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kiamonstaa

  • 3 years ago

3√(2x)+x√(8x)-5√(18x)

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  1. kiamonstaa
    • 3 years ago
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    show all work.

  2. helder_edwin
    • 3 years ago
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    \[ \large 3\sqrt{2x}+x\sqrt{8x}-5\sqrt{18x} \]

  3. marcoduuuh
    • 3 years ago
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    -12√2x+2x√2x

  4. helder_edwin
    • 3 years ago
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    first \[ \large \sqrt{8x}=\sqrt{2\cdot4x}=2\sqrt{2x} \]

  5. helder_edwin
    • 3 years ago
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    \[ \large \sqrt{18x}=\sqrt{2\cdot9x}=3\sqrt{2x} \]

  6. kiamonstaa
    • 3 years ago
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    okay what's next?

  7. helder_edwin
    • 3 years ago
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    plug in the main expresion.

  8. helder_edwin
    • 3 years ago
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    \[ \large 3\sqrt{2x}+x\sqrt{8x}-5\sqrt{18x}= \] \[ \large 3\sqrt{2x}+x[2\sqrt{2x}]-5[3\sqrt{2x}]= 3\sqrt{2x}+2x\sqrt{2x}-15\sqrt{2x} \]

  9. helder_edwin
    • 3 years ago
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    now add/substract like terms

  10. kiamonstaa
    • 3 years ago
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    ohh.....okay thanks

  11. helder_edwin
    • 3 years ago
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    u r welcome

  12. kiamonstaa
    • 3 years ago
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    is that it or am i missing something?

  13. helder_edwin
    • 3 years ago
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    what did u get?

  14. kiamonstaa
    • 3 years ago
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    \[12\sqrt{2x}+2x \sqrt{2x}\]

  15. kiamonstaa
    • 3 years ago
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    oh. ha nvm i got it.

  16. helder_edwin
    • 3 years ago
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    \[ \large =-12\sqrt{2x}+2x\sqrt{2x}=(2x-12)\sqrt{2x} \]

  17. kiamonstaa
    • 3 years ago
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    ohhh.

  18. marcoduuuh
    • 3 years ago
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    Which would be my answer. That's the final one without all the work. :p

  19. kiamonstaa
    • 3 years ago
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    lol yeaa

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