💡Okay, so I have this problem here: If ( k in mathbb{R} ), then " ( k > 3 )" is the ______ condition for "the equation ( dfrac{x^{2}}{k-3} - dfrac{y^{2}}{k+3} = 1 ) represents a hyperbola". I need to figure out whether ( k > 3 ) is a sufficient condition, necessary condition, both, or neither.Alright, let me start by recalling what defines a hyperbola. A hyperbola is a type of conic section, and its standard form is either ( dfrac{(x - h)^2}{a^2} - dfrac{(y - k)^2}{b^2} = 1 ) for a hyperbola that opens left and right, or ( dfrac{(y - k)^2}{b^2} - dfrac{(x - h)^2}{a^2} = 1 ) for one that opens up and down. In both cases, the denominators under the squared terms must be positive because ( a^2 ) and ( b^2 ) are always positive.Looking at the given equation ( dfrac{x^{2}}{k-3} - dfrac{y^{2}}{k+3} = 1 ), it resembles the standard form of a hyperbola that opens left and right. So, for this equation to represent a hyperbola, both denominators ( k - 3 ) and ( k + 3 ) must be positive. Let me write that down:1. ( k - 3 > 0 ) implies ( k > 3 ).2. ( k + 3 > 0 ) implies ( k > -3 ).So, for both denominators to be positive, ( k ) must satisfy both conditions. That means ( k > 3 ) because if ( k > 3 ), it automatically satisfies ( k > -3 ). Wait, but hold on. What if ( k - 3 ) is negative? Then the equation would not represent a hyperbola because the denominator under ( x^2 ) would be negative, making the entire term negative, which isn't the standard form of a hyperbola. Similarly, if ( k + 3 ) is negative, the denominator under ( y^2 ) would be negative, which also disrupts the standard form.But actually, in the standard form of a hyperbola, only one of the denominators needs to be positive because the hyperbola opens in the direction of the positive term. Hmm, maybe I was too hasty before.Let me think again. The equation ( dfrac{x^{2}}{A} - dfrac{y^{2}}{B} = 1 ) represents a hyperbola if either ( A > 0 ) and ( B > 0 ), or ( A < 0 ) and ( B < 0 ). Wait, is that right? Because if both ( A ) and ( B ) are positive, it's a hyperbola opening left and right. If both are negative, the equation would become ( dfrac{x^{2}}{A} - dfrac{y^{2}}{B} = 1 ), but since ( A ) and ( B ) are negative, it's equivalent to ( -dfrac{x^{2}}{|A|} + dfrac{y^{2}}{|B|} = 1 ), which is a hyperbola opening up and down.So, actually, for the equation to represent a hyperbola, we need either both denominators positive or both negative. That means:Either:1. ( k - 3 > 0 ) and ( k + 3 > 0 ), which implies ( k > 3 ).Or:2. ( k - 3 < 0 ) and ( k + 3 < 0 ), which implies ( k < -3 ).Therefore, the equation represents a hyperbola if ( k > 3 ) or ( k < -3 ). So, coming back to the original question: Is ( k > 3 ) a sufficient condition, necessary condition, both, or neither for the equation to represent a hyperbola?Well, if ( k > 3 ), then the equation definitely represents a hyperbola. So, ( k > 3 ) is a sufficient condition. However, it's not necessary because the equation can also represent a hyperbola when ( k < -3 ). Therefore, ( k > 3 ) is not required; it's just one way to ensure the equation is a hyperbola.To summarize:- Sufficient condition: If ( k > 3 ), then the equation is a hyperbola. So, it's sufficient.- Necessary condition: But the equation can be a hyperbola even if ( k > 3 ) is not true (i.e., when ( k < -3 )). So, it's not necessary.Hence, ( k > 3 ) is a sufficient but not necessary condition.**Final Answer**The answer is boxed{text{sufficient but not necessary}}.

💡Okay, so I have this quadratic trinomial, which is x squared plus 7x plus n, and it's given that it has a factor of (x + 5). I need to find the value of n. Hmm, let me think about how to approach this.First, I remember that if a quadratic trinomial has a factor of (x + a), then one of its roots is x = -a. So, in this case, if (x + 5) is a factor, then one of the roots should be x = -5. That makes sense because if I plug x = -5 into the quadratic, it should equal zero.So, let me write that down: if x = -5 is a root, then substituting x = -5 into the quadratic equation should give me zero. The quadratic is x² + 7x + n, so plugging in x = -5:(-5)² + 7*(-5) + n = 0Calculating each term step by step:(-5) squared is 25.7 times (-5) is -35.So, putting it all together:25 - 35 + n = 0Simplifying that:25 - 35 is -10, so:-10 + n = 0To solve for n, I can add 10 to both sides:n = 10Wait, so n is 10? Let me double-check that.Alternatively, I can think about factoring the quadratic. If (x + 5) is a factor, then the quadratic can be written as (x + 5)(x + m), where m is some number. Let me expand that:(x + 5)(x + m) = x² + (5 + m)x + 5mComparing this to the original quadratic, which is x² + 7x + n, the coefficients should be equal. So, the coefficient of x in the expanded form is (5 + m), which should equal 7. Therefore:5 + m = 7Solving for m:m = 7 - 5 = 2Now, the constant term in the expanded form is 5m, which should equal n. So:n = 5m = 5*2 = 10Okay, that confirms it. So, n is indeed 10.Wait, but let me check the options given. The options are A: -18, B: 2, C: 10, D: 12. So, 10 is option C. That must be the correct answer.Just to make sure I didn't make any mistakes, let me substitute n = 10 back into the original quadratic and see if (x + 5) is indeed a factor.The quadratic becomes x² + 7x + 10. Let me factor this:Looking for two numbers that multiply to 10 and add up to 7. Those numbers are 5 and 2 because 5*2=10 and 5+2=7. So, the factors are (x + 5)(x + 2). Yep, that's correct. So, (x + 5) is definitely a factor when n = 10.Therefore, I'm confident that the value of n is 10, which is option C.

💡First, I need to calculate the product of the three complex numbers: (1 + i), (2 + i), and (3 + i).I'll start by multiplying the first two complex numbers: (1 + i) and (2 + i). Using the distributive property, I'll multiply each term in the first complex number by each term in the second complex number.1 * 2 = 21 * i = ii * 2 = 2ii * i = i²Since i² is equal to -1, the product becomes:2 + i + 2i - 1 = (2 - 1) + (i + 2i) = 1 + 3iNow, I'll multiply this result by the third complex number, (3 + i):(1 + 3i) * (3 + i)Again, using the distributive property:1 * 3 = 31 * i = i3i * 3 = 9i3i * i = 3i²Since i² is -1, this becomes:3 + i + 9i - 3 = (3 - 3) + (i + 9i) = 0 + 10i = 10iTherefore, the final result of multiplying (1 + i)(2 + i)(3 + i) is 10i.

💡Okay, so I have this probability problem here, and I need to figure out the chance of drawing a white ball after adding one white ball to a bag that initially has three balls. The tricky part is that I don't know the colors of the initial three balls, and all possible combinations are equally likely. Hmm, let's break this down step by step.First, I need to consider all possible combinations of the initial three balls. Since each ball can either be white (W) or not white (N), there are 2 possibilities for each ball. So, for three balls, that's 2^3 = 8 possible combinations. Let me list them out:1. NNN2. NNW3. NWN4. WNN5. NWW6. WNW7. WWN8. WWWOkay, so these are all the possible initial configurations of the bag. Now, I add one white ball to each of these configurations. Let's see what each configuration becomes:1. NNN + W = NNNW (1 white out of 4 balls)2. NNW + W = NNWW (2 whites out of 4 balls)3. NWN + W = NWNW (2 whites out of 4 balls)4. WNN + W = WNNW (2 whites out of 4 balls)5. NWW + W = NWWW (3 whites out of 4 balls)6. WNW + W = WNWW (3 whites out of 4 balls)7. WWN + W = WWNW (3 whites out of 4 balls)8. WWW + W = WWWW (4 whites out of 4 balls)Now, for each of these new configurations, I need to calculate the probability of drawing a white ball. Let's do that:1. NNNW: 1 white out of 4 balls → Probability = 1/42. NNWW: 2 whites out of 4 balls → Probability = 2/4 = 1/23. NWNW: 2 whites out of 4 balls → Probability = 2/4 = 1/24. WNNW: 2 whites out of 4 balls → Probability = 2/4 = 1/25. NWWW: 3 whites out of 4 balls → Probability = 3/46. WNWW: 3 whites out of 4 balls → Probability = 3/47. WWNW: 3 whites out of 4 balls → Probability = 3/48. WWWW: 4 whites out of 4 balls → Probability = 4/4 = 1Alright, so now I have the probabilities for each initial configuration after adding the white ball. Since all initial hypotheses are equally likely, I need to find the average of these probabilities to get the overall probability of drawing a white ball.Let's add up all these probabilities:1/4 + 1/2 + 1/2 + 1/2 + 3/4 + 3/4 + 3/4 + 1To make it easier, I'll convert all fractions to have the same denominator, which is 4:1/4 + 2/4 + 2/4 + 2/4 + 3/4 + 3/4 + 3/4 + 4/4Now, add them up:1 + 2 + 2 + 2 + 3 + 3 + 3 + 4 = 20So, the total sum is 20/4 = 5.Now, since there are 8 initial configurations, I divide the total sum by 8:5 / 8Therefore, the probability of drawing a white ball is 5/8.Wait, let me double-check my calculations to make sure I didn't make a mistake. Adding up the numerators:1 + 2 + 2 + 2 + 3 + 3 + 3 + 4 = 20Yes, that's correct. And 20 divided by 4 is indeed 5. Then, 5 divided by 8 is 5/8. Okay, that seems right.I think I've covered all possible initial configurations, calculated the probabilities correctly after adding the white ball, and then averaged them out since each initial configuration is equally likely. So, I'm confident that the probability is 5/8.

💡Alright, so I need to figure out the perimeter of this pentagon FGHIJ. Let me try to visualize it based on the description. First, point F is at the origin. Okay, so F is at (0,0). Then, G is directly south of F at a distance of 2 units. Since south is the negative y-direction, G must be at (0, -2). Got that.Next, H is 2 units to the east from G. East is the positive x-direction, so starting from G at (0, -2), moving 2 units east would put H at (2, -2). That makes sense.Now, I is northeast of H, forming a right triangle with sides 1 unit each from H. Hmm, northeast means moving equally in the positive x and positive y directions. Since it's a right triangle with legs of 1 unit each, I think that means from H, we move 1 unit east and 1 unit north. So starting from H at (2, -2), moving 1 unit east would take us to (3, -2), and then 1 unit north would take us to (3, -1). So I is at (3, -1). Wait, but the problem says it's forming a right triangle with sides 1 unit each from H, conforming HI and a vertical leg not drawn. Maybe I misinterpreted. If HI is the hypotenuse of a right triangle with legs of 1 unit each, then HI would be sqrt(1^2 + 1^2) = sqrt(2). So the length of HI is sqrt(2). But where exactly is I located? If it's northeast of H, then it's moving equally in x and y directions. So from H at (2, -2), moving 1 unit northeast would mean moving 1/sqrt(2) in x and 1/sqrt(2) in y. But the problem says the legs are 1 unit each, so maybe it's moving 1 unit east and 1 unit north, making the coordinates (3, -1). So I think I is at (3, -1). Okay, moving on to J. J is placed so that it forms right triangles both with I (along IJ) and F (along FJ), each with the legs being 1.5 units. So both IJ and FJ are hypotenuses of right triangles with legs of 1.5 units each. That means IJ and FJ each have lengths of sqrt(1.5^2 + 1.5^2) = sqrt(4.5). But where exactly is J located? It needs to satisfy both conditions: forming a right triangle with I and with F. Let me think. Since F is at (0,0) and J is forming a right triangle with F, the legs are 1.5 units each. So from F, moving 1.5 units in some direction and 1.5 units in a perpendicular direction to reach J. Similarly, from I at (3, -1), moving 1.5 units in some direction and 1.5 units in a perpendicular direction to reach J. This seems like J is the intersection point of two circles: one centered at F with radius sqrt(4.5) and another centered at I with radius sqrt(4.5). But since both legs are 1.5 units, the coordinates of J can be determined based on the directions from F and I. Assuming that from F, moving east and north, and from I, moving west and north, since J needs to be somewhere in the first quadrant to form right triangles with both F and I. So from F at (0,0), moving 1.5 units east and 1.5 units north would place J at (1.5, 1.5). From I at (3, -1), moving 1.5 units west and 1.5 units north would also place J at (3 - 1.5, -1 + 1.5) = (1.5, 0.5). Wait, that's conflicting. Hmm, maybe I need to adjust the directions. If from F, moving east and north, and from I, moving west and north, but the distances are 1.5 units each. So from F, moving 1.5 east and 1.5 north gives (1.5, 1.5). From I, moving 1.5 west and 1.5 north gives (3 - 1.5, -1 + 1.5) = (1.5, 0.5). These two points don't coincide, so perhaps my assumption about the directions is wrong. Alternatively, maybe from I, moving northwest to reach J. So from I at (3, -1), moving northwest would mean moving in the negative x and positive y directions. If the legs are 1.5 units each, then moving 1.5 units west and 1.5 units north would take us to (3 - 1.5, -1 + 1.5) = (1.5, 0.5). From F at (0,0), moving northeast would mean moving 1.5 units east and 1.5 units north to (1.5, 1.5). Again, conflicting points. Wait, maybe the right triangles are not necessarily aligned with the axes. So the legs could be in any perpendicular directions, not necessarily along the x and y axes. That complicates things. Alternatively, perhaps J is located such that FJ and IJ are both diagonals of squares with side 1.5 units. So FJ would be sqrt(1.5^2 + 1.5^2) = sqrt(4.5), and similarly for IJ. But without knowing the exact directions, it's hard to pinpoint J's coordinates. Maybe I can use coordinate geometry to find J. Let's denote J as (x, y). From F at (0,0), the distance to J is sqrt(4.5), so:sqrt(x^2 + y^2) = sqrt(4.5) => x^2 + y^2 = 4.5From I at (3, -1), the distance to J is sqrt(4.5), so:sqrt((x - 3)^2 + (y + 1)^2) = sqrt(4.5) => (x - 3)^2 + (y + 1)^2 = 4.5Now we have two equations:1. x^2 + y^2 = 4.52. (x - 3)^2 + (y + 1)^2 = 4.5Subtracting equation 1 from equation 2:(x - 3)^2 + (y + 1)^2 - x^2 - y^2 = 0Expanding:(x^2 - 6x + 9) + (y^2 + 2y + 1) - x^2 - y^2 = 0Simplify:-6x + 9 + 2y + 1 = 0 => -6x + 2y + 10 = 0 => -3x + y + 5 = 0 => y = 3x - 5Now plug this into equation 1:x^2 + (3x - 5)^2 = 4.5Expand:x^2 + 9x^2 - 30x + 25 = 4.5 => 10x^2 - 30x + 25 - 4.5 = 0 => 10x^2 - 30x + 20.5 = 0Divide by 5:2x^2 - 6x + 4.1 = 0Multiply by 10 to eliminate decimal:20x^2 - 60x + 41 = 0Using quadratic formula:x = [60 ± sqrt(3600 - 3280)] / 40 = [60 ± sqrt(320)] / 40 = [60 ± 8*sqrt(5)] / 40 = [15 ± 2*sqrt(5)] / 10So x = (15 + 2√5)/10 ≈ (15 + 4.472)/10 ≈ 1.9472Or x = (15 - 2√5)/10 ≈ (15 - 4.472)/10 ≈ 1.0528Now find y:y = 3x - 5For x ≈ 1.9472:y ≈ 3*1.9472 - 5 ≈ 5.8416 - 5 ≈ 0.8416For x ≈ 1.0528:y ≈ 3*1.0528 - 5 ≈ 3.1584 - 5 ≈ -1.8416So J could be at approximately (1.9472, 0.8416) or (1.0528, -1.8416). Since the pentagon is likely convex and J is connected back to F, the lower y-coordinate might not make sense as it would be below G and H. So probably J is at (1.9472, 0.8416).But let's see, from I at (3, -1), moving to J at (1.9472, 0.8416), the distance should be sqrt(4.5). Let's check:sqrt((3 - 1.9472)^2 + (-1 - 0.8416)^2) ≈ sqrt((1.0528)^2 + (-1.8416)^2) ≈ sqrt(1.108 + 3.391) ≈ sqrt(4.5) ≈ 2.1213. That works.Similarly, from F at (0,0) to J at (1.9472, 0.8416):sqrt((1.9472)^2 + (0.8416)^2) ≈ sqrt(3.79 + 0.708) ≈ sqrt(4.5) ≈ 2.1213. That also works.So J is at approximately (1.9472, 0.8416). But for exact values, we can keep it as ( (15 + 2√5)/10 , 3*(15 + 2√5)/10 - 5 ). Simplifying y:y = (45 + 6√5)/10 - 5 = (45 + 6√5 - 50)/10 = (-5 + 6√5)/10 = (6√5 - 5)/10So J is at ( (15 + 2√5)/10 , (6√5 - 5)/10 )Now, to find the perimeter, we need the lengths of all sides: FG, GH, HI, IJ, and JF.FG is from F(0,0) to G(0,-2): distance is 2 units.GH is from G(0,-2) to H(2,-2): distance is 2 units.HI is from H(2,-2) to I(3,-1): distance is sqrt((3-2)^2 + (-1 - (-2))^2) = sqrt(1 + 1) = sqrt(2).IJ is from I(3,-1) to J( (15 + 2√5)/10 , (6√5 - 5)/10 ): distance is sqrt(4.5) as established earlier.JF is from J to F(0,0): distance is sqrt(4.5).So adding them up:FG + GH + HI + IJ + JF = 2 + 2 + sqrt(2) + sqrt(4.5) + sqrt(4.5)Simplify sqrt(4.5): sqrt(4.5) = sqrt(9/2) = (3)/sqrt(2) = (3√2)/2So sqrt(4.5) + sqrt(4.5) = 2*(3√2)/2 = 3√2Therefore, total perimeter = 2 + 2 + sqrt(2) + 3√2 = 4 + 4√2So the perimeter is 4 + 4√2 units.

💡Alright, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. There are two parts to this problem. Let me tackle them one by one.Starting with part (1): If 2a sin B equals the square root of 3 times b, find the measure of angle A. Hmm, okay. I remember the Law of Sines, which says that in any triangle, the ratio of a side to the sine of its opposite angle is constant. So, that would mean a/sin A = b/sin B = c/sin C. Maybe I can use that here.Given 2a sin B = sqrt(3) b. Let me rewrite that equation using the Law of Sines. Since a/sin A = b/sin B, that means a = (b sin A)/sin B. Let me substitute that into the equation.So, 2 * (b sin A / sin B) * sin B = sqrt(3) b. Wait, the sin B cancels out here, right? So, 2b sin A = sqrt(3) b. Then, I can divide both sides by b, assuming b isn't zero, which it can't be in a triangle. So, 2 sin A = sqrt(3). Therefore, sin A = sqrt(3)/2.Okay, so sin A equals sqrt(3)/2. I know that sin 60 degrees is sqrt(3)/2, and sin 120 degrees is also sqrt(3)/2. So, angle A could be either 60 degrees or 120 degrees. But in a triangle, the sum of angles is 180 degrees, so depending on the other angles, both could be possible. But the problem doesn't give any more information, so I think both are valid. So, angle A is either 60 degrees or 120 degrees.Wait, but in the problem statement, it just says "find the measure of angle A." So, I think both are possible answers. Let me just make sure I didn't make any mistakes in the steps. Starting from 2a sin B = sqrt(3) b, using Law of Sines, a = (b sin A)/sin B, substituting, getting 2 sin A = sqrt(3), so sin A = sqrt(3)/2. Yeah, that seems correct. So, angle A is either 60 degrees or 120 degrees.Alright, moving on to part (2): If the altitude on side BC is equal to a/2, find the maximum value of (c/b) + (b/c). Hmm, okay. So, the altitude on BC is a/2. Let me visualize the triangle. BC is side a, right? Because in triangle ABC, side opposite angle A is a, so BC is side a.So, the altitude from A to BC is a/2. The area of the triangle can be expressed in two ways: one using the base BC and the height, and another using sides b, c, and the sine of angle A.So, area = (1/2)*base*height = (1/2)*a*(a/2) = (1/2)*a*(a/2) = a²/4.Alternatively, area = (1/2)*b*c*sin A. So, equating the two expressions for area:(1/2)*b*c*sin A = a²/4.Multiplying both sides by 2: b*c*sin A = a²/2.So, a² = 2*b*c*sin A.Now, from the Law of Cosines, a² = b² + c² - 2*b*c*cos A.So, setting these equal: 2*b*c*sin A = b² + c² - 2*b*c*cos A.Let me rearrange this equation:b² + c² = 2*b*c*sin A + 2*b*c*cos A.Factor out 2*b*c: b² + c² = 2*b*c*(sin A + cos A).Now, the expression we need to maximize is (c/b) + (b/c). Let me denote this as E = (c/b) + (b/c).I can write E as (b² + c²)/(b*c). So, E = (b² + c²)/(b*c).From the earlier equation, b² + c² = 2*b*c*(sin A + cos A). So, substituting into E:E = [2*b*c*(sin A + cos A)] / (b*c) = 2*(sin A + cos A).So, E = 2*(sin A + cos A). Now, I need to find the maximum value of E.I recall that sin A + cos A can be rewritten using a trigonometric identity. Specifically, sin A + cos A = sqrt(2)*sin(A + 45°). Let me verify that.Yes, because sin A + cos A = sqrt(2)*sin(A + 45°). So, sin A + cos A = sqrt(2)*sin(A + 45°). Therefore, E = 2*sqrt(2)*sin(A + 45°).The maximum value of sin(theta) is 1, so the maximum value of E is 2*sqrt(2)*1 = 2*sqrt(2).But wait, is there any restriction on angle A? Since A is an angle in a triangle, it must be between 0 and 180 degrees. So, A + 45° must be between 45° and 225°. The sine function reaches its maximum at 90°, so when A + 45° = 90°, which means A = 45°, then sin(A + 45°) = 1.Therefore, the maximum value of E is 2*sqrt(2), achieved when angle A is 45 degrees.Let me just recap to make sure I didn't make any mistakes. Starting from the altitude, found the area in two ways, set them equal, used Law of Cosines to relate a² to b and c, then expressed (c/b + b/c) in terms of sin A + cos A, converted that to a single sine function, and found its maximum. Seems solid.So, for part (2), the maximum value is 2*sqrt(2).**Final Answer**(1) The measure of angle ( A ) is (boxed{dfrac{pi}{3}}) or (boxed{dfrac{2pi}{3}}).(2) The maximum value of ( dfrac{c}{b} + dfrac{b}{c} ) is (boxed{2sqrt{2}}).