answer：First, I recognize that the coordinates of point ( A ) relative to point ( B ) are given as ((-3, 4)). This means that point ( A ) is positioned 3 units to the left and 4 units above point ( B ).To find the coordinates of point ( B ) relative to point ( A ), I need to reverse the directions. If point ( A ) is now the origin, point ( B ) will be positioned in the opposite directions: 3 units to the right and 4 units below point ( A ).Therefore, the coordinates of point ( B ) when point ( A ) is the origin are ((3, -4)).

answer：Okay, so I have this quadratic equation: 3x² + ax + 12x + 16 = 0. The problem is asking for the sum of all values of 'a' that make this equation have only one solution. Hmm, okay. Let me try to figure this out step by step.First, I remember that a quadratic equation has only one solution when its discriminant is zero. The discriminant formula is b² - 4ac. If that equals zero, the quadratic has exactly one real root. So, I need to find the discriminant of this equation and set it equal to zero, then solve for 'a'.But before I can do that, I think I should combine like terms in the equation. I see that there are two terms with 'x': ax and 12x. So, I can combine those. Let me rewrite the equation:3x² + (a + 12)x + 16 = 0.Okay, so now it's in the standard quadratic form: Ax² + Bx + C = 0, where A is 3, B is (a + 12), and C is 16.Now, the discriminant D is B² - 4AC. Plugging in the values we have:D = (a + 12)² - 4 * 3 * 16.Let me compute that:First, expand (a + 12)². That would be a² + 24a + 144.Then, compute 4 * 3 * 16. 4 times 3 is 12, and 12 times 16 is 192.So, the discriminant D becomes:D = a² + 24a + 144 - 192.Simplify that:a² + 24a - 48.So, for the equation to have only one solution, D must be zero:a² + 24a - 48 = 0.Now, I have a quadratic equation in terms of 'a'. I need to solve for 'a'. I can use the quadratic formula, which is:a = [-b ± √(b² - 4ac)] / (2a).Wait, hold on, that might be confusing because the variable here is also 'a'. Let me make sure I don't mix up the variables. In the quadratic formula, the coefficients are usually denoted as A, B, C, so in this case, for the equation a² + 24a - 48 = 0, the coefficients are:A = 1 (coefficient of a²),B = 24 (coefficient of a),C = -48 (constant term).So, applying the quadratic formula:a = [-B ± √(B² - 4AC)] / (2A).Plugging in the values:a = [-24 ± √(24² - 4 * 1 * (-48))] / (2 * 1).Compute the discriminant inside the square root:24² is 576,4 * 1 * (-48) is -192,So, 576 - (-192) is 576 + 192, which is 768.So, the equation becomes:a = [-24 ± √768] / 2.Simplify √768. Let me see, 768 divided by 256 is 3, so √768 = √(256 * 3) = 16√3.So, now we have:a = [-24 ± 16√3] / 2.Simplify the fractions:-24 divided by 2 is -12,16√3 divided by 2 is 8√3.So, the two solutions are:a = -12 + 8√3,anda = -12 - 8√3.Now, the problem asks for the sum of these values of 'a'. So, let's add them together:(-12 + 8√3) + (-12 - 8√3).Combine like terms:-12 -12 = -24,8√3 - 8√3 = 0.So, the sum is -24.Wait, that's interesting. So, regardless of the values of 'a' that make the original equation have only one solution, their sum is -24. I think that makes sense because in a quadratic equation, the sum of the roots is -B/A. In this case, when we had the equation a² + 24a - 48 = 0, the sum of the roots (which are the values of 'a') is -24/1 = -24. So, that's another way to see it without having to compute the individual roots.So, either way, whether I compute the roots and add them or use the property of quadratic equations, I get the same result: -24.I think that's the answer. Let me just double-check my steps to make sure I didn't make any mistakes.1. Combined like terms: 3x² + (a + 12)x + 16 = 0. That seems correct.2. Calculated discriminant: (a + 12)² - 4*3*16 = a² + 24a + 144 - 192 = a² + 24a - 48. That looks right.3. Set discriminant to zero: a² + 24a - 48 = 0. Correct.4. Applied quadratic formula: a = [-24 ± √(576 + 192)] / 2 = [-24 ± √768]/2 = [-24 ± 16√3]/2 = -12 ± 8√3. That seems correct.5. Sum of the roots: (-12 + 8√3) + (-12 - 8√3) = -24. Correct.Alternatively, using the sum of roots formula: for equation a² + 24a - 48 = 0, sum of roots is -24/1 = -24. Yep, that's consistent.So, I feel confident that the sum of the values of 'a' is -24.**Final Answer**The sum of the values of ( a ) is boxed{-24}.

answer：First, I identify the total number of shirts the store initially has, which is 49.Next, I note that after selling some shirts, there are 28 shirts remaining.To find out how many shirts were sold, I subtract the number of shirts left from the initial number of shirts.So, 49 shirts minus 28 shirts equals 21 shirts sold.Therefore, the store sold 21 shirts.

answer：First, I'll denote the number of kilometers Natalia rode on Tuesday as ( T ).On Wednesday, she rode 50% fewer kilometers than on Tuesday, which means she rode ( frac{T}{2} ) kilometers.On Thursday, she rode as many kilometers as the sum of the kilometers from Monday and Wednesday, which is ( 40 + frac{T}{2} ) kilometers.The total distance she rode over the four days is the sum of the distances from Monday, Tuesday, Wednesday, and Thursday, which is given as 180 kilometers. So, I can write the equation:[40 , (text{Monday}) + T , (text{Tuesday}) + frac{T}{2} , (text{Wednesday}) + left(40 + frac{T}{2}right) , (text{Thursday}) = 180]Now, let's solve for ( T ):[40 + T + frac{T}{2} + 40 + frac{T}{2} = 180]Combine like terms:[80 + T + frac{T}{2} + frac{T}{2} = 180]Simplify the equation:[80 + 2T = 180]Subtract 80 from both sides:[2T = 100]Divide both sides by 2 to solve for ( T ):[T = 50]Therefore, Natalia rode (boxed{50}) kilometers on Tuesday.

answer：Okay, so I have these three parts to solve about unimodal functions. Let me start with part (1). First, I need to figure out if the functions f₁(x) = x - 2x² and f₂(x) = |log₂(x + 0.5)| are unimodal on the interval [0,1]. If they are, I have to find the peak point. If not, explain why.Starting with f₁(x) = x - 2x². Hmm, this looks like a quadratic function. Quadratic functions are usually parabolas, so they should have a single peak or valley. Since the coefficient of x² is negative (-2), it should open downward, meaning it has a maximum point, which would be the peak. So, f₁(x) should be unimodal.To find the peak point, I can take the derivative of f₁(x). The derivative f₁’(x) = 1 - 4x. Setting this equal to zero to find critical points: 1 - 4x = 0 => x = 1/4. Now, I need to check the behavior around x = 1/4. For x < 1/4, say x = 0, f₁’(0) = 1 > 0, so the function is increasing. For x > 1/4, say x = 1/2, f₁’(1/2) = 1 - 4*(1/2) = 1 - 2 = -1 < 0, so the function is decreasing. Therefore, f₁(x) is increasing on [0, 1/4] and decreasing on [1/4, 1], which makes it unimodal with the peak at x = 1/4.Now, moving on to f₂(x) = |log₂(x + 0.5)|. This function involves an absolute value of a logarithm. Let me analyze its behavior on [0,1].First, let's consider the expression inside the absolute value: log₂(x + 0.5). The logarithm function log₂(y) is defined for y > 0, which is true here since x + 0.5 ranges from 0.5 to 1.5 as x goes from 0 to 1.The logarithm log₂(y) is negative when y < 1 and positive when y > 1. So, for x + 0.5 < 1, which is x < 0.5, log₂(x + 0.5) is negative, and the absolute value will make it positive. For x + 0.5 ≥ 1, which is x ≥ 0.5, log₂(x + 0.5) is non-negative, so the absolute value doesn't change it.Therefore, f₂(x) can be written as:- For 0 ≤ x < 0.5: f₂(x) = -log₂(x + 0.5)- For 0.5 ≤ x ≤ 1: f₂(x) = log₂(x + 0.5)Now, let's find the derivative for each interval.For 0 ≤ x < 0.5:f₂(x) = -log₂(x + 0.5)Derivative: f₂’(x) = - [1 / ( (x + 0.5) ln 2 ) ] * (1) = -1 / [ (x + 0.5) ln 2 ]Since ln 2 is positive, and x + 0.5 is positive, f₂’(x) is negative. So, f₂(x) is decreasing on [0, 0.5).For 0.5 ≤ x ≤ 1:f₂(x) = log₂(x + 0.5)Derivative: f₂’(x) = 1 / [ (x + 0.5) ln 2 ]Again, ln 2 is positive, and x + 0.5 is positive, so f₂’(x) is positive. Thus, f₂(x) is increasing on [0.5, 1].So, f₂(x) is decreasing on [0, 0.5) and increasing on [0.5, 1]. That means it has a minimum at x = 0.5, not a maximum. Since a unimodal function should have a single peak (a maximum), but here we have a minimum, f₂(x) is not unimodal on [0,1]. It actually has a V-shape with the lowest point at x = 0.5.So, for part (1), only f₁(x) is unimodal with the peak at x = 1/4, and f₂(x) is not unimodal.Moving on to part (2). The function is f(x) = a x³ + x, where a < 0. We need to find the range of a such that f(x) is unimodal on [1,2].First, let's recall that a function is unimodal on [a,b] if it increases to a peak and then decreases. So, it should have exactly one critical point in (a,b), which is a maximum.To find critical points, take the derivative:f’(x) = 3a x² + 1Set f’(x) = 0:3a x² + 1 = 0 => x² = -1/(3a)Since a < 0, -1/(3a) is positive, so x² is positive, which gives real solutions. The critical points are x = sqrt(-1/(3a)) and x = -sqrt(-1/(3a)). But since we're considering the interval [1,2], we only care about positive x. So, the critical point is x = sqrt(-1/(3a)).Now, for f(x) to be unimodal on [1,2], this critical point must lie within (1,2). So, we need:1 < sqrt(-1/(3a)) < 2Let me solve this inequality step by step.First, sqrt(-1/(3a)) > 1:sqrt(-1/(3a)) > 1 => -1/(3a) > 1 => -1 > 3a => a > -1/3But since a < 0, this gives -1/3 < a < 0.Second, sqrt(-1/(3a)) < 2:sqrt(-1/(3a)) < 2 => -1/(3a) < 4 => -1 < 12a => a > -1/12Again, since a < 0, this gives -1/12 < a < 0.Combining both inequalities, we have:From the first inequality: a > -1/3From the second inequality: a > -1/12So, the stricter condition is a > -1/12. But also, a must be less than 0.Therefore, the range of a is (-1/12, 0). But wait, let me double-check.Wait, actually, the critical point x = sqrt(-1/(3a)) must lie in (1,2). So, when a is between -1/3 and -1/12, does x lie in (1,2)?Wait, let's compute sqrt(-1/(3a)):If a = -1/12, then sqrt(-1/(3*(-1/12))) = sqrt(1/(3*(1/12))) = sqrt(1/(1/4)) = sqrt(4) = 2.If a = -1/3, then sqrt(-1/(3*(-1/3))) = sqrt(1/1) = 1.So, when a is between -1/3 and -1/12, sqrt(-1/(3a)) is between 1 and 2. Therefore, the critical point lies in (1,2) when a is in (-1/3, -1/12).Wait, but earlier, when solving the inequalities, I got a > -1/12 and a > -1/3, which would mean a > -1/12. But actually, since a is negative, the interval is -1/3 < a < -1/12.Wait, let me think again.Given that a < 0, and we have:1 < sqrt(-1/(3a)) < 2So, sqrt(-1/(3a)) is between 1 and 2.Squaring all parts:1 < -1/(3a) < 4Multiply all parts by 3a, but since a is negative, the inequalities reverse when multiplying.So, 3a < -1 < 12aWait, that seems confusing. Let me approach it differently.Let me denote k = sqrt(-1/(3a)). Then, 1 < k < 2.So, 1 < k < 2 => 1 < sqrt(-1/(3a)) < 2Square all parts:1 < -1/(3a) < 4Multiply all parts by 3a, but since a is negative, the inequalities reverse.So, 3a < -1 < 12aWait, that can't be right because 3a < -1 and -1 < 12a.But 3a < -1 => a < -1/3And -1 < 12a => a > -1/12So, combining these, we have -1/12 < a < -1/3But wait, that's impossible because -1/12 is greater than -1/3. So, the interval would be empty.Wait, that can't be. I must have messed up the direction of the inequalities.Let me try another approach.From 1 < sqrt(-1/(3a)) < 2First, sqrt(-1/(3a)) > 1:=> -1/(3a) > 1Multiply both sides by 3a, but since a < 0, inequality reverses:-1 < 3a=> 3a > -1=> a > -1/3Second, sqrt(-1/(3a)) < 2:=> -1/(3a) < 4Multiply both sides by 3a, which is negative, so inequality reverses:-1 > 12a=> 12a < -1=> a < -1/12So, combining both results:a > -1/3 and a < -1/12Thus, the range is (-1/3, -1/12)Wait, that makes sense because when a is between -1/3 and -1/12, the critical point x = sqrt(-1/(3a)) is between 1 and 2.So, the range of a is (-1/3, -1/12). Therefore, a must be greater than -1/3 and less than -1/12.Okay, that seems correct.Now, part (3). Let f(x) be a unimodal function on [a,b]. Suppose m, n are in (a,b), m < n, and f(m) ≥ f(n). We need to prove that (a,n) is a peak-containing interval.First, let's recall that a unimodal function on [a,b] has a peak point x₀ in (a,b) such that f is increasing on [a, x₀] and decreasing on [x₀, b].Given that m < n and f(m) ≥ f(n), we need to show that the peak point x₀ is in (a,n). That is, the peak-containing interval is (a,n).Assume for contradiction that x₀ ≥ n. Then, since x₀ is the peak, f is decreasing on [x₀, b]. But since x₀ ≥ n, f is decreasing on [n, b]. So, for m < n, if f is decreasing on [n, b], then f(m) ≤ f(n) because m < n. But this contradicts the given condition that f(m) ≥ f(n). Therefore, our assumption that x₀ ≥ n must be false. Hence, x₀ < n.Therefore, the peak point x₀ is in (a,n), so (a,n) is a peak-containing interval.Wait, let me make sure I didn't skip any steps.We know f is increasing on [a, x₀] and decreasing on [x₀, b]. If x₀ were ≥ n, then on [n, b], f is decreasing. So, for any m < n, f(m) ≤ f(n) because f is decreasing after n. But we have f(m) ≥ f(n), which can't happen if f is decreasing after n. Therefore, x₀ must be less than n, so the peak is in (a,n).Yes, that makes sense.So, summarizing:1. f₁ is unimodal with peak at 1/4; f₂ is not unimodal.2. The range of a is (-1/3, -1/12).3. Proved that (a,n) is a peak-containing interval.**Final Answer**(1) ① is unimodal with peak at boxed{dfrac{1}{4}}; ② is not unimodal.(2) The range of a is boxed{left( -dfrac{1}{3}, -dfrac{1}{12} right)}.(3) Proved as above.

answer：First, I need to calculate the area of the equilateral triangle with a side length of 6 units. The formula for the area of an equilateral triangle is ( frac{s^2 sqrt{3}}{4} ). Plugging in the side length:[A = frac{6^2 sqrt{3}}{4} = frac{36 sqrt{3}}{4} = 9 sqrt{3}]Next, I'll determine the perimeter of the triangle by multiplying the side length by 3:[P = 3 times 6 = 18]Finally, I'll find the ratio of the area to the perimeter:[text{Ratio} = frac{9 sqrt{3}}{18} = frac{sqrt{3}}{2}]