💡Okay, so I have this problem about ellipses, and I need to figure out the standard equation of an ellipse that passes through a specific point and shares the same foci as another given ellipse. Hmm, let me try to break this down step by step.First, the given ellipse is (frac{y^2}{25} + frac{x^2}{9} = 1). I remember that the standard form of an ellipse is (frac{(y - k)^2}{a^2} + frac{(x - h)^2}{b^2} = 1) for a vertical ellipse, where (a > b), and the center is at ((h, k)). In this case, the center is at the origin since there are no shifts in the equation. So, (a^2 = 25) and (b^2 = 9), which means (a = 5) and (b = 3).Now, I need to find the foci of this given ellipse. I recall that for an ellipse, the distance from the center to each focus is given by (c), where (c^2 = a^2 - b^2). Plugging in the values, (c^2 = 25 - 9 = 16), so (c = 4). Since this is a vertical ellipse, the foci are located at ((0, pm 4)).The problem states that the desired ellipse shares these same foci, so the foci of our new ellipse are also at ((0, pm 4)). That means for our new ellipse, (c = 4) as well. Now, the standard form of our desired ellipse will also be vertical because the foci are on the y-axis. So, its equation should look like (frac{y^2}{A^2} + frac{x^2}{B^2} = 1), where (A > B). For this ellipse, the relationship between (A), (B), and (c) is (c^2 = A^2 - B^2). We know (c = 4), so (16 = A^2 - B^2). That's one equation.Additionally, the ellipse passes through the point ((sqrt{3}, -sqrt{5})). Plugging this point into the equation of the ellipse should satisfy it. So, substituting (x = sqrt{3}) and (y = -sqrt{5}) into (frac{y^2}{A^2} + frac{x^2}{B^2} = 1), we get:[frac{(-sqrt{5})^2}{A^2} + frac{(sqrt{3})^2}{B^2} = 1]Simplifying, that becomes:[frac{5}{A^2} + frac{3}{B^2} = 1]So now I have two equations:1. (A^2 - B^2 = 16)2. (frac{5}{A^2} + frac{3}{B^2} = 1)I need to solve this system of equations to find (A^2) and (B^2). Let me denote (A^2 = m) and (B^2 = n) to make it simpler. Then, the equations become:1. (m - n = 16)2. (frac{5}{m} + frac{3}{n} = 1)From the first equation, I can express (m = n + 16). Now, substitute this into the second equation:[frac{5}{n + 16} + frac{3}{n} = 1]To solve for (n), I'll find a common denominator for the fractions on the left side. The common denominator is (n(n + 16)). So, rewriting the equation:[frac{5n}{n(n + 16)} + frac{3(n + 16)}{n(n + 16)} = 1]Combine the numerators:[frac{5n + 3(n + 16)}{n(n + 16)} = 1]Simplify the numerator:[5n + 3n + 48 = 8n + 48]So, the equation becomes:[frac{8n + 48}{n(n + 16)} = 1]Multiply both sides by (n(n + 16)) to eliminate the denominator:[8n + 48 = n(n + 16)]Expand the right side:[8n + 48 = n^2 + 16n]Bring all terms to one side to form a quadratic equation:[n^2 + 16n - 8n - 48 = 0]Simplify:[n^2 + 8n - 48 = 0]Now, solve this quadratic equation. I can use the quadratic formula: (n = frac{-b pm sqrt{b^2 - 4ac}}{2a}), where (a = 1), (b = 8), and (c = -48).Calculate the discriminant:[b^2 - 4ac = 64 - 4(1)(-48) = 64 + 192 = 256]So,[n = frac{-8 pm sqrt{256}}{2} = frac{-8 pm 16}{2}]This gives two solutions:1. (n = frac{-8 + 16}{2} = frac{8}{2} = 4)2. (n = frac{-8 - 16}{2} = frac{-24}{2} = -12)Since (n = B^2) must be positive, we discard the negative solution. So, (n = 4).Now, substitute (n = 4) back into (m = n + 16):[m = 4 + 16 = 20]So, (A^2 = 20) and (B^2 = 4). Therefore, the standard equation of the ellipse is:[frac{y^2}{20} + frac{x^2}{4} = 1]Looking back at the options provided:A) (frac{x^2}{20} + frac{y^2}{4} = 1)B) (frac{x^2}{2sqrt{5}} + frac{y^2}{4} = 1)C) (frac{y^2}{20} + frac{x^2}{4} = 1)D) (frac{x^2}{4} + frac{y^2}{2sqrt{5}} = 1)The equation I found matches option C. So, the correct answer is C.**Final Answer**boxed{C}

💡Okay, so I've got this problem where I have three n x n matrices A, B, and C with complex entries. They satisfy two conditions: A squared equals B squared equals C squared, and B cubed equals ABC plus 2 times the identity matrix. I need to prove that A to the sixth power is the identity matrix.Alright, let's start by writing down what we know. We have:1. ( A^2 = B^2 = C^2 )2. ( B^3 = ABC + 2I )I need to show that ( A^6 = I ). Hmm, okay. So, maybe I can manipulate these equations to express A in terms of B and C or something like that.First, since ( A^2 = B^2 = C^2 ), let's denote this common matrix as D. So, ( D = A^2 = B^2 = C^2 ). That might simplify things a bit.So, now we have:1. ( A^2 = B^2 = C^2 = D )2. ( B^3 = ABC + 2I )I wonder if I can express C in terms of A or B. Let's see. Maybe if I can find a relationship between A and C, that would help.Looking at the second equation, ( B^3 = ABC + 2I ). Maybe I can solve for C? Let's try.From ( B^3 = ABC + 2I ), if I can isolate C, that would be helpful. Let's subtract ABC from both sides:( B^3 - ABC = 2I )Hmm, factoring out B on the left side:( B(B^2 - AC) = 2I )But since ( B^2 = D ), we can substitute that in:( B(D - AC) = 2I )Hmm, not sure if that helps directly. Maybe I can express D in terms of A or C.Wait, since ( D = A^2 ), and ( D = C^2 ), so ( A^2 = C^2 ). That means ( A ) and ( C ) are square roots of the same matrix D. Hmm, but matrices don't necessarily have unique square roots, so I can't directly say A equals C or something like that.But maybe I can find a relationship between A and C. Let's think about the equation ( B^3 = ABC + 2I ) again.If I can express B in terms of A and C, maybe I can substitute back into the equation.Alternatively, maybe I can find an expression for B^3 in terms of A and C, and then relate it to D.Wait, let's try to compute B^5. Since we have B^3, maybe we can find B^5 in terms of lower powers.So, ( B^5 = B^3 cdot B^2 ). We know ( B^3 = ABC + 2I ) and ( B^2 = D ). So,( B^5 = (ABC + 2I)D )But ( D = C^2 ), so:( B^5 = (ABC + 2I)C^2 )Let's distribute that:( B^5 = ABC cdot C^2 + 2I cdot C^2 )Simplify:( ABC cdot C^2 = AB C^3 )And ( 2I cdot C^2 = 2C^2 )So,( B^5 = AB C^3 + 2C^2 )Hmm, okay. Now, ( C^3 = C cdot C^2 = C cdot D ). So,( B^5 = AB cdot D cdot C + 2C^2 )But ( D = A^2 ), so:( B^5 = AB cdot A^2 cdot C + 2C^2 )Simplify ( AB cdot A^2 ):( AB cdot A^2 = A cdot B cdot A^2 = A^3 B )Wait, is that correct? Matrix multiplication is associative, so yes, ( (AB)A^2 = A(B A^2) ). But I'm not sure if that helps.Alternatively, maybe I can factor differently.Wait, let's see. Since ( D = A^2 ), and ( D = B^2 ), so ( B^2 = A^2 ). So, ( B^2 = A^2 ), which might imply that B and A commute? Not necessarily, but maybe.Alternatively, perhaps I can find a relationship between A and C.Wait, let's go back to the equation ( B^5 = AB C^3 + 2C^2 ). Maybe I can express C^3 in terms of something else.Since ( C^2 = D ), ( C^3 = C cdot D ). So, ( C^3 = C cdot D ). Therefore,( B^5 = AB cdot D cdot C + 2C^2 )But ( D = B^2 ), so:( B^5 = AB cdot B^2 cdot C + 2C^2 )Simplify ( AB cdot B^2 ):( AB cdot B^2 = A cdot B^3 )So,( B^5 = A cdot B^3 cdot C + 2C^2 )But from the given, ( B^3 = ABC + 2I ), so substitute that in:( B^5 = A(ABC + 2I)C + 2C^2 )Let's expand that:( B^5 = A cdot ABC cdot C + A cdot 2I cdot C + 2C^2 )Simplify each term:First term: ( A cdot ABC cdot C = A^2 B C^2 )Second term: ( A cdot 2I cdot C = 2A C )Third term: ( 2C^2 )So,( B^5 = A^2 B C^2 + 2AC + 2C^2 )But ( A^2 = D ), ( C^2 = D ), so:( B^5 = D cdot B cdot D + 2AC + 2D )Hmm, ( D cdot B cdot D ). Since ( D = B^2 ), this becomes ( B^2 cdot B cdot B^2 = B^5 ). Wait, that's interesting.So,( B^5 = B^5 + 2AC + 2D )Subtract ( B^5 ) from both sides:( 0 = 2AC + 2D )Divide both sides by 2:( 0 = AC + D )But ( D = A^2 ), so:( 0 = AC + A^2 )Factor out A:( 0 = A(C + A) )So, ( A(C + A) = 0 )Hmm, so either A is the zero matrix or ( C + A = 0 ). But A is an n x n matrix with complex entries, and if A were zero, then ( A^2 = 0 ), so ( B^2 = 0 ) and ( C^2 = 0 ). But then ( B^3 = ABC + 2I ) would imply ( 0 = 0 + 2I ), which is impossible because 2I is not zero. So, A cannot be zero. Therefore, ( C + A = 0 ), which implies ( C = -A ).Okay, so that's a useful relationship: ( C = -A ).Now, let's substitute ( C = -A ) back into our equations.First, since ( C = -A ), ( C^2 = (-A)^2 = A^2 ), which is consistent with ( A^2 = C^2 ).Now, let's look at the second equation ( B^3 = ABC + 2I ). Substitute ( C = -A ):( B^3 = AB(-A) + 2I = -ABA + 2I )So,( B^3 = -ABA + 2I )Hmm, okay. Let's see if we can find a relationship involving B^3.Also, since ( B^2 = A^2 ), we can write ( B^2 = A^2 ).Let me try to compute ( B^3 ) in another way. Since ( B^3 = B cdot B^2 = B cdot A^2 ). So,( B^3 = B A^2 )But from the equation above, ( B^3 = -ABA + 2I ). Therefore,( B A^2 = -ABA + 2I )Let's rearrange this:( B A^2 + ABA = 2I )Factor out A from the middle term:( B A^2 + A B A = 2I )Hmm, can I factor this differently? Maybe factor A on the left or right.Wait, let's see:( B A^2 + A B A = 2I )Let me factor A on the right:( B A^2 + A B A = A (B A) + A (B A) = 2A (B A) )Wait, no, that's not quite right. Let me see:Wait, ( B A^2 = B A cdot A ) and ( A B A = A cdot B cdot A ). So, they are similar but not the same.Alternatively, maybe I can factor A on the left or right.Wait, let's consider:( B A^2 + A B A = A (B A) + (A B) A )Hmm, not sure if that helps.Alternatively, maybe I can factor A on the left:( A (B A) + A (B A) = 2A (B A) ). Wait, no, that's not correct because the first term is ( B A^2 ), not ( A B A ).Wait, maybe I can write ( B A^2 = A (B A) ) if A and B commute, but I don't know if they do.Alternatively, perhaps I can consider that ( B A^2 = A (B A) ) if A and B commute, but since we don't know that, maybe that's not helpful.Wait, maybe I can use the fact that ( B^2 = A^2 ). So, ( B^2 = A^2 ), which might imply that B and A commute? Let's see.If ( B^2 = A^2 ), then ( B^2 A = A B^2 ). So, ( B^2 A = A B^2 ). But does that imply that B and A commute? Not necessarily. For example, if B and A are such that they don't commute, but their squares do.Hmm, maybe not. So, perhaps I need a different approach.Wait, let's go back to the equation ( B A^2 + A B A = 2I ). Maybe I can factor A on the right.Let me write it as:( B A^2 + A B A = A (B A) + A (B A) = 2 A (B A) )Wait, no, that's not correct because ( B A^2 ) is ( B A A ), which is ( (B A) A ), not ( A (B A) ). So, unless A and B commute, these are different.Hmm, maybe I can write ( B A^2 = (B A) A ) and ( A B A = A (B A) ). So, the equation becomes:( (B A) A + A (B A) = 2I )Let me denote ( X = B A ). Then, the equation becomes:( X A + A X = 2I )So,( X A + A X = 2I )Hmm, that's an interesting equation. It's a commutator-like equation, but not exactly a commutator.Wait, if I can find X such that ( X A + A X = 2I ), maybe I can solve for X.But X is ( B A ), so perhaps I can find an expression for X in terms of A.Alternatively, maybe I can find an expression for B in terms of A.Wait, let's see. Since ( B^2 = A^2 ), perhaps I can write B as ( B = A P ), where P is some matrix such that ( P^2 = I ). Because then ( B^2 = A P A P = A^2 P^2 = A^2 I = A^2 ), which matches ( B^2 = A^2 ).So, let's assume ( B = A P ), where ( P^2 = I ). Then, let's substitute this into our equation ( X A + A X = 2I ), where ( X = B A = A P A ).So, ( X = A P A ). Then, the equation becomes:( (A P A) A + A (A P A) = 2I )Simplify each term:First term: ( A P A cdot A = A P A^2 )Second term: ( A cdot A P A = A^2 P A )So, the equation is:( A P A^2 + A^2 P A = 2I )Factor out A on the left and right:( A (P A^2) + (A^2 P) A = 2I )Hmm, not sure if that helps. Maybe I can factor A^2.Wait, since ( A^2 = D ), which is the same as ( B^2 ), and ( C = -A ), so ( C^2 = A^2 = D ).Wait, maybe I can use the fact that ( C = -A ) in the original equation ( B^3 = ABC + 2I ). Let's try that.We have ( B^3 = ABC + 2I ), and ( C = -A ), so:( B^3 = AB(-A) + 2I = -ABA + 2I )So,( B^3 = -ABA + 2I )But earlier, we also have ( B^3 = B A^2 ). So,( B A^2 = -ABA + 2I )Which gives:( B A^2 + ABA = 2I )Which is the same equation we had before.Hmm, maybe I can use the assumption ( B = A P ) with ( P^2 = I ) to substitute into this equation.So, ( B = A P ), so ( B A^2 = A P A^2 ), and ( A B A = A (A P) A = A^2 P A ).So, the equation becomes:( A P A^2 + A^2 P A = 2I )Let me factor out A on the left and right:( A (P A^2) + (A^2 P) A = 2I )Hmm, maybe I can write ( P A^2 ) and ( A^2 P ) in terms of something else.Wait, since ( A^2 = D ), which is also ( B^2 ), and ( B = A P ), so ( B^2 = A P A P = A^2 P^2 = A^2 I = A^2 ), which is consistent.But I'm not sure if that helps directly.Wait, maybe I can consider that ( P ) commutes with ( A ). If ( P ) commutes with ( A ), then ( P A = A P ), which would make ( B = A P ) commute with A, but I don't know if that's necessarily true.Alternatively, maybe I can find ( P ) such that ( P A = A P ), but that's an assumption I can't make without justification.Hmm, maybe I need a different approach.Wait, let's go back to the equation ( B^5 = B^5 + 2AC + 2D ), which led us to ( AC + D = 0 ), and then ( AC = -D = -A^2 ). So, ( AC = -A^2 ).But since ( C = -A ), then ( AC = A(-A) = -A^2 ), which matches. So, that's consistent.So, we have ( C = -A ), and ( AC = -A^2 ).Now, let's try to compute ( B^3 ) again, using ( C = -A ):( B^3 = ABC + 2I = AB(-A) + 2I = -ABA + 2I )So,( B^3 = -ABA + 2I )But we also have ( B^3 = B A^2 ), so:( B A^2 = -ABA + 2I )Let me rearrange this:( B A^2 + ABA = 2I )Hmm, as before.Wait, maybe I can factor A on the left or right.Let me try to factor A on the left:( A (B A) + A (B A) = 2I )Wait, no, that's not correct because ( B A^2 ) is ( B A A ), which is ( (B A) A ), not ( A (B A) ).Alternatively, maybe I can factor A on the right:( B A^2 + A B A = A (B A) + A (B A) = 2 A (B A) )Wait, no, that's not correct because ( B A^2 ) is ( B A A ), which is ( (B A) A ), and ( A B A ) is ( A (B A) ). So, they are similar but not the same.Wait, unless ( B A = A B ), which would make ( B A^2 = A B A ), but we don't know if B and A commute.Hmm, maybe I can assume that B and A commute? Let's see.If ( B A = A B ), then ( B A^2 = A B A ), so the equation becomes:( B A^2 + A B A = 2 B A^2 = 2I )Which would imply ( B A^2 = I ). Then, ( B = A^{-2} ), assuming A is invertible.But wait, do we know if A is invertible? Let's check.From ( A^2 = B^2 = C^2 ), and since ( C = -A ), ( C^2 = A^2 ), which is consistent.But does ( A^2 ) being equal to ( B^2 ) imply that A is invertible? Not necessarily, unless ( A^2 ) is invertible, which would require that A is invertible.Wait, if ( A^2 ) is invertible, then A must be invertible because the product of invertible matrices is invertible, and if A were not invertible, then ( A^2 ) wouldn't be either.But we don't know if ( A^2 ) is invertible yet.Wait, let's see. From the equation ( B^3 = ABC + 2I ), if we can show that B is invertible, then perhaps A and C are invertible as well.Let me check if B is invertible. Suppose B is not invertible, then there exists a non-zero vector v such that Bv = 0. Then, from ( B^3 = ABC + 2I ), applying both sides to v:( B^3 v = (ABC + 2I) v )Left side: ( B^3 v = B^2 (B v) = B^2 0 = 0 )Right side: ( ABC v + 2I v = A B C v + 2v )But Bv = 0, so ( ABC v = A B (C v) = A 0 = 0 ). Therefore, right side is 0 + 2v = 2v.So, 0 = 2v, which implies v = 0, contradicting the assumption that v is non-zero. Therefore, B must be invertible.So, B is invertible. Therefore, since ( B^2 = A^2 ), A must also be invertible because if A were not invertible, ( A^2 ) wouldn't be, contradicting ( B^2 ) being invertible.Therefore, A is invertible, and so is C since ( C = -A ).Okay, so A is invertible. Therefore, we can write ( A^{-1} ).Now, going back to the equation ( B A^2 + A B A = 2I ), since A is invertible, maybe we can manipulate this equation.Let me factor A on the right:( B A^2 + A B A = A (B A) + A (B A) = 2 A (B A) )Wait, no, that's not correct because ( B A^2 ) is ( B A A ), which is ( (B A) A ), not ( A (B A) ).But since A is invertible, maybe I can multiply both sides by ( A^{-1} ) on the left or right.Let me try multiplying both sides by ( A^{-1} ) on the left:( A^{-1} B A^2 + A^{-1} A B A = A^{-1} 2I )Simplify:First term: ( A^{-1} B A^2 = (A^{-1} B A) A )Second term: ( A^{-1} A B A = B A )Right side: ( 2 A^{-1} )So, the equation becomes:( (A^{-1} B A) A + B A = 2 A^{-1} )Let me denote ( X = A^{-1} B A ). Then, the equation becomes:( X A + B A = 2 A^{-1} )But ( X = A^{-1} B A ), so ( X A = A^{-1} B A cdot A = A^{-1} B A^2 )Wait, maybe that's not helpful.Alternatively, let's write ( X A + B A = 2 A^{-1} ). Since ( X = A^{-1} B A ), we can write ( X A = A^{-1} B A cdot A = A^{-1} B A^2 ). Hmm, not sure.Wait, maybe I can factor A on the right:( X A + B A = (X + B) A = 2 A^{-1} )So,( (X + B) A = 2 A^{-1} )Multiply both sides by A on the right:( (X + B) A A = 2 A^{-1} A )Simplify:( (X + B) A^2 = 2I )But ( X = A^{-1} B A ), so:( (A^{-1} B A + B) A^2 = 2I )Factor B on the left:( B (A^{-1} A + I) A^2 = 2I )Simplify ( A^{-1} A = I ):( B (I + I) A^2 = 2I )So,( B (2I) A^2 = 2I )Simplify:( 2 B A^2 = 2I )Divide both sides by 2:( B A^2 = I )So, ( B A^2 = I )But we also have ( B^3 = -ABA + 2I ). Let's use ( B A^2 = I ) to express B in terms of A.From ( B A^2 = I ), we can write ( B = A^{-2} ). Because if ( B A^2 = I ), then multiplying both sides by ( A^{-2} ) on the right gives ( B = A^{-2} ).So, ( B = A^{-2} ).Now, let's substitute ( B = A^{-2} ) into the equation ( B^3 = -ABA + 2I ):( (A^{-2})^3 = -A cdot A^{-2} cdot A + 2I )Simplify left side:( A^{-6} )Right side:( -A cdot A^{-2} cdot A + 2I = -A^{-1} cdot A + 2I = -I + 2I = I )So,( A^{-6} = I )Multiply both sides by ( A^6 ):( I = A^6 )Therefore, ( A^6 = I ).So, we've shown that ( A^6 = I ), which is what we needed to prove.

💡Okay, so I have this geometry problem here. Let me try to understand it step by step. There are four points, P, Q, R, S, lying on a straight line in that order. So, the order is P first, then Q, then R, and finally S. The lengths of the segments PQ, PR, and PS are given as a, b, and c respectively. So, PQ is a, which is the distance from P to Q. PR is b, which is the distance from P to R. And PS is c, which is the distance from P to S. Since all these points are on a straight line, I can visualize this as P---Q---R---S, with the distances between them being a, (b - a), and (c - b) respectively. Because PR is from P to R, which is PQ plus QR, so QR would be (b - a). Similarly, PS is from P to S, which is PR plus RS, so RS would be (c - b).Now, the problem says that segments PQ and RS may be rotated about points Q and R respectively, such that points P and S coincide, forming a non-degenerate triangle. Hmm, okay. So, if we rotate PQ around Q and RS around R, the points P and S will meet at some point, forming a triangle. Since it's a non-degenerate triangle, the three points must not be colinear, meaning the triangle must have a positive area.I need to figure out which of the inequalities I, II, and III must be satisfied for this to happen. Let's list the inequalities again:I. a < c/2 II. b < a + c/2 III. b < c/2 So, I need to determine which of these inequalities must hold true for the triangle to be non-degenerate.First, let me try to visualize the rotation. When we rotate PQ around Q, point P will move in a circular path with center at Q and radius PQ, which is a. Similarly, when we rotate RS around R, point S will move in a circular path with center at R and radius RS, which is (c - b). The goal is to have these two rotations result in points P and S coinciding at some point, say T, forming triangle TQR.Wait, actually, if P and S coincide, then the triangle formed would be TQR, where T is the new position of both P and S after rotation. So, the triangle has sides TQ, TR, and QR. Since QR is fixed as (b - a), and TQ is equal to PQ, which is a, and TR is equal to RS, which is (c - b). But for this triangle to be non-degenerate, the sum of any two sides must be greater than the third side. So, applying the triangle inequality theorem:1. TQ + QR > TR 2. TQ + TR > QR 3. QR + TR > TQ Substituting the lengths:1. a + (b - a) > (c - b) Simplifying: b > c - b So, 2b > c Therefore, b > c/2 2. a + (c - b) > (b - a) Simplifying: a + c - b > b - a Bring like terms together: a + a + c > b + b So, 2a + c > 2b Dividing both sides by 2: a + c/2 > b Which is the same as b < a + c/2 3. (b - a) + (c - b) > a Simplifying: b - a + c - b > a So, c - a > a Therefore, c > 2a Which is the same as a < c/2 So, from the triangle inequalities, we derive three conditions:1. b > c/2 2. b < a + c/2 3. a < c/2 Now, let's look back at the given inequalities:I. a < c/2 II. b < a + c/2 III. b < c/2 From our derived conditions, condition 1 is b > c/2, which contradicts inequality III (b < c/2). Therefore, inequality III cannot be necessarily true; in fact, it's the opposite. So, III is not a required condition.Conditions 2 and 3 correspond directly to inequalities II and I, respectively. Therefore, both I and II must be satisfied for the triangle to be non-degenerate.Let me double-check my reasoning. If a is not less than c/2, then c would be less than or equal to 2a, which would mean that when we try to form the triangle, the side opposite to the largest angle might not satisfy the triangle inequality, leading to a degenerate triangle or no triangle at all. Similarly, if b is not less than a + c/2, then the sum of a and c/2 might not be greater than b, again violating the triangle inequality.Also, since we derived that b must be greater than c/2, inequality III, which states b < c/2, is not only unnecessary but also impossible if the triangle is to be non-degenerate. Therefore, III is not a valid condition.So, the correct answer should include inequalities I and II only, which corresponds to option C.**Final Answer**boxed{C}

💡First, I see that the value of ( w ) is given as 90. This is the starting point for solving the problem.Next, I'll use the equation for ( z ) which depends on ( w ). By substituting ( w = 90 ) into the equation ( z = 4w + 40 ), I can calculate the value of ( z ).Once I have ( z ), I can use the equation for ( y ) which depends on ( z ). Substituting the calculated value of ( z ) into ( y = 3z + 15 ) will give me the value of ( y ).Finally, with ( y ) determined, I can substitute it into the equation for ( x ) which is ( x = 2y + 6 ). This will provide the final value of ( x ).By following this step-by-step substitution method, I can systematically find the value of ( x ) based on the given equations.

💡First, I need to determine the total distance Susan traveled during her trip. She drove four different segments with varying speeds and distances.1. **First Segment:** - Distance: 40 miles - Speed: 15 mph - Time: ( frac{40}{15} = frac{8}{3} ) hours2. **Second Segment:** - Distance: 20 miles - Speed: 60 mph - Time: ( frac{20}{60} = frac{1}{3} ) hours3. **Third Segment:** - Distance: 30 miles - Speed: 50 mph - Time: ( frac{30}{50} = frac{3}{5} ) hours4. **Fourth Segment:** - Distance: 10 miles - Speed: 70 mph - Time: ( frac{10}{70} = frac{1}{7} ) hoursNext, I'll calculate the total distance by adding up all the individual distances:[text{Total Distance} = 40 + 20 + 30 + 10 = 100 text{ miles}]Then, I'll calculate the total time by adding up all the individual times:[text{Total Time} = frac{8}{3} + frac{1}{3} + frac{3}{5} + frac{1}{7}]To add these fractions, I'll find a common denominator, which is 105:[frac{8}{3} = frac{280}{105}, quad frac{1}{3} = frac{35}{105}, quad frac{3}{5} = frac{63}{105}, quad frac{1}{7} = frac{15}{105}][text{Total Time} = frac{280 + 35 + 63 + 15}{105} = frac{393}{105} approx 3.74 text{ hours}]Finally, I'll calculate the average speed by dividing the total distance by the total time:[text{Average Speed} = frac{100}{3.74} approx 26.74 text{ mph}]

💡Okay, so I have this problem where I need to prove that it's impossible to fill a k×k grid with numbers from 1 to k² such that the sum of each row and each column is a power of 2. Hmm, interesting. Let me try to break this down step by step.First, let me understand what a power of 2 means. A power of 2 is any number that can be expressed as 2 raised to some integer exponent, like 2, 4, 8, 16, and so on. So, each row and each column in the grid must add up to one of these numbers.Now, the grid has k rows and k columns, and we're filling it with numbers from 1 to k². That means each number from 1 to k² is used exactly once. So, the total sum of all the numbers in the grid is the sum from 1 to k², which is given by the formula:Total sum = (k²)(k² + 1)/2Okay, that's the total sum. Now, since each row must sum to a power of 2, let's denote the sum of each row as 2^a, where a is some integer. Since there are k rows, the total sum can also be expressed as k times 2^a. So, we have:k * 2^a = (k²)(k² + 1)/2Hmm, that's an equation I can work with. Let me rearrange it to solve for 2^a:2^a = (k²)(k² + 1)/(2k)Simplify that:2^a = (k(k² + 1))/2So, 2^a must equal k(k² + 1)/2. Now, since 2^a is a power of 2, the right-hand side must also be a power of 2. Let's analyze the right-hand side.First, let's consider whether k is even or odd. Maybe that will help.Case 1: k is odd.If k is odd, then k² is also odd because an odd number squared is odd. Then, k² + 1 is even because an odd number plus 1 is even. So, k² + 1 is divisible by 2. Therefore, k(k² + 1)/2 simplifies to:(k² + 1)/2 * kSince k is odd, let's denote k = 2m + 1 for some integer m. Then, k² = (2m + 1)² = 4m² + 4m + 1, so k² + 1 = 4m² + 4m + 2 = 2(2m² + 2m + 1). Therefore, (k² + 1)/2 = 2m² + 2m + 1, which is an integer.So, in this case, k(k² + 1)/2 = (2m + 1)(2m² + 2m + 1). Now, this product is (odd number)*(odd number) = odd number. But 2^a is always even for a ≥ 1. So, we have an odd number equal to an even number, which is impossible. Therefore, if k is odd, this leads to a contradiction.Case 2: k is even.If k is even, then k = 2m for some integer m. Then, k² = 4m², which is even, and k² + 1 = 4m² + 1, which is odd. So, k(k² + 1)/2 = (2m)(4m² + 1)/2 = m(4m² + 1). Now, 4m² + 1 is odd, so m(4m² + 1) is m times an odd number.Now, 2^a must equal m(4m² + 1). Since 2^a is a power of 2, m(4m² + 1) must also be a power of 2. Let's analyze this.Since 4m² + 1 is odd, m must be a power of 2 for their product to be a power of 2. Let me denote m = 2^n for some integer n ≥ 0. Then, 4m² + 1 = 4*(2^n)^2 + 1 = 4*4^n + 1 = 4^{n+1} + 1. So, m(4m² + 1) = 2^n*(4^{n+1} + 1).Now, 4^{n+1} + 1 is equal to (2^{2(n+1)}) + 1, which is 2^{2n+2} + 1. This is an odd number because 2^{2n+2} is even, and adding 1 makes it odd. Therefore, 2^n*(4^{n+1} + 1) is 2^n times an odd number. For this to be a power of 2, the odd number must be 1 because any other odd number would introduce a prime factor other than 2.So, 4^{n+1} + 1 = 1. Let's solve for n:4^{n+1} + 1 = 14^{n+1} = 0But 4^{n+1} is always positive for any integer n ≥ 0, so this equation has no solution. Therefore, our assumption that k is even also leads to a contradiction.Wait, did I make a mistake here? Let me double-check. If k is even, then m = k/2. So, m is an integer. Then, m(4m² + 1) must be a power of 2. Since 4m² + 1 is odd, m must be a power of 2, say m = 2^n. Then, 4m² + 1 = 4*(2^n)^2 + 1 = 4^{n+1} + 1. So, m(4m² + 1) = 2^n*(4^{n+1} + 1). For this to be a power of 2, 4^{n+1} + 1 must be a power of 2 as well. But 4^{n+1} is a power of 2, specifically 2^{2(n+1)}, so 4^{n+1} + 1 = 2^{2(n+1)} + 1. This is one more than a power of 2. The only time when 2^b + 1 is a power of 2 is when b = 0, because 2^0 + 1 = 2, which is 2^1. For b > 0, 2^b + 1 is not a power of 2 because it would be odd and greater than 1, hence not a power of 2.So, 2^{2(n+1)} + 1 is a power of 2 only if 2(n+1) = 0, which implies n+1 = 0, so n = -1. But n must be a non-negative integer because m = 2^n and m is a positive integer. Therefore, there is no solution in this case either.Therefore, whether k is odd or even, we reach a contradiction. Hence, it is impossible to fill a k×k grid with numbers from 1 to k² such that each row and column sums to a power of 2.Wait, but let me think again. Maybe I missed something. For example, what if k = 1? The problem states k > 1, so k = 1 is not considered. For k = 2, let's test it manually.For k = 2, the grid is 2x2, numbers 1, 2, 3, 4. The total sum is 10. Each row must sum to a power of 2. The possible powers of 2 less than or equal to 10 are 2, 4, 8. Since there are 2 rows, each row must sum to 5, but 5 is not a power of 2. Wait, that's a problem. Alternatively, maybe each row sums to 4 and 6? But 6 is not a power of 2 either. Hmm, maybe it's impossible for k=2 as well.Wait, but according to our earlier reasoning, for k=2, which is even, m = 1, so m(4m² + 1) = 1*(4 + 1) = 5, which is not a power of 2. So, indeed, it's impossible for k=2 as well.Another test case: k=3. The grid is 3x3, numbers 1 to 9. Total sum is 45. Each row must sum to a power of 2. The possible powers of 2 up to 45 are 2, 4, 8, 16, 32. Since there are 3 rows, the total sum must be 3*2^a. So, 3*2^a = 45, which implies 2^a = 15. But 15 is not a power of 2, so it's impossible.Similarly, for k=4, total sum is 1+2+...+16 = 136. Each row must sum to 2^a, and there are 4 rows, so 4*2^a = 136, which implies 2^a = 34. But 34 is not a power of 2, so again impossible.So, it seems that for any k > 1, the total sum cannot be expressed as k times a power of 2, which is necessary for the row sums to be powers of 2. Therefore, the initial assumption that such a grid exists must be false.I think this covers both cases, whether k is odd or even, leading to contradictions. Therefore, it's impossible to fill the grid as described.