Answer:
AC = 9
Step-by-step explanation:
[tex]cosine=\dfrac{adjacent}{hypotenuse}\\\\\text{We have:}\\\\\cos C=\dfrac{9\sqrt{130}}{130}\\\\adjacent=AC\\\\hypotenuse=\sqrt{130}\\\\\text{Substitute:}\\\\\dfrac{9\sqrt{130}}{130}=\dfrac{AC}{\sqrt{130}}\qquad\text{cross multiply}\\\\130AC=(9\sqrt{130})(\sqrt{130})\qquad\text{use}\ \sqrt{a}\cdot\sqrt{a}=a\\\\130AC=(9)(130)\qquad\text{divide both sides by 130}\\\\\dfrac{130AC}{130}=\dfrac{(9)(130)}{130}\\\\AC=9[/tex]
