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anonymous

  • 5 years ago

show me how to work 4y=12y^2

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  1. anonymous
    • 5 years ago
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    what have you done so far?

  2. amistre64
    • 5 years ago
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    -4y from each side; then factor out a y and see which y values make it equal zero

  3. anonymous
    • 5 years ago
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    did you just say subtract 4y from both sides?

  4. anonymous
    • 5 years ago
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    nothing i dont understand how to work it

  5. amistre64
    • 5 years ago
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    yes

  6. anonymous
    • 5 years ago
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    umm thats wrong.... more like divide.

  7. anonymous
    • 5 years ago
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    thats what i thought it would be division

  8. amistre64
    • 5 years ago
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    4y = 12y^2 ; -4y 0 = 12y^2 -4y ; factor out 4y 0 = 4y(3y - 1) ; y = 0 or 1/3

  9. anonymous
    • 5 years ago
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    ^2 means squared right that's why i typed it like that because 12y squared

  10. anonymous
    • 5 years ago
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    so 4y=12y squared

  11. amistre64
    • 5 years ago
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    am i still 'wrong' ? lol

  12. anonymous
    • 5 years ago
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    oh sorry yeah do it like he said and check using quadratic.

  13. anonymous
    • 5 years ago
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    ok i'm a 9th grader so i'm a bit confused lol

  14. amistre64
    • 5 years ago
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    you can also /4y if you want 1 = 3y ; y=1/3 but your missing something because... 4y CAN be a zero, which this way does not allow for

  15. amistre64
    • 5 years ago
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    if it can be zero; stay away from division

  16. anonymous
    • 5 years ago
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    \[4y=12y^2\]\[4y-4y=4y^2-4y\]\[0=4y^2-4y\]\[4(y^2-y)=0\]\[4(y(y+1))=0\]

  17. anonymous
    • 5 years ago
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    yes good thinking ^-^

  18. anonymous
    • 5 years ago
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    *12y^2

  19. amistre64
    • 5 years ago
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    when you know it can never be zero; then division is fine ;)

  20. anonymous
    • 5 years ago
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    ok ty

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