💡First, I need to determine the amount of alcohol in each vessel. For the first vessel with a capacity of 2 liters and a 25% alcohol concentration, the amount of alcohol is 0.25 multiplied by 2 liters, which equals 0.5 liters. Next, for the second vessel with a capacity of 6 liters and a 40% alcohol concentration, the amount of alcohol is 0.40 multiplied by 6 liters, resulting in 2.4 liters.Adding the alcohol from both vessels gives a total of 0.5 liters plus 2.4 liters, which equals 2.9 liters of alcohol.The total volume of the mixture is 10 liters, as the 8 liters poured from the two vessels are filled up to 10 liters with water. To find the new concentration, I divide the total alcohol (2.9 liters) by the total volume (10 liters) and then multiply by 100% to get the percentage.This calculation results in a new concentration of 29%.

💡Alright, so I'm trying to understand this problem about triangles inscribed in circles and their circumradii. The problem states: "An acute triangle is inscribed in a circle. Prove that the circle's radius is not less than the circumradius of the triangle. Is this statement true for an obtuse triangle?" First, I need to recall some basic concepts. An acute triangle is one where all three angles are less than 90 degrees. A circumradius of a triangle is the radius of the circumscribed circle that passes through all three vertices of the triangle. So, if a triangle is inscribed in a circle, that circle is the circumcircle of the triangle, and its radius is the circumradius.Wait, but the problem says "an acute triangle is inscribed in a circle." So, does that mean the circle is the circumcircle? Or is it a different circle? Hmm, the wording is a bit confusing. It says the triangle is inscribed in the circle, which usually means that the circle is the circumcircle. But then it asks to prove that the circle's radius is not less than the circumradius of the triangle. That seems a bit redundant because if the circle is the circumcircle, then its radius is exactly the circumradius. So maybe I'm misunderstanding something.Perhaps the circle is not necessarily the circumcircle. Maybe it's another circle that the triangle is inscribed in, and we need to show that the radius of this circle is at least as big as the circumradius of the triangle. That makes more sense. So, the triangle is inscribed in some circle, and we need to prove that the radius of this circle is not less than the triangle's circumradius.Okay, so let's think about that. If a triangle is inscribed in a circle, that circle is the circumcircle, right? So, maybe the problem is saying that if an acute triangle is inscribed in a circle, then the radius of that circle is not less than the circumradius of the triangle. But that would mean the circle is the circumcircle, and its radius is equal to the circumradius. So, the statement would be trivially true because it's saying the radius is not less than itself.Hmm, maybe I'm missing something. Perhaps the circle is not the circumcircle but another circle that contains the triangle. So, the triangle is inscribed in a larger circle, and we need to show that the radius of this larger circle is at least the circumradius of the triangle.Wait, that could make sense. So, if you have an acute triangle, and you inscribe it in some circle, then the radius of that circle must be at least the circumradius of the triangle. So, the circumradius is the smallest possible radius for a circle that can inscribe the triangle. Therefore, any other circle that can inscribe the triangle must have a radius equal to or larger than the circumradius.But why is this specifically about acute triangles? Maybe for acute triangles, the circumradius has a particular property that ensures this. For example, in an acute triangle, the circumradius is related to the sides in a certain way that might make this statement hold.Let me recall the formula for the circumradius of a triangle. The circumradius ( R ) of a triangle with sides ( a ), ( b ), and ( c ) is given by:[ R = frac{a b c}{4 Delta} ]where ( Delta ) is the area of the triangle. Alternatively, using the Law of Sines:[ R = frac{a}{2 sin A} = frac{b}{2 sin B} = frac{c}{2 sin C} ]Since the triangle is acute, all angles are less than 90 degrees, so all sines of the angles are positive and less than 1. That might be useful.Now, if the triangle is inscribed in a circle of radius ( R' ), then ( R' ) must satisfy certain conditions. For example, the sides of the triangle must be chords of the circle. The length of a chord is related to the radius and the central angle subtended by the chord. The formula for the length of a chord is:[ text{Length} = 2 R' sin theta ]where ( theta ) is half the central angle. Wait, actually, the chord length is:[ text{Length} = 2 R' sin left( frac{alpha}{2} right) ]where ( alpha ) is the central angle corresponding to the chord.But in a triangle inscribed in a circle, the central angles correspond to twice the inscribed angles. So, if the triangle has angles ( A ), ( B ), and ( C ), then the central angles would be ( 2A ), ( 2B ), and ( 2C ). Since the triangle is acute, each angle is less than 90 degrees, so each central angle is less than 180 degrees.Wait, but in that case, the central angles would each be less than 180 degrees, meaning that each side of the triangle is a chord subtending less than 180 degrees. That might be important.Now, if we have a circle with radius ( R' ) containing the triangle, then each side of the triangle must be less than or equal to the diameter of the circle. But since the triangle is acute, none of the sides can be the diameter, because that would make the triangle right-angled. So, all sides must be less than the diameter.But how does that relate to the circumradius? The circumradius ( R ) is related to the sides and the angles. Maybe we can compare ( R' ) and ( R ) using the chord lengths.Let me think about one side of the triangle. Suppose side ( a ) is opposite angle ( A ). In the circumcircle, the length of side ( a ) is:[ a = 2 R sin A ]In the other circle with radius ( R' ), the same side ( a ) is a chord subtending angle ( 2A ) at the center. So, the length of side ( a ) is also:[ a = 2 R' sin A ]Wait, that can't be right because the central angle would be ( 2A ), so the chord length should be:[ a = 2 R' sin left( frac{2A}{2} right) = 2 R' sin A ]So, actually, both expressions for ( a ) are the same:[ 2 R sin A = 2 R' sin A ]Which would imply ( R = R' ). But that contradicts the idea that ( R' ) could be larger or smaller. Hmm, maybe I'm making a mistake here.Wait, no. In the circumcircle, the central angle corresponding to side ( a ) is indeed ( 2A ), so the chord length is ( 2 R sin A ). In another circle, if the triangle is inscribed, the central angles might be different. Wait, no, the central angles would still correspond to twice the inscribed angles, right? Because the inscribed angle theorem states that an angle subtended by an arc at the center is twice the inscribed angle.So, regardless of the circle, the central angles would still be ( 2A ), ( 2B ), and ( 2C ). Therefore, the chord lengths would be:[ a = 2 R' sin A ][ b = 2 R' sin B ][ c = 2 R' sin C ]But in the circumcircle, we have:[ a = 2 R sin A ][ b = 2 R sin B ][ c = 2 R sin C ]So, comparing these, we get:[ 2 R' sin A = 2 R sin A ][ 2 R' sin B = 2 R sin B ][ 2 R' sin C = 2 R sin C ]Which again implies ( R' = R ). So, this suggests that the radius ( R' ) must equal the circumradius ( R ). But that contradicts the problem statement which asks to prove that ( R' ) is not less than ( R ). So, maybe my approach is flawed.Perhaps I'm assuming that the central angles are the same, but in reality, if the triangle is inscribed in a different circle, the central angles could be different. Wait, but the inscribed angles are fixed because the triangle is fixed. So, the central angles must still be twice the inscribed angles. Therefore, the chord lengths must be proportional to the sines of the angles, and thus the radius must be the same.Wait, that seems to suggest that the circumradius is unique for a given triangle. So, if a triangle is inscribed in a circle, that circle must be the circumcircle, and its radius is uniquely determined. Therefore, the radius cannot be less than the circumradius because it is exactly the circumradius.But the problem says "an acute triangle is inscribed in a circle. Prove that the circle's radius is not less than the circumradius of the triangle." If the circle is the circumcircle, then the radius is equal to the circumradius, so it's not less than. But if the circle is not the circumcircle, then how can the triangle be inscribed in it? Because the circumcircle is the smallest circle that can contain the triangle with all vertices on the circumference.Wait, maybe the problem is considering circles that are not necessarily the circumcircle but contain the triangle, meaning the triangle is inscribed in the circle, but the circle might be larger. But in that case, the radius would have to be at least the circumradius because the circumradius is the smallest possible radius for a circle containing the triangle with all vertices on the circumference.So, perhaps the problem is trying to say that if you have an acute triangle inscribed in any circle, then the radius of that circle is at least the circumradius of the triangle. Which makes sense because the circumradius is the minimal such radius.But then why specify that the triangle is acute? Maybe for non-acute triangles, this isn't necessarily true? Let me think about that.If the triangle is obtuse, then the circumradius is still defined, but the circumcircle still passes through all three vertices. However, in an obtuse triangle, the circumradius is larger than half the length of the longest side. Wait, actually, in any triangle, the circumradius is given by ( R = frac{a}{2 sin A} ), and for an obtuse triangle, one of the angles is greater than 90 degrees, so its sine is still positive but less than 1. Wait, no, sine of angles greater than 90 degrees is still positive but less than or equal to 1. Wait, actually, sine of 90 degrees is 1, and sine decreases after that. So, for an obtuse angle, say 120 degrees, ( sin 120^circ = frac{sqrt{3}}{2} ), which is less than 1.So, for an obtuse triangle, the circumradius would be ( R = frac{a}{2 sin A} ), where ( A ) is the obtuse angle. Since ( sin A < 1 ), ( R ) would be greater than ( frac{a}{2} ). But the longest side is opposite the largest angle, which is the obtuse angle. So, the longest side is ( a ), and ( R = frac{a}{2 sin A} > frac{a}{2} ).But if we consider a circle with radius ( frac{a}{2} ), which is the radius of the circle with diameter ( a ), then the triangle can be inscribed in that circle because the obtuse angle would lie inside the circle. Wait, actually, in an obtuse triangle, the circumradius is larger than ( frac{a}{2} ), but the circle with diameter ( a ) would have a radius of ( frac{a}{2} ), and the triangle would be inscribed in it with the obtuse angle inside the circle.Wait, but in that case, the radius ( frac{a}{2} ) is less than the circumradius ( R ). So, for an obtuse triangle, you can have a circle with a radius smaller than the circumradius that still inscribes the triangle. Therefore, the statement that the circle's radius is not less than the circumradius is not true for obtuse triangles.So, putting it all together, for an acute triangle, any circle that inscribes it must have a radius at least equal to the circumradius, because the circumradius is the minimal such radius. But for an obtuse triangle, you can have a circle with a smaller radius that still inscribes the triangle, specifically the circle with diameter equal to the longest side, which has a radius smaller than the circumradius.Therefore, the statement is true for acute triangles but false for obtuse triangles.

💡First, I need to determine the total volume of the bucket, which I'll denote as ( V ) liters.Tap A fills the bucket at a rate of 3 liters per minute. This means that in one minute, Tap A adds 3 liters to the bucket.Next, Tap B fills one-third of the bucket in 20 minutes. To find the rate at which Tap B fills the bucket, I'll calculate the time it takes to fill the entire bucket. Since one-third takes 20 minutes, the whole bucket would take ( 20 times 3 = 60 ) minutes. Therefore, Tap B's rate is ( frac{V}{60} ) liters per minute.When both taps are working together, they fill the bucket in 10 minutes. This means their combined rate is ( frac{V}{10} ) liters per minute.I can now set up an equation to represent the combined rates of Tap A and Tap B:[3 + frac{V}{60} = frac{V}{10}]To solve for ( V ), I'll first eliminate the fractions by finding a common denominator, which is 60:[3 times 60 + V = 6V]This simplifies to:[180 + V = 6V]Subtracting ( V ) from both sides gives:[180 = 5V]Finally, dividing both sides by 5:[V = frac{180}{5} = 36]Therefore, the total volume of the bucket is 36 liters.

💡First, I need to solve the equation ( x - frac{1}{2} = frac{7}{8} - frac{2}{3} ).To isolate ( x ), I'll add ( frac{1}{2} ) to both sides of the equation:[x = frac{7}{8} - frac{2}{3} + frac{1}{2}]Next, I need to find a common denominator for the fractions ( frac{7}{8} ), ( frac{2}{3} ), and ( frac{1}{2} ). The least common multiple of 8, 3, and 2 is 24. I'll convert each fraction to have this common denominator:[frac{7}{8} = frac{21}{24}, quad frac{2}{3} = frac{16}{24}, quad text{and} quad frac{1}{2} = frac{12}{24}]Now, substitute these values back into the equation:[x = frac{21}{24} - frac{16}{24} + frac{12}{24}]Combine the fractions:[x = left( frac{21}{24} - frac{16}{24} right) + frac{12}{24} = frac{5}{24} + frac{12}{24} = frac{17}{24}]Since ( frac{17}{24} ) is already in its simplest form, no further simplification is needed.Therefore, the value of ( x ) that satisfies the equation is ( frac{17}{24} ).

💡Okay, so I have this problem about an ellipse, and I need to find its equation and then prove something about a point D on a line intersecting the ellipse. Let me try to break this down step by step.First, part (1) asks for the equation of the ellipse C. The ellipse is given by the standard equation x²/a² + y²/b² = 1, where a > b > 0. The right focus is F, and there's a point P(1, 3/2) on the ellipse. Also, PF is perpendicular to the x-axis. Hmm, okay.So, PF is perpendicular to the x-axis. That means the line PF is vertical. Since P is at (1, 3/2), and PF is vertical, the focus F must have the same x-coordinate as P. So, F is at (1, 0) because it's the right focus. Wait, actually, the foci of an ellipse are located at (±c, 0) where c is the distance from the center to the focus, and c² = a² - b².But in this case, since PF is vertical, and P is at (1, 3/2), the focus F must be at (1, 0). So, that tells me that the distance from the center to the focus is 1, so c = 1. Therefore, c² = 1 = a² - b². So, a² - b² = 1.Also, since P(1, 3/2) lies on the ellipse, plugging into the ellipse equation: (1)²/a² + (3/2)²/b² = 1. That simplifies to 1/a² + 9/(4b²) = 1.So now I have two equations:1. a² - b² = 12. 1/a² + 9/(4b²) = 1I need to solve these two equations to find a² and b².Let me denote u = 1/a² and v = 1/b². Then the equations become:1. 1/u - 1/v = 12. u + (9/4)v = 1Wait, no. Let me think again. From equation 1: a² - b² = 1, so a² = b² + 1.From equation 2: 1/a² + 9/(4b²) = 1. Let's substitute a² from equation 1 into equation 2.So, 1/(b² + 1) + 9/(4b²) = 1.Let me compute this:Let me denote b² = t, so t > 0.Then, 1/(t + 1) + 9/(4t) = 1.Multiply both sides by 4t(t + 1):4t + 9(t + 1) = 4t(t + 1)Simplify:4t + 9t + 9 = 4t² + 4tCombine like terms:13t + 9 = 4t² + 4tBring all terms to one side:4t² + 4t - 13t - 9 = 0Simplify:4t² - 9t - 9 = 0Now, solve for t using quadratic formula:t = [9 ± sqrt(81 + 144)] / 8Because discriminant D = 81 + 4*4*9 = 81 + 144 = 225So sqrt(225) = 15Thus, t = [9 ± 15]/8So two solutions:t = (9 + 15)/8 = 24/8 = 3t = (9 - 15)/8 = (-6)/8 = -3/4But t = b² must be positive, so t = 3.Thus, b² = 3, so a² = b² + 1 = 4.Therefore, the equation of the ellipse is x²/4 + y²/3 = 1.Okay, that seems solid. Let me just double-check:Point P(1, 3/2): 1²/4 + (3/2)²/3 = 1/4 + (9/4)/3 = 1/4 + 3/4 = 1. Correct.Focus at (1,0): c² = a² - b² = 4 - 3 = 1, so c = 1. So foci at (±1, 0). Correct.Alright, part (1) is done.Now, part (2): A line l passes through F(1,0) and intersects the ellipse at points A and B (different from P). D is a point on line l. The slopes of PA, PD, PB are k1, k2, k3 respectively, and k1 + k3 = 2k2. Need to prove that the abscissa (x-coordinate) of D is a constant.Hmm, okay. So, I need to show that the x-coordinate of D is always the same, regardless of the line l.Let me think about how to approach this. Maybe parametrize the line l, find points A and B, express the slopes k1, k2, k3, set up the condition k1 + k3 = 2k2, and solve for the coordinates of D.First, let's parametrize line l. Since it passes through F(1,0), we can write it in the form y = m(x - 1), where m is the slope. Alternatively, if the slope is infinite (vertical line), it would be x = 1, but since P is on x=1, and A and B are different from P, so x=1 would only intersect at P, which is not allowed. So, line l can't be vertical, so we can write it as y = m(x - 1).Alternatively, sometimes it's easier to parametrize using a parameter t, but maybe slope-intercept is okay.So, line l: y = m(x - 1). Let's find points A and B where this line intersects the ellipse x²/4 + y²/3 = 1.Substitute y = m(x - 1) into the ellipse equation:x²/4 + [m²(x - 1)²]/3 = 1Multiply both sides by 12 to eliminate denominators:3x² + 4m²(x - 1)² = 12Expand (x - 1)²: x² - 2x + 1So:3x² + 4m²(x² - 2x + 1) = 12Expand:3x² + 4m²x² - 8m²x + 4m² = 12Combine like terms:(3 + 4m²)x² - 8m²x + (4m² - 12) = 0So, quadratic in x: Ax² + Bx + C = 0, whereA = 3 + 4m²B = -8m²C = 4m² - 12Now, since the line passes through F(1,0), which is a focus, but not necessarily a point on the ellipse. Wait, actually, F is inside the ellipse because for an ellipse, the foci are inside. So, the line passes through F and intersects the ellipse at A and B.Therefore, the quadratic equation will have two roots: x = 1 (since F is on the line, but F is inside the ellipse, so x=1 is not on the ellipse. Wait, actually, F is not on the ellipse, so x=1 is not a root.Wait, but when we plug x=1 into the quadratic equation:A(1)^2 + B(1) + C = (3 + 4m²) - 8m² + (4m² - 12) = 3 + 4m² - 8m² + 4m² - 12 = (3 - 12) + (4m² - 8m² + 4m²) = -9 + 0 = -9 ≠ 0So, x=1 is not a root. Therefore, the quadratic equation has two roots x1 and x2, which are the x-coordinates of points A and B.Let me denote x1 and x2 as the roots. Then, by Vieta's formula:x1 + x2 = -B/A = (8m²)/(3 + 4m²)x1 * x2 = C/A = (4m² - 12)/(3 + 4m²)Similarly, the corresponding y-coordinates are y1 = m(x1 - 1) and y2 = m(x2 - 1).Now, point D is on line l, so its coordinates can be written as (x0, m(x0 - 1)).We need to find the slopes k1, k2, k3.Slope k1 is the slope of PA: (y1 - 3/2)/(x1 - 1)Similarly, k3 is the slope of PB: (y2 - 3/2)/(x2 - 1)Slope k2 is the slope of PD: (m(x0 - 1) - 3/2)/(x0 - 1)Given that k1 + k3 = 2k2.So, let's write expressions for k1, k2, k3.First, k1 = [y1 - 3/2]/(x1 - 1) = [m(x1 - 1) - 3/2]/(x1 - 1) = m - (3/2)/(x1 - 1)Similarly, k3 = [y2 - 3/2]/(x2 - 1) = m - (3/2)/(x2 - 1)k2 = [m(x0 - 1) - 3/2]/(x0 - 1) = m - (3/2)/(x0 - 1)Given that k1 + k3 = 2k2.So, let's write:[m - (3/2)/(x1 - 1)] + [m - (3/2)/(x2 - 1)] = 2[m - (3/2)/(x0 - 1)]Simplify left side:2m - (3/2)[1/(x1 - 1) + 1/(x2 - 1)] = 2m - (3/2)[(x1 + x2 - 2)/(x1x2 - x1 - x2 + 1)]Wait, let me compute 1/(x1 - 1) + 1/(x2 - 1):= [ (x2 - 1) + (x1 - 1) ] / [(x1 - 1)(x2 - 1)]= (x1 + x2 - 2) / [x1x2 - x1 - x2 + 1]So, left side becomes:2m - (3/2)*(x1 + x2 - 2)/(x1x2 - x1 - x2 + 1)Right side is:2m - (3/2)*(2)/(x0 - 1)So, equate both sides:2m - (3/2)*(x1 + x2 - 2)/(x1x2 - x1 - x2 + 1) = 2m - (3/2)*(2)/(x0 - 1)Subtract 2m from both sides:- (3/2)*(x1 + x2 - 2)/(x1x2 - x1 - x2 + 1) = - (3/2)*(2)/(x0 - 1)Multiply both sides by (-2/3):(x1 + x2 - 2)/(x1x2 - x1 - x2 + 1) = 2/(x0 - 1)So, (x1 + x2 - 2)/(x1x2 - x1 - x2 + 1) = 2/(x0 - 1)Therefore, x0 - 1 = 2*(x1x2 - x1 - x2 + 1)/(x1 + x2 - 2)So, x0 = 1 + 2*(x1x2 - x1 - x2 + 1)/(x1 + x2 - 2)Now, let's compute x1 + x2 and x1x2 from earlier.From Vieta's formulas:x1 + x2 = 8m²/(3 + 4m²)x1x2 = (4m² - 12)/(3 + 4m²)So, let's compute numerator and denominator:Numerator: x1x2 - x1 - x2 + 1 = x1x2 - (x1 + x2) + 1= (4m² - 12)/(3 + 4m²) - (8m²)/(3 + 4m²) + 1= [4m² - 12 - 8m²]/(3 + 4m²) + 1= (-4m² - 12)/(3 + 4m²) + 1= [(-4m² - 12) + (3 + 4m²)]/(3 + 4m²)= (-4m² - 12 + 3 + 4m²)/(3 + 4m²)= (-9)/(3 + 4m²)Denominator: x1 + x2 - 2 = (8m²)/(3 + 4m²) - 2= (8m² - 2*(3 + 4m²))/(3 + 4m²)= (8m² - 6 - 8m²)/(3 + 4m²)= (-6)/(3 + 4m²)So, putting numerator and denominator together:Numerator: -9/(3 + 4m²)Denominator: -6/(3 + 4m²)So, their ratio is (-9)/(-6) = 3/2Thus, x0 = 1 + 2*(3/2) = 1 + 3 = 4So, x0 = 4, which is a constant, independent of m.Therefore, the abscissa (x-coordinate) of point D is always 4.Wait, that seems to be the case. Let me just verify the calculations.We had:Numerator: x1x2 - x1 - x2 + 1 = (-9)/(3 + 4m²)Denominator: x1 + x2 - 2 = (-6)/(3 + 4m²)So, their ratio is (-9)/(-6) = 3/2Thus, x0 = 1 + 2*(3/2) = 4.Yes, that seems correct.Alternatively, let me think if there's another way to see this. Maybe using properties of ellipses or harmonic division, but I think the algebraic approach worked.So, regardless of the slope m of line l, the x-coordinate of D is always 4. Therefore, it's a constant.**Final Answer**The equation of the ellipse is boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}, and the abscissa of point ( D ) is the constant value boxed{4}.

💡Okay, so I have this geometry problem here. It says: "If a line l is perpendicular to every line in plane α, then line l is perpendicular to plane α." Then it asks about the converse, inverse, and contrapositive of this proposition and how many of them are true. The options are A: 3, B: 2, C: 1, D: 0.First, I need to make sure I understand the original proposition. It says that if a line is perpendicular to every single line in a plane, then it must be perpendicular to the entire plane. Hmm, I think that's a standard result in geometry. If a line is perpendicular to every line in a plane, it means it's perpendicular to the plane itself. So the original statement is true.Now, let's recall what converse, inverse, and contrapositive mean. - **Converse**: This is formed by switching the hypothesis and conclusion of the original statement. So, the converse would be: "If a line l is perpendicular to plane α, then line l is perpendicular to every line in plane α."- **Inverse**: This is formed by negating both the hypothesis and the conclusion of the original statement. So, the inverse would be: "If a line l is not perpendicular to every line in plane α, then line l is not perpendicular to plane α."- **Contrapositive**: This is formed by both switching and negating the hypothesis and conclusion. So, the contrapositive would be: "If a line l is not perpendicular to plane α, then line l is not perpendicular to every line in plane α."Alright, so now I need to evaluate the truth value of each of these.Starting with the **converse**: "If a line l is perpendicular to plane α, then line l is perpendicular to every line in plane α." Hmm, I think this is also true. Because if a line is perpendicular to a plane, by definition, it should be perpendicular to every line in that plane. So, the converse is true.Next, the **inverse**: "If a line l is not perpendicular to every line in plane α, then line l is not perpendicular to plane α." Let me think about this. If a line isn't perpendicular to every line in the plane, does that mean it's not perpendicular to the plane? Well, if it's not perpendicular to even one line in the plane, then it can't be perpendicular to the entire plane. So, yes, the inverse seems to be true as well.Lastly, the **contrapositive**: "If a line l is not perpendicular to plane α, then line l is not perpendicular to every line in plane α." This is logically equivalent to the original statement. Since the original statement is true, the contrapositive must also be true. So, the contrapositive is true.Wait, so all three—the converse, inverse, and contrapositive—are true? That would mean the answer is A: 3. But let me double-check to make sure I'm not making a mistake.Thinking about the original statement: If a line is perpendicular to every line in a plane, then it's perpendicular to the plane. That's definitely true because being perpendicular to every line in the plane implies it's the normal vector of the plane.The converse: If a line is perpendicular to the plane, then it's perpendicular to every line in the plane. That also makes sense because the line would be orthogonal to every direction in the plane.The inverse: If a line isn't perpendicular to every line in the plane, then it's not perpendicular to the plane. This is the contrapositive of the converse, actually. Since the converse is true, its contrapositive (which is the inverse) should also be true.Wait, hold on. Is the inverse actually the contrapositive of the converse? Let me clarify.Original: If P, then Q.Converse: If Q, then P.Inverse: If not P, then not Q.Contrapositive: If not Q, then not P.So, the inverse is not the contrapositive of the converse. The inverse is just the negation of both parts. But in this case, if the original is true, the contrapositive is true, and if the converse is true, then its contrapositive (which would be the inverse) is also true.But wait, no. The inverse is not necessarily the contrapositive of the converse. Let me think again.Original: P → QConverse: Q → PInverse: ~P → ~QContrapositive: ~Q → ~PSo, the inverse is ~P → ~Q, which is not the same as the contrapositive of the converse. The contrapositive of the converse would be ~P → ~Q, which is actually the inverse. So, in this case, the inverse is the contrapositive of the converse.Since the converse is true, its contrapositive (which is the inverse) is also true. So, yes, the inverse is true.Therefore, all three—the converse, inverse, and contrapositive—are true. So, the number of true propositions is 3, which is option A.But wait, let me think again. Sometimes in logic, the inverse isn't always true even if the original statement is. But in this case, since the original statement is a definition, maybe the inverse holds.Wait, the original statement is a definition? Or is it a theorem?In geometry, the definition of a line being perpendicular to a plane is that it is perpendicular to every line in the plane. So, actually, the original statement is a definition, which is always true. Therefore, its converse, inverse, and contrapositive would all be true because they are logically equivalent in this case.Wait, no. If it's a definition, then the original statement is an equivalence. So, P if and only if Q. Therefore, both the original and the converse are true by definition.But the inverse and contrapositive are also related. Since P ↔ Q is true, then ~P ↔ ~Q is also true, which would make both the inverse and the contrapositive true as well.So, yes, in this case, since the original statement is an equivalence (a definition), all four related statements (original, converse, inverse, contrapositive) are true. But the question only asks about converse, inverse, and contrapositive, so that's three true statements.Therefore, the answer should be A: 3.But wait, I'm a bit confused because sometimes in logic, the inverse isn't necessarily true. But in this case, since the original is a definition, it's an equivalence, so all related conditionals hold.Yes, I think that's correct. So, all three are true.**Final Answer**boxed{A}