💡Okay, so I need to compute the product of four sine functions: sin 6°, sin 42°, sin 66°, and sin 78°. Hmm, that seems a bit complicated at first glance, but maybe there's a pattern or identity I can use to simplify it.First, I remember that sometimes products of sines can be simplified using product-to-sum formulas or by recognizing complementary angles. Let me see if any of these angles are complementary. Looking at the angles: 6°, 42°, 66°, and 78°. I know that complementary angles add up to 90°, so let's check:- 6° and 84° are complementary because 6 + 84 = 90.- 42° and 48° are complementary because 42 + 48 = 90.- 66° and 24° are complementary because 66 + 24 = 90.- 78° and 12° are complementary because 78 + 12 = 90.Hmm, so none of the given angles are directly complementary to each other, but maybe I can express some of them in terms of their complements. For example, sin 78° is equal to cos 12°, since sin θ = cos(90° - θ). Similarly, sin 66° is equal to cos 24°, and sin 42° is equal to cos 48°, but 48° isn't one of the angles here. Maybe this isn't the most straightforward approach.Alternatively, I remember that products of sines can sometimes be expressed using multiple-angle identities or by using complex exponentials, but that might be too advanced for now. Let me think about another approach.I recall that there's a formula for the product of sines in an arithmetic progression, but I'm not sure if these angles form such a progression. Let me check the differences:- 42° - 6° = 36°- 66° - 42° = 24°- 78° - 66° = 12°No, the differences aren't consistent, so that might not help either.Wait, maybe I can pair the angles in a way that allows me to use some identities. Let's try pairing sin 6° with sin 78° and sin 42° with sin 66°. Starting with sin 6° and sin 78°, I can use the identity sin A sin B = [cos(A - B) - cos(A + B)] / 2. Let's apply that:sin 6° sin 78° = [cos(6° - 78°) - cos(6° + 78°)] / 2= [cos(-72°) - cos(84°)] / 2Since cos(-θ) = cos θ, this simplifies to:= [cos 72° - cos 84°] / 2Similarly, for sin 42° and sin 66°, applying the same identity:sin 42° sin 66° = [cos(42° - 66°) - cos(42° + 66°)] / 2= [cos(-24°) - cos(108°)] / 2Again, cos(-θ) = cos θ, so:= [cos 24° - cos 108°] / 2Now, our original product becomes:[sin 6° sin 78°] * [sin 42° sin 66°] = ([cos 72° - cos 84°] / 2) * ([cos 24° - cos 108°] / 2)= (1/4) [cos 72° - cos 84°][cos 24° - cos 108°]This seems a bit messy, but maybe expanding the product will help. Let's multiply the two brackets:= (1/4) [cos 72° cos 24° - cos 72° cos 108° - cos 84° cos 24° + cos 84° cos 108°]Now, I have four terms involving products of cosines. I can use the identity cos A cos B = [cos(A + B) + cos(A - B)] / 2 for each term.Let's compute each term one by one:1. cos 72° cos 24° = [cos(72° + 24°) + cos(72° - 24°)] / 2= [cos 96° + cos 48°] / 22. cos 72° cos 108° = [cos(72° + 108°) + cos(72° - 108°)] / 2= [cos 180° + cos(-36°)] / 2= [cos 180° + cos 36°] / 2Since cos 180° = -1, this becomes:= [-1 + cos 36°] / 23. cos 84° cos 24° = [cos(84° + 24°) + cos(84° - 24°)] / 2= [cos 108° + cos 60°] / 2cos 60° = 0.5, so:= [cos 108° + 0.5] / 24. cos 84° cos 108° = [cos(84° + 108°) + cos(84° - 108°)] / 2= [cos 192° + cos(-24°)] / 2= [cos 192° + cos 24°] / 2Since cos 192° = cos(180° + 12°) = -cos 12°, this becomes:= [-cos 12° + cos 24°] / 2Now, substituting these back into our expression:= (1/4) [ ( [cos 96° + cos 48°] / 2 ) - ( [ -1 + cos 36° ] / 2 ) - ( [cos 108° + 0.5 ] / 2 ) + ( [ -cos 12° + cos 24° ] / 2 ) ]Let's simplify each term inside the brackets:First term: [cos 96° + cos 48°] / 2Second term: - [ -1 + cos 36° ] / 2 = [1 - cos 36°] / 2Third term: - [cos 108° + 0.5 ] / 2 = -cos 108° / 2 - 0.25Fourth term: [ -cos 12° + cos 24° ] / 2So combining all these:= (1/4) [ (cos 96° + cos 48°)/2 + (1 - cos 36°)/2 - cos 108°/2 - 0.25 + (-cos 12° + cos 24°)/2 ]Now, let's combine the fractions:= (1/4) [ (cos 96° + cos 48° + 1 - cos 36° - cos 108° - cos 12° + cos 24°) / 2 - 0.25 ]Wait, actually, I think I made a mistake in combining the terms. Let me re-express it step by step.First, let's list all the terms:1. (cos 96° + cos 48°) / 22. (1 - cos 36°) / 23. -cos 108° / 24. -0.255. (-cos 12° + cos 24°) / 2So, combining terms 1 to 5:= (1/4) [ (cos 96° + cos 48° + 1 - cos 36° - cos 108° - cos 12° + cos 24°) / 2 - 0.25 ]Wait, actually, terms 1, 2, 3, and 5 are all divided by 2, and term 4 is -0.25. So let's factor out the 1/2:= (1/4) [ (cos 96° + cos 48° + 1 - cos 36° - cos 108° - cos 12° + cos 24°) / 2 - 0.25 ]= (1/4) [ (cos 96° + cos 48° + 1 - cos 36° - cos 108° - cos 12° + cos 24° - 0.5) / 2 ]Wait, no, that's not quite right. Let me think again.Actually, term 4 is -0.25, which is -1/4. So when we combine all terms:= (1/4) [ (cos 96° + cos 48° + 1 - cos 36° - cos 108° - cos 12° + cos 24°) / 2 - 1/4 ]Now, let's combine the constants:1 - 1/4 = 3/4So:= (1/4) [ (cos 96° + cos 48° - cos 36° - cos 108° - cos 12° + cos 24°) / 2 + 3/4 ]This is getting quite complicated. Maybe there's a better approach. Let me think if there's a different identity or method I can use.I remember that sometimes products of sines can be related to roots of unity or using complex numbers, but that might be too advanced. Alternatively, maybe I can use multiple-angle identities or look for symmetries.Wait, another idea: perhaps using the identity for sin A sin(60° - A) sin(60° + A) = (sin 3A)/4. Let me check if this applies here.Looking at the angles: 6°, 42°, 66°, 78°. Let's see if any of these can be expressed as 60° ± something.- 6° = 60° - 54°- 42° = 60° - 18°- 66° = 60° + 6°- 78° = 60° + 18°Hmm, interesting. So 6° = 60° - 54°, 42° = 60° - 18°, 66° = 60° + 6°, and 78° = 60° + 18°. Wait, if I consider A = 6°, then:sin 6° sin(60° - 6°) sin(60° + 6°) = sin 6° sin 54° sin 66° = (sin 18°)/4But in our product, we have sin 6°, sin 42°, sin 66°, sin 78°. So sin 42° is sin(60° - 18°), and sin 78° is sin(60° + 18°). So maybe I can pair sin 6° with sin 66° and sin 42° with sin 78°, and apply the identity separately.Let's try that.First, consider sin 6° sin 66°. Using the identity sin A sin(60° + A) = [cos(A - (60° + A)) - cos(A + (60° + A))]/2, but that might not be the best way. Alternatively, using the identity I mentioned earlier:sin A sin(60° - A) sin(60° + A) = (sin 3A)/4But in this case, if I take A = 6°, then sin 6° sin 54° sin 66° = sin 18° /4. But we don't have sin 54° in our product. Hmm.Alternatively, maybe I can consider the product sin 6° sin 42° sin 66° sin 78° as [sin 6° sin 66°][sin 42° sin 78°]. Let's compute each pair separately.First pair: sin 6° sin 66°. Let's use the identity sin A sin B = [cos(A - B) - cos(A + B)] / 2.So:sin 6° sin 66° = [cos(6° - 66°) - cos(6° + 66°)] / 2= [cos(-60°) - cos(72°)] / 2= [cos 60° - cos 72°] / 2Since cos(-θ) = cos θ.cos 60° = 0.5, so:= [0.5 - cos 72°] / 2Similarly, for the second pair: sin 42° sin 78°.Using the same identity:sin 42° sin 78° = [cos(42° - 78°) - cos(42° + 78°)] / 2= [cos(-36°) - cos(120°)] / 2= [cos 36° - cos 120°] / 2cos 120° = -0.5, so:= [cos 36° - (-0.5)] / 2= [cos 36° + 0.5] / 2Now, our original product becomes:[sin 6° sin 66°][sin 42° sin 78°] = ([0.5 - cos 72°] / 2) * ([cos 36° + 0.5] / 2)= (1/4)(0.5 - cos 72°)(cos 36° + 0.5)Let me compute (0.5 - cos 72°)(cos 36° + 0.5). Let's expand this:= 0.5 * cos 36° + 0.5 * 0.5 - cos 72° * cos 36° - cos 72° * 0.5= 0.5 cos 36° + 0.25 - cos 72° cos 36° - 0.5 cos 72°Now, let's compute each term:1. 0.5 cos 36°2. 0.253. -cos 72° cos 36°4. -0.5 cos 72°Let me see if I can simplify terms 3 and 4. Maybe using the identity cos A cos B = [cos(A+B) + cos(A-B)] / 2.For term 3: cos 72° cos 36° = [cos(72° + 36°) + cos(72° - 36°)] / 2= [cos 108° + cos 36°] / 2Similarly, term 4: -0.5 cos 72° = -0.5 cos 72°So substituting back:= 0.5 cos 36° + 0.25 - [cos 108° + cos 36°] / 2 - 0.5 cos 72°Let's distribute the negative sign:= 0.5 cos 36° + 0.25 - 0.5 cos 108° - 0.5 cos 36° - 0.5 cos 72°Now, combine like terms:- 0.5 cos 36° + 0.5 cos 36° = 0So we're left with:= 0.25 - 0.5 cos 108° - 0.5 cos 72°Now, let's compute cos 108° and cos 72°. I know that cos 108° = cos(180° - 72°) = -cos 72°. So:cos 108° = -cos 72°Therefore, the expression becomes:= 0.25 - 0.5*(-cos 72°) - 0.5 cos 72°= 0.25 + 0.5 cos 72° - 0.5 cos 72°= 0.25Wow, that simplified nicely! So the entire expression inside the brackets is 0.25.Therefore, our original product is:= (1/4)(0.25)= 1/16So, sin 6° sin 42° sin 66° sin 78° = 1/16.I think that's the answer. Let me just double-check my steps to make sure I didn't make any mistakes.Starting from the product, I paired sin 6° with sin 66° and sin 42° with sin 78°, used the product-to-sum identities, expanded, and through simplification, everything canceled out except for 0.25, leading to the final result of 1/16. It seems correct.

💡Okay, so I need to find the domain of the function ( f(x) = sqrt{(ln x - 2)(x - ln x - 1)} ). Hmm, the domain of a function is basically all the real numbers for which the function is defined. Since this function involves a square root, the expression inside the square root must be greater than or equal to zero. So, I need to solve the inequality:[(ln x - 2)(x - ln x - 1) geq 0]Alright, let's break this down. First, I know that the natural logarithm function, ( ln x ), is only defined for ( x > 0 ). So, right away, I know that ( x ) must be positive. That gives me a starting point: ( x > 0 ).Now, the expression inside the square root is a product of two terms: ( (ln x - 2) ) and ( (x - ln x - 1) ). For their product to be non-negative, either both terms are non-negative or both are non-positive. So, I need to analyze the signs of each term.Let me consider each factor separately.First, ( ln x - 2 geq 0 ) implies ( ln x geq 2 ), which means ( x geq e^2 ) because ( e^2 ) is approximately 7.389. So, when ( x geq e^2 ), the first term is non-negative.Second, ( x - ln x - 1 geq 0 ). Let me denote this as ( g(x) = x - ln x - 1 ). I need to find when ( g(x) geq 0 ). To do this, maybe I should analyze the behavior of ( g(x) ).Let's compute the derivative of ( g(x) ) to find its critical points and determine where it's increasing or decreasing.[g'(x) = frac{d}{dx}(x - ln x - 1) = 1 - frac{1}{x}]Setting ( g'(x) = 0 ):[1 - frac{1}{x} = 0 implies frac{1}{x} = 1 implies x = 1]So, ( x = 1 ) is a critical point. Let's analyze the intervals around this point.For ( x < 1 ), say ( x = 0.5 ):[g'(0.5) = 1 - frac{1}{0.5} = 1 - 2 = -1 < 0]So, ( g(x) ) is decreasing on ( (0, 1) ).For ( x > 1 ), say ( x = 2 ):[g'(2) = 1 - frac{1}{2} = 0.5 > 0]So, ( g(x) ) is increasing on ( (1, infty) ).Therefore, ( g(x) ) has a minimum at ( x = 1 ). Let's compute ( g(1) ):[g(1) = 1 - ln 1 - 1 = 1 - 0 - 1 = 0]So, ( g(1) = 0 ). Since ( g(x) ) is decreasing before ( x = 1 ) and increasing after ( x = 1 ), and it reaches a minimum of 0 at ( x = 1 ), this means ( g(x) geq 0 ) for all ( x > 0 ).Wait, that can't be right. Because if ( g(x) ) is always non-negative, then ( x - ln x - 1 geq 0 ) for all ( x > 0 ). Is that true?Let me test a value less than 1, say ( x = 0.5 ):[g(0.5) = 0.5 - ln 0.5 - 1 = 0.5 - (-0.6931) - 1 ≈ 0.5 + 0.6931 - 1 ≈ 0.1931 > 0]And at ( x = 1 ), it's 0, and for ( x > 1 ), it's increasing. So, yes, ( g(x) geq 0 ) for all ( x > 0 ).So, the second factor ( (x - ln x - 1) ) is always non-negative for ( x > 0 ).Therefore, the sign of the product ( (ln x - 2)(x - ln x - 1) ) depends solely on the sign of ( ln x - 2 ).Since ( x - ln x - 1 geq 0 ) for all ( x > 0 ), the product will be non-negative when ( ln x - 2 geq 0 ), which is when ( x geq e^2 ).But wait, at ( x = 1 ), ( g(1) = 0 ), so the product becomes ( (ln 1 - 2)(0) = (-2)(0) = 0 ), which is also non-negative. So, ( x = 1 ) is also a valid point.Therefore, the expression inside the square root is non-negative when ( x geq e^2 ) or ( x = 1 ).But let me double-check this. Suppose ( x ) is between 0 and 1, say ( x = 0.5 ). Then:[ln 0.5 ≈ -0.6931][ln x - 2 ≈ -0.6931 - 2 = -2.6931 < 0][x - ln x - 1 ≈ 0.5 - (-0.6931) - 1 ≈ 0.5 + 0.6931 - 1 ≈ 0.1931 > 0]So, the product is negative times positive, which is negative. Hence, the square root is not defined here.For ( x ) between 1 and ( e^2 ), say ( x = 2 ):[ln 2 ≈ 0.6931][ln x - 2 ≈ 0.6931 - 2 = -1.3069 < 0][x - ln x - 1 ≈ 2 - 0.6931 - 1 ≈ 0.3069 > 0]Again, the product is negative times positive, which is negative. So, the square root is not defined here either.At ( x = e^2 ):[ln e^2 = 2][ln x - 2 = 0][x - ln x - 1 = e^2 - 2 - 1 = e^2 - 3 ≈ 7.389 - 3 ≈ 4.389 > 0]So, the product is 0 times positive, which is 0. Hence, the square root is defined.For ( x > e^2 ), say ( x = 10 ):[ln 10 ≈ 2.3026][ln x - 2 ≈ 2.3026 - 2 = 0.3026 > 0][x - ln x - 1 ≈ 10 - 2.3026 - 1 ≈ 6.6974 > 0]So, the product is positive times positive, which is positive. Hence, the square root is defined.Therefore, the expression inside the square root is non-negative when ( x = 1 ) or ( x geq e^2 ).So, the domain of ( f(x) ) is all ( x ) such that ( x = 1 ) or ( x geq e^2 ).But wait, let me check ( x = 1 ):[f(1) = sqrt{(ln 1 - 2)(1 - ln 1 - 1)} = sqrt{(0 - 2)(1 - 0 - 1)} = sqrt{(-2)(0)} = sqrt{0} = 0]So, it's defined at ( x = 1 ).Therefore, combining these, the domain is ( x = 1 ) and ( x geq e^2 ).But in interval notation, how do we write this? It's the union of the singleton set ( {1} ) and the interval ( [e^2, infty) ).So, the domain is ( {1} cup [e^2, infty) ).I think that's it. Let me just recap:1. The function is a square root, so the inside must be non-negative.2. The inside is a product of two terms: ( (ln x - 2) ) and ( (x - ln x - 1) ).3. Analyzed each term: - ( ln x - 2 geq 0 ) when ( x geq e^2 ). - ( x - ln x - 1 geq 0 ) for all ( x > 0 ), with equality only at ( x = 1 ).4. Therefore, the product is non-negative when ( x geq e^2 ) or ( x = 1 ).5. Hence, the domain is ( {1} cup [e^2, infty) ).I don't think I missed anything. Let me just visualize the graph of ( g(x) = x - ln x - 1 ) to confirm it's always non-negative. At ( x = 1 ), it's zero, and it's increasing for ( x > 1 ) and decreasing for ( x < 1 ), but since at ( x ) approaching 0, ( ln x ) goes to negative infinity, so ( x - ln x - 1 ) goes to positive infinity. So yes, it's always non-negative.Therefore, my conclusion is correct.

💡Okay, so I have this problem where I need to find a new function G by substituting x in F with 2x/(1+x²), and then simplify it. The function F is given as the logarithm of (1+x)/(1-x). The options are A) F, B) 2F, C) F squared, and D) F over 2. Hmm, let me try to figure this out step by step.First, let me write down what F is:F = log[(1 + x)/(1 - x)]Now, I need to replace every x in F with 2x/(1 + x²). So, wherever there's an x in F, I'll put 2x/(1 + x²) instead. That means:G = log[(1 + (2x/(1 + x²)))/(1 - (2x/(1 + x²)))]Okay, that looks a bit complicated, but maybe I can simplify the expression inside the logarithm. Let's work on the numerator and the denominator separately.Starting with the numerator:1 + (2x/(1 + x²)) = (1*(1 + x²) + 2x)/(1 + x²) = (1 + x² + 2x)/(1 + x²)Similarly, the denominator:1 - (2x/(1 + x²)) = (1*(1 + x²) - 2x)/(1 + x²) = (1 + x² - 2x)/(1 + x²)So now, G becomes:G = log[( (1 + x² + 2x)/(1 + x²) ) / ( (1 + x² - 2x)/(1 + x²) )]Since both the numerator and the denominator have the same denominator (1 + x²), when I divide them, the (1 + x²) terms will cancel out. So, simplifying that:G = log[(1 + x² + 2x)/(1 + x² - 2x)]Hmm, looking at the numerator and the denominator, I notice that they resemble perfect squares. Let me check:Numerator: 1 + x² + 2x = (1 + x)²Denominator: 1 + x² - 2x = (1 - x)²Yes, that's right! So, substituting back:G = log[( (1 + x)² ) / ( (1 - x)² )]Now, using the logarithm property that log(a²) = 2 log(a), I can take the exponent 2 out of the logarithm:G = log[(1 + x)/(1 - x)]² = 2 log[(1 + x)/(1 - x)]But wait, log[(1 + x)/(1 - x)] is exactly F. So, substituting back:G = 2FSo, after simplifying, G is equal to 2F. Therefore, the correct answer should be option B) 2F.Let me just double-check my steps to make sure I didn't make any mistakes. I substituted x with 2x/(1 + x²) correctly, simplified the numerator and denominator, recognized the perfect squares, applied the logarithm property, and ended up with 2F. Everything seems to add up, so I think I'm confident with this answer.**Final Answer**boxed{B}

💡Alright, so I have this problem where Xiao Zhang has 200 pencils and Xiao Li has 20 pens. Every time they exchange, Xiao Zhang gives Xiao Li 6 pencils, and Xiao Li gives Xiao Zhang 1 pen in return. I need to figure out after how many such exchanges the number of pencils Xiao Zhang has will be 11 times the number of pens Xiao Li has.Okay, let's break this down. First, let's think about what happens in each exchange. Every time they exchange, Xiao Zhang loses 6 pencils and gains 1 pen. Conversely, Xiao Li gains 6 pencils and loses 1 pen. So, over multiple exchanges, the number of pencils and pens each person has will change accordingly.Let's denote the number of exchanges as ( n ). After ( n ) exchanges, Xiao Zhang will have given away ( 6n ) pencils and received ( n ) pens. Similarly, Xiao Li will have received ( 6n ) pencils and given away ( n ) pens.So, the number of pencils Xiao Zhang has after ( n ) exchanges would be his initial number of pencils minus the pencils he gave away. That would be ( 200 - 6n ).On the other hand, the number of pens Xiao Li has after ( n ) exchanges would be her initial number of pens minus the pens she gave away. That would be ( 20 - n ).According to the problem, after a certain number of exchanges, the number of pencils Xiao Zhang has is 11 times the number of pens Xiao Li has. So, we can set up an equation:[ 200 - 6n = 11 times (20 - n) ]Now, let's solve this equation step by step.First, expand the right side:[ 200 - 6n = 220 - 11n ]Now, let's get all the terms involving ( n ) on one side and the constants on the other. I'll add ( 11n ) to both sides to eliminate the negative ( 11n ):[ 200 - 6n + 11n = 220 ]Simplify the left side:[ 200 + 5n = 220 ]Now, subtract 200 from both sides to isolate the term with ( n ):[ 5n = 20 ]Finally, divide both sides by 5 to solve for ( n ):[ n = 4 ]So, after 4 exchanges, Xiao Zhang will have 11 times as many pencils as Xiao Li has pens.Let me double-check this to make sure I didn't make a mistake. After 4 exchanges:- Xiao Zhang has ( 200 - 6 times 4 = 200 - 24 = 176 ) pencils.- Xiao Li has ( 20 - 4 = 16 ) pens.Is 176 equal to 11 times 16? Let's see: ( 11 times 16 = 176 ). Yes, that checks out.So, it looks like 4 exchanges is the correct answer.

💡Okay, so I have this problem about a geometric sequence. Let me try to understand what it's asking. It says that the sum of the first n terms is denoted by S_n. Then, it tells me that the ratio of S_8 to S_4 is 4, and I need to find the ratio of S_12 to S_4. The options are A: 3, B: 4, C: 13/4, and D: 13.Alright, first, let's recall what a geometric sequence is. In a geometric sequence, each term after the first is found by multiplying the previous term by a constant called the common ratio, usually denoted by r. So, if the first term is a, then the sequence is a, ar, ar^2, ar^3, and so on.The sum of the first n terms of a geometric sequence can be calculated using the formula:S_n = a * (1 - r^n) / (1 - r), when r ≠ 1.If r = 1, then every term is just a, and the sum S_n would be a * n. But since the ratio S_8/S_4 is given as 4, which is greater than 1, I can assume that r is not 1 because if r were 1, then S_8 would be 8a and S_4 would be 4a, so their ratio would be 2, not 4. So, r ≠ 1.So, I can use the formula S_n = a*(1 - r^n)/(1 - r).Given that S_8/S_4 = 4, let's write that out:S_8 / S_4 = [a*(1 - r^8)/(1 - r)] / [a*(1 - r^4)/(1 - r)] = (1 - r^8)/(1 - r^4) = 4.Simplify that:(1 - r^8)/(1 - r^4) = 4.Notice that 1 - r^8 can be factored as (1 - r^4)(1 + r^4). So,(1 - r^4)(1 + r^4)/(1 - r^4) = 1 + r^4 = 4.So, 1 + r^4 = 4.Therefore, r^4 = 3.So, r^4 = 3. That means r^2 = sqrt(3), but since r could be positive or negative, but since we're dealing with sums, the sign might matter. However, since the ratio S_8/S_4 is positive, and the sum S_4 is positive, r must be positive because if r were negative, the terms would alternate in sign, and the sum might not necessarily be positive. So, r is positive.So, r^4 = 3, which means r = 3^(1/4).Now, I need to find S_12/S_4.Let's compute S_12 and S_4 using the formula.First, S_4 = a*(1 - r^4)/(1 - r) = a*(1 - 3)/(1 - r) = a*(-2)/(1 - r).Similarly, S_8 = a*(1 - r^8)/(1 - r) = a*(1 - (r^4)^2)/(1 - r) = a*(1 - 9)/(1 - r) = a*(-8)/(1 - r).Wait, but from earlier, S_8/S_4 = 4, so let's check:S_8/S_4 = [a*(-8)/(1 - r)] / [a*(-2)/(1 - r)] = (-8)/(-2) = 4, which matches the given condition. So, that's consistent.Now, let's compute S_12.S_12 = a*(1 - r^12)/(1 - r).But r^12 = (r^4)^3 = 3^3 = 27.So, S_12 = a*(1 - 27)/(1 - r) = a*(-26)/(1 - r).Now, S_4 is a*(-2)/(1 - r), so S_12/S_4 = [a*(-26)/(1 - r)] / [a*(-2)/(1 - r)] = (-26)/(-2) = 13.So, the ratio S_12/S_4 is 13.Looking back at the options, D is 13, so that should be the answer.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. I used the formula for the sum of a geometric series correctly.2. I correctly factored 1 - r^8 as (1 - r^4)(1 + r^4).3. I solved for r^4 correctly, getting r^4 = 3.4. I computed S_4 and S_8 correctly, noting that r^4 = 3.5. I computed S_12 correctly by noting that r^12 = (r^4)^3 = 27.6. I then took the ratio S_12/S_4, which simplified to 13.Everything seems to check out. So, I'm confident that the answer is D: 13.

💡First, determine Ken's cycling speeds in both rain and snow conditions.When it's raining, Ken cycles 30 miles in 20 minutes. To find his speed in miles per hour:[text{Rain Speed} = frac{60}{20} times 30 = 3 times 30 = 90 text{ miles per hour}]When it's snowing, Ken cycles 10 miles in 20 minutes. To find his speed in miles per hour:[text{Snow Speed} = frac{60}{20} times 10 = 3 times 10 = 30 text{ miles per hour}]Ken cycles 1 hour each day. If it snows 4 times in a week, the distance covered during snow is:[4 times 30 = 120 text{ miles}]Let ( R ) be the number of times it rained. The distance covered during rain is:[R times 90 text{ miles}]The total distance cycled in the week is 390 miles:[R times 90 + 120 = 390]Solving for ( R ):[R times 90 = 390 - 120 R times 90 = 270 R = frac{270}{90} R = 3]