answer：Alright, so I have this problem about fencing a rectangular field that's on a slope. The field is supposed to be fenced on three sides, leaving one side of 50 feet uncovered. There are two types of fencing materials available: Material A costs 8 per foot and has a maximum span of 100 feet, while Material B costs 12 per foot and has no length restrictions. The area of the field is 1200 square feet. I need to figure out the optimal combination of these materials to minimize the cost of fencing.First, I need to understand the problem clearly. We have a rectangular field, and one of its sides is 50 feet, which is left uncovered. So, we need to fence the other three sides. The area is given as 1200 square feet, which should help me find the other dimension of the field.Since the area of a rectangle is length multiplied by width, and one of the sides is 50 feet, I can set up the equation:Area = length × width1200 = 50 × widthSo, width = 1200 / 50 = 24 feet.Okay, so the field is 50 feet by 24 feet. Now, we need to fence three sides. That would be two sides of 24 feet each and one side of 50 feet. So, the total fencing needed is:Total fencing = 2 × 24 + 50 = 48 + 50 = 98 feet.Now, I need to decide which fencing material to use for which side to minimize the cost. Material A is cheaper at 8 per foot but has a maximum span of 100 feet. Material B is more expensive at 12 per foot but has no length restrictions.I need to consider the cost implications of using each material for different sides. Let's think about the options:1. **Using Material A for all three sides:** - Total cost = 98 feet × 8/foot = 784.2. **Using Material B for the 50-foot side and Material A for the two 24-foot sides:** - Cost for Material B = 50 feet × 12/foot = 600. - Cost for Material A = 2 × 24 feet × 8/foot = 48 feet × 8/foot = 384. - Total cost = 600 + 384 = 984.3. **Using Material A for the 50-foot side and Material B for the two 24-foot sides:** - Cost for Material A = 50 feet × 8/foot = 400. - Cost for Material B = 2 × 24 feet × 12/foot = 48 feet × 12/foot = 576. - Total cost = 400 + 576 = 976.4. **Using Material B for all three sides:** - Total cost = 98 feet × 12/foot = 1176.Now, comparing all these options:- Option 1: 784- Option 2: 984- Option 3: 976- Option 4: 1176Clearly, Option 1 is the cheapest at 784. However, I need to make sure that Material A can be used for all three sides without violating its maximum span of 100 feet. Since the total fencing required is 98 feet, which is less than 100 feet, it's feasible to use Material A for the entire fencing.But wait, let me double-check if there's any restriction on using Material A for individual sides. The problem states that Material A has a maximum span of 100 feet, but it doesn't specify whether this is per piece or in total. If it's per piece, then each side must be less than or equal to 100 feet. In this case, the longest side is 50 feet, which is well within the 100-foot limit. So, using Material A for all sides is acceptable.Is there any other combination that could potentially be cheaper? For example, using Material A for some parts and Material B for others. But since Material A is cheaper, it makes sense to use it as much as possible. Given that the total fencing needed is 98 feet, which is under the 100-foot limit of Material A, using Material A exclusively is the most cost-effective solution.Therefore, the optimal combination is to use Material A for all three sides, resulting in a total cost of 784.

answer：First, I need to determine what number, when multiplied by 45, equals 35% of 900.To find 35% of 900, I'll calculate 0.35 multiplied by 900, which equals 315.Now, I have the equation 45 multiplied by an unknown number equals 315. To find the unknown number, I'll divide 315 by 45.315 divided by 45 equals 7.Therefore, the number we're looking for is 7.

answer：Okay, so I have this problem about a trapezoid. Let me try to visualize it first. A trapezoid has two parallel sides called bases, and the other two sides are non-parallel. The longer base is given as 115 units, and I need to find the shorter base. The problem mentions the line joining the midpoints of the diagonals. Hmm, I remember that in a trapezoid, the segment that connects the midpoints of the non-parallel sides (the legs) is called the midsegment, and its length is the average of the lengths of the two bases. But wait, this problem is talking about the midpoints of the diagonals, not the legs. Is there a similar property for the midpoints of the diagonals?Let me think. I recall that in a trapezoid, the segment connecting the midpoints of the diagonals is parallel to the bases and its length is half the difference of the lengths of the two bases. So, if I denote the longer base as ( B ) and the shorter base as ( b ), then the length of this segment should be ( frac{B - b}{2} ).The problem states that this length is 6. So, I can set up the equation:[frac{B - b}{2} = 6]Given that the longer base ( B ) is 115, I can substitute that into the equation:[frac{115 - b}{2} = 6]To solve for ( b ), I'll multiply both sides by 2:[115 - b = 12]Then, subtract 12 from 115:[b = 115 - 12 = 103]So, the shorter base should be 103 units long. Let me double-check to make sure I didn't make a mistake. If the longer base is 115 and the shorter is 103, the difference is 12. Half of that difference is 6, which matches the given length of the segment connecting the midpoints of the diagonals. That seems right.I think I got it. The shorter base is 103.

answer：First, I need to understand the structure of the club and how it changes each year. The club starts with 21 people, consisting of 7 leaders and 14 regular members.Each year, the 7 leaders leave the club, leaving only the regular members. Each regular member then recruits 3 new people, which means each regular member is effectively replaced by 4 people (themselves plus 3 new recruits). This results in the number of regular members quadrupling each year.After the recruitment, 7 new leaders are elected from outside the club, bringing the total number of people back to the previous year's total plus the new leaders.To find the total number of people in the club after five years, I can set up a recursive formula where each year's total is calculated based on the previous year's total. Starting with 21 people, I can compute the totals for each subsequent year up to five years.By applying this recursive formula year by year, I can determine the total number of people in the club after five years.

answer：Okay, so I have this problem here: Given that sin α = 3/5 and α is between π/2 and π, I need to find the value of sin 2α. Hmm, let me think about how to approach this.First, I remember that sin 2α is a double angle formula. I think the formula is sin 2α = 2 sin α cos α. Yeah, that sounds right. So, I know sin α is 3/5, but I don't know cos α yet. I need to find cos α to use the double angle formula.Since I know sin α, I can use the Pythagorean identity to find cos α. The identity is sin²α + cos²α = 1. So, if sin α is 3/5, then sin²α is (3/5)², which is 9/25. Plugging that into the identity, I get 9/25 + cos²α = 1. To solve for cos²α, I subtract 9/25 from both sides: cos²α = 1 - 9/25. That simplifies to 16/25.Now, taking the square root of both sides gives me cos α = ±√(16/25) = ±4/5. But wait, I need to figure out whether cos α is positive or negative. The problem says α is in the interval (π/2, π), which is the second quadrant. In the second quadrant, sine is positive and cosine is negative. Since sin α is positive, that makes sense, and cos α should be negative. So, cos α = -4/5.Alright, now I have both sin α and cos α. Let me plug them into the double angle formula. So, sin 2α = 2 * (3/5) * (-4/5). Let me calculate that. 2 times 3 is 6, and 5 times 5 is 25. So, 6/25 times -4 is -24/25. Therefore, sin 2α is -24/25.Wait, let me double-check my steps to make sure I didn't make any mistakes. First, I used the double angle formula correctly, right? Yes, sin 2α = 2 sin α cos α. Then, I found cos α using the Pythagorean identity, which is correct. I correctly identified that in the second quadrant, cosine is negative, so cos α is -4/5. Plugging those values into the formula, I multiplied 2 * 3/5 * -4/5, which gives me -24/25. That seems right.Just to be thorough, maybe I can think about the angle α. If sin α is 3/5, then α is an angle in the second quadrant where the opposite side is 3 and the hypotenuse is 5. Using the Pythagorean theorem, the adjacent side would be -4 (since it's in the second quadrant). So, cos α is -4/5, which matches what I found earlier.Another way to think about it is using the unit circle. In the second quadrant, the x-coordinate (which corresponds to cosine) is negative, and the y-coordinate (sine) is positive. So, with sin α = 3/5, the y-coordinate is 3/5, and the x-coordinate would be -4/5 to satisfy the equation x² + y² = 1. That makes sense.So, putting it all together, sin 2α is -24/25. I think that's the correct answer. I don't see any mistakes in my reasoning, so I feel confident about this solution.**Final Answer**The value of sin 2alpha is boxed{-dfrac{24}{25}}.

answer：First, I need to determine the total number of trees the gardener is planting. There are 5 maple trees, 4 oak trees, and 6 birch trees, making a total of 15 trees.Next, I want to find the probability that no two birch trees are next to each other. To do this, I'll consider the maple and oak trees as "non-birch" trees. There are 5 maple + 4 oak = 9 non-birch trees.These 9 non-birch trees create slots where the birch trees can be placed. Specifically, there are 9 + 1 = 10 slots available for the birch trees (including the positions before the first non-birch tree and after the last non-birch tree).I need to choose 6 out of these 10 slots to place the birch trees. The number of ways to do this is given by the combination formula ( binom{10}{6} ).Calculating ( binom{10}{6} ):[binom{10}{6} = frac{10!}{6! cdot 4!} = 210]Now, I'll calculate the total number of ways to arrange all 15 trees. This is the number of ways to choose 6 positions out of 15 for the birch trees, which is ( binom{15}{6} ).Calculating ( binom{15}{6} ):[binom{15}{6} = frac{15!}{6! cdot 9!} = 5005]The probability that no two birch trees are adjacent is the ratio of the favorable arrangements to the total arrangements:[frac{210}{5005}]Simplifying this fraction:[frac{210}{5005} = frac{2}{45}]Finally, adding the numerator and denominator of the simplified fraction:[2 + 45 = 47]