answer:(1) First, we will find the intersection point of l_{1} and l_{2}. To do this, we will rewrite l_{1} in the form y=sqrt{3}x + 2sqrt{3} and use the given parametric form of l_{2} to obtain the equations: begin{cases} y = sqrt{3}x + 2sqrt{3} y = -sqrt{3}x end{cases} Solving this system of equations, we find the intersection point P(-1, sqrt{3}). Converting this point to polar coordinates, we get P(2, frac{2π}{3}). (2) Now, we will find the value of the expression frac{1}{|OA|^{2}}+ frac{1}{|OB|^{2}}+ frac{1}{|OC|^{2}}. We'll use the polar coordinate system with the origin as the pole and the positive x-axis as the polar axis. Substituting x=ρcos θ and y=ρsin θ into the ellipse equation, we get: frac{1}{ρ^{2}}= frac{cos^{2}θ}{4}+sin^{2}θ Let A(ρ_{1},θ), B(ρ_{2},θ+120^{circ}), and C(ρ_{3},θ-120^{circ}). Using these points and the equation above, we have: begin{aligned} frac{1}{|OA|^{2}}+ frac{1}{|OB|^{2}}+ frac{1}{|OC|^{2}} &= frac{1}{ρ_{1}^{2}}+ frac{1}{ρ_{2}^{2}}+ frac{1}{ρ_{3}^{2}} &= frac{1}{4} left[ cos^{2}θ + cos^{2}(θ+120^{circ}) + cos^{2}(θ-120^{circ}) right] &quad + left[ sin^{2}θ + sin^{2}(θ+120^{circ}) + sin^{2}(θ-120^{circ}) right] &= frac{1}{4} left[ cos^{2}θ + frac{1}{4}(cos θ + sqrt{3} sin θ)^{2} + frac{1}{4}(cos θ - sqrt{3} sin θ)^{2} right] &quad + sin^{2}θ + frac{1}{4}(-sin θ + sqrt{3} cos θ)^{2} + frac{1}{4}(sin θ + sqrt{3} cos θ)^{2} &= boxed{frac{15}{8}} end{aligned}

answer:1. **Prime Factorization of 60**: [ 60 = 2^2 cdot 3 cdot 5 ] This factorization shows that 60 is composed of the square of 2, 3, and 5. 2. **Factors of 60**: The total number of factors of a number given its prime factorization p^a cdot q^b cdot r^c is (a+1)(b+1)(c+1). For 60, this gives: [ (2+1)(1+1)(1+1) = 3 cdot 2 cdot 2 = 12 ] These factors are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. 3. **Multiples of 6 among the Factors**: A multiple of 6 must be divisible by both 2 and 3. Therefore, we need to identify factors of 60 that include at least 2^1 cdot 3^1. From the list of factors, we identify those that are multiples of 6: - 6 = 2 cdot 3 - 12 = 2^2 cdot 3 - 30 = 2 cdot 3 cdot 5 - 60 = 2^2 cdot 3 cdot 5 4. **Counting the Multiples of 6**: We have identified 4 factors of 60 that are multiples of 6: 6, 12, 30, and 60. Thus, there are 4 positive factors of 60 that are also multiples of 6. The correct answer is 4. The final answer is boxed{C. 4}

answer:First, we convert 100111001_6 from base-6 to a decimal representation. This can be written as: 100111001_6 = 6^8 + 6^5 + 6^4 + 6^3 + 6 + 1 Letting x = 6, the number is: x^8 + x^5 + x^4 + x^3 + x + 1 Again, using the Rational Root Theorem, x+1 is a factor. The polynomial simplifies as: x^8 + x^5 + x^4 + x^3 + x + 1 = (x+1)(x^7 - x^6 + x^5 + x^3 + 1) The x^7 - x^6 + x^5 + x^3 + 1 term can be factored further by adding and subtracting terms and using binomial expansions, as well as grouping: (x+1)(x^5(x^2 - x + 1) + (x+1)(x^2 - x + 1)) = (x+1)(x^2 - x + 1)(x^5 + x + 1) Continue to factor the quintic polynomial: x^5 + x + 1 = x^5 - x^2 + x^2 + x + 1 = x^2(x^3 - 1) + x^2 + x + 1 = (x^2 + x + 1)(x^3 - x^2 + 1) Thus, the complete factorization is: (x+1)(x^2 - x + 1)(x^2 + x + 1)(x^3 - x^2 + 1) Substitute x = 6: 7 cdot 31 cdot 43 cdot 169 Here, 169 is 13^2 and hence not a prime. Checking the primes, the largest prime factor is boxed{43}.

answer:1. **Error Identification**: Ron reversed a. Let's denote the reversed number as a' with the erroneous product being 221: [ a' cdot b = 221 ] 2. **Factorize 221**: To find possible values of a' and b, we factorize 221: [ 221 = 13 times 17 ] This suggests possible pairs (a', b) as (13, 17) or (17, 13). 3. **Determine Correct a and b**: a' must be a two-digit number. The valid pairs having a' as a two-digit number are (13, 17) or (17, 13). Given either choice, we convert to identify a via reversing digits: - If a' = 13, then a = 31. - If a' = 17, then a = 71. 4. **Calculate the Correct Products**: Depending on the determination from the previous step, compute ab: - ab = 31 times 17 = 527 - ab = 71 times 13 = 923 Identifying the possible calculations, the individual values depend on selecting the pair which stresses the numeric relations correctly. 5. **Conclusion**: Assuming a' is the simpler digit reversal, 71 reversed to 17 seems credible, so the correct product of a and b should be 923. The final answer is boxed{textbf{(B)} 923}