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anonymous

  • one year ago

(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

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  1. anonymous
    • one year ago
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    Can you help me with this @misssunshinexxoxo

  2. anonymous
    • one year ago
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    @SolomonZelman?

  3. SolomonZelman
    • one year ago
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    \(\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~y^{-4}~x^3\right)^{-2} }\) like this?

  4. anonymous
    • one year ago
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    Yeah. Brb five minutes or less sorry making lunch.

  5. SolomonZelman
    • one year ago
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    :)

  6. SolomonZelman
    • one year ago
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    Ok, I will tell you another property..... \(\displaystyle \Large {\rm w}^\color{red}{\rm \LARGE a}\times{\rm w}^\color{blue}{\rm \LARGE b}={\rm w}^{\color{red}{\rm \LARGE a}+\color{blue}{\rm \LARGE b}}\)

  7. SolomonZelman
    • one year ago
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    this property, can be applied to the terms with an x. lets do a little replacement, just to visualize it better: \(\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~\cdot~y^{-4}~\cdot~x^3\right)^{-2} }\) \(\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~\cdot~x^3~\cdot~y^{-4}\right)^{-2} }\)

  8. SolomonZelman
    • one year ago
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    so, you can apply the property I told you just now to the first 2 terms in this expression.

  9. anonymous
    • one year ago
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    Back Okay.

  10. SolomonZelman
    • one year ago
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    ok, take your time to read, and apply the property i have provided...

  11. anonymous
    • one year ago
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    So we don't change the terms into positive exponents?

  12. SolomonZelman
    • one year ago
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    we will do that later, but for now we just want to simplify inside the parenthesis (using this new property i told you).....

  13. SolomonZelman
    • one year ago
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    \(\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~\cdot~x^3~\cdot~y^{-4}\right)^{-2} }\) (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, \(\displaystyle \Large {\rm x}^\color{red}{\rm \LARGE -2}\times{\rm w}^\color{blue}{\rm \LARGE 3}={\rm x}^{?}\)

  14. anonymous
    • one year ago
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    By the bases being different i can't add them together right. so i guess x would be still x^-2

  15. SolomonZelman
    • one year ago
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    you can apply it to the " x\(^{-2}\) • x\(^3\) " part (where the base is same, the base is x)

  16. anonymous
    • one year ago
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    we would just have x^1 which we don't count so we just have x.

  17. anonymous
    • one year ago
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    Sorry thought i already enter it forgot i had to click on post. >~< @SolomonZelman

  18. SolomonZelman
    • one year ago
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    yes, x^1 is correct.

  19. SolomonZelman
    • one year ago
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    And x^1 is the same exact thing as x (besides the fact that you waste extra space and time when writing ^1 there)

  20. SolomonZelman
    • one year ago
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    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: \(\LARGE\color{black}{ \displaystyle \large \left(x~\cdot~y^{-4}\right)^{-2} }\)

  21. SolomonZelman
    • one year ago
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    now, apply the property of \(a^{-b}\)=\(\dfrac{1}{a^b}\) to the " y\(^{-4}\) " part

  22. anonymous
    • one year ago
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    Okay y^-4 =1/y^4 @SolomonZelman

  23. SolomonZelman
    • one year ago
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    yes

  24. anonymous
    • one year ago
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    So how do i solve for -2 then?

  25. SolomonZelman
    • one year ago
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    our expression: \( \LARGE\color{black}{ \displaystyle \large \left(x~\cdot~y^{-4}\right)^{-2} }\) is going to become: \( \LARGE\color{black}{ \displaystyle \large \left(x~\cdot~\dfrac{1}{y^{4}}\right)^{-2} }\) \( \LARGE\color{black}{ \displaystyle \large \left(\dfrac{x}{y^{4}}\right)^{-2} }\)

  26. anonymous
    • one year ago
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    Okay, so we don't change two?

  27. SolomonZelman
    • one year ago
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    I will give you another propery (which can is derived from \(a^{-b}\)=1 /\(a^b\)) the property is: \(\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{\rm a}}{\color{blue}{\rm b}}\right)^{\Large-w} = \left(\dfrac{\color{blue}{\rm b}}{\color{red}{\rm a}}\right)^{\Large w} }\)

  28. SolomonZelman
    • one year ago
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    Text - correct: `which can *be* derived...`

  29. SolomonZelman
    • one year ago
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    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: \(\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{ x}}{\color{blue}{ y^4}}\right)^{\Large-2} = \left(\dfrac{\color{blue}{y^4}}{\color{red}{x}}\right)^{\Large 2} }\)

  30. SolomonZelman
    • one year ago
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    so far so good?

  31. anonymous
    • one year ago
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    Yeah just make the reciprocal and it turn the negative into a positive.

  32. anonymous
    • one year ago
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    Thanks for your help again. I appreciate it. @SolomonZelman

  33. SolomonZelman
    • one year ago
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    this is not it

  34. anonymous
    • one year ago
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    There's more DX

  35. SolomonZelman
    • one year ago
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    \(\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{ x}}{\color{blue}{ y^4}}\right)^{\Large-2} = \left(\dfrac{\color{blue}{y^4}}{\color{red}{x}}\right)^{\Large 2}=\dfrac{\left(\color{blue}{y^4}\right)^2}{\left(\color{red}{x}\right)^2} }\)

  36. SolomonZelman
    • one year ago
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    yes, this one more step...

  37. anonymous
    • one year ago
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    Alright.

  38. SolomonZelman
    • one year ago
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    \((\)y\(^4\)\()\)\(\large^{^2}\) = y\(^{4~\times~ 4}\) = y\(^?\)

  39. anonymous
    • one year ago
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    y^16

  40. SolomonZelman
    • one year ago
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    yes, y\(^{16}\)/x\(^2\)

  41. SolomonZelman
    • one year ago
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    everything makes sense?

  42. anonymous
    • one year ago
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    Yeah so far do i divide by x or i'm just writing it as y^16 over x^2

  43. SolomonZelman
    • one year ago
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    you had \(\color{black}{ \displaystyle \large \left(\dfrac{\color{blue}{y^4}}{\color{red}{x}}\right)^{\Large 2}=\dfrac{\left(\color{blue}{y^4}\right)^2}{\left(\color{red}{x}\right)^2} }\)

  44. anonymous
    • one year ago
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    so we multiply the exponent y^4 with 2 and we get 16/x^2 Yeah it makes sense :)

  45. SolomonZelman
    • one year ago
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    applying the rule: \(\color{black}{ \displaystyle \large \left(\dfrac{\color{blue}{\rm A}}{\color{red}{\rm B}}\right)^{\Large w}=\dfrac{\left(\color{blue}{\rm A}\right)^w}{\left(\color{red}{\rm B}\right)^w} }\)

  46. SolomonZelman
    • one year ago
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    and then after we applied this rule above, we simplify on the top.

  47. SolomonZelman
    • one year ago
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    Which gives the result of: y\(^{16}\)/x\(^2\)

  48. SolomonZelman
    • one year ago
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    and that is it. can't do anything with this....

  49. anonymous
    • one year ago
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    Alright thanks @SolomonZelman

  50. SolomonZelman
    • one year ago
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    yw, gtg.

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