answer:To determine how many two-digit numbers have at least one digit that is a 4, let's analyze the two-digit numbers systematically: Step 1: Count numbers with 4 as the tens digit - The tens digit ranges from 1 to 9 for two-digit numbers. - If the tens digit is 4, the possible numbers are 40, 41, 42, 43, 44, 45, 46, 47, 48, and 49. Since there are 10 possible numbers: [ 40, 41, 42, 43, 44, 45, 46, 47, 48, text{ and } 49 ] Thus, we have: [ text{Total numbers with tens digit 4} = 10 ] Step 2: Count numbers with 4 as the ones digit - The ones digit also ranges from 0 to 9. - If the ones digit is 4, the possible numbers are 14, 24, 34, 44, 54, 64, 74, 84, and 94. Since there are 9 possible numbers: [ 14, 24, 34, 44, 54, 64, 74, 84, text{ and } 94 ] Thus, we have: [ text{Total numbers with ones digit 4} = 9 ] Step 3: Adjust for overlap - There is an overlap between numbers counted in both Step 1 and Step 2, specifically the number 44. Therefore, we must subtract this overlap once to avoid double-counting: [ text{Overlap} = 1 ] Step 4: Calculate the total number Using the principle of inclusion and exclusion: [ text{Total} = (text{Numbers with 4 as tens digit}) + (text{Numbers with 4 as ones digit}) - (text{Overlap}) ] [ text{Total} = 10 + 9 - 1 = 18 ] # Conclusion: The number of two-digit numbers that have at least one digit that is a 4 is: [ boxed{18} ]

answer:Using the sine addition formula, sin(alpha + beta) = sin alpha cos beta + cos alpha sin beta, we can rewrite the expression as sin(10^{circ} + 50^{circ}) = sin 60^{circ}. Since sin 60^{circ} = dfrac{sqrt{3}}{2}, the correct answer is boxed{text{C}}.

answer:For semi-annual compounding: [A_1 = 8000 left(1 + frac{0.15}{2}right)^{2 times 3} = 8000 left(1 + 0.075right)^6 = 8000 times (1.075)^6 = 8000 times 1.50756784 = 12060.54] For monthly compounding: [A_2 = 8000 left(1 + frac{0.15}{12}right)^{12 times 3} = 8000 left(1 + 0.0125right)^{36} = 8000 times (1.0125)^36 = 8000 times 1.5314067 = 12251.25] The difference: [A_2 - A_1 = 12251.25 - 12060.54 = boxed{190.71}]

answer:Let's denote a_n = |A_{n-1}A_n|. Given that B_n lies on y = x^2, the height of the equilateral triangle A_{n-1}B_nA_n based on its base a_n is a_n frac{sqrt{3}}{2}. Since B_n lies on y=x^2, the y-coordinate of B_n is (x_{n-1} + frac{a_n}{2})^2. Consequently, we have: [ left(a_n frac{sqrt{3}}{2}right)^2 = left(x_{n-1} + frac{a_n}{2}right)^2 ] Expanding and simplifying, we find: [ frac{3}{4} a_n^2 = x_{n-1}^2 + a_n x_{n-1} + frac{a_n^2}{4} ] [ frac{1}{2} a_n^2 = x_{n-1}^2 + a_n x_{n-1} ] This implies a relationship involving x_{n-1}, which is dependent on x_{n-2} and so forth, leading back to x_0 = 0. From here, rather than solving the quadratic, observe that for large n, the terms x_{n-1} and a_n increase, and the relationship tends towards simpler forms. Here, as n increases, the term frac{a_n^2}{2} approximately equals a_n x_{n-1}, giving a_n approx 2x_{n-1}, an approximation that simplifies with larger n. Using this approximation iteratively, we seek n such that: [ A_0A_n = a_1 + a_2 + cdots + a_n approx frac{n(n+1)}{2} ] Solving frac{n^2 + n}{2} geq 50, we find n^2 + n - 100 geq 0. The positive root of this equation is approximately n approx 9.3. Thus, the smallest integer n geq 10. Conclusion: The smallest integer n such that A_0A_n geq 50 is n = boxed{10}.