💡Okay, so I have this math problem here, and I need to figure out which of the options A, B, C, or D equals 9. Let me take it step by step because I want to make sure I understand each part correctly.First, let me look at option A: (3 times 3 div 3 + 3). Hmm, okay, so I need to remember the order of operations, which is multiplication and division first, from left to right, and then addition and subtraction. So, starting with (3 times 3), that's 9. Then, I divide that by 3, so (9 div 3) is 3. Now, I add the last 3, so (3 + 3) equals 6. Wait, that's only 6, not 9. So option A doesn't work.Alright, moving on to option B: (3 div 3 + 3 times 3). Again, I'll follow the order of operations. First, (3 div 3) is 1, and (3 times 3) is 9. Now, I add those two results together: (1 + 9) equals 10. That's more than 9, so option B isn't the right answer either.Next is option C: (3 times 3 - 3 + 3). Let's break this down. First, (3 times 3) is 9. Then, I subtract 3 from that, which gives me (9 - 3 = 6). Finally, I add the last 3, so (6 + 3 = 9). Oh, that's exactly what we're looking for! So option C equals 9.Just to be thorough, let me check option D: (3 div 3 + 3 div 3). Following the order of operations, both divisions come first. So, (3 div 3) is 1, and the other (3 div 3) is also 1. Adding those together gives me (1 + 1 = 2). That's way too low, so option D isn't the answer.So, after going through each option carefully, it looks like option C is the one that equals 9.

💡Okay, so I have this problem where there's a square with a side length of 3 units. They're saying that n points are randomly selected inside this square. Out of these n points, m points are within a distance of less than 1 unit from any of the square's vertices. The question is asking for the experimental value of the circumference of π obtained using random simulation. Hmm, okay, let me try to break this down step by step.First, I need to visualize the square. It's a square with each side being 3 units long. So, the area of the square would be side length squared, which is 3*3 = 9 square units. That part seems straightforward.Now, the problem mentions that n points are randomly selected within this square. So, these points are uniformly distributed across the entire area of the square. That means each point has an equal chance of being anywhere inside the square.Next, among these n points, m points are at a distance less than 1 from the vertices of the square. So, for each vertex, there's a region around it where if a point falls within that region, it's considered to be within a distance of 1 from that vertex. Since the square has four vertices, there are four such regions.I think each of these regions is a quarter-circle with a radius of 1 unit. Because if you consider a circle of radius 1 around each vertex, the part of the circle that lies inside the square is a quarter-circle. So, each vertex contributes a quarter-circle area where points can be within 1 unit of that vertex.So, the total area covered by all four quarter-circles would be 4 times the area of a quarter-circle. The area of a full circle is πr², so a quarter-circle would be (π*1²)/4 = π/4. Therefore, four quarter-circles would have a total area of 4*(π/4) = π. So, the combined area where points are within 1 unit of any vertex is π square units.Now, since the points are randomly distributed, the probability that a point falls within this area is the ratio of the area of the four quarter-circles to the area of the entire square. So, the probability P is π/9, because the area of the square is 9.In terms of the points selected, the expected number of points within this area would be n*(π/9). But in reality, we have m points within this area. So, we can set up the equation m = n*(π/9). Solving for π, we get π = (9m)/n.Therefore, the experimental value of π obtained from this random simulation would be 9m divided by n.Wait, let me double-check that. The area where points are within 1 unit of any vertex is π, and the total area is 9. So, the ratio is π/9, which should be equal to the ratio of points m/n. So, π/9 = m/n, which implies π = 9m/n. Yeah, that seems correct.I think that's the reasoning. The key was recognizing that the regions around each vertex are quarter-circles, calculating their combined area, and then using the ratio of areas to estimate π based on the proportion of points falling within those regions.

💡First, I need to convert the number 1300000 into scientific notation.To do this, I'll move the decimal point to the right of the first non-zero digit, which is 1, resulting in 1.3.Next, I'll count how many places I moved the decimal point. In this case, I moved it 6 places to the left.Finally, I'll express the number in scientific notation by multiplying 1.3 by 10 raised to the power of 6.Therefore, 1300000 in scientific notation is 1.3 × 10⁶.

💡First, I recognize that there were initially 31 students in the fourth grade at the start of the year.During the year, 5 students left the school. To find out how many students remained after these departures, I subtract the number of students who left from the initial number of students:31 - 5 = 26 students remained.Next, 11 new students joined the school. To determine the total number of students at the end of the year, I add the number of new students to the remaining number:26 + 11 = 37 students.Therefore, there were 37 students in the fourth grade at the end of the year.

💡Okay, so I have this problem where there's a circle given by the equation ( x^2 + y^2 = 4 ) and a hyperbola given by ( x^2 - frac{y^2}{4} = 1 ) with the condition that ( x > 0 ). There's a line ( l ) that's tangent to the circle and intersects the hyperbola at two points, ( M ) and ( N ). I need to find the equation of the locus of the intersection point of the tangents to the hyperbola at points ( M ) and ( N ).First, I need to understand what each part means. The circle has a radius of 2, centered at the origin. The hyperbola is a standard hyperbola opening to the left and right, but since ( x > 0 ), we're only considering the right branch. The line ( l ) is tangent to the circle, meaning it touches the circle at exactly one point, and this line also intersects the hyperbola at two distinct points ( M ) and ( N ).I think I need to find the equation of the tangent line ( l ) to the circle first. The general equation of a tangent to a circle ( x^2 + y^2 = r^2 ) at a point ( (x_1, y_1) ) is ( xx_1 + yy_1 = r^2 ). Since our circle has radius 2, the tangent line at ( (x_1, y_1) ) would be ( xx_1 + yy_1 = 4 ).But wait, the line ( l ) is tangent to the circle, so it can be represented in this form. However, I don't know the point of tangency yet. Maybe I can represent the tangent line in a different form, like ( y = mx + c ), and use the condition that the distance from the center of the circle (which is the origin) to the line is equal to the radius.Yes, that sounds better. The distance from the origin to the line ( y = mx + c ) is ( frac{|c|}{sqrt{1 + m^2}}} ). Since the line is tangent to the circle, this distance must be equal to the radius, which is 2. So,[frac{|c|}{sqrt{1 + m^2}} = 2 implies |c| = 2sqrt{1 + m^2}]So, the equation of the tangent line can be written as ( y = mx pm 2sqrt{1 + m^2} ).Now, this line intersects the hyperbola ( x^2 - frac{y^2}{4} = 1 ). Let's substitute ( y ) from the line into the hyperbola equation.Substituting ( y = mx + c ) into the hyperbola:[x^2 - frac{(mx + c)^2}{4} = 1]Expanding this:[x^2 - frac{m^2x^2 + 2mcx + c^2}{4} = 1]Multiply through by 4 to eliminate the denominator:[4x^2 - (m^2x^2 + 2mcx + c^2) = 4]Simplify:[4x^2 - m^2x^2 - 2mcx - c^2 - 4 = 0]Combine like terms:[(4 - m^2)x^2 - 2mcx - (c^2 + 4) = 0]This is a quadratic equation in ( x ). Since the line intersects the hyperbola at two distinct points ( M ) and ( N ), the discriminant of this quadratic must be positive.The discriminant ( D ) is:[D = ( -2mc )^2 - 4(4 - m^2)( - (c^2 + 4) )]Simplify:[D = 4m^2c^2 - 4(4 - m^2)( -c^2 - 4 )]Factor out the 4:[D = 4[ m^2c^2 - (4 - m^2)( -c^2 - 4 ) ]]Let me compute the term inside the brackets:First, expand ( (4 - m^2)( -c^2 - 4 ) ):[(4 - m^2)( -c^2 - 4 ) = -4c^2 - 16 + m^2c^2 + 4m^2]So,[m^2c^2 - ( -4c^2 - 16 + m^2c^2 + 4m^2 ) = m^2c^2 + 4c^2 + 16 - m^2c^2 - 4m^2]Simplify:[4c^2 + 16 - 4m^2]So, the discriminant becomes:[D = 4(4c^2 + 16 - 4m^2) = 16c^2 + 64 - 16m^2]Since the discriminant must be positive:[16c^2 + 64 - 16m^2 > 0 implies c^2 + 4 - m^2 > 0]But from earlier, we have ( c = pm 2sqrt{1 + m^2} ). Let's substitute ( c^2 = 4(1 + m^2) ) into the inequality:[4(1 + m^2) + 4 - m^2 > 0 implies 4 + 4m^2 + 4 - m^2 > 0 implies 3m^2 + 8 > 0]Which is always true since ( m^2 ) is non-negative. So, the discriminant condition is satisfied for all real ( m ).Now, moving on. The points ( M ) and ( N ) lie on both the line ( l ) and the hyperbola. Let me denote their coordinates as ( M(x_1, y_1) ) and ( N(x_2, y_2) ).I need to find the equations of the tangents to the hyperbola at ( M ) and ( N ), and then find their intersection point. The locus of this intersection point as the tangent line ( l ) varies is what I need to find.First, let's recall the equation of the tangent to the hyperbola ( x^2 - frac{y^2}{4} = 1 ) at a point ( (x_0, y_0) ) on the hyperbola. The equation is:[xx_0 - frac{yy_0}{4} = 1]So, for point ( M(x_1, y_1) ), the tangent is:[xx_1 - frac{yy_1}{4} = 1]Similarly, for point ( N(x_2, y_2) ), the tangent is:[xx_2 - frac{yy_2}{4} = 1]Let me denote the intersection point of these two tangents as ( P(h, k) ). So, ( P ) lies on both tangents, meaning:1. ( h x_1 - frac{k y_1}{4} = 1 )2. ( h x_2 - frac{k y_2}{4} = 1 )So, we have a system of equations:[begin{cases}h x_1 - frac{k y_1}{4} = 1 h x_2 - frac{k y_2}{4} = 1end{cases}]This can be written in matrix form as:[begin{pmatrix}x_1 & -frac{y_1}{4} x_2 & -frac{y_2}{4}end{pmatrix}begin{pmatrix}h kend{pmatrix}=begin{pmatrix}1 1end{pmatrix}]To solve for ( h ) and ( k ), we can use Cramer's Rule. The determinant of the coefficient matrix is:[D = x_1 left(-frac{y_2}{4}right) - x_2 left(-frac{y_1}{4}right) = -frac{x_1 y_2}{4} + frac{x_2 y_1}{4} = frac{x_2 y_1 - x_1 y_2}{4}]Assuming ( D neq 0 ), which should be the case since ( M ) and ( N ) are distinct points.Now, the determinant for ( h ) is:[D_h = begin{vmatrix}1 & -frac{y_1}{4} 1 & -frac{y_2}{4}end{vmatrix}= 1 cdot left(-frac{y_2}{4}right) - 1 cdot left(-frac{y_1}{4}right) = -frac{y_2}{4} + frac{y_1}{4} = frac{y_1 - y_2}{4}]Similarly, the determinant for ( k ) is:[D_k = begin{vmatrix}x_1 & 1 x_2 & 1end{vmatrix}= x_1 cdot 1 - x_2 cdot 1 = x_1 - x_2]Therefore, using Cramer's Rule:[h = frac{D_h}{D} = frac{frac{y_1 - y_2}{4}}{frac{x_2 y_1 - x_1 y_2}{4}} = frac{y_1 - y_2}{x_2 y_1 - x_1 y_2}][k = frac{D_k}{D} = frac{x_1 - x_2}{frac{x_2 y_1 - x_1 y_2}{4}} = frac{4(x_1 - x_2)}{x_2 y_1 - x_1 y_2}]Hmm, this seems a bit complicated. Maybe there's a better way to approach this.Alternatively, since both ( M ) and ( N ) lie on the line ( l ), which is tangent to the circle, perhaps I can parameterize the tangent line.Let me consider the tangent line to the circle ( x^2 + y^2 = 4 ). A general tangent to this circle can be written as ( x cos theta + y sin theta = 2 ), where ( theta ) is the angle made with the x-axis.Yes, that's another way to represent the tangent line. So, maybe using parametric equations for the tangent line.Let me set ( l ) as ( x cos theta + y sin theta = 2 ). Then, this line intersects the hyperbola ( x^2 - frac{y^2}{4} = 1 ).Let me substitute ( y ) from the line equation into the hyperbola.From ( x cos theta + y sin theta = 2 ), solve for ( y ):[y = frac{2 - x cos theta}{sin theta}]Substitute into the hyperbola:[x^2 - frac{left( frac{2 - x cos theta}{sin theta} right)^2}{4} = 1]Simplify:[x^2 - frac{(2 - x cos theta)^2}{4 sin^2 theta} = 1]Multiply through by ( 4 sin^2 theta ):[4 sin^2 theta cdot x^2 - (2 - x cos theta)^2 = 4 sin^2 theta]Expand ( (2 - x cos theta)^2 ):[4 - 4 x cos theta + x^2 cos^2 theta]So, substituting back:[4 sin^2 theta cdot x^2 - (4 - 4 x cos theta + x^2 cos^2 theta) = 4 sin^2 theta]Simplify:[4 sin^2 theta x^2 - 4 + 4 x cos theta - x^2 cos^2 theta - 4 sin^2 theta = 0]Combine like terms:[(4 sin^2 theta - cos^2 theta) x^2 + 4 cos theta cdot x - (4 + 4 sin^2 theta) = 0]This is a quadratic in ( x ). Let me denote this as:[A x^2 + B x + C = 0]Where:- ( A = 4 sin^2 theta - cos^2 theta )- ( B = 4 cos theta )- ( C = -4(1 + sin^2 theta) )The solutions to this quadratic will give me the x-coordinates of points ( M ) and ( N ). Let me denote them as ( x_1 ) and ( x_2 ).From quadratic equation, the sum and product of roots are:- ( x_1 + x_2 = -B/A = - (4 cos theta) / (4 sin^2 theta - cos^2 theta) )- ( x_1 x_2 = C/A = (-4(1 + sin^2 theta)) / (4 sin^2 theta - cos^2 theta) )Similarly, the corresponding ( y )-coordinates can be found using the line equation ( y = frac{2 - x cos theta}{sin theta} ).So,- ( y_1 = frac{2 - x_1 cos theta}{sin theta} )- ( y_2 = frac{2 - x_2 cos theta}{sin theta} )Now, I need to find the equations of the tangents at ( M ) and ( N ) to the hyperbola.As before, the tangent at ( (x_0, y_0) ) is ( x x_0 - frac{y y_0}{4} = 1 ).So, for ( M(x_1, y_1) ), the tangent is:[x x_1 - frac{y y_1}{4} = 1]Similarly, for ( N(x_2, y_2) ), the tangent is:[x x_2 - frac{y y_2}{4} = 1]Let me denote the intersection point of these two tangents as ( P(h, k) ). So, substituting ( (h, k) ) into both equations:1. ( h x_1 - frac{k y_1}{4} = 1 )2. ( h x_2 - frac{k y_2}{4} = 1 )This is the same system as before. So, perhaps I can express ( h ) and ( k ) in terms of ( x_1, x_2, y_1, y_2 ).Alternatively, maybe I can find a relationship between ( h ) and ( k ) by eliminating ( x_1, x_2, y_1, y_2 ).But this seems complicated. Maybe I can find a parametric equation for ( P(h, k) ) in terms of ( theta ), since the tangent line is parameterized by ( theta ).Let me try that.From earlier, we have expressions for ( x_1 + x_2 ) and ( x_1 x_2 ), as well as ( y_1 ) and ( y_2 ) in terms of ( x_1 ) and ( x_2 ).Let me express ( y_1 ) and ( y_2 ) in terms of ( x_1 ) and ( x_2 ):[y_1 = frac{2 - x_1 cos theta}{sin theta}, quad y_2 = frac{2 - x_2 cos theta}{sin theta}]So, ( y_1 - y_2 = frac{(2 - x_1 cos theta) - (2 - x_2 cos theta)}{sin theta} = frac{(x_2 - x_1) cos theta}{sin theta} )Similarly, ( x_2 y_1 - x_1 y_2 = x_2 cdot frac{2 - x_1 cos theta}{sin theta} - x_1 cdot frac{2 - x_2 cos theta}{sin theta} )Simplify:[frac{x_2 (2 - x_1 cos theta) - x_1 (2 - x_2 cos theta)}{sin theta} = frac{2 x_2 - x_1 x_2 cos theta - 2 x_1 + x_1 x_2 cos theta}{sin theta} = frac{2(x_2 - x_1)}{sin theta}]So, ( x_2 y_1 - x_1 y_2 = frac{2(x_2 - x_1)}{sin theta} )From earlier, ( h = frac{y_1 - y_2}{x_2 y_1 - x_1 y_2} = frac{frac{(x_2 - x_1) cos theta}{sin theta}}{frac{2(x_2 - x_1)}{sin theta}} = frac{cos theta}{2} )Similarly, ( k = frac{4(x_1 - x_2)}{x_2 y_1 - x_1 y_2} = frac{4(x_1 - x_2)}{frac{2(x_2 - x_1)}{sin theta}} = frac{4(x_1 - x_2) sin theta}{2(x_2 - x_1)} = frac{4 sin theta}{2} cdot frac{(x_1 - x_2)}{(x_2 - x_1)} = -2 sin theta )Wait, let me check that again.From ( k = frac{4(x_1 - x_2)}{x_2 y_1 - x_1 y_2} ), and ( x_2 y_1 - x_1 y_2 = frac{2(x_2 - x_1)}{sin theta} ), so:[k = frac{4(x_1 - x_2)}{frac{2(x_2 - x_1)}{sin theta}} = frac{4(x_1 - x_2) sin theta}{2(x_2 - x_1)} = frac{4 sin theta (x_1 - x_2)}{2 (x_2 - x_1)} = frac{4 sin theta (-1)(x_2 - x_1)}{2 (x_2 - x_1)} = -2 sin theta]Yes, that's correct.So, we have:[h = frac{cos theta}{2}, quad k = -2 sin theta]So, ( h = frac{cos theta}{2} ) and ( k = -2 sin theta ). Let's express ( cos theta ) and ( sin theta ) in terms of ( h ) and ( k ):[cos theta = 2h, quad sin theta = -frac{k}{2}]Since ( cos^2 theta + sin^2 theta = 1 ), substitute:[(2h)^2 + left(-frac{k}{2}right)^2 = 1 implies 4h^2 + frac{k^2}{4} = 1]Multiply through by 4 to eliminate the fraction:[16h^2 + k^2 = 4]So, the equation relating ( h ) and ( k ) is ( 16h^2 + k^2 = 4 ). Therefore, the locus of the intersection point ( P(h, k) ) is the ellipse given by ( 16x^2 + y^2 = 4 ).But wait, we need to consider the condition ( x > 0 ) for the hyperbola. Since ( h = frac{cos theta}{2} ), and ( cos theta ) can range between -1 and 1, ( h ) can range between -0.5 and 0.5. However, since the hyperbola is only for ( x > 0 ), we need to see if ( h ) is positive.But in our case, ( h = frac{cos theta}{2} ), and ( cos theta ) can be positive or negative. However, the hyperbola only has points with ( x > 0 ), so the intersection points ( M ) and ( N ) must have positive ( x )-coordinates. Therefore, the line ( l ) must intersect the hyperbola in the right branch, which implies that the tangent line must be such that the solutions ( x_1 ) and ( x_2 ) are positive.Looking back at the quadratic equation in ( x ):[(4 sin^2 theta - cos^2 theta) x^2 + 4 cos theta cdot x - 4(1 + sin^2 theta) = 0]For both roots ( x_1 ) and ( x_2 ) to be positive, the following conditions must be satisfied:1. The quadratic must have real roots, which we already ensured by the discriminant being positive.2. The product of the roots ( x_1 x_2 = frac{C}{A} = frac{-4(1 + sin^2 theta)}{4 sin^2 theta - cos^2 theta} ) must be positive.3. The sum of the roots ( x_1 + x_2 = -frac{B}{A} = -frac{4 cos theta}{4 sin^2 theta - cos^2 theta} ) must be positive.Let me analyze the product ( x_1 x_2 ):[x_1 x_2 = frac{-4(1 + sin^2 theta)}{4 sin^2 theta - cos^2 theta}]For this to be positive, the numerator and denominator must have the same sign.Numerator: ( -4(1 + sin^2 theta) ) is always negative since ( 1 + sin^2 theta geq 1 ).Denominator: ( 4 sin^2 theta - cos^2 theta ). Let's see when this is negative.( 4 sin^2 theta - cos^2 theta < 0 implies 4 sin^2 theta < cos^2 theta implies 4 tan^2 theta < 1 implies tan^2 theta < frac{1}{4} implies |tan theta| < frac{1}{2} )So, the denominator is negative when ( |tan theta| < frac{1}{2} ), which makes the product ( x_1 x_2 ) positive since numerator is negative and denominator is negative.Similarly, if ( 4 sin^2 theta - cos^2 theta > 0 ), the denominator is positive, and the numerator is negative, so the product is negative, which would imply that one root is positive and one is negative. But since we're only considering ( x > 0 ), we need both roots to be positive, so the product must be positive, hence the denominator must be negative.Therefore, ( 4 sin^2 theta - cos^2 theta < 0 implies tan^2 theta < frac{1}{4} implies |theta| < arctanleft( frac{1}{2} right) approx 26.565^circ ).So, ( theta ) is restricted to ( (-arctan(1/2), arctan(1/2)) ).Now, considering the sum ( x_1 + x_2 = -frac{4 cos theta}{4 sin^2 theta - cos^2 theta} ).Since ( 4 sin^2 theta - cos^2 theta < 0 ), the denominator is negative. The numerator is ( -4 cos theta ). For the sum to be positive:[-frac{4 cos theta}{text{negative}} > 0 implies -4 cos theta > 0 implies cos theta < 0]But wait, ( theta ) is restricted to ( (-arctan(1/2), arctan(1/2)) ), so ( cos theta ) is positive because ( theta ) is between -26.565° and 26.565°, where cosine is positive.Therefore, ( -4 cos theta ) is negative, and the denominator is negative, so the sum ( x_1 + x_2 ) is positive.Wait, let me double-check:Sum ( x_1 + x_2 = -frac{4 cos theta}{4 sin^2 theta - cos^2 theta} )Denominator is negative, numerator is ( -4 cos theta ). Since ( cos theta > 0 ) in this range, numerator is negative. So, negative divided by negative is positive. So, yes, the sum is positive.Therefore, both roots ( x_1 ) and ( x_2 ) are positive when ( |theta| < arctan(1/2) ).So, going back, the locus of ( P(h, k) ) is ( 16h^2 + k^2 = 4 ), but with ( h = frac{cos theta}{2} ) and ( theta ) restricted such that ( |theta| < arctan(1/2) ).So, ( h = frac{cos theta}{2} ), and since ( |theta| < arctan(1/2) ), ( cos theta ) ranges from ( cos(arctan(1/2)) ) to 1.Compute ( cos(arctan(1/2)) ):Let ( phi = arctan(1/2) ), so ( tan phi = 1/2 ). Then, ( cos phi = frac{2}{sqrt{1 + (1/2)^2}} = frac{2}{sqrt{5/4}} = frac{2}{sqrt{5}/2} = frac{4}{sqrt{5}} ). Wait, that can't be right because ( cos phi ) must be less than 1.Wait, let's compute it correctly.If ( tan phi = 1/2 ), then in a right triangle, the opposite side is 1, adjacent side is 2, hypotenuse is ( sqrt{1 + 4} = sqrt{5} ). Therefore, ( cos phi = frac{2}{sqrt{5}} approx 0.894 ).So, ( cos theta ) ranges from ( frac{2}{sqrt{5}} ) to 1 as ( theta ) goes from ( arctan(1/2) ) to 0. Therefore, ( h = frac{cos theta}{2} ) ranges from ( frac{1}{sqrt{5}} ) to ( frac{1}{2} ).So, the locus is the ellipse ( 16x^2 + y^2 = 4 ) with ( x ) restricted to ( frac{1}{sqrt{5}} < x leq frac{1}{2} ).Wait, let me compute ( frac{2}{sqrt{5}} / 2 = frac{1}{sqrt{5}} approx 0.447 ), and ( 1/2 = 0.5 ). So, ( x ) ranges from approximately 0.447 to 0.5.Therefore, the equation of the locus is ( 16x^2 + y^2 = 4 ) with ( frac{1}{sqrt{5}} < x leq frac{1}{2} ).But let me write ( frac{1}{sqrt{5}} ) as ( frac{sqrt{5}}{5} ) for rationalization.So, the final answer is ( 16x^2 + y^2 = 4 ) with ( frac{sqrt{5}}{5} < x leq frac{1}{2} ).**Final Answer**The equation of the locus is boxed{16x^2 + y^2 = 4} with the condition that boxed{frac{sqrt{5}}{5} < x leq frac{1}{2}}.However, since the problem asks for the equation of the locus, and the condition is part of the locus, I think it's appropriate to present both together. So, the final boxed answer is:boxed{16x^2 + y^2 = 4, quad left( frac{sqrt{5}}{5} < x leq frac{1}{2} right)}

💡Okay, so I have this problem where I need to find the angle PQR given three points in 3D space: P(-3,1,7), Q(-4,1,3), and R(-5,0,4). Hmm, angle PQR means the angle at point Q, right? So, it's the angle between the lines QP and QR. First, I think I need to find the vectors QP and QR. To do that, I can subtract the coordinates of Q from P and R respectively. Let me write that down.Vector QP is P - Q, so that would be (-3 - (-4), 1 - 1, 7 - 3) which simplifies to (1, 0, 4). Similarly, vector QR is R - Q, which is (-5 - (-4), 0 - 1, 4 - 3) simplifying to (-1, -1, 1). Now, to find the angle between these two vectors, I remember that the dot product formula relates the cosine of the angle to the dot product of the vectors divided by the product of their magnitudes. The formula is:cos(theta) = (QP · QR) / (|QP| |QR|)Let me compute the dot product first. QP is (1, 0, 4) and QR is (-1, -1, 1). So, the dot product is (1)(-1) + (0)(-1) + (4)(1) = -1 + 0 + 4 = 3.Next, I need the magnitudes of QP and QR. The magnitude of QP is sqrt(1^2 + 0^2 + 4^2) = sqrt(1 + 0 + 16) = sqrt(17). The magnitude of QR is sqrt((-1)^2 + (-1)^2 + 1^2) = sqrt(1 + 1 + 1) = sqrt(3).So, plugging these into the formula, cos(theta) = 3 / (sqrt(17) * sqrt(3)) = 3 / sqrt(51). Hmm, sqrt(51) is approximately 7.1414, so 3 divided by that is roughly 0.4196. To find theta, I take the arccosine of 0.4196. Using a calculator, arccos(0.4196) is approximately 65 degrees. So, the angle PQR is about 65 degrees.Wait, let me double-check my calculations. The vectors QP and QR seem correct. The dot product is indeed 3, and the magnitudes are sqrt(17) and sqrt(3). So, 3 divided by sqrt(51) is correct. Converting that to degrees gives around 65 degrees. Yeah, that seems right.