💡Okay, so I have this problem about an ellipse. The equation is given as (x²/a²) + (y²/b²) = 1, where a > b > 0. The ellipse has two foci, F₁ and F₂, and one endpoint of the minor axis is B. The triangle formed by F₁, F₂, and B is an acute triangle, and I need to find the eccentricity of the ellipse. The options are A. √2/2, B. √3/2, C. (√5 - 1)/2, and D. 1/2.First, I should recall some basic properties of an ellipse. The standard form is (x²/a²) + (y²/b²) = 1, where a is the semi-major axis and b is the semi-minor axis. The distance between the center and each focus is c, where c² = a² - b². The eccentricity e is defined as e = c/a.Given that a > b, the major axis is along the x-axis, and the minor axis is along the y-axis. The foci F₁ and F₂ are located at (-c, 0) and (c, 0) respectively. The endpoint of the minor axis B is at (0, b).Now, the triangle F₁F₂B is an acute triangle. That means all three angles in the triangle are less than 90 degrees. To determine the conditions for this triangle to be acute, I can use the property that in any triangle, if the square of each side is less than the sum of the squares of the other two sides, then the triangle is acute.So, let's denote the sides of the triangle:- Side F₁F₂: This is the distance between the two foci, which is 2c.- Side F₁B: This is the distance from F₁ to B. Since F₁ is at (-c, 0) and B is at (0, b), the distance can be calculated using the distance formula: √[(0 - (-c))² + (b - 0)²] = √(c² + b²).- Side F₂B: Similarly, this is the distance from F₂ to B. F₂ is at (c, 0), so the distance is √[(0 - c)² + (b - 0)²] = √(c² + b²).So, the triangle has sides of lengths 2c, √(c² + b²), and √(c² + b²). That means it's an isosceles triangle with two sides equal.For the triangle to be acute, all three conditions must hold:1. (2c)² < (√(c² + b²))² + (√(c² + b²))²2. (√(c² + b²))² < (2c)² + (√(c² + b²))²3. (√(c² + b²))² < (2c)² + (√(c² + b²))²But since two sides are equal, the second and third conditions are actually the same. So, we only need to check the first condition and one of the others.Let's compute the first condition:(2c)² < (√(c² + b²))² + (√(c² + b²))²Simplify:4c² < (c² + b²) + (c² + b²)4c² < 2c² + 2b²Subtract 2c² from both sides:2c² < 2b²Divide both sides by 2:c² < b²But wait, in an ellipse, we know that c² = a² - b². So, substituting:a² - b² < b²a² < 2b²Divide both sides by b²:(a²)/(b²) < 2But (a²)/(b²) is equal to (a/b)², which is greater than 1 since a > b. However, this condition says it's less than 2. So, (a/b)² < 2, which implies that a/b < √2.But I also need to check the other conditions. Let's take the second condition:(√(c² + b²))² < (2c)² + (√(c² + b²))²Simplify:c² + b² < 4c² + c² + b²c² + b² < 5c² + b²Subtract c² + b² from both sides:0 < 4c²Which is always true since c > 0. So, the second condition doesn't give us any new information.Therefore, the only meaningful condition is c² < b², which translates to a² - b² < b², so a² < 2b².But I need to express this in terms of eccentricity. Since e = c/a, and c² = a² - b², let's express b² in terms of a and e.We have c = ae, so c² = a²e². Then, b² = a² - c² = a² - a²e² = a²(1 - e²).So, substituting into the inequality a² < 2b²:a² < 2 * a²(1 - e²)Divide both sides by a² (since a > 0):1 < 2(1 - e²)1 < 2 - 2e²Subtract 2 from both sides:-1 < -2e²Multiply both sides by (-1), which reverses the inequality:1 > 2e²Divide both sides by 2:1/2 > e²Take square roots:√(1/2) > eWhich simplifies to:e < √2 / 2 ≈ 0.707But wait, the options given are √2/2, √3/2, (√5 -1)/2, and 1/2. So, e must be less than √2/2. But we need to find the exact value of e.However, the problem states that the triangle is acute, so all angles are less than 90 degrees. We derived that c² < b², which gives us e < √2/2. But the options include √2/2 as an option, so perhaps the condition is e < √2/2, but the maximum possible e for the triangle to be acute is √2/2. However, since e must be less than √2/2, but the options don't include a value less than √2/2 except 1/2. Wait, 1/2 is less than √2/2, which is approximately 0.707. So, 1/2 is 0.5, which is less than √2/2.But wait, let's double-check. Maybe I made a mistake in the condition. Let me go back.We have the triangle with sides 2c, √(c² + b²), √(c² + b²). For it to be acute, all angles must be less than 90 degrees. The angle opposite the side 2c is the angle at B, which is the vertex at (0, b). The other two angles are at F₁ and F₂, which are equal because the triangle is isosceles.So, to ensure all angles are acute, we need:1. The angle at B is acute: (2c)² < (√(c² + b²))² + (√(c² + b²))²2. The angles at F₁ and F₂ are acute: (√(c² + b²))² < (2c)² + (√(c² + b²))²But as we saw, the second condition is always true, so the only real condition is the first one, which gives c² < b².But c² = a² - b², so a² - b² < b² ⇒ a² < 2b² ⇒ (a/b)² < 2 ⇒ a/b < √2.But we also know that in an ellipse, b = a√(1 - e²). So, let's express b in terms of a and e:b = a√(1 - e²)Then, a/b = 1 / √(1 - e²)So, the condition a/b < √2 becomes:1 / √(1 - e²) < √2Square both sides:1 / (1 - e²) < 2Multiply both sides by (1 - e²):1 < 2(1 - e²)1 < 2 - 2e²Subtract 2:-1 < -2e²Multiply by (-1), reverse inequality:1 > 2e²So, e² < 1/2 ⇒ e < √(1/2) = √2/2 ≈ 0.707So, the eccentricity must be less than √2/2. But the options are A. √2/2, B. √3/2, C. (√5 -1)/2, D. 1/2.Since e must be less than √2/2, the only possible answer is D. 1/2, because √2/2 is approximately 0.707, and 1/2 is 0.5, which is less than √2/2. The other options are larger than √2/2 except for D.Wait, but let me think again. The condition is e < √2/2, so the maximum possible e is approaching √2/2, but not reaching it. However, the options include √2/2 as an option. So, perhaps the answer is √2/2, but according to our derivation, e must be less than √2/2. So, maybe I made a mistake.Alternatively, perhaps I need to consider the triangle being acute, which requires that all angles are less than 90 degrees. So, not only the angle at B, but also the angles at F₁ and F₂ must be acute.Wait, but earlier I thought that the angles at F₁ and F₂ are automatically acute because the side opposite to them is √(c² + b²), and the other sides are 2c and √(c² + b²). So, for the angle at F₁, the sides adjacent are 2c and √(c² + b²), and the side opposite is √(c² + b²). So, to check if the angle at F₁ is acute, we need:(√(c² + b²))² < (2c)² + (√(c² + b²))²Which simplifies to:c² + b² < 4c² + c² + b²Which is:c² + b² < 5c² + b²Subtract c² + b²:0 < 4c²Which is always true. So, the angle at F₁ is always acute, regardless of the values. Similarly for F₂.Therefore, the only condition we need is that the angle at B is acute, which gives us c² < b², leading to e < √2/2.But the options include √2/2 as an option, so perhaps the answer is √2/2, but according to our condition, e must be less than √2/2. So, maybe the answer is D. 1/2, because it's the only option less than √2/2.Alternatively, perhaps I need to consider the triangle being acute in a stricter sense, meaning that all sides must satisfy the condition, not just one. But as we saw, the other conditions are automatically satisfied.Wait, let me think differently. Maybe I should use vectors or coordinates to compute the angles.Let's place the ellipse in the coordinate system. F₁ is at (-c, 0), F₂ at (c, 0), and B at (0, b).So, vectors:- From F₁ to F₂: (2c, 0)- From F₁ to B: (c, b)- From F₂ to B: (-c, b)To check if the triangle is acute, all dot products between the vectors should be positive because in an acute triangle, the cosine of each angle is positive, meaning the dot product is positive.So, let's compute the dot products:1. Between vectors F₁F₂ and F₁B:F₁F₂ vector is (2c, 0)F₁B vector is (c, b)Dot product: (2c)(c) + (0)(b) = 2c²Since c > 0, this is positive.2. Between vectors F₂F₁ and F₂B:F₂F₁ vector is (-2c, 0)F₂B vector is (-c, b)Dot product: (-2c)(-c) + (0)(b) = 2c²Positive again.3. Between vectors BF₁ and BF₂:BF₁ vector is (c, -b)BF₂ vector is (-c, -b)Dot product: (c)(-c) + (-b)(-b) = -c² + b²For the angle at B to be acute, this dot product must be positive:-c² + b² > 0 ⇒ b² > c² ⇒ same as before.So, the condition is b² > c², which is the same as c² < b², leading to e < √2/2.Therefore, the eccentricity must be less than √2/2. Among the options, only D. 1/2 satisfies this.Wait, but let me check if 1/2 is indeed the correct answer. Let's compute e = 1/2.If e = 1/2, then c = a/2. Then, b² = a² - c² = a² - (a²/4) = (3a²)/4 ⇒ b = (√3 a)/2.So, b² = (3a²)/4, and c² = a²/4. So, b² = 3c², which is greater than c², so the condition b² > c² is satisfied.Therefore, e = 1/2 is valid.But wait, what if e is exactly √2/2? Then, c = a√2/2, so c² = a²/2. Then, b² = a² - a²/2 = a²/2, so b² = c². Then, the dot product at B would be zero, meaning the angle is 90 degrees, making the triangle right-angled, not acute. Therefore, e must be less than √2/2.Hence, the correct answer is D. 1/2.

💡Okay, so I have this problem here: I need to find the equation of a line that passes through the center of a given circle and is perpendicular to another given line. The circle is defined by the equation ( x^2 + (y - 1)^2 = 4 ), and the line is given by ( 3x + 2y + 1 = 0 ). The options are A through D, each with different equations. First, I need to recall some basic concepts. The general equation of a circle is ( (x - h)^2 + (y - k)^2 = r^2 ), where ((h, k)) is the center and ( r ) is the radius. So, looking at the given circle equation ( x^2 + (y - 1)^2 = 4 ), I can see that the center is at ((0, 1)) because there's no ( x ) term, meaning ( h = 0 ), and the ( y ) term is ( (y - 1) ), so ( k = 1 ). The radius is ( sqrt{4} = 2 ), but I don't think the radius is directly needed here.Next, I need to find the line that is perpendicular to ( 3x + 2y + 1 = 0 ) and passes through the center of the circle, which is ((0, 1)). To find a line perpendicular to another, I remember that the slopes of perpendicular lines are negative reciprocals of each other. So, if I can find the slope of the given line ( 3x + 2y + 1 = 0 ), then I can find the slope of the line perpendicular to it.Let me rearrange the given line equation into slope-intercept form ( y = mx + b ) to find its slope. Starting with ( 3x + 2y + 1 = 0 ), I'll subtract ( 3x ) and 1 from both sides:( 2y = -3x - 1 )Now, divide both sides by 2:( y = (-3/2)x - 1/2 )So, the slope ( m ) of this line is ( -3/2 ). Therefore, the slope of the line perpendicular to this should be the negative reciprocal, which would be ( 2/3 ). Now that I have the slope of the desired line, which is ( 2/3 ), and it passes through the point ((0, 1)), I can use the point-slope form of a line equation to write the equation. The point-slope form is ( y - y_1 = m(x - x_1) ), where ( (x_1, y_1) ) is a point the line passes through, and ( m ) is the slope.Plugging in the values, I get:( y - 1 = (2/3)(x - 0) )Simplifying this:( y - 1 = (2/3)x )To convert this into standard form ( Ax + By + C = 0 ), I'll multiply both sides by 3 to eliminate the fraction:( 3(y - 1) = 2x )Expanding the left side:( 3y - 3 = 2x )Now, bringing all terms to one side:( -2x + 3y - 3 = 0 )But usually, we prefer the coefficient of ( x ) to be positive, so I'll multiply the entire equation by -1:( 2x - 3y + 3 = 0 )Looking at the options provided:A: ( 2x - 3y + 3 = 0 )B: ( 2x - 3y - 3 = 0 )C: ( 2x + 3y + 3 = 0 )D: ( 2x + 3y - 3 = 0 )So, the equation I derived is exactly option A. Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the slope: original line has slope ( -3/2 ), so perpendicular slope is ( 2/3 ). That seems right. Then using point-slope with point (0,1):( y - 1 = (2/3)x ). Converting to standard form: ( 2x - 3y + 3 = 0 ). Yes, that looks correct. Alternatively, maybe I should check if the line passes through (0,1). Plugging x=0 and y=1 into option A: ( 2(0) - 3(1) + 3 = 0 - 3 + 3 = 0 ). So, yes, it satisfies the equation. Just to be thorough, let me check if the slope is correct. The original line had slope ( -3/2 ), so the perpendicular slope should be ( 2/3 ). If I rearrange option A into slope-intercept form:( 2x - 3y + 3 = 0 )Subtract 2x and 3 from both sides:( -3y = -2x - 3 )Divide by -3:( y = (2/3)x + 1 )So, the slope is indeed ( 2/3 ), which is correct. And the y-intercept is 1, which matches the center of the circle. Therefore, I'm confident that option A is the correct answer.**Final Answer**boxed{A}

💡First, identify Glory's favorite number, which is given as 450.Next, calculate Misty's favorite number by dividing Glory's number by 3:450 ÷ 3 = 150.Then, determine Dawn's favorite number by multiplying Glory's number by 2:450 × 2 = 900.Finally, add all three numbers together to find the total sum:150 + 450 + 900 = 1500.

💡Okay, so I've got this equation here: (left[left(10^{3}right)left(10^{x}right)right]^{2}=10^{18}). I need to find the value of (x). Hmm, let me think about how to approach this step by step.First, I remember that when you multiply two exponents with the same base, you can add the exponents. So, (10^3 times 10^x) should be (10^{3+x}). Let me write that down:[left(10^{3} times 10^{x}right)^2 = 10^{18}][left(10^{3+x}right)^2 = 10^{18}]Okay, now I have (left(10^{3+x}right)^2). I also recall that when you raise a power to another power, you multiply the exponents. So, (left(10^{3+x}right)^2) becomes (10^{2(3+x)}). Let me write that:[10^{2(3+x)} = 10^{18}]Now, since the bases are the same (both are base 10), the exponents must be equal. That means:[2(3 + x) = 18]Alright, time to solve for (x). I'll start by expanding the left side:[2 times 3 + 2 times x = 18][6 + 2x = 18]Next, I'll subtract 6 from both sides to isolate the term with (x):[2x = 18 - 6][2x = 12]Finally, I'll divide both sides by 2 to solve for (x):[x = frac{12}{2}][x = 6]Let me double-check my steps to make sure I didn't make any mistakes. Starting from the original equation, I combined the exponents correctly, then applied the power rule, set the exponents equal, and solved the resulting equation. Everything seems to add up. So, I'm confident that (x = 6) is the correct answer.

💡Alright, so I have this problem about a sequence {a_n} where the sum of the first n terms is given by S_n = 2^{n+2} - 4. I need to find the general formula for a_n and then, using that, find the sum of the first n terms of another sequence {b_n}, where each term b_n is a_n multiplied by the logarithm base 2 of a_n. Starting with part (1), finding the general formula for a_n. I remember that the sum of the first n terms, S_n, is related to the sequence itself. Specifically, the nth term of the sequence can be found by subtracting the sum of the first (n-1) terms from the sum of the first n terms. So, a_n = S_n - S_{n-1}. Let me write that down:a_n = S_n - S_{n-1}Given that S_n = 2^{n+2} - 4, then S_{n-1} would be 2^{(n-1)+2} - 4, which simplifies to 2^{n+1} - 4. So, substituting these into the equation for a_n:a_n = (2^{n+2} - 4) - (2^{n+1} - 4)Simplifying this:a_n = 2^{n+2} - 4 - 2^{n+1} + 4The -4 and +4 cancel out, so:a_n = 2^{n+2} - 2^{n+1}I can factor out 2^{n+1}:a_n = 2^{n+1}(2 - 1) = 2^{n+1}(1) = 2^{n+1}Wait, so a_n is 2^{n+1}. Let me check if this makes sense for the first few terms.For n=1: a_1 = 2^{1+1} = 4. Let me check S_1: 2^{1+2} - 4 = 8 - 4 = 4. That matches.For n=2: a_2 = 2^{2+1} = 8. S_2 should be 2^{2+2} - 4 = 16 - 4 = 12. So, a_1 + a_2 = 4 + 8 = 12. That works.For n=3: a_3 = 2^{3+1} = 16. S_3 = 2^{3+2} - 4 = 32 - 4 = 28. Then a_1 + a_2 + a_3 = 4 + 8 + 16 = 28. Perfect.So, it seems that a_n = 2^{n+1} is correct for n ≥ 1.Moving on to part (2), where b_n = a_n * log_2(a_n). Since a_n = 2^{n+1}, then log_2(a_n) is log_2(2^{n+1}) which is just n+1. So, b_n = 2^{n+1} * (n+1). That simplifies to (n+1) * 2^{n+1}.So, b_n = (n+1) * 2^{n+1}. Now, I need to find the sum of the first n terms of {b_n}, which is T_n = b_1 + b_2 + ... + b_n.So, T_n = sum_{k=1}^{n} (k+1) * 2^{k+1}Let me write out the terms to see the pattern:For k=1: (1+1)*2^{1+1} = 2*4 = 8For k=2: (2+1)*2^{2+1} = 3*8 = 24For k=3: (3+1)*2^{3+1} = 4*16 = 64And so on, up to k=n: (n+1)*2^{n+1}So, the sum T_n is 8 + 24 + 64 + ... + (n+1)*2^{n+1}Hmm, this looks like a series where each term is a multiple of a geometric sequence. Maybe I can use the method of differences or find a recursive formula.I remember that for sums involving k*2^k, there's a standard technique. Let me recall that.Let me denote T_n = sum_{k=1}^{n} (k+1)*2^{k+1}I can factor out the 2^{k+1}:T_n = sum_{k=1}^{n} (k+1)*2^{k+1} = sum_{k=1}^{n} (k+1)*2^{k+1}Let me make a substitution: let m = k+1, so when k=1, m=2, and when k=n, m=n+1. So, T_n = sum_{m=2}^{n+1} m*2^{m}But that might complicate things. Alternatively, perhaps I can write this as:T_n = sum_{k=1}^{n} (k+1)*2^{k+1} = sum_{k=1}^{n} k*2^{k+1} + sum_{k=1}^{n} 2^{k+1}So, T_n = 2*sum_{k=1}^{n} k*2^{k} + sum_{k=1}^{n} 2^{k+1}Let me compute each part separately.First, compute sum_{k=1}^{n} 2^{k+1}. That's a geometric series.sum_{k=1}^{n} 2^{k+1} = 2^{2} + 2^{3} + ... + 2^{n+1} = sum_{m=2}^{n+1} 2^{m} = sum_{m=0}^{n+1} 2^{m} - 2^{0} - 2^{1} = (2^{n+2} - 1) - 1 - 2 = 2^{n+2} - 4Wait, that's interesting. Because S_n = 2^{n+2} - 4, which is the sum of a_n. Hmm, maybe there's a connection.But let me just note that sum_{k=1}^{n} 2^{k+1} = 2^{n+2} - 4.Now, the other part is sum_{k=1}^{n} k*2^{k}. I need a formula for that.I recall that sum_{k=0}^{n} k*r^{k} = r*(1 - (n+1)*r^{n} + n*r^{n+1}) / (1 - r)^2, for r ≠ 1.In this case, r=2, so let me apply that formula.sum_{k=0}^{n} k*2^{k} = 2*(1 - (n+1)*2^{n} + n*2^{n+1}) / (1 - 2)^2Simplify denominator: (1 - 2)^2 = 1So, sum_{k=0}^{n} k*2^{k} = 2*(1 - (n+1)*2^{n} + n*2^{n+1})Simplify numerator:1 - (n+1)*2^{n} + n*2^{n+1} = 1 - (n+1)*2^{n} + 2n*2^{n} = 1 + (2n - n -1)*2^{n} = 1 + (n -1)*2^{n}So, sum_{k=0}^{n} k*2^{k} = 2*(1 + (n -1)*2^{n}) = 2 + 2*(n -1)*2^{n} = 2 + (n -1)*2^{n+1}But we need sum_{k=1}^{n} k*2^{k} = sum_{k=0}^{n} k*2^{k} - 0*2^{0} = 2 + (n -1)*2^{n+1}So, sum_{k=1}^{n} k*2^{k} = 2 + (n -1)*2^{n+1}Therefore, going back to T_n:T_n = 2*sum_{k=1}^{n} k*2^{k} + sum_{k=1}^{n} 2^{k+1} = 2*(2 + (n -1)*2^{n+1}) + (2^{n+2} -4)Simplify:First, compute 2*(2 + (n -1)*2^{n+1}) = 4 + 2*(n -1)*2^{n+1} = 4 + (n -1)*2^{n+2}Then, add (2^{n+2} -4):So, T_n = 4 + (n -1)*2^{n+2} + 2^{n+2} -4The 4 and -4 cancel out:T_n = (n -1)*2^{n+2} + 2^{n+2} = (n -1 +1)*2^{n+2} = n*2^{n+2}Wait, that's a nice result. So, T_n = n*2^{n+2}Let me verify this with small n.For n=1: T_1 = b_1 = (1+1)*2^{1+1} = 2*4=8. According to the formula, T_1=1*2^{1+2}=1*8=8. Correct.For n=2: T_2 = b_1 + b_2 = 8 + 24=32. Formula: 2*2^{2+2}=2*16=32. Correct.For n=3: T_3=8+24+64=96. Formula:3*2^{3+2}=3*32=96. Correct.Looks good.So, summarizing:(1) a_n = 2^{n+1}(2) T_n = n*2^{n+2}**Final Answer**(1) The general term is boxed{2^{n+1}}.(2) The sum of the first n terms of {b_n} is boxed{n cdot 2^{n+2}}.

💡First, I recognize that both terms in the expression (10! - 9!) involve factorials, which are products of consecutive integers up to a certain number.I know that (10!) can be expressed as (10 times 9!), because (10! = 10 times 9 times 8 times dots times 1) and (9! = 9 times 8 times dots times 1).By factoring out (9!) from both terms, the expression becomes (9! times (10 - 1)), which simplifies to (9! times 9).Next, I calculate (9!) by multiplying the numbers from 1 to 9 together:[9! = 9 times 8 times 7 times 6 times 5 times 4 times 3 times 2 times 1 = 362880]Finally, I multiply (362880) by (9) to find the value of (10! - 9!):[362880 times 9 = 3265920]