Let ABCDEF be a regular hexagon Then A¯B¯+A¯C¯+A¯D¯+EA¯+FA¯=

FA¯+A¯C¯=F¯C¯=2A¯B¯ and EA¯+A¯D¯=E¯D¯=A¯B¯

Let ABCDEF be a regular hexagon Then A¯B¯+A¯C¯+A¯D¯+EA¯+FA¯=

FA¯+A¯C¯=F¯C¯=2A¯B¯ and EA¯+A¯D¯=E¯D¯=A¯B¯

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Let ABCDEF be a regular hexagon Then $\bar{A} \bar{B} + \bar{A} \bar{C} + \bar{A} \bar{D} + \bar{E A} + \bar{F A} =$

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$4 \bar{A } \bar{B}$

$2 \bar{A} \bar{B}$

$\bar{A} \bar{B}$

$3 \bar{A} \bar{B}$

answer is C.

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$\bar{F A} + \bar{A} \bar{C} = \bar{F} \bar{C} = 2 \bar{A} \bar{B} and \bar{E A} + \bar{A} \bar{D} = \bar{E} \bar{D} = \bar{A} \bar{B}$

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Let ABCDEF be a regular hexagon Then A¯B¯+A¯C¯+A¯D¯+EA¯+FA¯=

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Let ABCDEF be a regular hexagon Then $\bar{A} \bar{B} + \bar{A} \bar{C} + \bar{A} \bar{D} + \bar{E A} + \bar{F A} =$