answer：First, I'll define the number of white balls as ( W ) and the number of yellow balls as ( Y ). The total number of balls ordered is 64, so:[W + Y = 64]Due to the error, 20 extra yellow balls were dispatched, making the new ratio of white to yellow balls ( frac{8}{13} ):[frac{W}{Y + 20} = frac{8}{13}]From this ratio, I can express ( W ) in terms of ( Y ):[W = frac{8(Y + 20)}{13}]Substituting this expression for ( W ) back into the total number of balls equation:[frac{8(Y + 20)}{13} + Y = 64]To eliminate the fraction, I'll multiply the entire equation by 13:[8(Y + 20) + 13Y = 832]Expanding and combining like terms:[8Y + 160 + 13Y = 832][21Y + 160 = 832]Subtracting 160 from both sides:[21Y = 672]Dividing by 21 to find ( Y ):[Y = 32]Now, substituting ( Y = 32 ) back into the total number of balls equation to find ( W ):[W + 32 = 64][W = 32]Thus, the original number of white and yellow balls was both 32, resulting in an original ratio of ( 1:1 ).

answer：Okay, so I have this problem about "local odd functions." I'm not entirely sure what that means, but the problem defines it. It says that a function f(x) is called a "local odd function" if there exists a real number x in the domain such that f(-x) = -f(x). So, unlike a regular odd function, which satisfies f(-x) = -f(x) for all x in the domain, a local odd function only needs to satisfy this condition for at least one x in the domain.Alright, so there are two parts to this problem. Let me tackle them one by one.**Part (1):**The function given is f(x) = log₂(x + m). We need to find the range of real number m such that f(x) is a "local odd function" on the interval [-1, 1].First, I need to make sure that the function is defined on the interval [-1, 1]. Since it's a logarithm function, the argument x + m must be positive for all x in [-1, 1]. So, for x = -1, the smallest value in the interval, we have:-1 + m > 0 ⇒ m > 1.So, m must be greater than 1 for the function to be defined on the entire interval.Now, since f(x) is a "local odd function," there must exist some x in [-1, 1] such that f(-x) = -f(x). Let's write this condition out:log₂(-x + m) = -log₂(x + m).Hmm, let's see. If I exponentiate both sides with base 2 to eliminate the logarithms, I get:2^{log₂(-x + m)} = 2^{-log₂(x + m)}.Simplifying both sides, this becomes:(-x + m) = 1 / (x + m).So, we have:(-x + m) = 1 / (x + m).Let me write that as:m - x = 1 / (m + x).Now, let's multiply both sides by (m + x) to eliminate the denominator:(m - x)(m + x) = 1.This simplifies to:m² - x² = 1.So, m² = x² + 1.Since x is in the interval [-1, 1], x² ranges from 0 to 1. Therefore, m² ranges from 1 to 2. Taking square roots, m ranges from 1 to √2. However, earlier we found that m must be greater than 1 for the function to be defined on the entire interval. So, m must be in the interval (1, √2).Wait, let me double-check that. If m² = x² + 1, and x² can be as small as 0 and as large as 1, then m² can be as small as 1 and as large as 2. So, m can be between 1 and √2. But since m must be greater than 1, the range is (1, √2).Okay, that seems reasonable.**Part (2):**Now, the function is f(x) = 9^x - m * 3^{x+1} - 3. We need to find the range of m such that f(x) is a "local odd function" on the domain R.Again, since it's a "local odd function," there exists some real number x such that f(-x) = -f(x). Let's write that condition:9^{-x} - m * 3^{-x + 1} - 3 = -[9^x - m * 3^{x + 1} - 3].Let me simplify both sides.First, the left side:9^{-x} - m * 3^{-x + 1} - 3.The right side:-9^x + m * 3^{x + 1} + 3.So, putting them together:9^{-x} - m * 3^{-x + 1} - 3 = -9^x + m * 3^{x + 1} + 3.Let's bring all terms to one side:9^{-x} - m * 3^{-x + 1} - 3 + 9^x - m * 3^{x + 1} - 3 = 0.Wait, that's a bit messy. Let me rearrange the equation step by step.Starting from:9^{-x} - m * 3^{-x + 1} - 3 = -9^x + m * 3^{x + 1} + 3.Let me add 9^x to both sides and subtract m * 3^{-x + 1} and subtract 3 from both sides:9^{-x} + 9^x - m * 3^{-x + 1} - m * 3^{x + 1} - 6 = 0.Hmm, that's better.Now, notice that 9^x is (3^2)^x = 3^{2x}, and 9^{-x} = 3^{-2x}. Also, 3^{-x + 1} = 3 * 3^{-x}, and 3^{x + 1} = 3 * 3^x.So, let's rewrite the equation:3^{-2x} + 3^{2x} - 3m * 3^{-x} - 3m * 3^x - 6 = 0.Let me factor out the 3:(3^{-2x} + 3^{2x}) - 3m(3^{-x} + 3^x) - 6 = 0.Now, let me set t = 3^x + 3^{-x}. Since 3^x is always positive, t is always greater than or equal to 2 (by AM ≥ GM: (3^x + 3^{-x})/2 ≥ √(3^x * 3^{-x}) = 1, so t ≥ 2).Now, let's express 3^{2x} + 3^{-2x} in terms of t. Notice that:(3^x + 3^{-x})^2 = 3^{2x} + 2 + 3^{-2x}.Therefore, 3^{2x} + 3^{-2x} = t² - 2.Substituting back into the equation:(t² - 2) - 3m * t - 6 = 0.Simplify:t² - 3m t - 8 = 0.So, we have a quadratic equation in terms of t:t² - 3m t - 8 = 0.We can solve for t:t = [3m ± √(9m² + 32)] / 2.But since t must be greater than or equal to 2, we need to ensure that the solution t is at least 2.Let me denote the solutions:t = [3m + √(9m² + 32)] / 2 and t = [3m - √(9m² + 32)] / 2.Since √(9m² + 32) is always positive, the second solution [3m - √(9m² + 32)] / 2 could be negative or positive. Let's check.Compute [3m - √(9m² + 32)] / 2:Since √(9m² + 32) > √(9m²) = 3|m|.So, if m is positive, 3m - √(9m² + 32) < 0.If m is negative, 3m is negative, and √(9m² + 32) is positive, so 3m - √(9m² + 32) is negative.Therefore, the second solution is always negative, which cannot be since t ≥ 2. So, we discard the second solution.Thus, the only valid solution is:t = [3m + √(9m² + 32)] / 2.We need this t to be greater than or equal to 2:[3m + √(9m² + 32)] / 2 ≥ 2.Multiply both sides by 2:3m + √(9m² + 32) ≥ 4.Let me isolate the square root:√(9m² + 32) ≥ 4 - 3m.Now, since the square root is always non-negative, the right side must also be non-negative:4 - 3m ≥ 0 ⇒ m ≤ 4/3.So, m must be less than or equal to 4/3.Now, squaring both sides (since both sides are non-negative):9m² + 32 ≥ (4 - 3m)².Compute the right side:(4 - 3m)² = 16 - 24m + 9m².So, we have:9m² + 32 ≥ 16 - 24m + 9m².Subtract 9m² from both sides:32 ≥ 16 - 24m.Subtract 16:16 ≥ -24m.Divide both sides by -24 (remembering to reverse the inequality sign):m ≥ -16/24 ⇒ m ≥ -2/3.So, combining the two conditions:m ≤ 4/3 and m ≥ -2/3.Therefore, m must be in the interval [-2/3, 4/3].Wait, but let me double-check this. When I squared both sides, I assumed that 4 - 3m is non-negative, which led to m ≤ 4/3. But when m > 4/3, 4 - 3m becomes negative, and the inequality √(9m² + 32) ≥ 4 - 3m is automatically true because the left side is positive and the right side is negative. So, for m > 4/3, the inequality holds without any additional constraints.But earlier, we found that t must be ≥ 2, which led us to m ≥ -2/3. So, combining these, for m > 4/3, the inequality holds, and for m ≤ 4/3, we have m ≥ -2/3.Therefore, the range of m is m ≥ -2/3.Wait, that seems contradictory. Let me think again.When m > 4/3, the inequality √(9m² + 32) ≥ 4 - 3m is automatically true because the right side is negative. So, for m > 4/3, the condition t ≥ 2 is automatically satisfied because t is always ≥ 2, regardless of m.But when m ≤ 4/3, we have the condition that t ≥ 2, which led us to m ≥ -2/3.Therefore, combining both cases:- For m > 4/3: condition holds.- For m ≤ 4/3: m must be ≥ -2/3.So, overall, m must be ≥ -2/3.Therefore, the range of m is [-2/3, ∞).Wait, but earlier, when m > 4/3, t is still ≥ 2, so the equation t² - 3m t - 8 = 0 has a solution t ≥ 2, which means that the function is a "local odd function" for m > 4/3 as well.So, actually, m can be any real number greater than or equal to -2/3.Therefore, the range of m is m ≥ -2/3.So, in conclusion:Part (1): m ∈ (1, √2)Part (2): m ∈ [-2/3, ∞)I think that's it.

answer：First, I need to determine how many points Lefty scored. Let's denote Lefty's points as ( L ).Righty scores half as many points as Lefty, so Righty's points are ( frac{L}{2} ).The third teammate scores six times as many points as Righty, which is ( 6 times frac{L}{2} = 3L ).The total points scored by the team is the sum of the points scored by Lefty, Righty, and the third teammate:[L + frac{L}{2} + 3L = 4.5L]The average points scored per player is given as 30. Since there are three players, the total points can also be calculated by multiplying the average by the number of players:[3 times 30 = 90]Setting the two expressions for total points equal to each other:[4.5L = 90]Solving for ( L ):[L = frac{90}{4.5} = 20]Therefore, Lefty scored 20 points.

answer：Okay, so I'm trying to figure out how Anya, Vanya, and Sanya managed to pay for their bus fare without having any copper coins, but each of them ended up spending five kopecks. Let me break this down step by step.First, I need to understand what the problem is saying. They got on a bus, and they didn't have any copper coins. Copper coins are typically lower denomination coins, like pennies or something similar. So, they didn't have small coins, but they still managed to pay for their fare, spending five kopecks each. That means each of them paid five kopecks, but they didn't have the exact change, especially in small denominations.I guess the key here is that they didn't have copper coins, which are usually the smaller denominations. So, they couldn't make exact change individually. But somehow, together, they managed to pay the fare. Maybe they combined their money in some way to make the exact amount needed.Let me think about how they could do that. If they didn't have copper coins, they probably had higher denomination coins. Maybe they used larger coins to pay the total fare and then somehow got the change back. But how?If each of them needed to spend five kopecks, the total fare would be 15 kopecks. So, maybe they pooled their money together to make 15 kopecks, and then one of them paid the fare, and the others got their change back. But since they didn't have copper coins, they couldn't make exact change individually.Wait, but if they pooled their money, they could have one person pay the total fare and then distribute the change. For example, if Anya, Vanya, and Sanya each had, say, 10 kopecks, they could give 10 kopecks each to one person, making 30 kopecks. Then that person could pay the 15 kopecks fare and give back 15 kopecks as change. But they only needed to spend 5 kopecks each, not 10.Hmm, maybe they didn't have 10 kopecks each. Maybe they had larger coins, like 15 kopecks each. If each had 15 kopecks, they could give that to one person, making 45 kopecks. Then that person could pay the 15 kopecks fare and give back 30 kopecks as change. But then each person would have spent 15 kopecks, not 5.Wait, that doesn't add up. They only needed to spend 5 kopecks each, so the total fare is 15 kopecks. If they each had 5 kopecks, they could just pay individually, but they didn't have copper coins. So, maybe they had larger coins, like 10 kopecks each. If each had 10 kopecks, they could give 10 kopecks to one person, making 30 kopecks. Then that person could pay the 15 kopecks fare and give back 15 kopecks as change. But then each person would have spent 10 kopecks, not 5.This is confusing. Maybe they didn't have 10 kopecks each. Maybe they had different denominations. Let's say Anya had 15 kopecks, Vanya had 15 kopecks, and Sanya had 15 kopecks. They could give all their money to one person, making 45 kopecks. Then that person could pay the 15 kopecks fare and give back 30 kopecks as change. But again, that would mean each person spent 15 kopecks, not 5.Wait, maybe they didn't all have the same amount. Maybe Anya had 5 kopecks, Vanya had 10 kopecks, and Sanya had 15 kopecks. Then, Anya could pay her 5 kopecks, Vanya could pay her 10 kopecks, and Sanya could pay her 15 kopecks. But that would make the total fare 30 kopecks, not 15.No, that's not right. The total fare should be 15 kopecks. So, maybe they each had 5 kopecks, but they didn't have copper coins, so they couldn't make exact change. Maybe they used a higher denomination to pay and then got change back.Wait, if they each had 10 kopecks, they could give 10 kopecks each to one person, making 30 kopecks. Then that person could pay the 15 kopecks fare and give back 15 kopecks as change. But then each person would have spent 10 kopecks, not 5.I'm going in circles here. Maybe the key is that they didn't have copper coins, so they couldn't make exact change, but they could use larger coins to pay and then get change back. So, if they each had, say, 15 kopecks, they could give that to one person, making 45 kopecks. Then that person could pay the 15 kopecks fare and give back 30 kopecks as change. But then each person would have spent 15 kopecks, not 5.Wait, maybe they didn't all have the same amount. Maybe Anya had 5 kopecks, Vanya had 10 kopecks, and Sanya had 15 kopecks. Then, Anya could pay her 5 kopecks, Vanya could pay her 10 kopecks, and Sanya could pay her 15 kopecks. But that would make the total fare 30 kopecks, not 15.No, that's not right. The total fare should be 15 kopecks. So, maybe they each had 5 kopecks, but they didn't have copper coins, so they couldn't make exact change. Maybe they used a higher denomination to pay and then got change back.Wait, if they each had 10 kopecks, they could give 10 kopecks each to one person, making 30 kopecks. Then that person could pay the 15 kopecks fare and give back 15 kopecks as change. But then each person would have spent 10 kopecks, not 5.I'm stuck. Maybe I need to think differently. Maybe they didn't pay individually but collectively. So, they gave money to one person who paid the fare and then gave them change. But since they didn't have copper coins, they had to use larger denominations.Let me try this. Suppose Anya, Vanya, and Sanya each had 15 kopecks. They give all their money to one person, making 45 kopecks. That person pays the 15 kopecks fare and gets 30 kopecks back as change. Then, they distribute the 30 kopecks back to themselves, each getting 10 kopecks. So, each person spent 5 kopecks (15 given - 10 received = 5 spent). That makes sense!So, the solution is that they pooled their money together, one person paid the fare, and then the change was distributed back to them, allowing each to have effectively spent 5 kopecks without needing copper coins.

answer：Okay, so I have this problem here where a projectile is launched, and its height in feet is given by the equation y = -16t² + 64t. I need to find out at what times t the projectile will reach 25 feet and 49 feet in height. The answers should be decimals rounded to the nearest tenth. Hmm, okay, let me think about how to approach this.First, I remember that this is a quadratic equation in terms of t. The general form of a quadratic equation is y = at² + bt + c. In this case, a is -16, b is 64, and c is 0 because there's no constant term. So, the equation is y = -16t² + 64t.Now, I need to find the times t when the height y is 25 feet and 49 feet. That means I need to solve for t when y = 25 and when y = 49. For each of these, I'll set up the equation and solve the quadratic for t.Let me start with y = 25 feet. So, substituting 25 into the equation:25 = -16t² + 64tHmm, okay, so I can rearrange this equation to standard quadratic form. That means moving all terms to one side so that it equals zero. Let me subtract 25 from both sides:0 = -16t² + 64t - 25Alternatively, I can write it as:-16t² + 64t - 25 = 0But usually, it's easier to work with quadratics where the coefficient of t² is positive. So, I can multiply both sides of the equation by -1 to make that happen:16t² - 64t + 25 = 0Alright, now I have a standard quadratic equation: 16t² - 64t + 25 = 0. To solve for t, I can use the quadratic formula, which is t = [-b ± √(b² - 4ac)] / (2a). In this equation, a = 16, b = -64, and c = 25.Plugging these values into the quadratic formula:t = [-(-64) ± √((-64)² - 4 * 16 * 25)] / (2 * 16)Simplifying the numerator first:-(-64) is 64.Now, let's compute the discriminant, which is the part under the square root:(-64)² = 40964 * 16 * 25 = 1600So, the discriminant is 4096 - 1600 = 2496Therefore, the equation becomes:t = [64 ± √2496] / 32Now, I need to compute √2496. Let me see, 2496 is not a perfect square, so I'll need to approximate it. Let me try to find a number that when squared gives close to 2496.I know that 49² = 2401 and 50² = 2500. So, √2496 is just a bit less than 50. Let me compute 49.96²:49.96 * 49.96 = (50 - 0.04)² = 50² - 2*50*0.04 + 0.04² = 2500 - 4 + 0.0016 = 2496.0016Wow, that's really close to 2496. So, √2496 ≈ 49.96So, plugging that back into the equation:t = [64 ± 49.96] / 32Now, let's compute both possibilities:First, with the plus sign:t = (64 + 49.96) / 32 = 113.96 / 32 ≈ 3.561Second, with the minus sign:t = (64 - 49.96) / 32 = 14.04 / 32 ≈ 0.439So, the two times when the projectile is at 25 feet are approximately 0.439 seconds and 3.561 seconds. Since the problem asks for the times rounded to the nearest tenth, I need to round these numbers.0.439 rounded to the nearest tenth is 0.4 (since the hundredth digit is 3, which is less than 5, we don't round up).3.561 rounded to the nearest tenth is 3.6 (since the hundredth digit is 6, which is 5 or more, we round up the tenth digit from 5 to 6).Okay, so for y = 25 feet, the projectile is at that height at approximately 0.4 seconds and 3.6 seconds.Now, moving on to y = 49 feet. Let's do the same process.Starting with the equation:49 = -16t² + 64tAgain, rearranging to standard quadratic form:0 = -16t² + 64t - 49Multiplying both sides by -1 to make the coefficient of t² positive:16t² - 64t + 49 = 0So, now we have another quadratic equation: 16t² - 64t + 49 = 0. Again, using the quadratic formula:t = [-b ± √(b² - 4ac)] / (2a)Here, a = 16, b = -64, and c = 49.Plugging these into the formula:t = [-(-64) ± √((-64)² - 4 * 16 * 49)] / (2 * 16)Simplifying:t = [64 ± √(4096 - 3136)] / 32Calculating the discriminant:4096 - 3136 = 960So, √960. Let me see, 30² = 900 and 31² = 961. So, √960 is just a bit less than 31. Let me compute 30.98²:30.98 * 30.98 = (31 - 0.02)² = 31² - 2*31*0.02 + 0.02² = 961 - 1.24 + 0.0004 = 959.7604That's very close to 960. So, √960 ≈ 30.98So, plugging back into the equation:t = [64 ± 30.98] / 32Calculating both possibilities:First, with the plus sign:t = (64 + 30.98) / 32 = 94.98 / 32 ≈ 2.968Second, with the minus sign:t = (64 - 30.98) / 32 = 33.02 / 32 ≈ 1.032So, the two times when the projectile is at 49 feet are approximately 1.032 seconds and 2.968 seconds. Rounding these to the nearest tenth:1.032 rounded to the nearest tenth is 1.0 (since the hundredth digit is 3, which is less than 5, we don't round up).2.968 rounded to the nearest tenth is 3.0 (since the hundredth digit is 6, which is 5 or more, we round up the tenth digit from 9 to 10, which makes it 3.0).Wait, hold on, 2.968 is approximately 3.0 when rounded to the nearest tenth? Let me double-check that. The tenths place is 9, the hundredths place is 6. Since 6 is greater than or equal to 5, we round up the tenths place. But 9 + 1 is 10, so we carry over, making it 3.0. Yes, that's correct.So, for y = 49 feet, the projectile is at that height at approximately 1.0 seconds and 3.0 seconds.Let me just recap to make sure I didn't make any mistakes. For y = 25, I set up the equation, rearranged it, used the quadratic formula, calculated the discriminant, approximated the square roots, solved for t, and rounded. The same process for y = 49. The calculations seem correct, and the rounding makes sense.I also remember that in projectile motion, the projectile goes up, reaches a maximum height, and then comes back down. So, for each height below the maximum, there are two times: one on the way up and one on the way down. That's why we have two solutions for each y value.Just to be thorough, maybe I should check if 25 and 49 feet are indeed below the maximum height of the projectile. Let me find the maximum height to confirm.The maximum height occurs at the vertex of the parabola. The time at which the maximum height is reached is given by t = -b/(2a). In the original equation, a = -16 and b = 64.So, t = -64/(2*(-16)) = -64/(-32) = 2 seconds.So, at t = 2 seconds, the projectile is at its maximum height. Let me compute y at t = 2:y = -16*(2)² + 64*(2) = -16*4 + 128 = -64 + 128 = 64 feet.So, the maximum height is 64 feet. Therefore, both 25 feet and 49 feet are below the maximum height, which makes sense why there are two times for each.Just to double-check, let me plug t = 0.4 into the original equation and see if y is approximately 25.y = -16*(0.4)² + 64*(0.4) = -16*(0.16) + 25.6 = -2.56 + 25.6 = 23.04 feet.Hmm, that's not exactly 25 feet. Wait, did I do something wrong? Let me check my calculations.Wait, 0.4 squared is 0.16, multiplied by -16 is -2.56. 64 times 0.4 is 25.6. So, -2.56 + 25.6 is indeed 23.04. That's not 25. So, maybe my rounding was too aggressive?Wait, the exact value was approximately 0.439 seconds. Let me compute y at t = 0.439.y = -16*(0.439)² + 64*(0.439)First, compute (0.439)^2: 0.439 * 0.439 ≈ 0.192Then, -16 * 0.192 ≈ -3.07264 * 0.439 ≈ 28.1So, y ≈ -3.072 + 28.1 ≈ 25.028 feet, which is approximately 25 feet. Okay, so the exact value at t ≈ 0.439 is about 25 feet. So, rounding 0.439 to 0.4 is a bit off, but since the question asks for rounding to the nearest tenth, 0.4 is correct.Similarly, let's check t = 3.6:y = -16*(3.6)^2 + 64*(3.6)3.6 squared is 12.96-16 * 12.96 = -207.3664 * 3.6 = 230.4So, y = -207.36 + 230.4 = 23.04 feet. Again, that's not 25. Wait, that's the same as before. Hmm, that seems odd.Wait, no, actually, at t = 3.6, which is on the way down, the projectile should be descending past 25 feet again. But according to the calculation, it's 23.04 feet, which is below 25. That suggests that maybe my rounding is causing some discrepancy.Wait, let me compute y at t = 3.561, which was the exact value before rounding.y = -16*(3.561)^2 + 64*(3.561)First, compute (3.561)^2: 3.561 * 3.561 ≈ 12.68Then, -16 * 12.68 ≈ -202.8864 * 3.561 ≈ 228.224So, y ≈ -202.88 + 228.224 ≈ 25.344 feet, which is approximately 25.34 feet. So, that's close to 25.34, which rounds to 25.3, but since we're rounding t to the nearest tenth, 3.6 is acceptable because at t = 3.6, it's slightly below 25, but the exact time when it's exactly 25 is 3.561, which is closer to 3.6 than to 3.5.Similarly, for t = 0.439, it's approximately 25.028, which is just above 25, so rounding to 0.4 is okay.I think the slight discrepancy is because when we round t to the nearest tenth, the corresponding y value might not be exactly 25, but it's the closest tenth. So, it's acceptable.Similarly, let me check y at t = 1.0 and t = 3.0 for 49 feet.First, t = 1.0:y = -16*(1.0)^2 + 64*(1.0) = -16 + 64 = 48 feet.Hmm, that's 48 feet, not 49. Wait, but the exact value was t ≈ 1.032.Let me compute y at t = 1.032:y = -16*(1.032)^2 + 64*(1.032)First, (1.032)^2 ≈ 1.065-16 * 1.065 ≈ -17.0464 * 1.032 ≈ 66.048So, y ≈ -17.04 + 66.048 ≈ 49.008 feet, which is approximately 49.01 feet. So, that's very close to 49 feet. So, rounding t to 1.0 is acceptable because at t = 1.032, it's just over 49, and the nearest tenth is 1.0.Similarly, at t = 3.0:y = -16*(3.0)^2 + 64*(3.0) = -16*9 + 192 = -144 + 192 = 48 feet.Again, 48 feet, which is just below 49. But the exact time when y = 49 is t ≈ 2.968, which is approximately 3.0 when rounded to the nearest tenth.So, even though at t = 3.0, y is 48 feet, the exact time when y = 49 is just before 3.0, at 2.968, which is closer to 3.0 than to 2.9. Therefore, rounding to 3.0 is correct.I think that makes sense. So, the rounding is based on the t value, not the y value. So, even if the y value at the rounded t is slightly off, the t value itself is correctly rounded.Therefore, my final answers are:For y = 25 feet: t ≈ 0.4 seconds and t ≈ 3.6 seconds.For y = 49 feet: t ≈ 1.0 seconds and t ≈ 3.0 seconds.I think that's all. I don't see any mistakes in my calculations, and the checks confirm that the rounded times are appropriate.

answer：Okay, so I have this problem with four propositions about lines and planes in space. I need to figure out how many of them are true. Let me go through each one step by step.**Proposition (1): Two lines perpendicular to the same plane are parallel.**Hmm, I remember that in three-dimensional geometry, if two lines are both perpendicular to the same plane, they must be parallel to each other. This is because the direction of each line is exactly the same as the normal vector of the plane. So, they can't intersect and must be parallel. I think this one is true.**Proposition (2): If lines a and b are skew lines, then through any point P in space, there can always be drawn a line that intersects both line a and line b.**Skew lines are lines that don't lie in the same plane and don't intersect. So, if I have two skew lines, a and b, and I pick any point P in space, can I always draw a line through P that intersects both a and b? Wait, I'm not sure about this. If P is in a certain position, maybe it's not possible. For example, if P is on one of the lines, say line a, then the line through P can only intersect a at P, but can it intersect b? Since a and b are skew, they don't intersect, so the line through P can't intersect both unless it's specifically constructed. But actually, I think there's a concept called a transversal line that intersects both skew lines. However, not every point P can have such a line. If P is in a specific position relative to the plane containing a and b, maybe it's possible, but if P is in a different position, it might not be. I think this proposition is false because there might be some points P where you can't draw such a line. Maybe when P is on a certain side or in a certain plane, it's not possible.**Proposition (3): If a line is parallel to a plane, then it is parallel to any line within the plane.**Okay, so if a line is parallel to a plane, does that mean it's parallel to every line in that plane? I don't think so. Being parallel to a plane means the line doesn't intersect the plane and lies in a direction that's parallel to the plane. But within the plane, there can be lines in various directions. For example, imagine a horizontal line above a horizontal plane. The line is parallel to the plane, but within the plane, there are vertical lines, diagonal lines, etc. The horizontal line isn't parallel to all of them; it's only parallel to lines that have the same direction. So, this proposition is false.**Proposition (4): If a line is perpendicular to a plane, then this line is perpendicular to any line within the plane.**Alright, if a line is perpendicular to a plane, it means it's perpendicular to every line in that plane. I remember that the definition of a line being perpendicular to a plane is that it's perpendicular to every line in the plane. So, this should be true.**Summary:**1. True2. False3. False4. TrueSo, out of the four propositions, two are true. Therefore, the answer should be B: 2.**Final Answer**boxed{B}