💡First, I need to determine the amount of alcohol in the original 15-liter mixture. Since the mixture is 25% alcohol, I can calculate this by multiplying 15 liters by 25%.Next, I'll remove 2 liters of alcohol from the mixture. This will reduce the total amount of alcohol and also decrease the overall volume of the mixture.After removing the alcohol, I'll add 3 liters of water to the mixture. This will increase the total volume of the mixture but will not change the amount of alcohol remaining.Finally, to find the percentage of alcohol in the final mixture, I'll divide the remaining amount of alcohol by the new total volume of the mixture and then multiply by 100 to get the percentage.

💡Okay, so I have this problem where I'm given a point (1, -2) on the terminal side of an angle α, and I need to figure out which of the given trigonometric statements are true. The options are about tan α, cos α, and sin α. Hmm, let me think about how to approach this.First, I remember that in trigonometry, when you have a point (x, y) on the terminal side of an angle in standard position, you can use that to find the trigonometric functions. I think the key here is to find the radius, which is the distance from the origin to the point (1, -2). That should help me find all the necessary trigonometric ratios.So, to find the radius r, I can use the Pythagorean theorem. The formula is r = sqrt(x² + y²). Plugging in the values, x is 1 and y is -2. So, r = sqrt(1² + (-2)²) = sqrt(1 + 4) = sqrt(5). Okay, so r is sqrt(5). That seems straightforward.Now, let's tackle each trigonometric function one by one.Starting with tan α. I recall that tan α is the ratio of the opposite side to the adjacent side in a right triangle, which translates to y/x in the coordinate system. So, tan α = y/x. Given the point (1, -2), y is -2 and x is 1. Therefore, tan α = -2/1 = -2. Hmm, looking at the options, option C says tan α = -2, which matches my calculation. So, option C seems correct. But wait, option A says tan α = -1/2. That would be the case if it were x/y instead of y/x. So, option A is incorrect because it swaps the numerator and denominator.Next, let's look at cos α. Cosine is the adjacent side over the hypotenuse, which corresponds to x/r in the coordinate system. So, cos α = x/r. We have x = 1 and r = sqrt(5). Therefore, cos α = 1/sqrt(5). To rationalize the denominator, we can multiply numerator and denominator by sqrt(5), which gives us sqrt(5)/5. So, cos α = sqrt(5)/5. Checking the options, option B states exactly that, so option B is correct.Now, moving on to sin α. Sine is the opposite side over the hypotenuse, which is y/r. Given y = -2 and r = sqrt(5), sin α = -2/sqrt(5). Rationalizing the denominator, we get -2sqrt(5)/5. Looking at option D, it says sin α = sqrt(5)/5. That's positive, but our calculation shows it's negative. So, option D is incorrect because it doesn't account for the negative value from the y-coordinate.Wait a second, let me double-check my calculations to make sure I didn't make a mistake. For tan α, y is -2 and x is 1, so -2/1 is indeed -2. For cos α, x is 1 and r is sqrt(5), so 1/sqrt(5) is sqrt(5)/5 when rationalized. And for sin α, y is -2, so -2/sqrt(5) is -2sqrt(5)/5. Yeah, that seems right.I also remember that the signs of the trigonometric functions depend on the quadrant in which the angle is located. Since the point (1, -2) is in the fourth quadrant (because x is positive and y is negative), tan α should be negative, which matches our result for tan α. Cos α is positive in the fourth quadrant, which aligns with our positive value for cos α. Sin α is negative in the fourth quadrant, which matches our negative value for sin α.So, summarizing my findings:- Option A: Incorrect, because tan α is -2, not -1/2.- Option B: Correct, cos α is sqrt(5)/5.- Option C: Correct, tan α is -2.- Option D: Incorrect, sin α is -2sqrt(5)/5, not sqrt(5)/5.Therefore, the correct options are B and C.

💡Alright, so I have this problem here: Given that ( a ) and ( b ) are real numbers, and ( sqrt{2a + 6} + |b - sqrt{2}| = 0 ), I need to find the absolute value of ( a + b ). Hmm, okay, let me try to figure this out step by step.First, I remember that square roots and absolute values are always non-negative. That means ( sqrt{2a + 6} ) is greater than or equal to zero, and ( |b - sqrt{2}| ) is also greater than or equal to zero. Since their sum is zero, each of them must individually be zero. Otherwise, if one was positive, the other would have to be negative to make the sum zero, but that's not possible because they can't be negative. So, both ( sqrt{2a + 6} ) and ( |b - sqrt{2}| ) must equal zero.Starting with the square root part: ( sqrt{2a + 6} = 0 ). To solve for ( a ), I can square both sides to get rid of the square root. That gives me ( 2a + 6 = 0 ). Now, solving for ( a ), I subtract 6 from both sides: ( 2a = -6 ). Then, divide both sides by 2: ( a = -3 ). Okay, so ( a ) is -3.Next, let's tackle the absolute value part: ( |b - sqrt{2}| = 0 ). I know that the absolute value of a number is zero only if the number itself is zero. So, ( b - sqrt{2} = 0 ). Solving for ( b ), I add ( sqrt{2} ) to both sides: ( b = sqrt{2} ). Got it, so ( b ) is ( sqrt{2} ).Now, I need to find the absolute value of ( a + b ). Let's compute ( a + b ) first. Substituting the values we found: ( a + b = -3 + sqrt{2} ). Hmm, ( sqrt{2} ) is approximately 1.414, so ( -3 + 1.414 ) is roughly ( -1.586 ). Since we're looking for the absolute value, we take the positive version of this number. So, ( |a + b| = | -3 + sqrt{2} | = 3 - sqrt{2} ).Wait a second, let me double-check that. If ( a + b = -3 + sqrt{2} ), then the absolute value would indeed be ( 3 - sqrt{2} ) because ( sqrt{2} ) is less than 3, making ( -3 + sqrt{2} ) negative. Taking the absolute value flips the sign, so it becomes positive ( 3 - sqrt{2} ).Just to make sure I didn't make any mistakes, let me go through the steps again. Starting with the equation ( sqrt{2a + 6} + |b - sqrt{2}| = 0 ), both terms must be zero. Solving ( sqrt{2a + 6} = 0 ) gives ( a = -3 ), and ( |b - sqrt{2}| = 0 ) gives ( b = sqrt{2} ). Adding them together, ( a + b = -3 + sqrt{2} ), and the absolute value is ( 3 - sqrt{2} ). Yeah, that seems correct.I think I've got it. The absolute value of ( a + b ) is ( 3 - sqrt{2} ).

💡First, I need to calculate (2overrightarrow{a}) by multiplying each component of vector (overrightarrow{a}) by 2.Next, I will subtract vector (overrightarrow{b}) from the result by subtracting each corresponding component.Finally, I will compare the resulting vector with the given options to determine the correct answer.

💡Alright, so I have this problem about an ellipse and a line, and I need to do two things: first, write the parametric equation for the ellipse and the general equation for the line, and second, find the coordinates of a specific point P on the ellipse that's a certain distance from another point A. Let me try to break this down step by step.Starting with part (1): writing the parametric equation for the ellipse C and the general equation for the line l.The ellipse is given by the equation ( frac{x^2}{4} + frac{y^2}{3} = 1 ). I remember that the standard form of an ellipse is ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ), where a is the semi-major axis and b is the semi-minor axis. So in this case, ( a^2 = 4 ) which means ( a = 2 ), and ( b^2 = 3 ) so ( b = sqrt{3} ).For parametric equations of an ellipse, I think they are usually written as:[begin{cases}x = a cos theta y = b sin thetaend{cases}]where ( theta ) is the parameter. So plugging in the values of a and b, the parametric equations for this ellipse should be:[begin{cases}x = 2 cos theta y = sqrt{3} sin thetaend{cases}]Okay, that seems straightforward. Now, for the line l, which is given in parametric form:[begin{cases}x = -3 + sqrt{3} t y = 2sqrt{3} + tend{cases}]I need to write this in the general (Cartesian) form. To do that, I should eliminate the parameter t. From the second equation, I can solve for t:[y = 2sqrt{3} + t implies t = y - 2sqrt{3}]Now, substitute this expression for t into the first equation:[x = -3 + sqrt{3} t = -3 + sqrt{3}(y - 2sqrt{3})]Let me simplify that:[x = -3 + sqrt{3} y - 2 times 3 = -3 + sqrt{3} y - 6 = sqrt{3} y - 9]So, bringing all terms to one side:[x - sqrt{3} y + 9 = 0]That should be the general equation of the line l.Alright, part (1) seems done. Now, moving on to part (2): Let A(1,0), and we need to find the coordinates of point P on ellipse C such that the distance from P to A is ( frac{3}{2} ).First, let's denote point P as ( (x, y) ). Since P lies on the ellipse, it must satisfy the ellipse equation:[frac{x^2}{4} + frac{y^2}{3} = 1]Also, the distance between P and A is given by:[sqrt{(x - 1)^2 + (y - 0)^2} = frac{3}{2}]Squaring both sides to eliminate the square root:[(x - 1)^2 + y^2 = left( frac{3}{2} right)^2 = frac{9}{4}]So now I have two equations:1. ( frac{x^2}{4} + frac{y^2}{3} = 1 )2. ( (x - 1)^2 + y^2 = frac{9}{4} )I need to solve these simultaneously to find the coordinates of P.Let me expand the second equation:[(x - 1)^2 + y^2 = x^2 - 2x + 1 + y^2 = frac{9}{4}]So,[x^2 - 2x + 1 + y^2 = frac{9}{4}]Let me rearrange this:[x^2 + y^2 - 2x + 1 = frac{9}{4}]Subtracting ( frac{9}{4} ) from both sides:[x^2 + y^2 - 2x + 1 - frac{9}{4} = 0]Simplify the constants:[x^2 + y^2 - 2x - frac{5}{4} = 0]Now, let's see if I can express this in terms of the ellipse equation. From the ellipse equation, I can solve for ( y^2 ):[frac{x^2}{4} + frac{y^2}{3} = 1 implies y^2 = 3 left(1 - frac{x^2}{4}right) = 3 - frac{3x^2}{4}]So, substitute ( y^2 = 3 - frac{3x^2}{4} ) into the equation from the distance:[x^2 + left(3 - frac{3x^2}{4}right) - 2x - frac{5}{4} = 0]Let me simplify this step by step.First, expand the terms:[x^2 + 3 - frac{3x^2}{4} - 2x - frac{5}{4} = 0]Combine like terms. Let's handle the ( x^2 ) terms:[x^2 - frac{3x^2}{4} = frac{4x^2}{4} - frac{3x^2}{4} = frac{x^2}{4}]Now, the constants:[3 - frac{5}{4} = frac{12}{4} - frac{5}{4} = frac{7}{4}]So putting it all together:[frac{x^2}{4} - 2x + frac{7}{4} = 0]Multiply every term by 4 to eliminate denominators:[x^2 - 8x + 7 = 0]Now, this is a quadratic equation in x. Let's solve for x using the quadratic formula:[x = frac{8 pm sqrt{64 - 28}}{2} = frac{8 pm sqrt{36}}{2} = frac{8 pm 6}{2}]So, two solutions:1. ( x = frac{8 + 6}{2} = frac{14}{2} = 7 )2. ( x = frac{8 - 6}{2} = frac{2}{2} = 1 )Now, let's check these x-values in the ellipse equation to find the corresponding y-values.First, for x = 7:[frac{7^2}{4} + frac{y^2}{3} = 1 implies frac{49}{4} + frac{y^2}{3} = 1 implies frac{y^2}{3} = 1 - frac{49}{4} = -frac{45}{4}]This gives ( y^2 = -frac{135}{4} ), which is not possible since y^2 cannot be negative. So, x = 7 is not a valid solution.Next, for x = 1:[frac{1^2}{4} + frac{y^2}{3} = 1 implies frac{1}{4} + frac{y^2}{3} = 1 implies frac{y^2}{3} = frac{3}{4} implies y^2 = frac{9}{4} implies y = pm frac{3}{2}]So, the coordinates of point P are (1, ( frac{3}{2} )) and (1, ( -frac{3}{2} )).Wait a second, but the problem says "the coordinates of point P". It doesn't specify if there are multiple points, so I think both are valid solutions.But let me double-check if these points are indeed at a distance of ( frac{3}{2} ) from A(1,0).For P(1, ( frac{3}{2} )):Distance = ( sqrt{(1 - 1)^2 + left( frac{3}{2} - 0 right)^2} = sqrt{0 + frac{9}{4}} = frac{3}{2} ). That's correct.For P(1, ( -frac{3}{2} )):Distance = ( sqrt{(1 - 1)^2 + left( -frac{3}{2} - 0 right)^2} = sqrt{0 + frac{9}{4}} = frac{3}{2} ). Also correct.So both points are valid. Therefore, the coordinates of point P are (1, ( frac{3}{2} )) and (1, ( -frac{3}{2} )).Wait, but in the parametric form of the ellipse, we have ( x = 2 cos theta ) and ( y = sqrt{3} sin theta ). Let me see if these points fit into that.For P(1, ( frac{3}{2} )):( x = 1 = 2 cos theta implies cos theta = frac{1}{2} implies theta = frac{pi}{3} ) or ( theta = frac{5pi}{3} )Then, ( y = sqrt{3} sin theta )For ( theta = frac{pi}{3} ), ( y = sqrt{3} times frac{sqrt{3}}{2} = frac{3}{2} )For ( theta = frac{5pi}{3} ), ( y = sqrt{3} times left( -frac{sqrt{3}}{2} right) = -frac{3}{2} )So, that matches the points we found.Therefore, everything checks out. The coordinates of point P are indeed (1, ( frac{3}{2} )) and (1, ( -frac{3}{2} )).**Final Answer**The coordinates of point ( P ) are (boxed{(1, pmfrac{3}{2})}).

💡Okay, so I need to prove this equation using mathematical induction. The equation is 1 + 3 + 5 + … + (2n - 1) = n², where n is a positive integer. I remember that mathematical induction has two main steps: the base case and the inductive step. First, the base case. I think I need to check if the equation holds true for the smallest value of n, which is usually 1. Let me plug in n = 1 into both sides of the equation. On the left side, I just have 1, and on the right side, it's 1², which is also 1. So, 1 = 1, which is true. That means the base case is verified.Now, moving on to the inductive step. I need to assume that the equation holds for some arbitrary positive integer k. So, I'll assume that 1 + 3 + 5 + … + (2k - 1) = k². This is called the induction hypothesis. The next part is where I need to show that if the equation holds for n = k, then it must also hold for n = k + 1. That means I need to prove that 1 + 3 + 5 + … + (2k - 1) + [2(k + 1) - 1] = (k + 1)².Let me break that down. The left side of the equation for n = k + 1 is the sum up to (2(k + 1) - 1). Simplifying that term, 2(k + 1) - 1 is 2k + 2 - 1, which is 2k + 1. So, the left side becomes 1 + 3 + 5 + … + (2k - 1) + (2k + 1).From my induction hypothesis, I know that 1 + 3 + 5 + … + (2k - 1) equals k². So, I can substitute that into the equation. That gives me k² + (2k + 1).Now, I need to simplify k² + 2k + 1. I remember that this is a perfect square trinomial. Specifically, k² + 2k + 1 factors into (k + 1)². So, k² + 2k + 1 = (k + 1)².Therefore, I've shown that if the equation holds for n = k, it also holds for n = k + 1. Since the base case is true and the inductive step holds, by mathematical induction, the equation 1 + 3 + 5 + … + (2n - 1) = n² is true for all positive integers n.Wait, but the question is asking specifically about the transformation during the inductive step. It gives four options:A: 1 + 3 + 5 + … + (2k + 1) = k²B: 1 + 3 + 5 + … + (2k + 1) = (k + 1)²C: 1 + 3 + 5 + … + (2k + 1) = (k + 2)²D: 1 + 3 + 5 + … + (2k + 1) = (k + 3)²From my previous reasoning, when n = k + 1, the sum becomes k² + (2k + 1) = (k + 1)². So, the equation should be transformed into 1 + 3 + 5 + … + (2k + 1) = (k + 1)², which is option B.I think that makes sense because when we add the next odd number, which is 2k + 1, to the sum up to k terms, we get the square of (k + 1). So, the correct transformation is option B.