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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)=
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3 votes
3^2-3^2÷(2^5-2^4-2^3-2^2-2^1+2^0)=

User Schummar
by
8.1k points

2 Answers

2 votes

Answer:

6

Explanation:

It simply 6

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