answer:By using the fundamental trigonometric identity for the same angle, we know that sin^2 alpha + cos^2 alpha = 1. Therefore, the correct choice is: boxed{C}

answer:1. **Identifying the Triangle Type**: Given side lengths 5, 12, 13, we recognize triangle ABC as a right triangle because 5^2 + 12^2 = 13^2. 2. **Folding Point A to Point C**: The crease formed by folding such that point A falls on point C will be a line segment that is equidistant from A and C. This is the perpendicular bisector of segment AC. 3. **Finding Midpoint E of AC**: Assuming A is at the origin (0,0) and C is on the x-axis at (12,0), the midpoint E of AC is at left(frac{12}{2}, 0right) = (6, 0). 4. **Equation of the Perpendicular Bisector**: Since AC is horizontal, the perpendicular bisector of AC is a vertical line passing through E. The equation of this vertical line is x = 6. 5. **Intersection with BC**: If B is at (0,5), the slope of BC is frac{5}{12}. The equation of BC is y = frac{5}{12}x + 5. Setting x = 6 in the equation of BC gives y = frac{5}{12}(6) + 5 = 7.5. The intersection point G on BC is (6, 7.5). 6. **Length of the Crease (EG)**: The length of the crease is the distance between E and G. Using the distance formula: [ EG = sqrt{(6 - 6)^2 + (7.5 - 0)^2} = sqrt{0 + 56.25} = 7.5 ] Thus, the length of the crease is 7.5. The final answer is boxed{textbf{(C)} 7.5}

answer:1. **Define the Function:** Given the function ( f(n) = frac{6^n}{n!} ), we need to evaluate it for different values of ( n ) to determine the smallest positive integer ( C ) such that ( f(n) leq C ) for all positive integers ( n ). 2. **Evaluate for Small ( n ):** - For ( n = 1 ): [ f(1) = frac{6^1}{1!} = 6 ] - For ( n = 2 ): [ f(2) = frac{6^2}{2!} = frac{36}{2} = 18 ] - For ( n = 3 ): [ f(3) = frac{6^3}{3!} = frac{216}{6} = 36 ] - For ( n = 4 ): [ f(4) = frac{6^4}{4!} = frac{1296}{24} = 54 ] - For ( n = 5 ): [ f(5) = frac{6^5}{5!} = frac{7776}{120} = 64.8 ] 3. **Analyze for ( n geq 6 ):** - For ( n = 6 ): [ f(6) = frac{6^6}{6!} = frac{46656}{720} = 64.8 ] - We observe that ( f(6) leq 64.8 ). 4. **General Expression for ( n geq 6 ):** For ( n geq 6 ), [ f(n) = frac{6^n}{n!} ] We need to evaluate whether ( f(n) ) decreases for ( n geq 6 ). This can be observed through the inequality, [ frac{6^{n+1}}{(n+1)!} = frac{6^n cdot 6}{n! cdot (n+1)} = frac{6}{n+1} cdot frac{6^n}{n!} = frac{6}{n+1} cdot f(n) ] Since ( frac{6}{n+1} ) is less than 1 for ( n geq 6 ), it shows that ( f(n) ) is decreasing for ( n geq 6 ). 5. **Conclusion:** The largest ( f(n) ) for ( n = 1, 2, 3, 4, 5 ) is 64.8, and for ( n geq 6 ), ( f(n) leq 64.8 ). Thus, the smallest positive integer ( C ) such that ( f(n) leq C ) for all ( n in mathbb{Z}^+ ) is: [ boxed{65} ]

answer:Given the complex number 2-bi where b in mathbb{R}, we are told that the sum of its real part and its imaginary part equals zero. This means we can set up the following equation based on the given information: [ text{Real part} + text{Imaginary part} = 0 ] For the complex number 2-bi, the real part is 2 and the imaginary part is -b (since the imaginary part of a complex number a+bi is b). Therefore, we substitute these into our equation: [ 2 + (-b) = 0 ] Simplifying the equation gives us: [ 2 - b = 0 ] To find the value of b, we solve for b: [ 2 = b ] Thus, the value of b that satisfies the given condition is 2. Looking at the options provided: A: 2 B: frac{2}{3} C: -frac{2}{3} D: -2 We see that the correct answer is A: 2. Therefore, the final answer, encapsulated as requested, is: [ boxed{A} ]