💡Alright, so I need to prove that in any triangle, at least one of the internal angles is not greater than 60 degrees. And I have to use the method of contradiction to do this. Okay, let's break this down step by step.First, I remember that in a triangle, the sum of all three internal angles is always 180 degrees. That's a fundamental property of triangles. So, if I can somehow relate the angles to this sum, I might be able to make progress.Now, the proposition says that at least one angle is not greater than 60 degrees. In other words, one angle is 60 degrees or less. To prove this using contradiction, I need to assume the opposite and show that this leads to a contradiction.So, what is the opposite of "at least one angle is not greater than 60 degrees"? It would be "all angles are greater than 60 degrees." That makes sense because if I assume that all three angles are greater than 60 degrees, then none of them are 60 degrees or less, which directly contradicts the original statement.Alright, so let's assume that all three internal angles of a triangle are greater than 60 degrees. Let's denote the angles as A, B, and C. So, A > 60°, B > 60°, and C > 60°.If I add these inequalities together, I get A + B + C > 60° + 60° + 60°, which simplifies to A + B + C > 180°. But wait, we know from the triangle angle sum property that A + B + C = 180°. So, this leads to a contradiction because 180° cannot be greater than itself.This contradiction implies that our initial assumption—that all three angles are greater than 60 degrees—must be false. Therefore, the original statement must be true: in any triangle, at least one of the internal angles is not greater than 60 degrees.Let me double-check to make sure I didn't make any mistakes. I assumed the opposite of what I wanted to prove, which is a valid approach for contradiction. Then, I used the known property that the sum of angles in a triangle is 180 degrees. By assuming all angles are greater than 60, their sum would exceed 180, which is impossible. So, the contradiction holds, and the proof seems solid.I think I've covered all the necessary steps. I started with the assumption, applied the known property, derived a contradiction, and concluded that the original statement must be true. This makes sense logically, and I don't see any flaws in the reasoning.Just to be thorough, let me consider if there's any other way this could be approached. Maybe by looking at specific types of triangles, like equilateral, isosceles, or scalene, but I think the general proof using contradiction covers all cases. In an equilateral triangle, all angles are exactly 60 degrees, so that satisfies the condition. In other triangles, if one angle is larger than 60, the others have to compensate by being smaller to keep the total at 180 degrees.Yeah, I think I'm confident with this proof. It logically follows and uses the properties of triangles effectively.

💡Alright, so I have this problem here with three variables: a, b, and c. I need to figure out their order from greatest to least and choose the correct option among A, B, C, D. Let's take it step by step.First, let's write down what each variable is:- ( a = log_{pi} e )- ( b = 2^{cos frac{7pi}{3}} )- ( c = log_{3} sin frac{17pi}{6} )Okay, so I need to evaluate each of these expressions and then compare them.Starting with ( a = log_{pi} e ). Hmm, logarithms can sometimes be tricky, but I remember that ( log_b a ) is the exponent to which we must raise b to get a. So, ( log_{pi} e ) is the exponent that we need to raise ( pi ) to in order to get e. Since ( pi ) is approximately 3.1416 and e is approximately 2.7183, and both are greater than 1, I know that ( log_{pi} e ) will be a number between 0 and 1 because ( e < pi ). So, ( a ) is somewhere between 0 and 1. Maybe around 0.8 or something? I might need to calculate it more precisely later, but for now, I'll note that ( 0 < a < 1 ).Next, ( b = 2^{cos frac{7pi}{3}} ). Let's tackle the exponent first: ( cos frac{7pi}{3} ). I know that the cosine function has a period of ( 2pi ), so I can subtract ( 2pi ) from ( frac{7pi}{3} ) to find an equivalent angle within the first rotation. Let's compute that:( frac{7pi}{3} - 2pi = frac{7pi}{3} - frac{6pi}{3} = frac{pi}{3} ).So, ( cos frac{7pi}{3} = cos frac{pi}{3} ). I remember that ( cos frac{pi}{3} = frac{1}{2} ). Therefore, ( b = 2^{frac{1}{2}} = sqrt{2} ). Okay, so ( b ) is approximately 1.4142. That's definitely greater than 1.Now, onto ( c = log_{3} sin frac{17pi}{6} ). Again, let's simplify the angle inside the sine function. The sine function also has a period of ( 2pi ), so let's subtract multiples of ( 2pi ) to get an angle between 0 and ( 2pi ).( frac{17pi}{6} ) is more than ( 2pi ) because ( 2pi = frac{12pi}{6} ). So, subtracting ( 2pi ):( frac{17pi}{6} - 2pi = frac{17pi}{6} - frac{12pi}{6} = frac{5pi}{6} ).But wait, ( frac{5pi}{6} ) is still greater than ( pi ), so maybe I need to subtract another ( 2pi )? No, wait, ( frac{5pi}{6} ) is less than ( 2pi ), so that's fine. But actually, ( frac{17pi}{6} ) is equivalent to ( frac{17pi}{6} - 2pi = frac{5pi}{6} ). So, ( sin frac{17pi}{6} = sin frac{5pi}{6} ).I remember that ( sin frac{5pi}{6} = sin (pi - frac{pi}{6}) = sin frac{pi}{6} = frac{1}{2} ). Wait, but hold on. ( frac{5pi}{6} ) is in the second quadrant where sine is positive, so it's ( frac{1}{2} ). However, ( frac{17pi}{6} ) is actually in the fourth quadrant because ( frac{17pi}{6} = 2pi + frac{5pi}{6} ), right? Wait, no, that's not correct. Let me think again.Actually, ( frac{17pi}{6} ) is equal to ( 2pi + frac{5pi}{6} ), which is more than ( 2pi ). So, to find the reference angle, we can subtract ( 2pi ):( frac{17pi}{6} - 2pi = frac{17pi}{6} - frac{12pi}{6} = frac{5pi}{6} ).But since ( frac{17pi}{6} ) is in the fourth quadrant (because after subtracting ( 2pi ), it's ( frac{5pi}{6} ), which is in the second quadrant, but wait, no, ( frac{5pi}{6} ) is in the second quadrant, but ( frac{17pi}{6} ) is actually ( 2pi + frac{5pi}{6} ), which is the same as ( frac{5pi}{6} ) in terms of sine, but with a sign based on the quadrant.Wait, I'm getting confused. Let me visualize the unit circle. ( 2pi ) is a full rotation, so ( frac{17pi}{6} ) is ( 2pi + frac{5pi}{6} ), which is the same as ( frac{5pi}{6} ) but in the next rotation. So, ( sin frac{17pi}{6} = sin frac{5pi}{6} ). But ( frac{5pi}{6} ) is in the second quadrant where sine is positive, so ( sin frac{5pi}{6} = frac{1}{2} ). Therefore, ( sin frac{17pi}{6} = frac{1}{2} ).Wait, but hold on, isn't ( frac{17pi}{6} ) actually in the fourth quadrant? Because ( 2pi = frac{12pi}{6} ), so ( frac{17pi}{6} - 2pi = frac{5pi}{6} ), but ( frac{5pi}{6} ) is in the second quadrant. Hmm, maybe I'm overcomplicating this. Let me just use the identity that ( sin (2pi + theta) = sin theta ). So, ( sin frac{17pi}{6} = sin frac{5pi}{6} = frac{1}{2} ).Wait, but actually, ( frac{17pi}{6} ) is ( 2pi + frac{5pi}{6} ), which is the same as ( frac{5pi}{6} ) in terms of sine, but since it's in the fourth quadrant, sine should be negative. Wait, no, because ( sin theta = sin (pi - theta) ), but in this case, ( frac{5pi}{6} ) is in the second quadrant, so ( sin frac{5pi}{6} = sin (pi - frac{pi}{6}) = sin frac{pi}{6} = frac{1}{2} ). But since ( frac{17pi}{6} ) is in the fourth quadrant, sine should be negative. So, is it ( -frac{1}{2} )?Wait, I'm getting conflicting thoughts here. Let me double-check. The angle ( frac{17pi}{6} ) is equal to ( 2pi + frac{5pi}{6} ). So, it's like starting at 0, going around the circle once (which is ( 2pi )), and then an additional ( frac{5pi}{6} ). So, ( frac{5pi}{6} ) is in the second quadrant, but since we've gone around the circle once, it's actually in the same position as ( frac{5pi}{6} ), but in the next rotation. Wait, no, the sine function is periodic with period ( 2pi ), so ( sin frac{17pi}{6} = sin frac{5pi}{6} ). But ( frac{5pi}{6} ) is in the second quadrant where sine is positive, so ( sin frac{5pi}{6} = frac{1}{2} ). Therefore, ( sin frac{17pi}{6} = frac{1}{2} ).Wait, but that doesn't make sense because ( frac{17pi}{6} ) is more than ( 2pi ), so it's like ( 2pi + frac{5pi}{6} ), which is the same as ( frac{5pi}{6} ) in terms of sine, but in the next cycle. So, actually, the sine should be positive because it's in the second quadrant. Hmm, I think I'm confusing myself. Maybe I should just calculate it numerically.Let me compute ( frac{17pi}{6} ) in degrees to better understand. Since ( pi ) radians is 180 degrees, so ( frac{17pi}{6} ) is ( frac{17}{6} times 180 = 17 times 30 = 510 ) degrees. 510 degrees is more than 360 degrees, so subtracting 360 gives 150 degrees. So, ( sin 510^circ = sin 150^circ ). And ( sin 150^circ = sin (180^circ - 30^circ) = sin 30^circ = frac{1}{2} ). So, yes, ( sin frac{17pi}{6} = frac{1}{2} ).Wait, but 510 degrees is in the second quadrant, so sine is positive. So, ( sin frac{17pi}{6} = frac{1}{2} ). Therefore, ( c = log_{3} frac{1}{2} ). Hmm, logarithms of numbers less than 1 are negative because the logarithm function is negative for arguments between 0 and 1. So, ( log_{3} frac{1}{2} ) is negative. Specifically, ( log_{3} frac{1}{2} = -log_{3} 2 ). Since ( log_{3} 2 ) is approximately 0.6309, so ( c approx -0.6309 ).Okay, so now I have:- ( a ) is between 0 and 1, approximately 0.8- ( b ) is approximately 1.4142- ( c ) is approximately -0.6309So, putting them in order from greatest to least: ( b > a > c ). Therefore, the correct option is A.Wait, let me double-check my calculations to make sure I didn't make any mistakes.For ( a = log_{pi} e ), since ( pi ) is about 3.14 and e is about 2.718, and since ( pi > e ), ( log_{pi} e ) should be less than 1, which makes sense. Using the change of base formula, ( log_{pi} e = frac{ln e}{ln pi} = frac{1}{ln pi} approx frac{1}{1.144} approx 0.873 ). So, ( a approx 0.873 ).For ( b = 2^{cos frac{7pi}{3}} ), we simplified ( frac{7pi}{3} ) to ( frac{pi}{3} ), so ( cos frac{pi}{3} = 0.5 ), hence ( b = 2^{0.5} = sqrt{2} approx 1.4142 ).For ( c = log_{3} sin frac{17pi}{6} ), we found that ( sin frac{17pi}{6} = frac{1}{2} ), so ( c = log_{3} frac{1}{2} = -log_{3} 2 approx -0.6309 ).So, indeed, ( b approx 1.4142 > a approx 0.873 > c approx -0.6309 ). Therefore, the correct order is ( b > a > c ), which corresponds to option A.I think that's solid. I don't see any mistakes in my reasoning now.

💡First, I'll assume the fuel tank has a capacity of 12 units to simplify calculations, as 12 is a common multiple of the denominators 4, 3, and 2 mentioned in the problem.1. **Initial Fill with Brand Z:** - The tank starts with 12 units of brand Z gasoline.2. **First Refill with Brand Y:** - When the tank is 1/4 empty, it has 9 units of brand Z left. - It is then filled with 3 units of brand Y. - **Current Composition:** 9 units Z, 3 units Y.3. **Second Refill with Brand W:** - When the tank is 1/3 empty again, it has 8 units of gasoline left (6 units Z and 2 units Y). - It is then filled with 4 units of brand W. - **Current Composition:** 6 units Z, 2 units Y, 4 units W.4. **Third Refill with Brand X:** - When the tank is half empty again, it has 6 units of gasoline left (4 units Z, 1 unit Y, and 1 unit W). - It is then filled with 6 units of brand X. - **Current Composition:** 4 units Z, 1 unit Y, 1 unit W, 6 units X.5. **Fourth Refill with Brand Y:** - When the tank is half empty again, it has 6 units of gasoline left (2 units Z, 0.5 units Y, 0.5 units W, and 3 units X). - It is then filled with 6 units of brand Y. - **Current Composition:** 2 units Z, 0.5 units Y, 0.5 units W, 3 units X, 6 units Y.6. **Final Refill with Brand Z:** - When the tank is 3/4 empty, it has 3 units of gasoline left (1 unit Z, 0.25 units Y, 0.25 units W, and 1.5 units X). - It is then filled with 9 units of brand Z. - **Final Composition:** 10 units Z, 0.25 units Y, 0.25 units W, 1.5 units X.7. **Calculating the Percentage of Brand Z:** - **Total Gasoline in Tank:** 12 units. - **Amount of Brand Z:** 10 units. - **Percentage of Brand Z:** (10 units / 12 units) × 100 = 83.33%.Therefore, the percentage of brand Z gasoline in the tank is 83.33%.

💡Okay, so I have this equation here: 3 times (5 plus kx) equals 15x plus 15. I need to find the value of k such that there are infinitely many solutions for x. Hmm, infinitely many solutions usually means that the equation is an identity, right? Like, both sides are the same no matter what x is. So, if I can make the equation true for any x, then that would mean there are infinitely many solutions.Let me write down the equation again to make it clearer:3(5 + kx) = 15x + 15.First, I should probably expand the left side. So, distributing the 3, I get:3 * 5 + 3 * kx = 15x + 15.That simplifies to:15 + 3kx = 15x + 15.Now, looking at this, I can subtract 15 from both sides to simplify further:15 + 3kx - 15 = 15x + 15 - 15.Which simplifies to:3kx = 15x.Okay, so now I have 3kx equals 15x. For this equation to hold true for infinitely many x, the coefficients of x on both sides must be equal. That means 3k must equal 15. So, setting them equal:3k = 15.To solve for k, I divide both sides by 3:k = 15 / 3.Which simplifies to:k = 5.Wait, let me double-check that. If k is 5, then plugging back into the original equation:3(5 + 5x) = 15x + 15.Expanding the left side:15 + 15x = 15x + 15.Yes, that's the same as the right side, so it's an identity. That means any x would satisfy the equation, which is exactly what we want for infinitely many solutions.I think that makes sense. So, k has to be 5 to make the equation hold true for any x. If k were any other number, then the equation would either have one solution or no solution, but not infinitely many.Just to be thorough, let me consider what happens if k is not 5. Suppose k is 4. Then the equation becomes:3(5 + 4x) = 15x + 15.Expanding:15 + 12x = 15x + 15.Subtracting 15 from both sides:12x = 15x.Subtracting 12x:0 = 3x.Which implies x = 0. So, only one solution. Similarly, if k is 6, then:3(5 + 6x) = 15x + 15.Expanding:15 + 18x = 15x + 15.Subtracting 15:18x = 15x.Subtracting 15x:3x = 0.Again, x = 0. So, in both cases where k is not 5, we only get one solution. Therefore, k must be exactly 5 to have infinitely many solutions.I think that's solid. I don't see any mistakes in my reasoning. So, yeah, k equals 5 is the answer.

💡First, I need to find the prime factors of 6241. To do this, I'll start by checking if 6241 is a perfect square. Calculating the square root of 6241 gives approximately 79, but 79 squared is 6241, so 79 is a prime factor.Next, I'll divide 6241 by 79 to find the other factor. Dividing 6241 by 79 results in 79 again, indicating that 79 is the only prime factor of 6241.Therefore, the largest prime factor of 6241 is 79.

💡First, I need to determine how many distinct 4-digit numbers can be formed using the digits of 2033 that are greater than 3000. To do this, I'll consider the possible leading digits, which must be either 3 or 2 to ensure the number is greater than 3000.**Case 1: Leading digit is 3**- The remaining digits available are 2, 0, and 3.- Since the digit 3 is repeated, I'll calculate the permutations considering the repetition: [ frac{3!}{2!1!} = frac{6}{2} = 3 ]- This gives me 3 valid numbers when the leading digit is 3.**Case 2: Leading digit is 2**- The remaining digits available are 0, 3, and 3.- Again, since the digit 3 is repeated, the number of permutations is: [ frac{3!}{2!1!} = frac{6}{2} = 3 ]- This gives me 3 valid numbers when the leading digit is 2.**Conclusion**- Adding the numbers from both cases: [ 3 + 3 = 6 ]- Therefore, there are 6 valid 4-digit numbers that can be formed using the digits of 2033 and are greater than 3000.