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anonymous

  • one year ago

***WILL MEDAL AND FAN*** Express answer in exact form. Find the area of the larger segment whose chord is 8" long in a circle with an 8" radius. (Hint: A chord divides a circle into two segments. In problem 1, you found the area of the smaller segment.)

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  1. anonymous
    • one year ago
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    I practically already know (How) to get the answer, I just need to make it look like this::

  2. anonymous
    • one year ago
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    |dw:1437655413905:dw|

  3. anonymous
    • one year ago
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    And it says to drag numbers out of the box and place the apropiately in the boxes

  4. anonymous
    • one year ago
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    Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers.

  5. anonymous
    • one year ago
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    That's what it tells me to do

  6. phi
    • one year ago
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    Did you find the area of the "small" segment?

  7. anonymous
    • one year ago
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    yES

  8. anonymous
    • one year ago
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    dO YOU WANT IT

  9. anonymous
    • one year ago
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    Woops. Caps lock again

  10. phi
    • one year ago
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    yes

  11. anonymous
    • one year ago
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    Okay. Gimmee a sec

  12. anonymous
    • one year ago
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    |dw:1437655965362:dw|

  13. anonymous
    • one year ago
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    There you go.

  14. phi
    • one year ago
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    ok. the area on the other side of the chord |dw:1437656246763:dw|

  15. phi
    • one year ago
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    the area of the entire circle \( \pi \ r^2\) = \( 64 \pi\)

  16. phi
    • one year ago
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    thus you want to do area of circle - area of segment \[ 64 \pi - \left( \frac{64 \pi}{6} - 16 \sqrt{3}\right) \]

  17. anonymous
    • one year ago
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    Okay

  18. phi
    • one year ago
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    I would simplify that

  19. phi
    • one year ago
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    \[ 64 \pi - \left( \frac{64 \pi}{6} - 16 \sqrt{3}\right) \\ 64 \pi -\frac{64 \pi}{6} + 16 \sqrt{3}\]

  20. phi
    • one year ago
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    multiply the first term by 6/6 \[ \frac{6}{6} \cdot 64 \pi - \frac{64 \pi}{6} + 16 \sqrt{3}\\ \frac{6\cdot 64 \pi - 64 \pi}{6} + 16 \sqrt{3} \] can you finish?

  21. phi
    • one year ago
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    to make it easier, I would "factor out" 64 pi so it looks like this \[ \frac{(6 -1) 64 \pi}{6} + 16 \sqrt{3} \]

  22. anonymous
    • one year ago
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    So we use distribution?

  23. phi
    • one year ago
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    distribution would "undo" the factoring out. Instead, simplify (6-1)

  24. anonymous
    • one year ago
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    So 5

  25. phi
    • one year ago
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    yes. also, we can divide top and bottom by 2

  26. anonymous
    • one year ago
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    That would make the 5 uneven

  27. phi
    • one year ago
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    you have \[ \frac{5 \cdot 64 \pi}{6} + 16 \sqrt{3} \] you can divide the 6 by 2 and the 64 by 2

  28. anonymous
    • one year ago
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    Oh that makes sense

  29. anonymous
    • one year ago
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    So 32 and 3?

  30. phi
    • one year ago
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    yes, so we now have \[ \frac{5 \cdot 32 \pi}{3} + 16 \sqrt{3} \] 32 is 2*2*2*2*2 so we can't divide any more. multiply 5*32 to get 160 and the final version is \[ \frac{160}{3} \pi+ 16 \sqrt{3} \]

  31. phi
    • one year ago
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    now you have to fill in your boxes with those numbers

  32. anonymous
    • one year ago
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    Okay. Thanks that really helps

  33. anonymous
    • one year ago
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    And btw,

  34. anonymous
    • one year ago
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    Is there any way to possibly save conversations for future reference?

  35. phi
    • one year ago
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    You could save the link ... I don't think OS will delete this post. Or maybe take screen shots?

  36. anonymous
    • one year ago
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    Yeah, but this really isn't my personal computer though

  37. anonymous
    • one year ago
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    Maybe I'll just save the link to my Google Docs account then

  38. phi
    • one year ago
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    yes, save the link.

  39. anonymous
    • one year ago
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    I will

  40. anonymous
    • one year ago
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    Thanks for the medal

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