answer：Okay, so I have this problem where I need to define a function ( f(x) ) on the interval ( (0, +infty) ) that satisfies the inequalities ( 9f(x) < xf'(x) < 10f(x) ) and also ( f(x) > 0 ). Then, I need to find the range of ( dfrac{f(2)}{f(1)} ). Hmm, that seems a bit tricky, but let me try to break it down step by step.First, I notice that the inequalities involve both ( f(x) ) and its derivative ( f'(x) ). This makes me think that maybe I can rewrite the inequality in terms of a differential equation or something similar. Let me write down the given inequalities:1. ( 9f(x) < xf'(x) )2. ( xf'(x) < 10f(x) )So, combining these, I have ( 9f(x) < xf'(x) < 10f(x) ). That means ( xf'(x) ) is sandwiched between ( 9f(x) ) and ( 10f(x) ). Maybe I can manipulate these inequalities to find a relationship between ( f(x) ) and ( x ).Let me consider the first inequality: ( 9f(x) < xf'(x) ). If I divide both sides by ( x ), I get ( dfrac{9}{x}f(x) < f'(x) ). Similarly, for the second inequality: ( xf'(x) < 10f(x) ), dividing both sides by ( x ) gives ( f'(x) < dfrac{10}{x}f(x) ).So now I have:1. ( f'(x) > dfrac{9}{x}f(x) )2. ( f'(x) < dfrac{10}{x}f(x) )This looks like a pair of differential inequalities. Maybe I can solve them to find bounds on ( f(x) ).Let me consider the first inequality: ( f'(x) > dfrac{9}{x}f(x) ). This is a linear differential inequality. I can rewrite it as:( dfrac{f'(x)}{f(x)} > dfrac{9}{x} )Integrating both sides from 1 to 2 (since I need to find ( f(2)/f(1) )) might help. Let me try that.Integrate the left side:( int_{1}^{2} dfrac{f'(x)}{f(x)} dx > int_{1}^{2} dfrac{9}{x} dx )The left integral is ( ln(f(2)) - ln(f(1)) = lnleft( dfrac{f(2)}{f(1)} right) ).The right integral is ( 9 ln(2) - 9 ln(1) = 9 ln(2) ) since ( ln(1) = 0 ).So, we have:( lnleft( dfrac{f(2)}{f(1)} right) > 9 ln(2) )Exponentiating both sides:( dfrac{f(2)}{f(1)} > e^{9 ln(2)} = 2^9 = 512 )Okay, so that gives a lower bound: ( dfrac{f(2)}{f(1)} > 512 ).Now, let's do the same for the second inequality: ( f'(x) < dfrac{10}{x}f(x) ).Rewriting:( dfrac{f'(x)}{f(x)} < dfrac{10}{x} )Integrate both sides from 1 to 2:( int_{1}^{2} dfrac{f'(x)}{f(x)} dx < int_{1}^{2} dfrac{10}{x} dx )Again, the left integral is ( lnleft( dfrac{f(2)}{f(1)} right) ).The right integral is ( 10 ln(2) - 10 ln(1) = 10 ln(2) ).So:( lnleft( dfrac{f(2)}{f(1)} right) < 10 ln(2) )Exponentiating both sides:( dfrac{f(2)}{f(1)} < e^{10 ln(2)} = 2^{10} = 1024 )So, combining both results, we have:( 512 < dfrac{f(2)}{f(1)} < 1024 )Therefore, the range of ( dfrac{f(2)}{f(1)} ) is ( (512, 1024) ).Wait, but let me double-check my steps to make sure I didn't make a mistake. I transformed the inequalities into differential inequalities, integrated them, and found the bounds. That seems correct. I also made sure to exponentiate correctly, so the bounds are ( 2^9 ) and ( 2^{10} ), which are 512 and 1024, respectively.I think that makes sense. The function ( f(x) ) is growing faster than a function proportional to ( x^9 ) but slower than one proportional to ( x^{10} ). So, when evaluating the ratio ( f(2)/f(1) ), it should be between ( 2^9 ) and ( 2^{10} ). Yeah, that seems right.I don't see any errors in my reasoning, so I feel confident about this answer.**Final Answer**The range of ( dfrac{f(2)}{f(1)} ) is boxed{(512, 1024)}.

answer：Okay, so I've got this problem where I need to find the probability that the distance between two randomly selected points on opposite sides of a unit square is less than a given length ( p ), where ( 1 < p < sqrt{2} ). Hmm, let me try to visualize this.First, let's think about the unit square. It has sides of length 1. If I pick two opposite sides, say the top and bottom, then I can represent the positions of the points on these sides using coordinates. Let me denote the point on the bottom side as ( (x, 0) ) and the point on the top side as ( (y, 1) ). Here, ( x ) and ( y ) are both between 0 and 1 since they're on the unit square.Now, the distance between these two points can be calculated using the distance formula. So, the distance ( d ) between ( (x, 0) ) and ( (y, 1) ) is:[d = sqrt{(x - y)^2 + (1 - 0)^2} = sqrt{(x - y)^2 + 1}]We want this distance ( d ) to be less than ( p ). So, setting up the inequality:[sqrt{(x - y)^2 + 1} < p]To simplify this, I'll square both sides to eliminate the square root:[(x - y)^2 + 1 < p^2]Subtracting 1 from both sides gives:[(x - y)^2 < p^2 - 1]Taking the square root of both sides, we get:[|x - y| < sqrt{p^2 - 1}]So, this inequality tells me that the difference between ( x ) and ( y ) must be less than ( sqrt{p^2 - 1} ). Since ( x ) and ( y ) are both between 0 and 1, I can visualize this on a coordinate plane where ( x ) is on the horizontal axis and ( y ) is on the vertical axis. The region where ( |x - y| < sqrt{p^2 - 1} ) would form a band around the line ( y = x ).But wait, since ( x ) and ( y ) are both between 0 and 1, the entire sample space is a unit square in the ( xy )-plane. The area where ( |x - y| < sqrt{p^2 - 1} ) is the region between the lines ( y = x + sqrt{p^2 - 1} ) and ( y = x - sqrt{p^2 - 1} ).Now, to find the probability, I need to calculate the area of this region and then divide it by the total area of the unit square, which is 1. So, the probability ( P ) is just the area of the region where ( |x - y| < sqrt{p^2 - 1} ).Let me sketch this out mentally. The lines ( y = x + sqrt{p^2 - 1} ) and ( y = x - sqrt{p^2 - 1} ) are parallel to the line ( y = x ) and are offset by ( sqrt{p^2 - 1} ). Since ( p ) is between 1 and ( sqrt{2} ), ( sqrt{p^2 - 1} ) will be between 0 and 1. That means these lines will intersect the unit square.The region between these two lines within the unit square is a band. The width of this band is ( 2sqrt{p^2 - 1} ). However, because the lines intersect the square, the area isn't just a simple rectangle. Instead, it's a sort of "strip" across the square.To calculate the area, I can subtract the areas of the two triangles that lie outside the band from the total area of the square. Each of these triangles has a base and height of ( 1 - sqrt{p^2 - 1} ). So, the area of one triangle is:[text{Area of one triangle} = frac{1}{2} times (1 - sqrt{p^2 - 1}) times (1 - sqrt{p^2 - 1}) = frac{1}{2}(1 - sqrt{p^2 - 1})^2]Since there are two such triangles, the total area outside the band is:[text{Total area outside} = 2 times frac{1}{2}(1 - sqrt{p^2 - 1})^2 = (1 - sqrt{p^2 - 1})^2]Therefore, the area inside the band is:[text{Area inside} = text{Total area} - text{Total area outside} = 1 - (1 - sqrt{p^2 - 1})^2]Expanding the squared term:[(1 - sqrt{p^2 - 1})^2 = 1 - 2sqrt{p^2 - 1} + (p^2 - 1)]So, substituting back:[text{Area inside} = 1 - [1 - 2sqrt{p^2 - 1} + p^2 - 1] = 1 - 1 + 2sqrt{p^2 - 1} - p^2 + 1]Simplifying:[text{Area inside} = 2sqrt{p^2 - 1} - p^2 + 1]Since the total area is 1, the probability ( P ) is just this area:[P = 2sqrt{p^2 - 1} - p^2 + 1]Let me double-check this result. When ( p = 1 ), the distance must be less than 1, but since the minimum distance between two opposite sides is 1 (the side length), the probability should be 0. Plugging ( p = 1 ) into the formula:[P = 2sqrt{1 - 1} - 1 + 1 = 0 - 1 + 1 = 0]That checks out. Now, when ( p = sqrt{2} ), the maximum possible distance between two points on opposite sides is ( sqrt{2} ) (the diagonal of the square). So, the probability should be 1. Plugging ( p = sqrt{2} ):[P = 2sqrt{2 - 1} - 2 + 1 = 2(1) - 2 + 1 = 2 - 2 + 1 = 1]That also checks out. So, the formula seems to make sense at the boundary conditions.I think this makes sense. The probability increases as ( p ) increases, which is intuitive because a larger ( p ) allows for more possible points within the required distance. The formula accounts for the area within the band, which is the region where the distance condition is satisfied.So, after all this, I believe the probability that the distance between the two randomly selected points is less than ( p ) is:[2sqrt{p^2 - 1} - p^2 + 1]**Final Answer**The probability is boxed{2sqrt{p^2 - 1} - p^2 + 1}.

answer：Okay, so I have this problem here about a sequence defined by the formula ( a_n = n^2 - 2lambda n + 1 ). The question is asking for the range of ( lambda ) such that the sequence is increasing. The options are A: ( (-infty, 1) ), B: ( (-infty, 1] ), C: ( (-infty, frac{3}{2}) ), and D: ( (-infty, frac{3}{2}] ).Alright, let me start by understanding what it means for a sequence to be increasing. A sequence ( {a_n} ) is increasing if each term is greater than the previous one. In other words, for all ( n ), ( a_{n+1} > a_n ). So, I need to ensure that this inequality holds for every natural number ( n ).Given the formula ( a_n = n^2 - 2lambda n + 1 ), let's write out ( a_{n+1} ) as well. That would be ( a_{n+1} = (n+1)^2 - 2lambda(n+1) + 1 ).Now, let's compute ( a_{n+1} - a_n ) to see what condition on ( lambda ) makes this difference positive.First, expand ( a_{n+1} ):( a_{n+1} = (n^2 + 2n + 1) - 2lambda n - 2lambda + 1 )Simplify that:( a_{n+1} = n^2 + 2n + 1 - 2lambda n - 2lambda + 1 )Combine like terms:( a_{n+1} = n^2 + 2n - 2lambda n + (1 + 1 - 2lambda) )Which simplifies to:( a_{n+1} = n^2 + (2 - 2lambda)n + (2 - 2lambda) )Now, subtract ( a_n ) from ( a_{n+1} ):( a_{n+1} - a_n = [n^2 + (2 - 2lambda)n + (2 - 2lambda)] - [n^2 - 2lambda n + 1] )Let's compute each term:- The ( n^2 ) terms cancel out.- The linear term: ( (2 - 2lambda)n - (-2lambda n) = (2 - 2lambda + 2lambda)n = 2n )- The constant term: ( (2 - 2lambda) - 1 = 1 - 2lambda )So, putting it all together:( a_{n+1} - a_n = 2n + (1 - 2lambda) )For the sequence to be increasing, this difference must be positive for all ( n ). So, we have:( 2n + (1 - 2lambda) > 0 ) for all ( n geq 1 ).Let's solve this inequality for ( lambda ):( 2n + 1 - 2lambda > 0 )Subtract ( 2n + 1 ) from both sides:( -2lambda > -2n - 1 )Multiply both sides by ( -1 ), which reverses the inequality:( 2lambda < 2n + 1 )Divide both sides by 2:( lambda < n + frac{1}{2} )So, ( lambda ) must be less than ( n + frac{1}{2} ) for all ( n geq 1 ). But since ( n ) starts at 1 and increases, the smallest value of ( n + frac{1}{2} ) is when ( n = 1 ), which is ( 1 + frac{1}{2} = frac{3}{2} ).Therefore, to satisfy ( lambda < n + frac{1}{2} ) for all ( n geq 1 ), ( lambda ) must be less than the smallest such value, which is ( frac{3}{2} ). So, ( lambda < frac{3}{2} ).Looking back at the options, that corresponds to option C: ( (-infty, frac{3}{2}) ).Wait, but let me double-check. If ( lambda = frac{3}{2} ), what happens? Let's plug it back into the difference:( a_{n+1} - a_n = 2n + 1 - 2lambda = 2n + 1 - 3 = 2n - 2 ).For ( n = 1 ), this would be ( 2(1) - 2 = 0 ). So, ( a_2 - a_1 = 0 ), meaning ( a_2 = a_1 ). But the sequence is supposed to be increasing, so ( a_{n+1} > a_n ) must hold strictly. Therefore, ( lambda ) cannot be equal to ( frac{3}{2} ); it must be strictly less than ( frac{3}{2} ).Hence, the correct range is ( (-infty, frac{3}{2}) ), which is option C.**Final Answer**boxed{C}

answer：Alright, so I have this problem here where I need to figure out which quadrants the set of points satisfying both inequalities ( y > -frac{1}{2}x ) and ( y > 3x + 6 ) are contained within. The options are Quadrants I and II, II and IV, I and III, III and IV, or I and IV. Let me try to break this down step by step.First, I remember that the coordinate plane is divided into four quadrants: I, II, III, and IV. Quadrant I is where both x and y are positive, Quadrant II is where x is negative and y is positive, Quadrant III is where both x and y are negative, and Quadrant IV is where x is positive and y is negative.Now, I need to analyze the two inequalities given. Let's start with the first one: ( y > -frac{1}{2}x ). This is a linear inequality, and I know that to graph it, I can first graph the line ( y = -frac{1}{2}x ) and then determine which side of the line satisfies the inequality.The line ( y = -frac{1}{2}x ) has a slope of -1/2 and passes through the origin (0,0). Since the inequality is ( y > -frac{1}{2}x ), the region that satisfies this inequality is above the line. So, above the line ( y = -frac{1}{2}x ) would be the area where y is greater than -1/2 times x. Now, thinking about quadrants, since the line passes through the origin and has a negative slope, it goes from Quadrant II to Quadrant IV. The region above this line would cover parts of Quadrant I and Quadrant II because above the line in Quadrant I would be where both x and y are positive, and above the line in Quadrant II would be where x is negative and y is positive.Next, let's look at the second inequality: ( y > 3x + 6 ). Again, this is a linear inequality. I can graph the line ( y = 3x + 6 ) and then determine which side satisfies the inequality.The line ( y = 3x + 6 ) has a slope of 3 and a y-intercept at (0,6). This means it crosses the y-axis at 6 and rises steeply to the right. Since the inequality is ( y > 3x + 6 ), the region that satisfies this inequality is above the line.Thinking about quadrants, the line ( y = 3x + 6 ) crosses the y-axis at 6, which is in Quadrant I, and it extends into Quadrant III as x becomes more negative. However, since the inequality is above the line, in Quadrant I, the region above the line would still be where y is positive and x is positive. In Quadrant III, even though the line extends there, the region above the line would actually be where y is negative because the line is going downward into Quadrant III. But wait, the inequality is ( y > 3x + 6 ), so in Quadrant III, where both x and y are negative, it's possible that some points satisfy this inequality, but I need to check.To better understand the regions, I think it would help to find where these two lines intersect each other. That way, I can see the point where both inequalities switch from one quadrant to another.So, let's set the two equations equal to each other to find their intersection point:( -frac{1}{2}x = 3x + 6 )Let me solve for x:( -frac{1}{2}x - 3x = 6 )Combining like terms:( -frac{7}{2}x = 6 )Multiply both sides by -2/7:( x = 6 times (-frac{2}{7}) = -frac{12}{7} )So, x is -12/7. Now, let's find y by plugging this back into one of the equations. I'll use ( y = -frac{1}{2}x ):( y = -frac{1}{2} times (-frac{12}{7}) = frac{6}{7} )So, the intersection point is at (-12/7, 6/7). This point is in Quadrant II because x is negative and y is positive.Now, knowing that the two lines intersect in Quadrant II, I can analyze the regions defined by both inequalities.For the first inequality ( y > -frac{1}{2}x ), the region above the line includes Quadrant I and Quadrant II. For the second inequality ( y > 3x + 6 ), the region above the line includes parts of Quadrant I and Quadrant II as well, but it also extends into Quadrant III. However, since the intersection point is in Quadrant II, the overlapping region where both inequalities are satisfied must be in Quadrant II.But wait, I should also consider if there's any part of the overlapping region in Quadrant I. Since both inequalities have regions in Quadrant I, it's possible that there is an overlapping region there as well.Let me think about specific points. For example, take the point (0,7). Plugging into both inequalities:For ( y > -frac{1}{2}x ): 7 > 0, which is true.For ( y > 3x + 6 ): 7 > 6, which is also true.So, (0,7) is in Quadrant I and satisfies both inequalities.Another point, say (1,10):For ( y > -frac{1}{2}x ): 10 > -0.5, which is true.For ( y > 3x + 6 ): 10 > 9, which is true.So, (1,10) is also in Quadrant I and satisfies both inequalities.What about a point in Quadrant II, say (-2, 3):For ( y > -frac{1}{2}x ): 3 > 1, which is true.For ( y > 3x + 6 ): 3 > 0, which is true.So, (-2,3) is in Quadrant II and satisfies both inequalities.Now, what about a point in Quadrant III, say (-3, -3):For ( y > -frac{1}{2}x ): -3 > 1.5, which is false.So, this point doesn't satisfy the first inequality.Another point in Quadrant III, say (-4, -2):For ( y > -frac{1}{2}x ): -2 > 2, which is false.Again, doesn't satisfy the first inequality.What about a point in Quadrant IV, say (2, -1):For ( y > -frac{1}{2}x ): -1 > -1, which is not true because it's equal, not greater.So, this point doesn't satisfy the first inequality.Another point in Quadrant IV, say (3, -2):For ( y > -frac{1}{2}x ): -2 > -1.5, which is false.So, this point also doesn't satisfy the first inequality.From these examples, it seems that the overlapping region where both inequalities are satisfied is in Quadrants I and II.Wait, but earlier I thought that the second inequality might extend into Quadrant III, but the intersection point is in Quadrant II, and the overlapping region seems to be in Quadrants I and II.Let me double-check by considering the regions more carefully.The first inequality ( y > -frac{1}{2}x ) is above a line that goes from Quadrant II to Quadrant IV. So, above this line would be Quadrant I and Quadrant II.The second inequality ( y > 3x + 6 ) is above a line that goes from Quadrant II to Quadrant I and into Quadrant III. However, above this line in Quadrant III would require y to be greater than a negative value, but since y is negative in Quadrant III, it's possible for some points to satisfy ( y > 3x + 6 ) there, but only if 3x + 6 is negative enough.But when we consider the intersection point at (-12/7, 6/7), which is in Quadrant II, the overlapping region above both lines would start from that point and extend upwards into Quadrant I.In Quadrant III, even though the line ( y = 3x + 6 ) extends there, the region above it would require y to be greater than a negative number, but since y is negative in Quadrant III, it's possible only if 3x + 6 is less than y, which is negative. However, for x negative enough, 3x + 6 could be negative, but we need to check if there are points in Quadrant III that satisfy both inequalities.Let me pick a point in Quadrant III, say (-3, -1):For ( y > -frac{1}{2}x ): -1 > 1.5, which is false.So, this point doesn't satisfy the first inequality.Another point (-4, -2):For ( y > -frac{1}{2}x ): -2 > 2, which is false.Again, doesn't satisfy the first inequality.It seems that in Quadrant III, the first inequality ( y > -frac{1}{2}x ) is not satisfied because y is negative and -1/2 x would be positive when x is negative, making y > positive number impossible in Quadrant III.Therefore, the overlapping region where both inequalities are satisfied is only in Quadrants I and II.So, the correct answer should be Quadrants I and II, which is option A.

answer：Okay, so I have this problem about triangle ABC. The sides opposite angles A, B, and C are labeled as a, b, and c respectively. The problem gives me that c squared equals a times cosine B plus b times cosine A, and it also tells me that a and b are both 3. I need to find the perimeter of triangle ABC.Alright, let's start by writing down what we know. We have a triangle with sides a, b, c opposite angles A, B, C. Given that a = b = 3, so it's an isosceles triangle because two sides are equal. That might be useful later on.The equation given is c² = a cos B + b cos A. Since a and b are both 3, I can substitute those in. So that becomes c² = 3 cos B + 3 cos A. Maybe I can factor out the 3? So c² = 3 (cos A + cos B). Hmm, that might be helpful.Now, I remember the Law of Cosines, which relates the sides of a triangle to the cosine of one of its angles. The Law of Cosines says that for any triangle, c² = a² + b² - 2ab cos C. Similarly, we can write expressions for cos A and cos B.Let me write those down. For angle A, the Law of Cosines gives:cos A = (b² + c² - a²) / (2bc)And for angle B, it's:cos B = (a² + c² - b²) / (2ac)Since a and b are both 3, let's substitute those values into these expressions.First, cos A:cos A = (3² + c² - 3²) / (2 * 3 * c) = (9 + c² - 9) / (6c) = c² / (6c) = c / 6Similarly, cos B:cos B = (3² + c² - 3²) / (2 * 3 * c) = (9 + c² - 9) / (6c) = c² / (6c) = c / 6Wait, that's interesting. Both cos A and cos B are equal to c/6. So, going back to the original equation, c² = 3 (cos A + cos B) = 3 (c/6 + c/6) = 3 (2c/6) = 3 (c/3) = c.So, c² = c. Let me write that as c² - c = 0, which factors to c(c - 1) = 0. So, c = 0 or c = 1. But c is a side length of a triangle, so it can't be zero. Therefore, c = 1.Now, since a = 3, b = 3, and c = 1, the perimeter is just a + b + c = 3 + 3 + 1 = 7.Wait, let me double-check that. If c = 1, then sides are 3, 3, 1. Does that make sense? Let me check the triangle inequality. The sum of any two sides should be greater than the third side.3 + 3 > 1: 6 > 1, which is true.3 + 1 > 3: 4 > 3, which is true.3 + 1 > 3: Same as above, true.So, yes, the sides 3, 3, 1 do form a valid triangle.Let me also verify the original equation with c = 1. So, c² = 1² = 1.a cos B + b cos A: Since a = b = 3, and earlier we found cos A = cos B = c/6 = 1/6.So, 3*(1/6) + 3*(1/6) = (3/6) + (3/6) = 0.5 + 0.5 = 1. Which equals c². So, that checks out.Therefore, I think the perimeter is indeed 7.**Final Answer**The perimeter of triangle ABC is boxed{7}.

answer：Alright, so I have this problem where I'm given two vectors, vector a and vector b. Vector a is (x, 3/2, 3) and vector b is (-1, y, 2). The question says that these two vectors are collinear, and I need to find the sum of x and y, which is x + y. Okay, first, I need to remember what it means for two vectors to be collinear. From what I recall, collinear vectors are vectors that lie on the same line or parallel lines. In other words, one vector is a scalar multiple of the other. So, if vector a is collinear with vector b, there must be some scalar, let's call it k, such that vector a equals k times vector b. So, mathematically, that would mean:a = k * bBreaking that down component-wise, since both vectors are in three dimensions, each corresponding component of vector a should be equal to k times the corresponding component of vector b. Let me write that out:- The first component of a is x, and the first component of b is -1. So, x = k * (-1)- The second component of a is 3/2, and the second component of b is y. So, 3/2 = k * y- The third component of a is 3, and the third component of b is 2. So, 3 = k * 2Alright, so now I have three equations:1. x = -k2. 3/2 = k * y3. 3 = 2kI think the best approach here is to solve for k first using the third equation because it only involves k and a known value. Let's do that.From equation 3:3 = 2kSo, solving for k:k = 3 / 2k = 1.5Okay, so k is 1.5. Now, let's plug this value of k into the other equations to find x and y.Starting with equation 1:x = -kx = -1.5So, x is -1.5.Now, moving on to equation 2:3/2 = k * yWe know k is 1.5, so:3/2 = 1.5 * yBut 3/2 is equal to 1.5, so:1.5 = 1.5 * yTo solve for y, divide both sides by 1.5:y = 1.5 / 1.5y = 1So, y is 1.Now, the question asks for x + y. Let's add them together:x + y = -1.5 + 1 = -0.5Hmm, that gives me -0.5. Wait, is that correct? Let me double-check my calculations to make sure I didn't make a mistake.Starting again, vector a is (x, 3/2, 3) and vector b is (-1, y, 2). They are collinear, so a = k * b.So, component-wise:x = k * (-1)3/2 = k * y3 = k * 2From the third equation, 3 = 2k, so k = 3/2, which is 1.5. That seems right.Then, x = -k = -1.5. That looks correct.For the second component, 3/2 = k * y, which is 1.5 * y. So, 1.5 = 1.5 * y, which implies y = 1. That also seems correct.Adding x and y: -1.5 + 1 = -0.5. So, x + y is -0.5.Wait, but in fractions, -0.5 is the same as -1/2. Maybe I should express it as a fraction instead of a decimal for consistency with the given values.So, x + y = -1/2.Let me just make sure there isn't another way to approach this problem that might give a different result. Sometimes, with vectors, there can be different interpretations, but in this case, since they are collinear, the scalar multiple approach seems straightforward.Alternatively, I could check if the ratios of the corresponding components are equal, which is another way to determine collinearity.So, for vectors a and b to be collinear, the ratios of their corresponding components should be equal. That is:x / (-1) = (3/2) / y = 3 / 2Let's see if that holds.From the third component ratio, 3 / 2 is equal to k, which we already found as 1.5.So, x / (-1) should equal 1.5, which gives x = -1.5, as before.Similarly, (3/2) / y should equal 1.5, so:(3/2) / y = 3/2Which implies that y = 1, as before.So, both methods give the same result, which reinforces that x + y is indeed -1/2.I think I'm confident with this answer now. It's always good to check your work, especially when dealing with vector problems, because it's easy to mix up components or make a calculation error.Just to recap:1. Recognize that collinear vectors are scalar multiples of each other.2. Set up the equations component-wise.3. Solve for the scalar k using the component with known values.4. Use k to find the unknown components x and y.5. Sum x and y to get the final answer.6. Double-check using an alternative method to ensure consistency.This systematic approach helps prevent mistakes and ensures that the solution is correct.Another thing I might consider is plugging the values back into the original vectors to see if they are indeed scalar multiples.So, vector a would be (-1.5, 1.5, 3), and vector b is (-1, 1, 2).If I multiply vector b by k = 1.5, I should get vector a.Let's see:-1 * 1.5 = -1.5 (matches the first component of a)1 * 1.5 = 1.5 (matches the second component of a)2 * 1.5 = 3 (matches the third component of a)Perfect, that confirms that vector a is indeed 1.5 times vector b, so they are collinear.Therefore, x + y = -1.5 + 1 = -0.5, which is -1/2.I think that's solid. No mistakes detected in the process, and the verification step confirms the result.