answer:To solve the problem of determining the minimal number of balls that must be drawn to ensure 12 balls of the same number, we follow these steps: 1. **Understanding the initial condition:** We have a set of balls numbered from 1 to 2016. 2. **Calculating the maximum number of balls that can be drawn without getting 12 balls of the same number:** For each number ( k ) from 1 to 2016, we can draw at most 11 balls without reaching the 12th ball, which would ensure 12 balls of that number. Let's calculate this as follows: - For ( k = 1 ), we can have 1 ball (since there is only one ball labeled 1). - For ( k = 2 ), we can have 2 balls. - For ( k = 3 ), we can have 3 balls. - ... - For ( k = 11 ), we can have 11 balls. - For ( k geq 12 ), we can only have 11 balls each to avoid having 12 balls of the same number. Summarize this in a formula: [ text{Sum} = 1 + 2 + 3 + cdots + 11 + 11 times 2005 ] 3. **Summing up the contributions:** - Sum of the first 11 terms: [ sum_{k=1}^{11} k = frac{11 times (11 + 1)}{2} = 66 ] - For each remaining number from 12 to 2016 (which is 2005 numbers), we draw 11 balls: [ 11 times 2005 = 22055 ] - Total sum: [ 66 + 22055 = 22121 ] 4. **Final step: Ensuring 12 balls of the same number:** If we draw 22121 balls, it is still possible not to have 12 balls of any particular number. However, drawing one more ball would force us to have 12 balls of at least one number. Therefore, the minimum number of balls that must be drawn to ensure 12 balls of the same number is: ( boxed{22122} )

answer:Given the equation y = -20t^2 + 30t + 60, we need to find the time t when y = 0. This involves solving the quadratic equation: [ -20t^2 + 30t + 60 = 0 ] To make calculations simpler, divide the entire equation by -10: [ 2t^2 - 3t - 6 = 0 ] Using the quadratic formula (t = frac{-b pm sqrt{b^2 - 4ac}}{2a}), where (a = 2), (b = -3), and (c = -6), we get: [ t = frac{-(-3) pm sqrt{(-3)^2 - 4 cdot 2 cdot (-6)}}{2 cdot 2} ] [ t = frac{3 pm sqrt{9 + 48}}{4} ] [ t = frac{3 pm sqrt{57}}{4} ] Since time cannot be negative, we choose the positive root: [ t = frac{3 + sqrt{57}}{4} ] So, the time when the ball hits the ground is ( t = boxed{frac{3 + sqrt{57}}{4}} ) seconds.

answer:Given the vector mathbf{u} = (a, b, c), the magnitude |mathbf{u}| is defined by: [ |mathbf{u}| = sqrt{a^2 + b^2 + c^2} ] Since it is given that |mathbf{u}| = 6, we use the definition of magnitude: [ sqrt{a^2 + b^2 + c^2} = 6 ] Squaring both sides, we get: [ a^2 + b^2 + c^2 = 36 ] The dot product of mathbf{u} with itself is: [ mathbf{u} cdot mathbf{u} = a^2 + b^2 + c^2 ] Thus, substituting from the magnitude squared, we find: [ mathbf{u} cdot mathbf{u} = 36 ] Therefore, mathbf{u} cdot mathbf{u} = boxed{36}.

answer:Given the equation 3cos 2alpha -8cos alpha =5, we start by applying the double angle formula for cosine, which gives us: [ 3(2cos^2alpha - 1) - 8cosalpha = 5 ] Expanding and rearranging the terms, we get: [ 6cos^2alpha - 3 - 8cosalpha - 5 = 0 ] Simplifying, this becomes: [ 6cos^2alpha - 8cosalpha - 8 = 0 ] Dividing the entire equation by 2 for simplicity, we obtain: [ 3cos^2alpha - 4cosalpha - 4 = 0 ] This is a quadratic equation in terms of cosalpha. Solving this quadratic equation, we find that cosalpha could be 2 or -frac{2}{3}. However, since cosalpha = 2 is not possible within the domain of alpha in (0, pi), we discard it and consider cosalpha = -frac{2}{3}. Given alpha in (0, pi), and specifically alpha in (frac{pi}{2}, pi) due to the negative cosine value, we find sinalpha using the Pythagorean identity: [ sinalpha = sqrt{1 - cos^2alpha} = sqrt{1 - left(-frac{2}{3}right)^2} = sqrt{1 - frac{4}{9}} = sqrt{frac{5}{9}} = frac{sqrt{5}}{3} ] Therefore, the correct answer is: [ boxed{text{A. } frac{sqrt{5}}{3}} ]