answer:Starting from the equation: [4 overrightarrow{OP} - 3 overrightarrow{OQ} + 6 overrightarrow{OR} + k overrightarrow{OS} = mathbf{0}.] We can rewrite this as: [4 overrightarrow{OP} - 3 overrightarrow{OQ} = -6 overrightarrow{OR} - k overrightarrow{OS}.] Let T be the point such that [overrightarrow{OT} = 4 overrightarrow{OP} - 3 overrightarrow{OQ} = -6 overrightarrow{OR} - k overrightarrow{OS}.] Since 4 - 3 = 1, T lies on line PQ. If -6 - k = 1, then T would also lie on line RS, which forces P, Q, R, and S to be coplanar. Solving -6 - k = 1 gives us k = -7. Therefore, the scalar k that makes points P, Q, R, and S coplanar is [boxed{-7}.]

answer:Since the universal set U={xinmathbb{N}^{*}|xleq 4}={1,2,3,4}, and A={1,4}, B={2,4}, Therefore, Acap B={4}, Thus, the complement of Acap B in U is {1,2,3}. Hence, the correct option is: boxed{text{A}}. **Analysis:** Given the universal set U={xinmathbb{N}^{*}|xleq 4}, and the sets A={1,4}, B={2,4}. By applying the properties and operations of the complement set, we find Acap B and then its complement to obtain the answer.

answer:Using the angle addition formulas and trigonometric identities, we start with: [ tan x + tan y = frac{sin x + sin y}{cos x + cos y} = frac{frac{85}{65}}{frac{60}{65}} = frac{85}{60} = frac{17}{12}. ] We also know: [ sin(x + y) = 2 sinleft(frac{x + y}{2}right) cosleft(frac{x - y}{2}right), quad cos(x + y) = 2 cosleft(frac{x + y}{2}right) cosleft(frac{x - y}{2}right). ] By dividing these, we obtain: [ tanleft(frac{x + y}{2}right) = frac{sinleft(frac{x + y}{2}right)}{cosleft(frac{x + y}{2}right)} = frac{frac{85}{65}}{frac{60}{65}} = frac{17}{12}. ] Now, using the identity tan(x+y) = frac{2tanleft(frac{x+y}{2}right)}{1-tan^2left(frac{x+y}{2}right)}, we find: [ tan(x+y) = frac{2 cdot frac{17}{12}}{1-left(frac{17}{12}right)^2} = frac{frac{34}{12}}{1 - frac{289}{144}} = frac{frac{34}{12}}{frac{-145}{144}} = -frac{34 times 144}{12 times 145} = -frac{34 times 12}{145}. ] Upon simplification, [ tan(x + y) = -frac{34 times 12}{145} = -frac{408}{145}. ] Thus, tan x + tan y = boxed{-frac{408}{145}}.

answer:The key for solving this problem is understanding the independence of each coin flip. Each flip of a fair coin has no memory of previous outcomes, and each flip is an independent event. 1. The probability of flipping tails on any single flip of a fair coin is frac{1}{2}. 2. Since the coin flips are independent, the probability of two specific outcomes happening in sequence (tails followed by another tails) is the product of their individual probabilities. Probability of tails on the ninth flip: frac{1}{2} Probability of tails on the tenth flip: frac{1}{2} Combined probability of both events: frac{1}{2} times frac{1}{2} = frac{1}{4} Therefore, the probability that Molly's next two flips will both be tails is boxed{frac{1}{4}}.