answer:First, let's calculate how many days Olaf and his men can survive with the 250 gallons of water they have. Since each man needs 1/2 a gallon of water per day, and there are 25 men, the total daily water consumption for the crew is: 25 men * 1/2 gallon/man/day = 12.5 gallons/day Now, let's find out how many days the 250 gallons of water will last: 250 gallons / 12.5 gallons/day = 20 days Now that we know they have enough water for 20 days, we can calculate how many miles they can travel in that time, given that the boat can go 200 miles per day: 20 days * 200 miles/day = 4000 miles Therefore, Olaf needs to travel boxed{4000} miles.

answer:1. **Identify Circle Touching Points and Line Tangents**: Let's denote the circles by (C) and (C') with centers (O) and (O') respectively. The circles (C) and (C') are touching externally. Let the common tangent touch (C) at point (A) and (C') at point (B). 2. **Determine Point (P)**: Point (P) is defined as the midpoint of line segment (AB). 3. **Special Case (Circles of Equal Size)**: - If the circles (C) and (C') are of equal radius, by symmetry, the midpoint (P) of the segment (AB) will lie on the line connecting the centers (O) and (O') (common tangent line). - Since (PA = AO), triangle (APO) is isosceles with (angle APO = 45^circ). - Similarly, triangles are isosceles for the other circle giving (angle BPO' = 45^circ). - Hence, [ angle OPO' = angle APO + angle BPO' = 45^circ + 45^circ = 90^circ. ] 4. **General Case (Circles of Unequal Size)**: - Let the lines (AB) and (OO') meet at point (X). Let points (C) and (C') touch externally at (Y). - Assign (angle AXO = k). - In triangle (AXO) we have: [ angle AOX = 90^circ - k. ] - Thus, in triangle (OBA) (where (O) is the center of the circle passing through (A)), we find: [ angle OBA = frac{90^circ - k}{2}. ] - Similarly, in triangle (BO'X) we get: [ angle BO'X = 90^circ - k. ] - In triangle (O'YB) (where (O') is the center of the circle passing through (B)): [ angle O'YB = frac{90^circ + k}{2}. ] - Therefore, triangle analysis ensures that: [ angle AYB = 90^circ. ] - Consequently, point (Y) lies on the circle with diameter (AB), meaning (PY = PA). - This deduces (P) lies on the common tangent line. 5. **Calculate Angles Around (P)**: - Let's determine (angle POY): [ angle POY = frac{angle AOY}{2} = frac{90^circ + k}{2}. ] - Thus, (angle OPY) becomes: [ angle OPY = frac{90^circ - k}{2}. ] - Similarly, at point (P) around line (XO'), we get: [ angle XO'P = frac{angle XO'B}{2} = frac{90^circ - k}{2}. ] - Therefore, (angle O'PY) is: [ angle O'PY = frac{90^circ + k}{2}. ] 6. **Combine Angle Measures**: - Hence, the total angle (angle OPO') at point (P) simplifies to: [ angle OPO' = angle OPY + angle O'PY = frac{90^circ - k}{2} + frac{90^circ + k}{2} = 90^circ. ] # Conclusion: [ boxed{90^circ} ]

answer:Let's denote the number of men in the first group as M. We know that the amount of work done is constant, so we can set up a proportion based on the work formula: Work = Men × Time × Rate. For the first group: Work = M men × 18 days × 6 hours/day For the second group: Work = 15 men × 12 days × 6 hours/day Since the work done by both groups is the same, we can set the two equations equal to each other: M men × 18 days × 6 hours/day = 15 men × 12 days × 6 hours/day Now we can cancel out the common factors on both sides of the equation: M men × 18 days = 15 men × 12 days Now we can solve for M: M × 18 = 15 × 12 M = (15 × 12) / 18 M = 180 / 18 M = 10 So, there were boxed{10} men in the first group.

answer:(1) According to the given information, we have a^{2}+b^{2}=4. The equation of the hyperbola is frac{x^{2}}{a^{2}}-frac{y^{2}}{4-a^{2}}=1 (0 < a^{2} < 4). Substituting point (3, sqrt{7}) into the equation above, we get: frac{9}{a^{2}}-frac{7}{4-a^{2}}=1. Solving the equation, we obtain a^{2}=18 (text{neglected}) or a^{2}=2. Hence, the equation of the hyperbola is boxed{frac{x^{2}}{2}-frac{y^{2}}{2}=1}. (2) According to the given information, let the equation of line l be y=kx+2. Substituting the equation of line l into the equation of hyperbola C and simplifying, we obtain: (1-k^{2})x^{2}-4kx-6=0. Since line l intersects hyperbola C at two distinct points E and F, we have: begin{cases} 1-k^{2} neq 0 triangle = (-4k)^{2} + 4 times 6(1-k)^{2} > 0 end{cases} iff begin{cases} k neq pm 1 -sqrt{3} < k < sqrt{3} end{cases}. So, k in (-sqrt{3},-1) cup (1, sqrt{3}). Let E(x_{1},y_{1}), F(x_{2},y_{2}), then we have: x_{1}+x_{2}=frac{4k}{1-k^{2}}, x_{1}x_{2}=-frac{6}{1-k^{2}}. Then, |EF|=sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}=sqrt{(1+k^{2})(x_{1}-x_{2})^{2}}=sqrt{1+k^{2}} cdot sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=sqrt{1+k^{2}} cdot frac{2sqrt{2}sqrt{3-k^{2}}}{|1-k^{2}|}. The distance d from origin O to line l is d=frac{2}{sqrt{1+k^{2}}}. So, the area of triangle OEF is S_{triangle OEF}=frac{1}{2}d cdot |EF|=frac{1}{2} cdot frac{2}{sqrt{1+k^{2}}} cdot sqrt{1+k^{2}} cdot frac{2sqrt{2}sqrt{3-k^{2}}}{|1-k^{2}|}=boxed{frac{2sqrt{2}sqrt{3-k^{2}}}{|1-k^{2}|}}. If S_{triangle OEF}=2sqrt{2}, then frac{2sqrt{2}sqrt{3-k^{2}}}{|1-k^{2}|}=2sqrt{2} iff k^{4}-k^{2}-2=0. Solving for k, we get k=pm sqrt{2}. Both values satisfy the condition. Therefore, there are two lines l that meet the requirements, and their equations are boxed{y=sqrt{2}x+2} and boxed{y=-sqrt{2}x+2}.