answer:To find the number of trailing zeros in the product 1!2!3!4!cdots49!50!, we need to count the number of times 5 appears as a factor in these factorials. - Numbers divisible by 5 up to 50: 5, 10, 15, ..., 50. These contribute one factor of 5 each. The count is lfloor 50/5 rfloor = 10. - Numbers divisible by 25 up to 50: 25, 50. These contribute an additional factor of 5 each (since 25 = 5^2 and 50 = 2 times 25). The count is lfloor 50/25 rfloor = 2. Adding these, the total count of factors of 5 is 10 + 2 = 12. Thus, the number of trailing zeros in the product of factorials from 1! to 50! is 12. The remainder of 12 when divided by 100 is boxed{12}.

answer:Let's denote the quiz score of the student who dropped the class as ( x ). The total score of all 16 students before one dropped the class is ( 16 times 62.5 ). When the student dropped the class, there were 15 students left, and the new average became 63.0. So the total score of the remaining 15 students is ( 15 times 63 ). The difference between the total score of the 16 students and the total score of the 15 students is the score of the student who dropped the class. So we have: ( 16 times 62.5 - 15 times 63 = x ) ( 1000 - 945 = x ) ( x = 55 ) Therefore, the quiz score of the student who dropped the class is boxed{55} .

answer:1. **Label the Hexagon and Identify Symmetry:** Let the hexagon be labeled as ( ABCDEF ) with ( AB = AF = 1 ), ( BC = EF = 2 ), and ( CD = DE = 3 ). The hexagon is symmetric about the line ( AD ), and ( AD ) is a diameter of the circle. 2. **Apply Ptolemy's Theorem:** Ptolemy's theorem states that for a cyclic quadrilateral ( ABCD ), the sum of the products of its two pairs of opposite sides is equal to the product of its diagonals: [ AC cdot BD = AB cdot CD + AD cdot BC ] 3. **Identify the Diagonals and Sides:** In our hexagon, we need to consider the diagonals ( AC ) and ( BD ). Since ( AD ) is a diameter, ( AD = 2r ). 4. **Calculate the Lengths of the Diagonals:** Using the properties of the circle and the symmetry of the hexagon, we can find the lengths of the diagonals ( AC ) and ( BD ) using the cosine rule in the triangles formed by the sides and the radii of the circle. For ( AC ): [ AC = sqrt{4r^2 - 9} ] For ( BD ): [ BD = sqrt{4r^2 - 1} ] 5. **Substitute into Ptolemy's Theorem:** Substitute the lengths of the diagonals and the sides into Ptolemy's theorem: [ sqrt{4r^2 - 9} cdot sqrt{4r^2 - 1} = 1 cdot 3 + 2r cdot 2 ] 6. **Simplify the Equation:** Simplify the left-hand side using the product of square roots: [ sqrt{(4r^2 - 9)(4r^2 - 1)} = 3 + 4r ] [ sqrt{16r^4 - 4r^2 - 36r^2 + 9} = 3 + 4r ] [ sqrt{16r^4 - 40r^2 + 9} = 3 + 4r ] 7. **Square Both Sides:** To eliminate the square root, square both sides of the equation: [ 16r^4 - 40r^2 + 9 = (3 + 4r)^2 ] [ 16r^4 - 40r^2 + 9 = 9 + 24r + 16r^2 ] 8. **Combine Like Terms:** Move all terms to one side of the equation to set it to zero: [ 16r^4 - 40r^2 + 9 - 9 - 24r - 16r^2 = 0 ] [ 16r^4 - 56r^2 - 24r = 0 ] 9. **Factor the Equation:** Factor out the common term: [ 8r(2r^3 - 7r - 3) = 0 ] 10. **Solve for ( r ):** Since ( r neq 0 ) (as it is the radius of the circle), we have: [ 2r^3 - 7r - 3 = 0 ] (blacksquare)

answer:Elizabeth starts with 10 reusable water bottles. After losing 2 at school, she has 10 - 2 = 8 water bottles left. Then, after one is stolen at dance practice, she has 8 - 1 = 7 water bottles remaining. To secure her remaining water bottles, Elizabeth places 3 stickers on each of them. Therefore, the total number of stickers she uses is calculated as follows: [7 text{ water bottles} times 3 text{ stickers per bottle} = 21 text{ stickers}] Thus, Elizabeth uses a total of boxed{21} stickers on her water bottles.