If $2a+b=10$ and $2ab=9$, then one of the values of $2a-bis$:

We use the standard algebraic identity relating the square of a sum and the square of a difference:
$$(2a-b{)}^2=(2a+b{)}^2-4(2a)(b)$$
$$(2a-b{)}^2=(2a+b{)}^2-4(2ab)$$
Given values are $2a+b=10$ and $2ab=9$. Substituting these values into the equation:
$$(2a-b{)}^2=(10{)}^2-4(9)$$
$$(2a-b{)}^2=100-36$$
$$(2a-b{)}^2=64$$
Taking the square root on both sides:
$$2a-b=\pm \sqrt{64}=\pm 8$$
Thus, one of the possible values for $2a-b$ is $8$.