answer：Okay, so I have this problem here where I need to find the value of ( k ) such that two given lines are perpendicular. Hmm, I remember that for two lines to be perpendicular, their direction vectors must have a dot product of zero. Let me try to recall how that works.First, let me write down the equations of the lines again to make sure I have them right. The first line is given by the vector equation:[begin{pmatrix} 3 1 -4 end{pmatrix} + t begin{pmatrix} k 1 2 end{pmatrix}]And the second line is:[begin{pmatrix} 0 -2 3 end{pmatrix} + u begin{pmatrix} -4 2 -1 end{pmatrix}]So, the direction vector of the first line is ( begin{pmatrix} k 1 2 end{pmatrix} ) and the direction vector of the second line is ( begin{pmatrix} -4 2 -1 end{pmatrix} ). Got that down.Now, to find if they are perpendicular, I need to compute the dot product of these two direction vectors and set it equal to zero. The dot product formula for two vectors ( mathbf{a} = begin{pmatrix} a_1 a_2 a_3 end{pmatrix} ) and ( mathbf{b} = begin{pmatrix} b_1 b_2 b_3 end{pmatrix} ) is:[mathbf{a} cdot mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3]So applying that to our direction vectors:[(k)(-4) + (1)(2) + (2)(-1) = 0]Let me compute each term step by step. First, ( k times -4 ) is ( -4k ). Then, ( 1 times 2 ) is ( 2 ). Next, ( 2 times -1 ) is ( -2 ). So putting it all together:[-4k + 2 - 2 = 0]Wait, that simplifies to:[-4k + 0 = 0]Which further simplifies to:[-4k = 0]To solve for ( k ), I can divide both sides by -4:[k = 0]Hmm, so ( k ) must be zero for the lines to be perpendicular. Let me double-check my calculations to make sure I didn't make a mistake.Starting again, the dot product:[(k)(-4) + (1)(2) + (2)(-1)]That's:[-4k + 2 - 2]Which is indeed:[-4k + 0 = 0]So, yes, ( k = 0 ) is the correct value. I think I did that correctly. Just to visualize, if ( k ) is zero, the direction vector of the first line becomes ( begin{pmatrix} 0 1 2 end{pmatrix} ). Let me compute the dot product again with ( k = 0 ):[(0)(-4) + (1)(2) + (2)(-1) = 0 + 2 - 2 = 0]Yep, that works out. So, the lines are indeed perpendicular when ( k = 0 ).I guess another way to think about it is that the direction vectors must form a right angle, which geometrically means their projections onto each axis must satisfy the condition that their combined effects cancel out in terms of the dot product. But I think the algebraic approach I took is sufficient here.Just to recap, the steps were:1. Identify the direction vectors of both lines.2. Compute their dot product.3. Set the dot product equal to zero because the lines are perpendicular.4. Solve the resulting equation for ( k ).Everything seems to check out. I don't think I missed anything here. It's a straightforward problem once you remember the condition for perpendicularity of lines in vector form.**Final Answer**The value of ( k ) is boxed{0}.

answer：First, I recognize that the set S contains exactly 10 numbers and has an original average of A. This means the total sum of all numbers in the set is 10A.When one of the numbers is increased by 6, the new sum of the set becomes 10A + 6. The problem states that the new average is 6.8, so the new total sum can also be expressed as 10 * 6.8.Setting up the equation:10A + 6 = 68Solving for A:10A = 68 - 610A = 62A = 62 / 10A = 6.2Therefore, the original average of set S was 6.2.

answer：First, I need to simplify the complex number ( z = frac{3}{1 + 2i} ) to express it in the standard form ( a + bi ).To do this, I'll multiply both the numerator and the denominator by the complex conjugate of the denominator, which is ( 1 - 2i ). This step helps eliminate the imaginary unit from the denominator.After multiplying, the expression becomes:[z = frac{3(1 - 2i)}{(1 + 2i)(1 - 2i)}]Next, I'll simplify the denominator using the difference of squares formula:[(1 + 2i)(1 - 2i) = 1^2 - (2i)^2 = 1 - 4i^2 = 1 - 4(-1) = 1 + 4 = 5]Now, I'll simplify the numerator:[3(1 - 2i) = 3 - 6i]Putting it all together, the simplified form of ( z ) is:[z = frac{3 - 6i}{5} = frac{3}{5} - frac{6}{5}i]From this, I can see that the real part ( a ) is ( frac{3}{5} ) and the imaginary part ( b ) is ( -frac{6}{5} ).In the complex plane, the real part corresponds to the x-coordinate, and the imaginary part corresponds to the y-coordinate. Therefore, the point ( left( frac{3}{5}, -frac{6}{5} right) ) lies in the fourth quadrant, where the x-coordinate is positive and the y-coordinate is negative.Thus, the correct answer is the fourth quadrant.

answer：First, I'll define the total number of passengers on the ship as ( P ).According to the problem, 20% of the passengers held round-trip tickets and took their cars aboard the ship. This means that ( 0.20P ) passengers had round-trip tickets and brought their cars.Next, it's stated that 20% of the passengers with round-trip tickets did not take their cars aboard the ship. This implies that 80% of the passengers with round-trip tickets did take their cars aboard the ship.Let ( R ) represent the total number of passengers with round-trip tickets. Since 80% of ( R ) equals ( 0.20P ), I can set up the equation:[0.80R = 0.20P]To find ( R ), I'll divide both sides of the equation by 0.80:[R = frac{0.20P}{0.80} = 0.25P]This calculation shows that 25% of the ship's passengers held round-trip tickets.

answer：First, I need to understand the relationship between the speeds of the new and old boats. The new boat travels 30% faster than the old boat. This means if the old boat's speed is ( v ), the new boat's speed is ( 1.30v ).Next, I'll determine the time it takes for the old boat to travel 150 miles. Using the formula ( text{Time} = frac{text{Distance}}{text{Speed}} ), the time ( t ) is ( frac{150}{v} ).Since both boats travel for the same amount of time, I can use this time to calculate the distance the new boat travels. Using the formula ( text{Distance} = text{Speed} times text{Time} ), the distance ( D ) for the new boat is ( 1.30v times frac{150}{v} ).Simplifying this, the ( v ) cancels out, leaving ( D = 1.30 times 150 = 195 ) miles.Therefore, the new boat can travel 195 miles in the same time it takes the old boat to travel 150 miles.

answer：Okay, so I have this problem here: the function f(x) = x² + a x - 1 has a minimum value of -2 on the interval [0, 3], and I need to find the real number a. Hmm, okay, let's see how to approach this.First, I remember that for quadratic functions, the graph is a parabola. Since the coefficient of x² is positive (it's 1), the parabola opens upwards, which means it has a minimum point, called the vertex. So, the vertex will give me the minimum value of the function. But wait, the minimum is given on the interval [0, 3], so I need to check where this minimum occurs.The vertex of a parabola given by f(x) = ax² + bx + c is at x = -b/(2a). In this case, my function is f(x) = x² + a x - 1, so a is 1, b is a, and c is -1. So, the x-coordinate of the vertex is x = -a/(2*1) = -a/2.Now, I need to see if this vertex lies within the interval [0, 3]. If it does, then the minimum value is at the vertex. If it doesn't, then the minimum will be at one of the endpoints, either x=0 or x=3.So, let's consider three cases:1. The vertex is inside the interval [0, 3].2. The vertex is to the left of the interval (x < 0).3. The vertex is to the right of the interval (x > 3).But since the parabola opens upwards, if the vertex is outside the interval, the minimum on the interval will be at the nearest endpoint. So, if the vertex is to the left, the minimum is at x=0; if it's to the right, the minimum is at x=3.Given that the minimum value is -2, I need to check each case.First, let's check if the vertex is inside [0, 3]. That means -a/2 must be between 0 and 3.So, 0 ≤ -a/2 ≤ 3.Multiplying all parts by -2 (and remembering to reverse the inequalities when multiplying by a negative):0 ≥ a ≥ -6.So, if a is between -6 and 0, the vertex is inside the interval. Otherwise, it's outside.Now, let's compute the minimum value at the vertex. The y-coordinate of the vertex is f(-a/2).So, f(-a/2) = (-a/2)² + a*(-a/2) - 1.Calculating that:(-a/2)² = a²/4a*(-a/2) = -a²/2So, f(-a/2) = a²/4 - a²/2 - 1.Simplify:a²/4 - 2a²/4 = -a²/4So, f(-a/2) = -a²/4 - 1.We are told that the minimum value is -2, so:-a²/4 - 1 = -2Multiply both sides by 4:-a² - 4 = -8Add 4 to both sides:-a² = -4Multiply both sides by -1:a² = 4So, a = ±2.But earlier, we found that for the vertex to be inside [0, 3], a must be between -6 and 0. So, a = -2 is valid, but a = 2 is not because 2 is greater than 0, which is outside the range where the vertex is inside the interval.So, a = -2 is a possible solution. But let's check the other cases to make sure.Case 2: Vertex is to the left of the interval (x < 0). That happens when -a/2 < 0, which implies a > 0.In this case, the minimum on [0, 3] is at x=0.Compute f(0) = 0 + 0 - 1 = -1.But we are told the minimum is -2, so this case doesn't work because -1 is not equal to -2.Case 3: Vertex is to the right of the interval (x > 3). That happens when -a/2 > 3, which implies a < -6.In this case, the minimum on [0, 3] is at x=3.Compute f(3) = 9 + 3a - 1 = 8 + 3a.Set this equal to -2:8 + 3a = -2Subtract 8:3a = -10Divide by 3:a = -10/3 ≈ -3.333...But wait, we had earlier that for the vertex to be to the right of the interval, a must be less than -6. But -10/3 is approximately -3.333, which is greater than -6. So, this is a contradiction. Therefore, this case doesn't work either.Wait, that doesn't make sense. If a = -10/3, which is approximately -3.333, then -a/2 = 10/6 ≈ 1.666, which is within [0, 3]. So, actually, a = -10/3 would place the vertex inside the interval, not to the right. So, my earlier reasoning was flawed.I think I made a mistake in determining the conditions for the vertex being inside or outside the interval.Let me re-examine that.The vertex is at x = -a/2.If x = -a/2 is in [0, 3], then 0 ≤ -a/2 ≤ 3.Multiplying by -2 (and reversing inequalities):0 ≥ a ≥ -6.So, a is between -6 and 0.If a > 0, then -a/2 < 0, so vertex is to the left.If a < -6, then -a/2 > 3, so vertex is to the right.So, when a = -10/3 ≈ -3.333, which is greater than -6, so -a/2 = 10/6 ≈ 1.666, which is within [0, 3]. Therefore, the vertex is inside the interval, so the minimum is at the vertex, not at x=3.So, in this case, when I set f(3) = -2, I got a = -10/3, but that actually places the vertex inside the interval, so that solution is invalid because the minimum would actually be at the vertex, not at x=3.Therefore, the only valid solution is a = -2.Wait, but earlier, when I set the vertex value to -2, I got a² = 4, so a = ±2, but only a = -2 is valid because a must be between -6 and 0.So, a = -2 is the solution.But let me double-check.Compute f(-a/2) when a = -2.x = -(-2)/2 = 1.f(1) = 1 + (-2)(1) -1 = 1 -2 -1 = -2.Yes, that works.Now, check if a = -2 is within the range where the vertex is inside [0, 3]. Since a = -2 is between -6 and 0, yes, it is.So, the minimum value is indeed at the vertex, and it's -2.Therefore, the value of a is -2.Wait, but earlier, when I considered a = -10/3, I thought it was a solution, but realized it was invalid because it actually places the vertex inside the interval. So, that case doesn't give a valid solution.Therefore, the only valid solution is a = -2.But let me check f(3) when a = -2.f(3) = 9 + (-2)(3) -1 = 9 -6 -1 = 2.Which is greater than -2, so the minimum is indeed at the vertex.Similarly, f(0) = -1, which is greater than -2.So, everything checks out.Therefore, the value of a is -2.Wait, but in the initial problem, the user wrote the function as f(x) = x² + a x -1, and the minimum is -2 on [0,3]. So, according to my calculations, a = -2.But in the initial thought process, I thought a = -10/3, but realized that was incorrect because it placed the vertex inside the interval. So, a = -2 is the correct answer.Wait, but let me double-check the calculations when I set the vertex value to -2.f(-a/2) = (-a/2)^2 + a*(-a/2) -1 = a²/4 - a²/2 -1 = -a²/4 -1.Set this equal to -2:-a²/4 -1 = -2Add 1 to both sides:-a²/4 = -1Multiply both sides by -4:a² = 4So, a = ±2.But since a must be between -6 and 0, a = -2 is the only solution.Yes, that's correct.Therefore, the value of a is -2.