answer:Given non-zero numbers ( a, b, c ) such that any two of the three equations ( a x^{11} + b x^{4} + c = 0 ), ( b x^{11} + c x^{4} + a = 0 ), ( c x^{11} + a x^{4} + b = 0 ) share a common root. We want to prove that all three equations have a common root. 1. Since no root can be zero (due to non-zero constant terms (c, a, b)), let ( p ) be a common root of the first two equations: [ a p^{11} + b p^{4} + c = 0 ] [ b p^{11} + c p^{4} + a = 0 ] 2. To eliminate ( p^{11} ) and obtain a relation between the equations, we multiply the first equation by ( b ) and the second by ( a ): [ b(a p^{11} + b p^{4} + c) = 0 ] [ a(b p^{11} + c p^{4} + a) = 0 ] 3. This yields two new equations: [ a b p^{11} + b^2 p^{4} + b c = 0 ] [ a b p^{11} + a c p^{4} + a^2 = 0 ] 4. Subtract the second equation from the first to eliminate ( a b p^{11} ): [ b^2 p^{4} + b c - a c p^{4} - a^2 = 0 ] [ p^{4} (b^2 - a c) + b c - a^2 = 0 ] 5. Consequently, we get: [ p^{4} = frac{a^2 - b c}{b^2 - a c} ] 6. Now, consider the common root ( p ) and the third equation: [ c p^{11} + a p^{4} + b = 0 ] 7. Similar to the previous derivation, let’s consider another combination: [ b(b p^{11} + c p^{4} + a) - c(a p^{11} + b p^{4} + c) = 0 ] 8. Simplify to: [ b^2 p^{11} + b c p^{4} + a b - c a p^{11} - b c p^{4} - c^2 = 0 ] [ p^{11} (b^2 - a c) + a b - c^2 = 0 ] [ p^{11} = frac{c^2 - a b}{b^2 - a c} ] 9. We now have two equations: [ p^{4} = frac{a^2 - b c}{b^2 - a c} ] [ p^{11} = frac{c^2 - a b}{b^2 - a c} ] 10. For consistency, the absolute values of these expressions must satisfy both: - Either: [ |p|^{4} = left|frac{a^2 - b c}{b^2 - a c}right| leq 1 ] and: [ |p|^{11} = left|frac{c^2 - a b}{b^2 - a c}right| geq 1 ] - Or the reverse: [ |p|^{4} = left|frac{a^2 - b c}{b^2 - a c}right| geq 1 ] and: [ |p|^{11} = left|frac{c^2 - a b}{b^2 - a c}right| leq 1 ] 11. This intermediate condition implies ( |p| = 1 ) and consequently: [ left|frac{a^2 - b c}{b^2 - a c}right| = 1 ] and: [ left|frac{c^2 - a b}{b^2 - a c}right| = 1 ] 12. Therefore, if the expressions involving ( a, b, c ) result in ( p =1 ) (or similarly ( p = -1 )), these would cover all three given equations as the common root. Hence, we conclude that all three equations: [ a x^{11} + b x^{4} + c = 0, quad b x^{11} + c x^{4} + a = 0, quad c x^{11} + a x^{4} + b = 0 ] have a shared common root by symmetry of ( p= 1 ) or ( p = -1 ). [ boxed{text{All three equations share a common root}} ]

answer:Mike picked 7 apples, Nancy picked 3 apples, and Keith picked 6 apples. To find the total number of apples picked, you add the number of apples each person picked: Mike's apples + Nancy's apples + Keith's apples = Total apples 7 + 3 + 6 = 16 So, in total, boxed{16} apples were picked.

answer:Based on the given information, vectors overrightarrow{a} and overrightarrow{b} are not collinear, and we have overrightarrow{AB}= overrightarrow{a}+2 overrightarrow{b}, overrightarrow{BC}=-4 overrightarrow{a}- overrightarrow{b}, overrightarrow{CD}=-5 overrightarrow{a}-3 overrightarrow{b}. To find vector overrightarrow{AD}, we can sum up the vectors from A to D: overrightarrow{AD} = overrightarrow{AB} + overrightarrow{BC} + overrightarrow{CD} = overrightarrow{a}+2 overrightarrow{b} - 4 overrightarrow{a} - overrightarrow{b} - 5 overrightarrow{a} - 3 overrightarrow{b}. Simplifying the above expression, we get: overrightarrow{AD} = -8 overrightarrow{a} - 2 overrightarrow{b}. Analyzing the result, we find: overrightarrow{AD} = 2 (-4 overrightarrow{a} - overrightarrow{b}) = 2 overrightarrow{BC}. This means that line segment AD is parallel to line segment BC because overrightarrow{AD} is a scalar multiple of overrightarrow{BC}. Now we should check if vectors overrightarrow{AB} and overrightarrow{CD} are collinear. Since overrightarrow{AB} and overrightarrow{CD} have different coefficients for overrightarrow{a} and overrightarrow{b}, they are not collinear, indicating that lines AB and CD are not parallel. Given that opposite sides AD and BC are parallel and the other pair of opposite sides AB and CD are not parallel, we can conclude that quadrilateral ABCD is a trapezoid. The final answer is: boxed{A: text{Trapezoid}}

answer:1. Setting ( y = 0 ), we derive: [ g(g(x)) = xg(0) + g(x) ] Particularly, ( g(g(0)) = g(0) ). 2. Setting ( x = g(0) ) and ( y = 0 ), we find: [ g(g(g(0))) = g(0)^2 + g(g(0)) ] Since ( g(g(g(0))) = g(g(0)) = g(0) ), it follows that ( g(0) = g(0)^2 + g(0) ), leading to ( g(0) = 0 ). 3. With ( g(0) = 0 ), we simplify to ( g(g(x)) = g(x) ). Setting ( x = 2 ) in the original equation: [ g(y + 2) = 2g(y) + 2 ] Using ( g(x) + 1 ) when substituting ( y = g(x) ), we get: [ g(g(x) + 2) = 2g(x) + 2 = g(x) + 2 ] For nonzero ( x ), setting ( y = frac{1}{x} ), we find: [ g left( frac{1}{x} right) = frac{1}{x} ] Thus, ( g(x) = x ) for all ( x ). Therefore, ( m = 1 ) and ( t = frac{1}{3} ), so ( m times t = boxed{frac{1}{3}} ).