answer:To determine which set of numbers cannot form the lengths of the sides of a right triangle, we apply the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. We examine each option: **Option A: sqrt{3}, 2, sqrt{5}** We check if the sum of the squares of the two smaller numbers equals the square of the largest number: [ (sqrt{3})^2 + 2^2 stackrel{?}{=} (sqrt{5})^2 ] [ 3 + 4 stackrel{?}{=} 5 ] [ 7 neq 5 ] Since this equation does not hold, set A cannot form a right triangle. **Option B: 3, 4, 5** We apply the same principle: [ 3^2 + 4^2 stackrel{?}{=} 5^2 ] [ 9 + 16 = 25 ] [ 25 = 25 ] This equation holds, so set B can form a right triangle. **Option C: 0.6, 0.8, 1** Again, we check: [ 0.6^2 + 0.8^2 stackrel{?}{=} 1^2 ] [ 0.36 + 0.64 = 1 ] [ 1 = 1 ] This equation holds, so set C can form a right triangle. **Option D: 130, 120, 50** Finally, we check: [ 50^2 + 120^2 stackrel{?}{=} 130^2 ] [ 2500 + 14400 = 16900 ] [ 16900 = 16900 ] This equation holds, so set D can form a right triangle. Since only set A does not satisfy the condition to form a right triangle, the correct answer is: [ boxed{text{A}} ]

answer:**Analysis** This question examines the properties of geometric sequences, involving the monotonicity of the sequence. It is a medium-level question. By verifying each option through the monotonicity and the general formula of a geometric sequence, we can find the solution. **Solution** Given that {a_n} is a geometric sequence with positive terms and q is its common ratio, and K_n is the product of its first n terms, From K_6 = K_7, we can deduce that a_7 = 1, so option B is correct; From K_5 < K_6, we can deduce that a_6 > 1, thus q= frac{a_7}{a_6} in (0,1), so option A is correct; Since {a_n} is a geometric sequence with positive terms and q in (0,1), the sequence is monotonically decreasing, Therefore, K_9 < K_5, making option C incorrect; Combining K_5 < K_6, K_6 = K_7 > K_8, we can conclude that option D is correct. Hence, the answer is boxed{text{C}}.

answer:Let P(x_1, y_1) and Q(x, y), Since the equation of the right directrix is x=3, The coordinates of point H are (3, y). Given |HQ| = lambda |PH| (lambda geq 1), So overrightarrow{HP} = frac{-1}{1+lambda} overrightarrow{PQ}, Therefore, by the section formula, we can get x_1 = frac{3(1+lambda)-x}{lambda}, y_1 = y, Substituting into the equation of the ellipse, we get the trajectory equation of point Q as frac{[x-3(1+lambda)]^2}{3lambda^2} + frac{y^2}{2} = 1 Therefore, the eccentricity e = frac{sqrt{3lambda^2-2}}{sqrt{3}lambda} = sqrt{1- frac{2}{3lambda^2}} in (frac{sqrt{3}}{3}, 1). Hence, the correct choice is boxed{C}. First, determine the relationship between the coordinates of P and Q, use the equation of the ellipse to obtain the trajectory equation of point Q, and then the range of the eccentricity can be determined. This problem examines the comprehensive issue of lines and conic sections, simple properties of ellipses, and is a challenging problem type in college entrance exams, requiring strong comprehensive ability and significant calculation, classified as difficult.

answer:1. **Initial Information:** The initial value of the number on the computer screen is given as 123. Every minute, this number increases by 102. 2. **Objective:** The objective is to determine if Fedya can manipulate the order of the digits of the number such that it never becomes a four-digit number. 3. **Number Sequence Pattern:** Starting from 123, each minute the new number can be calculated as: [ 123 + 102n quad text{where } n text{ is the number of minutes passed} ] We analyze the sequence of numbers: [ 123, 225, 327, 429, 531, 633, 735, 837, 939, 1041, ldots ] Notice that after 8 increments, the number reaches 939. Adding 102 again: [ 939 + 102 = 1041 ] So, at the 9th increment, the number becomes four digits. 4. **Digit Manipulation Strategy:** Fedya can rearrange the digits of the number whenever needed. The critical observation is: - The last digit of a number increases in such a way that it cycles through all possible digits from 3 to 9, back down through 1 and 2 as the last digits, then combining the hundreds column increment. 5. **Critical Handling of the Last Digit:** - Whenever the last digit of the incremented number becomes 1 or 2, Fedya can rearrange the number to bring 1 or 2 at the first position. - Let’s inspect this with an example where Fedya uses the property of the number: If the number is about to reach 939: [ text{Number: } 939 Rightarrow 1041 quad (text{before it becomes 4-digit}) ] Fedya rearranges the digits before it reaches this stage. Hence backtracking to: [ 903 Rightarrow 1005 ] Rearranges number to handle the cycle. 6. **Strategic Step to Avoid Growing Beyond Three Digits:** The increment progression towards the 4th digit can be controlled: [ text{From 939:} ] Fedya can manipulate as: - Make the last digit sum as 1 or 2 firmly by rearranging without adding it beyond. Observing through various possible cycles of the number increments: - Each digit checks (1, 2, etc., on which configuration should handle): - On the effective digit pattern when last is 1, safely becomes: [ 231 - Rightarrow manipulation on last position to prioritizing leq 2 ] 7. **Conclusion:** By constantly monitoring and rearranging the digits to keep the last digit to be (leq 2) and pushing it away from surpassing a 4th digit excessively, Fedya can control and assure the number remains within a three-digit cycle, procedurally avoiding a four-digit start. Hence, Fedya can always rearrange the digits appropriately: [ boxed{text{Yes, Fedya can keep the number from becoming a four-digit number.}} ]