answer:To solve the given problem, we follow these steps closely aligned with the provided solution: 1. **Identify the Focus and Directrix**: From the equation of the parabola C:y^{2}=2pxleft(p > 0right), we can determine the focus F and the point G on the directrix that intersects the x-axis. The focus of a parabola in this form is at F(frac{p}{2}, 0), and since the directrix is equidistant from the focus on the opposite side, G is at G(-frac{p}{2}, 0). 2. **Equation of Line AB**: A line with a slope of frac{4}{3} passing through point F(frac{p}{2}, 0) can be represented as y=frac{4}{3}(x-frac{p}{2}). 3. **Intersection Points**: To find where this line intersects the parabola, we solve the system of equations: [ left{ begin{array}{l} y=frac{4}{3}(x-frac{p}{2}) {y}^{2}=2px end{array} right. ] Substituting y from the first equation into the second and simplifying, we get: [ left(frac{4}{3}(x-frac{p}{2})right)^2 = 2px ] This simplifies to: [ 8x^{2}-17px+2p^{2}=0 ] Solving this quadratic equation for x, we find x=frac{p}{8} or x=2p. 4. **Coordinates of A and B**: Substituting x=2p into the equation of line AB, we find A is at (2p, 2p). Similarly, substituting x=frac{p}{8}, we find B is at left(frac{p}{8}, -frac{p}{2}right). 5. **Distances |GA| and |GB|**: To find |GA|, we use the distance formula: [ |GA| = sqrt{left(2p+frac{p}{2}right)^2 + (2p)^2} = frac{sqrt{41}}{2}p ] Similarly, for |GB|, we have: [ |GB| = sqrt{left(frac{p}{8}+frac{p}{2}right)^2 + left(-frac{1}{2}pright)^2} = frac{sqrt{41}}{8}p ] 6. **Ratio frac{|GA|}{|GB|}**: Finally, we calculate the ratio: [ frac{|GA|}{|GB|} = frac{frac{sqrt{41}}{2}p}{frac{sqrt{41}}{8}p} = 4 ] Therefore, the answer is boxed{4}.

answer:1. **Determine the radius**: Given the diameter is ( 8 ) meters, the radius ( r ) is half of the diameter, which is ( r = frac{8}{2} = 4 ) meters. 2. **Apply the area formula**: The area ( A ) of a circle is calculated by the formula ( A = pi r^2 ). 3. **Calculate the area**: Substituting the radius into the formula gives ( A = pi (4^2) = 16pi ). Conclusion: The area of the circle is ( boxed{16 pi} ) square meters.

answer:We are given the recurrence relation for (a_n): [a_{0} = sin x, quad a_{n} = (-1)^{leftlfloor frac{n}{2} rightrfloor} sqrt{1-a_{n-1}^2}] Here, (leftlfloor frac{n}{2} rightrfloor) represents the floor function, which denotes the integer part of (frac{n}{2}). Let's examine the behavior of the sequence (a_n) by observing a few initial cases. 1. For (n=0): [a_{0} = sin x] 2. For (n=1): [ leftlfloor frac{1}{2} rightrfloor = 0 quad Rightarrow quad a_{1} = (-1)^{0} sqrt{1-a_{0}^2} = cos x ] 3. For (n=2): [ leftlfloor frac{2}{2} rightrfloor = 1 quad Rightarrow quad a_{2} = (-1)^{1} sqrt{1-a_{1}^2} = -sin x ] 4. For (n=3): [ leftlfloor frac{3}{2} rightrfloor = 1 quad Rightarrow quad a_{3} = (-1)^{1} sqrt{1-a_{2}^2} = -cos x ] Notice that we have a cyclic pattern repeating every 4 steps. We can generalize this as follows: [ a_{n} = begin{cases} sin x & text{if } n = 4k cos x & text{if } n = 4k + 1 -sin x & text{if } n = 4k + 2 -cos x & text{if } n = 4k + 3 end{cases} ] Given (n = 1989), we can find the equivalence class of 1989 modulo 4: [ 1989 div 4 = 497 quad text{remainder } 1 ] So, (1989 equiv 1 pmod{4}), meaning (1989 = 4 cdot 497 + 1). Hence, [ a_{1989} = a_{4 cdot 497 + 1} = cos x ] # Conclusion: [ boxed{cos x text{ (D)}} ]

answer:We know from the property of determinants that the determinant of the product of matrices is equal to the product of their determinants. Thus, we apply this property to the given matrices mathbf{A}, mathbf{B}, and mathbf{C}. [ det (mathbf{A} mathbf{B} mathbf{C}) = (det mathbf{A})(det mathbf{B})(det mathbf{C}) ] Given: [ det mathbf{A} = 3, quad det mathbf{B} = 5, quad det mathbf{C} = 4 ] We substitute these values into our equation: [ det (mathbf{A} mathbf{B} mathbf{C}) = (3)(5)(4) = 60 ] Thus, the determinant of the product mathbf{A} mathbf{B} mathbf{C} is boxed{60}.