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anonymous

  • one year ago

I really need help with this problem. it says to multiply and simplify if possible. (7 3sqrt4)(6 3sqrt2)

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  1. sweetburger
    • one year ago
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    \[(7 (3\sqrt{4})) ( 6 (3\sqrt{2}))\] like this?

  2. anonymous
    • one year ago
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    yes

  3. anonymous
    • one year ago
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    Yes here it means like to solve as sqrt of 4 is equwl to 2 and then mutiplt 2 with 7 and 3.. solve another side same like that.. it will give you answer in points..

  4. sweetburger
    • one year ago
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    basically what @Rahammii said

  5. anonymous
    • one year ago
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    And then multiply both answers of brcket with eachother

  6. anonymous
    • one year ago
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    Gimee medal plzzz

  7. sweetburger
    • one year ago
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    u actually wouldnt have to take anything out from teh square root cus its multiplication...

  8. anonymous
    • one year ago
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    Hmm yes.. right on..

  9. anonymous
    • one year ago
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    |dw:1433253936674:dw|

  10. sweetburger
    • one year ago
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    OHHHHHH THAT CHANGES THINGS

  11. anonymous
    • one year ago
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    Okay i know its answer too

  12. sweetburger
    • one year ago
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    ya this can be rewritten simply

  13. anonymous
    • one year ago
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    Taking power 1/3 to both brackets.. then simplify it..

  14. sweetburger
    • one year ago
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    4^(1/3) x 2^(1/3) = blank then take this answer and multiply it by 42

  15. anonymous
    • one year ago
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    @sweetburger.. it can be solve like method i provided..

  16. sweetburger
    • one year ago
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    @Rahammii funny that is the method u provided just i took it an easy way

  17. anonymous
    • one year ago
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    Oh okay okay ..

  18. anonymous
    • one year ago
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    so it would be 2x42?

  19. sweetburger
    • one year ago
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    ya bosss

  20. anonymous
    • one year ago
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    then what would I do with the 7 and the 6?

  21. sweetburger
    • one year ago
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    no the 7x6 = 42 thats where your 42 came from boss

  22. sweetburger
    • one year ago
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    sorry about the confusion there :/

  23. anonymous
    • one year ago
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    Ohh ok

  24. anonymous
    • one year ago
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    thank you for helping me :)

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