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Jamierox4ev3r

  • one year ago

Reviewing some math basics before I start college. I'm unsure of a few things, so please check my work as I go. Thank you. First question: 2c. Simplify the following expression. Write your answer without negative exponents.

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  1. Jamierox4ev3r
    • one year ago
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    Give me some time to type out the expression, thanks.

  2. Jamierox4ev3r
    • one year ago
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    \[\left( \frac{ 3x ^{\frac{ 3 }{ 2 }}y ^{3} }{ x ^{2} y ^{-\frac{ 1 }{ 2 }}} \right)^{-2}\]

  3. Jamierox4ev3r
    • one year ago
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    oh my goodness that was a heck of an expression to get out. So far, what I've done is this: \[\left( \frac{ x ^{2}y ^{-\frac{ 1 }{ 2 }} }{ 3x ^{\frac{ 3 }{ 2 }} y ^{3}} \right)^{2}\] basically, I got ride of the negative 2 exponent by flipping the fraction around. If this was an incorrect thing to do, someone let me know now. My recollection of algebra rules is a bit foggy

  4. Jamierox4ev3r
    • one year ago
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    any idea? @Nnesha

  5. Nnesha
    • one year ago
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    wait lol :P

  6. Jamierox4ev3r
    • one year ago
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    alright :P thanks

  7. Nnesha
    • one year ago
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    alright back

  8. Nnesha
    • one year ago
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    sorry abt that i need to buy a new laptop >.< i'm just gonna use draw tool |dw:1440364115001:dw| but i'm sure we will get the same answer :=) yes

  9. Jamierox4ev3r
    • one year ago
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    lol nice xD \(\LaTeX\) is complicated stuffs

  10. Nnesha
    • one year ago
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    that's the same thing as this \[\left( \frac{ x ^{2}y ^{-\frac{ 1 }{ 2 }} }{ 3x ^{\frac{ 3 }{ 2 }} y ^{3}} \right)^{2}\] so yes your first step is right wanna try next step :=)

  11. Nnesha
    • one year ago
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    this is the 3rd time i was about to use equation tool but my laptop turned off so i'm like ugghh

  12. Jamierox4ev3r
    • one year ago
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    Sure. Here is what I think I should do next Distribute the 2 exponent. So I would get this: \[ \frac{ x ^{4} y ^{-1}}{ 9x ^{3}y ^{6} }\\]

  13. Jamierox4ev3r
    • one year ago
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    oh man xD let me try that again \(\Large\frac{x^{4} y^{-1}}{9x^{3}y^{6}}\)

  14. Nnesha
    • one year ago
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    yes that's right ! doing great so far!

  15. Jamierox4ev3r
    • one year ago
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    There we go! and :oo I did correct things! Yay me cx

  16. Jamierox4ev3r
    • one year ago
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    Kinda confused from there, but if I recall correctly, then I would simplify as follows: \(\Large\frac{xy^{-7}}{9}\)

  17. Jamierox4ev3r
    • one year ago
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    would that be a correct way of simplifying?

  18. Nnesha
    • one year ago
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    perfect!

  19. Nnesha
    • one year ago
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    one more step!

  20. Jamierox4ev3r
    • one year ago
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    wao! alright I wasn't even sure if that was correct XD So last of all, I assume that since we don't want any negative exponents for the final simplified form, we would put the y in the denominator in order to eliminate the negative exponent. It would look like this: \(\Large\frac{x}{9y^{7}}\)

  21. Jamierox4ev3r
    • one year ago
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    and I'm assuming that I have arrived at the final answer

  22. Jamierox4ev3r
    • one year ago
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    ...I think

  23. Nnesha
    • one year ago
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    YAY!!

  24. Nnesha
    • one year ago
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    that's it!

  25. Jamierox4ev3r
    • one year ago
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    It's right? :o

  26. Jamierox4ev3r
    • one year ago
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    YES!

  27. Jamierox4ev3r
    • one year ago
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    oh my gosh, I didn't think I remembered any of this algebra, it's been almost 3 long years since I've done this stuff. Thank you so much <33

  28. Nnesha
    • one year ago
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    eh i didn't do anything :/ you posted this question to tell me how SMART you are :(

  29. Nnesha
    • one year ago
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    hahh jk you gonna get A+ good luck wth ur college stuff!

  30. Jamierox4ev3r
    • one year ago
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    Thank you so so much XD I didn't know if I was heading in the right direction to be honest, in my textbook, there's no answer key for any of the review. So I just tried things and crossed my fingers lol. Thanks for helping me make sure that I was doing the right thing.

  31. Nnesha
    • one year ago
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    my pleasure! <3

  32. triciaal
    • one year ago
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    all the best with college. be confident relax and ask for help when you really need it .

  33. Jamierox4ev3r
    • one year ago
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    @triciaal thank you so much! I really appreciate it. I'm pretty nervous to begin college honestly, but reviewing early concepts is a good way to prep. for the higher math courses that I'm going to be taking in the extremely near future. To be precise, in two days! Thank you for you advice, I will make sure I ask for help whenever I need it and do my best to always remain confident in my abilities.

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