File:Calculations of Mass and Radius for a Low Density Black Hole 1.JPG

This image shows some kind of formula that could be converted to TeX. Storing formulas as images makes it harder to change them. TeX also helps making sure that they all use the same font and size.
A replacement has been proposed:

${\displaystyle {\begin{aligned}&\rho ={\frac {m}{{\frac {4}{3}}*\pi *r^{3}}}\\&{\frac {m}{\left({\frac {2Gm}{c^{2}}}\right)^{3}*{\frac {4}{3}}*\pi }}=1000Kg/M^{3}\\&{\frac {m}{{\frac {8G^{3}m^{3}}{c^{6}}}*{\frac {4\pi }{3}}}}=1000Kg/M^{3}\\&{\frac {3mc^{6}}{32*\pi *m^{3}G^{3}}}=1000Kg/M^{3}\\&3c^{6}=32000*\pi *m^{2}G^{3}*Kg/M^{3}\\&m={\sqrt {\frac {3c^{6}}{32000*\pi *G^{3}*Kg/M^{3}}}}={\frac {c^{3}/100}{\sqrt {3.2/3*\pi *G^{3}*Kg/M^{3}}}}\\&m\approx {\frac {2.7*10^{23}M^{3}/{\text{sec}}^{3}}{\sqrt {10/3*3*10^{-31}M^{9}*{\text{sec}}^{-6}*Kg^{-3}*Kg/M^{3}}}}={\frac {2.7*10^{27}M^{3}/{\text{sec}}^{3}}{\sqrt {10*10^{-31}M^{6}/{\text{sec}}^{6}*Kg^{-2}}}}\\&m\approx {\frac {2.7*10^{23}}{10^{-15}}}Kg=2.7*10^{38}Kg\\&r={\frac {2Gm}{c^{2}}}=4*10^{8}M\end{aligned}}}$

In your article, replace the image with:

<math>\begin{align}
&\rho=\frac{m}{\frac{4}{3}*\pi*r^3}\\
&\frac{m}{\left(\frac{2Gm}{c^2}\right)^3*\frac{4}{3}*\pi}=1000Kg/M^3\\
&\frac{m}{\frac{8G^3m^3}{c^6}*\frac{4\pi}{3} }=1000Kg/M^3\\
&\frac{3mc^6}{32*\pi*m^3G^3}=1000Kg/M^3\\
&3c^6=32000*\pi*m^2G^3*Kg/M^3\\
&m=\sqrt{\frac{3c^6}{32000*\pi*G^3*Kg/M^3} }=\frac{c^3/100}{\sqrt{3.2/3*\pi*G^3*Kg/M^3} }\\
&m\approx\frac{2.7*10^{23}M^3/\text{sec}^3}{\sqrt{10/3*3*10^{-31}M^9*\text{sec}^{-6}*Kg^{-3}*Kg/M^3} } = \frac{2.7*10^{27}M^3/\text{sec}^3}{\sqrt{10*10^{-31}M^6/\text{sec}^6*Kg^{-2} } }\\
&m\approx\frac{2.7*10^{23} }{10^{-15} }Kg=2.7*10^{38}Kg\\
&r=\frac{2Gm}{c^2}=4*10^8M
\end{align}</math>