💡Okay, so I have this problem here that asks about the relationship between two statements: "alpha = frac{pi}{6} + 2kpi, where k is an integer," and "sin alpha = frac{1}{2}." The question is asking whether the first statement is a sufficient condition, necessary condition, both, or neither for the second statement. The options are A, B, C, and D, with A being "Sufficient but not necessary condition," B being "Necessary but not sufficient condition," C being "Necessary and sufficient condition," and D being "Neither sufficient nor necessary condition."Alright, let's break this down step by step. First, I need to understand what each of these terms means: sufficient condition, necessary condition, and so on. A sufficient condition means that if the condition is met, then the statement must be true. However, it doesn't necessarily mean that the condition is the only way for the statement to be true. On the other hand, a necessary condition means that for the statement to be true, the condition must be met, but meeting the condition doesn't guarantee the statement is true. If something is both necessary and sufficient, then meeting the condition is exactly what makes the statement true and vice versa.So, applying this to the problem: the first statement is "alpha = frac{pi}{6} + 2kpi, where k is an integer." The second statement is "sin alpha = frac{1}{2}." I need to figure out if the first statement is a sufficient condition, necessary condition, both, or neither for the second statement.Let me start by recalling some trigonometry. The sine function, sin alpha, equals frac{1}{2} at specific angles. I remember that sin frac{pi}{6} = frac{1}{2}, and also sin frac{5pi}{6} = frac{1}{2}. Moreover, since the sine function is periodic with a period of 2pi, these solutions repeat every 2pi. So, the general solutions for sin alpha = frac{1}{2} are alpha = frac{pi}{6} + 2kpi and alpha = frac{5pi}{6} + 2kpi, where k is any integer.Now, the first statement given is only one of these solutions: alpha = frac{pi}{6} + 2kpi. It doesn't include the other solution, alpha = frac{5pi}{6} + 2kpi. So, if I take the first statement as true, then sin alpha must be frac{1}{2} because it's one of the solutions. That means the first statement is a sufficient condition for the second statement because it guarantees that sin alpha = frac{1}{2}.However, is it a necessary condition? That is, is the first statement the only way for sin alpha to be frac{1}{2}? No, because as I mentioned earlier, there's another set of solutions, alpha = frac{5pi}{6} + 2kpi, which also satisfy sin alpha = frac{1}{2}. Therefore, the first statement isn't necessary because there are other values of alpha that make sin alpha = frac{1}{2} without satisfying alpha = frac{pi}{6} + 2kpi.So, putting it all together, the first statement is sufficient but not necessary for the second statement. That would correspond to option A.Wait, let me double-check to make sure I haven't missed anything. If alpha = frac{pi}{6} + 2kpi, then yes, sin alpha will definitely be frac{1}{2}. So, it's sufficient. But since there are other angles where sin alpha = frac{1}{2}, it's not necessary for alpha to be exactly frac{pi}{6} + 2kpi. Therefore, my initial conclusion seems correct.I think I've covered all the bases here. I considered both the sufficiency and necessity, checked the general solutions for the sine function, and compared them to the given statement. Everything seems to align with option A.

💡To find the product of (0.overline{6}) and 8, I first need to convert the repeating decimal (0.overline{6}) into a fraction.Let (x = 0.overline{6}). Multiplying both sides by 10 gives (10x = 6.overline{6}). Subtracting the original equation from this new equation eliminates the repeating part: (10x - x = 6.overline{6} - 0.overline{6}), which simplifies to (9x = 6). Solving for (x) yields (x = frac{6}{9} = frac{2}{3}).Now that I have (0.overline{6}) expressed as (frac{2}{3}), I can multiply this fraction by 8:[frac{2}{3} times 8 = frac{16}{3}]Converting (frac{16}{3}) into a mixed number gives (5 frac{1}{3}).Therefore, the product of (0.overline{6}) and 8 is (5 frac{1}{3}).

💡First, I need to calculate the distance Bob traveled during each segment of his road trip. To do this, I'll use the formula:[ text{Distance} = text{Speed} times text{Time} ]I'll start by converting the time for each segment from hours and minutes to decimal hours.1. **First Segment:** - **Time:** 1 hour and 45 minutes - **Conversion:** ( 1 + frac{45}{60} = 1.75 ) hours - **Speed:** 65 mph - **Distance:** ( 65 times 1.75 = 113.75 ) miles2. **Second Segment:** - **Time:** 2 hours and 20 minutes - **Conversion:** ( 2 + frac{20}{60} approx 2.333 ) hours - **Speed:** 45 mph - **Distance:** ( 45 times 2.333 approx 104.985 ) miles3. **Third Segment:** - **Time:** 1 hour and 30 minutes - **Conversion:** ( 1 + frac{30}{60} = 1.5 ) hours - **Speed:** 35 mph - **Distance:** ( 35 times 1.5 = 52.5 ) miles4. **Fourth Segment:** - **Time:** 2 hours and 15 minutes - **Conversion:** ( 2 + frac{15}{60} = 2.25 ) hours - **Speed:** 50 mph - **Distance:** ( 50 times 2.25 = 112.5 ) milesAfter calculating the distance for each segment, I'll sum them up to find the total distance Bob traveled:[ text{Total Distance} = 113.75 + 104.985 + 52.5 + 112.5 = 383.735 text{ miles} ]

💡To determine the product of two prime numbers that sum to 93, we start by noting that the sum of two primes is odd. Since the only even prime number is 2, one of the primes must be 2, and the other must be 91. However, 91 is not a prime number because it is divisible by 7 and 13. Next, we try the next smallest prime numbers. We find that 93 - 89 = 4, but 4 is not a prime. Continuing this process, we find that the correct pair of primes is 2 and 89. The product of these two primes is:[ 2 times 89 = 178 ]Thus, the product of the two prime numbers is (boxed{C}).

💡Okay, so I have this problem where I need to find the number of intersection points between the graphs of ( y = 3log x ) and ( y = log(3x) + log 3 ). Hmm, let me think about how to approach this.First, I remember that to find intersection points, I need to set the two equations equal to each other and solve for ( x ). So, let me write that down:[ 3log x = log(3x) + log 3 ]Alright, now I need to simplify this equation. I recall some logarithmic properties that might help here. One property is that ( log(ab) = log a + log b ). Let me apply that to the right side of the equation.So, ( log(3x) ) can be written as ( log 3 + log x ). That means the right side becomes:[ log 3 + log x + log 3 ]Combining like terms, that's:[ 2log 3 + log x ]Now, my equation looks like this:[ 3log x = 2log 3 + log x ]Hmm, okay, I can subtract ( log x ) from both sides to get:[ 2log x = 2log 3 ]Dividing both sides by 2 gives:[ log x = log 3 ]If ( log x = log 3 ), then ( x = 3 ). So, that suggests that there is an intersection point at ( x = 3 ).But wait, I should make sure there aren't any other solutions. Sometimes, when dealing with logarithmic equations, there might be multiple solutions or sometimes no solution. Let me think about the behavior of both functions to see if they could intersect more than once.The first function is ( y = 3log x ). Since the coefficient of ( log x ) is 3, this function will grow faster than ( log x ) as ( x ) increases. The second function is ( y = log(3x) + log 3 ). Let me simplify that:[ y = log(3x) + log 3 = log 3 + log x + log 3 = 2log 3 + log x ]So, the second function is essentially ( y = log x + 2log 3 ). Comparing this to the first function, which is ( y = 3log x ), I can see that for large values of ( x ), ( 3log x ) will grow much faster than ( log x + 2log 3 ). This suggests that after ( x = 3 ), the first function will be above the second function and won't intersect again.What about for values of ( x ) less than 3? Let me test a value, say ( x = 1 ):For ( x = 1 ):- ( y = 3log 1 = 0 )- ( y = log(3*1) + log 3 = log 3 + log 3 = 2log 3 ) which is approximately ( 2*0.477 = 0.954 )So, at ( x = 1 ), the second function is higher than the first function.At ( x = 3 ), both functions equal ( 3log 3 approx 3*0.477 = 1.431 ).Now, let me check a value between 1 and 3, say ( x = 2 ):For ( x = 2 ):- ( y = 3log 2 approx 3*0.301 = 0.903 )- ( y = log(3*2) + log 3 = log 6 + log 3 approx 0.778 + 0.477 = 1.255 )So, at ( x = 2 ), the second function is still higher than the first function.Wait, so from ( x = 1 ) to ( x = 3 ), the second function starts higher and meets the first function at ( x = 3 ). But does the first function ever cross the second function before ( x = 3 )?Let me check a value less than 1, say ( x = 0.5 ):For ( x = 0.5 ):- ( y = 3log 0.5 approx 3*(-0.301) = -0.903 )- ( y = log(3*0.5) + log 3 = log 1.5 + log 3 approx 0.176 + 0.477 = 0.653 )So, at ( x = 0.5 ), the second function is still higher.Hmm, so it seems like the second function is always above the first function for ( x < 3 ) and they meet exactly at ( x = 3 ). After that, the first function grows faster and stays above the second function.Therefore, it looks like there's only one intersection point at ( x = 3 ).But just to be thorough, let me consider the derivatives of both functions to see if they could potentially intersect more than once.The derivative of ( y = 3log x ) is ( y' = frac{3}{x} ).The derivative of ( y = log(3x) + log 3 ) is ( y' = frac{1}{x} ).So, the slope of the first function is always three times the slope of the second function. This means that the first function is steeper for all ( x > 0 ). Since they intersect at ( x = 3 ), and the first function is steeper beyond that point, they won't intersect again.Therefore, I'm confident that there's only one intersection point.

💡Okay, so I have this problem where I need to find two positive integers, m and n, such that their product is 5000. But there's a catch: neither m nor n can be divisible by 10. Then, I need to find the sum of m and n. Hmm, let me think about how to approach this.First, I know that 5000 is the product of m and n, so I can write that as:[ mn = 5000 ]Now, I need to factorize 5000 to understand its prime components. Let me do that step by step. Starting with 5000, I can divide by 10 to get 500, and then divide by 10 again to get 50, and once more by 10 to get 5. So, that's three 10s, which is (10^3). But 10 itself is (2 times 5), so:[ 5000 = 10^3 times 5 = (2 times 5)^3 times 5 = 2^3 times 5^3 times 5 = 2^3 times 5^4 ]So, the prime factorization of 5000 is (2^3 times 5^4). That means m and n must be factors of 5000, and their product should cover all these prime factors.But the problem says neither m nor n can be divisible by 10. Divisibility by 10 means that a number must have both 2 and 5 as factors. So, if neither m nor n can be divisible by 10, that means one of them can have only the 2s and the other can have only the 5s. Because if one had both, it would be divisible by 10, which is not allowed.So, I need to split the prime factors between m and n such that m gets all the 2s and n gets all the 5s, or vice versa. Let me try both possibilities.First, let's assign all the 2s to m and all the 5s to n:- m would be (2^3 = 8)- n would be (5^4 = 625)Let me check if this works:- (8 times 625 = 5000), which is correct.- Is 8 divisible by 10? No, because it doesn't have a 5 as a factor.- Is 625 divisible by 10? No, because it doesn't have a 2 as a factor.Great, that works. Now, let me check the other possibility where m gets all the 5s and n gets all the 2s:- m would be (5^4 = 625)- n would be (2^3 = 8)Checking this:- (625 times 8 = 5000), which is correct.- 625 is not divisible by 10, as it lacks a 2.- 8 is not divisible by 10, as it lacks a 5.So both assignments work, but they result in the same pair of numbers, just swapped. Therefore, the sum (m + n) would be the same in both cases.Calculating the sum:[ m + n = 8 + 625 = 633 ]Wait, let me make sure there are no other possibilities. Could there be another way to split the factors without having one number take all the 2s and the other all the 5s? For example, could m have some 2s and some 5s, but not enough to make it divisible by 10?But if m had both 2s and 5s, even one of each, it would be divisible by 10, which is against the problem's condition. So, the only way to ensure neither m nor n is divisible by 10 is to have one take all the 2s and the other take all the 5s.Therefore, the only possible pairs are (8, 625) and (625, 8), both leading to the same sum of 633.I think that's the only solution. Let me just recap:1. Factorize 5000 into primes: (2^3 times 5^4).2. Since neither m nor n can be divisible by 10, they must not share both 2 and 5 as factors.3. Assign all 2s to one number and all 5s to the other.4. Calculate the numbers: 8 and 625.5. Sum them to get 633.I don't see any other way to split the factors without violating the divisibility condition. So, I'm confident that 633 is the correct answer.