answer:(1) Let N(x, y). By definition of the companion point, we have the system: begin{cases} x = frac{x_0}{a} y = frac{y_0}{b} end{cases}, which implies begin{cases} x_0 = ax y_0 = by end{cases}. Since M is on the ellipse, we have: frac{x_0^2}{a^2} + frac{y_0^2}{b^2} = 1 quad (a>b>0), thus frac{(ax)^2}{a^2} + frac{(by)^2}{b^2} = 1 quad (a>b>0), resulting in the equation x^2 + y^2 = 1. (2) From frac{1}{2} = frac{1}{a}, we get a=2. Also, frac{1}{a^2} + frac{9}{4b^2}=1, so b = sqrt{3}. Since point M(x_0, y_0) is on the ellipse, frac{x_0^2}{4} + frac{y_0^2}{3} = 1, hence y_0^2 = 3 - frac{3}{4}x_0^2, and 0 le x_0^2 le 4. The dot product overrightarrow{OM} cdot overrightarrow{ON} is (x_0, y_0) cdot left(frac{x_0}{2}, frac{y_0}{sqrt{3}}right) = frac{x_0^2}{2} + frac{y_0^2}{sqrt{3}} = left(frac{2-sqrt{3}}{4}right)x_0^2 + sqrt{3}, and since frac{2-sqrt{3}}{4} > 0, the range of overrightarrow{OM} cdot overrightarrow{ON} is left[sqrt{3}, 2right]. (3) Let A(x_1, y_1) and B(x_2, y_2), then the companion points are Pleft(frac{x_1}{2}, frac{y_1}{sqrt{3}}right) and Qleft(frac{x_2}{2}, frac{y_2}{sqrt{3}}right). - When the line l has a slope, let its equation be y=kx+m. From the system begin{cases} y = kx + m frac{x^2}{4} + frac{y^2}{3} = 1 end{cases}, we obtain (3+4k^2)x^2 + 8kmx + 4(m^2-3) = 0. Therefore, we have begin{cases} Delta = 48(3+4k^2-m^2) > 0 x_1 + x_2 = -frac{8km}{3+4k^2} x_1 x_2 = frac{4(m^2-3)}{3+4k^2} end{cases}. The circle with PQ as its diameter passes through O, implying 3x_1x_2 + 4y_1y_2 = 0. After rearranging, (3+4k^2)x_1x_2 + 4mk(x_1+x_2) + 4m^2 = 0. Substituting the first set of equations, we get 3+4k^2=2m^2. Since 3+4k^2>0, then m^2>0, so Delta=48m^2>0. The distance from O to the line y=kx+m is d = frac{|m|}{sqrt{1+k^2}}, and |AB| = sqrt{1+k^2} cdot sqrt{(x_1+x_2)^2-4x_1x_2} = sqrt{1+k^2} cdot frac{4sqrt{3}|m|}{3+4k^2}, therefore boxed{S_{triangle OAB} = frac{1}{2}|AB|d = sqrt{3}}. - When the line l has no slope, let its equation be x=m (-2 < m < 2). Combining this with the ellipse equation, we get y^2 = frac{3(4-m^2)}{4}. Substituting into 3x_1x_2 + 4y_1y_2 = 0 yields 3m^2 = 4, and thus m^2 = 2. This gives y^2 = frac{3}{2}, and S_{triangle OAB} = frac{1}{2}|AB| times d = sqrt{3}. In summary, the area of triangle OAB is a fixed value boxed{sqrt{3}}.

answer:1. **Calculate the total cost of the pizza:** The plain pizza cost 12 and the extra cheese on one-third (i.e., 4 slices) costs an additional 3 dollars. Therefore, the total cost of the pizza is: [ 12 + 3 = 15 text{ dollars} ] 2. **Determine the cost per slice:** The pizza has 12 slices, and the total cost is 15 dollars, making the cost per slice: [ frac{15}{12} = 1.25 text{ dollars per slice} ] 3. **Calculate the number of slices each person ate:** Liam ate all 4 extra cheese slices and 4 plain slices, totaling 8 slices. Ellen ate the remaining 4 plain slices. 4. **Calculate the cost for Ellen:** Ellen ate 4 plain slices. Since the plain pizza costs 12 dollars for 12 slices, each plain slice costs: [ frac{12}{12} = 1 text{ dollar per slice} ] Therefore, Ellen paid: [ 4 times 1 = 4 text{ dollars} ] 5. **Calculate the cost for Liam:** Liam ate 4 extra cheese slices and 4 plain slices. By the cost calculation: [ 8 times 1.25 = 10 text{ dollars} ] 6. **Calculate the difference in payment:** The difference in the amount paid by Liam and Ellen is: [ 10 - 4 = 6 text{ dollars} ] Therefore, Liam paid 6 more dollars than Ellen. The final answer is boxed{textbf{(C) } 6}.

answer:A full circle represents 360 degrees. If the manufacturing department's sector is 162 degrees, we can find the percentage of the circle this represents by dividing the manufacturing department's degrees by the total degrees in a circle and then multiplying by 100 to convert it to a percentage. So, the calculation would be: (162 degrees / 360 degrees) * 100 = 45% Therefore, boxed{45%} of Megatek's employees are in the manufacturing department.

answer:1. **Visualizing the Problem** We start by visualizing the given problem. The steel cube has edges of length (3 , mathrm{cm}) and the cone has a diameter (8 , mathrm{cm}) and height (24 , mathrm{cm}). We are given that one of the interior diagonals of the cube coincides with the axis of the cone. 2. **Understanding Vertices Positions** Consider that the cube is oriented such that one vertex (A) of the cube is at the bottom, directly above the vertex of the cone. Its opposite vertex (Q) is at the top. Additionally, vertices (B), (C), and (D) touch the sides of the cone, forming an equilateral triangle. 3. **Diagonal Calculation** Each interior diagonal of the cube, such as (AQ), can be calculated using the Pythagorean theorem in 3D: [ AQ = sqrt{3^2 + 3^2 + 3^2} = sqrt{27} = 3sqrt{3} ] 4. **Length of Face Diagonal** The diagonal across each face of the cube, such as (BQ), (BC), (BD), etc., is: [ BQ = sqrt{3^2 + 3^2} = sqrt{18} = 3sqrt{2} ] 5. **Geometric Relations of the Cone and Cube** Since vertices (B), (C), and (D) form an equilateral triangle within the plane, determine the height of this plane above the vertex (T) of the cone. This requires determining the segment in the base of the cone perpendicular to one side of the equilateral triangle within the placed cube. Axis (TG = GQ). Given that the radius of the cone is (4 , mathrm{cm}) (i.e., half of 8 cm), calculate the vertical distance from the cone’s tip (T) to where the equilateral triangle (BCD) is formed, called (G). 6. **Distance from the Axis of the Cone to Points B, C, D** Using the geometry, ( text{BG} ) being the median: [ BG = sqrt{ left(B Q^2 - frac{ BC^2 }{4} right)} ] Considering (BC = 3sqrt{2}) has been calculated as the face diagonal: [ BC = 3sqrt{2} implies BG = frac{3sqrt{2}}{sqrt{3}} = sqrt{6} ] 7. **Height (TG)** calculation: Since (TG = x = frac{sqrt{6}}{4}): [ TG = 24 left(frac{sqrt{6}}{4}right) = 6sqrt{6} ] 8. **Distance (AG)** calculation: To find ( AG), note: [ AQ = 3sqrt{3}, QG = sqrt{(3sqrt{2})^2 - (sqrt{6})^2} = sqrt{18 - 6} = sqrt{12} = 2sqrt{3} ] Thus: : AG = AQ - QG = 3sqrt{3} - 2sqrt{3} = sqrt{3} ] 9. **Final Calculation** Finally, we determine the distance from the tip (T) to vertex (A). [ TA = TG - AG = 6sqrt{6} - sqrt{3} ] # Conclusion: [ boxed{6 sqrt{6} - sqrt{3}} ]