Recent content by khatche4

  1. K

    Solving Equations: 2 Equations, 1 Unknown Variable

    I am given 2 equations (for a calculus problem)... y(x2 +4)=5 and x2 +y2 -2y=0. I know how to solve for y in terms of x for the first equation, but I am seriously stumped on how to get passed: y2 -2y=-x2
  2. K

    Area of the Region Inside the Curve.

    Yeah. It's due in the morning... Thanks for your help, though. It wasn't the -2, though, because I changed it to check...
  3. K

    Area of the Region Inside the Curve.

    When I use: \int[(1+\sqrt{1-x^2})-(\frac{5}{x^2+4}]^2 dx a=-2 and b=3 (my professor is letting us use our calculator) When I put it in, it tells me that it's an unreal answer? (And I know we aren't supposed to get them).
  4. K

    Area of the Region Inside the Curve.

    One more question, I think... Because a half circle is involved, will I use \frac{\pi r^2}{2} for the equation of the circle?
  5. K

    Area of the Region Inside the Curve.

    Thank you VERY much! I think that will definitely help... I'll post my answer (if I get it done) on here. Thanks!
  6. K

    Area of the Region Inside the Curve.

    Ok, so, now I'm trying to figure out the integral. Going from: with a=-2 and b=3 \int((x^2+y^2-2y)-(y(x^2+4))dx obviously I have to solve for y still, correct? That's why I'm confused, because when solving for y... The circle has a y^2-2y and how would I solve for y in that case?
  7. K

    Area of the Region Inside the Curve.

    Oh my gosh... I'm dumb as a box of rocks. That totally makes sense. So, I get y=-2, 1, and 3... But I'll be using -2 and 3, correct? And it is y, right? Hah. I really appreciate your help.. I'm still going to need some more, I'm sure... But I seriously do appreciate it.
  8. K

    Area of the Region Inside the Curve.

    I already tried that, and then it wasn't easily factored... So that's when I got stuck.
  9. K

    Area of the Region Inside the Curve.

    (\frac{5}{x^2+4})2-2(\frac{5}{x^2+4}=-x2 which then gives me I have (\frac{5}{x^2+4})2-(\frac{10}{x^2+4})=-x2 But when I graphed it, I don't really think that's right...
  10. K

    Area of the Region Inside the Curve.

    **I totally meant in terms of x... Wow. That was silly.
  11. K

    Area of the Region Inside the Curve.

    I know that, I just don't know how to solve for x in the situation where there is a "y^2-2y) situation...
  12. K

    Area of the Region Inside the Curve.

    How would you solve for x when in the equation for the circle we're given "y^2-2y"? I know I should know this, but I don't... :(
  13. K

    Area of the Region Inside the Curve.

    So, I will set, for example, \sqrt{2y-y^2}=\sqrt{\frac{5}{y}-4}? Then solve it from there, to find a and b?
  14. K

    Area of the Region Inside the Curve.

    Would it be possible if I could get help getting started? I don't have any work yet, but I am really trying to figure it out.. Problem: Determine the area of the region inside the curve x^2+y^2-2y=0 and outside y(x^2+4)=5. It has a graph and I'll upload a picture if needed. Basically...
  15. K

    Solving System of Equations: Partial Fraction Decomposition Help

    Hello. I'm trying to solve a system of equations problem (for my Partial Fraction Decomposition problem)... 1 = -4 + 2B + 2C + D + E 3 = 4 - 2B + 2C - D + E 3 = -25 + 20B + 10C + 4D + 2E 5 = 25 - 20B + 10C - 4D + 2E So that boils down to.. 1 = 2C + E 3 = 2C + E 3 = 10C + 2E 5 =...
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