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Bunuel
\(\frac{7^9 + 7^8}{8} =\)

(A) 1/8
(B) 7/8
(C) 77/8
(D) 78
(E) 79


I think there is a formatting problem. The answer should be 7^8.
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Bunuel
\(\frac{7^9 + 7^8}{8} =\)

(A) 1/8
(B) 7/8
(C) 77/8
(D) 78
(E) 79


I think there is a formatting problem. The answer should be 7^8.

Edited. Thank you.
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