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bara.[barb+barc)xx(bara+barb+barc)] is e...

`bara.[barb+barc)xx(bara+barb+barc)]` is equal to

A

`[bara" "barb" "barc]`

B

`2[bara" "barb" "barc]`

C

`3[bara" "barb" "barc]`

D

0

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The correct Answer is:
D
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