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mathhasproblems

  • 3 years ago

√c (3 + 2√5) and √6 (7√3 + 6) Please help???

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  1. satellite73
    • 3 years ago
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    is that a c in front?

  2. LoveYou*69
    • 3 years ago
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    its asking for the square root of 3 then(3+2√5) and then the other question

  3. LoveYou*69
    • 3 years ago
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    @ satellite

  4. mathhasproblems
    • 3 years ago
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    The left :/

  5. LoveYou*69
    • 3 years ago
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    ):

  6. FoolForMath
    • 3 years ago
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    What are we suppose to do here?

  7. LoveYou*69
    • 3 years ago
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    Multiply and Simplify

  8. LoveYou*69
    • 3 years ago
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    Multiply/Distribute in this case, and simplify

  9. LoveYou*69
    • 3 years ago
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    I got the first one, just need help with the second

  10. Pippa
    • 3 years ago
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    Y dont you retype the question?

  11. Pippa
    • 3 years ago
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    It isnt so clear

  12. myininaya
    • 3 years ago
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    \[\sqrt{3} (3+2 \sqrt{5}) \sqrt{6}(7 \sqrt{3}+6)?\]

  13. myininaya
    • 3 years ago
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    we are finding product right?

  14. LoveYou*69
    • 3 years ago
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    \[√6(7√3 + 6)\] thats the one i still need help on

  15. Pippa
    • 3 years ago
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    myinaya it itwo separate equations

  16. LoveYou*69
    • 3 years ago
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    distributing and simplifying

  17. myininaya
    • 3 years ago
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    \[\sqrt{6}(7 \sqrt{3}+6)\]

  18. myininaya
    • 3 years ago
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    \[\sqrt{6} \cdot 7 \sqrt{3}+\sqrt{6} \cdot 6 \]

  19. myininaya
    • 3 years ago
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    i use distributive property there

  20. myininaya
    • 3 years ago
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    a(b+c)=ab+ac

  21. myininaya
    • 3 years ago
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    \[7 \sqrt{6} \sqrt{3}+6 \sqrt{6}\] \[7 \sqrt{2 \cdot 3} \sqrt{3}+6 \sqrt{6}\]

  22. myininaya
    • 3 years ago
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    \[7 \sqrt{2} \sqrt{3} \sqrt{3}+6 \sqrt{6}\]

  23. satellite73
    • 3 years ago
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    is the "c" a "3"??

  24. myininaya
    • 3 years ago
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    \[7 \sqrt{2}(3)+6 \sqrt{6}=7(3) \sqrt{2}+6 \sqrt{6}=21 \sqrt{2}+6 \sqrt{6}\]

  25. FoolForMath
    • 3 years ago
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    \[ \sqrt{6}(7 \sqrt{3}+6) = 21\sqrt{2} +6\sqrt{6} \]

  26. LoveYou*69
    • 3 years ago
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    Thank you guys so much!!

  27. FoolForMath
    • 3 years ago
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    Distributive property holds because they are rationals.

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