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anonymous

  • 5 years ago

Solve for k: ((16-2k)/3)^2+(2((16-2k)/3))+k=40

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  1. anonymous
    • 5 years ago
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    (256-64k+4k^2)/9 + (32-4k)/3 + k =40

  2. anonymous
    • 5 years ago
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    then put them all over 9

  3. anonymous
    • 5 years ago
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    (256-64k+4k^2 + 96-12k+9k-360)/9 = 0

  4. anonymous
    • 5 years ago
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    Wait how'd you get the first step?

  5. anonymous
    • 5 years ago
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    Never mind 16 squared?

  6. anonymous
    • 5 years ago
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    I expanded the first part squared and multiplied the second part by 2

  7. anonymous
    • 5 years ago
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    Do you know what to do from there?

  8. anonymous
    • 5 years ago
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    i think so

  9. anonymous
    • 5 years ago
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    just say so if you need more help

  10. anonymous
    • 5 years ago
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    okay i got to -64k+4k^2-12k+9k=-352, how do i get all the k's alone?

  11. anonymous
    • 5 years ago
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    you can group all the ks together and get 4k^2 + 67k +352 = 0

  12. anonymous
    • 5 years ago
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    then use the quadratic equation with a=4, b=67, c=352

  13. anonymous
    • 5 years ago
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    -67, sorry

  14. anonymous
    • 5 years ago
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    whats the quadratic equation?

  15. anonymous
    • 5 years ago
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    \[k=(-b \pm \sqrt{b ^{2}-4ac})/2a\]

  16. anonymous
    • 5 years ago
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    gotcha!

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