Question

$$8 + [ = 1$$

Answer

$$t=-(1-Bo*x)/(8*e^2*IM*l^2*n^2*h*m^2*s*u*b*r*sin(g))$$

Solution


Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[{\imath }^{3}{l}^{2}{n}^{2}th{e}^{2}{m}^{2}s(\sin{g})ubr\times 8+Box=1\]
Isolate \({\imath }^{2}\).
\[{\imath }^{2}\imath {l}^{2}{n}^{2}th{e}^{2}{m}^{2}s(\sin{g})ubr\times 8+Box=1\]
Use Square Rule: \({i}^{2}=-1\).
\[-1\times \imath {l}^{2}{n}^{2}th{e}^{2}{m}^{2}s(\sin{g})ubr\times 8+Box=1\]
Simplify  \(1\times \imath {l}^{2}{n}^{2}th{e}^{2}{m}^{2}s(\sin{g})ubr\times 8\)  to  \(8{l}^{2}{n}^{2}th{m}^{2}subr\imath {e}^{2}\sin{g}\).
\[-8{l}^{2}{n}^{2}th{m}^{2}subr\imath {e}^{2}\sin{g}+Box=1\]
Regroup terms.
\[-8{e}^{2}\imath {l}^{2}{n}^{2}th{m}^{2}subr\sin{g}+Box=1\]
Regroup terms.
\[Box-8{e}^{2}\imath {l}^{2}{n}^{2}th{m}^{2}subr\sin{g}=1\]
Subtract \(Box\) from both sides.
\[-8{e}^{2}\imath {l}^{2}{n}^{2}th{m}^{2}subr\sin{g}=1-Box\]
Divide both sides by \(-8\).
\[{e}^{2}\imath {l}^{2}{n}^{2}th{m}^{2}subr\sin{g}=-\frac{1-Box}{8}\]
Divide both sides by \({e}^{2}\).
\[\imath {l}^{2}{n}^{2}th{m}^{2}subr\sin{g}=-\frac{\frac{1-Box}{8}}{{e}^{2}}\]
Simplify  \(\frac{\frac{1-Box}{8}}{{e}^{2}}\)  to  \(\frac{1-Box}{8{e}^{2}}\).
\[\imath {l}^{2}{n}^{2}th{m}^{2}subr\sin{g}=-\frac{1-Box}{8{e}^{2}}\]
Divide both sides by \(\imath \).
\[{l}^{2}{n}^{2}th{m}^{2}subr\sin{g}=-\frac{\frac{1-Box}{8{e}^{2}}}{\imath }\]
Simplify  \(\frac{\frac{1-Box}{8{e}^{2}}}{\imath }\)  to  \(\frac{1-Box}{8{e}^{2}\imath }\).
\[{l}^{2}{n}^{2}th{m}^{2}subr\sin{g}=-\frac{1-Box}{8{e}^{2}\imath }\]
Divide both sides by \({l}^{2}\).
\[{n}^{2}th{m}^{2}subr\sin{g}=-\frac{\frac{1-Box}{8{e}^{2}\imath }}{{l}^{2}}\]
Simplify  \(\frac{\frac{1-Box}{8{e}^{2}\imath }}{{l}^{2}}\)  to  \(\frac{1-Box}{8{e}^{2}\imath {l}^{2}}\).
\[{n}^{2}th{m}^{2}subr\sin{g}=-\frac{1-Box}{8{e}^{2}\imath {l}^{2}}\]
Divide both sides by \({n}^{2}\).
\[th{m}^{2}subr\sin{g}=-\frac{\frac{1-Box}{8{e}^{2}\imath {l}^{2}}}{{n}^{2}}\]
Simplify  \(\frac{\frac{1-Box}{8{e}^{2}\imath {l}^{2}}}{{n}^{2}}\)  to  \(\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}}\).
\[th{m}^{2}subr\sin{g}=-\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}}\]
Divide both sides by \(h\).
\[t{m}^{2}subr\sin{g}=-\frac{\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}}}{h}\]
Simplify  \(\frac{\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}}}{h}\)  to  \(\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h}\).
\[t{m}^{2}subr\sin{g}=-\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h}\]
Divide both sides by \({m}^{2}\).
\[tsubr\sin{g}=-\frac{\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h}}{{m}^{2}}\]
Simplify  \(\frac{\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h}}{{m}^{2}}\)  to  \(\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}}\).
\[tsubr\sin{g}=-\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}}\]
Divide both sides by \(s\).
\[tubr\sin{g}=-\frac{\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}}}{s}\]
Simplify  \(\frac{\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}}}{s}\)  to  \(\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}s}\).
\[tubr\sin{g}=-\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}s}\]
Divide both sides by \(u\).
\[tbr\sin{g}=-\frac{\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}s}}{u}\]
Simplify  \(\frac{\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}s}}{u}\)  to  \(\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}su}\).
\[tbr\sin{g}=-\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}su}\]
Divide both sides by \(b\).
\[tr\sin{g}=-\frac{\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}su}}{b}\]
Simplify  \(\frac{\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}su}}{b}\)  to  \(\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}sub}\).
\[tr\sin{g}=-\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}sub}\]
Divide both sides by \(r\).
\[t\sin{g}=-\frac{\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}sub}}{r}\]
Simplify  \(\frac{\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}sub}}{r}\)  to  \(\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}subr}\).
\[t\sin{g}=-\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}subr}\]
Divide both sides by \(\sin{g}\).
\[t=-\frac{\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}subr}}{\sin{g}}\]
Simplify  \(\frac{\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}subr}}{\sin{g}}\)  to  \(\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}subr\sin{g}}\).
\[t=-\frac{1-Box}{8{e}^{2}\imath {l}^{2}{n}^{2}h{m}^{2}subr\sin{g}}\]