answer:To maximize the area of the projection of a rectangular parallelepiped (a box) onto the horizontal plane, we start by understanding the nature of the projection of this box onto any plane. The projection of a rectangular parallelepiped onto a plane results in either a parallelogram or a polygon with up to six sides, depending on the orientation of the box. 1. **Identify projections of a parallelepiped:** The projection of a parallelepiped (box) can be considered as several parallelograms forming a polygon. If we denote the projected vertices of the parallelepiped as (A', B', C', D', E', F'), then we can form a hexagon (A'B'C'D'E'F'), where (A', B', C', D', E', F') are the projections of the vertices (A, B, C, D, E, F) of the original parallelepiped. Note that this hexagon may degenerate into a quadrilateral if the parallelogram edges align perfectly. 2. **Form multiple parallelograms:** Let this hexagon consist of three key parallelograms, for example, the parallelograms (A'B'C'O', C'D'E'O'), and (E'F'A'O'), where (O') is an interior point of the hexagon and represents the projection of a vertex from the original three-dimensional figure. 3. **Understanding parallelogram properties:** The area of the hexagon can be split into sums of areas of the parallelograms projected. Each of these areas is maximized when considering that the projection area is proportional to the cosine of the angle between the plane of projection and the plane of the parallelogram in three-dimensional space. 4. **Maximizing the projection area:** By geometry, the area of a projected parallelogram is maximized when its plane is parallel to the projection plane. So, for a horizontal projection plane, the parallelogram must be aligned such that one of its faces (which appears as a parallelogram in projection) lies parallel to the horizontal plane. 5. **Choosing optimal positioning:** To maximize the total projected area on the horizontal plane, the optimal arrangement would be that at least one face of the parallelepiped lies completely parallel to the horizontal plane. In other words, the box should be placed such that the base lies flat on the horizontal plane, ensuring all projections of horizontal faces result in their true shape with maximal area. # Conclusion: The parallelepiped should be arranged in space such that at least one of its faces lies completely parallel to the horizontal plane. This configuration maximizes the area of the projection onto the horizontal plane. [ boxed{text{The parallelepiped should be oriented such that one face lies parallel to the horizontal plane.}} ]

answer:Solution: In option A, the correct expression should be ( frac {n}{m})^{7}=n^{7}m^{-7}; For option B, the left side of the equation is positive, while the right side is negative; For option C, when x=y=1, the equation does not hold; Option D is correct. Therefore, the answer is boxed{D}. By applying the rules of exponentiation and the conversion between radicals and fractional exponents, in option A, the correct expression should be ( frac {n}{m})^{7}=n^{7}m^{-7}; For option B, the left side of the equation is positive, while the right side is negative; For option C, when x=y=1, the equation does not hold, which can be determined by the process of elimination. This question tests the conversion between radicals and fractional exponents, and the rules of exponentiation, focusing on computational skills.

answer:To find an approximate solution of the quadratic equation x^{2}-x=1.4, we observe the given table and compare the values of x^{2}-x with 1.4 for different values of x. - For x=1.1, we have x^{2}-x=0.11 - For x=1.2, we have x^{2}-x=0.24 - For x=1.3, we have x^{2}-x=0.39 - For x=1.4, we have x^{2}-x=0.56 - For x=1.5, we have x^{2}-x=0.75 - For x=1.6, we have x^{2}-x=0.96 - For x=1.7, we have x^{2}-x=1.19 - For x=1.8, we have x^{2}-x=1.44 - For x=1.9, we have x^{2}-x=1.71 By comparing these values with 1.4, we notice that when x=1.8, x^{2}-x=1.44 is the closest to 1.4. Therefore, the value of x that makes x^{2}-x closest to 1.4 is 1.8. Hence, the approximate solution of the quadratic equation x^{2}-x=1.4 is boxed{D}.

answer:1. **Factorize the given equation:** Given ( a^3 - b^3 = 22x^3 ) and ( a - b = 2x ), start by factorizing the left side using the difference of cubes formula: [ a^3 - b^3 = (a-b)(a^2 + ab + b^2). ] Substituting ( a - b = 2x ), we get: [ 2x(a^2 + ab + b^2) = 22x^3. ] 2. **Simplify the equation:** Assuming ( x neq 0 ) (since ( a neq b )), divide both sides by ( 2x ): [ a^2 + ab + b^2 = 11x^2. ] Also, ( (a-b)^2 = (2x)^2 = 4x^2 ), so: [ a^2 - 2ab + b^2 = 4x^2. ] Adding ( 3ab ) to both sides: [ a^2 + ab + b^2 = 4x^2 + 3ab. ] Equating to the earlier expression gives: [ 4x^2 + 3ab = 11x^2, ] [ 3ab = 11x^2 - 4x^2 = 7x^2, ] [ ab = frac{7x^2}{3}. ] 3. **Solve for ( a ) in terms of ( x ):** From ( a - b = 2x ) and ( ab = frac{7x^2}{3} ): [ b = a - 2x, ] [ a(a - 2x) = frac{7x^2}{3}, ] [ a^2 - 2ax - frac{7x^2}{3} = 0. ] Solving this quadratic equation for ( a ): [ a = frac{2x pm sqrt{(2x)^2 + 4 cdot frac{7x^2}{3}}}{2}, ] [ a = frac{2x pm sqrt{4x^2 + frac{28x^2}{3}}}{2}, ] [ a = frac{2x pm xsqrt{frac{40}{3}}}{2}, ] [ a = x + frac{xsqrt{frac{40}{3}}}{2} quad text{or} quad a = x - frac{xsqrt{frac{40}{3}}}{2}. ] **Conclusion:** The possible values of ( a ) are ( x + frac{xsqrt{frac{40}{3}}}{2} ) or ( x - frac{xsqrt{frac{40}{3}}}{2} ). Therefore, the correct answer is: [ x left(1 + frac{sqrt{frac{40{3}}}{2}right) text{ or } x left(1 - frac{sqrt{frac{40}{3}}}{2}right)}. ] boxed{The final answer is **(boxed{(C)})**.}