$\frac{3}{8}$ $\frac{3}{x}$
2.49915/5x =(1/200)+1/(100+x)
2.42x1.23
\int_{-\infty}^{0}\frac{dx}{(x+1)^2}
5=-2*x+3
Find h(-4) and one value of x which h (x) = -3
-2x+3=0
-7 - (-2)=
0.3x+0.4y=100 , 0.5x+0.2y=120
\begin{aligned}
& 2y-3x = -27
\\\\
& 5y+3x=6
\end{aligned}