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3^2-3^2÷(2^5-2^4-2^3-2^2-2^1+2^0)=

3^2-3^2÷(2^5-2^4-2^3-2^2-2^1+2^0)=
asked 79.9k views
3 votes
3^2-3^2÷(2^5-2^4-2^3-2^2-2^1+2^0)=

asked
User Schummar
by
8.1k points

2 Answers

2 votes

Answer:

6

Explanation:

It simply 6

answered
User Kirill Leontev
by
8.5k points
6 votes
6 because calculator
answered
User Anton Bogdanov
by
8.1k points
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