Asked by bob
                Simply the equation:
2x^4y^0z^2/(yz^2)^-2 x 2zx^2y^-4
Can you help?
I got the denominator to be
y^-6 x 2z^5 x x^2
When I solved it, I got
x^2y^6/z^3
            
        2x^4y^0z^2/(yz^2)^-2 x 2zx^2y^-4
Can you help?
I got the denominator to be
y^-6 x 2z^5 x x^2
When I solved it, I got
x^2y^6/z^3
Answers
                    Answered by
            Reiny
            
    ok, first of all you can't "solve" this because you need an equation to solve. You don't have one
secondly, you are using x as a variable, but then seem to use x as a multiplication sign, very confusing.
thirdly, you lack of brackets make the question ambiguous.
I will assume you mean:
simplify the expression 2x^4y^0z^2/( (yz^2)^-2 x 2zx^2y^-4)
= 2x^4z^2/((y^-2 z^-4)(2z x^2 y^-4))
= 2x^4z^2/(2y^-6 x^2 z^-3)
= x^2 y^6 z^5
    
secondly, you are using x as a variable, but then seem to use x as a multiplication sign, very confusing.
thirdly, you lack of brackets make the question ambiguous.
I will assume you mean:
simplify the expression 2x^4y^0z^2/( (yz^2)^-2 x 2zx^2y^-4)
= 2x^4z^2/((y^-2 z^-4)(2z x^2 y^-4))
= 2x^4z^2/(2y^-6 x^2 z^-3)
= x^2 y^6 z^5
                    Answered by
            bob
            
    I didn't ask for the problem to be solved, just simplified.
I know I shouldn't have used an x for a multiplication sign, sorry. My bad.
Thank you. I found my error.
    
I know I shouldn't have used an x for a multiplication sign, sorry. My bad.
Thank you. I found my error.
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