💡Okay, so I have this problem about a curve C given by parametric equations, and I need to find its standard and polar equations, and then find the distance between two points A and B on this curve. Let me try to break this down step by step.First, part (I) asks for the standard equation and polar equation of curve C. The parametric equations given are:x = 2 cos φ y = 2 + 2 sin φHmm, these look familiar. I remember that parametric equations involving sine and cosine usually represent circles or ellipses. Since both x and y are expressed in terms of sine and cosine with the same coefficient (2), this is likely a circle.To find the standard equation, I should eliminate the parameter φ. I know that cos²φ + sin²φ = 1. So, let's express cos φ and sin φ from the parametric equations.From x = 2 cos φ, we can write cos φ = x / 2. From y = 2 + 2 sin φ, subtract 2 from both sides to get y - 2 = 2 sin φ, so sin φ = (y - 2) / 2.Now, plug these into the Pythagorean identity:cos²φ + sin²φ = 1 (x/2)² + ((y - 2)/2)² = 1 (x²)/4 + ((y - 2)²)/4 = 1 Multiply both sides by 4 to eliminate the denominators: x² + (y - 2)² = 4So, the standard equation of curve C is x² + (y - 2)² = 4. That makes sense—it's a circle with center at (0, 2) and radius 2.Now, for the polar equation. I remember that polar coordinates relate to Cartesian coordinates via x = ρ cos θ and y = ρ sin θ. So, let's substitute these into the standard equation.Starting with x² + (y - 2)² = 4, substitute x and y:(ρ cos θ)² + (ρ sin θ - 2)² = 4 Expand both terms: ρ² cos²θ + (ρ sin θ - 2)² = 4 First term is ρ² cos²θ. Second term: (ρ sin θ - 2)² = ρ² sin²θ - 4ρ sin θ + 4.So, putting it all together: ρ² cos²θ + ρ² sin²θ - 4ρ sin θ + 4 = 4 Combine like terms: ρ² (cos²θ + sin²θ) - 4ρ sin θ + 4 = 4 Since cos²θ + sin²θ = 1: ρ² - 4ρ sin θ + 4 = 4 Subtract 4 from both sides: ρ² - 4ρ sin θ = 0 Factor out ρ: ρ(ρ - 4 sin θ) = 0So, either ρ = 0 or ρ = 4 sin θ. Since ρ = 0 is just the origin, which is a single point, the polar equation of the curve is ρ = 4 sin θ. That seems right because it's a circle above the origin.Okay, part (I) seems done. Now, part (II) asks for the distance |AB| where points A and B are on curve C with polar coordinates (ρ₁, π/3) and (ρ₂, 5π/6) respectively.First, I need to find the Cartesian coordinates of points A and B. Since they are given in polar coordinates, I can convert them using x = ρ cos θ and y = ρ sin θ.But wait, I also know that these points lie on curve C, whose polar equation is ρ = 4 sin θ. So, for point A, θ = π/3, so ρ₁ = 4 sin(π/3). Similarly, for point B, θ = 5π/6, so ρ₂ = 4 sin(5π/6).Let me compute ρ₁ and ρ₂ first.For point A: ρ₁ = 4 sin(π/3) I remember that sin(π/3) is √3/2, so ρ₁ = 4*(√3/2) = 2√3.For point B: ρ₂ = 4 sin(5π/6) Sin(5π/6) is also √3/2 because 5π/6 is in the second quadrant where sine is positive. So, ρ₂ = 4*(√3/2) = 2√3 as well.Wait, both ρ₁ and ρ₂ are 2√3? Interesting. So, points A and B both have the same radius but different angles.Now, let's find their Cartesian coordinates.For point A: x₁ = ρ₁ cos(π/3) = 2√3 * cos(π/3) Cos(π/3) is 1/2, so x₁ = 2√3*(1/2) = √3 y₁ = ρ₁ sin(π/3) = 2√3 * (√3/2) = 2√3*(√3)/2 = (2*3)/2 = 3 So, point A is (√3, 3)For point B: x₂ = ρ₂ cos(5π/6) = 2√3 * cos(5π/6) Cos(5π/6) is -√3/2, so x₂ = 2√3*(-√3/2) = - (2√3*√3)/2 = - (6)/2 = -3 Wait, hold on, that can't be right. Let me recalculate.Wait, 2√3 * (-√3/2) = (2√3)*( -√3)/2 = (2*(-3))/2 = (-6)/2 = -3. Hmm, okay, that seems correct.y₂ = ρ₂ sin(5π/6) = 2√3 * sin(5π/6) Sin(5π/6) is 1/2, so y₂ = 2√3*(1/2) = √3 Wait, that doesn't seem right. Wait, sin(5π/6) is actually 1/2? Wait, no, sin(π/6) is 1/2, and sin(5π/6) is also 1/2 because it's in the second quadrant. So yes, y₂ = 2√3*(1/2) = √3.Wait, but hold on, point B is at (x₂, y₂) = (-3, √3). But let me check that again because when I calculated x₂, I got -3, which seems a bit large because the circle has a radius of 2. Wait, is that possible?Wait, the circle is centered at (0, 2) with radius 2, so the maximum x-coordinate would be 2, right? So, if point B has x-coordinate -3, that's outside the circle. That can't be right. I must have made a mistake.Wait, let's go back. The polar equation is ρ = 4 sin θ. So, for θ = 5π/6, ρ = 4 sin(5π/6) = 4*(1/2) = 2. Wait, hold on, earlier I thought sin(5π/6) was √3/2, but actually, sin(5π/6) is 1/2. Let me confirm.Yes, sin(π/6) = 1/2, and sin(5π/6) = sin(π - π/6) = sin(π/6) = 1/2. So, ρ₂ = 4*(1/2) = 2, not 2√3. I think I messed up earlier.Wait, so I think I made a mistake when calculating ρ₁ and ρ₂. Let's recalculate.For point A: θ = π/3 ρ₁ = 4 sin(π/3) = 4*(√3/2) = 2√3 That's correct.For point B: θ = 5π/6 ρ₂ = 4 sin(5π/6) = 4*(1/2) = 2 Ah, okay, so ρ₂ is 2, not 2√3. That makes more sense because the radius of the circle is 2, so ρ can't be more than 4, but in this case, it's 2.So, point B has ρ = 2, θ = 5π/6.Now, let's find the Cartesian coordinates again.For point A: x₁ = ρ₁ cos(π/3) = 2√3 * cos(π/3) = 2√3*(1/2) = √3 y₁ = ρ₁ sin(π/3) = 2√3*(√3/2) = (2√3*√3)/2 = (6)/2 = 3 So, point A is (√3, 3)For point B: x₂ = ρ₂ cos(5π/6) = 2 * cos(5π/6) Cos(5π/6) is -√3/2, so x₂ = 2*(-√3/2) = -√3 y₂ = ρ₂ sin(5π/6) = 2 * sin(5π/6) = 2*(1/2) = 1 So, point B is (-√3, 1)Okay, that makes more sense. Now, both points are on the circle x² + (y - 2)² = 4. Let's verify.For point A: (√3)² + (3 - 2)² = 3 + 1 = 4. Correct. For point B: (-√3)² + (1 - 2)² = 3 + 1 = 4. Correct.Good, so points A and B are correctly found.Now, to find the distance |AB|, I can use the distance formula between points A(√3, 3) and B(-√3, 1).The distance formula is sqrt[(x₂ - x₁)² + (y₂ - y₁)²]Plugging in the coordinates:Distance = sqrt[(-√3 - √3)² + (1 - 3)²] Simplify:First, (-√3 - √3) = -2√3 Second, (1 - 3) = -2So, distance = sqrt[(-2√3)² + (-2)²] = sqrt[(4*3) + 4] = sqrt[12 + 4] = sqrt[16] = 4So, the distance |AB| is 4.Alternatively, since both points are on the circle, and the angle between them at the origin is π/3 to 5π/6, which is a difference of 5π/6 - π/3 = 5π/6 - 2π/6 = 3π/6 = π/2. So, the angle between OA and OB is π/2, which means triangle AOB is a right-angled triangle with OA and OB as legs, and AB as the hypotenuse.Given that OA = ρ₁ = 2√3 and OB = ρ₂ = 2, then by Pythagoras, AB² = OA² + OB² = (2√3)² + 2² = 12 + 4 = 16, so AB = 4. That's another way to see it.So, both methods confirm that |AB| is 4.I think that's it. I went through the steps, made a mistake initially with calculating ρ₂, but corrected it, and now everything checks out.

💡Okay, so I have this problem here: If ( r > 0 ) and ( p, q > 0 ) such that ( p^2r > q^2r ), then which of the following must be true? The options are:1. ( -p > -q )2. ( -p > q )3. ( 1 > -frac{q}{p} )4. ( 1 < frac{q}{p} )5. None of theseAnd the choices are labeled A to E, with E being "None of these." Alright, let's try to figure this out step by step.First, let's parse the given information. We know that ( r ) is positive, and both ( p ) and ( q ) are positive numbers. The inequality given is ( p^2r > q^2r ). Since ( r ) is positive, I can divide both sides of the inequality by ( r ) without changing the direction of the inequality. That would give me ( p^2 > q^2 ).Now, ( p^2 > q^2 ) implies that ( p ) is greater than ( q ) because both ( p ) and ( q ) are positive. Taking square roots on both sides (since they are positive, the square root function is increasing), we get ( p > q ).Alright, so we've established that ( p > q ). Now, let's look at each of the options one by one to see which must be true.1. ( -p > -q ): Hmm, this seems like it's related to ( p > q ). If I multiply both sides of ( p > q ) by -1, the inequality direction reverses, right? So, ( -p < -q ). That means ( -p > -q ) is false. So, option 1 is not true.2. ( -p > q ): This is interesting. Since ( p ) and ( q ) are both positive, ( -p ) is negative, and ( q ) is positive. So, a negative number cannot be greater than a positive number. Therefore, ( -p > q ) is false. So, option 2 is not true.3. ( 1 > -frac{q}{p} ): Let's see. Since ( p ) and ( q ) are positive, ( frac{q}{p} ) is positive, and thus ( -frac{q}{p} ) is negative. So, ( 1 > ) a negative number is always true because 1 is greater than any negative number. Therefore, option 3 is true.4. ( 1 < frac{q}{p} ): From ( p > q ), dividing both sides by ( p ) (which is positive, so inequality direction remains the same), we get ( frac{p}{p} > frac{q}{p} ), which simplifies to ( 1 > frac{q}{p} ). So, ( 1 < frac{q}{p} ) is false. Therefore, option 4 is not true.So, after evaluating all the options, only option 3 is true. Therefore, the correct answer should be option C, which corresponds to ( 1 > -frac{q}{p} ).Wait, but let me double-check to make sure I didn't make any mistakes. Starting from ( p^2r > q^2r ), dividing both sides by ( r ) gives ( p^2 > q^2 ), which leads to ( p > q ) since both are positive. Then, for option 1, flipping the inequality when multiplying by -1 gives ( -p < -q ), so ( -p > -q ) is indeed false. For option 2, ( -p ) is negative and ( q ) is positive, so ( -p > q ) is impossible. For option 3, ( -frac{q}{p} ) is negative, and 1 is greater than any negative number, so that's true. For option 4, since ( p > q ), ( frac{q}{p} < 1 ), so ( 1 < frac{q}{p} ) is false. Yep, everything checks out. So, I'm confident that option C is the correct answer.

💡Alright, so I have this problem about a moving point M whose distance to a fixed point F(1,0) and a fixed line x=4 has a ratio of 1/2. The trajectory of M is called curve C. There are two parts: first, finding the equation of curve C, and second, dealing with some properties of lines intersecting this curve.Starting with part (1). I remember that when a point's distance to a fixed point and a fixed line has a constant ratio, it's called a conic section. The ratio is called the eccentricity. If the eccentricity is less than 1, it's an ellipse; if it's equal to 1, it's a parabola; and if it's greater than 1, it's a hyperbola. Here, the ratio is 1/2, which is less than 1, so it should be an ellipse.But let me try to derive it step by step. Let’s denote the moving point as M(x, y). The distance from M to F(1,0) is sqrt[(x - 1)^2 + y^2]. The distance from M to the line x=4 is |x - 4|. According to the problem, the ratio of these distances is 1/2. So, we can write:sqrt[(x - 1)^2 + y^2] / |x - 4| = 1/2To eliminate the square root, I'll square both sides:[(x - 1)^2 + y^2] / (x - 4)^2 = 1/4Multiplying both sides by (x - 4)^2:(x - 1)^2 + y^2 = (1/4)(x - 4)^2Now, expanding both sides:Left side: (x^2 - 2x + 1) + y^2Right side: (1/4)(x^2 - 8x + 16) = (1/4)x^2 - 2x + 4So, putting it all together:x^2 - 2x + 1 + y^2 = (1/4)x^2 - 2x + 4Let me subtract the right side from both sides to bring everything to the left:x^2 - 2x + 1 + y^2 - (1/4)x^2 + 2x - 4 = 0Simplify term by term:x^2 - (1/4)x^2 = (3/4)x^2-2x + 2x = 01 - 4 = -3So, we have:(3/4)x^2 + y^2 - 3 = 0Multiply both sides by 4 to eliminate the fraction:3x^2 + 4y^2 - 12 = 0Bring the constant to the other side:3x^2 + 4y^2 = 12Divide both sides by 12 to get the standard form:(x^2)/4 + (y^2)/3 = 1So, that's the equation of curve C. It's an ellipse centered at the origin, with semi-major axis 2 along the x-axis and semi-minor axis sqrt(3) along the y-axis.Moving on to part (2). We have point P(4,0). A line l with a non-zero slope passes through F(1,0) and intersects curve C at points A and B. We need to find the sum of the slopes of lines PA and PB, which are k1 and k2 respectively.First, let's visualize this. Curve C is an ellipse, and point P is at (4,0), which is outside the ellipse since the ellipse goes up to x=2. The line l passes through F(1,0) and intersects the ellipse at A and B. Then, from P, we draw lines to A and B, and we need to find the sum of their slopes.I think the key here is to parametrize the line l. Since it passes through F(1,0) and has a non-zero slope, we can write its equation as y = m(x - 1), where m is the slope. But since the slope is non-zero, m ≠ 0.Alternatively, sometimes it's easier to use a parameter t for the slope, but let me stick with m for now.So, the equation of line l is y = m(x - 1). We need to find the points where this line intersects the ellipse C.Substituting y = m(x - 1) into the ellipse equation:(x^2)/4 + [m(x - 1)]^2 / 3 = 1Let me expand this:(x^2)/4 + (m^2(x^2 - 2x + 1))/3 = 1Multiply through by 12 to eliminate denominators:3x^2 + 4m^2(x^2 - 2x + 1) = 12Expand the terms:3x^2 + 4m^2x^2 - 8m^2x + 4m^2 = 12Combine like terms:(3 + 4m^2)x^2 - 8m^2x + (4m^2 - 12) = 0This is a quadratic equation in x. Let me write it as:(4m^2 + 3)x^2 - 8m^2x + (4m^2 - 12) = 0Let me denote this as Ax^2 + Bx + C = 0, where:A = 4m^2 + 3B = -8m^2C = 4m^2 - 12The solutions to this quadratic will give the x-coordinates of points A and B. Let me denote the roots as x1 and x2. Then, from Vieta's formulas:x1 + x2 = -B/A = (8m^2)/(4m^2 + 3)x1 * x2 = C/A = (4m^2 - 12)/(4m^2 + 3)Similarly, the y-coordinates of A and B can be found using y = m(x - 1):y1 = m(x1 - 1)y2 = m(x2 - 1)So, points A and B are (x1, m(x1 - 1)) and (x2, m(x2 - 1)).Now, we need to find the slopes k1 and k2 of lines PA and PB. Point P is (4,0), so the slope from P to A is:k1 = [m(x1 - 1) - 0]/[x1 - 4] = m(x1 - 1)/(x1 - 4)Similarly, k2 = m(x2 - 1)/(x2 - 4)We need to find k1 + k2.So, let's compute k1 + k2:k1 + k2 = m(x1 - 1)/(x1 - 4) + m(x2 - 1)/(x2 - 4)Factor out m:= m [ (x1 - 1)/(x1 - 4) + (x2 - 1)/(x2 - 4) ]Let me compute the expression inside the brackets:Let’s denote S = (x1 - 1)/(x1 - 4) + (x2 - 1)/(x2 - 4)To combine these fractions, find a common denominator:= [ (x1 - 1)(x2 - 4) + (x2 - 1)(x1 - 4) ] / [ (x1 - 4)(x2 - 4) ]Expand the numerator:= [x1x2 - 4x1 - x2 + 4 + x1x2 - 4x2 - x1 + 4] / [ (x1 - 4)(x2 - 4) ]Combine like terms:Numerator:x1x2 + x1x2 = 2x1x2-4x1 - x1 = -5x1-4x2 - x2 = -5x24 + 4 = 8So, numerator = 2x1x2 -5x1 -5x2 +8Denominator = (x1 - 4)(x2 - 4) = x1x2 -4x1 -4x2 +16So, S = [2x1x2 -5x1 -5x2 +8] / [x1x2 -4x1 -4x2 +16]Now, let's express this in terms of x1 + x2 and x1x2, which we have from Vieta's formulas.We know:x1 + x2 = 8m^2 / (4m^2 + 3)x1x2 = (4m^2 - 12)/(4m^2 + 3)Let me compute the numerator and denominator separately.First, numerator:2x1x2 -5(x1 + x2) +8= 2*(4m^2 -12)/(4m^2 +3) -5*(8m^2)/(4m^2 +3) +8Let me combine these terms over the common denominator (4m^2 +3):= [2*(4m^2 -12) -5*8m^2 +8*(4m^2 +3)] / (4m^2 +3)Compute numerator:2*(4m^2 -12) = 8m^2 -24-5*8m^2 = -40m^28*(4m^2 +3) =32m^2 +24Add them together:8m^2 -24 -40m^2 +32m^2 +24Combine like terms:(8m^2 -40m^2 +32m^2) + (-24 +24)= 0m^2 + 0 = 0So, numerator is 0.Denominator:x1x2 -4(x1 +x2) +16= (4m^2 -12)/(4m^2 +3) -4*(8m^2)/(4m^2 +3) +16Again, combine over common denominator:= [ (4m^2 -12) -4*8m^2 +16*(4m^2 +3) ] / (4m^2 +3)Compute numerator:4m^2 -12 -32m^2 +64m^2 +48Combine like terms:(4m^2 -32m^2 +64m^2) + (-12 +48)= 36m^2 +36Factor:= 36(m^2 +1)So, denominator = 36(m^2 +1)/(4m^2 +3)Therefore, S = 0 / [36(m^2 +1)/(4m^2 +3)] = 0Thus, S = 0, so k1 + k2 = m * 0 = 0Wait, that's interesting. So regardless of the slope m, the sum of the slopes k1 + k2 is zero.Let me verify this because it's a surprising result. Maybe I made a mistake in the algebra.Looking back, when I computed the numerator of S, it turned out to be zero. That seems correct because all the terms canceled out. The denominator was non-zero because 36(m^2 +1) is always positive, and 4m^2 +3 is also positive. So, S is indeed zero.Therefore, k1 + k2 = m * 0 = 0So, the sum of the slopes is zero.Alternatively, maybe there's a geometric interpretation. Since P is at (4,0), which is the same x-coordinate as the directrix x=4. Maybe there's some symmetry here. The line l passes through F(1,0), which is the focus, and intersects the ellipse at A and B. Then, connecting P to A and B, the slopes might have some symmetric properties that cause their sum to be zero.But regardless, the algebra shows that k1 + k2 = 0.So, the final answer for part (2) is 0.**Final Answer**(1) The equation of curve ( C ) is (boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}).(2) The value of ( k_1 + k_2 ) is (boxed{0}).

💡First, I note that Jessie's initial weight is 69 kilograms.She lost 35 kilograms through jogging. Subtracting this from her initial weight gives:69 kg - 35 kg = 34 kgNext, she lost an additional 10 kilograms through her diet plan. Subtracting this from her weight after jogging results in:34 kg - 10 kg = 24 kgFinally, she gained 5 kilograms through a strength training program. Adding this to her weight after the diet plan provides:24 kg + 5 kg = 29 kgTherefore, Jessie's current weight, after considering all the changes, is 29 kilograms.

💡Okay, so I have this problem about a triangular pyramid, also known as a tetrahedron, O-ABC. The given information is that angle BOC is 90 degrees, OA is perpendicular to the plane BOC, and the lengths of AB, BC, and AC are sqrt(10), sqrt(13), and sqrt(5) respectively. All the points O, A, B, C lie on the surface of a sphere S, and I need to find the surface area of this sphere.Hmm, let me try to visualize this. So, we have a tetrahedron with vertex O connected to A, B, and C. The angle between BO and CO is 90 degrees, which means that BO and CO are perpendicular to each other. Also, OA is perpendicular to the plane BOC. That suggests that OA is like a vertical edge from point O to the plane containing B and C.Since OA is perpendicular to the plane BOC, and angle BOC is 90 degrees, it seems like the edges OA, OB, and OC are all mutually perpendicular. That is, OA, OB, and OC form a set of perpendicular edges meeting at O. So, if I consider O as the origin in a 3D coordinate system, then points A, B, and C can be placed along the x, y, and z axes respectively.Let me denote the lengths of OA, OB, and OC as a, b, and c respectively. So, OA = a, OB = b, and OC = c. Since OA is perpendicular to the plane BOC, which contains points B and C, then OA is perpendicular to both OB and OC. So, OA, OB, and OC are all mutually perpendicular.Given that AB = sqrt(10), BC = sqrt(13), and AC = sqrt(5), I can express these distances in terms of a, b, and c.First, let's find AB. Since A is at (a, 0, 0) and B is at (0, b, 0), the distance AB can be calculated using the distance formula in 3D:AB = sqrt[(a - 0)^2 + (0 - b)^2 + (0 - 0)^2] = sqrt(a^2 + b^2) = sqrt(10)So, sqrt(a^2 + b^2) = sqrt(10) => a^2 + b^2 = 10. (1)Similarly, let's find AC. Point A is at (a, 0, 0) and point C is at (0, 0, c). So, distance AC is:AC = sqrt[(a - 0)^2 + (0 - 0)^2 + (0 - c)^2] = sqrt(a^2 + c^2) = sqrt(5)So, sqrt(a^2 + c^2) = sqrt(5) => a^2 + c^2 = 5. (2)Now, let's find BC. Point B is at (0, b, 0) and point C is at (0, 0, c). So, distance BC is:BC = sqrt[(0 - 0)^2 + (b - 0)^2 + (0 - c)^2] = sqrt(b^2 + c^2) = sqrt(13)So, sqrt(b^2 + c^2) = sqrt(13) => b^2 + c^2 = 13. (3)Now, I have three equations:1. a^2 + b^2 = 102. a^2 + c^2 = 53. b^2 + c^2 = 13I need to solve for a^2, b^2, and c^2.Let me subtract equation (2) from equation (1):(a^2 + b^2) - (a^2 + c^2) = 10 - 5 => b^2 - c^2 = 5. (4)From equation (3), b^2 + c^2 = 13. (3)Now, I can add equations (4) and (3):(b^2 - c^2) + (b^2 + c^2) = 5 + 13 => 2b^2 = 18 => b^2 = 9 => b = 3.Now, substitute b^2 = 9 into equation (1):a^2 + 9 = 10 => a^2 = 1 => a = 1.Similarly, substitute b^2 = 9 into equation (3):9 + c^2 = 13 => c^2 = 4 => c = 2.So, we have a = 1, b = 3, c = 2.Therefore, the lengths of OA, OB, and OC are 1, 3, and 2 respectively.Now, since points O, A, B, C lie on the surface of sphere S, the sphere is the circumscribed sphere (circum sphere) of the tetrahedron O-ABC.To find the surface area of the sphere, we need to find its radius first. The surface area is 4πr², where r is the radius of the sphere.For a tetrahedron with mutually perpendicular edges at a vertex (like OA, OB, OC here), the circumradius R can be found using the formula:R = (1/2) * sqrt(a² + b² + c²)This formula comes from the fact that the tetrahedron can be inscribed in a rectangular box with side lengths a, b, c, and the circumradius of the tetrahedron is half the space diagonal of the box.So, let's compute a² + b² + c²:a² = 1, b² = 9, c² = 4.So, a² + b² + c² = 1 + 9 + 4 = 14.Therefore, R = (1/2) * sqrt(14).Hence, the radius of the sphere is sqrt(14)/2.Now, the surface area is 4πr² = 4π*(sqrt(14)/2)².Let's compute that:(sqrt(14)/2)² = (14)/4 = 7/2.So, 4π*(7/2) = 4*(7/2)*π = 14π.Therefore, the surface area of the sphere is 14π.Wait, let me just double-check my steps to make sure I didn't make a mistake.1. I set up the coordinates correctly, assuming O is the origin, and A, B, C lie along the axes.2. Calculated AB, AC, BC distances correctly using the distance formula.3. Set up the three equations correctly and solved for a², b², c².4. Found a=1, b=3, c=2.5. Recognized that the tetrahedron can be inscribed in a rectangular box with sides a, b, c, so the circumradius is half the space diagonal.6. Calculated the space diagonal as sqrt(a² + b² + c²) = sqrt(14), so radius is sqrt(14)/2.7. Calculated the surface area as 4π*(sqrt(14)/2)² = 14π.Everything seems to check out. I think this is correct.**Final Answer**The surface area of sphere ( S ) is boxed{14pi}.

💡First, determine the original number of girls by subtracting the number of boys from the total number of students.18 students - 8 boys = 10 girlsOn the day of the trip, the number of girls and boys was the same, which means there were 8 girls present.Finally, calculate the number of girls who were not able to join the trip by subtracting the number of girls present from the original number of girls.10 girls - 8 girls = 2 girls