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

Given BD∥AE and △BCD∼△ACE , what is the m∠CAE?

Let ${\displaystyle y=f\left(x\right)}$ be a function D=(-12,13) and range ${\displaystyle R=(-\mathrm{\infty },\mathrm{\infty }).}$ Find the domain D and range R for the following functions and enter your answers using interval notation. Keep in mind order of operations. Be sure your intervals are in the correct order, and enter exact answers only (no approximations).
a) ${\displaystyle y=f\left(\frac{1}{2}x\right)}$
b) ${\displaystyle y=5f\left(2x\right)-2}$

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If you double the net force on an object, youll

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A point mass m having charge q is connected with massless spring to a rigid wall on a horizontal surface.A horizontal uniform electric field E is switched on and mass is displaced through a distance x from equilibrium position.
Work done by electric force is $W=qE.x$ as electric force is constant.
But F(external) on spring mas system is equal to kx. So $W=kx.x=k{x}^2$ (equation 1) as external force i.e, electric force is constant.
We know that work done is stored in the form of potential energy and we know that potential energy stored in the spring is $1/2k{x}^2$.
So $W=1/2k{x}^2$ (let it be equation 2)
But the above 2 equations are not equal. From where did the 1/2 come from in the 2nd equation?

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Calculate all the values of s such that $$(s-<!--\; -\; -->\frac{21}{8}{)}^2=\frac{361}{44}$$.

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What angle is subtended at the center of a circle of radius 2 km by an arc of length 9 m?

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How to derive this simple equality?
Let us define
$L_i\triangleq <!--\; \triangleq \; -->\log <!--\; \; -->\left({\displaystyle \frac{Prob(x_i=+1)}{Prob(x_i=-<!--\; -\; -->1)}}\right)$
$E\{x_i\}\triangleq <!--\; \triangleq \; -->Prob(x_i=+1)-<!--\; -\; -->Prob(x_i=-<!--\; -\; -->1)$
I need to show that
$E\{x_i\}=\tanh <!--\; \; -->(L_i/2)$
I can write
$\begin{array}{}\text{(1a)}& E\{x_i\}& =e^{log(Prob(x_i=+1)-<!--\; -\; -->Prob(x_i=-<!--\; -\; -->1))}\text{(1b)}& & =e^{L_i}\end{array}$
if I use the following:
$\tanh <!--\; \; -->(z)={\displaystyle \frac{\sinh <!--\; \; -->(z)}{\cosh <!--\; \; -->(z)}}={\displaystyle \frac{e^z-<!--\; -\; -->e^{-<!--\; -\; -->z}}{e^z+e^{-<!--\; -\; -->z}}}={\displaystyle \frac{e^{2z}-<!--\; -\; -->1}{e^{2z}+1}}$
and put $z=L_i/2$, then
$\begin{array}{}\text{(2)}& \tanh <!--\; \; -->(L_i/2)={\displaystyle \frac{e^{L_i}-<!--\; -\; -->1}{e^{l_i}+1}}\end{array}$
how do I arrive at (2) using (1a)

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What is the general solution of the differential equation ${\displaystyle y\prime -2xy=x^3}$?

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