answer:We need to place 7 indistinguishable balls into 6 distinguishable boxes (representing the maximum number of digits in base 5 that can reach 5^6), where no box (digit place) contains more than 4 balls (since the highest digit in base 5 is 4). The number of ways to place 7 objects into 6 boxes can be calculated using the stars and bars theorem: {7 + 6 - 1 choose 6 - 1} = {12 choose 5} = 792. We must remove configurations where a "box" has more than 4 balls. We calculate: - Configurations with at least 5 balls in one box: - Choose 1 box for 5 balls: {6 choose 1} ways. - Distribute the remaining 2 balls into any of the 6 boxes: {2 + 6 - 1 choose 6 - 1} = {7 choose 5} = 21 ways. - Total invalid configurations: 6 * 21 = 126 ways. Thus, the number of valid configurations is m = 792 - 126 = 666. To find star(m), calculate the sum of the digits of 666: star(666) = 6 + 6 + 6 = boxed{18}.

answer:- Total two-person sub-committees from 8 members: There are binom{8}{2} ways to choose any two members from 8. [ binom{8}{2} = frac{8 cdot 7}{2} = 28 ] - No male (both are female) committees: There are binom{3}{2} ways to choose the two members from the 3 women only. [ binom{3}{2} = frac{3 cdot 2}{2} = 3 ] - Sub-committees with at least one male: Subtract the all-female committees from the total committees. [ 28 - 3 = 25 ] Thus, there are boxed{25} possible two-person sub-committees that include at least one male.

answer:The line of sight is highest when it is tangent to the parabola, and it is in the upper right direction. Let the equation of the tangent line be y=kx-2 (k>0). By combining this with the equation of the parabola, we get 2x^2-kx+2=0. Delta=k^2-16=0 k=4 (the negative value is discarded). Therefore, the tangent line is y=4x-2. Substituting x=3 into this equation yields y=10. Point B satisfies the condition as long as it is below this tangent line. Therefore, a<10. Hence, the correct choice is boxed{D}.

answer:Given: Perimeter of the circumscribed trapezoid is ( 12 ). 1. **Understand the sum of the bases and the legs:** According to the properties of a circumscribed trapezoid, the sum of the lengths of the bases (let's denote them as ( a ) and ( b )) is equal to the sum of the lengths of the legs (let's denote them as ( c ) and ( d )). 2. **Compute the sum of the bases and the legs:** From the problem statement: [ text{Perimeter} = text{sum of all sides} = a + b + c + d ] Given that the perimeter is ( 12 ), we have: [ a + b + c + d = 12 ] Since the sum of the bases is equal to the sum of the legs: [ a + b = c + d ] Let ( S ) denote the sum of the bases (or the legs), we have: [ S + S = 12 implies 2S = 12 implies S = 6 ] 3. **Determine the sum of the bases:** Given ( S ) as the sum of the bases: [ a + b = 6 ] 4. **Calculate the midline of the trapezoid:** The midline (or median) of a trapezoid is computed as the average of the lengths of the bases. Therefore: [ text{Midline} = frac{a + b}{2} = frac{6}{2} = 3 ] 5. **Conclusion:** The length of the midline of the trapezoid is: [ boxed{3} ]