💡Okay, so I have this problem here about propositions and their converses. The original proposition is: "If (a + b geq 2), then at least one of (a) or (b) is not less than 1." I need to figure out if this proposition and its converse are true or false. The options are A, B, C, D, where A says the original is true and the converse is false, and so on.First, let me understand the original proposition. It's saying that if the sum of two numbers (a) and (b) is at least 2, then at least one of them is not less than 1. In other words, if (a + b) is 2 or more, then either (a geq 1) or (b geq 1) (or both). I think this is similar to the pigeonhole principle. If both (a) and (b) were less than 1, then their sum would be less than 2. So, if their sum is at least 2, it can't be that both are less than 1. Therefore, at least one must be 1 or more. That makes sense, so I think the original proposition is true.Now, the converse. The converse of a proposition swaps the hypothesis and the conclusion. So, the converse would be: "If at least one of (a) or (b) is not less than 1, then (a + b geq 2)." Hmm, let's test this. Suppose (a = 1) and (b = 0.5). Then (a) is not less than 1, but (a + b = 1.5), which is less than 2. So, the converse doesn't hold in this case. Therefore, the converse is false.Wait, let me think of another example. If (a = 1) and (b = 1), then (a + b = 2), which satisfies the converse. But the previous example already shows that the converse isn't always true. So, the converse is false.Therefore, the original proposition is true, and the converse is false. So, the answer should be option A.But just to be thorough, let me consider another case for the converse. Suppose (a = 1.5) and (b = 0.4). Then (a) is not less than 1, but (a + b = 1.9), which is still less than 2. So, again, the converse fails. What if both (a) and (b) are exactly 1? Then (a + b = 2), which satisfies the converse. But since there are cases where the converse doesn't hold, like when one is exactly 1 and the other is less than 1, the converse isn't universally true.So, yeah, I'm confident now. The original is true, the converse is false. So, option A is correct.

💡Alright, so I have this quadratic equation: (x^2 + (k + 1)x + 3k - 6 = 0). The problem has two parts. First, I need to prove that this equation always has two real roots, regardless of the value of (k). Second, if one of the roots is not less than 7, I need to find the range of values for (k).Starting with part (1): Proving the equation always has two real roots. I remember that for a quadratic equation (ax^2 + bx + c = 0), the discriminant (Delta = b^2 - 4ac) determines the nature of the roots. If (Delta > 0), there are two distinct real roots; if (Delta = 0), there's exactly one real root (a repeated root); and if (Delta < 0), there are no real roots, only complex ones.So, for our equation, (a = 1), (b = (k + 1)), and (c = 3k - 6). Let me compute the discriminant:[Delta = (k + 1)^2 - 4 times 1 times (3k - 6)]Expanding ((k + 1)^2):[(k + 1)^2 = k^2 + 2k + 1]Now, expanding the second term:[4 times 1 times (3k - 6) = 12k - 24]So, putting it all together:[Delta = k^2 + 2k + 1 - 12k + 24]Combining like terms:[Delta = k^2 - 10k + 25]Hmm, this looks like a perfect square. Let me check:[k^2 - 10k + 25 = (k - 5)^2]Yes, that's correct. So, the discriminant is ((k - 5)^2). Since any square is always non-negative, ((k - 5)^2 geq 0). Therefore, the discriminant is always non-negative, which means the quadratic equation always has two real roots (they could be equal if the discriminant is zero, but they are still real).So, part (1) is proven.Moving on to part (2): If the equation has one root not less than 7, find the range of values for (k).First, let's recall that the roots of a quadratic equation (ax^2 + bx + c = 0) can be found using the quadratic formula:[x = frac{-b pm sqrt{Delta}}{2a}]But since we already have the discriminant as ((k - 5)^2), which is a perfect square, the roots will be:[x = frac{-(k + 1) pm (k - 5)}{2}]Let me compute both roots:First root with the plus sign:[x_1 = frac{-(k + 1) + (k - 5)}{2} = frac{-k - 1 + k - 5}{2} = frac{-6}{2} = -3]Second root with the minus sign:[x_2 = frac{-(k + 1) - (k - 5)}{2} = frac{-k - 1 - k + 5}{2} = frac{-2k + 4}{2} = -k + 2]So, the two roots are (x_1 = -3) and (x_2 = -k + 2).Now, the problem states that one root is not less than 7. That means at least one of the roots is greater than or equal to 7.Looking at the roots:- (x_1 = -3) is always less than 7, so it can't be the root that's not less than 7.- Therefore, the other root (x_2 = -k + 2) must be greater than or equal to 7.So, we set up the inequality:[-k + 2 geq 7]Solving for (k):[-k + 2 geq 7 -k geq 7 - 2 -k geq 5 ]Multiplying both sides by -1 (remember to reverse the inequality sign when multiplying by a negative):[k leq -5]So, the range of values for (k) is all real numbers less than or equal to -5.Let me double-check this result. If (k = -5), then (x_2 = -(-5) + 2 = 5 + 2 = 7), which is exactly 7, satisfying the condition. If (k < -5), say (k = -6), then (x_2 = -(-6) + 2 = 6 + 2 = 8), which is greater than 7. And if (k > -5), say (k = 0), then (x_2 = -0 + 2 = 2), which is less than 7. So, the inequality (k leq -5) correctly identifies the values of (k) for which one root is not less than 7.Therefore, the range of (k) is (k leq -5).

💡Okay, so I have this triangle ABC where the sides opposite angles A, B, and C are a, b, and c respectively. I know that a is greater than b, a is 5, c is 6, and sin B is 3/5. I need to find the values of b and sin A for part (1), and then for part (2), I need to find sin(2A + π/4).First, let me try to visualize the triangle. Since a is opposite angle A and a is 5, and c is 6 opposite angle C, so side c is longer than side a. Hmm, but a is greater than b, so side b must be less than 5. That might be useful later.Given that sin B is 3/5, I can find cos B using the Pythagorean identity. Since sin² B + cos² B = 1, cos B would be sqrt(1 - (3/5)²) = sqrt(1 - 9/25) = sqrt(16/25) = 4/5. So cos B is 4/5.Now, to find side b, I can use the Law of Cosines. The Law of Cosines states that b² = a² + c² - 2ac cos B. Plugging in the known values: b² = 5² + 6² - 2*5*6*(4/5). Let me compute that step by step.5 squared is 25, 6 squared is 36, so 25 + 36 is 61. Then, 2*5*6 is 60, and 60*(4/5) is 48. So, subtracting 48 from 61 gives 13. Therefore, b squared is 13, so b is sqrt(13). That seems straightforward.Now, moving on to finding sin A. For this, I can use the Law of Sines, which states that a/sin A = b/sin B. We know a is 5, b is sqrt(13), and sin B is 3/5. So, setting up the equation: 5/sin A = sqrt(13)/(3/5). Let me solve for sin A.First, cross-multiplying gives 5*(3/5) = sqrt(13)*sin A. Simplifying the left side: 5*(3/5) is 3. So, 3 = sqrt(13)*sin A. Therefore, sin A is 3/sqrt(13). To rationalize the denominator, multiply numerator and denominator by sqrt(13): (3*sqrt(13))/13. So, sin A is 3*sqrt(13)/13.Wait, but I should check if this makes sense. Since a is 5 and c is 6, and a > b, so side a is 5, which is less than c, which is 6. So, angle A should be less than angle C. Hmm, but I don't know angle C yet. Maybe I can find it later, but for now, sin A seems reasonable.Moving on to part (2), I need to find sin(2A + π/4). To do this, I can use the sine addition formula: sin(2A + π/4) = sin(2A)cos(π/4) + cos(2A)sin(π/4). I know that cos(π/4) and sin(π/4) are both sqrt(2)/2, so I can factor that out later.First, I need to find sin(2A) and cos(2A). To find these, I can use double-angle identities. Sin(2A) is 2 sin A cos A, and cos(2A) is 1 - 2 sin² A. I already have sin A as 3*sqrt(13)/13, so I need to find cos A.Using the Pythagorean identity again, cos² A = 1 - sin² A. So, sin² A is (9*13)/169 = 117/169. Therefore, cos² A is 1 - 117/169 = (169 - 117)/169 = 52/169. Taking the square root, cos A is sqrt(52)/13. Simplify sqrt(52): sqrt(4*13) = 2*sqrt(13). So, cos A is 2*sqrt(13)/13.Now, compute sin(2A): 2 sin A cos A = 2*(3*sqrt(13)/13)*(2*sqrt(13)/13). Let's calculate that: 2*3*2 = 12, sqrt(13)*sqrt(13) = 13, and denominator is 13*13 = 169. So, sin(2A) is (12*13)/169 = 156/169. Wait, that can't be right because 12*13 is 156, but 156/169 simplifies to 12/13. Wait, no, 12*13 is 156, but 156 divided by 169 is 12/13 because 156 = 12*13 and 169 = 13*13. So, sin(2A) is 12/13.Next, compute cos(2A): 1 - 2 sin² A. We already have sin² A as 117/169, so 2 sin² A is 234/169. Therefore, cos(2A) is 1 - 234/169 = (169 - 234)/169 = (-65)/169 = -5/13.Now, plug these into the sine addition formula: sin(2A + π/4) = sin(2A)cos(π/4) + cos(2A)sin(π/4). Both cos(π/4) and sin(π/4) are sqrt(2)/2, so:sin(2A + π/4) = (12/13)*(sqrt(2)/2) + (-5/13)*(sqrt(2)/2).Factor out sqrt(2)/2: [12/13 - 5/13] * sqrt(2)/2 = (7/13) * sqrt(2)/2 = 7*sqrt(2)/26.Wait, let me double-check that. 12/13 - 5/13 is indeed 7/13. Then, multiplying by sqrt(2)/2 gives 7*sqrt(2)/26. That seems correct.So, summarizing:1. b is sqrt(13) and sin A is 3*sqrt(13)/13.2. sin(2A + π/4) is 7*sqrt(2)/26.I think that covers everything. I should probably check my steps again to make sure I didn't make any calculation errors.Starting with part (1):- Found cos B correctly: 4/5.- Applied Law of Cosines: b² = 25 + 36 - 48 = 13, so b = sqrt(13). Correct.- Law of Sines: 5/sin A = sqrt(13)/(3/5). Cross-multiplied: 5*(3/5) = 3 = sqrt(13) sin A. So sin A = 3/sqrt(13) = 3*sqrt(13)/13. Correct.For part (2):- Found cos A using Pythagorean identity: sqrt(1 - (9*13)/169) = sqrt(52/169) = 2*sqrt(13)/13. Correct.- Sin(2A) = 2*(3*sqrt(13)/13)*(2*sqrt(13)/13) = 12/13. Correct.- Cos(2A) = 1 - 2*(9*13)/169 = 1 - 234/169 = -65/169 = -5/13. Correct.- Applied sine addition formula: (12/13 - 5/13)*sqrt(2)/2 = 7*sqrt(2)/26. Correct.Everything seems to check out. I think I'm confident with these answers.

💡First, I need to determine the proportion of employees who are no older than 45 years old in the total population. There are 120 employees in this age group out of 200 total employees.Next, I'll calculate the proportion by dividing the number of employees no older than 45 by the total number of employees: 120 divided by 200 equals 0.6.To find out how many employees from this age group should be included in the sample of 25, I'll multiply the total sample size by the proportion: 25 multiplied by 0.6 equals 15.Therefore, 15 employees who are no older than 45 years old should be drawn for the sample.

💡To determine the digit ( B ) that makes the number ( 947B ) divisible by 5, I recall that a number is divisible by 5 if its last digit is either 0 or 5.First, I consider the possibility that ( B = 0 ). Substituting 0 into the number gives ( 9470 ). Checking the divisibility, since the last digit is 0, ( 9470 ) is divisible by 5.Next, I examine the case where ( B = 5 ). Substituting 5 into the number results in ( 9475 ). Again, since the last digit is 5, ( 9475 ) is also divisible by 5.Therefore, the valid digits for ( B ) are 0 and 5.

💡First, I identify the total number of people in the group, which is 49.Next, I note the number of Chinese individuals, which is 22, and the number of Australians, which is 11.To find the number of Americans, I subtract the number of Chinese and Australians from the total number of people.So, 49 minus 22 equals 27, and then 27 minus 11 equals 16.Therefore, there are 16 Americans in the group.