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  1. itssimplyme

    Can you double check these questions. I haven't looked at the second question, but the first...

    Can you double check these questions. I haven't looked at the second question, but the first seems too complicated to be right. I'll reply as soon as you get back to me on these
  2. itssimplyme

    1. Sorry about the late reply. deep in my own studies right now. Alright, let's see here...

    1. Sorry about the late reply. deep in my own studies right now. Alright, let's see here.... maximum of that.... -that's y= 1/√cosx + sinx + 5 , I'm assuming. So the maximum will be at a point where the gradient is 0. that is, dy/dx=0 So we have to work out dy/dx of this equation.
  3. itssimplyme

    I'm born in '94 NOT '92 (Which it insists on saying, for some odd reason....

    I'm born in '94 NOT '92 (Which it insists on saying, for some odd reason....
  4. itssimplyme

    ....wha- ZOMG< WTH?! T^T

    ....wha- ZOMG< WTH?! T^T
  5. itssimplyme

    Mathematics: Post your doubts here!

    q3ii-: This should be alright if you know the formula. So you want the volume between two points. If you sketched the curve properly, you should get your limits: 0 and 2. Now use the formulae π∫(y^2)dx. You know y= (x-2)^2. So substitute that into your formula: 1. π∫((x-2)^2)^2 dx =...
  6. itssimplyme

    Does anyone know how to solve ln x = -2?

    Wait, I think you're onto something there...:confused:
  7. itssimplyme

    Does anyone know how to solve ln x = -2?

    I thought natural logs under zero were undefined, but this came up in a past paper question. Or at least, it's one of the steps you have to work through.
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