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Mathematics: Post your doubts here!

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Can someone please help me finish off this question? I sort of got stuck..

The question is as below:
24cwi8k.png


I solved and reached:
2hz3hw3.png


Now the last step I'm supposed to:
2h2ghv7.png
which should be quite simple, don't know what I'm missing.

Can someone please help me out?
 
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When you expand ((1+x^2)^(2)) you get is x^4 + 2x^2 + 1

Now from part a you got the expansion like ; 32 + 80x^2 + 80x^4

Now multiply each brackets.

(x^4 + 2x^2 + 1)(32 + 80x^2 + 80x^4) = 32x^4 + 80x^6 + 80x^8 + 64x^2 + 160x^4 + 160x^6 + 32 + 80x^2 + 80x^4

Now we need coff. of x^4, hence, add 'em, 32 + 160 +80 = 272 :)
 
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Do you know how to do the question using vector move step?
the vector move step is pretty easy, you have to look at A and B look at how many x steps it takes and how many y steps it takes and in what direction for you to move from A to B or vice versa for example, look at A to B, it moves from 2 to -2 in x direction, so use this:
Final position - initial position
(-2) - (2) = -4 so 4 steps back in x direction
and for y direction:
8 - 14 = -6 so six steps down.
Now you can find the co ordinates for D using equation of line and stuff and then use this translation to find the co ordinates for C.
 
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I dont understand the last step, the one right before the final answer..
How does the denominator finalize to become 1? I know that the numerator is the answer we're trying to reach, okay. But the denominator should be 1, now how does 1 + (tant) (tan 1/4pi) equate to 1? (tan 1/4pi) is equal to 1 not zero, so how did tant get cancelled off? Shouldn't it be, tant - (tan 1/4pi) / (tan 1/4pi) + tant ?
I don't think it's possible to substitute 1 with (tan 1/4pi) only in the numerator..

Please do explain.
 
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Dude, first off, you really need to cool off and let go of that khuari a bit. Give the lady some respect. Secondly, what you are asking is pretty simple. Look, you need to find the angle between a curve and a tangent, right? That would require you to know two things: The gradient of the tangent to the curve and the other is the point on the curve to which you need angle from. Next you differentiate the equation of the curve and find the gradient of that point by substitution and now all you need to know is the fact that a gradient is just simple the ratio of the height of an point to the width of it, that means,

Rise/Run. Which is just like the tan formula, Tan@ = perp/base so if you take the tan inverse of the gradient of that point and then take the tan inverse of the gradient of the tangent, you will have two angles, subtract the larger angle from the smaller angle and you have the angle between a tangent and a point on the graph. Lastly, I would humbly ask you to remove that comment from the thread, it looks extremely degrading on such an intellectual forum.
I dont find anything "degrading on this intellectual forum".But anyways i dont bother all this.Nice explanation :) .But my question is that the gradient of tangent and curve will be the same at that point ryt?.So how are u supposed to find the angle my friend?
 
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