answer：Okay, so I've got this problem about a water channel, and I need to find the width at the bottom. Let me see... The cross-section is a trapezium, which I think is the same as a trapezoid in some countries. So, a trapezium has two parallel sides, right? In this case, the top and the bottom of the channel.The problem says the top width is 14 meters, and the area of the cross-section is 880 square meters. The depth, which I assume is the height of the trapezium, is 80 meters. I need to find the width at the bottom.Alright, so I remember the formula for the area of a trapezium is something like the average of the two parallel sides multiplied by the height. Let me write that down to make sure I have it right.Area = (1/2) * (sum of the two parallel sides) * heightSo, in this case, the area is 880, the height is 80, and one of the sides is 14. Let's call the bottom width 'b'. Plugging these into the formula:880 = (1/2) * (14 + b) * 80Hmm, okay, let's simplify this step by step. First, I can simplify the multiplication by 1/2 and 80. Let me calculate 1/2 of 80 first.1/2 * 80 = 40So now the equation becomes:880 = 40 * (14 + b)Alright, now I need to solve for 'b'. Let's divide both sides by 40 to isolate the term with 'b'.880 / 40 = (14 + b)Calculating 880 divided by 40... Let's see, 40 times 22 is 880, right? So, 880 / 40 = 22So now we have:22 = 14 + bTo find 'b', I just need to subtract 14 from both sides.22 - 14 = bWhich gives:8 = bSo, the width at the bottom of the channel is 8 meters.Wait, let me double-check that to make sure I didn't make any mistakes. Starting from the area formula:Area = (1/2) * (top + bottom) * heightPlugging in the values:880 = (1/2) * (14 + 8) * 80Calculating inside the parentheses first:14 + 8 = 22Then multiply by 1/2:(1/2) * 22 = 11Now multiply by the height, 80:11 * 80 = 880Yes, that matches the given area. So, my calculation seems correct. The width at the bottom is indeed 8 meters.I think that's it. It makes sense because the area formula worked out when I plugged the numbers back in, so I must have done it right.

answer：Alright, so I'm trying to figure out how many moles of hydrochloric acid (HCl) are formed when I combine 3 moles of sodium chloride (NaCl) and 3 moles of nitric acid (HNO3). Okay, let's start by recalling what I know about chemical reactions.First, I remember that when an acid reacts with a salt, it can produce another acid and another salt. In this case, sodium chloride is a salt, and nitric acid is a strong acid. So, maybe they can react to form hydrochloric acid and some other product.Let me write down the possible chemical equation for this reaction. If NaCl reacts with HNO3, the products might be NaNO3 (sodium nitrate) and HCl (hydrochloric acid). So, the unbalanced equation would look like this:NaCl + HNO3 → NaNO3 + HClNow, I need to balance this equation to make sure the number of atoms of each element is the same on both sides. Let's see:On the left side, I have:- 1 Na atom in NaCl- 1 Cl atom in NaCl- 1 H atom in HNO3- 1 N atom in HNO3- 3 O atoms in HNO3On the right side, I have:- 1 Na atom in NaNO3- 1 N atom in NaNO3- 3 O atoms in NaNO3- 1 H atom in HCl- 1 Cl atom in HClIt looks like the equation is already balanced as written. So, the balanced chemical equation is:NaCl + HNO3 → NaNO3 + HClOkay, now that I have the balanced equation, I can use stoichiometry to find out how many moles of HCl are produced when 3 moles of NaCl and 3 moles of HNO3 react.From the balanced equation, the mole ratio of NaCl to HCl is 1:1, and the mole ratio of HNO3 to HCl is also 1:1. That means 1 mole of NaCl reacts with 1 mole of HNO3 to produce 1 mole of HCl.Since I have 3 moles of NaCl and 3 moles of HNO3, and the mole ratio is 1:1:1, I would expect 3 moles of HCl to be produced.But wait, I should double-check if there's any limiting reagent here. Both reactants are present in equal moles (3 moles each), and the reaction requires them in a 1:1 ratio. So, neither is in excess; both are completely consumed in the reaction. That means the theoretical yield of HCl is indeed 3 moles.However, I'm a bit confused because I've heard that nitric acid is a strong oxidizing agent, and sometimes reactions involving nitric acid can be more complex, especially with salts like NaCl. Maybe there's more to this reaction than just a simple acid-base reaction?Let me think about this. Nitric acid is a strong acid, and sodium chloride is a neutral salt. In a typical acid-base reaction, the acid donates protons (H+ ions) to the base. But in this case, NaCl is not a base; it's a neutral salt. So, maybe the reaction doesn't proceed as straightforwardly as I thought.Perhaps the reaction between NaCl and HNO3 is not just a simple exchange of ions but involves some redox chemistry. Nitric acid can act as an oxidizing agent, and sodium chloride can act as a reducing agent under certain conditions. This might lead to the formation of different products, such as chlorine gas (Cl2) or other compounds, instead of just HCl.If that's the case, then the stoichiometry might change, and the amount of HCl produced could be different. I need to look into this further.Let me try to write a possible redox reaction between NaCl and HNO3. In such reactions, nitric acid is often reduced, and chloride ions are oxidized. For example, in a reaction where nitric acid reacts with sodium chloride, the products could include sodium nitrate (NaNO3), water (H2O), and chlorine gas (Cl2).Here's a possible unbalanced equation:NaCl + HNO3 → NaNO3 + H2O + Cl2↑Now, let's balance this equation. First, I'll balance the sodium (Na) atoms:NaCl → NaNO3So, I need to balance Na: 1 Na on both sides.Next, balance the nitrogen (N) atoms:HNO3 → NaNO3There is 1 N on both sides, so that's balanced.Now, balance the oxygen (O) atoms. On the left, there are 3 O in HNO3, and on the right, there are 3 O in NaNO3 and 1 O in H2O, totaling 4 O. To balance, I'll need to adjust the coefficients.Let's try:NaCl + HNO3 → NaNO3 + H2O + Cl2↑To balance O, I can adjust the coefficients of H2O. Let's see:If I have 1 HNO3 on the left, it provides 3 O. On the right, NaNO3 has 3 O, and H2O has 1 O, totaling 4 O. To balance, I need to have 4 O on the left. Since HNO3 provides 3 O, I need another O somewhere. Maybe I need to add H2O on the left?Wait, that might complicate things. Alternatively, I can adjust the coefficients to make the O atoms balance.Let's try:NaCl + 3 HNO3 → NaNO3 + H2O + Cl2↑Now, on the left, we have 3 HNO3, which provides 9 O atoms. On the right, NaNO3 has 3 O, H2O has 1 O, and Cl2 has none. That's a total of 4 O on the right, which doesn't match the 9 O on the left. Hmm, that's not balanced.Maybe I need to adjust the coefficients differently. Let's try:NaCl + 4 HNO3 → NaNO3 + 2 H2O + Cl2↑Now, on the left, 4 HNO3 provides 12 O atoms. On the right, NaNO3 has 3 O, 2 H2O have 2 O, totaling 5 O. Still not balanced.This is getting tricky. Maybe I need to use a different approach. Let's try to balance the equation step by step.First, write the skeleton equation:NaCl + HNO3 → NaNO3 + H2O + Cl2↑Now, balance the Na atoms:1 Na on both sides, so that's balanced.Next, balance the Cl atoms:1 Cl on the left (in NaCl) and 2 Cl on the right (in Cl2). To balance, I'll need 2 NaCl on the left:2 NaCl + HNO3 → NaNO3 + H2O + Cl2↑Now, balance the N atoms:1 N in HNO3 and 1 N in NaNO3. So, that's balanced.Next, balance the O atoms:On the left, HNO3 has 3 O. On the right, NaNO3 has 3 O and H2O has 1 O, totaling 4 O. To balance, I'll need to adjust the coefficients.Let's try:2 NaCl + 4 HNO3 → 2 NaNO3 + 2 H2O + Cl2↑Now, on the left, 4 HNO3 provides 12 O atoms. On the right, 2 NaNO3 have 6 O, and 2 H2O have 2 O, totaling 8 O. Still not balanced.Maybe I need to adjust the coefficients further. Let's try:2 NaCl + 6 HNO3 → 2 NaNO3 + 3 H2O + Cl2↑Now, on the left, 6 HNO3 provides 18 O atoms. On the right, 2 NaNO3 have 6 O, and 3 H2O have 3 O, totaling 9 O. Still not balanced.This is getting complicated. Maybe I need to use a different method, like the oxidation number method, to balance the redox reaction.Let's identify the oxidation numbers:In NaCl:- Na: +1- Cl: -1In HNO3:- H: +1- N: +5- O: -2In NaNO3:- Na: +1- N: +5- O: -2In H2O:- H: +1- O: -2In Cl2:- Cl: 0So, the changes in oxidation numbers are:- Cl: -1 → 0 (oxidation)- N: +5 → +5 (no change)Wait, that means there's no change in oxidation number for nitrogen. So, maybe this isn't a redox reaction after all. That simplifies things.If there's no redox reaction, then the initial assumption that NaCl and HNO3 simply exchange ions to form NaNO3 and HCl might be correct.So, going back to the original balanced equation:NaCl + HNO3 → NaNO3 + HClWith a 1:1:1:1 mole ratio.Given that I have 3 moles of NaCl and 3 moles of HNO3, and the reaction requires them in a 1:1 ratio, the limiting reagent is neither; both are completely consumed.Therefore, the theoretical yield of HCl is 3 moles.But earlier, I was confused about the possibility of a redox reaction. It seems like in this case, since there's no change in oxidation numbers for nitrogen, it's not a redox reaction, and the simple acid-base reaction holds.So, to summarize:- Balanced equation: NaCl + HNO3 → NaNO3 + HCl- Mole ratio: 1:1:1:1- Moles of NaCl: 3- Moles of HNO3: 3- Limiting reagent: Neither; both are fully consumed- Moles of HCl produced: 3Therefore, 3 moles of hydrochloric acid are formed when 3 moles of sodium chloride and 3 moles of nitric acid are combined.

answer：First, I need to determine the individual work rates of A and B. A can complete the work in 20 days, so A's work rate is ( frac{1}{20} ) of the work per day. Similarly, B can complete the work in 12 days, so B's work rate is ( frac{1}{12} ) of the work per day.When A and B work together, their combined work rate is the sum of their individual rates:[frac{1}{20} + frac{1}{12} = frac{3}{60} + frac{5}{60} = frac{8}{60} = frac{2}{15} text{ of the work per day}]Let ( x ) be the number of days A and B work together. During these ( x ) days, they complete:[frac{2}{15}x text{ of the work}]After A leaves, B works alone for 3 more days. In these 3 days, B completes:[frac{1}{12} times 3 = frac{1}{4} text{ of the work}]The total work completed by both A and B together plus the work done by B alone should equal the entire work:[frac{2}{15}x + frac{1}{4} = 1]To solve for ( x ), I'll first subtract ( frac{1}{4} ) from both sides:[frac{2}{15}x = 1 - frac{1}{4} = frac{3}{4}]Next, I'll solve for ( x ) by multiplying both sides by ( frac{15}{2} ):[x = frac{3}{4} times frac{15}{2} = frac{45}{8} = 5.625 text{ days}]Since we typically express the number of days in whole numbers, I'll round down to the nearest whole number:[x = 5 text{ days}]

answer：Okay, so I have this arithmetic sequence problem here. It says that in the arithmetic sequence {a_n}, we have a_5 = 3 and a_6 = -2. I need to find the sum from a_3 to a_8. Hmm, arithmetic sequences, right? They have a common difference between consecutive terms. Let me try to recall how these work.First, since it's an arithmetic sequence, the difference between any two consecutive terms is constant. Let me denote this common difference as d. So, the formula for the nth term of an arithmetic sequence is a_n = a_1 + (n-1)d, where a_1 is the first term.Given that a_5 = 3 and a_6 = -2, I can set up two equations using the formula. Let's write them out:a_5 = a_1 + 4d = 3 a_6 = a_1 + 5d = -2Now, if I subtract the first equation from the second, I can eliminate a_1 and solve for d:(a_1 + 5d) - (a_1 + 4d) = -2 - 3 Simplifying that, I get:a_1 + 5d - a_1 - 4d = -5 Which simplifies to:d = -5So, the common difference d is -5. That means each term is 5 less than the previous term.Now, I need to find the sum from a_3 to a_8. Let me note that this is the sum of six terms: a_3, a_4, a_5, a_6, a_7, a_8. To find this sum, I can use the formula for the sum of an arithmetic series, which is S = n/2 * (first term + last term), where n is the number of terms.But before I can use that, I need to know the values of a_3 and a_8. Alternatively, I could express all these terms in terms of a_5 and a_6, which I already know.Wait, since I know a_5 and a_6, maybe I can express a_3 and a_8 in terms of these. Let me think.Since a_5 is the 5th term, a_3 is two terms before that. So, a_3 = a_5 - 2d. Similarly, a_8 is three terms after a_5, so a_8 = a_5 + 3d.Let me compute a_3:a_3 = a_5 - 2d = 3 - 2*(-5) = 3 + 10 = 13And a_8:a_8 = a_5 + 3d = 3 + 3*(-5) = 3 - 15 = -12So, the first term in the sum is 13, and the last term is -12. The number of terms is 6, as I thought earlier.Now, applying the sum formula:S = n/2 * (first term + last term) S = 6/2 * (13 + (-12)) S = 3 * (1) S = 3Wait, that seems straightforward. Alternatively, I could think about pairing terms. Since it's an arithmetic sequence, the sum of terms equidistant from the ends are equal. So, a_3 + a_8 = a_4 + a_7 = a_5 + a_6.Given that, since there are six terms, there are three such pairs. Each pair sums to a_5 + a_6, which is 3 + (-2) = 1. Therefore, the total sum is 3 * 1 = 3.Either way, whether I calculate each term individually or use the pairing method, I get the same result. So, the sum from a_3 to a_8 is 3.I think that makes sense. Let me just verify by calculating each term:a_5 = 3 a_6 = -2 a_7 = a_6 + d = -2 + (-5) = -7 a_8 = a_7 + d = -7 + (-5) = -12 a_4 = a_5 - d = 3 - (-5) = 8 a_3 = a_4 - d = 8 - (-5) = 13So, the terms are: 13, 8, 3, -2, -7, -12.Adding them up: 13 + 8 = 21; 21 + 3 = 24; 24 + (-2) = 22; 22 + (-7) = 15; 15 + (-12) = 3.Yep, that adds up to 3. So, I'm confident that the sum is indeed 3.**Final Answer**The sum (a_3 + a_4 + ldots + a_8) is boxed{3}.

answer：Okay, so I have this problem about an even function f(x) defined on all real numbers. It says that f(0) is 0, and when x is greater than 0, f(x) is log base 1/2 of x. There are two parts: first, I need to find the expression for f(x), and second, solve the inequality f(x² - 1) > -2. Hmm, let's start with part (1).Alright, since f(x) is an even function, that means f(-x) = f(x) for all x. So, if I know what f(x) is when x is positive, I can figure out what it is when x is negative by just mirroring it. Also, f(0) is given as 0. So, for x > 0, f(x) is log base 1/2 of x. Let me write that down.For x > 0: f(x) = log_{1/2}(x)Since it's even, for x < 0: f(x) = f(-x) = log_{1/2}(-x)And at x = 0: f(0) = 0So, putting that all together, the function f(x) is piecewise defined. Let me write that out:f(x) = { log_{1/2}(x) if x > 0, 0 if x = 0, log_{1/2}(-x) if x < 0 }Okay, that seems straightforward. Now, moving on to part (2): solving the inequality f(x² - 1) > -2.First, I need to understand what f(x² - 1) is. Since f is an even function, f(x² - 1) is the same as f(|x² - 1|). Because x² - 1 can be positive or negative, but since f is even, it doesn't matter. So, f(x² - 1) = f(|x² - 1|).So, the inequality becomes f(|x² - 1|) > -2.Now, I need to figure out when f(y) > -2, where y = |x² - 1|. Since y is always non-negative, and f(y) is defined as log_{1/2}(y) for y > 0, and f(0) = 0.Wait, so f(y) = log_{1/2}(y) when y > 0, and f(0) = 0. So, let's consider y = |x² - 1|.So, f(y) > -2 translates to log_{1/2}(y) > -2, because y is non-negative, and when y = 0, f(y) = 0, which is greater than -2. So, we have two cases:Case 1: y = 0. Then f(y) = 0 > -2, which is true.Case 2: y > 0. Then log_{1/2}(y) > -2.So, let's solve log_{1/2}(y) > -2.Remember that log base 1/2 is a decreasing function. So, when the log is greater than -2, the argument y must be less than (1/2)^{-2}.Calculating (1/2)^{-2} is the same as 2^2, which is 4. So, log_{1/2}(y) > -2 implies y < 4.But y is |x² - 1|, so |x² - 1| < 4.So, now we have |x² - 1| < 4.Let me solve this inequality.First, |x² - 1| < 4 means that -4 < x² - 1 < 4.Adding 1 to all parts: -4 + 1 < x² < 4 + 1, so -3 < x² < 5.But x² is always non-negative, so -3 < x² is automatically true. So, the inequality reduces to x² < 5.Therefore, x² < 5 implies that |x| < sqrt(5), so x is between -sqrt(5) and sqrt(5).But wait, let me double-check.Starting from |x² - 1| < 4.Which is equivalent to -4 < x² - 1 < 4.Adding 1: -3 < x² < 5.Since x² is always ≥ 0, the left inequality is always true, so we only need x² < 5.So, x² < 5 implies x ∈ (-sqrt(5), sqrt(5)).But hold on, is that the only solution? Let me think.Wait, when y = |x² - 1|, y can be 0 or positive. So, in the case when y = 0, which is when x² - 1 = 0, so x² = 1, so x = ±1. Then f(y) = f(0) = 0, which is greater than -2.So, x = ±1 are included in the solution set.But when y > 0, we have y < 4, which gives |x² - 1| < 4, leading to x² < 5, so x ∈ (-sqrt(5), sqrt(5)).So, combining both cases, the solution is x ∈ (-sqrt(5), sqrt(5)).Wait, but hold on, when y = |x² - 1|, and y < 4, that gives x² - 1 < 4 and x² - 1 > -4.So, x² - 1 < 4 gives x² < 5, and x² - 1 > -4 gives x² > -3, which is always true since x² is non-negative.So, the only restriction is x² < 5, so x ∈ (-sqrt(5), sqrt(5)).Therefore, the solution set is all real numbers x such that |x| < sqrt(5).So, in interval notation, that's (-sqrt(5), sqrt(5)).Wait, but let me confirm if I considered all cases correctly.When y = |x² - 1|, y is always non-negative. So, f(y) is log_{1/2}(y) when y > 0, and 0 when y = 0.So, f(y) > -2.Case 1: y = 0. Then f(y) = 0 > -2, which is true. So, x² - 1 = 0, so x = ±1.Case 2: y > 0. Then log_{1/2}(y) > -2. As log_{1/2}(y) is decreasing, this implies y < (1/2)^{-2} = 4.So, y = |x² - 1| < 4.So, |x² - 1| < 4 implies -4 < x² - 1 < 4, which gives -3 < x² < 5. Since x² is always ≥ 0, x² < 5, so x ∈ (-sqrt(5), sqrt(5)).But wait, when y = |x² - 1|, it's possible that y could be greater than or equal to 4, but in that case, f(y) would be log_{1/2}(y) ≤ -2, which doesn't satisfy the inequality. So, we only consider y < 4.Therefore, combining both cases, the solution is x ∈ (-sqrt(5), sqrt(5)).But hold on, when y = 0, x = ±1, which are within (-sqrt(5), sqrt(5)) since sqrt(5) is approximately 2.236, and 1 is less than that. So, the entire interval includes x = ±1.Therefore, the solution set is indeed (-sqrt(5), sqrt(5)).I think that's it. So, to recap:1. The function f(x) is even, so f(x) = log_{1/2}(|x|) for x ≠ 0, and f(0) = 0.2. For the inequality f(x² - 1) > -2, we transformed it to f(|x² - 1|) > -2, which led us to |x² - 1| < 4, resulting in x ∈ (-sqrt(5), sqrt(5)).I don't see any mistakes in this reasoning, so I think that's the correct solution.

answer：First, I recognize that the ratio of red balls to blue balls is 3:7. This means for every 3 red balls, there are 7 blue balls.Next, I calculate the number of tennis balls among the red balls. Since 70% of the red balls are tennis balls, I multiply the number of red balls by 0.7.Similarly, I calculate the number of tennis balls among the blue balls. Given that 30% of the blue balls are tennis balls, I multiply the number of blue balls by 0.3.To find the total number of tennis balls, I add the number of tennis balls from the red balls and the blue balls together.Finally, I express the total number of tennis balls in terms of the number of red balls, resulting in the total number of tennis balls being 1.4 times the number of red balls.