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question:Given sin ^{2}(frac{π}{4}+alpha )=frac{2}{3}, find the value of sin 2alpha.

answer:Given sin ^{2}left(frac{pi}{4}+alpha right)=frac{2}{3}, we aim to find the value of sin 2alpha. Step 1: Use the identity sin^{2}x = frac{1 - cos 2x}{2} to express sin^{2}left(frac{pi}{4}+alpharight). sin^{2}left(frac{pi}{4}+alpharight) = frac{1 - cosleft(frac{pi}{2} + 2alpharight)}{2} Step 2: Recognize that cosleft(frac{pi}{2} + 2alpharight) = -sin 2alpha and substitute it into the equation. frac{1 - (-sin 2alpha)}{2} = frac{2}{3} Step 3: Simplify the equation to solve for sin 2alpha. frac{1 + sin 2alpha}{2} = frac{2}{3} Step 4: Multiply both sides by 2 and subtract 1 to isolate sin 2alpha. 1 + sin 2alpha = frac{4}{3} sin 2alpha = frac{4}{3} - 1 sin 2alpha = frac{1}{3} Therefore, the value of sin 2alpha is boxed{frac{1}{3}}.

question:w many miles can a car travel in 30 minutes if it travels frac{2}{3} as fast as a train going 90 miles per hour?

answer:First, calculate the car's speed: frac{2}{3}times90text{ miles per hour} = 2times30=60text{ miles per hour}. Now, compute the distance the car travels in 30 minutes: frac{60 text{ miles}}{60text{ minutes}} times 30text{ minutes} = 1 times 30 = boxed{30text{ miles}}.

question:Given alpha and beta are distinct planes, and m and n are distinct lines, there are four propositions: 1. If m parallel n and m perp alpha, then n perp alpha. 2. If m parallel alpha and alpha cap beta = n, then m parallel n. 3. If m perp alpha, n perp beta, and alpha perp beta, then m perp n. 4. If n subset alpha, m subset beta, and alpha parallel beta, then m parallel n. The number of correct propositions is ( ) A: 1 B: 2 C: 3 D: 4

answer:If m parallel n and m perp alpha, according to the second theorem of line-perpendicular-to-plane, we can conclude m perp alpha, hence proposition 1 is correct. If m parallel alpha and alpha cap beta = n, then m and n could either be parallel or skew, hence proposition 2 is incorrect. If m perp alpha and alpha perp beta, then m is either parallel to beta or contained in beta. Given n perp beta, then m perp n, hence proposition 3 is incorrect. If n subset alpha, m subset beta, and alpha parallel beta, then m and n could either be parallel or skew, hence proposition 4 is incorrect. Therefore, the correct choice is boxed{text{A}}.

question:(1) The area of the figure formed by the curve y=sin x where x in left[frac{pi}{3}, 0right], the lines x=frac{pi}{3}, x=0, and the xaxis is _______. (2) In a geometric sequence {a_n}, if a_{11}=1, then a_{1} cdot a_{2} cdot a_{3} cdot ldots cdot a_{n} = a_{1} cdot a_{2} cdot a_{3} cdot ldots cdot a_{21n} holds for n < 21 and n in mathbb{N}^*. By analogy, in an arithmetic sequence {b_n}, if b_{10}=0, then there exists an equation b_{1}+b_{2}+ ldots +b_{n}= _______. (3) One day, Xiao Zhao, Xiao Zhang, Xiao Li, and Xiao Liu went to the cinema together. After arriving, they found that movies A, B, C, D, and E were being shown that day. They discussed and decided to watch one of them together. Xiao Zhao said: Anything but B is fine; Xiao Zhang said: B, C, D, E are all fine; Xiao Li said: I like D, but anything but C is fine; Xiao Liu said: Everything but E is fine. Based on this, the movie that all four of them can watch together is _______. (4) Given that the line y=b intersects the functions f(x)=2x+3 and g(x)=ax+ln x at points A and B respectively, if the minimum value of |AB| is 2, then a+b= _______.

answer:(1) **Analysis** This question tests the calculation of area using definite integrals, which is a basic question. Note that the area is calculated as int_{a}^{b}|f(x)|dx. **Solution** Since int_{- frac{pi}{3}}^{0}sin xdx=(-cos x)|_{- frac{pi}{3}}^{0}=-1-(- frac{1}{2})=- frac{1}{2}, Therefore, S= frac{1}{2}. Hence, the answer is boxed{frac{1}{2}}. (2) **Analysis** This question tests the analogy thinking, which is a basic question. Note that in a geometric sequence, the product becomes a sum in an arithmetic sequence, and for the geometric sequence, 21=11 times 2 - 1, so for the arithmetic sequence, it should be 10 times 2 - 1 = 19. Also, pay attention to the condition about n mentioned in the parentheses. **Solution** By analogy, The product in a geometric sequence becomes a sum in an arithmetic sequence, For the geometric sequence, 21=11 times 2 - 1, For the arithmetic sequence, it should be 10 times 2 - 1 = 19, Therefore, the answer is b_{1}+b_{2}+...+b_{19-n} for n < 19 and n in mathbb{N}^*. Hence, the answer is boxed{b_{1}+b_{2}+...+b_{19-n}}. (3) **Analysis** This question tests the application of the intersection of sets, which is a basic question. Write down the sets of movies that Xiao Zhao, Xiao Zhang, Xiao Li, and Xiao Liu can watch, then find the intersection. **Solution** The sets of movies that Xiao Zhao, Xiao Zhang, Xiao Li, and Xiao Liu can watch are {A,C,D,E}, {B,C,D,E}, {A,B,D,E}, {A,B,C,D}, Finding the intersection gives {D}. Hence, the answer is boxed{D}. (4) **Analysis** This question tests the use of derivatives to study the tangent lines of curves and minimum value problems, which is a basic question. According to the analysis of the combination of numbers and shapes, to minimize |AB|, the key is that the tangent line of the curve g(x) at point B is parallel to the line y=2x+3. Further, based on the condition, establish the equations for a and b, solve for a and b, and thus obtain a+b. **Solution** Let B(x_0,b), to minimize |AB|, The tangent line of the curve g(x) at point B is parallel to the line y=2x+3, g'(x)=a+ frac{1}{x}, hence a+ frac{1}{{x}_{0}}=2, solving gives {x}_{0}= frac{1}{2-a}, Therefore, a cdot frac{1}{2-a}+ln frac{1}{2-a}=b (1), Also, it is easy to find the coordinates of point A as left( frac{b-3}{2},bright), Therefore, |AB|= frac{1}{2-a}- frac{b-3}{2}=2 (2), Solving (1) and (2) gives a=1, b=1, Therefore, a+b=2, Hence, the answer is boxed{2}.