Alan

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There are two possibilities:  n = 0 or n = -1

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Just divide 32 by 8.

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$$(x+\frac{1}{x})^N=x^N+Nx^{N-1}x^{-1}+\frac{N(N-1)}{2!}x^{N-2}x^{-2}+...$$

The fifth term will therefore contain x^N-4x^-4 or x^N-8.  For this to be independent of x we must have N - 8 = 0 or N = 8

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It looks like nobody actually answered your question SquareRoot!  Here's an (admittedly oversimplified) explanation:

Suppose you switch on a torch and point it to your left.  At the same time you switch on another torch and point it to your right.  15 billion years later the light beams will be 30 billion light-years apart!

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The following link will tell you about the endorelational properties "reflexive", "symmetric" and "antisymmetric", which you can then use to answer your question.

Use asin (press 2nd then asin on the calculator on the home page here):

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\mathtt{0.37}}\right)} = {\mathtt{21.715\: \!617\: \!283\: \!264^{\circ}}}$$

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Ok.  In that case just replace tanθ = 440/h  by sinθ = 440/h where h is now the line of sight distance.

cosθ*dθ/dt = -440/h²dh/dt    when θ = 45°  then cosθ = 1/√2,  h = 440*√2 so

dh/dt = 440^2*2*(1/√2)*0.05/440 = (22*√2) m/s ≈ 31.1 m/s

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I don't understand what you are asking Melody; that is the speed at which the plane is moving away from the observer!

Or perhaps you are asking how it would be done if the plane were immediately above the observer.  In this case, take a small interval of time δt during which the angle changes by δθ.  Now we have  tan(δθ) = vδt/440.

When δθ is small, tan(δθ) = δθ, so δθ = vδt/440  or v = 440*δθ/δt.

Given the rate of change of angle, δθ/δt, we get the plane's speed.

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