answer:1. Let's consider the function f(x) = 2x^3 - 2x defined on the interval [-1, 1]. 2. We will use the arithmetic and geometric mean inequality (AM-GM inequality), which states that for non-negative numbers a and b: frac{a + b}{2} geq sqrt{ab} 3. Apply this inequality to the numbers a = x^2 and b = 1 - x^2 (noting that x^2 geq 0 and 1 - x^2 geq 0 for x in [-1, 1]): frac{x^2 + (1 - x^2)}{2} geq sqrt{x^2 cdot (1 - x^2)} 4. Simplifying the left-hand side, we get: frac{1}{2} geq |x|sqrt{1 - x^2} 5. This implies: sqrt{1 - x^2} geq 2|x|(1 - x^2) = 2|x|(1 - x^2) = |f(x)| 6. Inequality (5) implies that the values of f(x) on the interval [-1, 1] are bounded within the unit circle with radius 1 and centered at the origin (0, 0). Therefore, the maximum distance between any two points on the graph of f(x) is at most the diameter of this circle, which is 2. 7. Indeed, this maximum distance of 2 can be attained, for instance, by the points A = (-1, 0) and D = (1, 0). 8. To verify that the maximum distance is 2, consider the equality condition in the AM-GM inequality: equality holds if and only if x^2 = 1 - x^2, which yields x = pm frac{1}{sqrt{2}}. 9. This gives two more points: B = left(-frac{1}{sqrt{2}}, frac{1}{sqrt{2}}right) and C = left(frac{1}{sqrt{2}}, -frac{1}{sqrt{2}}right). 10. Both pairs of points, (A, D) and (B, C), lie on the unit circle and are on the graph of f(x). Moreover, they are diametrically opposite each other, confirming the maximum distance is indeed 2. Conclusion. boxed{A=(-1, 0) text{ and } D=(1, 0) text{ or } B=left(-frac{1}{sqrt{2}}, frac{1}{sqrt{2}}right) text{ and } C=left(frac{1}{sqrt{2}}, -frac{1}{sqrt{2}}right)}.

answer:1. **Original cube dimensions:** Let edge length of the original cube be (a). Therefore, the volume (V_{text{cube}} = a^3). 2. **New rectangular solid dimensions:** Changing dimensions according to the problem, the dimensions are ((a-2)), (a), and ((a+2)). Volume of the new rectangular solid (V_{text{new}} = (a-2) cdot a cdot (a+2)). 3. **Expanding and simplifying the new volume:** [ V_{text{new}} = (a-2) cdot (a+2) cdot a = (a^2 - 4) cdot a = a^3 - 4a ] 4. **Equation setup based on volume difference:** The volume difference is (8): [ a^3 - (a^3 - 4a) = 8 ] Simplifying, we have: [ a^3 - a^3 + 4a = 8 implies 4a = 8 implies a = 2 ] 5. **Determining the cube's original volume:** Using (a = 2): [ V_{text{cube}} = 2^3 = 8 ] 6. **Conclusion and boxed answer:** The original volume of the cube was (8). boxed{- The correct answer is **(B) 8**}

answer:We are asked to find the number of 5-digit numbers such that the product of their digits equals 180. 1. **Factorize 180 into prime factors**: [ 180 = 2^2 cdot 3^2 cdot 5 ] 2. **List all possible factorizations of 180 into five single-digit numbers**: We need to find sets of 5 digits (a, b, c, d, e) such that (a cdot b cdot c cdot d cdot e = 180), where each (a, b, c, d, e in {1, 2, 3, 4, 5, 6, 7, 8, 9}). The possible valid combinations (ignoring permutations) are: [ begin{aligned} 1. & quad 1 cdot 1 cdot 4 cdot 5 cdot 9 2. & quad 1 cdot 1 cdot 5 cdot 6 cdot 6 3. & quad 1 cdot 2 cdot 2 cdot 5 cdot 9 4. & quad 1 cdot 2 cdot 3 cdot 5 cdot 6 5. & quad 1 cdot 3 cdot 3 cdot 4 cdot 5 6. & quad 2 cdot 2 cdot 3 cdot 3 cdot 5 end{aligned} ] 3. **Calculate the number of permutations for each combination**: - For (1 cdot 1 cdot 4 cdot 5 cdot 9): - This set has 2 identical digits (1 and 1), so the number of permutations is: [ frac{5!}{2!} = frac{120}{2} = 60 ] - For (1 cdot 1 cdot 5 cdot 6 cdot 6): - This set has 2 identical digits (1 and 1) and another pair of identical digits (6 and 6), so the number of permutations is: [ frac{5!}{2! cdot 2!} = frac{120}{4} = 30 ] - For (1 cdot 2 cdot 2 cdot 5 cdot 9): - This set has 2 identical digits (2 and 2), so the number of permutations is: [ frac{5!}{2!} = frac{120}{2} = 60 ] - For (1 cdot 2 cdot 3 cdot 5 cdot 6): - All digits are distinct, so the number of permutations is: [ 5! = 120 ] - For (1 cdot 3 cdot 3 cdot 4 cdot 5): - This set has 2 identical digits (3 and 3), so the number of permutations is: [ frac{5!}{2!} = frac{120}{2} = 60 ] - For (2 cdot 2 cdot 3 cdot 3 cdot 5): - This set has 2 pairs of identical digits (2 and 2, 3 and 3), so the number of permutations is: [ frac{5!}{2! cdot 2!} = frac{120}{4} = 30 ] 4. **Sum all the permutations**: [ 60 + 30 + 60 + 120 + 60 + 30 = 360 ] # Conclusion: [ boxed{360} ]

answer:From the given, the focus of the parabola y^{2}=4x is F(1,0). Since PM+PQ=PM+PF-1, Therefore, when PM+PF is minimized, the sum of the lengths of PQ and PM is minimized. The minimum value of PM+PF is the distance MF= sqrt{34}. Hence, the answer is sqrt{34}-1. Because PM+PQ=PM+PF-1, when PM+PF is minimized, the sum of the lengths of PQ and PM is minimized, and the minimum value of PM+PF is MF= sqrt{34}, which can be calculated. This problem uses the parabola as a medium to examine the geometric properties of the parabola and tests the students' ability to analyze and solve problems. Therefore, the minimum value of the sum of the lengths of PQ and PM is boxed{sqrt{34}-1}.