💡Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. The problem gives me an equation: b² = a² + c² - ac. I need to figure out what angle B is. The options are π/3, π/6, 2π/3, and 5π/6. Hmm, okay.First, I remember that in triangles, the Law of Cosines is really useful for relating the sides of a triangle to one of its angles. The Law of Cosines formula is: b² = a² + c² - 2ac cos(B)So, comparing this to the equation given in the problem, which is b² = a² + c² - ac, I can see that both equations have a² + c², but the problem's equation subtracts ac instead of 2ac cos(B). Let me write both equations down to compare:From the problem: b² = a² + c² - ac From the Law of Cosines: b² = a² + c² - 2ac cos(B)Since both equal b², I can set them equal to each other:a² + c² - ac = a² + c² - 2ac cos(B)Hmm, okay, so if I subtract a² + c² from both sides, I get:-ac = -2ac cos(B)Now, I can multiply both sides by -1 to make it simpler:ac = 2ac cos(B)Assuming that a and c are not zero (which makes sense because they are sides of a triangle and must have positive lengths), I can divide both sides by ac:1 = 2 cos(B)So, dividing both sides by 2:cos(B) = 1/2Now, I need to find the angle B where the cosine of B is 1/2. I remember that cos(π/3) is 1/2, which is 60 degrees. So, B must be π/3.Wait, but let me double-check. Cosine is positive in the first and fourth quadrants, but since we're dealing with a triangle, angles are between 0 and π (0 and 180 degrees). So, the only angle in that range where cosine is 1/2 is π/3. Is there any chance it could be another angle? Let me think. Cos(π/3) is 1/2, and cos(2π/3) is -1/2. So, no, 2π/3 would give a negative cosine, which doesn't match our equation. Similarly, π/6 has a cosine of √3/2, which is about 0.866, not 1/2. And 5π/6 is in the second quadrant with a cosine of -√3/2, which is also not 1/2. So, yeah, π/3 is the only one that fits.Therefore, angle B is π/3, which is option A.**Final Answer**boxed{A}

💡Alright, so I have this problem here about random variables. Let me try to understand it step by step. Okay, the problem says that the mathematical expectation (which I think is the same as the mean) of a random variable X is 5. So, E(X) = 5. Got that. And the variance of X is 9. Variance measures how spread out the numbers are, right? So, Var(X) = 9.Now, there's another random variable Y defined as Y = 2X - 5. I need to find the mathematical expectation (mu) and the standard deviation (sigma) of Y, and then add them together to get mu + sigma.Let me recall some properties of expectation and variance. For expectation, if I have a linear transformation like Y = aX + b, then the expectation of Y is E(Y) = aE(X) + b. That seems straightforward. So, in this case, a is 2 and b is -5. Therefore, E(Y) should be 2*E(X) - 5.Given that E(X) is 5, plugging that in, E(Y) = 2*5 - 5 = 10 - 5 = 5. So, mu is 5. That part seems simple enough.Now, moving on to variance. I remember that variance has a different rule when it comes to linear transformations. Specifically, Var(Y) = a^2 * Var(X). The constant term b doesn't affect the variance because it just shifts all the values by a constant, which doesn't change their spread. So, in this case, a is 2, and Var(X) is 9. Therefore, Var(Y) = 2^2 * 9 = 4 * 9 = 36.But the problem asks for the standard deviation, which is the square root of the variance. So, sigma = sqrt(Var(Y)) = sqrt(36) = 6.Now, I need to find mu + sigma. Since mu is 5 and sigma is 6, adding them together gives 5 + 6 = 11.Wait, let me double-check my calculations to make sure I didn't make any mistakes. For expectation: Y = 2X - 5, so E(Y) = 2E(X) - 5 = 2*5 - 5 = 10 - 5 = 5. Yep, that's correct.For variance: Var(Y) = (2)^2 * Var(X) = 4 * 9 = 36. Standard deviation is sqrt(36) = 6. That seems right too.Adding them together: 5 + 6 = 11. Okay, I think that's the correct answer.I guess the key here was remembering how expectation and variance behave under linear transformations. For expectation, it's linear, so constants just get added or subtracted, and coefficients multiply. For variance, it's a bit different because variance depends on the spread, so only the squared coefficient affects it, and constants don't matter.So, in summary, mu is 5 and sigma is 6, so mu + sigma is 11.

💡First, I need to determine how many points Vanessa scored in the game. The team's total score is 75 points, and there are eight other players who each averaged 4.5 points.To find the total points scored by the eight players, I'll multiply the number of players by their average points:8 players × 4.5 points/player = 36 pointsNow, I'll subtract the points scored by the other players from the team's total to find Vanessa's score:75 total points - 36 points = 39 pointsTherefore, Vanessa scored 39 points in the game.

💡First, I need to determine the total cost of the pen in cents. Since the pen costs 6 dollars and 1 dollar equals 100 cents, the pen costs 600 cents.Joshua has 5 dollars, which is equivalent to 500 cents. He needs an additional 32 cents to purchase the pen, making the total amount he needs 632 cents.To find out how many cents Joshua borrowed from his neighbor, I subtract the amount he already has from the total amount needed: 632 cents minus 500 cents equals 132 cents.Therefore, Joshua borrowed 132 cents from his neighbor.

💡Okay, so I have this problem about an infinite geometric sequence. Let me try to understand it step by step. The sequence is denoted as {a_n}, and the sum of its terms is S_n. The first term, a_1, is given as 1. The common ratio is a bit tricky—it's given as "a - 3/2". Hmm, that's interesting because the common ratio usually is a constant, but here it's expressed in terms of 'a', which is also the limit of the sum S_n as n approaches infinity. So, the limit of S_n as n goes to infinity is equal to 'a'. Alright, so I need to find the value of 'a'. Let me recall the formula for the sum of an infinite geometric series. If the first term is a_1 and the common ratio is r, then the sum S is given by S = a_1 / (1 - r), provided that |r| < 1. In this case, the first term a_1 is 1, and the common ratio r is (a - 3/2). So, substituting these into the formula, the sum S should be 1 / (1 - (a - 3/2)). But wait, the problem says that the limit of S_n as n approaches infinity is equal to 'a'. So, that means the sum S is equal to 'a'. Therefore, I can set up the equation:a = 1 / (1 - (a - 3/2))Let me simplify the denominator first. 1 - (a - 3/2) is the same as 1 - a + 3/2. Combining the constants, 1 + 3/2 is 5/2. So, the denominator becomes (5/2 - a). Therefore, the equation becomes:a = 1 / (5/2 - a)Hmm, okay. Let me write that down:a = 1 / (5/2 - a)To solve for 'a', I can cross-multiply. So, multiplying both sides by (5/2 - a) gives:a * (5/2 - a) = 1Expanding the left side:(5/2)a - a^2 = 1Let me rearrange this equation to form a quadratic equation:-a^2 + (5/2)a - 1 = 0It's usually better to have the coefficient of a^2 positive, so I'll multiply the entire equation by -1:a^2 - (5/2)a + 1 = 0To eliminate the fraction, I can multiply every term by 2:2a^2 - 5a + 2 = 0Now, I have a quadratic equation: 2a^2 - 5a + 2 = 0. Let me try to factor this. I'm looking for two numbers that multiply to (2 * 2) = 4 and add up to -5. Hmm, let's see. 2a^2 can be factored as 2a and a, and the constants 2 can be factored as -2 and -1. Let me check:(2a - 1)(a - 2) = 2a^2 - 4a - a + 2 = 2a^2 - 5a + 2Yes, that works. So, the factors are (2a - 1) and (a - 2). Setting each factor equal to zero gives the solutions:2a - 1 = 0 => 2a = 1 => a = 1/2a - 2 = 0 => a = 2So, the solutions are a = 1/2 and a = 2. But wait, I need to check if these solutions are valid in the context of the problem. Remember, for the sum of an infinite geometric series to converge, the absolute value of the common ratio must be less than 1. The common ratio here is (a - 3/2). So, let's compute |a - 3/2| for each solution and see if it's less than 1.First, for a = 1/2:|1/2 - 3/2| = |-1| = 1Hmm, that's equal to 1, which doesn't satisfy |r| < 1. So, a = 1/2 is not a valid solution because the series wouldn't converge.Next, for a = 2:|2 - 3/2| = |1/2| = 1/2That's less than 1, so this solution is valid.Therefore, the value of 'a' must be 2.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the beginning, the sum S is equal to 1 / (1 - r), where r is (a - 3/2). Then, setting S equal to 'a' gives the equation a = 1 / (1 - (a - 3/2)). Simplifying that denominator: 1 - a + 3/2 = 5/2 - a. So, a = 1 / (5/2 - a). Cross-multiplying gives a*(5/2 - a) = 1, which expands to (5/2)a - a^2 = 1. Rearranging, -a^2 + (5/2)a - 1 = 0, which becomes 2a^2 - 5a + 2 = 0 after multiplying by -2. Factoring that quadratic gives (2a - 1)(a - 2) = 0, so a = 1/2 or a = 2. Checking the common ratio condition, |a - 3/2| < 1. For a = 1/2, |1/2 - 3/2| = 1, which is not less than 1. For a = 2, |2 - 3/2| = 1/2, which is less than 1. So, yes, a = 2 is the correct solution.I think that covers all the steps. I didn't skip anything, and the logic seems sound. So, I'm confident that the answer is 2.**Final Answer**The value of ( a ) is boxed{2}.

💡Okay, so I need to figure out if 30 divides ( p^2 - 1 ) for any prime number ( p geq 7 ). Hmm, let me break this down step by step.First, I remember that ( p^2 - 1 ) can be factored as ( (p - 1)(p + 1) ). Since ( p ) is a prime number greater than or equal to 7, it must be odd. That means both ( p - 1 ) and ( p + 1 ) are even numbers. In fact, they are consecutive even numbers, so one of them must be divisible by 4. Therefore, their product ( (p - 1)(p + 1) ) is divisible by ( 2 times 4 = 8 ).Next, I need to check if this product is divisible by 3 and 5 because 30 is ( 2 times 3 times 5 ). Let's start with 3. For any prime ( p geq 7 ), ( p ) cannot be divisible by 3. So, ( p ) must leave a remainder of either 1 or 2 when divided by 3. If ( p equiv 1 mod 3 ), then ( p - 1 ) is divisible by 3. If ( p equiv 2 mod 3 ), then ( p + 1 ) is divisible by 3. So, in either case, ( (p - 1)(p + 1) ) is divisible by 3.Now, let's check divisibility by 5. Similarly, ( p ) cannot be divisible by 5 since it's a prime number greater than 5. So, ( p ) can leave a remainder of 1, 2, 3, or 4 when divided by 5. If ( p equiv 1 mod 5 ), then ( p - 1 ) is divisible by 5. If ( p equiv 4 mod 5 ), then ( p + 1 ) is divisible by 5. If ( p equiv 2 mod 5 ), then ( p + 3 ) would be divisible by 5, but that's not directly relevant here. Wait, actually, I think I made a mistake there. Let me correct that.If ( p equiv 1 mod 5 ), then ( p - 1 equiv 0 mod 5 ). If ( p equiv 4 mod 5 ), then ( p + 1 equiv 0 mod 5 ). For ( p equiv 2 mod 5 ), ( p + 3 equiv 0 mod 5 ), but since we're only considering ( p - 1 ) and ( p + 1 ), this case doesn't directly apply. Similarly, if ( p equiv 3 mod 5 ), then ( p - 2 equiv 0 mod 5 ), which also isn't directly relevant. Hmm, so actually, only when ( p equiv 1 ) or ( 4 mod 5 ) does ( p^2 - 1 ) become divisible by 5. Does that mean it's not always divisible by 5?Wait, no. Because for any prime ( p geq 7 ), ( p ) must be congruent to 1, 2, 3, or 4 modulo 5. If ( p equiv 1 ) or ( 4 mod 5 ), then ( p^2 - 1 ) is divisible by 5. If ( p equiv 2 ) or ( 3 mod 5 ), then ( p^2 equiv 4 mod 5 ), so ( p^2 - 1 equiv 3 mod 5 ), which is not divisible by 5. Oh, so does that mean ( p^2 - 1 ) isn't always divisible by 5?But wait, I thought earlier that since ( p ) is a prime greater than 5, it must be congruent to 1 or 4 modulo 5, but that's not necessarily true. For example, 7 is congruent to 2 modulo 5, and 11 is congruent to 1 modulo 5. So, primes can be congruent to 1, 2, 3, or 4 modulo 5. Therefore, ( p^2 - 1 ) is only divisible by 5 when ( p equiv 1 ) or ( 4 mod 5 ).So, putting it all together, ( p^2 - 1 ) is always divisible by 8 and 3, but only sometimes divisible by 5, depending on the value of ( p ) modulo 5. Therefore, 30 divides ( p^2 - 1 ) only when ( p equiv 1 ) or ( 4 mod 5 ). Wait, but the question is asking if 30 divides ( p^2 - 1 ) for any prime ( p geq 7 ). So, does it mean that there exists at least one prime ( p geq 7 ) such that 30 divides ( p^2 - 1 )? Or does it mean that for all primes ( p geq 7 ), 30 divides ( p^2 - 1 )?Looking back at the question: "Determine if 30 divides ( p^2 - 1 ) for any prime number ( p geq 7 )." The wording is a bit ambiguous, but I think it's asking whether 30 divides ( p^2 - 1 ) for any prime ( p geq 7 ), meaning whether there exists such a prime where 30 divides ( p^2 - 1 ). If that's the case, then yes, because for primes like 11, which is congruent to 1 modulo 5, ( 11^2 - 1 = 120 ), and 120 is divisible by 30. Similarly, 19 is congruent to 4 modulo 5, and ( 19^2 - 1 = 360 ), which is also divisible by 30.However, if the question is interpreted as whether 30 always divides ( p^2 - 1 ) for any prime ( p geq 7 ), then the answer would be no because, as I saw earlier, for primes like 7, which is congruent to 2 modulo 5, ( 7^2 - 1 = 48 ), which is not divisible by 30.But the options given are:A) alwaysB) neverC) sometimes onlyD) only if ( p = 7 )E) none of theseSo, if the question is asking whether 30 divides ( p^2 - 1 ) for any prime ( p geq 7 ), meaning whether there exists such primes, then the answer is sometimes only, which is option C. However, if it's asking whether 30 always divides ( p^2 - 1 ) for any prime ( p geq 7 ), then the answer would be no, but that's not one of the options. Wait, option B is "never," which would be incorrect because we have examples where it does divide.Alternatively, maybe the question is asking whether 30 divides ( p^2 - 1 ) for all primes ( p geq 7 ). In that case, the answer would be no, but since "never" is an option, that would be incorrect because it does happen sometimes.Wait, I'm getting confused. Let me re-examine the question: "Determine if 30 divides ( p^2 - 1 ) for any prime number ( p geq 7 )." The phrase "for any" can be ambiguous. It could mean "for all" or "for any given." If it's "for all," then the answer is no. If it's "for any given," meaning whether it's possible, then the answer is yes, sometimes.Given the options, I think the intended interpretation is whether 30 always divides ( p^2 - 1 ) for any prime ( p geq 7 ). If that's the case, then the answer is no, but since "never" is an option, that would be incorrect because it does happen sometimes. Therefore, the correct answer is C) sometimes only.But wait, in my initial analysis, I thought that ( p^2 - 1 ) is always divisible by 8 and 3, and sometimes divisible by 5, depending on ( p ) modulo 5. Therefore, 30 divides ( p^2 - 1 ) if and only if ( p equiv 1 ) or ( 4 mod 5 ). So, it's not always, but it does happen for some primes ( p geq 7 ). Therefore, the answer is C) sometimes only.However, looking back at the initial problem statement, the user provided a thought process that concluded the answer is A) always. But based on my analysis, that's not correct because it's not always divisible by 5. Therefore, the correct answer should be C) sometimes only.Wait, but let me double-check with specific examples. Let's take ( p = 7 ): ( 7^2 - 1 = 48 ), which is not divisible by 30. ( p = 11 ): ( 11^2 - 1 = 120 ), which is divisible by 30. ( p = 13 ): ( 13^2 - 1 = 168 ), which is not divisible by 30. ( p = 17 ): ( 17^2 - 1 = 288 ), which is not divisible by 30. ( p = 19 ): ( 19^2 - 1 = 360 ), which is divisible by 30. So, it alternates depending on whether ( p equiv 1 ) or ( 4 mod 5 ). Therefore, it's not always, but it does happen for some primes ( p geq 7 ). So, the answer is C) sometimes only.But wait, the initial thought process concluded A) always, which seems incorrect. Maybe I made a mistake in my analysis. Let me check again.Wait, I think I confused myself earlier. Let's clarify:- ( p^2 - 1 = (p - 1)(p + 1) )- Since ( p ) is odd, both ( p - 1 ) and ( p + 1 ) are even, so their product is divisible by 4.- Additionally, one of them is divisible by 2, so overall divisible by 8.- For divisibility by 3: among any three consecutive numbers, one is divisible by 3. Since ( p ) is not divisible by 3, either ( p - 1 ) or ( p + 1 ) must be divisible by 3.- For divisibility by 5: similar logic, but since ( p ) is a prime greater than 5, it's not divisible by 5. So, ( p ) can be congruent to 1, 2, 3, or 4 modulo 5. If ( p equiv 1 mod 5 ), then ( p - 1 equiv 0 mod 5 ). If ( p equiv 4 mod 5 ), then ( p + 1 equiv 0 mod 5 ). If ( p equiv 2 mod 5 ), then ( p + 3 equiv 0 mod 5 ), but that's not directly relevant. Similarly, if ( p equiv 3 mod 5 ), then ( p - 2 equiv 0 mod 5 ). So, only when ( p equiv 1 ) or ( 4 mod 5 ) does ( p^2 - 1 ) become divisible by 5.Therefore, ( p^2 - 1 ) is always divisible by 8 and 3, hence by 24, but only sometimes divisible by 5. Therefore, 30 divides ( p^2 - 1 ) only when ( p equiv 1 ) or ( 4 mod 5 ). So, it's not always, but it does happen for some primes ( p geq 7 ). Therefore, the answer is C) sometimes only.Wait, but in the initial thought process, the conclusion was A) always, which contradicts my analysis. Maybe the initial thought process was incorrect. Let me verify with specific examples:- ( p = 7 ): ( 7^2 - 1 = 48 ), not divisible by 30.- ( p = 11 ): ( 11^2 - 1 = 120 ), divisible by 30.- ( p = 13 ): ( 13^2 - 1 = 168 ), not divisible by 30.- ( p = 17 ): ( 17^2 - 1 = 288 ), not divisible by 30.- ( p = 19 ): ( 19^2 - 1 = 360 ), divisible by 30.- ( p = 23 ): ( 23^2 - 1 = 528 ), not divisible by 30.- ( p = 29 ): ( 29^2 - 1 = 840 ), divisible by 30.So, it alternates depending on whether ( p equiv 1 ) or ( 4 mod 5 ). Therefore, it's not always, but it does happen for some primes ( p geq 7 ). Hence, the correct answer is C) sometimes only.