Question

Could someone show me how to do the problem: (4b^2n^3/zn^2)^1?

Answer 1

\[\left(\begin{matrix}4b^2n^3 \\ zn^2\end{matrix}\right)^{-1}\]

Answer 2

\(\bf \left(\cfrac{4b^2n^3}{zn^2}\right)^1\\
\textit{any number by itself, is really raised to 1 }\\ so\quad anyvalue = anyvalue^1\\\quad\\\quad\\
\textit{so }\quad \left(\cfrac{4b^2n^3}{zn^2}\right)^1 \implies \left(\cfrac{4b^2n^3}{zn^2}\right) \implies \cfrac{4b^2n^3}{zn^2}\)

Answer 3

and from there, you cancel out like-terms, above and below \(\bf \cfrac{4b^2n^3}{zn^2} \implies \cfrac{4b\cdot b\cdot n\cdot n\cdot n}{z\cdot n\cdot n}\)

see any like-terms above and below?

Answer 4

N^3 and N^2, but wouldn't you have to flip the whole equation, since you got rid of the negative 1?

Answer 5

negative one? I didn't see any :(.... wait... shoot... I just saw it :(

Answer 6

but anyhow, yes, you do have to flip the rational :)

Answer 7

Okay, so you would get:

\[\left( \frac{ zn^2 }{ 4b^2n^3 } \right)\]

Answer 8

yes it would

Answer 9

Okay, and then subtract n^2 from n^3?

Answer 10

\(\bf \left(\cfrac{4b^2n^3}{zn^2}\right)^{-1}\\\quad \\

\textit{so }\quad \left(\cfrac{4b^2n^3}{zn^2}\right)^{-1}
\implies \left(\cfrac{zn^2}{4b^2n^3}\right) \implies \cfrac{zn^2}{4b^2n^3}\\\quad \\

\cfrac{zn^2}{4b^2n^3} \implies \cfrac{z\cdot n\cdot n}{4\cdot b\cdot b\cdot n\cdot n\cdot n}
\)

Answer 11

you would cancel out, like-terms, if one above, looks the same as the one below, then, they cancel

Answer 12

\(\bf \cfrac{zn^2}{4b^2n^3} \implies \cfrac{z\cdot \cancel{n}\cdot \cancel{n}}{4\cdot b\cdot b\cdot n\cdot \cancel{n}\cdot \cancel{n}}\)

Answer 13

Okay, so the answer would be:

\[ \left(\begin{matrix}zn \\ 4b^2\end{matrix}\right)\]?

Answer 14

yeap

Answer 15

hhh... wait... hmmm

Answer 16

notice, the numerator, all n's cancel out there

Answer 17

Oh, sorry for replying so late. The answer would be z/4b^2n?

Answer 18

yeap \(\bf \cfrac{zn^2}{4b^2n^3} \implies \cfrac{z\cdot \cancel{n}\cdot \cancel{n}}{4\cdot b\cdot b\cdot n\cdot \cancel{n}\cdot \cancel{n}} \implies \cfrac{z}{4b^2n}\)

Answer 19

Thanks!