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![\bf \begin{cases} c(x)=4x-2\\ d(x)=x^2+5x \end{cases}\qquad (c\circ d)(x)\iff c[\ d(x)\ ] \\\\\\ c[\ d(x)\ ]=4\boxed{d(x)}-2\implies c[\ d(x)\ ]=4\left( \boxed{x^2+5x} \right)-2 \\\\\\ c[\ d(x)\ ]=4x^2+20x-2](/tpl/images/0366/8944/d66b6.png)
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If c(x) = 4x-2 and d(x) = x^2+5x, what is (cod)(x)...
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