💡Okay, so I have this problem where I need to translate the graph of the function ( y = sin(2x + frac{pi}{3}) ) to the right by ( a ) units, and after translating, the new graph should be ( y = sin(2x) ). I need to find the value of ( a ) where ( 0 < a < pi ).Hmm, let me recall how function translations work. If I have a function ( y = f(x) ) and I translate it to the right by ( a ) units, the new function becomes ( y = f(x - a) ). So, applying this to my function, translating ( y = sin(2x + frac{pi}{3}) ) to the right by ( a ) units should give me ( y = sin(2(x - a) + frac{pi}{3}) ).Let me write that down:Original function: ( y = sin(2x + frac{pi}{3}) )After translating right by ( a ) units: ( y = sin(2(x - a) + frac{pi}{3}) )Simplify the argument inside the sine function:( 2(x - a) + frac{pi}{3} = 2x - 2a + frac{pi}{3} )So, the translated function is ( y = sin(2x - 2a + frac{pi}{3}) ).But the problem states that after translating, the equation becomes ( y = sin(2x) ). That means:( sin(2x - 2a + frac{pi}{3}) = sin(2x) )For these two sine functions to be equal for all ( x ), their arguments must differ by an integer multiple of ( 2pi ) or be supplementary angles differing by an integer multiple of ( 2pi ). However, since we're dealing with a translation, which is a horizontal shift, the phase shift should be such that the functions align perfectly. So, the simplest case is that the arguments differ by an integer multiple of ( 2pi ).But let me think again. If I have ( sin(theta) = sin(phi) ), then ( theta = phi + 2pi n ) or ( theta = pi - phi + 2pi n ) for some integer ( n ). However, in this case, since we are translating the graph, we need the functions to coincide exactly, which would require that the phase shift is such that the entire graph is shifted without any reflection or other transformations.Therefore, the phase shift must satisfy:( 2x - 2a + frac{pi}{3} = 2x + 2pi n )for some integer ( n ). Let's subtract ( 2x ) from both sides:( -2a + frac{pi}{3} = 2pi n )Now, solving for ( a ):( -2a = 2pi n - frac{pi}{3} )( 2a = frac{pi}{3} - 2pi n )( a = frac{pi}{6} - pi n )Now, since ( a ) must be between ( 0 ) and ( pi ), let's find the integer ( n ) such that ( a ) falls within this interval.Let's try ( n = 0 ):( a = frac{pi}{6} - 0 = frac{pi}{6} )Which is approximately 0.523 radians, and since ( 0 < frac{pi}{6} < pi ), this works.Let's check ( n = 1 ):( a = frac{pi}{6} - pi = -frac{5pi}{6} )Which is negative, so it doesn't satisfy ( 0 < a < pi ).Similarly, ( n = -1 ):( a = frac{pi}{6} + pi = frac{7pi}{6} )Which is greater than ( pi ), so it doesn't satisfy the condition either.Therefore, the only valid solution is ( a = frac{pi}{6} ).Let me verify this. If I translate the original function ( y = sin(2x + frac{pi}{3}) ) to the right by ( frac{pi}{6} ) units, the new function becomes:( y = sin(2(x - frac{pi}{6}) + frac{pi}{3}) )Simplify inside the sine:( 2x - 2 cdot frac{pi}{6} + frac{pi}{3} = 2x - frac{pi}{3} + frac{pi}{3} = 2x )So, the translated function is indeed ( y = sin(2x) ), which matches the given condition.Therefore, the value of ( a ) is ( frac{pi}{6} ).

💡Okay, so I have this problem here: The least common multiple of x, 12, and 15 is 180. I need to find the greatest possible value of x. Hmm, let's break this down step by step.First, I remember that the least common multiple, or LCM, of a set of numbers is the smallest number that all of them can divide into without leaving a remainder. So, if the LCM of x, 12, and 15 is 180, that means 180 is the smallest number that x, 12, and 15 can all divide into.I think a good starting point is to factor each number into its prime factors. Let me try that. Starting with 12: I know that 12 can be broken down into 2 times 6, and 6 is 2 times 3. So, 12 is 2 squared times 3. In mathematical terms, that's 2² × 3.Next, 15: That's 3 times 5. So, 15 is 3 × 5.Now, the LCM of these numbers is given as 180. Let me factor 180 as well. 180 divided by 2 is 90, divided by 2 again is 45. 45 divided by 3 is 15, divided by 3 again is 5, and then 5 is a prime number. So, putting that together, 180 is 2 squared times 3 squared times 5. So, 2² × 3² × 5.Okay, so now I have the prime factors:- 12 = 2² × 3- 15 = 3 × 5- LCM (12, 15, x) = 180 = 2² × 3² × 5I remember that the LCM of multiple numbers is found by taking the highest power of each prime number that appears in the factorization of any of the numbers. So, for example, for the prime number 2, the highest power is 2² in both 12 and 180. For 3, the highest power is 3² in 180. For 5, it's 5¹.So, if I already have 12 and 15, their LCM is 60 because:- 12 = 2² × 3- 15 = 3 × 5- LCM(12, 15) = 2² × 3 × 5 = 60But in this problem, the LCM of x, 12, and 15 is 180, which is higher than 60. That means x must be contributing an additional factor that isn't already covered by 12 and 15. Specifically, since 60 is 2² × 3 × 5, and 180 is 2² × 3² × 5, the extra factor is 3. So, x must have an additional 3 to make the LCM go up to 180.But wait, x could have more factors, right? As long as they don't introduce any new prime factors beyond what's already in 180. So, x can have factors of 2, 3, and 5, but not any other primes because that would make the LCM larger than 180, which isn't allowed.So, to maximize x, I want x to be as large as possible without making the LCM exceed 180. That means x can include all the prime factors of 180, but not more than what's already there.Let me think about the exponents. For 2, the highest power in 180 is 2². Since 12 already has 2², x can have at most 2². For 3, 180 has 3², and 12 and 15 only have 3¹ each, so x can have 3² to contribute to the LCM. For 5, 15 already has 5¹, so x can have at most 5¹.Therefore, the maximum x can be is 2² × 3² × 5¹. Let me calculate that:2² is 4, 3² is 9, and 5¹ is 5. Multiplying them together: 4 × 9 = 36, and 36 × 5 = 180.Wait, so x can be 180? Let me check if that makes sense. If x is 180, then the LCM of 180, 12, and 15 should be 180. Let's verify:- 180 is already a multiple of 12 because 180 ÷ 12 = 15, which is an integer.- 180 is also a multiple of 15 because 180 ÷ 15 = 12, which is an integer.So, yes, the LCM of 180, 12, and 15 is indeed 180. Therefore, x can be 180.But hold on, is 180 the only possible value? Or are there smaller values of x that also result in an LCM of 180? For example, if x is 90, which is 2 × 3² × 5, would that work?Let me check: LCM of 90, 12, and 15.- 90 = 2 × 3² × 5- 12 = 2² × 3- 15 = 3 × 5The LCM would be the highest powers of each prime: 2², 3², and 5¹, which is 180. So, yes, 90 also works.Similarly, if x is 60, which is 2² × 3 × 5, the LCM of 60, 12, and 15 is still 60, which is less than 180. So, 60 wouldn't work because the LCM needs to be 180.Wait, so x can be 90 or 180. But the question asks for the greatest possible value of x. So, between 90 and 180, 180 is larger. Therefore, 180 is the maximum value x can take.But let me think again. Is there any number larger than 180 that can have an LCM of 180 with 12 and 15? For example, if x is 360, which is 2³ × 3² × 5. Then, the LCM of 360, 12, and 15 would be 360 because 360 has a higher power of 2 (2³) than 12 (2²). But the problem states that the LCM is 180, so 360 can't be a solution because it would make the LCM larger than 180.Therefore, x cannot be larger than 180 because that would exceed the given LCM. So, 180 is indeed the maximum possible value for x.To summarize, by factorizing the numbers and understanding how LCM works, I determined that x must include the necessary prime factors to reach the LCM of 180 without introducing any new prime factors or higher exponents than those present in 180. The highest such x is 180 itself.

💡Okay, so I've got this problem here where I'm told that f(x) is an odd function and g(x) is an even function. They also give me that f(x) + g(x) equals 1 divided by (x - 1). I need to find f(3). Hmm, okay, let's break this down step by step.First, I remember that odd functions have the property where f(-x) = -f(x). So, if I plug in -x into an odd function, it's the same as flipping the sign of the function. On the other hand, even functions have the property where g(-x) = g(x). So, plugging in -x into an even function doesn't change its value. That seems important.Given that f(x) + g(x) = 1/(x - 1), I can write that equation for x and also for -x. Let me try that. If I replace x with -x in the equation, I get f(-x) + g(-x) = 1/(-x - 1). But since f is odd and g is even, this becomes -f(x) + g(x) = 1/(-x - 1). So now I have two equations:1. f(x) + g(x) = 1/(x - 1)2. -f(x) + g(x) = 1/(-x - 1)Hmm, okay, so now I have a system of two equations with two unknowns: f(x) and g(x). Maybe I can solve this system to find expressions for f(x) and g(x). Let me write them down again:Equation 1: f(x) + g(x) = 1/(x - 1)Equation 2: -f(x) + g(x) = 1/(-x - 1)If I subtract Equation 2 from Equation 1, I can eliminate g(x). Let's try that:[f(x) + g(x)] - [-f(x) + g(x)] = [1/(x - 1)] - [1/(-x - 1)]Simplifying the left side:f(x) + g(x) + f(x) - g(x) = 2f(x)And the right side:1/(x - 1) - 1/(-x - 1) = 1/(x - 1) + 1/(x + 1)Because subtracting a negative is like adding. So, 1/(x - 1) + 1/(x + 1). Let me combine these fractions. The common denominator would be (x - 1)(x + 1). So:[ (x + 1) + (x - 1) ] / (x - 1)(x + 1) = (2x) / (x² - 1)So, putting it all together:2f(x) = 2x / (x² - 1)Divide both sides by 2:f(x) = x / (x² - 1)Okay, so that's f(x). Now, since I need to find f(3), let's plug in x = 3 into this expression.f(3) = 3 / (3² - 1) = 3 / (9 - 1) = 3 / 8So, f(3) is 3/8. Let me double-check my steps to make sure I didn't make a mistake.1. I used the properties of odd and even functions correctly. f(-x) = -f(x) and g(-x) = g(x). That seems right.2. I set up the two equations correctly by substituting x and -x into the original equation.3. Subtracting the second equation from the first to eliminate g(x) was a good move. That gave me 2f(x) on the left.4. On the right side, I correctly simplified 1/(x - 1) - 1/(-x - 1) to 1/(x - 1) + 1/(x + 1), then combined them into a single fraction.5. Simplifying the numerator: (x + 1) + (x - 1) = 2x, which is correct.6. The denominator is (x - 1)(x + 1) = x² - 1, which is correct.7. So, 2f(x) = 2x / (x² - 1), which simplifies to f(x) = x / (x² - 1). That seems right.8. Plugging in x = 3: 3 / (9 - 1) = 3/8. Yep, that checks out.Wait, let me just make sure that f(x) is indeed an odd function as given. If f(x) = x / (x² - 1), then f(-x) should be -f(x). Let's test that:f(-x) = (-x) / ((-x)² - 1) = (-x) / (x² - 1) = - (x / (x² - 1)) = -f(x). Perfect, so f(x) is indeed odd. That's a good consistency check.Similarly, let's find g(x) to make sure everything lines up. From Equation 1: f(x) + g(x) = 1/(x - 1). So, g(x) = 1/(x - 1) - f(x) = 1/(x - 1) - x/(x² - 1).Let me combine these terms. The common denominator is (x - 1)(x + 1). So:[ (x + 1) - x ] / (x² - 1) = (x + 1 - x) / (x² - 1) = 1 / (x² - 1)So, g(x) = 1 / (x² - 1). Let's check if this is even. g(-x) = 1 / ((-x)² - 1) = 1 / (x² - 1) = g(x). Yep, so g(x) is even as required.Alright, so everything seems consistent. I found f(x) correctly, checked that it's odd, found g(x) and checked that it's even, and then computed f(3) as 3/8.I think that's solid. So, the answer should be 3/8, which is option C.**Final Answer**boxed{C}

💡To determine the value of the real number ( a ) such that the complex number ( (a^{2} - 3a + 2) + (a - 1)i ) is purely imaginary, we need to ensure that the real part of the complex number is zero and the imaginary part is non-zero.1. **Set the real part equal to zero:** [ a^{2} - 3a + 2 = 0 ] Solving this quadratic equation: [ a^{2} - 3a + 2 = 0 implies (a - 1)(a - 2) = 0 implies a = 1 text{ or } a = 2 ]2. **Ensure the imaginary part is non-zero:** [ a - 1 neq 0 implies a neq 1 ]3. **Combine the results:** The value ( a = 1 ) is excluded because it makes the imaginary part zero. Therefore, the valid solution is: [ a = 2 ]

💡To find the fraction equivalent of the repeating decimal (0.overline{85}), I'll start by letting (x = 0.overline{85}).Since the repeating part has two digits, I'll multiply both sides of the equation by 100 to shift the decimal point two places to the right:[100x = 85.858585ldots]Next, I'll subtract the original equation (x = 0.858585ldots) from this new equation to eliminate the repeating part:[100x - x = 85.858585ldots - 0.858585ldots][99x = 85]Finally, I'll solve for (x) by dividing both sides by 99:[x = frac{85}{99}]Therefore, the fraction that equals (0.overline{85}) is (frac{85}{99}).

💡First, I need to understand the pattern of numeric values assigned to the letters. The pattern is: 1, 2, 1, 0, -1, -2, -1, 0, 1, 2, and then it repeats every 10 letters.Next, I'll determine the position of each letter in the word "mathematics" within the alphabet. For example, 'm' is the 13th letter, 'a' is the 1st, and so on.Using the positions, I'll calculate the numeric value for each letter by finding the position modulo 10. This will give me the corresponding value from the pattern. For instance, 'm' at position 13 modulo 10 is 3, which corresponds to a value of 1.After obtaining the numeric values for all letters in "mathematics," I'll sum them up to find the total.Finally, I'll present the final sum as the answer.