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anonymous

  • one year ago

Solve this equation :)

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  1. anonymous
    • one year ago
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    |dw:1439764528624:dw|

  2. rishavraj
    • one year ago
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    is it \[\sqrt[3]{70 - 2x} - 10 = -6\]

  3. anonymous
    • one year ago
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    yes

  4. rishavraj
    • one year ago
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    okay then add 10 on both sides.......

  5. anonymous
    • one year ago
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    4

  6. rishavraj
    • one year ago
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    and also \[(\sqrt[x]{a})^x = a\]

  7. rishavraj
    • one year ago
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    so u got \[\sqrt[3]{70 - 2x} = 4\]

  8. anonymous
    • one year ago
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    yes

  9. rishavraj
    • one year ago
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    now \[(\sqrt[3]{70 - 2x})^3 = 4^3 \]

  10. anonymous
    • one year ago
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    ok i got 4^3=64

  11. rishavraj
    • one year ago
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    okay then 70 - 2x = 64

  12. rishavraj
    • one year ago
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    wht do u think .???whts the next step

  13. anonymous
    • one year ago
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    subtract 70

  14. rishavraj
    • one year ago
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    yupp

  15. anonymous
    • one year ago
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    -6 :)

  16. rishavraj
    • one year ago
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    i mean -2x = -6

  17. anonymous
    • one year ago
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    x=3

  18. anonymous
    • one year ago
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    Would that be extraneous or not?

  19. rishavraj
    • one year ago
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    yup :))

  20. anonymous
    • one year ago
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    wouldnt it be none extraneous?

  21. anonymous
    • one year ago
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    *non lol

  22. rishavraj
    • one year ago
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    plug x = 3 in main equation and see whether it satisfies or not`....

  23. anonymous
    • one year ago
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    yes, it does:) I plugged it into the 70-2x=64 equation not the orginal one

  24. rishavraj
    • one year ago
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    okay so its not extraneous

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spraguer (Moderator)
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