answer:1. **Understanding the Problem:** We are given a wooden model of a cube, and we need to establish that it is possible to make a hole through which an identical cube can pass. 2. **Projection Analysis:** Let's consider two projections of the cube. - Projection (a): The projecting lines are parallel to the diagonals of the cube. - Projection (b): The projecting lines are parallel to the edges of the cube. 3. **Properties of Diagonals:** For projection (a), observe that the diagonals of the faces perpendicular to the direction of the projecting lines are not distorted. Therefore, their lengths are preserved. 4. **Constructing the Other Projection:** Using the preserved diagonals in projection (a), we can find the sides of the square by considering the properties of a cube's diagonal. The diagonal of a square with side length ( a ) is ( asqrt{2} ). 5. **Projection Alignment:** Next, we show that the entire second projection fits completely inside the first projection: - Consider the square inside projection (a). - Draw a circle of radius ( frac{1}{2} ) of the diagonal of this square, centered at the origin ( O ). - This circle will fit within the hexagon inscribed into the projection (a). 6. **Mathematical Confirmation:** To further confirm, calculate the radius of the circle formed by the square’s diagonal: [ text{Diagonal of square} = asqrt{2} implies text{Radius} = frac{asqrt{2}}{2}. ] For the circle inscribed in the hexagon: [ text{Radius} = frac{asqrt{3}}{2}. ] Since ( sqrt{2} < sqrt{3} ), the inscribed circle fits within the hexagon. 7. **Hole Creation:** If we drill a hole along the dotted lines in projection (a) along the direction of the projecting lines, we create an aperture through which a congruent cube can pass. **Conclusion:** By drilling the hole at the specified location, it's possible to tunnel through the model cube such that a congruent cube can pass through. Therefore, the assertion holds true. [ boxed{} ]

answer:Pair the terms as follows: [left(frac{1}{x} + frac{1}{x + 15}right) + left(frac{1}{x + 3} + frac{1}{x + 13}right) - left(frac{1}{x + 5} + frac{1}{x + 11}right) - left(frac{1}{x + 7} + frac{1}{x + 9}right) = 0.] Then, [frac{2x + 15}{x^2 + 15x} + frac{2x + 16}{x^2 + 16x + 39} - frac{2x + 16}{x^2 + 16x + 55} - frac{2x + 16}{x^2 + 16x + 63} = 0.] Dividing by 2, we get: [frac{x + 7.5}{x^2 + 15x} + frac{x + 8}{x^2 + 16x + 39} - frac{x + 8}{x^2 + 16x + 55} - frac{x + 8}{x^2 + 16x + 63} = 0.] Let y = x + 8: [frac{y - 0.5}{y^2 - 49} + frac{y}{y^2 - 17} - frac{y}{y^2 - 1} - frac{y}{y^2 - 9} = 0.] Dividing by y (assuming y neq 0), [frac{1}{y^2 - 49} + frac{1}{y^2 - 17} - frac{1}{y^2 - 1} - frac{1}{y^2 - 9} = 0.] Let z = y^2, [frac{1}{z - 49} + frac{1}{z - 17} - frac{1}{z - 1} - frac{1}{z - 9} = 0.] This can be rewritten and solved similarly to the initial solution, leading to a quadratic in z: [z^2 - az + b = 0,] where a, b solve accordingly and yield z = 19 pm 6 sqrt{5}, y = pm sqrt{19 pm 6 sqrt{5}}, and x = -8 pm sqrt{19 pm 6 sqrt{5}}. Conclusion: Thus, a + b + c + d = 8 + 19 + 6 + 5 = boxed{38}.

answer:Let's denote the manufacturer's suggested retail price as x. The pet store regularly sells pet food at a discount of 10% to 30%. The additional discount during the sale is 20% off the already discounted price. The lowest possible price of the container of pet food is achieved when the store gives the maximum regular discount of 30% and then an additional 20% off that discounted price. First, we calculate the price after the maximum regular discount of 30%: Price after 30% discount = x - 0.30x = 0.70x Then, we calculate the price after the additional 20% discount on the already discounted price: Price after additional 20% discount = 0.70x - 0.20(0.70x) = 0.70x - 0.14x = 0.56x We are given that the lowest possible price after all discounts is 16.80: 0.56x = 16.80 Now, we solve for x to find the manufacturer's suggested retail price: x = 16.80 / 0.56 x = 30 Therefore, the manufacturer's suggested retail price is boxed{30} .

answer:We know that overrightarrow{Z_1 Z_2} = overrightarrow{OZ_2} - overrightarrow{OZ_1}. Therefore, the complex number corresponding to overrightarrow{Z_1 Z_2} is z_2 - z_1. Substituting the given expressions for z_1 and z_2, we obtain: z_2 - z_1 = (-a^2 + a + 2) + (a^2 + a - 6)i. Since this is a purely imaginary number, its real part must be zero: -a^2 + a + 2 = 0. Solving this quadratic equation, we find that a = -1 or a = 2. However, the imaginary part of z_2 - z_1 must not be zero for it to be purely imaginary. That is, a^2 + a - 6 neq 0. Substituting a = -1 and a = 2 into this inequality, we find that only a = -1 satisfies it. Thus, boxed{a = -1}.