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anonymous

  • 4 years ago

What is the area under the curves of y=3, 2y=5(x)^1/2 and 2y+x=6

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  1. anonymous
    • 4 years ago
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    is there a picture attached?

  2. anonymous
    • 4 years ago
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    Ah well it's just that graphing this is going to be a pain.. So let's see what we can do

  3. anonymous
    • 4 years ago
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    no but I can attach one

  4. anonymous
    • 4 years ago
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    if u can, please do so haha

  5. anonymous
    • 4 years ago
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    k here it is

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  6. anonymous
    • 4 years ago
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    omg lol more of these webworks... i'm so sick of them haha

  7. anonymous
    • 4 years ago
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    sorry

  8. anonymous
    • 4 years ago
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    nah it's fine.. actually give the wolframalpha link, they also list the intercepts. it's easier.

  9. anonymous
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=Graph+y%3D3%2C+2y%3D5x^%281%2F2%29%2C+2y%2Bx%3D6

  10. anonymous
    • 4 years ago
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    all right great thanks. Here we go..

  11. anonymous
    • 4 years ago
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    So here I'm going to split the graph in two at the point where x+2y = 6 and 2y = 5sqrt(x) intersect

  12. anonymous
    • 4 years ago
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    okay we did that

  13. anonymous
    • 4 years ago
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    That's the upper limit of our first integral

  14. anonymous
    • 4 years ago
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    okay

  15. anonymous
    • 4 years ago
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    Sorry my browser's having trouble, anyway the upper intersect was 1. Now the lower intersect, which occurs when y =3 and 2y+x=6 intersect.

  16. anonymous
    • 4 years ago
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    6+x=6, x = 0. So the first half of that area is: Integral (upper curve - lower curve, 0,1)

  17. anonymous
    • 4 years ago
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    how did you get one I got 1.36 doesnt it turn into a quadratic

  18. anonymous
    • 4 years ago
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    x+5sqrt(x) = 6? Well I mean take a look, what number plus 5 times it's square root is 6?

  19. anonymous
    • 4 years ago
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    okay i get it i never thought of it like that

  20. anonymous
    • 4 years ago
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    Keep in mind, we're still calculating the first half. So now, the upper curve minus the lower curve (well they're lines, but whatever) is: 3 - (6-x)/2

  21. anonymous
    • 4 years ago
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    So so far, the first half of our area is: Integral (3 - (6-x)/2, 0,1)

  22. anonymous
    • 4 years ago
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    Now let's find the 2nd half: This time the lower limit is 1, because we are starting at one, and the upper limit is the intersection of y=3 and y = 5/2sqrt(x)

  23. anonymous
    • 4 years ago
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    (5/2)*sqrt(x) = 3 sqrt(x) = 6/5 => x = 36/25

  24. anonymous
    • 4 years ago
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    So 2nd integral is again, upper curve minus the lower curve from 1 to 36/25, or Integral (3 - (5/2)*sqrt(x), 1, 36/25)

  25. anonymous
    • 4 years ago
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    Calculate each integral and then add them to get 107/300 (lol I cheated, used wolf to compute)

  26. anonymous
    • 4 years ago
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    so it is .356666667

  27. anonymous
    • 4 years ago
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    Yea that looks right.

  28. anonymous
    • 4 years ago
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    k i will try it and tell you if it works

  29. anonymous
    • 4 years ago
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    it works thanks sooooo much you have no idea

  30. anonymous
    • 4 years ago
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    np.

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