answer:To find the angles in the triangle with sides 3, 3, and sqrt{8}, use the Law of Cosines: For the angle opposite the side sqrt{8}, we use: [ cos C = frac{3^2 + 3^2 - (sqrt{8})^2}{2 times 3 times 3} = frac{9 + 9 - 8}{18} = frac{10}{18} = frac{5}{9}. ] Thus, [ C = cos^{-1}left(frac{5}{9}right). ] Using a calculator or trigonometric tables: [ C approx cos^{-1}(0.5556) approx 56.44^circ. ] Since the other two sides are equal (sides of 3 and 3), the angles opposite these sides are equal as well, call them A and B. Since the sum of angles in a triangle is 180^circ, [ A + B + 56.44^circ = 180^circ. ] Then, [ A + B = 123.56^circ. ] Since (A = B), [ 2A = 123.56^circ Rightarrow A = 61.78^circ. ] Thus the angles of the triangle are 56.44^circ, 61.78^circ, and 61.78^circ. Therefore, the final answer is [ boxed{56.44^circ, 61.78^circ, 61.78^circ}. ]

answer:Given the difference between the sum of interior angles and the sum of exterior angles of a polygon is 720^{circ}, and knowing that the sum of exterior angles of any polygon is always 360^{circ}, we can set up the following equation to find the sum of interior angles: text{Sum of interior angles} = 720^{circ} + 360^{circ} = 1080^{circ} Let n represent the number of sides of the polygon. The formula for the sum of interior angles of a polygon is (n-2) times 180^{circ}. Therefore, we can equate this to 1080^{circ} to solve for n: (n-2) times 180^{circ} = 1080^{circ} Solving for n involves dividing both sides of the equation by 180^{circ}: n - 2 = frac{1080^{circ}}{180^{circ}} = 6 Adding 2 to both sides to solve for n: n = 6 + 2 = 8 Therefore, the polygon is an octagon, which corresponds to option C. Thus, the correct answer is encapsulated as: boxed{C}

answer:Given the ellipse Gamma:frac{x^2}{a^2}+frac{y^2}{b^2}=1 (where a>b>0), its left focus is at F(-c,0), and its right vertex is at A(a,0). The distance between A and F is |AF|=a+c. Given that triangle AMF is a right isosceles triangle, we can deduce the coordinates of point M. Since AF is the hypotenuse and triangle AMF is isosceles, the legs AM and MF are equal. Therefore, the x-coordinate of M is halfway between A and F, and the y-coordinate of M is equal to the length of AM (or MF). This gives us Mleft(frac{a-c}{2}, frac{a+c}{2}right). Since M lies on the ellipse, we substitute its coordinates into the equation of the ellipse to get: [ frac{(a-c)^{2}}{4a^{2}}+frac{(a+c)^{2}}{4b^{2}}=1. ] Using the relationship between a, b, and c in an ellipse, where c^2=a^2-b^2 and the eccentricity e=frac{c}{a}, we can rewrite the equation as: [ frac{1}{4}(1-e)^{2}+frac{1}{4}cdot frac{1+e}{1-e}=1. ] Simplifying this equation, we obtain: [ e^{3}-3e^{2}-2e+2=0. ] Factoring this cubic equation, we get: [ (e+1)(e^{2}-4e+2)=0. ] Since the eccentricity e of an ellipse is between 0 and 1 (0<e<1), we solve the quadratic equation e^{2}-4e+2=0 to find the value of e. The solutions to this equation are e=2pmsqrt{2}. However, since 0<e<1, the only valid solution is e=2-sqrt{2}. Therefore, the eccentricity of the ellipse Gamma is boxed{2-sqrt{2}}, which corresponds to option C.

answer:Let each side of the equilateral triangle be of length (s). The formula for the area of an equilateral triangle is (frac{sqrt{3}}{4}s^2). The distance from the center (centroid) to the midpoint of any side of an equilateral triangle, which acts as the radius (r) of the circle that Fido can reach, is (frac{s}{sqrt{3}}). 1. **Calculate the area of the triangle**: [ text{Area of triangle} = frac{sqrt{3}}{4}s^2 ] 2. **Calculate the area Fido can reach**: [ text{Area reachable} = pi left(frac{s}{sqrt{3}}right)^2 = frac{pi s^2}{3} ] 3. **Calculate the fraction of the area reachable**: [ text{Fraction reachable} = frac{frac{pi s^2}{3}}{frac{sqrt{3}}{4}s^2} = frac{4pi s^2}{3sqrt{3}s^2} = frac{4pi}{3sqrt{3}} = frac{4sqrt{3}}{9}pi ] Thus, (a = 3) and (b = 9), and (ab = 3 cdot 9 = boxed{27}).