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$\frac{20b}{{\left(4b^3\right)}^3}$

Geometric proof of ${\displaystyle \frac{{\sin 60}^{\circ }}{{\sin 40}^{\circ }}=4{\sin 20}^{\circ }{\sin 80}^{\circ }}$

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Replace the Cartesian equations with equivalent polar equations. $$x^2+(y-<!--\; -\; -->2{)}^2=4$$

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For each of the following differential equations, determine the general or particular solution: ${\displaystyle 2x{y}^{\prime }+y=y^2\log x}$

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If an ideal cylindrical inductor/electromagnet is charged and then discharged, will the energy lost by charging the inductor/electromagnet be equal to the energy gained by discharging it or will some of it be lost to radiation? Based on what I know about magnetism, there should be no loss.

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Is it allowed to have the zeroth-component of a four-velocity be negative?
This is referring to $V^0$ for a curved space metric with signature $-+++$.

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differential equations of second order
How may I solve this differential equations:
$$y^{″}+4y=12x^2-<!--\; -\; -->16x\cos <!--\; \; -->(2x)?$$

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Show that $\stackrel{\to <!--\; \to \; -->}{a}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{b}=\frac{1}{4}|\stackrel{\to <!--\; \to \; -->}{a}+\stackrel{\to <!--\; \to \; -->}{b}{|}^2-<!--\; -\; -->\frac{1}{4}|\stackrel{\to <!--\; \to \; -->}{a}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{b}{|}^2$
I have tried:
$$\begin{gathered} \stackrel{\to <!--\; \to \; -->}{a}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{b}=\frac{1}{4}|\stackrel{\to <!--\; \to \; -->}{a}+\stackrel{\to <!--\; \to \; -->}{b}{|}^2-<!--\; -\; -->\frac{1}{4}|\stackrel{\to <!--\; \to \; -->}{a}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{b}{|}^2 \\ \stackrel{\to <!--\; \to \; -->}{a}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{b}=\frac{1}{4}(|\stackrel{\to <!--\; \to \; -->}{a}{|}^2+2|\stackrel{\to <!--\; \to \; -->}{a}||\stackrel{\to <!--\; \to \; -->}{b}|+|\stackrel{\to <!--\; \to \; -->}{b}{|}^2)-<!--\; -\; -->\frac{1}{4}(|\stackrel{\to <!--\; \to \; -->}{a}{|}^2-<!--\; -\; -->2|\stackrel{\to <!--\; \to \; -->}{a}||\stackrel{\to <!--\; \to \; -->}{b}|+|\stackrel{\to <!--\; \to \; -->}{b}{|}^2) \\ \stackrel{\to <!--\; \to \; -->}{a}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{b}=|\stackrel{\to <!--\; \to \; -->}{a}||\stackrel{\to <!--\; \to \; -->}{b}| \end{gathered}$$
However it seems that the equation only holds true when $\stackrel{\to <!--\; \to \; -->}{a}$ and $\stackrel{\to <!--\; \to \; -->}{b}$ are collinear.
Is that true?

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What is the z-score of sample X, if $n=36,\mathrm{\mu <!--\; \mu \; -->}=13.7,\text{St.Dev}=21,{\mathrm{\mu <!--\; \mu \; -->}}_X=9.9$?

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