Respuesta :

[tex]\bf c^2=a^2+b^2\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ ------\\ a=m^2-n^2\\ b=2mn \end{cases}\\\\\\ c^2=(m^2-n^2)^2+(\underline{2mn})^2 \\\\\\ c^2=m^4-2m^2n^2+n^4+\underline{2^2m^2n^2} \\\\\\ c^2=m^4-2m^2n^2+n^4+{4m^2n^2} \\\\\\ c^2=m^4+2m^2n^2+n^4\impliedby \textit{perfect square trinomial} \\\\\\ c^2=(m^2+n^2)^2\implies c=\sqrt{(m^2+n^2)^2}\implies c=m^2+n^2[/tex]

Answer:

It's d

Step-by-step explanation:

I got it correct on the review