Answer
Due to Le Châtelier's principle, the $H_3O^+$ produced by the strong acid causes the equilibrium of the ionization of the weak acid to shift to the left and ionize even less than normally.
Therefore, the amount of hydronium ion produced by the weak acid is even smaller than the same by the strong acid, making valid the assumption that it can be neglected.
Work Step by Step
This is the equilibrium of the weak acid ionization:
$$HA(aq) + H_2O(l) \leftrightharpoons A^-(aq) + H_3O^+(aq)$$
When the strong acid reacts, it produces a large amount of $H_3O^+$
Since product is being added in large scale, Le Chatelier's principle says that the equilibrium is going to shift to the left, decreasing the production of hydronium ion by this weak acid.
In a mixture of acids, we calculate the total $H_3O^+$ by adding the amount given by each acid:
$$[H_3O^+]_{total} = [H_3O^+]_{strong \space acid} + [H_3O^+]_{weak \space acid}$$
Since $ [H_3O^+]_{strong \space acid} \gt\gt [H_3O^+]_{weak \space acid}$:
$[H_3O^+]_{weak \space acid} \approx 0$
$$[H_3O^+]_{total} \approx [H_3O^+]_{strong \space acid} $$
