answer:1. **Assumption and Graph Construction**: Suppose it is possible to set the heights of the columns such that each wire intersects exactly two other wires, once from below and once from above. Consider the vertices of the regular pentagon as A, B, C, D, E. 2. **Wire Intersection Between AB and CD**: Assume that at the intersection of wires AB and CD, the wire AB passes above CD. According to the condition, at the intersection of AB and DE, AB must pass below DE. Let us denote this scenario with the help of an illustration. Assume AB intersects CD from above, which implies: [ h_A + h_B > h_C + h_D ] Then, AB should intersect DE from below, which implies: [ h_A + h_B < h_D + h_E ] These two conditions should hold simultaneously. 3. **Further Analysis of Intersection Points**: Imagining the configuration of the wires, each wire from vertex A to B intersects another wire in triangle CDE at some point P. Here, AB intersects CDE inside the plane formed by C, D, and E at a point P. Now, AE is considered. Because AB intersects CD inside triangle CDE at point P, the segment AE must be on the same side of P if it maintains the same perspective intersection nature: [ h_A > h_C, h_D, h_E ] 4. **Validation with Other Segments**: Next, consider the wire DC. If AB intersects all possible connecting segments of CDE similarly, then for DC to intersect AB from below, it must intersect AE in a similar manner, implying a violation of the cross-intersection condition. 5. **Conclusion on Overall Configuration**: The established condition requires at least one point to have an intersection such that: [ h_A < h_C < h_D < h_E text{ or some height permutation} ] This results in continuous inequality violations as the setup renders each wire segment adhering to a repeating intersection in one direction, which contradicts our initial assumption. Hence, it is not possible to arrange the heights of all vertices such that each wire intersects every other wire exactly twice (once from below and once from above) because this eventually leads to contradictory conditions for the intersections across the segments' configurations. # Conclusion: Thus, the arrangement of column heights cannot be chosen to satisfy the given intersecting wire conditions. [ boxed{} ]

answer:To find the positive real number x such that lfloor x rfloor cdot x = 70, we start by considering the properties of the floor function, which gives us the inequality lfloor x rfloor leq x < lfloor x rfloor + 1. This means that the value of x is at least lfloor x rfloor but less than lfloor x rfloor + 1. Given the equation lfloor x rfloor cdot x = 70, we need to find a value of lfloor x rfloor that, when multiplied by a number in the range lfloor x rfloor leq x < lfloor x rfloor + 1, gives a product of 70. We can test integer values of lfloor x rfloor to find a suitable candidate. For lfloor x rfloor = 8, we have: - The lower bound of the product is 8 cdot 8 = 64. - The upper bound, without reaching lfloor x rfloor + 1 = 9, would be less than 9 cdot 9 = 81. Since 64 leq 70 < 81, lfloor x rfloor = 8 is a suitable choice because 70 falls within the range defined by multiplying 8 by a number between 8 and 9. Given lfloor x rfloor = 8, we substitute into the equation to find x: [ begin{align*} lfloor x rfloor cdot x &= 70 8x &= 70 x &= frac{70}{8} x &= 8.75 end{align*} ] Therefore, the positive real number x such that lfloor x rfloor cdot x = 70 is boxed{8.75}.

answer:(1) The inequality f(x) < 4-|x-1| can be rewritten as |3x+2|+|x-1| < 4. When x < − frac{2}{3}, the inequality becomes -3x-2-x+1 < 4, which yields − frac{5}{4} < x < − frac{2}{3}. When − frac{2}{3}leqslant xleqslant 1, the inequality becomes 3x+2-x+1 < 4, which yields − frac{2}{3}leqslant x < frac{1}{2}. When x > 1, the inequality becomes 3x+2+x-1 < 4, which has no solution. Thus, the solution set of the original inequality is boxed{− frac{5}{4} < x < frac{1}{2}}. (2) We have frac{1}{m}+ frac{1}{n}=( frac{1}{m}+ frac{1}{n})(m+n)=1+1+ frac{n}{m}+ frac{m}{n}geqslant 4, with equality holding if and only if m=n= frac{1}{2}. Let g(x)=|x-a|-f(x)=|x-a|-|3x+2|. We can express g(x) as a piecewise function: g(x) = begin{cases} 2x+2, & x < - frac{2}{3} -4x-2, & - frac{2}{3}leqslant xleqslant a -2x-2-a, & x > a end{cases} The maximum value of g(x) occurs at x=− frac{2}{3}, and is given by g{(x)}_{max}= frac{2}{3}+a. In order for the inequality to always hold true, we need g{(x)}_{max}= frac{2}{3}+aleqslant 4. Solving this inequality yields boxed{0 < aleqslant frac{10}{3}}.

answer:1. **Divide (x^{10}) by (x - 1):** Using the Remainder Theorem, substitute (x = 1) into (x^{10}) to find the remainder: [ (1)^{10} = 1 ] Thus, the remainder (r_3 = 1). 2. **Determine the quotient (q_3(x)):** By the division of (x^{10}) by (x-1), the quotient (q_3(x)) becomes a polynomial reduced in the degree by 1, which is (x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1). 3. **Divide (q_3(x)) by (x - 1) again:** Substitute (x = 1) into (q_3(x)) to get the remainder: [ q_3(1) = 1^9 + 1^8 + 1^7 + 1^6 + 1^5 + 1^4 + 1^3 + 1^2 + 1 + 1 = 10 ] Therefore, the remainder (r_4 = 10). 4. **Conclusion:** The value of (r_4) is (10). boxed{The correct answer is (textbf{(D) } 10).}