answer:**Analysis** Given the problem, the diameter of the sphere is the space diagonal of the rectangular solid. By calculating the space diagonal using the given dimensions of the rectangular solid, we can find the radius of the sphere and then use the formula for the volume of a sphere to solve the problem. **Solution** From the problem, the diameter of the sphere circumscribing the rectangular solid is sqrt{1^2 + 2^2 + 3^2} = sqrt{14}, Therefore, the radius of the sphere is dfrac{sqrt{14}}{2}, Thus, the volume of the sphere is dfrac{4}{3} times pi times left( dfrac{sqrt{14}}{2} right)^3 = dfrac{7sqrt{14}}{3}pi. Hence, the correct answer is boxed{C}.

answer:Since f(x)=-4x^{2}+4ax-4a-a^{2}=-4(x- dfrac {a}{2})^{2}-4a, the axis of symmetry is x= dfrac {a}{2}, When a < 0, dfrac {a}{2} < 0, thus f(x) is a decreasing function in the interval [0,1], Its maximum value is f(0)=-a^{2}-4a=-5, Therefore, a=-5, or a=1 (which is not consistent with the problem, discard), Therefore, a=-5; When a=0, f(x)=-4x^{2}, which is not consistent with the problem, discard; When 0 < a < 2, 0 < dfrac {a}{2} < 1, the maximum value of f(x) in the interval [0,1] is f( dfrac {a}{2})=-4a=-5, Therefore, a= dfrac {5}{4}; When ageqslant 2, dfrac {a}{2}geqslant 1, f(x) is an increasing function in the interval [0,1], Its maximum value is f(1)=-4+4a-4a-a^{2}=-5, Therefore, a=±1 (which is not consistent with the problem, discard); In summary, the value of a is boxed{dfrac {5}{4}} or boxed{-5}.

answer:1. **Define variables**: Let c be the number of cows and h be the number of chickens. 2. **Equations for legs and heads**: - A genetically modified cow has 5 legs. - Each chicken still has 2 legs. - Total legs = 5c + 2h. - Total heads = c + h as each animal has one head. 3. **Problem statement translation into the equation**: The problem states that the number of legs is 20 more than twice the number of heads: [ 5c + 2h = 20 + 2(c + h) ] 4. **Simplify the equation**: Expand and simplify: [ 5c + 2h = 20 + 2c + 2h ] Simplify by subtracting 2h from both sides: [ 5c = 20 + 2c ] Isolate c: [ 3c = 20 ] Solve for c: [ c = frac{20}{3} approx 6.67 ] 5. **Find nearest whole number (assuming realistic scenarios of whole animals)**: Since c cannot be a fraction in real terms, round down to nearest whole number: [ c = 6 ] Conclusion: The number of genetically modified cows on the farm is 6. The final answer is boxed{C) 6}

answer:1. **Understanding the Problem:** We are given ( n ) distinct positive integers ( x_1, x_2, ldots, x_n ) and we need to find the maximal value of the function: [ f(x_1, x_2, ldots, x_n) = frac{text{GCD}(x_1, S_1) + text{GCD}(x_2, S_2) + cdots + text{GCD}(x_n, S_n)}{x_1 + x_2 + cdots + x_n} ] where ( S_i ) is the sum of all ( x_j ) except ( x_i ). 2. **Case ( n = 2 ):** Let ( x_1 ) and ( x_2 ) be two distinct positive integers. Then: [ S_1 = x_2 quad text{and} quad S_2 = x_1 ] Therefore: [ f(x_1, x_2) = frac{text{GCD}(x_1, x_2) + text{GCD}(x_2, x_1)}{x_1 + x_2} = frac{2 cdot text{GCD}(x_1, x_2)}{x_1 + x_2} ] Let ( d = text{GCD}(x_1, x_2) ). Then ( x_1 = da ) and ( x_2 = db ) where ( a ) and ( b ) are coprime integers. Thus: [ f(x_1, x_2) = frac{2d}{da + db} = frac{2}{a + b} ] Since ( x_1 ) and ( x_2 ) are distinct, ( a ) and ( b ) are distinct positive integers, and ( a + b geq 3 ). Therefore: [ f(x_1, x_2) leq frac{2}{3} ] The equality holds when ( a = 1 ) and ( b = 2 ) (or vice versa), i.e., ( x_1 = 1 ) and ( x_2 = 2 ). 3. **Case ( n geq 3 ):** For ( n geq 3 ), we need to show that ( f(x_1, x_2, ldots, x_n) leq 1 ) and find when equality holds. Note that: [ text{GCD}(x_i, S_i) leq x_i ] for every ( i ). Therefore: [ text{GCD}(x_1, S_1) + text{GCD}(x_2, S_2) + cdots + text{GCD}(x_n, S_n) leq x_1 + x_2 + cdots + x_n ] This implies: [ f(x_1, x_2, ldots, x_n) leq 1 ] To achieve equality, we need ( text{GCD}(x_i, S_i) = x_i ) for every ( i ), which means ( x_i mid S_i ). This implies ( x_i mid (x_1 + x_2 + cdots + x_n) ) for every ( i ). 4. **Constructing a Sequence:** If ( (x_1, x_2, ldots, x_{n-1}) ) satisfies the condition, then ( (x_1, x_2, ldots, x_{n-1}, x_1 + x_2 + cdots + x_{n-1}) ) also satisfies the condition. Starting with ( (1, 2, 3) ) for ( n = 3 ), we can use induction to show that the sequence ( (1, 2, 3, 6, 12, ldots) ) satisfies the condition for all ( n geq 3 ). Conclusion: The maximal value of ( f(x_1, x_2, ldots, x_n) ) is: - ( frac{2}{3} ) for ( n = 2 ) - ( 1 ) for ( n geq 3 ) The final answer is ( boxed{1} ) for ( n geq 3 ) and ( boxed{frac{2}{3}} ) for ( n = 2 ).