3^3 is 27 but 27^0 is still 1. 3^0 = 1 also and 3^1 = 3 so you get 3+1 = 4. But it's been a while since I've mathed so I'm probably forgetting something with regards to order of operations and exponents, lol.
It's true there is is an order of precedence for mathematical operators. It's also true that parentheses override normal operator precedence, form the innermost set of parentheses out. I forgot that detail earlier.
Zero ??
2
Oh wait!
4
It is 66 ??
4?
3^3 is 27 but 27^0 is still 1. 3^0 = 1 also and 3^1 = 3 so you get 3+1 = 4. But it's been a while since I've mathed so I'm probably forgetting something with regards to order of operations and exponents, lol.
Correct, bOOm !
It's true there is is an order of precedence for mathematical operators. It's also true that parentheses override normal operator precedence, form the innermost set of parentheses out. I forgot that detail earlier.
2
I was wrong. My final answer is 4.
Correct, bOOm !
1+3=4
Correct?
(3^3)^0 + 3^(3^0) = 27^0 + 3^1 = 1 + 3 = 4.
Correct, bOOm !
(3^3)^0 + 3^(3^0) = 27^0+ 3^1 = 1 + 3 = 4.
4
I used Calculator.