answer：Alright, so I've got this function f(x) = |x² - 1| + x² + kx, and I need to solve two parts. Let me start with part (1).**Problem (1):** Find the range of values for the real number k such that for any x in the interval (0, +∞), f(x) ≥ 0 holds.Hmm, okay. So I need to ensure that f(x) is always non-negative for all positive x. Let me write down the function again:f(x) = |x² - 1| + x² + kx.First, I should consider the absolute value term |x² - 1|. This will change its behavior depending on whether x² is greater than or less than 1. So, I can split the function into two cases: when x is in (0, 1] and when x is in (1, +∞).**Case 1: x ∈ (0, 1]**In this interval, x² ≤ 1, so |x² - 1| = 1 - x². Therefore, the function becomes:f(x) = (1 - x²) + x² + kx = 1 + kx.So, f(x) = 1 + kx. I need this to be ≥ 0 for all x in (0, 1].So, 1 + kx ≥ 0 for all x ∈ (0, 1].Let me solve for k:kx ≥ -1.Since x is positive, I can divide both sides by x:k ≥ -1/x.But x is in (0, 1], so 1/x is ≥ 1. Therefore, -1/x is ≤ -1.So, to satisfy k ≥ -1/x for all x in (0, 1], k must be ≥ the maximum of -1/x in that interval.The maximum of -1/x occurs at x = 1, which is -1. So, k must be ≥ -1.**Case 2: x ∈ (1, +∞)**Here, x² > 1, so |x² - 1| = x² - 1. Therefore, the function becomes:f(x) = (x² - 1) + x² + kx = 2x² + kx - 1.I need 2x² + kx - 1 ≥ 0 for all x > 1.This is a quadratic in x. Let me consider it as a function of x:g(x) = 2x² + kx - 1.I need g(x) ≥ 0 for all x > 1.Since the coefficient of x² is positive (2), the parabola opens upwards. So, the minimum of g(x) occurs at the vertex.The vertex of a quadratic ax² + bx + c is at x = -b/(2a). So here, x = -k/(4).But wait, x must be greater than 1. So, the vertex is at x = -k/4.If the vertex is to the left of x = 1, then the function is increasing for x > 1, so the minimum value on x > 1 would be at x = 1.If the vertex is to the right of x = 1, then the minimum is at the vertex.So, let's check where the vertex is.x_vertex = -k/4.We need to see if x_vertex ≤ 1 or x_vertex > 1.Case 2a: If x_vertex ≤ 1, i.e., -k/4 ≤ 1 ⇒ -k ≤ 4 ⇒ k ≥ -4.In this case, the minimum of g(x) on x > 1 is at x = 1.So, compute g(1):g(1) = 2(1)² + k(1) - 1 = 2 + k - 1 = k + 1.We need g(1) ≥ 0 ⇒ k + 1 ≥ 0 ⇒ k ≥ -1.Case 2b: If x_vertex > 1, i.e., -k/4 > 1 ⇒ -k > 4 ⇒ k < -4.In this case, the minimum of g(x) is at x = -k/4.Compute g(-k/4):g(-k/4) = 2*(-k/4)² + k*(-k/4) - 1.Let me compute each term:First term: 2*(k²/16) = k²/8.Second term: -k²/4.Third term: -1.So, g(-k/4) = (k²/8) - (k²/4) - 1 = (k²/8 - 2k²/8) - 1 = (-k²/8) - 1.We need this to be ≥ 0:(-k²/8) - 1 ≥ 0 ⇒ -k²/8 ≥ 1 ⇒ -k² ≥ 8 ⇒ k² ≤ -8.But k² is always non-negative, so k² ≤ -8 is impossible. Therefore, there's no solution in this case.Thus, Case 2b doesn't yield any valid k. So, only Case 2a is relevant, which requires k ≥ -1.Combining both cases:From Case 1: k ≥ -1.From Case 2: k ≥ -1.Therefore, the range of k is k ≥ -1.Wait, but let me double-check. In Case 2, if k < -4, the minimum is at x = -k/4, but that leads to an impossible condition, so k must be ≥ -4? Wait, no, in Case 2a, we have k ≥ -4, but in Case 2b, k < -4, but in that case, the minimum is negative, which is not allowed. So, actually, k must be ≥ -1 to satisfy both cases.Wait, perhaps I made a miscalculation.Wait, in Case 2a, when k ≥ -4, the minimum is at x=1, which requires k ≥ -1.But if k is between -4 and -1, then k ≥ -4 but k < -1, then g(1) = k + 1 < 0, which would violate the condition. So, actually, for Case 2a, to have g(1) ≥ 0, k must be ≥ -1.Therefore, regardless of whether k is ≥ -4 or not, the condition is k ≥ -1.So, overall, the range of k is k ≥ -1.**Problem (2):** If the function f(x) has two distinct zero points x₁, x₂ in the interval (0, 2), find:(i) the range of values for the real number k;(ii) the range of values for 1/x₁ + 1/x₂.Alright, part (2) is more involved. Let me tackle it step by step.First, I need to find k such that f(x) has two distinct zeros in (0, 2). So, f(x) = 0 has two solutions in (0, 2).Again, since f(x) is defined piecewise due to the absolute value, I'll split into two cases: x ∈ (0, 1] and x ∈ (1, 2).**Case 1: x ∈ (0, 1]**Here, f(x) = 1 + kx.Set f(x) = 0:1 + kx = 0 ⇒ x = -1/k.But x must be in (0, 1]. So, -1/k must be in (0, 1].Which implies:-1/k > 0 ⇒ k < 0.And,-1/k ≤ 1 ⇒ -1 ≤ k ⇒ k ≥ -1.So, for x ∈ (0, 1], the equation f(x) = 0 has a solution x = -1/k only if k ∈ (-1, 0).If k = -1, then x = 1, which is the boundary. If k = 0, f(x) = 1, which never zero.**Case 2: x ∈ (1, 2)**Here, f(x) = 2x² + kx - 1.Set f(x) = 0:2x² + kx - 1 = 0.This is a quadratic equation. Let me denote it as:2x² + kx - 1 = 0.Let me compute its discriminant:D = k² + 8.Since D is always positive (k² ≥ 0, so D ≥ 8), there are two real roots.The roots are:x = [-k ± √(k² + 8)] / (4).Now, since x must be in (1, 2), let's analyze the roots.First, note that √(k² + 8) > |k|.So, let's compute both roots:x₁ = [-k + √(k² + 8)] / 4,x₂ = [-k - √(k² + 8)] / 4.Now, since √(k² + 8) > |k|, the second root x₂ is negative because:If k is positive: -k - √(k² + 8) is negative.If k is negative: Let's say k = -m where m > 0. Then,x₂ = [m - √(m² + 8)] / 4.But √(m² + 8) > m, so x₂ is negative.Therefore, only the first root x₁ is positive. So, in (1, 2), f(x) = 0 has at most one solution.So, to have two distinct zeros in (0, 2), f(x) must have one zero in (0, 1] and one zero in (1, 2). So, both cases must contribute one solution each.From Case 1: We need k ∈ (-1, 0) to have a solution in (0, 1].From Case 2: We need the root x₁ = [-k + √(k² + 8)] / 4 to be in (1, 2).So, let's find the conditions on k such that x₁ ∈ (1, 2).So:1 < [-k + √(k² + 8)] / 4 < 2.Multiply all parts by 4:4 < -k + √(k² + 8) < 8.Let me solve the left inequality first:4 < -k + √(k² + 8).Let me denote √(k² + 8) as S for simplicity.So, 4 < -k + S.Rearranged: S > k + 4.But S = √(k² + 8).So, √(k² + 8) > k + 4.Since S is always positive, and k is in (-1, 0), let's see:k + 4 is positive because k > -1, so k + 4 > 3.So, we can square both sides:k² + 8 > (k + 4)².Compute RHS: (k + 4)² = k² + 8k + 16.So,k² + 8 > k² + 8k + 16.Subtract k² from both sides:8 > 8k + 16.Subtract 16:-8 > 8k.Divide by 8:-1 > k.Which is k < -1.But from Case 1, k must be in (-1, 0). So, this seems conflicting.Wait, that suggests that the inequality 4 < -k + √(k² + 8) leads to k < -1, but in Case 1, we need k ∈ (-1, 0). So, is there a contradiction?Wait, perhaps I made a mistake in the direction of the inequality when squaring.Wait, when I have √(k² + 8) > k + 4, and both sides are positive, squaring is valid.But let me check for k ∈ (-1, 0):Let me pick k = -0.5.Compute S = √(0.25 + 8) = √(8.25) ≈ 2.872.Compute -k + S = 0.5 + 2.872 ≈ 3.372.Which is greater than 4? No, 3.372 < 4. So, the inequality 4 < -k + S is not satisfied for k = -0.5.Wait, so perhaps the left inequality 4 < -k + S is not satisfied for k ∈ (-1, 0). Therefore, maybe there's no solution?But that can't be, because we need two zeros.Wait, perhaps I need to re-examine.Wait, the quadratic equation in Case 2 is 2x² + kx - 1 = 0, and we have x₁ = [-k + √(k² + 8)] / 4.We need x₁ ∈ (1, 2). So, let's compute x₁ for k approaching -1 and k approaching 0.When k approaches -1 from the right (k → -1⁺):x₁ = [1 + √(1 + 8)] / 4 = [1 + 3] / 4 = 1.When k approaches 0 from the left (k → 0⁻):x₁ = [0 + √(0 + 8)] / 4 = √8 / 4 ≈ 2.828 / 4 ≈ 0.707.Wait, that's less than 1. So, as k increases from -1 to 0, x₁ decreases from 1 to approximately 0.707.But we need x₁ ∈ (1, 2). So, if x₁ is decreasing from 1 to 0.707 as k increases from -1 to 0, then x₁ is always ≤ 1 in this interval. Therefore, x₁ cannot be in (1, 2) when k ∈ (-1, 0).Wait, that suggests that there is no solution in (1, 2) when k ∈ (-1, 0). But that contradicts the problem statement which says that f(x) has two distinct zeros in (0, 2). So, perhaps I made a mistake in my analysis.Wait, let me check the quadratic equation again.Wait, in Case 2, x ∈ (1, 2), f(x) = 2x² + kx - 1.Wait, when k is negative, let's say k = -m where m > 0, then the equation becomes 2x² - mx - 1 = 0.The roots are x = [m ± √(m² + 8)] / 4.So, the positive root is [m + √(m² + 8)] / 4.Wait, when m increases, the root increases.Wait, when m = 1 (k = -1), the root is [1 + 3]/4 = 1.When m increases beyond 1, say m = 2 (k = -2), the root is [2 + √(4 + 8)] / 4 = [2 + √12] / 4 ≈ [2 + 3.464]/4 ≈ 5.464/4 ≈ 1.366, which is in (1, 2).Similarly, for m = 3 (k = -3), the root is [3 + √(9 + 8)] / 4 = [3 + √17]/4 ≈ [3 + 4.123]/4 ≈ 7.123/4 ≈ 1.781, still in (1, 2).Wait, but when m = 4 (k = -4), the root is [4 + √(16 + 8)] / 4 = [4 + √24]/4 ≈ [4 + 4.899]/4 ≈ 8.899/4 ≈ 2.224, which is greater than 2.So, for m > 4, the root is greater than 2, which is outside our interval.So, to have x₁ ∈ (1, 2), we need m ∈ (1, 4), i.e., k ∈ (-4, -1).Wait, but earlier, in Case 1, we had k ∈ (-1, 0) to have a solution in (0, 1]. But now, in Case 2, to have a solution in (1, 2), we need k ∈ (-4, -1).So, to have both solutions, k must satisfy both conditions, i.e., k ∈ (-4, -1) ∩ (-1, 0) = (-4, -1).Wait, but k ∈ (-4, -1) is the range where the quadratic in Case 2 has a solution in (1, 2). But in Case 1, for k ∈ (-1, 0), we have a solution in (0, 1]. So, for k ∈ (-4, -1), in Case 1, k is less than -1, so x = -1/k would be in (0, 1) because k < -1 ⇒ -1/k ∈ (0, 1).Wait, let me check:If k ∈ (-4, -1), then -1/k ∈ (1/4, 1).So, x = -1/k is in (1/4, 1), which is within (0, 1].So, for k ∈ (-4, -1), f(x) has one zero in (0, 1] and one zero in (1, 2), totaling two zeros in (0, 2).But wait, earlier, when I considered k ∈ (-1, 0), the root in Case 2 was less than 1, which is not in (1, 2). So, to have a root in (1, 2), k must be in (-4, -1).Therefore, the range of k for part (2)(i) is (-4, -1).Wait, but let me verify with k = -2:f(x) = |x² - 1| + x² - 2x.In (0, 1], f(x) = 1 - 2x. Setting to zero: 1 - 2x = 0 ⇒ x = 0.5, which is in (0, 1].In (1, 2), f(x) = 2x² - 2x - 1. Setting to zero: 2x² - 2x - 1 = 0 ⇒ x = [2 ± √(4 + 8)] / 4 = [2 ± √12]/4 = [2 ± 2√3]/4 = [1 ± √3]/2.The positive root is [1 + √3]/2 ≈ (1 + 1.732)/2 ≈ 1.366, which is in (1, 2). So, yes, two zeros.Similarly, for k = -4:f(x) = |x² - 1| + x² - 4x.In (0, 1], f(x) = 1 - 4x. Setting to zero: x = 1/4, which is in (0, 1].In (1, 2), f(x) = 2x² - 4x - 1. Setting to zero: 2x² - 4x - 1 = 0 ⇒ x = [4 ± √(16 + 8)] / 4 = [4 ± √24]/4 = [4 ± 2√6]/4 = [2 ± √6]/2.The positive root is [2 + √6]/2 ≈ (2 + 2.449)/2 ≈ 2.224, which is just above 2, so not in (1, 2). So, at k = -4, the root in (1, 2) is exactly at x = 2.224, which is outside (1, 2). Therefore, k must be greater than -4 to have the root in (1, 2).Similarly, for k approaching -4 from above, say k = -3.9:x₁ = [3.9 + √(3.9² + 8)] / 4 ≈ [3.9 + √(15.21 + 8)] / 4 ≈ [3.9 + √23.21]/4 ≈ [3.9 + 4.818]/4 ≈ 8.718/4 ≈ 2.1795, which is still greater than 2.Wait, so perhaps my earlier conclusion was wrong. Let me re-examine.Wait, when k = -4, the root is [4 + √(16 + 8)] / 4 = [4 + √24]/4 = [4 + 2√6]/4 = [2 + √6]/2 ≈ 2.224, which is greater than 2.So, to have x₁ < 2, we need:[-k + √(k² + 8)] / 4 < 2.Multiply both sides by 4:-k + √(k² + 8) < 8.Rearranged: √(k² + 8) < 8 + k.Since √(k² + 8) is positive, and 8 + k must also be positive. So, 8 + k > 0 ⇒ k > -8.But since k ∈ (-4, -1), this is satisfied.Now, square both sides:k² + 8 < (8 + k)² = 64 + 16k + k².Subtract k² from both sides:8 < 64 + 16k.Subtract 64:-56 < 16k.Divide by 16:-3.5 < k.So, k > -3.5.Therefore, combining with earlier, k ∈ (-3.5, -1).Wait, but earlier, when k = -4, the root was at x ≈ 2.224, which is outside (1, 2). So, to have x₁ < 2, we need k > -3.5.Similarly, to have x₁ > 1, we need:[-k + √(k² + 8)] / 4 > 1 ⇒ -k + √(k² + 8) > 4.As before, let's solve this.√(k² + 8) > k + 4.But since k is negative, let me set k = -m where m > 0.So, √(m² + 8) > -m + 4.Wait, but -m + 4 is positive only if m < 4.But m > 0, so for m < 4, -m + 4 > 0.So, √(m² + 8) > 4 - m.Square both sides:m² + 8 > (4 - m)² = 16 - 8m + m².Subtract m²:8 > 16 - 8m.Subtract 16:-8 > -8m.Divide by -8 (inequality sign reverses):1 < m.So, m > 1 ⇒ k = -m < -1.Therefore, combining with the previous condition, k ∈ (-3.5, -1).So, the range of k is (-3.5, -1).Wait, but earlier, when k = -3.5, let's check:k = -3.5, so m = 3.5.x₁ = [3.5 + √(12.25 + 8)] / 4 = [3.5 + √20.25]/4 = [3.5 + 4.5]/4 = 8/4 = 2.So, at k = -3.5, x₁ = 2, which is the boundary. Since we need x₁ ∈ (1, 2), k must be greater than -3.5.Similarly, when k approaches -1 from below, x₁ approaches 1 from above.So, the range of k is (-3.5, -1).But wait, 3.5 is 7/2, so -3.5 is -7/2.So, k ∈ (-7/2, -1).Therefore, part (2)(i) answer is k ∈ (-7/2, -1).Now, part (2)(ii): the range of values for 1/x₁ + 1/x₂.Given that x₁ is in (0, 1] and x₂ is in (1, 2).From Case 1: x₁ = -1/k, where k ∈ (-7/2, -1). So, x₁ = -1/k ∈ (1/7, 1).From Case 2: x₂ = [-k + √(k² + 8)] / 4.So, 1/x₁ + 1/x₂ = k + [4 / (-k + √(k² + 8))].Let me simplify this expression.Let me denote S = √(k² + 8).So, 1/x₂ = 4 / (-k + S).Let me rationalize the denominator:4 / (-k + S) = 4(-k - S) / [(-k + S)(-k - S)] = 4(-k - S) / (k² - S²).But S² = k² + 8, so k² - S² = -8.Thus,4(-k - S)/(-8) = (-4k - 4S)/(-8) = (4k + 4S)/8 = (k + S)/2.Therefore, 1/x₂ = (k + S)/2.So, 1/x₁ + 1/x₂ = k + (k + S)/2 = (2k + k + S)/2 = (3k + S)/2.Wait, that seems a bit complicated. Let me check my steps.Wait, 1/x₂ = 4 / (-k + S).Multiply numerator and denominator by (-k - S):4*(-k - S) / [(-k + S)(-k - S)] = 4*(-k - S) / (k² - S²).But S² = k² + 8, so denominator is k² - (k² + 8) = -8.Thus,4*(-k - S)/(-8) = (-4k - 4S)/(-8) = (4k + 4S)/8 = (k + S)/2.Yes, correct.So, 1/x₂ = (k + S)/2.Therefore, 1/x₁ + 1/x₂ = k + (k + S)/2 = (2k + k + S)/2 = (3k + S)/2.But S = √(k² + 8).So, 1/x₁ + 1/x₂ = (3k + √(k² + 8))/2.Now, we need to find the range of this expression as k varies in (-7/2, -1).Let me denote this expression as T(k) = (3k + √(k² + 8))/2.We need to find the range of T(k) for k ∈ (-7/2, -1).Let me analyze T(k).First, note that as k increases from -7/2 to -1, how does T(k) behave?Compute T(k) at k = -7/2:T(-7/2) = [3*(-7/2) + √{(49/4) + 8}]/2 = (-21/2 + √(49/4 + 32/4))/2 = (-21/2 + √(81/4))/2 = (-21/2 + 9/2)/2 = (-12/2)/2 = (-6)/2 = -3.Wait, that can't be right because 1/x₁ + 1/x₂ should be positive since x₁ and x₂ are positive.Wait, perhaps I made a miscalculation.Wait, let's compute T(-7/2):3k = 3*(-7/2) = -21/2.√(k² + 8) = √{(49/4) + 8} = √{(49/4) + (32/4)} = √(81/4) = 9/2.So, T(-7/2) = (-21/2 + 9/2)/2 = (-12/2)/2 = (-6)/2 = -3.But 1/x₁ + 1/x₂ is the sum of reciprocals of positive numbers, so it must be positive. Therefore, I must have made a mistake in the sign.Wait, let me re-examine the expression.From earlier:1/x₂ = (k + S)/2.But k is negative, S is positive.So, (k + S)/2 could be positive or negative depending on whether S > |k|.But S = √(k² + 8) > |k|, since 8 > 0.So, S > |k| ⇒ k + S > 0 because S > |k|.Therefore, 1/x₂ is positive.Similarly, 1/x₁ = k, but k is negative, so 1/x₁ is negative.Wait, but 1/x₁ is k, which is negative, but x₁ is positive, so 1/x₁ is positive. Wait, no, x₁ = -1/k, so 1/x₁ = -k, which is positive because k is negative.Wait, I think I messed up earlier.From Case 1: x₁ = -1/k, so 1/x₁ = -k.Since k is negative, -k is positive.Similarly, from Case 2: x₂ = [-k + S]/4, so 1/x₂ = 4/(-k + S).But as we saw earlier, 1/x₂ = (k + S)/2.Wait, but let me re-express T(k):T(k) = 1/x₁ + 1/x₂ = (-k) + (k + S)/2.So, T(k) = -k + (k + S)/2 = (-2k + k + S)/2 = (-k + S)/2.Ah, that's different from what I had earlier. I think I made a mistake in the earlier step.So, correct expression:T(k) = (-k) + (k + S)/2 = (-2k + k + S)/2 = (-k + S)/2.Yes, that makes more sense.So, T(k) = (S - k)/2, where S = √(k² + 8).So, T(k) = (√(k² + 8) - k)/2.Now, let's compute T(k) at k = -7/2 and k = -1.At k = -7/2:T(-7/2) = (√{(49/4) + 8} - (-7/2))/2 = (√(81/4) + 7/2)/2 = (9/2 + 7/2)/2 = (16/2)/2 = 8/2 = 4.At k = -1:T(-1) = (√(1 + 8) - (-1))/2 = (3 + 1)/2 = 4/2 = 2.So, as k increases from -7/2 to -1, T(k) decreases from 4 to 2.Therefore, the range of T(k) is (2, 4).Wait, but let me check the behavior in between.Let me pick k = -2:T(-2) = (√(4 + 8) - (-2))/2 = (√12 + 2)/2 ≈ (3.464 + 2)/2 ≈ 5.464/2 ≈ 2.732.Which is between 2 and 4.Similarly, k = -3:T(-3) = (√(9 + 8) - (-3))/2 = (√17 + 3)/2 ≈ (4.123 + 3)/2 ≈ 7.123/2 ≈ 3.5615.Which is also between 2 and 4.Therefore, as k increases from -7/2 to -1, T(k) decreases from 4 to 2.Thus, the range of 1/x₁ + 1/x₂ is (2, 4).So, the final answers are:(1) k ≥ -1.(2)(i) k ∈ (-7/2, -1).(2)(ii) The range is (2, 4).

answer：Alright, so I have this problem about two parallelograms, ABCD and A'B'C'D', in space. Points A'', B'', C'', D'' divide the segments AA', BB', CC', DD' in the same ratio. I need to figure out what can be said about the quadrilateral A''B''C''D''.Hmm, okay. Let me try to visualize this. We have two parallelograms, which are like four-sided figures with opposite sides equal and parallel. They're in space, so they might not be in the same plane. Each vertex of the first parallelogram is connected to the corresponding vertex of the second parallelogram, forming segments AA', BB', CC', and DD'. Then, points A'', B'', C'', D'' are somewhere along these segments, dividing them in the same ratio. So, if A'' divides AA' in a ratio of, say, 1:2, then B'', C'', and D'' also divide BB', CC', and DD' in the same 1:2 ratio.I think the key here is to figure out the properties of quadrilateral A''B''C''D''. Is it also a parallelogram? Or maybe something else? Let me try to break it down step by step.First, let's recall what defines a parallelogram. A quadrilateral is a parallelogram if both pairs of opposite sides are parallel and equal in length. So, if I can show that A''B'' is parallel and equal to C''D'' and that B''C'' is parallel and equal to A''D'', then A''B''C''D'' is a parallelogram.To do this, maybe I can use vectors. Let me assign position vectors to the points. Let’s denote the position vectors of points A, B, C, D as vectors a, b, c, d respectively. Similarly, the position vectors of A', B', C', D' will be a', b', c', d'.Since A'', B'', C'', D'' divide AA', BB', CC', DD' in the same ratio, let's say the ratio is k:1, where k is some constant. So, the position vector of A'' can be expressed as:a'' = (k * a' + a) / (k + 1)Similarly,b'' = (k * b' + b) / (k + 1)c'' = (k * c' + c) / (k + 1)d'' = (k * d' + d) / (k + 1)Okay, so now I have expressions for all four points of quadrilateral A''B''C''D''. Let me compute the vectors representing the sides of this quadrilateral.Vector A''B'' = b'' - a'' = [(k * b' + b) / (k + 1)] - [(k * a' + a) / (k + 1)] = [k(b' - a') + (b - a)] / (k + 1)Similarly, vector C''D'' = d'' - c'' = [(k * d' + d) / (k + 1)] - [(k * c' + c) / (k + 1)] = [k(d' - c') + (d - c)] / (k + 1)Now, since ABCD is a parallelogram, we know that b - a = d - c. Similarly, since A'B'C'D' is a parallelogram, b' - a' = d' - c'.So, substituting these into the expressions above:Vector A''B'' = [k(b' - a') + (b - a)] / (k + 1) = [k(d' - c') + (d - c)] / (k + 1) = Vector C''D''Therefore, A''B'' is equal and parallel to C''D''. Similarly, I can compute vectors B''C'' and A''D''.Vector B''C'' = c'' - b'' = [(k * c' + c) / (k + 1)] - [(k * b' + b) / (k + 1)] = [k(c' - b') + (c - b)] / (k + 1)Vector A''D'' = d'' - a'' = [(k * d' + d) / (k + 1)] - [(k * a' + a) / (k + 1)] = [k(d' - a') + (d - a)] / (k + 1)Again, since ABCD is a parallelogram, c - b = a - d, but wait, that might not be directly helpful. Let me think. Actually, in a parallelogram, opposite sides are equal and parallel, so c - b = d - a. Similarly, in A'B'C'D', c' - b' = d' - a'.So, substituting these:Vector B''C'' = [k(c' - b') + (c - b)] / (k + 1) = [k(d' - a') + (d - a)] / (k + 1) = Vector A''D''Therefore, B''C'' is equal and parallel to A''D''. Since both pairs of opposite sides of quadrilateral A''B''C''D'' are equal and parallel, it must be a parallelogram.Wait, but the problem says the parallelograms are in space. Does that affect anything? I mean, in space, the parallelograms could be skewed or something. But since we're connecting corresponding points with segments and taking points along those segments in the same ratio, the resulting figure should still be a parallelogram, right?Let me think of a specific example. Suppose ABCD is a square on the XY-plane, and A'B'C'D' is another square, maybe translated along the Z-axis. Then, points A'', B'', C'', D'' would lie along the vertical segments connecting A to A', etc., at the same height. So, the quadrilateral A''B''C''D'' would also be a square, hence a parallelogram.Alternatively, if ABCD and A'B'C'D' are not in the same plane and not necessarily aligned, but still parallelograms, then connecting corresponding points with segments and taking points along those segments in the same ratio should still result in a parallelogram. Because the affine combination (which is what dividing the segments in a ratio is) preserves the parallelogram structure.So, yeah, I think regardless of the orientation in space, as long as the original figures are parallelograms and the points divide the connecting segments in the same ratio, the resulting quadrilateral is also a parallelogram.I don't see any reason why it wouldn't be. Maybe if the ratio was different for each segment, it could be something else, but since it's the same ratio, the structure is preserved.So, in conclusion, quadrilateral A''B''C''D'' is a parallelogram.

answer：First, I need to calculate the diagonal of the square base. Since each side of the square is 12 cm, the diagonal ( d ) can be found using the Pythagorean theorem:[d = sqrt{12^2 + 12^2} = sqrt{288} = 12sqrt{2} text{ cm}]Next, I determine the length of a slant edge. The slant edge forms a right triangle with half of the diagonal and the altitude of the pyramid. Half the diagonal is:[frac{d}{2} = frac{12sqrt{2}}{2} = 6sqrt{2} text{ cm}]Using the Pythagorean theorem again, the length of the slant edge ( s ) is:[s = sqrt{(6sqrt{2})^2 + 15^2} = sqrt{72 + 225} = sqrt{297}]Finally, I calculate the total length of all the edges. There are four base edges and four slant edges:[4 times 12 + 4 times sqrt{297} approx 48 + 4 times 17.233 = 48 + 68.932 = 116.932]Rounding to the nearest whole number, the sum of the lengths of the pyramid's edges is:[boxed{117}]

answer：Okay, so Amber went to the store and bought some stuff: 10 rolls of toilet paper, 7 rolls of paper towels, and 3 boxes of tissues. I need to figure out how much one box of tissues costs. First, let's break down what we know. The total cost for everything she bought is 35. We also know the cost per roll for toilet paper and paper towels. Specifically, each roll of toilet paper costs 1.50, and each roll of paper towels costs 2. Alright, so to find out how much the tissues cost, I guess I need to first calculate how much she spent on toilet paper and paper towels separately, and then subtract that from the total to find out how much was spent on tissues. After that, since she bought 3 boxes, I can divide that amount by 3 to get the cost per box.Let's start with the toilet paper. She bought 10 rolls at 1.50 each. So, 10 times 1.50 should give me the total cost for toilet paper. Let me calculate that: 10 times 1.50 is 15. Okay, so she spent 15 on toilet paper.Next, the paper towels. She bought 7 rolls at 2 each. So, 7 times 2 is... let's see, 7 times 2 is 14. Got it, she spent 14 on paper towels.Now, adding those two amounts together: 15 for toilet paper plus 14 for paper towels equals 29. So, she spent 29 on toilet paper and paper towels combined.Since the total cost for everything was 35, I can subtract the 29 she spent on the other items to find out how much she spent on tissues. So, 35 minus 29 is 6. That means she spent 6 on tissues.But wait, she bought 3 boxes of tissues. To find out how much one box costs, I need to divide the total cost of tissues by the number of boxes. So, 6 divided by 3 boxes is 2 per box.Hmm, let me double-check that to make sure I didn't make a mistake. Toilet paper: 10 rolls at 1.50 is 15. Paper towels: 7 rolls at 2 is 14. Together, that's 29. Total spent is 35, so 35 minus 29 is 6 for tissues. Divided by 3 boxes, that's 2 per box. Yeah, that seems right.I think I've got it. One box of tissues costs 2.

answer：Alright, so I've got this limit problem to solve:[lim _{n rightarrow infty} frac{(2 n+1)^{3}+(3 n+2)^{3}}{(2 n+3)^{3}-(n-7)^{3}}]Hmm, okay. Limits involving sequences can sometimes be tricky, but I remember that when dealing with polynomials, especially as ( n ) approaches infinity, the highest degree terms dominate. So maybe I can simplify this expression by focusing on the highest degree parts.First, let me look at the numerator: ( (2n + 1)^3 + (3n + 2)^3 ). If I expand these, I'll get terms involving ( n^3 ), ( n^2 ), ( n ), and constants. Similarly, the denominator is ( (2n + 3)^3 - (n - 7)^3 ), which will also expand to similar terms.But expanding everything might be too tedious. Instead, I recall that for large ( n ), the lower-degree terms become negligible compared to the highest degree terms. So, perhaps I can approximate each term by its leading term.Let's see:- ( (2n + 1)^3 ) is approximately ( (2n)^3 = 8n^3 ) for large ( n ).- ( (3n + 2)^3 ) is approximately ( (3n)^3 = 27n^3 ) for large ( n ).- ( (2n + 3)^3 ) is approximately ( (2n)^3 = 8n^3 ) for large ( n ).- ( (n - 7)^3 ) is approximately ( n^3 ) for large ( n ).So, substituting these approximations into the original expression, the numerator becomes approximately ( 8n^3 + 27n^3 = 35n^3 ), and the denominator becomes approximately ( 8n^3 - n^3 = 7n^3 ).Therefore, the limit simplifies to:[lim _{n rightarrow infty} frac{35n^3}{7n^3} = lim _{n rightarrow infty} 5 = 5]Wait, but I feel like I might be missing something here. Maybe I should check by expanding the terms more carefully to ensure that the lower-degree terms don't affect the limit.Let's try expanding ( (2n + 1)^3 ):[(2n + 1)^3 = 8n^3 + 12n^2 + 6n + 1]Similarly, ( (3n + 2)^3 ):[(3n + 2)^3 = 27n^3 + 54n^2 + 36n + 8]Adding these together:[8n^3 + 12n^2 + 6n + 1 + 27n^3 + 54n^2 + 36n + 8 = 35n^3 + 66n^2 + 42n + 9]Now, the denominator:( (2n + 3)^3 ):[(2n + 3)^3 = 8n^3 + 36n^2 + 54n + 27]( (n - 7)^3 ):[(n - 7)^3 = n^3 - 21n^2 + 147n - 343]Subtracting these:[8n^3 + 36n^2 + 54n + 27 - (n^3 - 21n^2 + 147n - 343) = 7n^3 + 57n^2 - 93n + 370]So now, the limit becomes:[lim _{n rightarrow infty} frac{35n^3 + 66n^2 + 42n + 9}{7n^3 + 57n^2 - 93n + 370}]To evaluate this, I can divide both numerator and denominator by ( n^3 ):[lim _{n rightarrow infty} frac{35 + frac{66}{n} + frac{42}{n^2} + frac{9}{n^3}}{7 + frac{57}{n} - frac{93}{n^2} + frac{370}{n^3}}]As ( n ) approaches infinity, the terms with ( frac{1}{n} ), ( frac{1}{n^2} ), and ( frac{1}{n^3} ) all approach zero. So, the expression simplifies to:[frac{35}{7} = 5]Okay, so my initial approximation was correct. The lower-degree terms don't affect the limit because they become insignificant as ( n ) grows large. Therefore, the limit is indeed 5.I think I've got it right. But just to be thorough, maybe I can test with a large value of ( n ) to see if the ratio approaches 5.Let's pick ( n = 1000 ):Numerator:( (2*1000 + 1)^3 + (3*1000 + 2)^3 = 2001^3 + 3002^3 )Denominator:( (2*1000 + 3)^3 - (1000 - 7)^3 = 2003^3 - 993^3 )Calculating these:Numerator:2001^3 ≈ 8,012,006,0013002^3 ≈ 27,054,018,008Total numerator ≈ 35,066,024,009Denominator:2003^3 ≈ 8,048,024,027993^3 ≈ 977,294,597Total denominator ≈ 8,048,024,027 - 977,294,597 ≈ 7,070,729,430Now, the ratio:35,066,024,009 / 7,070,729,430 ≈ 4.956Hmm, that's close to 5 but not exactly 5. Maybe because ( n = 1000 ) isn't large enough? Let's try ( n = 10,000 ):Numerator:20001^3 + 30002^3 ≈ 8.0012006001e+11 + 2.70054018008e+12 ≈ 3.50066024009e+12Denominator:20003^3 - 9993^3 ≈ 8.0048024027e+11 - 9.977294597e+10 ≈ 7.007072943e+11Ratio:3.50066024009e+12 / 7.007072943e+11 ≈ 4.995Closer to 5. So as ( n ) increases, the ratio approaches 5. Therefore, my conclusion is correct.I think I've thoroughly checked it. The limit is 5.

answer：First, I need to determine the total distance traveled during the trip. The car travels at two different speeds during different parts of the trip.For the first 4 hours, the car averages 55 mph. To find the distance covered in this part, I multiply the speed by the time:55 mph × 4 hours = 220 miles.For the remaining 2 hours, the car averages 70 mph. The distance covered in this part is:70 mph × 2 hours = 140 miles.Next, I add the distances from both parts to find the total distance traveled:220 miles + 140 miles = 360 miles.The total time for the trip is 6 hours. To find the average speed, I divide the total distance by the total time:360 miles / 6 hours = 60 mph.Therefore, the average speed for the entire trip is 60 mph.