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calculusxy

  • one year ago

MEDAL!!! Question will be posted below.

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  1. calculusxy
    • one year ago
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    \[\huge v^2 - 20 = 0 \]

  2. calculusxy
    • one year ago
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    Solve with quadratic formula

  3. Nnesha
    • one year ago
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    ok have you tried anything yet ?

  4. calculusxy
    • one year ago
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    yes gimme a moment

  5. Nnesha
    • one year ago
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    okay

  6. anonymous
    • one year ago
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    \[\frac{ -b \pm \sqrt{b^2-4ac}}{ 2a } \]

  7. calculusxy
    • one year ago
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    \[\frac{ -0 \pm \sqrt{0 - 4(-20)} }{ 2 }\]

  8. anonymous
    • one year ago
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    yes

  9. calculusxy
    • one year ago
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    i thought the new equation would be like: \[v^2 + 0a - 20 = 0\]

  10. calculusxy
    • one year ago
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    \[\frac{ -0 \pm 8.9 }{ 2 }\]

  11. anonymous
    • one year ago
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    0v would be better, but yes. v^2 +0v -20 = 0

  12. Nnesha
    • one year ago
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    v^2 -20 is same as `v^2 -0v -20`

  13. Nnesha
    • one year ago
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    well would be great if get the final answer in radical

  14. Nnesha
    • one year ago
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    don't take square of 20 factor it out

  15. Nnesha
    • one year ago
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    square root*

  16. calculusxy
    • one year ago
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    the answer key for this in decimal form...

  17. Nnesha
    • one year ago
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    oh okay then.

  18. calculusxy
    • one year ago
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    \[\frac{ -0 \pm 8.9 }{ 2 } = 4.45 \]

  19. anonymous
    • one year ago
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    remember it's \[\pm \]

  20. calculusxy
    • one year ago
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    okay so -4.45 or 4.45

  21. anonymous
    • one year ago
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    yes

  22. Nnesha
    • one year ago
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    ^ye right it's \[\rm a \pm b ~= a-b ~~ and ~a+b\]

  23. calculusxy
    • one year ago
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    is my answer correct?

  24. anonymous
    • one year ago
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    yes it is. if your key says 4.4721... it's only because you cut off some decimals from 8.9

  25. Nnesha
    • one year ago
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    i didn't get 4.4 `5`

  26. anonymous
    • one year ago
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    but you did solve it correctly.

  27. calculusxy
    • one year ago
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    ok thank you :)

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