. Total area = (3 times 2 = 6) square units. - **Polygon E:** Consists of 4 squares. Total area = (4 times 4 = 16) square units. After comparing the areas: - **Polygon A:** 10 square units. - **Polygon B:** 12 square units. - **Polygon C:** 12 square units. - **Polygon D:** 6 square units. - **Polygon E:** 16 square units. Therefore, Polygon E has the largest area of all, which is (16) square units. boxed{The final answer is (boxed{textbf{(E)} 16 text{ sq units}}).}

answer:If you mistakenly multiplied a decimal by 12 and got 84.6, we can find the original decimal by dividing 84.6 by 12. So, the calculation would be: 84.6 ÷ 12 = 7.05 Therefore, the original decimal should be boxed{7.05} .

answer:1. **Initial Assumption**: We start by assuming that it is indeed possible to place three arcs of great circles on a sphere, each measuring (300^circ), in such a way that no two of them share any points, whether endpoints or otherwise. 2. **Establish Intersection Points**: Let’s consider these arcs and the great circles they are a part of. Suppose the points of intersection of these great circles are labeled as (A, B, C) and their respective diametrically opposite points (A', B', C'). 3. **Characteristics of the Arcs**: Since each arc is (300^circ), it means that each arc is more than half of a great circle, i.e., more than (180^circ). Thus, each arc must pass through at least one point from each pair ((A, A'), (B, B'), (C, C')). 4. **Connectivity Requirement**: For our arcs not to intersect, each arc must pass through exactly two points that are not diametrically opposite. Without loss of generality, assume that the arcs pass through the points as follows: - One arc through (A) and (B') - Another arc through (B) and (C') - The third arc through (C) and (A') 5. **An Arc's Angles**: Define the arc (overset{frown}{XY}) to be the minor arc between points (X) and (Y) on a great circle. Knowing that an arc through (A) and (B') is non-intersecting means the segment (A') to (B) must be less than (60^circ) because: [ 360^circ - 300^circ = 60^circ ] This corresponds to: [ angle A'B < 60^circ ] 6. **Inequality Relations**: Given (angle A'B < 60^circ), then the measure ( angle AB) on the sphere would be: [ angle AB = 180^circ - angle A'B > 120^circ ] By symmetry and similar corresponding angles, we also have: [ angle BC > 120^circ quad text{and} quad angle CA > 120^circ ] 7. **Spherical Triangle Angle Sum**: The sum of the interior angles of any spherical triangle formed by vertexes (O), (A), (B), and (C) (where (O) is the sphere’s center) exceeds (360^circ), which contradicts the geometric constraints of a spherical triangle on a unit sphere. 8. **Conclusion**: Thus, the assumption that these arcs can be positioned without intersections is false. No such configuration exists that allows three (300^circ) arcs of great circles on a sphere to not share any points or endpoints. [ boxed{} ]