answer:1. **Initial Setup**: - Total sprinters: (300) - Lanes (sprinters per race): (8) 2. **Race Process**: - Typically, each race eliminates (6) sprinters, advancing (2), except every third race which advances (2) sprinters and eliminates (6). 3. **Total Eliminations Needed**: - To find the champion, (299) sprinters must be eliminated. 4. **Number of Races Calculation**: - Let (n) denote the number of regular races, and (m) denote the number of double-advancement races. - We know (n + m) races total, and (6n + 6m = 299) from the eliminations. - Additionally, every third race is a double-advancement race, implying (m = frac{n+m}{3}). - Solving these equations, we substitute (m = frac{n + m}{3}) into the elimination equation: [ 6n + 6left(frac{n + m}{3}right) = 299 implies 8n + 2m = 299 ] - Solving (3m = n + m) gives (2m = n), substituting (n = 2m) into (8n + 2m = 299) gives: [ 18m = 299 implies m approx 16.61 ] - Since (m) must be whole, we take (m = 16), then (n = 2 times 16 = 32). - Total races: (n + m = 32 + 16 = 48). 5. **Conclusion**: Thus, the total number of races required to determine the ultimate champion sprinter is 48. The final answer is boxed{textbf{(C)} 48}

answer:Let the isosceles trapezoid have legs of length x and a shorter base of length y. Dropping perpendiculars from the endpoints of the shorter base to the longer base forms two right-angled triangles. The base angle is arcsin(0.6), giving the vertical sides of these triangles as 0.6x and the horizontal side as 0.8x. From the property that tangents from a point to a circle are equal, we have 2y + 2(0.8x) = 2x. Simplifying gives 2y + 1.6x = 2x, or 2y = 0.4x. Also, the sum of y and the two horizontal sides must equal the longer base, thus y + 1.6x = 20. From 2y = 0.4x, we have y = 0.2x. Substituting into the base equation gives 0.2x + 1.6x = 20, leading to 1.8x = 20 and therefore x = frac{20}{1.8} approx 11.11. Substituting back to find y, we get y = 0.2 times 11.11 approx 2.22. The height of the trapezoid, which is also the vertical side of the triangle, is 0.6 times 11.11 approx 6.67. The area of the trapezoid can be calculated as frac{1}{2}(y + 20)(height) = frac{1}{2}(2.22 + 20)(6.67) approx frac{1}{2}(22.22)(6.67) approx 74.14. Thus, the area is boxed{74.14}.

answer:Since the point (1, 0) lies on both the curves of functions g(x) and f(x): For g(x), we find the derivative: g'(x) = frac{d}{dx}(x^2 - 1) = 2x, Evaluating at x=1, we have: g'(1) = 2. Therefore, the equation of the tangent line at the point (1,0) is: y = 2(x - 1). For f(x), we find the derivative: f'(x) = frac{d}{dx}(t ln x) = frac{t}{x}. Evaluating at x=1, we get: f'(1) = t. Since the slopes must be equal for the two functions to have the same tangent line at that point, we can equate the derivatives from f(x) and g(x) at x=1: 2 = t. Thus, the value of t is boxed{2}, which corresponds to option C.

answer:The radius of each circle equals the z-coordinate of its center since the circles are tangent to both the x-axis and y-axis and the axes lie in the plane z=0. For the first circle centered at (3,3,3), the radius is 3. For the second circle centered at (15,12,3), the radius is also 3. The distance between the centers of the circles is given by [ sqrt{(15 - 3)^2 + (12 - 3)^2 + (3 - 3)^2} = sqrt{12^2 + 9^2 + 0^2} = sqrt{144 + 81} = sqrt{225} = 15. ] The distance between the closest points of the two circles is the distance between the centers minus the sum of the radii: [ 15 - 3 - 3 = boxed{9}. ]