$$ ဆိုရင် အလည်မှာဖြစ်သွားမှာဖြစ်ပြီး ရိုးရိုး $ ဆိုရင်တော့ ရှေ့မှာပဲရှိနေမှာပါ။
Precipitation of barium sulfate: \({SO4^2- + Ba^2+ -> BaSO4 v}\)
_Precipitation of barium sulfate:_ \\({SO4^2- + Ba^2+ -> BaSO4 v}\\)When $a \ne 0$, there are two solutions to $(ax^2 + bx + c = 0)$ and they are $$ x = {-b \pm \sqrt{b^2-4ac} \over 2a} $$
When $a \ne 0$, there are two solutions to $(ax^2 + bx + c = 0)$ and they are
$$ x = {-b \pm \sqrt{b^2-4ac} \over 2a} $$$$\int*{a}^{b} x^2 dx$$
$$\int*{a}^{b} x^2 dx$$$$ \frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{u^{s-1}}{e^{u}-1}\mathrm{d}u $$
$$
\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{u^{s-1}}{e^{u}-1}\mathrm{d}u
$$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$
$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$$ \frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{u^{s-1}}{e^{u}-1}\mathrm{d}u $$
$$
\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{u^{s-1}}{e^{u}-1}\mathrm{d}u
$$$$ \begin{equation*} \left\lbrace \begin{aligned} x \equiv & a_1 \bmod m_{1}\ x \equiv & a_2 \bmod m_{2}\ & \vdots\ x \equiv & a_n \bmod m_{n}\ \end{aligned} \right. \end{equation*} $$
$$
\begin{equation*}
\left\lbrace
\begin{aligned}
x \equiv & a_1 \bmod m_{1}\\
x \equiv & a_2 \bmod m_{2}\\
& \vdots\\
x \equiv & a_n \bmod m_{n}\\
\end{aligned}
\right.
\end{equation*}
$$$$ \begin{align*} a & = b \ X &\sim {\sf Norm}(10, 3) \ 5 & \le 10 \end{align*} $$
$$
\begin{align*}
a & = b \\
X &\sim {\sf Norm}(10, 3) \\
5 & \le 10
\end{align*}
$$$\lim\limits_{n\rightarrow\infty}{\left(1+\frac{1}{n}\right)^n}$
$\lim\limits_{n\rightarrow\infty}{\left(1+\frac{1}{n}\right)^n}$$$\mathop {\lim }\limits_{x \to c} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to c} \frac{{f’\left( x \right)}}{{g’\left( x \right)}}$$
$$\mathop {\lim }\limits\_{x \to c} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits\_{x \to c} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}$$