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question:A power boat and a raft both left dock ( A ) on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed relative to the river. After reaching dock ( B ) downriver, the power boat immediately turned around and traveled back upriver. This time, the power boat met the raft on the river 12 hours after leaving dock ( A ). Determine how many hours it took the power boat to travel from ( A ) to ( B ). A) ( frac{10r}{6p r} ) B) ( frac{12r}{6p r} ) C) ( frac{14r}{6p r} ) D) ( frac{16r}{6p r} )

answer:Let: - ( t ) be the time in hours for the power boat to travel from ( A ) to ( B ). - ( r ) be the speed of the river current (and the raft's speed). - ( p ) be the relative speed of the power boat with respect to the river. The distance from ( A ) to ( B ) is ( (p + r)t ), given the power boat's downstream speed. The raft travels a distance of ( 12r ) in 12 hours. The power boat travels a distance of ( (p - r)(12 - t) ) upstream from ( B ) to meet the raft. Setting up the equation based on distances: [ (p + r)t + (p - r)(12 - t) = 12r ] Expanding and solving for ( t ): [ pt + rt + 12p - pt - 12r + rt = 12r ] [ 12p - 12r + 2rt = 12r ] [ 12p = 24r - 2rt ] [ 2rt - 12p = -24r ] [ t(2r - 12p) = -24r ] [ t = frac{-24r}{2r - 12p} ] For the equation to be valid and feasible, we need to assume the values such that ( 2r neq 12p ) to avoid division by zero. However, assuming typical speeds, we simplify further: [ t = frac{24r}{12p - 2r} ] [ t = frac{24r}{2(6p - r)} ] [ t = frac{12r}{6p - r} ] Conclusion: [ t = frac{12r{6p - r}} ] The final answer is boxed{textbf{B}}

question:Suppose x is a purely imaginary number, y is a real number, and 2x 1 + i = y (3 y)i, then x + y is equal to ( ) A: 1 + frac{5}{2}i B: 1 + frac{5}{2}i C: 1 frac{5}{2}i D: 1 frac{5}{2}i

answer:Given that x is a purely imaginary number, let's denote it as x = ai where a neq 0. We are also given that y is a real number and the equation 2x - 1 + i = y - (3 - y)i holds true. We can use the necessary and sufficient condition for complex numbers equality to find the answer. First, substitute x = ai into the equation: [ 2ai - 1 + i = y - (3 - y)i ] Next, rewrite the equation by grouping the real and imaginary parts: [ -1 + (2a + 1)i = y + (y - 3)i ] Now, equate the real and imaginary parts: [ begin{cases} y = -1 y - 3 = 2a + 1 end{cases} ] Solving this system of equations, we get: [ begin{cases} a = -frac{5}{2} y = -1 end{cases} ] Finally, find x + y: [ x + y = -frac{5}{2}i - 1 ] Therefore, the correct answer is boxed{-1 - frac{5}{2}i}, which corresponds to option D.

question:des of a triangle are 13 cm . 14 cm . and 15 cm . respectively . calculate the area of triangle .

answer:To calculate the area of a triangle with sides of lengths 13 cm, 14 cm, and 15 cm, we can use Heron's formula. Heron's formula states that the area of a triangle with sides of lengths a, b, and c is: Area = √(s * (s - a) * (s - b) * (s - c)) where s is the semi-perimeter of the triangle, calculated as: s = (a + b + c) / 2 First, let's calculate the semi-perimeter: s = (13 + 14 + 15) / 2 s = 42 / 2 s = 21 cm Now, we can use Heron's formula to find the area: Area = √(21 * (21 - 13) * (21 - 14) * (21 - 15)) Area = √(21 * 8 * 7 * 6) Area = √(7056) Area = 84 cm² Therefore, the area of the triangle is boxed{84} square centimeters.

question:f C_{n}^{1}+3C_{n}^{2}+3^{2}C_{n}^{3}+cdots +3^{n2}C_{n}^{n1}+3^{n1}=85, then the value of n is ( ). A: 3 B: 4 C: 5 D: 6

answer:**Analysis** This problem tests the application of the binomial theorem, and it is a short-answer question. **Solution** Given: C_{n}^{1}+3C_{n}^{2}+3^{2}C_{n}^{3}+cdots+3^{n-2}C_{n}^{n-1}+3^{n-1}= frac{1}{3}[(3+1)^{n}-1]=85, Solving this equation, we find n=4. Therefore, the correct choice is boxed{B}.

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