Are the last 2's exponents? That would mean you want to know
[8(x^3y^4)^2]^-2
The +2 and -2 operations on (x^3y^4) would leave it unchanged, and leave you with
(1/64)(x^3y^4)
please help me simplify
(8(x^3y^4)2)-2
6 answers
The problem to solve is:
(8(x^3y^4)2)-2
Multiply 8 and
8*(x^3*y^4)2 evaluates to 64
Abort (8*(x^3*y^4)2)-2 evaluates to 64-64
The final answer is 64-64
(8(x^3y^4)2)-2
Multiply 8 and
8*(x^3*y^4)2 evaluates to 64
Abort (8*(x^3*y^4)2)-2 evaluates to 64-64
The final answer is 64-64
The 2's are not exponents
Apologies to you and James B, I'm home from work now and just looked at the problem and the 2's are exponents.
Apologies, I'm home from work now and was able to look at the problem for myself and the 2's are exponents.
If the 2's are exponents, I stand by my original answer.
3 answers