answer：Okay, so I have this trigonometry problem here. It says that tan α is 1/2 and tan β is -1/3. I need to find the value of this expression: (3 sin α cos β - sin β cos α) divided by (cos α cos β + 2 sin α sin β). The options are A: 7/8, B: 11/8, C: 7/4, D: 11/4. Hmm, let me think about how to approach this.First, I remember that when dealing with expressions involving sine and cosine, especially when given tangents, it might be helpful to express everything in terms of tangent. Because tangent is sine over cosine, so maybe I can rewrite the numerator and the denominator in terms of tan α and tan β.Looking at the numerator: 3 sin α cos β - sin β cos α. I notice that this looks a bit like the sine of a difference formula, which is sin(α - β) = sin α cos β - cos α sin β. But here, instead of just 1 times sin α cos β, it's 3 times that term. So maybe I can factor out something or manipulate it to use the tangent values.Similarly, the denominator is cos α cos β + 2 sin α sin β. That reminds me of the cosine of a difference formula, which is cos(α - β) = cos α cos β + sin α sin β. But here, the second term is multiplied by 2. So again, maybe I can factor or adjust it to use the tangent values.Wait, another thought: since both numerator and denominator have combinations of sin and cos, perhaps I can divide both numerator and denominator by cos α cos β. That way, I can express everything in terms of tan α and tan β, which are given.Let me try that. So, if I divide each term in the numerator and denominator by cos α cos β, I get:Numerator: (3 sin α cos β)/(cos α cos β) - (sin β cos α)/(cos α cos β)= 3 (sin α / cos α) - (sin β / cos β)= 3 tan α - tan βDenominator: (cos α cos β)/(cos α cos β) + (2 sin α sin β)/(cos α cos β)= 1 + 2 (sin α / cos α)(sin β / cos β)= 1 + 2 tan α tan βSo now, the entire expression simplifies to (3 tan α - tan β) / (1 + 2 tan α tan β). That's much better because I know tan α and tan β.Given tan α = 1/2 and tan β = -1/3, let's plug these values into the expression.First, compute the numerator: 3 tan α - tan β= 3*(1/2) - (-1/3)= 3/2 + 1/3To add these, I need a common denominator, which is 6.3/2 = 9/6 and 1/3 = 2/6So, 9/6 + 2/6 = 11/6Now, compute the denominator: 1 + 2 tan α tan β= 1 + 2*(1/2)*(-1/3)First, multiply 2, 1/2, and -1/3:2*(1/2) = 1, then 1*(-1/3) = -1/3So, the denominator is 1 - 1/3 = 2/3Now, the expression is (11/6) divided by (2/3). Dividing by a fraction is the same as multiplying by its reciprocal, so:11/6 * 3/2 = (11*3)/(6*2) = 33/12Simplify 33/12 by dividing numerator and denominator by 3:33 ÷ 3 = 1112 ÷ 3 = 4So, it simplifies to 11/4.Looking back at the options, D is 11/4. So, the answer should be D.Wait, let me double-check my steps to make sure I didn't make a mistake.1. I converted the original expression by dividing numerator and denominator by cos α cos β. That seems correct because cos α and cos β are not zero (since tan α and tan β are defined).2. Then, I expressed everything in terms of tan α and tan β, which are given. That seems right.3. Plugging in tan α = 1/2 and tan β = -1/3:Numerator: 3*(1/2) - (-1/3) = 3/2 + 1/3 = 11/6. That looks correct.Denominator: 1 + 2*(1/2)*(-1/3) = 1 - 1/3 = 2/3. That also seems correct.4. Dividing 11/6 by 2/3: 11/6 * 3/2 = 33/12 = 11/4. Yep, that's right.So, I think the answer is definitely D: 11/4.**Final Answer**boxed{D}

answer：First, I need to solve the inequality (3 - 2x < 0).I'll start by isolating the term with (x). Subtracting 3 from both sides gives:[-2x < -3]Next, I'll divide both sides by -2. It's important to remember that dividing or multiplying both sides of an inequality by a negative number reverses the inequality sign. So, dividing by -2 changes the inequality from (<) to (>):[x > frac{3}{2}]This means the solution set (M) consists of all real numbers greater than (frac{3}{2}).Now, I'll check the membership of 0 and 2 in the set (M):- (0) is not greater than (frac{3}{2}), so (0 notin M).- (2) is greater than (frac{3}{2}), so (2 in M).Based on this analysis, the correct option is B.

answer：Okay, so I've got this problem here: If sin theta and cos theta are the two roots of the equation x^{2} - ax + a = 0 with respect to x, then find cos(theta - frac{3pi}{2}) + sin(frac{3pi}{2} + theta).Hmm, let's break this down step by step. First, I remember that for a quadratic equation of the form x^2 - ax + a = 0, the sum of the roots is equal to a and the product of the roots is also equal to a. So, since the roots are sin theta and cos theta, that means:1. Sum of roots: sin theta + cos theta = a2. Product of roots: sin theta cdot cos theta = aWait, both the sum and the product are equal to a? That seems interesting. Maybe I can use this to find the value of a or something about theta.I also remember that (sin theta + cos theta)^2 = sin^2 theta + 2 sin theta cos theta + cos^2 theta. Since sin^2 theta + cos^2 theta = 1, this simplifies to 1 + 2 sin theta cos theta. Given that sin theta + cos theta = a, squaring both sides gives a^2 = 1 + 2 sin theta cos theta. But we also know that sin theta cos theta = a, so substituting that in, we get:a^2 = 1 + 2aSo, rearranging this equation: a^2 - 2a - 1 = 0. This is a quadratic equation in terms of a. Let's solve for a using the quadratic formula:a = frac{2 pm sqrt{(2)^2 - 4(1)(-1)}}{2(1)} = frac{2 pm sqrt{4 + 4}}{2} = frac{2 pm sqrt{8}}{2} = frac{2 pm 2sqrt{2}}{2} = 1 pm sqrt{2}So, a can be either 1 + sqrt{2} or 1 - sqrt{2}. But wait, the original quadratic equation is x^2 - ax + a = 0. For the roots to be real, the discriminant must be non-negative. The discriminant is D = a^2 - 4a. Let's compute this for both values of a:1. For a = 1 + sqrt{2}:D = (1 + sqrt{2})^2 - 4(1 + sqrt{2}) = 1 + 2sqrt{2} + 2 - 4 - 4sqrt{2} = (3 + 2sqrt{2}) - 4 - 4sqrt{2} = -1 - 2sqrt{2}Hmm, that's negative, which means the roots would be complex. But in the problem statement, the roots are given as sin theta and cos theta, which are real numbers. So, a = 1 + sqrt{2} is not valid.2. For a = 1 - sqrt{2}:D = (1 - sqrt{2})^2 - 4(1 - sqrt{2}) = 1 - 2sqrt{2} + 2 - 4 + 4sqrt{2} = (3 - 2sqrt{2}) - 4 + 4sqrt{2} = -1 + 2sqrt{2}This is positive because sqrt{2} approx 1.414, so 2sqrt{2} approx 2.828, and -1 + 2.828 approx 1.828 > 0. So, a = 1 - sqrt{2} is valid.Alright, so we've determined that a = 1 - sqrt{2}. Therefore, sin theta + cos theta = 1 - sqrt{2}.Now, moving on to the expression we need to evaluate: cos(theta - frac{3pi}{2}) + sin(frac{3pi}{2} + theta).Let me try to simplify this expression using trigonometric identities. First, let's recall some angle addition formulas:1. cos(A - B) = cos A cos B + sin A sin B2. sin(A + B) = sin A cos B + cos A sin BBut before diving into these, maybe there's a simpler way by recognizing the angles involved.Looking at cos(theta - frac{3pi}{2}):frac{3pi}{2} is 270 degrees. So, theta - frac{3pi}{2} is like shifting theta by 270 degrees. Alternatively, we can think of this as cos(theta - 2pi + frac{pi}{2}) because 2pi is a full circle, so subtracting 2pi brings it back to the same angle. So, cos(theta - frac{3pi}{2}) = cos(theta + frac{pi}{2}).Similarly, for sin(frac{3pi}{2} + theta), we can think of it as sin(pi + frac{pi}{2} + theta) = sin(pi + (frac{pi}{2} + theta)). But let's use the identities:1. cos(theta - frac{3pi}{2}) = cos(theta + frac{pi}{2}) because cos has a period of 2pi, so subtracting 2pi doesn't change the value.2. sin(frac{3pi}{2} + theta) = sin(pi + frac{pi}{2} + theta) = sin(pi + (theta + frac{pi}{2}))Wait, maybe a better approach is to use co-function identities or shift identities.I remember that cos(theta - frac{3pi}{2}) can be rewritten using the identity cos(theta - frac{3pi}{2}) = sin(theta - frac{pi}{2}) because cosine is sine shifted by frac{pi}{2}. Let me verify that:cos(theta - frac{3pi}{2}) = cos(theta + frac{pi}{2} - 2pi) = cos(theta + frac{pi}{2}) because cosine is periodic with period 2pi. And cos(theta + frac{pi}{2}) = -sin theta because cosine of an angle plus frac{pi}{2} is equal to negative sine of that angle. Similarly, sin(theta + frac{pi}{2}) = cos theta.Wait, let me make sure:cos(theta + frac{pi}{2}) = cos theta cos frac{pi}{2} - sin theta sin frac{pi}{2} = cos theta cdot 0 - sin theta cdot 1 = -sin thetaYes, that's correct. So, cos(theta - frac{3pi}{2}) = cos(theta + frac{pi}{2}) = -sin theta.Now, for the second term: sin(frac{3pi}{2} + theta). Let's see:sin(frac{3pi}{2} + theta) = sin(pi + frac{pi}{2} + theta) = sin(pi + (theta + frac{pi}{2}))Using the identity sin(pi + x) = -sin x, so:sin(pi + (theta + frac{pi}{2})) = -sin(theta + frac{pi}{2})And we know that sin(theta + frac{pi}{2}) = cos theta, so:sin(frac{3pi}{2} + theta) = -cos thetaPutting it all together:cos(theta - frac{3pi}{2}) + sin(frac{3pi}{2} + theta) = -sin theta - cos theta = -(sin theta + cos theta)But earlier, we found that sin theta + cos theta = a = 1 - sqrt{2}. Therefore:-(sin theta + cos theta) = -(1 - sqrt{2}) = sqrt{2} - 1So, the value of the expression is sqrt{2} - 1.Wait, let me double-check my steps to make sure I didn't make a mistake.1. I used the sum and product of roots correctly to find a = 1 - sqrt{2}.2. Then, I simplified the trigonometric expression by recognizing the angle shifts and using co-function identities.3. I correctly applied the identities to find that the expression simplifies to -(sin theta + cos theta).4. Substituting the value of a gives the final result.Everything seems to check out. I think I got it right.

answer：Okay, so I have this arithmetic sequence: 3, 5, 7, 9, and so on. I need to find the 50th term in this sequence. Hmm, arithmetic sequences... I remember they have a common difference between consecutive terms. Let me think about how to approach this.First, I should identify the first term of the sequence. Looking at the given numbers, the first term is clearly 3. That’s straightforward. Now, what's the common difference? Well, if I subtract the first term from the second term, that should give me the common difference. So, 5 minus 3 is 2. Okay, so the common difference, which I think is usually denoted as 'd', is 2.Now, I recall there's a formula to find the nth term of an arithmetic sequence. I think it's something like a_n equals a_1 plus (n minus 1) times d. Let me write that down to make sure: a_n = a_1 + (n - 1)d. Yeah, that sounds right. So, in this case, a_1 is 3, d is 2, and n is 50 because I need the 50th term.Let me plug those numbers into the formula. So, a_50 equals 3 plus (50 minus 1) times 2. Let me calculate that step by step. First, 50 minus 1 is 49. Then, 49 times 2 is 98. Finally, adding that to the first term, 3 plus 98 equals 101. So, the 50th term should be 101.Wait, let me double-check my calculations to make sure I didn't make a mistake. Starting from the beginning: first term is 3, common difference is 2. So, the sequence increases by 2 each time. The second term is 5, third is 7, fourth is 9, and so on. If I continue this pattern, each term is indeed 2 more than the previous one.Using the formula again: a_n = a_1 + (n - 1)d. Plugging in the values: a_50 = 3 + (50 - 1)*2. 50 minus 1 is 49, 49 times 2 is 98, and 3 plus 98 is 101. Yep, that seems consistent. I don't think I made any errors in my calculations.Just to be extra sure, maybe I can think about how many steps it takes to get from the first term to the 50th term. Since the first term is 3, the second term is 3 plus 2, the third term is 3 plus 2 times 2, and so on. So, the nth term would be 3 plus 2 times (n - 1). For the 50th term, that's 3 plus 2 times 49, which is 3 plus 98, giving me 101 again. Okay, that confirms it.I think I'm confident now that the 50th term of this arithmetic sequence is 101.

answer：Okay, so I need to find the area of a triangle, and I've been given the base and the height. The base is 3 meters, and the height is 6 meters. Hmm, I remember that the area of a triangle has something to do with multiplying the base and the height, but I'm not entirely sure about the exact formula. Let me think about this.First, I think the formula involves dividing by something after multiplying the base and height. Maybe it's divided by 2? Yeah, that sounds familiar. So, if I recall correctly, the formula should be something like:Area = (base × height) / 2Okay, so plugging in the numbers I have:Base = 3 metersHeight = 6 metersSo, multiplying them together: 3 × 6 = 18. Then, dividing by 2: 18 / 2 = 9.Wait, does that make sense? Let me visualize a triangle. If the base is 3 meters and the height is 6 meters, the area should be half of the area of a rectangle with the same base and height, right? Because a triangle is essentially half of a parallelogram.So, if the area of the rectangle would be 3 × 6 = 18 square meters, then the area of the triangle should indeed be half of that, which is 9 square meters.But just to be sure, maybe I should double-check the formula. I think it's always good to verify, especially when dealing with geometry. Let me recall: for any triangle, the area is given by (base × height) divided by 2. Yeah, that seems right.Another way to think about it is that the area of a triangle depends on its base and height, and since a triangle can be seen as half of a rectangle or parallelogram, the division by 2 makes sense to account for that.So, applying the formula again:Area = (3 m × 6 m) / 2 = 18 m² / 2 = 9 m²Yep, that checks out. I think I've got it right. The area of the triangle with a base of 3 meters and a height of 6 meters is 9 square meters.

answer：First, I need to determine the total number of each type of fruit required by all six bakeries on a weekly basis.For strawberries, I'll add the weekly requirements from each bakery:2 (from the first bakery) + 4 (second) + 12 (third) + 8 (fourth) + 15 (fifth) + 5 (sixth) = 46 sacks per week.Next, I'll calculate the total for blueberries:3 (first) + 2 (second) + 10 (third) + 4 (fourth) + 6 (fifth) + 9 (sixth) = 34 sacks per week.Then, for raspberries:5 (first) + 8 (second) + 7 (third) + 3 (fourth) + 12 (fifth) + 11 (sixth) = 46 sacks per week.Now, to find out the total number of sacks needed for 10 weeks, I'll multiply each weekly total by 10.For strawberries: 46 sacks/week * 10 weeks = 460 sacks.For blueberries: 34 sacks/week * 10 weeks = 340 sacks.For raspberries: 46 sacks/week * 10 weeks = 460 sacks.Finally, I'll summarize the totals for each type of fruit.