answer:To determine the range of values for k that allows the quadratic equation x^{2}-2x+k=0 to have two real roots, we need to ensure that the discriminant Delta of the equation is non-negative. The discriminant Delta for a quadratic equation ax^{2}+bx+c=0 is given by Delta=b^{2}-4ac. For our equation, a=1, b=-2, and c=k, so we have: [ Delta = (-2)^{2}-4(1)(k) = 4-4k ] For the equation to have two real roots, the discriminant must be greater than or equal to zero: [ 4-4k geqslant 0 ] Solving this inequality for k: [ 4 geqslant 4k implies 1 geqslant k implies k leqslant 1 ] Therefore, the range of values for k that allows the equation to have two real roots is encapsulated as: [ boxed{k leqslant 1} ]

answer:To determine the range of values for x for which the algebraic expression frac{sqrt{3-x}}{x+1} is meaningful, we need to ensure two conditions are met: 1. The expression under the square root, 3-x, must be greater than or equal to zero to avoid taking the square root of a negative number. This gives us the inequality: [3-x geqslant 0] Solving for x, we subtract 3 from both sides and multiply by -1 (remembering to flip the inequality sign): [x leqslant 3] 2. The denominator, x+1, must not be equal to zero to avoid division by zero. This gives us the equation: [x+1 neq 0] Solving for x, we subtract 1 from both sides: [x neq -1] Combining these two conditions, we find that x must be less than or equal to 3 but cannot be equal to -1. Therefore, the range of values for x is: [x leqslant 3 quad text{and} quad x neq -1] Thus, the final answer, encapsulating it as required, is: [boxed{x leqslant 3 quad text{and} quad x neq -1}]

answer:Solution: (Ⅰ) The function relationship is f(x)=(x-30)^{2}(x-10-a), where 20 leqslant x leqslant 25, (Ⅱ) The derivative of f(x) is f'(x)=2(x-30)⋅(x-10-a)+(x-30)^{2}=(3x-2a-50)(x-30). Setting f'(x)=0, we get x= dfrac{2a+50}{3} or x=30, Since 3 leqslant a leqslant 7, we have dfrac{56}{3} leqslant dfrac{2a+50}{3} leqslant dfrac{64}{3}, Therefore, ① If dfrac{2a+50}{3} leqslant 20, which is when 3 leqslant a leqslant 5, f'(x) leqslant 0 for x in [20,25], Thus, f(x) is a decreasing function on [20,25]. Hence, the maximum value of f(x), f(x)_{max}=f(20)=(30-20)^{2}(20-10-a)=100(10-a)=1000-10a, ② If 5 < a leqslant 7, then dfrac{2a+50}{3} in [20,25] When x in [20, dfrac{2a+50}{3}], f'(x) > 0, and f(x) is an increasing function on [20, dfrac{2a+50}{3}]; When x in [dfrac{2a+50}{3},25], f'(x) < 0, and f(x) is a decreasing function on [dfrac{2a+50}{3},25]. Therefore, the maximum value of f(x), f(x)_{max}=f(dfrac{2a+50}{3})=(30- dfrac{2a+50}{3})^{2}(dfrac{2a+50}{3}-10-a)=(dfrac{2a-40}{3})^{2}(dfrac{20-a}{3})=-dfrac{4(a-20)^{3}}{27}, Thus, when 3 leqslant a leqslant 5, the maximum profit g(a) is achieved at a selling price of 20 dollars, with a maximum profit of 1000-10a; When 5 < a leqslant 7, the maximum profit g(a) is achieved at a selling price of dfrac{2a+50}{3} dollars, with a maximum profit of -dfrac{4(a-20)^{3}}{27}. Therefore, the final answers are: - When 3 leqslant a leqslant 5, the selling price for maximum profit is boxed{20} dollars, and the maximum profit is boxed{1000-10a}. - When 5 < a leqslant 7, the selling price for maximum profit is boxed{dfrac{2a+50}{3}} dollars, and the maximum profit is boxed{-dfrac{4(a-20)^{3}}{27}}.

answer:To calculate the maximum number of each type of plane that can fit into the hangar, we need to consider the length of each plane and the required safety gap of 10 feet between each plane. For small planes (50 feet long): Each small plane requires 50 feet of space, plus a 10-foot safety gap. So, each small plane effectively occupies 60 feet of space (50 feet + 10 feet). The hangar is 900 feet long, so the maximum number of small planes that can fit is: 900 feet / 60 feet per small plane = 15 small planes For medium planes (75 feet long): Each medium plane requires 75 feet of space, plus a 10-foot safety gap. So, each medium plane effectively occupies 85 feet of space (75 feet + 10 feet). The hangar is 900 feet long, so the maximum number of medium planes that can fit is: 900 feet / 85 feet per medium plane = 10.5882 medium planes Since we can't have a fraction of a plane, we round down to the nearest whole number: 10 medium planes For large planes (110 feet long): Each large plane requires 110 feet of space, plus a 10-foot safety gap. So, each large plane effectively occupies 120 feet of space (110 feet + 10 feet). The hangar is 900 feet long, so the maximum number of large planes that can fit is: 900 feet / 120 feet per large plane = 7.5 large planes Again, we can't have a fraction of a plane, so we round down to the nearest whole number: 7 large planes In summary: - The maximum number of small planes that can fit is 15. - The maximum number of medium planes that can fit is 10. - The maximum number of large planes that can fit is boxed{7} .