answer:1. **Analyze Distance Relationship**: Given DQ=7, EQ=9, FQ=11, explore if any triangle formed with these distances is a right triangle: [ DQ^2 + EQ^2 = 7^2 + 9^2 = 49 + 81 = 130 neq 11^2 = 121. ] Therefore, triangle DQE is not a right triangle, and neither triangle DEF forms direct recognizable angles. 2. **Explore Triangle Relationship**: Since no right triangle can be directly formed, consider relationships that could relate to locating Q. Rotate triangle DQF about D by 60^circ to point Q'. 3. **Investigate Transformations and Triangle Calculations**: With Q inside equilateral triangle DEF, and given angle DQE is not 90^circ, we need a different approach for area calculations. 4. **Applying Law of Cosines in triangle DQF**: [ DF^2 = DQ^2 + FQ^2 - 2 cdot DQ cdot FQ cdot cos(angle DQF), ] As angle DQF = 60^circ in equilateral triangle: [ DF^2 = 49 + 121 - 2 cdot 7 cdot 11 cdot frac{1}{2} = 170 - 77 = 93. ] Since DF should equal DE = EF, all sides are sqrt{93}. 5. **Area Calculation**: Calculate the area of triangle DEF: [ text{Area} = frac{sqrt{3}}{4}(DF^2) = frac{sqrt{3}}{4} times 93 = frac{93sqrt{3}}{4}. ] Approximating sqrt{3} approx 1.732, the area becomes: [ frac{93 cdot 1.732}{4} approx 40.317. ] Thus, the area of triangle DEF to the nearest integer is 40. The final answer is boxed{B. 40}

answer:Given the hyperbola C:frac{x^2}{a^2}-frac{y^2}{b^2}=1 with a>0 and b>0, and the information about the left vertex A and the right focus F, where the focal length is 12, we proceed as follows: 1. **Focal Length Calculation**: The focal length of the hyperbola is given as the distance between the two foci, which is 2c=12. Therefore, we calculate the value of c as: [ 2c = 12 implies c = frac{12}{2} = 6. ] 2. **Relation Between MF and AF**: Given that MF perp AF and |MF| = 2|AF|, we can use the properties of the hyperbola to express MF and AF in terms of a, b, and c. For a hyperbola, the distance from a point on the hyperbola to a focus (MF in this case) can be expressed as frac{b^2}{a} when perpendicular to the line connecting the vertex and the focus. Also, |AF| = a + c. Therefore, we have: [ |MF| = frac{b^2}{a} quad text{and} quad |AF| = a + c. ] Given that |MF| = 2|AF|, we substitute the expressions for |MF| and |AF|: [ frac{b^2}{a} = 2(a + c). ] 3. **Solving for a**: We know that for a hyperbola, c^2 = a^2 + b^2. Substituting c = 6 and rearranging the equation frac{b^2}{a} = 2(a + c), we get: [ b^2 = 2a^2 + 2ac. ] Substituting b^2 = c^2 - a^2 into the equation, we get: [ c^2 - a^2 = 2a^2 + 2ac. ] Substituting c = 6 into the equation, we have: [ 36 - a^2 = 2a^2 + 12a. ] Rearranging the equation, we get: [ 3a^2 + 12a - 36 = 0. ] Solving this quadratic equation for a, we find that a = 2. Therefore, the correct answer is boxed{text{B}}.

answer:1. Start with z = g(w), which gives: [ z = frac{2w}{1+w} ] 2. To solve for w in terms of z: [ z(1+w) = 2w implies z + zw = 2w implies zw - 2w + z = 0 implies w(z-2) = -z ] [ w = frac{-z}{z-2} ] 3. Evaluate the options: - For different functions, apply g and transformations of z: - gleft(frac{1}{z}right) = frac{frac{2}{z}}{1+frac{1}{z}} = frac{2}{z+1} - -g(z) = -left(frac{2z}{1+z}right) = frac{-2z}{1+z} - g(-z) = frac{-2z}{1-z} - -g(-z) = -frac{-2z}{1-z} = frac{2z}{1-z} 4. By comparison, the expression for w: [ w = frac{-z}{z-2} ] None of the calculated options match directly; hence, introducing new options based on the solution obtained: - frac{-z}{z-2} corresponds directly to a different function or manipulation of z not listed originally. Conclusion: Given the reevaluation, the correctly matching expression for w is text{None of the original options. frac{-z}{z-2}}, and additional options should be introduced. The final answer is (E) boxed{frac{-z}{z-2}}

answer:Given the function ( g ), defined on the set of integers, satisfying the conditions: 1. ( g(1) > 1 ) 2. ( g(x+y) + x g(y) + y g(x) = g(x) g(y) + x + y + xy ) for any ( x, y in mathbb{Z} ) 3. ( 3 g(x) = g(x+1) + 2x - 1 ) for any ( x in mathbb{Z} ) We need to find ( g(5) ). 1. Let ( g(1) = a ). Using condition 3, when ( x = 1 ): [ 3g(1) = g(2) + 2 cdot 1 - 1 Rightarrow 3a = g(2) + 1 Rightarrow g(2) = 3a - 1 ] 2. Using condition 3 again, when ( x = 2 ): [ 3g(2) = g(3) + 2 cdot 2 - 1 Rightarrow 3(3a - 1) = g(3) + 4 - 1 Rightarrow 9a - 3 = g(3) + 3 Rightarrow g(3) = 9a - 6 ] 3. Using condition 3 again, when ( x = 3 ): [ 3g(3) = g(4) + 2 cdot 3 - 1 Rightarrow 3(9a - 6) = g(4) + 6 - 1 Rightarrow 27a - 18 = g(4) + 5 Rightarrow g(4) = 27a - 23 ] 4. Using condition 3 again, when ( x = 4 ): [ 3g(4) = g(5) + 2 cdot 4 - 1 Rightarrow 3(27a - 23) = g(5) + 8 - 1 Rightarrow 81a - 69 = g(5) + 7 Rightarrow g(5) = 81a - 76 ] 5. To determine ( a ), we use condition 2. Set ( x = 4 ) and ( y = 1 ): [ g(4+1) + 4g(1) + 1g(4) = g(4)g(1) + 4 + 1 + 4 cdot 1 Rightarrow g(5) + 4a + g(4) = g(4)a + 9 ] Substitute ( g(4) = 27a - 23 ) and ( g(5) = 81a - 76 ): [ 81a - 76 + 4a + 27a - 23 = (27a - 23)a + 9 Rightarrow 112a - 99 = 27a^2 - 23a + 9 ] 6. Simplify the resulting equation: [ 27a^2 - 23a + 9 = 112a - 99 Rightarrow 27a^2 - 23a + 9 - 112a + 99 = 0 Rightarrow 27a^2 - 135a + 108 = 0 ] 7. Solve the quadratic equation ( 27a^2 - 135a + 108 = 0 ): [ a^2 - 5a + 4 = 0 quad text{(Divide by 27)} ] 8. Factorize the quadratic equation: [ (a - 4)(a - 1) = 0 ] Hence, solutions are ( a = 4 ) or ( a = 1 ). 9. Given that ( g(1) = a > 1 ), we discard ( a = 1 ). Therefore, ( a = 4 ). 10. Finally, substitute ( a = 4 ) into ( g(5) = 81a - 76 ): [ g(5) = 81 cdot 4 - 76 = 324 - 76 = 248 ] Conclusively, ( boxed{248} )