answer：Alright, so I have this problem here with two parts. Let me try to tackle them step by step. First, part (I) asks for the general equation of line ( l ) and curve ( C ). Starting with line ( l ). The parametric equations are given as:[begin{cases}x = t cos phi y = -1 + t sin phiend{cases}]where ( t ) is the parameter. I remember that parametric equations can often be converted to Cartesian form by eliminating the parameter. So, I need to eliminate ( t ) from these equations.From the first equation, ( x = t cos phi ), I can solve for ( t ):[t = frac{x}{cos phi}]Now, plug this into the second equation:[y = -1 + left( frac{x}{cos phi} right) sin phi]Simplify that:[y = -1 + x tan phi]So, the equation of line ( l ) is ( y = x tan phi - 1 ). That seems straightforward.Now, moving on to curve ( C ). Its polar equation is given as:[rho = 2 sin left( theta + frac{pi}{3} right)]I need to convert this into Cartesian coordinates. I recall that polar coordinates can be converted using ( x = rho cos theta ) and ( y = rho sin theta ). Also, there's a trigonometric identity for sine of a sum:[sin (A + B) = sin A cos B + cos A sin B]Applying this to the equation:[rho = 2 left( sin theta cos frac{pi}{3} + cos theta sin frac{pi}{3} right)]We know that ( cos frac{pi}{3} = frac{1}{2} ) and ( sin frac{pi}{3} = frac{sqrt{3}}{2} ), so substituting these in:[rho = 2 left( sin theta cdot frac{1}{2} + cos theta cdot frac{sqrt{3}}{2} right ) = 2 left( frac{1}{2} sin theta + frac{sqrt{3}}{2} cos theta right )]Simplify:[rho = sin theta + sqrt{3} cos theta]Now, multiply both sides by ( rho ):[rho^2 = rho sin theta + sqrt{3} rho cos theta]But ( rho^2 = x^2 + y^2 ), ( rho sin theta = y ), and ( rho cos theta = x ). Substituting these in:[x^2 + y^2 = y + sqrt{3} x]Rearranging terms:[x^2 + y^2 - sqrt{3} x - y = 0]So, that's the Cartesian equation of curve ( C ).Alright, part (I) seems done. Now onto part (II). It says: In the Cartesian coordinate system, draw a perpendicular line to line ( l ) from point ( B(0,1) ), with the foot of the perpendicular being ( H ). We need to find the parametric equation of the trajectory of point ( H ) with ( phi ) as the parameter and identify the curve.Hmm, okay. So, point ( B ) is at (0,1). We need to drop a perpendicular from ( B ) to line ( l ), and the foot of this perpendicular is ( H ). As ( phi ) changes, ( H ) moves, tracing some curve. We need to find the parametric equations for ( H )'s coordinates in terms of ( phi ), and then figure out what curve this is.First, let's recall that the slope of line ( l ) is ( tan phi ). Therefore, the slope of a line perpendicular to ( l ) would be the negative reciprocal, which is ( -cot phi ).So, the line perpendicular to ( l ) passing through ( B(0,1) ) has the equation:[y = -cot phi cdot x + 1]Now, we have two lines: line ( l ) with equation ( y = tan phi cdot x - 1 ) and the perpendicular line ( y = -cot phi cdot x + 1 ). The point ( H ) is their intersection. So, let's solve these two equations simultaneously.Set them equal:[tan phi cdot x - 1 = -cot phi cdot x + 1]Bring all terms to one side:[tan phi cdot x + cot phi cdot x - 1 - 1 = 0]Factor out ( x ):[x (tan phi + cot phi) - 2 = 0]Solve for ( x ):[x = frac{2}{tan phi + cot phi}]Hmm, let's simplify ( tan phi + cot phi ). Recall that ( tan phi = frac{sin phi}{cos phi} ) and ( cot phi = frac{cos phi}{sin phi} ). So:[tan phi + cot phi = frac{sin phi}{cos phi} + frac{cos phi}{sin phi} = frac{sin^2 phi + cos^2 phi}{sin phi cos phi} = frac{1}{sin phi cos phi}]Because ( sin^2 phi + cos^2 phi = 1 ). So, substituting back:[x = frac{2}{frac{1}{sin phi cos phi}} = 2 sin phi cos phi = sin 2phi]Wait, that's an interesting simplification. So, ( x = sin 2phi ). Now, let's find ( y ). Let's plug this ( x ) back into one of the equations, say the equation of line ( l ):[y = tan phi cdot x - 1 = tan phi cdot sin 2phi - 1]But ( sin 2phi = 2 sin phi cos phi ), so:[y = tan phi cdot 2 sin phi cos phi - 1 = 2 sin^2 phi - 1]Because ( tan phi = frac{sin phi}{cos phi} ), so multiplying by ( sin phi cos phi ) gives ( 2 sin^2 phi ).So, ( y = 2 sin^2 phi - 1 ). Hmm, I recognize this as the double-angle identity for cosine:[cos 2phi = 1 - 2 sin^2 phi implies 2 sin^2 phi = 1 - cos 2phi implies 2 sin^2 phi - 1 = -cos 2phi]So, ( y = -cos 2phi ).Therefore, the parametric equations for ( H ) are:[x = sin 2phi y = -cos 2phi]Hmm, that looks like the parametric equations for a circle. Let me check:If we square both ( x ) and ( y ) and add them:[x^2 + y^2 = sin^2 2phi + cos^2 2phi = 1]So, indeed, ( x^2 + y^2 = 1 ), which is the equation of a unit circle centered at the origin.Wait, but in my earlier steps, I had ( x = sin 2phi ) and ( y = -cos 2phi ). So, it's a unit circle, but parameterized differently. So, the trajectory of ( H ) is a unit circle.But just to make sure, let me retrace my steps. I found the equations for ( x ) and ( y ) in terms of ( phi ), substituted back, and ended up with ( x^2 + y^2 = 1 ). So, regardless of the parameterization, it's a unit circle.Therefore, the parametric equations are:[begin{cases}x = sin 2phi y = -cos 2phiend{cases}]and the trajectory is a unit circle.Wait, but in the initial solution, the parametric equations were expressed in terms of ( tan phi ). Let me see if that's consistent with what I have.In the initial solution, they had:[x = frac{2 tan phi}{1 + tan^2 phi} y = frac{tan^2 phi - 1}{1 + tan^2 phi}]Hmm, let me see if that's equivalent to my parametric equations.Recall that ( sin 2phi = frac{2 tan phi}{1 + tan^2 phi} ) and ( cos 2phi = frac{1 - tan^2 phi}{1 + tan^2 phi} ). So, indeed, ( x = sin 2phi ) and ( y = -cos 2phi ) can be written as:[x = frac{2 tan phi}{1 + tan^2 phi} y = frac{tan^2 phi - 1}{1 + tan^2 phi}]Which is exactly what the initial solution had. So, both forms are equivalent, just expressed differently.Therefore, the parametric equations are correct, and the trajectory is a unit circle.I think that covers everything. I converted the parametric equations to Cartesian form, found the intersection point ( H ) by solving the perpendicular line, and then identified the trajectory as a unit circle. It all seems consistent.**Final Answer**The trajectory of point ( H ) is a unit circle, so the final answer is boxed{text{unit circle}}.

answer：First, I need to determine the number of hard drives sold by the hardware store. I'll start by calculating the total earnings from each type of item sold.The store sold 10 graphics cards at 600 each, so the total earnings from graphics cards are:10 × 600 = 6,000.Next, the store sold 8 CPUs at 200 each, so the total earnings from CPUs are:8 × 200 = 1,600.Additionally, the store sold 4 pairs of RAM at 60 per pair, so the total earnings from RAM are:4 × 60 = 240.Let ( H ) represent the number of hard drives sold. Each hard drive costs 80, so the total earnings from hard drives are:80H.The total earnings from all items sold is 8,960. I can set up the following equation to represent this:6,000 + 1,600 + 240 + 80H = 8,960.Combining the known earnings:6,000 + 1,600 + 240 = 7,840.Now, the equation simplifies to:7,840 + 80H = 8,960.To find ( H ), I'll subtract 7,840 from both sides:80H = 8,960 - 7,840,80H = 1,120.Finally, I'll divide both sides by 80 to solve for ( H ):H = 1,120 / 80,H = 14.Therefore, the store sold 14 hard drives.

answer：First, I need to find the value of ( a ) such that the real part and the imaginary part of the complex number ( z = a^2 - 2 + (3a - 4)i ) are equal. This means setting the real part equal to the imaginary part:[a^2 - 2 = 3a - 4]Next, I'll solve this equation for ( a ). Rearranging the terms:[a^2 - 3a + 2 = 0]This is a quadratic equation, which can be factored as:[(a - 1)(a - 2) = 0]So, the solutions are ( a = 1 ) and ( a = 2 ).Now, I need to check which of these solutions places the complex number ( z ) in the third quadrant. In the third quadrant, both the real and imaginary parts must be negative.For ( a = 1 ):[z = (1)^2 - 2 + (3(1) - 4)i = -1 - 1i]Both the real and imaginary parts are negative, so ( z ) is in the third quadrant.For ( a = 2 ):[z = (2)^2 - 2 + (3(2) - 4)i = 2 + 2i]Both the real and imaginary parts are positive, so ( z ) is in the first quadrant, not the third.Therefore, the only valid solution is ( a = 1 ).

answer：First, I need to expand the expression ((x - 2)(3x^2 - 2x + 5)). I'll distribute each term in the first polynomial to each term in the second polynomial.[(x - 2)(3x^2 - 2x + 5) = x cdot 3x^2 + x cdot (-2x) + x cdot 5 - 2 cdot 3x^2 - 2 cdot (-2x) - 2 cdot 5]Simplifying each term:[= 3x^3 - 2x^2 + 5x - 6x^2 + 4x - 10]Next, I'll combine like terms:[= 3x^3 - 8x^2 + 9x - 10]Now, I'll expand the second part of the expression, (4(x^3 + x^2 - 3x)):[4(x^3 + x^2 - 3x) = 4x^3 + 4x^2 - 12x]Finally, I'll add the two expanded polynomials together:[3x^3 - 8x^2 + 9x - 10 + 4x^3 + 4x^2 - 12x]Combining like terms:[= 7x^3 - 4x^2 - 3x - 10]The simplified expression has four nonzero terms.

answer：Okay, so I have this problem here: I need to calculate the sum of the geometric series ( 1 + left(frac{1}{4}right) + left(frac{1}{4}right)^2 + left(frac{1}{4}right)^3 + dots ). Hmm, let's see. I remember that a geometric series is one where each term is a constant multiple of the previous term. That constant is called the common ratio, right?So, looking at the series, the first term is 1. Then the next term is ( frac{1}{4} ), which is ( 1 times frac{1}{4} ). The term after that is ( left(frac{1}{4}right)^2 ), which is ( frac{1}{4} times frac{1}{4} ), and so on. So it seems like each term is multiplied by ( frac{1}{4} ) to get the next term. That means the common ratio ( r ) is ( frac{1}{4} ).Now, I recall that for an infinite geometric series, the sum can be found using a specific formula, but only if the absolute value of the common ratio is less than 1. In this case, ( |r| = frac{1}{4} ), which is definitely less than 1, so we can use the formula.The formula for the sum ( S ) of an infinite geometric series is:[ S = frac{a}{1 - r} ]Where ( a ) is the first term and ( r ) is the common ratio. Plugging in the values we have:[ S = frac{1}{1 - frac{1}{4}} ]Let me compute the denominator first. ( 1 - frac{1}{4} ) is equal to ( frac{3}{4} ). So now the formula becomes:[ S = frac{1}{frac{3}{4}} ]Dividing by a fraction is the same as multiplying by its reciprocal, so:[ S = 1 times frac{4}{3} = frac{4}{3} ]So, the sum of the series should be ( frac{4}{3} ). Let me double-check to make sure I didn't make any mistakes. The first term is 1, the ratio is ( frac{1}{4} ), and since ( |r| < 1 ), the formula applies. Calculating ( 1 - frac{1}{4} ) gives ( frac{3}{4} ), and the reciprocal of that is ( frac{4}{3} ). Yeah, that seems right.I think I've got it. The sum of the series is ( frac{4}{3} ).

answer：Okay, so I need to figure out the relationship between the areas of the faces of a tetrahedron where three edges meeting at one vertex are mutually perpendicular. Hmm, let me start by recalling what a tetrahedron is. It's a three-dimensional shape with four triangular faces. In this case, the tetrahedron is O-ABC, meaning one vertex is O, and the other three are A, B, and C.The problem mentions that OA, OB, and OC are all mutually perpendicular. That means each pair of these edges forms a right angle. So, if I imagine point O as the origin in a 3D coordinate system, then points A, B, and C would lie along the x, y, and z-axes, respectively. That makes sense because in a coordinate system, the axes are mutually perpendicular.Now, S is the area of the face opposite to vertex O. That face would be triangle ABC. The other areas, S₁, S₂, and S₃, are the areas of the side faces: triangles AOB, BOC, and COA, respectively. So, S₁ is the area of triangle AOB, S₂ is the area of triangle BOC, and S₃ is the area of triangle COA.I need to find a relationship between S and S₁, S₂, S₃. The problem is asking for an extension of the Pythagorean theorem into three dimensions. In the 2D case, for a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Maybe there's an analogous relationship here with areas instead of lengths.Let me think about how areas relate in this tetrahedron. Since OA, OB, and OC are mutually perpendicular, the triangles AOB, BOC, and COA are all right-angled triangles. That means their areas can be expressed in terms of the lengths of OA, OB, and OC.Let’s denote the lengths of OA, OB, and OC as a, b, and c respectively. Then, the areas of the side faces would be:- S₁ (area of triangle AOB) = (1/2) * OA * OB = (1/2) * a * b- S₂ (area of triangle BOC) = (1/2) * OB * OC = (1/2) * b * c- S₃ (area of triangle COA) = (1/2) * OC * OA = (1/2) * c * aNow, the face opposite to O is triangle ABC. To find its area, I need to figure out the lengths of sides AB, BC, and CA. Since OA, OB, and OC are mutually perpendicular, triangle ABC is not a right-angled triangle, but it's a triangle in 3D space.Let me compute the lengths of AB, BC, and CA using the Pythagorean theorem in 3D. - AB is the distance between points A and B. Since OA and OB are perpendicular, AB can be found using the Pythagorean theorem: AB = sqrt(OA² + OB²) = sqrt(a² + b²)- Similarly, BC = sqrt(OB² + OC²) = sqrt(b² + c²)- And CA = sqrt(OC² + OA²) = sqrt(c² + a²)So, triangle ABC has sides of length sqrt(a² + b²), sqrt(b² + c²), and sqrt(c² + a²). Hmm, that seems a bit complicated. Maybe there's a better way to find the area of triangle ABC without directly computing the lengths of its sides.Wait, perhaps I can use vectors or the cross product to find the area of triangle ABC. Since points A, B, and C are in 3D space with O as the origin, I can represent them as vectors OA, OB, and OC. Then, vectors AB and AC can be expressed as OB - OA and OC - OA, respectively.But actually, since OA, OB, and OC are mutually perpendicular, the volume of the tetrahedron can be found using the scalar triple product, which is (1/6) * |OA · (OB × OC)|. But I'm not sure if that's directly helpful for finding the area of face ABC.Alternatively, maybe I can find the area of triangle ABC using Heron's formula. Heron's formula states that the area of a triangle with sides of length x, y, z is sqrt[s(s - x)(s - y)(s - z)], where s is the semi-perimeter: (x + y + z)/2.But that might get messy because the sides are sqrt(a² + b²), sqrt(b² + c²), and sqrt(c² + a²). Let me see:Let’s denote:x = sqrt(a² + b²)y = sqrt(b² + c²)z = sqrt(c² + a²)Then, the semi-perimeter s = (x + y + z)/2.But computing s(s - x)(s - y)(s - z) seems complicated. Maybe there's a smarter way.Wait, another approach: since OA, OB, and OC are mutually perpendicular, the face ABC is actually the hypotenuse face of the tetrahedron. Maybe there's a formula similar to the Pythagorean theorem that relates the areas of the faces.In 2D, for a right-angled triangle, c² = a² + b². In 3D, perhaps the square of the area of the hypotenuse face is equal to the sum of the squares of the areas of the other three faces.Let me test this idea. If S is the area of triangle ABC, and S₁, S₂, S₃ are the areas of the other faces, then maybe S² = S₁² + S₂² + S₃².But is that true? Let me check with specific values.Suppose OA = a = 3, OB = b = 4, OC = c = 12. These are just arbitrary numbers, but let's see.Then, S₁ = (1/2)*3*4 = 6S₂ = (1/2)*4*12 = 24S₃ = (1/2)*12*3 = 18Now, let's compute the sides of triangle ABC:AB = sqrt(3² + 4²) = 5BC = sqrt(4² + 12²) = sqrt(16 + 144) = sqrt(160) = 4*sqrt(10)CA = sqrt(12² + 3²) = sqrt(144 + 9) = sqrt(153) = 3*sqrt(17)Now, let's compute the area of triangle ABC. Hmm, this might be tricky. Maybe using Heron's formula:First, compute the semi-perimeter:s = (5 + 4*sqrt(10) + 3*sqrt(17))/2That's already complicated. Maybe another approach.Alternatively, since OA, OB, OC are mutually perpendicular, the area of triangle ABC can be found using the formula for the area of a triangle in 3D space:Area = (1/2) * ||AB × AC||Where AB and AC are vectors.Let me compute vectors AB and AC.Vector AB = B - A = (0, 4, 0) - (3, 0, 0) = (-3, 4, 0)Vector AC = C - A = (0, 0, 12) - (3, 0, 0) = (-3, 0, 12)Now, compute the cross product AB × AC:|i   j   k||-3  4   0||-3  0  12|= i*(4*12 - 0*0) - j*(-3*12 - 0*(-3)) + k*(-3*0 - 4*(-3))= i*(48) - j*(-36) + k*(12)= (48, 36, 12)The magnitude of this vector is sqrt(48² + 36² + 12²) = sqrt(2304 + 1296 + 144) = sqrt(3744)Simplify sqrt(3744):3744 = 16 * 234 = 16 * 9 * 26 = 16 * 9 * 2 * 13 = 16 * 9 * 26Wait, sqrt(3744) = sqrt(16 * 234) = 4*sqrt(234)234 = 9 * 26, so sqrt(234) = 3*sqrt(26)Therefore, sqrt(3744) = 4*3*sqrt(26) = 12*sqrt(26)So, the area of triangle ABC is (1/2)*12*sqrt(26) = 6*sqrt(26)Now, let's compute S²:S² = (6*sqrt(26))² = 36 * 26 = 936Now, compute S₁² + S₂² + S₃²:S₁² = 6² = 36S₂² = 24² = 576S₃² = 18² = 324Sum: 36 + 576 + 324 = 936Hey, that's the same as S²! So, in this specific case, S² = S₁² + S₂² + S₃².That seems to confirm the relationship. Let me try another example to be sure.Let’s take OA = 1, OB = 1, OC = 1.Then, S₁ = (1/2)*1*1 = 0.5S₂ = (1/2)*1*1 = 0.5S₃ = (1/2)*1*1 = 0.5Compute the sides of triangle ABC:AB = sqrt(1² + 1²) = sqrt(2)BC = sqrt(1² + 1²) = sqrt(2)CA = sqrt(1² + 1²) = sqrt(2)So, triangle ABC is an equilateral triangle with sides sqrt(2). The area of an equilateral triangle is (sqrt(3)/4)*(side length)².So, area S = (sqrt(3)/4)*(sqrt(2))² = (sqrt(3)/4)*2 = sqrt(3)/2Compute S²: (sqrt(3)/2)² = 3/4Compute S₁² + S₂² + S₃²: 0.5² + 0.5² + 0.5² = 0.25 + 0.25 + 0.25 = 0.75 = 3/4Again, S² = S₁² + S₂² + S₃². So, this seems consistent.Therefore, it appears that in a tetrahedron where three edges meeting at one vertex are mutually perpendicular, the square of the area of the opposite face is equal to the sum of the squares of the areas of the other three faces.This is analogous to the Pythagorean theorem in three dimensions, extending it from lengths in 2D to areas in 3D.So, the relationship is S² = S₁² + S₂² + S₃².