answer:1. **Identify the sides of the triangle**: Let a = b - 2d, a + d = b - d, and a + 2d = b. Here b is the hypotenuse. 2. **Apply the Pythagorean Theorem**: The triangle is right-angled, so [ (a)^2 + (a+d)^2 = (a+2d)^2 ] [ b^2 - 4bd + 4d^2 + b^2 - 2bd + d^2 = b^2 ] 3. **Simplify and solve for b and d**: [ 2b^2 - 6bd + 5d^2 = b^2 ] [ b^2 - 6bd + 5d^2 = 0 ] Solve using the quadratic formula: [ d = frac{6b pm sqrt{(6b)^2 - 4 cdot 5 cdot b^2}}{2 cdot 5} = frac{6b pm sqrt{16b^2}}{10} = frac{6b pm 4b}{10} ] [ d = b quad text{or} quad d = frac{b}{5} ] 4. **Check the answer choices**: Possible values: - a = b-2d, b-d, b - For d = b: a = -b, not valid as side lengths are positive. - For d = b/5: a = 3b/5, 4b/5, b 5. **Finding which choices fit**: - textbf{(B)} 69: Could fit if 69 = b, a = 41.4, a + d = 55.2; not valid as not integers. - textbf{(C)} 48: Check if 48 = b, d = 9.6, non-integer d. - textbf{(D)} 120: Fits with 120 = b, d = frac{120}{5} = 24; yielding sides 72, 96, 120. This fits a scaled (3,4,5) triangle structure. 120 Conclusion: We verified that with sides as 72, 96, 120, the conditions and the Pythagorean theorem hold, making the proposed problem valid and the solution exact. The final answer is boxed{textbf{(D)} 120}

answer:Let's denote the number of men in the second group as M. We know that the amount of work done is constant in both cases. The work done is directly proportional to the number of men and the number of days they work. So we can write the relationship as: Work = Number of Men × Number of Days For the first group of 25 men working for 96 days, the work done can be represented as: Work = 25 men × 96 days For the second group of M men working for 60 days, the work done can be represented as: Work = M men × 60 days Since the amount of work done is the same in both cases, we can set the two equations equal to each other: 25 men × 96 days = M men × 60 days Now, we can solve for M: M = (25 men × 96 days) / 60 days M = (2400 men-days) / 60 days M = 40 men So, there are boxed{40} men in the second group.

answer:To solve the given problem, we will break down the expression into its components and solve each part step by step. 1. Calculate tan 45^{circ}: - We know that tan 45^{circ} = 1. 2. Evaluate (frac{1}{2})^{-1}: - The expression (frac{1}{2})^{-1} means we are finding the reciprocal of frac{1}{2}, which is 2. 3. Find the absolute value of |-2|: - The absolute value of |-2| is 2, because the absolute value of a number is its distance from 0 on the number line, without considering direction. Now, we add all these values together: tan 45^{circ} + (frac{1}{2})^{-1} + |-2| = 1 + 2 + 2 Adding these numbers: 1 + 2 + 2 = 5 Therefore, the final answer is boxed{5}.

answer:1. **Lemma:** Given segments on the number line such that the sum of their lengths is less than sqrt{2}, we can label the integers on the number line on intervals of length 1 such that no part of a segment is labeled. **Proof of Lemma:** Assume the contrary. We set a starting point for the integer 0, and from that starting point, we translate the intervals a length 1. Note that our intervals are still distinct in each translated position. Then the sum of all the segment lengths that have been overlapped with an integer on the number line must be 1, so the segments have length sum at least 1, a contradiction. 2. Assume that there are k total segments. Let them have lengths l_1, l_2, cdots, l_k. For a given orientation O of the coordinate plane, let x_1(O), cdots, x_k(O) and y_1(O), cdots, y_k(O) denote the lengths of the x and y components of the segments, respectively. Let [ S_x(O) = sum_{i=1}^{k} x_i(O) quad text{and} quad S_y(O) = sum_{i=1}^{k} y_i(O). ] 3. By the Cauchy-Schwarz inequality, we have: [ x_i(O) + y_i(O) le sqrt{2(x_i(O)^2 + y_i(O)^2)} = l_i sqrt{2}. ] Therefore, [ S_x(O) + S_y(O) le sum_{i=1}^{k} l_i sqrt{2} < sqrt{2} cdot sqrt{2} = 2. ] 4. Note that we can also write x_i(O) = l_i |cos(theta + alpha_i)| for fixed alpha_i. Similarly, y_i(O) = l_i |sin(theta + alpha_i)| = l_i |cos(theta + 3pi/2 + alpha_i)|. 5. If for a given O, S_x(O) < 1 and S_y(O) < 1, then by the lemma, we can fix the x and y coordinate lines to not overlap with any of the segments. 6. If S_x(O) < 1 and S_y(O) > 1, since S_x(O) is a continuous function of theta, and S_y(O) is the same function of theta, there exists a value of theta such that S_x(O) = 1. This means that since S_x(O) + S_y(O) < 2, we can again use continuity to find a value of theta such that S_x(O) < 1 and S_y(O) < 1, and apply the lemma to fix the x-coordinate lines and y-coordinate lines for the desired result. blacksquare