💡Alright, so I've got this problem here about a function f(x) = 2 sin x cos x + 2√3 cos²x - √3. It's divided into two parts, (Ⅰ) and (Ⅱ). Let me try to tackle them one by one.Starting with part (Ⅰ): I need to find the smallest positive period and the intervals where the function y = f(-3x) + 1 is monotonically decreasing. Hmm, okay. First, maybe I should simplify f(x) to make it easier to work with.Looking at f(x), I see 2 sin x cos x. I remember that 2 sin x cos x is equal to sin 2x. So that part simplifies to sin 2x. Then there's 2√3 cos²x. I recall that cos²x can be written using the double-angle identity: cos²x = (1 + cos 2x)/2. So substituting that in, 2√3 cos²x becomes 2√3*(1 + cos 2x)/2, which simplifies to √3*(1 + cos 2x). So putting it all together, f(x) becomes sin 2x + √3*(1 + cos 2x) - √3. The √3 and -√3 cancel out, so we're left with sin 2x + √3 cos 2x. Hmm, that looks like a single sine or cosine function. Maybe I can write this as a single sine function with a phase shift.I remember that a sin θ + b cos θ can be written as R sin(θ + φ), where R = √(a² + b²) and tan φ = b/a. In this case, a is 1 (coefficient of sin 2x) and b is √3 (coefficient of cos 2x). So R would be √(1 + 3) = √4 = 2. Then tan φ = √3/1 = √3, so φ is π/3. Therefore, f(x) = 2 sin(2x + π/3). Okay, so f(x) simplifies to 2 sin(2x + π/3). Now, the function we're interested in is y = f(-3x) + 1. Let's substitute -3x into f(x):f(-3x) = 2 sin(2*(-3x) + π/3) = 2 sin(-6x + π/3). But sin(-θ) = -sin θ, so this becomes 2*(-sin(6x - π/3)) = -2 sin(6x - π/3). Therefore, y = -2 sin(6x - π/3) + 1.Now, to find the smallest positive period of y. The period of sin(kx + c) is 2π/|k|. Here, k is 6, so the period is 2π/6 = π/3. So the smallest positive period is π/3.Next, we need to find the intervals where y is monotonically decreasing. Since y is a sine function with a negative coefficient, it's going to be decreasing where the sine function is increasing. Let's think about the derivative.The derivative of y with respect to x is dy/dx = -2 * 6 cos(6x - π/3) = -12 cos(6x - π/3). For y to be decreasing, dy/dx should be negative. So we need -12 cos(6x - π/3) < 0, which simplifies to cos(6x - π/3) > 0.When is cos θ positive? It's positive in the intervals (-π/2 + 2πk, π/2 + 2πk) for any integer k. So, let's set up the inequality:-π/2 + 2πk < 6x - π/3 < π/2 + 2πk.Let's solve for x:Add π/3 to all parts:-π/2 + π/3 + 2πk < 6x < π/2 + π/3 + 2πk.Compute -π/2 + π/3: that's (-3π/6 + 2π/6) = -π/6.Similarly, π/2 + π/3 is (3π/6 + 2π/6) = 5π/6.So:-π/6 + 2πk < 6x < 5π/6 + 2πk.Divide all parts by 6:-π/36 + (π/3)k < x < 5π/36 + (π/3)k.Therefore, the function y is decreasing on intervals [ -π/36 + (π/3)k, 5π/36 + (π/3)k ] for any integer k.But since we're looking for intervals of monotonic decrease, we can express this as:x ∈ [ (π/3)k - π/36, (π/3)k + 5π/36 ] for k ∈ ℤ.So that's part (Ⅰ) done.Moving on to part (Ⅱ): We have triangle ABC with sides a, b, c opposite angles A, B, C respectively. Given that A is an acute angle and f( (A/2) - π/6 ) = √3. Also, a = 7, and sin B + sin C = 13√3 /14. We need to find the area of triangle ABC.First, let's use the given f( (A/2) - π/6 ) = √3. From part (Ⅰ), we know f(x) = 2 sin(2x + π/3). So let's substitute x = (A/2) - π/6 into f(x):f( (A/2) - π/6 ) = 2 sin( 2*( (A/2) - π/6 ) + π/3 ) = 2 sin( A - π/3 + π/3 ) = 2 sin A.So, 2 sin A = √3. Therefore, sin A = √3 / 2. Since A is acute, A must be π/3 (60 degrees).Now, using the Law of Sines, which states that a / sin A = b / sin B = c / sin C = 2R, where R is the circumradius. Given a = 7 and sin A = √3 / 2, we can write:7 / (√3 / 2) = 14 / √3 = 2R. So R = 7 / √3.But we also know that sin B + sin C = 13√3 /14. Let's express sin B and sin C in terms of sides b and c.From the Law of Sines, sin B = b / (2R) and sin C = c / (2R). Therefore, sin B + sin C = (b + c) / (2R).Substituting the known values:(b + c) / (2*(7 / √3)) = 13√3 /14.Simplify the denominator: 2*(7 / √3) = 14 / √3.So, (b + c) / (14 / √3) = 13√3 /14.Multiply both sides by (14 / √3):b + c = (13√3 /14) * (14 / √3) = 13.So, b + c = 13.Now, we can use the Law of Cosines to relate sides a, b, c with angle A. The Law of Cosines states:a² = b² + c² - 2bc cos A.We know a = 7, A = π/3, so cos A = 1/2.Substituting:49 = b² + c² - 2bc*(1/2) = b² + c² - bc.We also know that b + c = 13. Let's square both sides:(b + c)² = b² + 2bc + c² = 169.So, b² + c² = 169 - 2bc.Substitute this into the earlier equation:49 = (169 - 2bc) - bc = 169 - 3bc.So, 49 = 169 - 3bc.Subtract 169 from both sides:49 - 169 = -3bc => -120 = -3bc => bc = 40.Now, the area of triangle ABC can be found using the formula:Area = (1/2) * bc * sin A.We have bc = 40 and sin A = √3 / 2.So, Area = (1/2) * 40 * (√3 / 2) = (20) * (√3 / 2) = 10√3.So, the area is 10√3.**Final Answer**The area of triangle ABC is boxed{10sqrt{3}}.

💡Alright, so I've got this problem here about real numbers a, b, and c. It says, "For real numbers a, b, and c, 'a > b' is ( ) 'ac² > bc²'." And then it gives me four options: A, B, C, D. I need to figure out whether 'a > b' is a sufficient condition, necessary condition, both, or neither for 'ac² > bc²'.Okay, let me start by recalling what these terms mean. A sufficient condition means that if 'a > b' is true, then 'ac² > bc²' must also be true. A necessary condition means that for 'ac² > bc²' to be true, 'a > b' must also be true. If it's both, then 'a > b' is both necessary and sufficient for 'ac² > bc²'. And if it's neither, then 'a > b' doesn't necessarily lead to 'ac² > bc²' and vice versa.So, first, let's think about whether 'a > b' implies 'ac² > bc²'. If I have 'a > b', and I multiply both sides by c², does that preserve the inequality? Well, c² is always non-negative because any real number squared is non-negative. So, c² is either positive or zero.If c² is positive, then multiplying both sides of 'a > b' by c² will preserve the inequality, so 'ac² > bc²' will hold. But if c² is zero, then both sides become zero, so 'ac² > bc²' becomes '0 > 0', which is not true. So, in cases where c is not zero, 'a > b' does imply 'ac² > bc²', but if c is zero, it doesn't.So, 'a > b' is not always sufficient to conclude 'ac² > bc²' because it depends on the value of c. Therefore, 'a > b' is not a sufficient condition in all cases.Now, let's consider the other direction: if 'ac² > bc²', does that imply 'a > b'? Well, if 'ac² > bc²', and c² is positive, then we can divide both sides by c² (which is positive, so the inequality direction remains the same) and get 'a > b'. But if c² is zero, then 'ac² > bc²' becomes '0 > 0', which is false. So, in cases where c is not zero, 'ac² > bc²' does imply 'a > b', but if c is zero, it doesn't.Wait, but if 'ac² > bc²' is true, then c² must be positive because if c² were zero, 'ac² > bc²' would be false. So, actually, if 'ac² > bc²' is true, then c² must be positive, and therefore, 'a > b' must also be true. So, in that case, 'ac² > bc²' implies 'a > b'.But if c is zero, 'ac² > bc²' is false, so it doesn't give us any information about 'a > b'. So, 'ac² > bc²' being true necessarily requires that 'a > b' is true, but 'a > b' doesn't necessarily lead to 'ac² > bc²' unless c is not zero.Therefore, 'a > b' is a necessary condition for 'ac² > bc²' because whenever 'ac² > bc²' is true, 'a > b' must also be true. However, it's not a sufficient condition because 'a > b' doesn't guarantee 'ac² > bc²' unless we know that c is not zero.So, putting it all together, 'a > b' is a necessary but not sufficient condition for 'ac² > bc²'. Therefore, the correct answer should be option B.But let me double-check to make sure I haven't missed anything. If c is zero, 'ac² > bc²' is false regardless of a and b. So, if 'ac² > bc²' is true, c can't be zero, and thus 'a > b' must be true. That makes 'a > b' necessary. However, 'a > b' doesn't ensure 'ac² > bc²' because if c is zero, 'ac² > bc²' is false even if 'a > b' is true. So, 'a > b' isn't sufficient.Yeah, I think that's right. So, the answer is B: Necessary but not sufficient condition.

💡First, I need to convert each binary number to its decimal equivalent.For (1001101_2), I'll calculate each bit's value:- (1 times 2^6 = 64)- (0 times 2^5 = 0)- (0 times 2^4 = 0)- (1 times 2^3 = 8)- (1 times 2^2 = 4)- (0 times 2^1 = 0)- (1 times 2^0 = 1)Adding these up: (64 + 0 + 0 + 8 + 4 + 0 + 1 = 77).Next, for (111000_2):- (1 times 2^5 = 32)- (1 times 2^4 = 16)- (1 times 2^3 = 8)- (0 times 2^2 = 0)- (0 times 2^1 = 0)- (0 times 2^0 = 0)Adding these: (32 + 16 + 8 + 0 + 0 + 0 = 56).Then, for (1010_2):- (1 times 2^3 = 8)- (0 times 2^2 = 0)- (1 times 2^1 = 2)- (0 times 2^0 = 0)Adding these: (8 + 0 + 2 + 0 = 10).Now, I'll sum the decimal equivalents:- (77 + 56 = 133)- (133 + 10 = 143)Therefore, the final sum is 143.

💡Okay, so I have this problem about an arithmetic sequence. It says that the sum of the first 9 terms is 27, and the 10th term is 8. I need to find the value of the 100th term. Hmm, arithmetic sequences... I remember they have a common difference between consecutive terms. Let me think about how to approach this.First, I recall that the sum of the first n terms of an arithmetic sequence can be calculated using the formula:[ S_n = frac{n}{2} times (2a_1 + (n-1)d) ]Where ( S_n ) is the sum of the first n terms, ( a_1 ) is the first term, and ( d ) is the common difference.Given that the sum of the first 9 terms is 27, I can plug these values into the formula:[ 27 = frac{9}{2} times (2a_1 + 8d) ]Let me simplify this equation. Multiplying both sides by 2 to eliminate the fraction:[ 54 = 9 times (2a_1 + 8d) ]Now, divide both sides by 9:[ 6 = 2a_1 + 8d ]I can simplify this further by dividing both sides by 2:[ 3 = a_1 + 4d ]Okay, so that's my first equation: ( a_1 + 4d = 3 ).Next, the problem states that the 10th term ( a_{10} = 8 ). The general formula for the nth term of an arithmetic sequence is:[ a_n = a_1 + (n-1)d ]So, for the 10th term:[ a_{10} = a_1 + 9d = 8 ]That's my second equation: ( a_1 + 9d = 8 ).Now, I have a system of two equations:1. ( a_1 + 4d = 3 )2. ( a_1 + 9d = 8 )I need to solve for ( a_1 ) and ( d ). Let me subtract the first equation from the second to eliminate ( a_1 ):[ (a_1 + 9d) - (a_1 + 4d) = 8 - 3 ][ 5d = 5 ][ d = 1 ]So, the common difference ( d ) is 1. Now, I can substitute ( d = 1 ) back into the first equation to find ( a_1 ):[ a_1 + 4(1) = 3 ][ a_1 + 4 = 3 ][ a_1 = 3 - 4 ][ a_1 = -1 ]Alright, so the first term ( a_1 ) is -1, and the common difference ( d ) is 1. Now, I need to find the 100th term, ( a_{100} ).Using the general formula again:[ a_{100} = a_1 + 99d ][ a_{100} = -1 + 99(1) ][ a_{100} = -1 + 99 ][ a_{100} = 98 ]So, the 100th term is 98. Let me just double-check my calculations to make sure I didn't make any mistakes.Starting with the sum of the first 9 terms:[ S_9 = frac{9}{2} times (2(-1) + 8(1)) ][ S_9 = frac{9}{2} times (-2 + 8) ][ S_9 = frac{9}{2} times 6 ][ S_9 = 9 times 3 ][ S_9 = 27 ]That checks out. Now, the 10th term:[ a_{10} = -1 + 9(1) ][ a_{10} = -1 + 9 ][ a_{10} = 8 ]That also checks out. So, everything seems consistent. Therefore, the 100th term should indeed be 98.

💡To find the sum of the (x)-coordinates where (g(x) = 2), I need to analyze each line segment of the graph (y = g(x)) and determine where it intersects the horizontal line (y = 2).1. **First Segment**: From ((-5, -3)) to ((-3, 0)) - Calculate the slope: (m = frac{0 - (-3)}{-3 - (-5)} = frac{3}{2}) - Equation of the line: (y = frac{3}{2}x + frac{15}{2}) - Set (y = 2) and solve for (x): (frac{3}{2}x + frac{15}{2} = 2 implies 3x + 15 = 4 implies 3x = -11 implies x = -frac{11}{3})2. **Second Segment**: From ((-1, -1)) to ((1, 3)) - Calculate the slope: (m = frac{3 - (-1)}{1 - (-1)} = 2) - Equation of the line: (y = 2x + 1) - Set (y = 2) and solve for (x): (2x + 1 = 2 implies 2x = 1 implies x = frac{1}{2})3. **Third Segment**: From ((3, 2)) to ((5, 6)) - Calculate the slope: (m = frac{6 - 2}{5 - 3} = 2) - Equation of the line: (y = 2x - 4) - Set (y = 2) and solve for (x): (2x - 4 = 2 implies 2x = 6 implies x = 3)After finding the (x)-coordinates where (g(x) = 2), I sum them up:[-frac{11}{3} + frac{1}{2} + 3 = frac{-22 + 3 + 18}{6} = frac{-1}{6}]

💡Okay, so I have this problem where I'm given that ( a > 0 ), ( bc > a^2 ), and ( a^2 - 2ab + c^2 = 0 ). I need to compare the magnitudes of ( a ), ( b ), and ( c ). Hmm, let's break this down step by step.First, I know that ( a ) is positive. That's straightforward. The second condition is ( bc > a^2 ), which tells me that the product of ( b ) and ( c ) is greater than ( a ) squared. The third condition is a quadratic equation: ( a^2 - 2ab + c^2 = 0 ). Maybe I can manipulate this equation to find relationships between ( a ), ( b ), and ( c ).Let me rewrite the quadratic equation:[ a^2 - 2ab + c^2 = 0 ]Hmm, this looks a bit like a quadratic in terms of ( a ) or ( c ). Let me see if I can factor it or complete the square. If I try to complete the square for ( a ), I can write:[ a^2 - 2ab + b^2 + c^2 - b^2 = 0 ][ (a - b)^2 + c^2 - b^2 = 0 ][ (a - b)^2 = b^2 - c^2 ]Wait, that might not be helpful. Maybe I should treat it as a quadratic in ( a ):[ a^2 - 2ab + c^2 = 0 ]Using the quadratic formula to solve for ( a ):[ a = frac{2b pm sqrt{(2b)^2 - 4 cdot 1 cdot c^2}}{2} ][ a = frac{2b pm sqrt{4b^2 - 4c^2}}{2} ][ a = b pm sqrt{b^2 - c^2} ]Since ( a > 0 ), the expression under the square root must be non-negative:[ b^2 - c^2 geq 0 ][ b^2 geq c^2 ][ |b| geq |c| ]But since ( a ) is positive, and from the quadratic equation, ( a = b pm sqrt{b^2 - c^2} ), we need to ensure that ( a ) is positive. If we take the negative sign:[ a = b - sqrt{b^2 - c^2} ]This would require that ( b > sqrt{b^2 - c^2} ), which is always true because ( sqrt{b^2 - c^2} < b ) when ( c neq 0 ). So, ( a = b - sqrt{b^2 - c^2} ) is positive.Alternatively, taking the positive sign:[ a = b + sqrt{b^2 - c^2} ]This would make ( a ) larger than ( b ), but let's see if that's possible given the other condition ( bc > a^2 ).Wait, if ( a = b + sqrt{b^2 - c^2} ), then ( a > b ). But then ( bc > a^2 ) would imply ( bc > (b + sqrt{b^2 - c^2})^2 ). That seems complicated, and I'm not sure if it's necessary. Maybe the negative sign is the right choice.So, let's stick with ( a = b - sqrt{b^2 - c^2} ). Now, let's see what this tells us about the relationship between ( a ), ( b ), and ( c ).Since ( sqrt{b^2 - c^2} ) is positive, ( a ) is less than ( b ). So, ( a < b ).Now, let's look at the condition ( bc > a^2 ). Since ( a < b ), and ( a > 0 ), ( b ) must be positive as well because if ( b ) were negative, ( a ) would be less than a negative number, which contradicts ( a > 0 ). So, ( b > 0 ).Given that ( bc > a^2 ) and ( a < b ), we can try to find a relationship between ( c ) and ( a ) or ( c ) and ( b ).From ( a = b - sqrt{b^2 - c^2} ), let's solve for ( c ):[ a = b - sqrt{b^2 - c^2} ][ sqrt{b^2 - c^2} = b - a ][ b^2 - c^2 = (b - a)^2 ][ b^2 - c^2 = b^2 - 2ab + a^2 ][ -c^2 = -2ab + a^2 ][ c^2 = 2ab - a^2 ][ c^2 = a(2b - a) ]Since ( c^2 ) is positive, ( 2b - a ) must be positive:[ 2b - a > 0 ][ 2b > a ][ b > frac{a}{2} ]Which is already implied since ( a < b ).Now, let's go back to the condition ( bc > a^2 ). Since ( b > 0 ) and ( c > 0 ) (because ( bc > a^2 > 0 )), we can write:[ c > frac{a^2}{b} ]From the equation ( c^2 = a(2b - a) ), we can express ( c ) as:[ c = sqrt{a(2b - a)} ]So, we have:[ sqrt{a(2b - a)} > frac{a^2}{b} ]Let's square both sides to eliminate the square root:[ a(2b - a) > frac{a^4}{b^2} ]Divide both sides by ( a ) (since ( a > 0 )):[ 2b - a > frac{a^3}{b^2} ]Multiply both sides by ( b^2 ):[ 2b^3 - a b^2 > a^3 ][ 2b^3 - a b^2 - a^3 > 0 ]Let's factor this expression. Maybe factor out ( a ):[ a(2b^3/a - b^2 - a^2) > 0 ]But that doesn't seem helpful. Alternatively, let's consider that ( a < b ), so ( a = kb ) where ( 0 < k < 1 ). Let's substitute ( a = kb ) into the inequality:[ 2b^3 - (kb) b^2 - (kb)^3 > 0 ][ 2b^3 - k b^3 - k^3 b^3 > 0 ][ b^3 (2 - k - k^3) > 0 ]Since ( b^3 > 0 ), we have:[ 2 - k - k^3 > 0 ]We need to find the values of ( k ) such that ( 2 - k - k^3 > 0 ). Let's analyze the function ( f(k) = 2 - k - k^3 ).Find ( f(k) = 0 ):[ 2 - k - k^3 = 0 ][ k^3 + k - 2 = 0 ]Let's try ( k = 1 ):[ 1 + 1 - 2 = 0 ]So, ( k = 1 ) is a root. Let's factor it:[ (k - 1)(k^2 + k + 2) = 0 ]The quadratic ( k^2 + k + 2 ) has discriminant ( 1 - 8 = -7 ), so no real roots. Thus, the only real root is ( k = 1 ).Since ( f(k) = 2 - k - k^3 ), and for ( k < 1 ), let's test ( k = 0 ):[ f(0) = 2 > 0 ]For ( k = 1 ):[ f(1) = 0 ]For ( k > 1 ), say ( k = 2 ):[ f(2) = 2 - 2 - 8 = -8 < 0 ]So, ( f(k) > 0 ) when ( k < 1 ). Since ( k = a/b < 1 ), this inequality holds. Therefore, our earlier steps are consistent.Now, going back, we have ( c = sqrt{a(2b - a)} ). Let's see how ( c ) compares to ( a ) and ( b ).Since ( c = sqrt{a(2b - a)} ), and ( a < b ), let's see if ( c ) is greater than ( a ):[ c > a ][ sqrt{a(2b - a)} > a ][ a(2b - a) > a^2 ][ 2ab - a^2 > a^2 ][ 2ab > 2a^2 ][ b > a ]Which is true, so ( c > a ).Now, how does ( c ) compare to ( b )? Let's check if ( c < b ):[ c < b ][ sqrt{a(2b - a)} < b ][ a(2b - a) < b^2 ][ 2ab - a^2 < b^2 ][ 2ab - a^2 - b^2 < 0 ][ -a^2 + 2ab - b^2 < 0 ][ -(a^2 - 2ab + b^2) < 0 ][ -(a - b)^2 < 0 ]Which is always true because ( (a - b)^2 ) is non-negative, so the negative of it is non-positive. Therefore, ( c < b ).So, putting it all together:- ( a < b )- ( c > a )- ( c < b )Therefore, the order is ( b > c > a ).Let me double-check to make sure I didn't make any mistakes. Starting from the quadratic equation, I solved for ( a ) and found that ( a = b - sqrt{b^2 - c^2} ), which implies ( a < b ). Then, using the condition ( bc > a^2 ), I derived that ( c > a ). Finally, by comparing ( c ) and ( b ), I concluded that ( c < b ). So, the order should be ( b > c > a ). That seems consistent.