answer:To solve the problem involving vector addition and subtraction, let's break down the given expression step by step, adhering closely to the provided solution: Given: overrightarrow{BA} + overrightarrow{CB} - overrightarrow{CD} + 2overrightarrow{AD} We start by recognizing that overrightarrow{CB} - overrightarrow{CD} is equivalent to moving from point C to point B and then from point D to point C, effectively moving from D to B, which can be represented by overrightarrow{DB}. So, the expression simplifies as follows: overrightarrow{BA} + overrightarrow{DB} + 2overrightarrow{AD} Next, we realize that overrightarrow{DB} is the opposite of overrightarrow{BD}, and thus can be written as -overrightarrow{BD}. Incorporating this, we have: overrightarrow{BA} - overrightarrow{BD} + 2overrightarrow{AD} Now, we notice overrightarrow{BA} - overrightarrow{BD} can be thought of moving from B to A and then from D to B, which effectively is moving from D to A, represented by overrightarrow{DA}. Hence, our expression further simplifies to: overrightarrow{DA} + 2overrightarrow{AD} Since overrightarrow{DA} is simply the reverse of overrightarrow{AD}, adding 2overrightarrow{AD} to it is equivalent to just adding overrightarrow{AD} to itself (because overrightarrow{DA} = -overrightarrow{AD}), leading to: overrightarrow{AD} Therefore, the final simplified form of the given vector expression is: boxed{overrightarrow{AD}}

answer:Tom reads for 10 hours over 5 days, which means he reads for 10 / 5 = 2 hours per day. He reads 50 pages per hour, so in 2 hours, he reads 50 * 2 = 100 pages per day. If he reads 100 pages per day for 7 days, he will read 100 * 7 = boxed{700} pages in 7 days.

answer:First, expand and simplify both sides of the equation: [ 27x^2 + 45x + 60 = 9x^2 - 60x ] Subtract 9x^2 and add 60x to both sides to bring all terms to one side: [ 27x^2 + 45x + 60 - 9x^2 + 60x = 0 ] [ 18x^2 + 105x + 60 = 0 ] Next, factorize the quadratic equation. Divide the entire equation by 3 to simplify: [ 6x^2 + 35x + 20 = 0 ] Using the quadratic formula, x = frac{-b pm sqrt{b^2 - 4ac}}{2a}, where a = 6, b = 35, and c = 20: [ x = frac{-35 pm sqrt{1225 - 480}}{12} ] [ x = frac{-35 pm sqrt{745}}{12} ] Since 745 is not a perfect square, we approximate sqrt{745} approx 27.29. Therefore, we have: [ x = frac{-35 + 27.29}{12} quad text{and} quad x = frac{-35 - 27.29}{12} ] [ x approx frac{-7.71}{12} approx -0.6425 quad text{and} quad x approx frac{-62.29}{12} approx -5.1908 ] Between x approx -0.6425 and x approx -5.1908, the largest value is x = boxed{-frac{7.71}{12}} when rounded.

answer:Based on the problem, we know that xi can take the values of 0, 1, 2, or 3. - For xi = 0, it means that the marksman missed the first three shots and takes the fourth shot regardless of whether it hits, thus the outcome contributes nothing to the remaining bullets. Therefore, the probability P(xi = 0) is 0.4^3, as the probability of missing each shot is 0.4 (the complementary probability of hitting with a 0.6 probability). - For xi = 1, it means that the marksman missed the first two shots but hit the target on the third shot. Therefore, the probability P(xi = 1) is 0.6 times 0.4^2. - For xi = 2, it means that the marksman missed the first shot and hit the target on the second shot. Therefore, the probability P(xi = 2) is 0.6 times 0.4. - For xi = 3, it means that the marksman hit the target on the first shot. Therefore, the probability P(xi = 3) is 0.6. The expectation Exi is calculated as the sum of the products of each possible value of xi and its corresponding probability: Exi = 0 cdot P(xi = 0) + 1 cdot P(xi = 1) + 2 cdot P(xi = 2) + 3 cdot P(xi = 3) Now substitute the probabilities: Exi = 0 cdot 0.4^3 + 1 cdot (0.6 times 0.4^2) + 2 cdot (0.6 times 0.4) + 3 cdot 0.6 Compute each term: Exi = 0 cdot 0.064 + 1 cdot (0.6 cdot 0.16) + 2 cdot (0.6 cdot 0.4) + 3 cdot 0.6 Exi = 0 + 0.096 + 0.48 + 1.8 Exi = 2.376 Therefore, the expected number of remaining bullets xi after the first hit is boxed{2.376}.