$\dfrac{15}{8}$
2
$1 + \dfrac{1}{2}\sqrt{2}$
$\dfrac{7}{4}$
$\sqrt{3}$

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$\text{In }\triangle ABC, \tan A=\frac34. \text{ In } \triangle ADE, \tan A=\frac{x}{2.5}. \quad\therefore \frac34 = \frac{x}{2.5}.$ Solving yields $x=\boxed{\frac{15}{8}}$