#### Answer

10

#### Work Step by Step

Since ACB is an inscribed angle and is thus equal to half of the measure of 180 degrees, it follows that ACB is a right angle. Thus, we know:
$AC^2 + CB^2 = AB^2$
We now solve for AC and CB. We call the length of CM k and the length CN j. Thus, we find:
$(2k)^2 + j^2 = 73$
And:
$k^2 + (2j)^2 = 52$
Thus, we obtain:
$4(52 - 4j^2) + j^2 = 73 \\ -15j^2 = -135 \\ j = 3$
This means:
$k = \sqrt{52 - 4(9)} = 4$
Since k is half of AC and j is half of CB, we find:
$AB = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$