What is the power of y? Is it 2? Are you supposed to multiply it all out?
It is not clear what the "problem" is. You cannot solve for x, y and z with one equation. In fact, you have not even written an equation.
You have already done a lot of factoring, but 4y^2 -1 and 5z^3 -3 can be factored some more.
Can someone help me on how to do this problem
2xy(3x^2)(4y^-1)(5z^-3)
4 answers
well it says to simplify.and that's all it gives me.
its either one of the following answers
a.6x^3/20z^3
b.6x^3y^2/20z^2
c.120x^3/z^3
d.120x^3z^3
its either one of the following answers
a.6x^3/20z^3
b.6x^3y^2/20z^2
c.120x^3/z^3
d.120x^3z^3
its (c), 120x^3/z^3.
thank you.
how did you get the answer?
how did you get the answer?