Alan

If  you mean a series expansion of something like ln(1 + x) then you can use a Taylor series to get:

$$ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...$$

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$$\sum_{n=0}^{n=\infty}n^3z^{2n}=\frac{z(z^2+4z+1)}{(z-1)^4}$$

This has a singularity at z = 5, so I guess the radius of convergence of your expression is when x²/5 = 1 or x = ±√5

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Edit: Just noticed that Chris has answered the same question elsewhere!

Work is a form of energy transfer.  The energy transferred to the water by the pump is used, first, to raise the water to the surface (potential energy) and then any remaining energy that has been transferred appears as kinetic energy as the water shoots out (presumably from a hose) at the top.

Not sure if this answers your question!

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(a) Work per minute = Potential energy per minute = mgh

Work each minute = 800*9.8*14 J = 109760 J

(b) Work per minute = kinetic energy per minute

Work each minute = (1/2)*800*18² J = 129600 J

(c) Power = rate of doing work = (109760 + 129600)/60  ≈ 3989 W  (because 1 W = 1 J/s)

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4x² - 4x - 8 - 4x²+5x → x - 8

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$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\frac{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\right)}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{10}}}}\right)}}\right)} = {\mathtt{54.435\: \!733\: \!877\: \!82^{\circ}}}$$

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This uses what is known as Leibnitz's rule for differentiation of integrals, which in its general form, where both limits are functions of the variable with respect to which everything is being differentiated, is:

In the original question here only the 2nd term on the RHS above is non-zero

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Yes, in this case, if you resolve it parallel and perpendicular to the direction of the force, and only use the component in the direction of the force to calculate the work done.

However, I would recommend not doing this, but stick to resolving the force instead, as it is less likely to lead to confusion in more complicated situations.

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Just use the on-site calculator:

$$\left({\mathtt{30}}{!}\right) {mod} \left({\mathtt{899}}\right) = {\mathtt{464}}$$

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Oops!  Missed the 4 equations bit!

In general there won't be a unique solution, though it's possible that all four equations are consistent.  Can't tell without seeing the equations.

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