💡Okay, so I have this question about deductive reasoning and syllogisms. There are four options, A through D, each presenting a different syllogistic argument. I need to figure out which one is both in the form of a syllogism and correct. Hmm, let's start by recalling what a syllogism is.A syllogism is a form of deductive reasoning consisting of two premises and a conclusion. The structure is usually: Major premise, Minor premise, Conclusion. The major premise is a general statement, the minor premise is a specific statement, and the conclusion connects the two. The conclusion must logically follow from the premises.Alright, let's look at each option one by one.**Option A:**- Major premise: Non-repeating infinite decimals are irrational numbers.- Minor premise: π is an irrational number.- Conclusion: π is a non-repeating infinite decimal.Hmm, so the major premise is a general statement about non-repeating infinite decimals being irrational. The minor premise is about π being irrational. The conclusion is that π is a non-repeating infinite decimal. Wait, does this follow? If non-repeating decimals are irrational, and π is irrational, does that necessarily mean π is a non-repeating decimal? I think this might be a case of affirming the consequent. The major premise is "If A, then B," and the minor premise is "B," so concluding "A." That's not a valid syllogism because there could be other ways to be irrational without being a non-repeating decimal. So, A might not be correct.**Option B:**- Major premise: Non-repeating infinite decimals are irrational numbers.- Minor premise: π is a non-repeating infinite decimal.- Conclusion: π is an irrational number.Okay, this seems more straightforward. The major premise is a general rule, the minor premise applies that rule to π, and the conclusion follows directly. Since non-repeating decimals are irrational, and π is a non-repeating decimal, then π must be irrational. This seems valid. It follows the structure of a syllogism correctly.**Option C:**- Major premise: π is a non-repeating infinite decimal.- Minor premise: Non-repeating infinite decimals are irrational numbers.- Conclusion: π is an irrational number.Wait, this looks similar to B, but the major and minor premises are swapped. In syllogisms, the major premise is supposed to be the general statement, and the minor premise is the specific case. Here, the major premise is about π, which is specific, and the minor premise is the general rule. That might not be a standard syllogism because the order is reversed. However, the conclusion still logically follows. But I think the form is not correct because the major premise should be the general statement. So, maybe C is not considered a correct syllogism in form, even though the conclusion is true.**Option D:**- Major premise: π is a non-repeating infinite decimal.- Minor premise: π is an irrational number.- Conclusion: Non-repeating infinite decimals are irrational numbers.This seems different. The major premise is about π being a non-repeating decimal, the minor premise is about π being irrational, and the conclusion is a general statement about all non-repeating decimals being irrational. Wait, that's going from specific to general, which is more like induction rather than deduction. In a syllogism, you go from general to specific. So, D is probably not a correct syllogism because it's moving in the wrong direction.So, summarizing my thoughts:- A: Affirming the consequent, invalid.- B: Correct syllogism, valid.- C: Premises are reversed, form is incorrect.- D: Inductive reasoning, not a syllogism.Therefore, the correct answer should be B.

💡First, I need to recall the formula for the projection of one vector onto another. The projection of vector (overrightarrow{a}) onto vector (overrightarrow{b}) is given by the dot product of (overrightarrow{a}) and (overrightarrow{b}) divided by the magnitude of (overrightarrow{b}).Next, I'll calculate the dot product of (overrightarrow{a}) and (overrightarrow{b}). This is done by multiplying the corresponding components of the vectors and then summing them up:[overrightarrow{a} cdot overrightarrow{b} = (2)(-2) + (3)(1) = -4 + 3 = -1]After finding the dot product, I'll determine the magnitude of vector (overrightarrow{b}). The magnitude is calculated using the square root of the sum of the squares of its components:[|overrightarrow{b}| = sqrt{(-2)^2 + (1)^2} = sqrt{4 + 1} = sqrt{5}]Finally, I'll use these values in the projection formula:[text{Projection of } overrightarrow{a} text{ onto } overrightarrow{b} = frac{overrightarrow{a} cdot overrightarrow{b}}{|overrightarrow{b}|} = frac{-1}{sqrt{5}} = -frac{sqrt{5}}{5}]

💡Okay, so I have this problem where I need to find the largest possible determinant of a matrix whose columns are a unit vector u, and two given vectors v and w. The vectors v and w are given as:v = [3, 2, -2] and w = [2, -1, 4].First, I remember that the determinant of a matrix with columns u, v, w is the same as the scalar triple product of u, v, and w. That is, determinant = u · (v × w). So, to find the determinant, I need to compute the cross product of v and w first, and then take the dot product of that result with u.Let me write down the cross product formula. For two vectors a = [a1, a2, a3] and b = [b1, b2, b3], the cross product a × b is given by:[a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1].So, applying this to vectors v and w:v × w = [v2w3 - v3w2, v3w1 - v1w3, v1w2 - v2w1].Plugging in the values:v2 = 2, v3 = -2, w3 = 4, w2 = -1.First component: 2*4 - (-2)*(-1) = 8 - 2 = 6.Second component: (-2)*2 - 3*4 = -4 - 12 = -16.Third component: 3*(-1) - 2*2 = -3 - 4 = -7.So, v × w = [6, -16, -7].Now, the determinant is u · (v × w). Since u is a unit vector, the dot product u · (v × w) is equal to the magnitude of (v × w) times the cosine of the angle between u and (v × w). To maximize this determinant, we need to maximize the dot product. The maximum value occurs when u is in the same direction as (v × w), because then the cosine of the angle is 1.Therefore, the maximum determinant is just the magnitude of (v × w).Calculating the magnitude:||v × w|| = sqrt(6^2 + (-16)^2 + (-7)^2) = sqrt(36 + 256 + 49).Adding those up: 36 + 256 = 292, 292 + 49 = 341.So, sqrt(341) is the magnitude.Therefore, the largest possible determinant is sqrt(341).Wait, let me double-check my cross product calculation to make sure I didn't make any mistakes.First component: v2w3 - v3w2 = 2*4 - (-2)*(-1) = 8 - 2 = 6. That seems correct.Second component: v3w1 - v1w3 = (-2)*2 - 3*4 = -4 - 12 = -16. That also looks right.Third component: v1w2 - v2w1 = 3*(-1) - 2*2 = -3 - 4 = -7. Correct again.So, the cross product is indeed [6, -16, -7].Calculating the magnitude:sqrt(6^2 + (-16)^2 + (-7)^2) = sqrt(36 + 256 + 49) = sqrt(341). Yep, that's correct.So, I think my reasoning is solid here. The maximum determinant occurs when u is aligned with the cross product vector, giving the maximum possible value of the scalar triple product, which is the magnitude of the cross product.I don't think I missed anything. The key steps were computing the cross product, then recognizing that the determinant is the dot product with u, and since u is a unit vector, the maximum occurs when u is in the direction of the cross product vector.So, the final answer should be sqrt(341).**Final Answer**The largest possible determinant of the matrix is boxed{sqrt{341}}.

💡First, I need to divide the coefficients of the terms. The coefficient in the numerator is 12, and in the denominator, it is 6. Dividing 12 by 6 gives me 2.Next, I'll handle the variable ( x ). In the numerator, the exponent of ( x ) is 5, and in the denominator, it is 1. Subtracting the exponents (5 - 1) gives me 4. So, ( x^5 div x = x^4 ).Finally, I'll address the variable ( y ). Both the numerator and the denominator have ( y ), so they cancel each other out.Combining these results, the simplified expression is ( 2x^4 ).

💡First, I need to calculate (frac{1}{6}) of 72. This means I will multiply 72 by (frac{1}{6}).Next, I take the result from the first step and find (frac{1}{3}) of that value. This involves multiplying the previous result by (frac{1}{3}).Finally, I take the result from the second step and determine (frac{1}{2}) of that value by multiplying it by (frac{1}{2}).By following these sequential multiplications, I will arrive at the final answer.

💡Okay, so I have this problem here: I need to find the smallest positive integer that leaves a remainder of 2 when divided by 5 and a remainder of 3 when divided by 7. Hmm, let me think about how to approach this.First, I know that when a number leaves a remainder when divided by another number, it's related to congruences in modular arithmetic. So, if a number x leaves a remainder of 2 when divided by 5, that means x is congruent to 2 modulo 5. Similarly, if it leaves a remainder of 3 when divided by 7, then x is congruent to 3 modulo 7.So, mathematically, I can write these two conditions as:1. x ≡ 2 mod 52. x ≡ 3 mod 7Now, I need to find the smallest positive integer x that satisfies both of these conditions. I remember that the Chinese Remainder Theorem can be used to solve systems of congruences like this, especially when the moduli are coprime. In this case, 5 and 7 are coprime because their greatest common divisor is 1. So, the Chinese Remainder Theorem should apply here.Let me recall how the Chinese Remainder Theorem works. It states that if we have two congruences:x ≡ a mod mx ≡ b mod nwhere m and n are coprime, then there exists a unique solution modulo m*n. So, in this case, the solution will be unique modulo 35 (since 5*7=35).Alright, so I need to find x such that:x ≡ 2 mod 5x ≡ 3 mod 7To solve this, I can express x in terms of one of the congruences and substitute into the other. Let's start with the first congruence:x ≡ 2 mod 5This means that x can be written as:x = 5k + 2where k is some integer.Now, I can substitute this expression for x into the second congruence:5k + 2 ≡ 3 mod 7Let me simplify this equation:5k + 2 ≡ 3 mod 7Subtract 2 from both sides:5k ≡ 1 mod 7So, now I have:5k ≡ 1 mod 7I need to solve for k. This means I need to find an integer k such that when 5k is divided by 7, the remainder is 1. In other words, 5k - 1 is divisible by 7.One way to solve this is to find the multiplicative inverse of 5 modulo 7. The multiplicative inverse of a number a modulo m is a number b such that:a*b ≡ 1 mod mSo, I need to find a number b such that:5*b ≡ 1 mod 7Let me try different values of b:- If b = 1: 5*1 = 5 ≡ 5 mod 7 → Not 1- If b = 2: 5*2 = 10 ≡ 3 mod 7 → Not 1- If b = 3: 5*3 = 15 ≡ 1 mod 7 → Yes! That works.So, the multiplicative inverse of 5 modulo 7 is 3. Therefore, I can multiply both sides of the equation 5k ≡ 1 mod 7 by 3 to solve for k:3*5k ≡ 3*1 mod 715k ≡ 3 mod 7But 15 mod 7 is 1, because 15 divided by 7 is 2 with a remainder of 1. So, this simplifies to:1k ≡ 3 mod 7Which means:k ≡ 3 mod 7So, k can be written as:k = 7m + 3where m is some integer.Now, substitute this back into the expression for x:x = 5k + 2x = 5*(7m + 3) + 2x = 35m + 15 + 2x = 35m + 17So, the general solution for x is x = 35m + 17, where m is any integer. Since we're looking for the smallest positive integer, we can set m = 0:x = 35*0 + 17 = 17Let me verify this solution to make sure it works.First, check x ≡ 2 mod 5:17 divided by 5 is 3 with a remainder of 2. So, 17 ≡ 2 mod 5. That's correct.Second, check x ≡ 3 mod 7:17 divided by 7 is 2 with a remainder of 3. So, 17 ≡ 3 mod 7. That's also correct.Therefore, 17 is indeed the smallest positive integer that satisfies both conditions.I think that's the solution. But just to be thorough, let me check the next possible value by setting m = 1:x = 35*1 + 17 = 52Check 52:52 divided by 5 is 10 with a remainder of 2. Good.52 divided by 7 is 7 with a remainder of 3. Also good.So, the solutions are 17, 52, 87, etc., each 35 apart. Since we're looking for the smallest positive integer, 17 is the answer.I don't think I made any mistakes in my reasoning. I used the Chinese Remainder Theorem correctly, found the multiplicative inverse, solved for k, substituted back, and verified the solution. Everything checks out.