💡Okay, so I have this problem about an equilateral triangle ABC. Points D and E are the midpoints of sides CA and CB, respectively. There's an ellipse with foci at A and B that passes through D and E, and a hyperbola with the same foci. I need to find out which relationship between the eccentricities e₁ (for the ellipse) and e₂ (for the hyperbola) is incorrect. The options are:A. e₂ + e₁ = 2 B. e₂ - e₁ = 2 C. e₂e₁ = 2 D. e₂ > 2Alright, let me start by recalling some properties of ellipses and hyperbolas.For an ellipse, the sum of the distances from any point on the ellipse to the two foci is constant and equal to the major axis length, 2a. The eccentricity e₁ is given by e₁ = c/a, where c is the distance from the center to a focus, and a is the semi-major axis.For a hyperbola, the difference of the distances from any point on the hyperbola to the two foci is constant and equal to the real axis length, 2a. The eccentricity e₂ is given by e₂ = c/a, where c is the distance from the center to a focus, and a is the semi-real axis.Given that ABC is an equilateral triangle, all sides are equal. Let's denote the length of each side as s. So, AB = BC = CA = s.Points D and E are midpoints, so AD = AE = s/2.First, let's analyze the ellipse.**Ellipse Analysis:**The ellipse has foci at A and B. It passes through points D and E.For an ellipse, the sum of distances from any point on the ellipse to the two foci is constant. So, for point D, which is on the ellipse, the sum AD + BD should be equal to 2a.Let me compute AD and BD.AD is the distance from A to D. Since D is the midpoint of CA, and ABC is equilateral, AD is equal to s/2.BD is the distance from B to D. Let me visualize triangle ABC. Since ABC is equilateral, all angles are 60 degrees.Point D is the midpoint of CA, so CD = DA = s/2.To find BD, I can use the Law of Cosines in triangle BCD.Wait, triangle BCD? Let me see. In triangle ABC, point D is the midpoint of CA, so CD = s/2. So, triangle BCD has sides BC = s, CD = s/2, and angle at C is 60 degrees.So, applying the Law of Cosines to triangle BCD:BD² = BC² + CD² - 2*BC*CD*cos(angle C)BD² = s² + (s/2)² - 2*s*(s/2)*cos(60°)cos(60°) is 0.5, so:BD² = s² + s²/4 - 2*s*(s/2)*0.5 BD² = (5s²/4) - (s²/2) BD² = (5s²/4 - 2s²/4) BD² = 3s²/4 BD = (s√3)/2So, BD = (s√3)/2.Therefore, the sum AD + BD = (s/2) + (s√3)/2 = s(1 + √3)/2.This sum is equal to 2a for the ellipse, so:2a = s(1 + √3)/2 a = s(1 + √3)/4Now, the distance between the foci of the ellipse is 2c, where c is the distance from the center to each focus. Since the foci are at A and B, which are separated by distance AB = s, so 2c = s, hence c = s/2.Eccentricity e₁ = c/a = (s/2) / [s(1 + √3)/4] = (s/2) * [4 / (s(1 + √3))] = (2) / (1 + √3)Simplify this:Multiply numerator and denominator by (1 - √3):e₁ = [2*(1 - √3)] / [(1 + √3)(1 - √3)] = [2*(1 - √3)] / (1 - 3) = [2*(1 - √3)] / (-2) = -(1 - √3) = √3 - 1So, e₁ = √3 - 1 ≈ 0.732**Hyperbola Analysis:**Now, the hyperbola has the same foci at A and B, and passes through points D and E.For a hyperbola, the difference of distances from any point on the hyperbola to the two foci is constant and equal to 2a.So, for point D on the hyperbola, |BD - AD| = 2a.We already computed AD = s/2 and BD = (s√3)/2.So, |BD - AD| = |(s√3)/2 - s/2| = s/2 |√3 - 1| = s(√3 - 1)/2Therefore, 2a = s(√3 - 1)/2 a = s(√3 - 1)/4The distance between the foci is still 2c = s, so c = s/2.Eccentricity e₂ = c/a = (s/2) / [s(√3 - 1)/4] = (s/2) * [4 / (s(√3 - 1))] = 2 / (√3 - 1)Simplify this:Multiply numerator and denominator by (√3 + 1):e₂ = [2*(√3 + 1)] / [(√3 - 1)(√3 + 1)] = [2*(√3 + 1)] / (3 - 1) = [2*(√3 + 1)] / 2 = √3 + 1So, e₂ = √3 + 1 ≈ 2.732**Checking the Options:**Now, let's check each option with e₁ ≈ 0.732 and e₂ ≈ 2.732.A. e₂ + e₁ = 2 Compute e₂ + e₁ ≈ 2.732 + 0.732 ≈ 3.464 ≠ 2 So, this is incorrect.B. e₂ - e₁ = 2 Compute e₂ - e₁ ≈ 2.732 - 0.732 ≈ 2 This is correct.C. e₂e₁ = 2 Compute e₂e₁ ≈ 2.732 * 0.732 ≈ 2 This is correct.D. e₂ > 2 Since e₂ ≈ 2.732 > 2, this is correct.So, the incorrect relationship is option A.**Final Answer**boxed{A}

💡Alright, so I have this problem where there are nine integers written on a board, and each of these integers is between 1 and 5. I need to figure out the sum of all these numbers based on some given conditions. Let's break down the problem step by step.First, the problem states that seven of the numbers are at least 2. That means there are seven numbers that are 2 or higher. So, out of the nine numbers, two of them could potentially be 1. But I need to check the other conditions to see if that's the case.Next, it says six of the numbers are greater than 2. So, six numbers are 3, 4, or 5. This means that out of the nine numbers, three of them could be 1 or 2. But wait, earlier it said seven numbers are at least 2, which would mean two numbers could be 1. So, combining these two pieces of information, if six numbers are greater than 2, then the remaining three numbers must be 2 or lower. But since seven numbers are at least 2, that means only two numbers can be 1. Therefore, out of the nine numbers, two are 1, and the remaining seven are 2 or higher.Now, moving on, it says three of the numbers are at least 4. So, three numbers are 4 or 5. And finally, it mentions that one of the numbers is at least 5. So, there's exactly one number that is 5.Let me try to summarize what I have so far:- Total numbers: 9- Numbers that are at least 2: 7- Numbers greater than 2: 6- Numbers at least 4: 3- Numbers at least 5: 1From this, I can deduce the following:1. Since one number is at least 5, that must be a 5 because the numbers are between 1 and 5.2. Three numbers are at least 4, which includes the number that is 5. So, there are two more numbers that are 4.3. Six numbers are greater than 2, which includes the three numbers that are 4 or 5. Therefore, there are three more numbers that are 3.4. Seven numbers are at least 2, which includes the six numbers that are greater than 2. So, there is one more number that is exactly 2.5. Finally, the remaining two numbers must be 1 because we've accounted for seven numbers already (three 3s, two 4s, one 5, and one 2), and there are nine numbers in total.So, putting it all together, the numbers on the board are:- Two 1s- One 2- Three 3s- Two 4s- One 5Now, to find the sum of all these numbers, I can calculate it as follows:- Sum of two 1s: 1 + 1 = 2- Sum of one 2: 2- Sum of three 3s: 3 + 3 + 3 = 9- Sum of two 4s: 4 + 4 = 8- Sum of one 5: 5Adding these up: 2 + 2 + 9 + 8 + 5 = 26So, the total sum of all the numbers on the board is 26.

💡Alright, so I need to find the radius of a circle when the area is given as 153.93804002589985 square meters. Hmm, okay, let's think about how to approach this. I remember that the area of a circle is calculated using the formula A = πr², where A is the area and r is the radius. So, if I have the area, I can rearrange this formula to solve for the radius.First, let's write down the formula again to make sure I have it right:A = πr²Given that A is 153.93804002589985, I can plug that into the formula:153.93804002589985 = πr²Now, I need to solve for r. To do that, I should isolate r on one side of the equation. Let's divide both sides of the equation by π to get rid of the π on the right side:153.93804002589985 / π = r²Okay, so now I have r squared equals the area divided by π. To find r, I need to take the square root of both sides. That should give me the radius.r = √(153.93804002589985 / π)Now, I need to calculate this value. I know that π is approximately 3.141592653589793. Let me plug that in:r = √(153.93804002589985 / 3.141592653589793)Let me do the division first:153.93804002589985 ÷ 3.141592653589793 ≈ 48.99999999999999Wow, that's really close to 49. So, r² ≈ 49Now, to find r, I take the square root of 49:r = √49 = 7So, the radius of the circle is approximately 7 meters.Wait a second, let me double-check my calculations to make sure I didn't make any mistakes. I'll start by recalculating the division:153.93804002589985 ÷ 3.141592653589793Let me use a calculator for this:153.93804002589985 ÷ 3.141592653589793 ≈ 48.99999999999999Yes, that's correct. It's essentially 49. So, r² = 49, and r = 7.Just to be thorough, let me plug the radius back into the area formula to see if I get the original area:A = πr² = π(7)² = π*49 ≈ 3.141592653589793*49 ≈ 153.93804002589985Yes, that matches the given area. So, my calculation seems correct.I think I've got it right. The radius of the circle is 7 meters.

💡Okay, so I need to evaluate the logarithm log base 5 of 1/125. Hmm, logarithms can sometimes be tricky for me, but I think I remember some basic properties that might help here.First, I recall that a logarithm log_b(a) asks the question: "To what power do we need to raise b to get a?" So, in this case, log_5(1/125) is asking, "5 raised to what power gives us 1/125?" Alright, so I need to find an exponent x such that 5^x = 1/125. Let me think about 125. I know that 5 cubed is 125 because 5 * 5 is 25, and 25 * 5 is 125. So, 5^3 = 125. But here, we have 1/125, which is the reciprocal of 125. I remember that taking the reciprocal of a number is the same as raising it to the power of -1. So, 1/125 is the same as 125^(-1). But since 125 is 5^3, I can substitute that in. So, 125^(-1) becomes (5^3)^(-1). Now, when you raise a power to another power, you multiply the exponents. So, (5^3)^(-1) is equal to 5^(3 * -1), which simplifies to 5^(-3). So, now I have 5^x = 5^(-3). Since the bases are the same and the equation holds true, the exponents must be equal. Therefore, x must be -3. Wait, let me double-check that. If I plug x = -3 back into the original equation, does it hold? 5^(-3) is indeed 1/(5^3), which is 1/125. Yes, that matches the argument of the logarithm. So, it seems like I've arrived at the correct answer. The logarithm base 5 of 1/125 is -3. I think I got it right!

💡Okay, so I have these two problems to solve based on the substitution method and Vieta's theorem. Let me try to work through them step by step.Starting with problem (1): I need to solve the equation (x^4 - x^2 - 6 = 0). The hint says to use substitution, like in Reading Material 1 where they let (y = x^2). So, I think I should do the same here. If I let (y = x^2), then (x^4) becomes (y^2). That should simplify the equation.So substituting, the equation becomes (y^2 - y - 6 = 0). Now, I need to solve this quadratic equation for (y). Let me see if I can factor it. Looking for two numbers that multiply to -6 and add to -1. Hmm, 2 and -3 work because 2 * (-3) = -6 and 2 + (-3) = -1. So, the equation factors as ((y - 3)(y + 2) = 0).Setting each factor equal to zero gives (y - 3 = 0) or (y + 2 = 0), so (y = 3) or (y = -2). But (y = x^2), and since (x^2) can't be negative, (y = -2) doesn't give a real solution. So, we only consider (y = 3).Now, substituting back, (x^2 = 3), so (x = pmsqrt{3}). Therefore, the solutions are (x = sqrt{3}) and (x = -sqrt{3}).Moving on to problem (2): We have real numbers (a) and (b) that satisfy the equation (a^4 - 3a^2 + 1 = 0) and (b^4 - 3b^2 + 1 = 0), with (a neq b). We need to find (a^4 + b^4).This seems a bit trickier. Let me think. Since both (a) and (b) satisfy the same quartic equation, maybe I can consider their squares. Let me set (u = a^2) and (v = b^2). Then, substituting into the equation, we get (u^2 - 3u + 1 = 0) and (v^2 - 3v + 1 = 0). So, (u) and (v) are roots of the quadratic equation (x^2 - 3x + 1 = 0).Since (a neq b), and assuming (a) and (b) are real, (u) and (v) must be distinct roots of the quadratic. Using Vieta's theorem, for the equation (x^2 - 3x + 1 = 0), the sum of the roots (u + v = 3) and the product (uv = 1).Now, we need to find (a^4 + b^4). But (a^4 = (a^2)^2 = u^2) and similarly (b^4 = v^2). So, (a^4 + b^4 = u^2 + v^2).I remember that (u^2 + v^2) can be expressed in terms of ((u + v)^2 - 2uv). Let me verify that:[u^2 + v^2 = (u + v)^2 - 2uv]Yes, that's correct. So substituting the known values from Vieta's theorem:[u^2 + v^2 = (3)^2 - 2(1) = 9 - 2 = 7]Therefore, (a^4 + b^4 = 7).Wait, let me make sure I didn't skip any steps. I transformed the original quartic equation into a quadratic by substitution, applied Vieta's theorem to find the sum and product of the roots, and then used an algebraic identity to find the sum of the squares of the roots. That seems solid.I don't think I made any mistakes here. The substitution method worked well, and using Vieta's theorem simplified the process of finding the required sum without having to solve for (a) and (b) explicitly. It's a neat application of these concepts.**Final Answer**(1) The solutions are (boxed{pm sqrt{3}}).(2) The value of (a^4 + b^4) is (boxed{7}).

💡First, I need to recall the conditions for a point to be located in the third quadrant of the Cartesian coordinate system. In the third quadrant, both the ( x )-coordinate and ( y )-coordinate of a point must be negative.Given the point ( P(-1, -2 + m) ), I observe that the ( x )-coordinate is (-1), which is already negative. This satisfies the first condition for being in the third quadrant.Next, I need to ensure that the ( y )-coordinate, which is (-2 + m), is also negative. To find the range of ( m ) that satisfies this condition, I set up the inequality:[-2 + m < 0]Solving for ( m ), I add 2 to both sides of the inequality:[m < 2]This means that for the point ( P(-1, -2 + m) ) to be in the third quadrant, the value of ( m ) must be less than 2.Finally, I compare this result with the provided answer choices and conclude that the correct answer is: