Respuesta :

dalendrk
[tex]c(x)=\dfrac{5}{x-2}\\the\ domain:\\D_c:x-2\neq0\to x\neq2\\\\d(x)=x+3\\the\ domain:\\D_d:x\in\mathbb{R}\\\\(cd)(x)=\dfrac{5}{x-2}\cdot(x+3)=\dfrac{5(x+3)}{x-2}\\the\ domain:\\D_{cd}:x\neq2[/tex]

Answer: The answer would be B

Step-by-step explanation:

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