answer:Given the ellipse C:frac{{x}^{2}}{{a}^{2}}+frac{{y}^{2}}{{b}^{2}}=1 where a>b>0, and the foci F_{1} and F_{2}, with point P on the ellipse such that PF_{2}bot F_{1}F_{2}. A line is drawn perpendicularly from P to F_{1}P, intersecting the x-axis at point A, and it is given that |AF_{2}|=frac{1}{2}c. The eccentricity of the ellipse is denoted as e. First, by applying the projection theorem to the given scenario, we can express the relationship between the distances as follows: [{|PF}_{2}|^{2}=|F_{1}F_{2}|cdot |AF_{2}|.] Given that |F_{1}F_{2}|=2c and |AF_{2}|=frac{1}{2}c, we substitute these values into the equation to find: [{|PF}_{2}|^{2}=2ctimes frac{1}{2}c = c^{2}.] Taking the square root of both sides gives us |PF_{2}|=c. Next, we consider the right triangle triangle PF_{1}F_{2}, where |PF_{1}| can be found using the Pythagorean theorem: [|PF_{1}|=sqrt{{c}^{2}+{(2c)}^{2}}=sqrt{5}c.] According to the definition of an ellipse, the sum of the distances from any point on the ellipse to the foci is constant, which is 2a for this ellipse. Therefore, we have: [sqrt{5}c+c=2a.] Solving for a gives us a=frac{(sqrt{5}+1)c}{2}. The eccentricity e of the ellipse is defined as e=frac{c}{a}. Substituting the expression for a we found earlier gives us: [e=frac{c}{frac{(sqrt{5}+1)c}{2}}=frac{2}{sqrt{5}+1}.] Multiplying the numerator and the denominator by sqrt{5}-1 to rationalize the denominator, we get: [e=frac{2(sqrt{5}-1)}{(sqrt{5}+1)(sqrt{5}-1)}=frac{sqrt{5}-1}{2}.] Squaring both sides to find e^{2}, we obtain: [e^{2}=left(frac{sqrt{5}-1}{2}right)^{2}=frac{5-2sqrt{5}+1}{4}=frac{3-sqrt{5}}{2}.] Therefore, the correct answer is boxed{A}.

answer:First, let's complete the square for g(x): [ g(x) = 3x^2 + 6x - 9 = 3(x^2 + 2x) - 9 = 3((x+1)^2 - 1) - 9 = 3(x+1)^2 - 12 ] The vertex form of the parabola is now clear: g(x) = 3(x+1)^2 - 12, and the vertex is at x = -1. This parabola opens upwards (since the coefficient of (x+1)^2 is positive), making g(x) increasing when x > -1 and decreasing when x < -1. Since we need to include x = -1 in the domain and make g(x) invertible, the two possible domains for g(x) to be invertible are: - decreasing from (-infty, -1] - increasing from [-1, +infty) To satisfy the point x = -1 being part of the interval and g(x) being invertible, we choose boxed{(-infty, -1]}.

answer:We start by simplifying the expression sqrt{4! cdot 4!}: 1. Recall that 4! = 4 times 3 times 2 times 1 = 24. 2. Thus, 4! cdot 4! = 24 times 24 = 576. 3. Compute the square root of 576: [ sqrt{576} = 24 ] Therefore, the value of sqrt{4! cdot 4!}, expressed as a positive integer, is boxed{24}.

answer:The decimal representation of frac{n}{570} terminates if and only if the denominator in its simplest form has only 2 and 5 as prime factors, after canceling out common factors with the numerator. The prime factorization of 570 is 2 cdot 3 cdot 5 cdot 19. To ensure the fraction frac{n}{570} terminates, the common factors must eliminate the primes 3 and 19. Therefore, we need to find n such that: 1. n is divisible by both 3 and 19 to cancel these primes from the denominator. Since 3 and 19 are coprime, n must be divisible by their product, 3 cdot 19 = 57. So we are seeking multiples of 57 within the range from 1 to 569. To find how many multiples of 57 are there between 1 and 569, calculate: - frac{569}{57} = 9.982... The floor of 9.982 is 9. Consequently, there are 9 multiples of 57 within this range. Thus, there are boxed{9} integer values of n for which frac{n}{570} has a terminating decimal representation.