anonymous
  • anonymous
(x^-2y^-4x^3)^-2 First remove the parenthesis now we use the ab = 1/ab property x^-2 = 1/x^2 now we have a positive exponent x^2. Next y^-4 = 1/y^4 now we have three positive exponents now we need to make the negative 2 positive. -2= 1/-2 now we have x^2y^4x^3 add the like terms x^2+ x^3 = x^5 (x^5y^4)^2
Mathematics
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Can you help me with this @misssunshinexxoxo
anonymous
  • anonymous
SolomonZelman
  • SolomonZelman
\(\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~y^{-4}~x^3\right)^{-2} }\) like this?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Yeah. Brb five minutes or less sorry making lunch.
SolomonZelman
  • SolomonZelman
:)
SolomonZelman
  • SolomonZelman
Ok, I will tell you another property..... \(\displaystyle \Large {\rm w}^\color{red}{\rm \LARGE a}\times{\rm w}^\color{blue}{\rm \LARGE b}={\rm w}^{\color{red}{\rm \LARGE a}+\color{blue}{\rm \LARGE b}}\)
SolomonZelman
  • SolomonZelman
this property, can be applied to the terms with an x. lets do a little replacement, just to visualize it better: \(\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~\cdot~y^{-4}~\cdot~x^3\right)^{-2} }\) \(\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~\cdot~x^3~\cdot~y^{-4}\right)^{-2} }\)
SolomonZelman
  • SolomonZelman
so, you can apply the property I told you just now to the first 2 terms in this expression.
anonymous
  • anonymous
Back Okay.
SolomonZelman
  • SolomonZelman
ok, take your time to read, and apply the property i have provided...
anonymous
  • anonymous
So we don't change the terms into positive exponents?
SolomonZelman
  • SolomonZelman
we will do that later, but for now we just want to simplify inside the parenthesis (using this new property i told you).....
SolomonZelman
  • SolomonZelman
\(\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~\cdot~x^3~\cdot~y^{-4}\right)^{-2} }\) (this is your expression - the order of how you multiply doesn't matter) ---------------------------------------------------- Knowing that: \(\displaystyle \Large {\rm w}^\color{red}{\rm \LARGE a}\times{\rm w}^\color{blue}{\rm \LARGE b}={\rm w}^{\color{red}{\rm \LARGE a}+\color{blue}{\rm \LARGE b}}\) then, \(\displaystyle \Large {\rm x}^\color{red}{\rm \LARGE -2}\times{\rm w}^\color{blue}{\rm \LARGE 3}={\rm x}^{?}\)
anonymous
  • anonymous
By the bases being different i can't add them together right. so i guess x would be still x^-2
SolomonZelman
  • SolomonZelman
you can apply it to the " x\(^{-2}\) • x\(^3\) " part (where the base is same, the base is x)
anonymous
  • anonymous
we would just have x^1 which we don't count so we just have x.
anonymous
  • anonymous
Sorry thought i already enter it forgot i had to click on post. >~< @SolomonZelman
SolomonZelman
  • SolomonZelman
yes, x^1 is correct.
SolomonZelman
  • SolomonZelman
And x^1 is the same exact thing as x (besides the fact that you waste extra space and time when writing ^1 there)
SolomonZelman
  • SolomonZelman
we had: \(\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~\cdot~x^3~\cdot~y^{-4}\right)^{-2} }\) our new expression (that is equivalent to the previous one, of course) is: \(\LARGE\color{black}{ \displaystyle \large \left(x~\cdot~y^{-4}\right)^{-2} }\)
SolomonZelman
  • SolomonZelman
now, apply the property of \(a^{-b}\)=\(\dfrac{1}{a^b}\) to the " y\(^{-4}\) " part
anonymous
  • anonymous
Okay y^-4 =1/y^4 @SolomonZelman
SolomonZelman
  • SolomonZelman
yes
anonymous
  • anonymous
So how do i solve for -2 then?
SolomonZelman
  • SolomonZelman
our expression: \( \LARGE\color{black}{ \displaystyle \large \left(x~\cdot~y^{-4}\right)^{-2} }\) is going to become: \( \LARGE\color{black}{ \displaystyle \large \left(x~\cdot~\dfrac{1}{y^{4}}\right)^{-2} }\) \( \LARGE\color{black}{ \displaystyle \large \left(\dfrac{x}{y^{4}}\right)^{-2} }\)
anonymous
  • anonymous
Okay, so we don't change two?
SolomonZelman
  • SolomonZelman
I will give you another propery (which can is derived from \(a^{-b}\)=1 /\(a^b\)) the property is: \(\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{\rm a}}{\color{blue}{\rm b}}\right)^{\Large-w} = \left(\dfrac{\color{blue}{\rm b}}{\color{red}{\rm a}}\right)^{\Large w} }\)
SolomonZelman
  • SolomonZelman
Text - correct: `which can *be* derived...`
SolomonZelman
  • SolomonZelman
Knowing that: \(\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{\rm a}}{\color{blue}{\rm b}}\right)^{\Large-w} = \left(\dfrac{\color{blue}{\rm b}}{\color{red}{\rm a}}\right)^{\Large w} }\) then, therefore: \(\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{ x}}{\color{blue}{ y^4}}\right)^{\Large-2} = \left(\dfrac{\color{blue}{y^4}}{\color{red}{x}}\right)^{\Large 2} }\)
SolomonZelman
  • SolomonZelman
so far so good?
anonymous
  • anonymous
Yeah just make the reciprocal and it turn the negative into a positive.
anonymous
  • anonymous
Thanks for your help again. I appreciate it. @SolomonZelman
SolomonZelman
  • SolomonZelman
this is not it
anonymous
  • anonymous
There's more DX
SolomonZelman
  • SolomonZelman
\(\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{ x}}{\color{blue}{ y^4}}\right)^{\Large-2} = \left(\dfrac{\color{blue}{y^4}}{\color{red}{x}}\right)^{\Large 2}=\dfrac{\left(\color{blue}{y^4}\right)^2}{\left(\color{red}{x}\right)^2} }\)
SolomonZelman
  • SolomonZelman
yes, this one more step...
anonymous
  • anonymous
Alright.
SolomonZelman
  • SolomonZelman
\((\)y\(^4\)\()\)\(\large^{^2}\) = y\(^{4~\times~ 4}\) = y\(^?\)
anonymous
  • anonymous
y^16
SolomonZelman
  • SolomonZelman
yes, y\(^{16}\)/x\(^2\)
SolomonZelman
  • SolomonZelman
everything makes sense?
anonymous
  • anonymous
Yeah so far do i divide by x or i'm just writing it as y^16 over x^2
SolomonZelman
  • SolomonZelman
you had \(\color{black}{ \displaystyle \large \left(\dfrac{\color{blue}{y^4}}{\color{red}{x}}\right)^{\Large 2}=\dfrac{\left(\color{blue}{y^4}\right)^2}{\left(\color{red}{x}\right)^2} }\)
anonymous
  • anonymous
so we multiply the exponent y^4 with 2 and we get 16/x^2 Yeah it makes sense :)
SolomonZelman
  • SolomonZelman
applying the rule: \(\color{black}{ \displaystyle \large \left(\dfrac{\color{blue}{\rm A}}{\color{red}{\rm B}}\right)^{\Large w}=\dfrac{\left(\color{blue}{\rm A}\right)^w}{\left(\color{red}{\rm B}\right)^w} }\)
SolomonZelman
  • SolomonZelman
and then after we applied this rule above, we simplify on the top.
SolomonZelman
  • SolomonZelman
Which gives the result of: y\(^{16}\)/x\(^2\)
SolomonZelman
  • SolomonZelman
and that is it. can't do anything with this....
anonymous
  • anonymous
Alright thanks @SolomonZelman
SolomonZelman
  • SolomonZelman
yw, gtg.

Looking for something else?

Not the answer you are looking for? Search for more explanations.