answer:```markdown **(a) Proof that the subset cannot have 7 elements:** 1. First, let's assume there is a set with 7 elements. Given the conditions, the median of this set is 10. 2. Therefore, 4 of these elements must be greater than or equal to the median, which is 10. 3. The remaining 3 elements must each be 1 or more. 4. Hence, the sum of all the elements in the set must be at least: [ 3 cdot 1 + 4 cdot 10 = 3 + 40 = 43 ] 5. However, we also know that the arithmetic mean of the set is 6. Therefore, the total sum of the elements should be: [ 6 cdot 7 = 42 ] 6. Therefore, we have a contradiction since (43 neq 42). This means a set with 7 elements satisfying the given conditions cannot exist. **(b) Proof that the subset must have at least 2n + 1 elements where n ge 4.** 7. Let the total number of elements in the set be (2n + 1) where (n) is a natural number. 8. Since the median is 10, there must be (n+1) elements that are at least 10. 9. The remaining (n) elements must be at least 1. 10. Thus, the sum of all elements in the set would be at least: [ n cdot 1 + 10 cdot (n+1) = n + 10n + 10 = 11n + 10 ] 11. On the other hand, if the arithmetic mean of the set is 6, then the sum of the elements should be: [ 6 cdot (2n + 1) = 12n + 6 ] 12. Therefore, we get the inequality: [ 12n + 6 geq 11n + 10 ] 13. Simplifying this inequality: [ 12n + 6 geq 11n + 10 implies n geq 4 ] 14. Consequently, the set must have at least (2 cdot 4 + 1 = 9) elements. 15. As an example, consider the set: [ {1, 1, 1, 1, 10, 10, 10, 10, 10} ] This set has 9 elements, four of which are 1 and five of which are 10, respecting the given conditions. # Conclusion [ boxed{a) text{no; } b) text{9}} ] ```

answer:To find two numbers whose sum is equal to 60 and whose greatest common divisor (GCD) and least common multiple (LCM) sum up to 84, let's use the properties of GCD and LCM. 1. Let the two numbers be ( a ) and ( b ). 2. We are given the following conditions: - ( a + b = 60 ) - ( text{GCD}(a, b) + text{LCM}(a, b) = 84 ) 3. We also know the relationship between GCD and LCM for two numbers: [ a cdot b = text{GCD}(a, b) cdot text{LCM}(a, b) ] 4. Let ( d = text{GCD}(a, b) ) and ( a = d cdot m ) and ( b = d cdot n ), where ( m ) and ( n ) are coprime (i.e., ( text{GCD}(m, n) = 1 )). 5. Therefore, given ( a + b = 60 ), we have: [ d cdot m + d cdot n = 60 ] [ d(m + n) = 60 ] 6. Also, we know from the properties of GCD and LCM: [ text{GCD}(a, b) + text{LCM}(a, b) = 84 ] Since ([a, b] = frac{a cdot b}{text{GCD}(a, b)}): [ d + frac{d^2 cdot m cdot n}{d} = 84 ] Simplifies to: [ d + d cdot m cdot n = 84 ] [ d(1 + m cdot n) = 84 ] 7. Now, solve for ( d ): [ d(1 + m cdot n) = 84 quad text{(equation 1)} ] [ d(m + n) = 60 quad text{(equation 2)} ] 8. From equation 2, solve for ( d ): [ d = frac{60}{m + n} ] 9. Substitute ( d ) in equation 1: [ frac{60}{m + n}(1 + m cdot n) = 84 ] [ 60 (1 + m cdot n) = 84(m + n) ] [ 60 + 60 m n = 84 m + 84 n ] [ 60 m n - 84 m - 84 n = -60 ] [ 60 m n - 84 m - 84 n + 60 = 0 ] [ 60 m n - 84 m - 84 n + 60 = 0 ] 10. Solve the equation by finding possible ( m ) and ( n ) that are coprime: - Start testing smaller values: if ( m = 3 ) and ( n = 2 ): [ d = frac{60}{3 + 2} = frac{60}{5} = 12 ] Hence, ( a = 12 cdot 3 = 36 ) and ( b = 12 cdot 2 = 24 ). 11. Verify: - Indeed, ( a + b = 36 + 24 = 60 ) - ( text{GCD}(36, 24) = 12 ) - ( text{LCM}(36, 24) = frac{36 cdot 24}{12} = 72 ) - ( 12 + 72 = 84 ) Therefore, the two numbers are: [ boxed{24 text{ and } 36} ]

answer:The domain of the function is (0,+infty). The derivative of y= frac {1}{4}x^{2}- frac {1}{2}ln x is y′= frac {1}{2}x- frac {1}{2x}. Setting frac {1}{2}x- frac {1}{2x}= frac {3}{4}, we get x=2. Therefore, the tangent point is (2,1- frac {1}{2}ln 2). The distance d from it to the line y= frac {3}{4}x-1, which is 3x-4y-4=0, is d= frac {|6-4+2ln 2-4|}{ sqrt {9+16}}= frac {2-2ln 2}{5}. Thus, the minimum distance from point P to the line y= frac {3}{4}x-1 is frac {2-2ln 2}{5}. Therefore, the answer is: boxed{frac {2-2ln 2}{5}}. First, we find the coordinates of the tangent point based on the geometric meaning of the derivative. To find the minimum distance from P to the line y= frac {3}{4}x-1, we calculate the distance from the tangent point to the line. Finally, we use the formula for the distance from a point to a line to solve it. This problem mainly examines the use of derivatives to study the equation of the tangent line at a point on the curve, as well as the formula for the distance from a point to a line, testing computational skills, and is considered a medium-level question.

answer:Let's call the speed at which Ted runs T mph and the speed at which Frank runs F mph. According to the problem, Ted runs two-thirds as fast as Frank, so we can write the following equation: T = (2/3) * F We also know that in two hours, Frank runs eight miles farther than Ted. This can be expressed as: 2 * F = 2 * T + 8 Now we can substitute the first equation into the second equation to solve for T: 2 * F = 2 * ((2/3) * F) + 8 Simplify the equation: 2 * F = (4/3) * F + 8 To solve for F, we need to get all the terms with F on one side of the equation: 2 * F - (4/3) * F = 8 To combine the terms, we need a common denominator, which is 3: (6/3) * F - (4/3) * F = 8 (2/3) * F = 8 Now we can solve for F: F = 8 / (2/3) F = 8 * (3/2) F = 12 Now that we have Frank's speed, we can find Ted's speed using the first equation: T = (2/3) * F T = (2/3) * 12 T = 8 So, Ted runs at a speed of boxed{8} mph.