answer:Since lfloor x rfloor represents the greatest integer less than or equal to x, we look at the number 5.7. The greatest integer less than or equal to 5.7 is 5. Therefore, lfloor 5.7 rfloor = boxed{5}.

answer:Assume the park is oriented so one side runs along the x-axis. Let's establish Alice's starting point as the origin (0,0) on a coordinate plane. If the sides of the hexagon are each 3 km, then: - Alice walks the first side from (0,0) to (3,0) (3 text{ km}), - Continues along the second side, transitioning to (3, 3sqrt{3}) (3sqrt{3} text{ km, since it's a shift from } (3,0) text{ to } (6, sqrt{3}) text{ and then perpendicular}), - Keeps going along the third side to (-3, 3sqrt{3}) (6 text{ km along the x-axis, totaling } 3 + 3sqrt{3} + 6). Now, Alice has walked 3 + 3sqrt{3} + 6 = 9 + 3sqrt{3} km. She has 10 - (9 + 3sqrt{3}) approx 1 - 5.2 approx -4.2 km left, so she turns around and walks back 4.2 km towards (3, 3sqrt{3}). The rough estimate for the final coordinates after retracing is between (0, 3sqrt{3}) text{ and } (-3, 3sqrt{3}). We calculate the final coordinates precisely, but given the distance walked back, a better approximation is near (0, 3sqrt{3}). Compute the Euclidean distance from the origin using these estimated final coordinates: [ text{Distance} = sqrt{(0-0)^2 + (3sqrt{3}-0)^2} = sqrt{0 + 27} = sqrt{27} = 3sqrt{3} text{ km}.] Convert 3sqrt{3} with decimal expansion sqrt{27} approx sqrt{27}, which is approximately sqrt{28} text{ km for approximate comparison}. In options given, closest is boxed{sqrt{27}} (not listed exactly but best estimate method).

answer:: 1. **Given Information:** - We are dealing with an isosceles triangle ( ABC ) where ( AB = BC ). - The angle at the base is ( alpha ). 2. **Find the ratio of the area of the triangle to the area of its circumscribed circle.** - Let ( R ) be the radius of the circumscribed circle around the triangle. - The area of the circumscribed circle is given by: [ S_{kp} = pi R^2 ] 3. **Express triangle dimensions in terms of ( x ) and ( alpha ):** - Let the sides ( AB ) and ( BC ) be ( x ). - The angle ( angle ABC ) (vertex angle) becomes: [ angle ABC = 180^circ - 2alpha ] - The angle ( angle OBC ) is: [ angle OBC = 90^circ - alpha ] 4. **Locate points and angles in the right triangle ( BEO ):** - ( BO ) is the radius ( R ) of the circumscribed circle. - Using properties of the circle and sine rule in ( triangle BEO ): [ BO = frac{BE}{sin alpha} ] - Since ( BE = frac{x}{2} ) (median of the isosceles triangle): [ BO = frac{frac{x}{2}}{sin alpha} = frac{x}{2 sin alpha} ] 5. **Write the area of the circumscribed circle in terms of ( x ) and ( alpha ):** - Using the previous result: [ S_{kp} = pi R^2 = pi BO^2 = pi left( frac{x}{2 sin alpha} right)^2 = frac{pi x^2}{4 sin^2 alpha} ] 6. **Calculate the area of ( triangle ABC ):** - Area formula for any triangle given two sides and included angle: [ S_{triangle ABC} = frac{1}{2} times AB times BC times sin( angle ABC ) ] - Substituting the values: [ S_{triangle ABC} = frac{1}{2} times x times x times sin(180^circ - 2 alpha) ] - Since ( sin(180^circ - 2 alpha) = sin 2 alpha ): [ S_{triangle ABC} = frac{x^2}{2} sin 2 alpha ] 7. **Form the ratio of the triangle's area to the circle's area:** [ frac{S_{triangle ABC}}{S_{kp}} = frac{frac{x^2}{2} sin 2 alpha}{frac{pi x^2}{4 sin^2 alpha}} ] 8. **Simplify the expression:** [ = frac{frac{x^2}{2} sin 2 alpha cdot 4 sin^2 alpha}{pi x^2} ] [ = frac{2 sin 2 alpha cdot sin^2 alpha}{pi} ] # Conclusion: The ratio of the area of the triangle to the area of its circumscribed circle is: [ boxed{frac{2 sin 2 alpha cdot sin^2 alpha}{pi}} ]

answer:1. **Analyze the quadratic expression**: The function given is (f(x) = frac{1}{lfloor x^2 - 8x + 20 rfloor}). First, observe the discriminant of (x^2 - 8x + 20): [ text{Discriminant} = 8^2 - 4 cdot 1 cdot 20 = 64 - 80 = -16 ] Since the discriminant is negative, the quadratic has no real roots and (x^2 - 8x + 20 > 0) for all real (x). 2. **Determine key intervals**: The function (f(x)) is undefined where (lfloor x^2 - 8x + 20 rfloor = 0). Therefore, we need to find when (0 leq x^2 - 8x + 20 < 1). - To find when (x^2 - 8x + 20 = 1), solve (x^2 - 8x + 19 = 0): [ x = frac{-(-8) pm sqrt{(-8)^2 - 4 cdot 1 cdot 19}}{2 cdot 1} = frac{8 pm sqrt{64 - 76}}{2} = frac{8 pm sqrt{-12}}{2} ] This equation has no real solutions, implying (x^2 - 8x + 20) does not reach 1. 3. **Conclude on the domain**: Since (x^2 - 8x + 20) never reaches values between 0 and 1, (lfloor x^2 - 8x + 20 rfloor neq 0) for all real (x). Therefore, (f(x)) is defined for all real numbers. [ boxed{(-infty, infty)} ]