When $k \ge 10$, $k!$ must be multiple of $100$ whose last two digits will be $0$. Therefore, the desired results is the same as the tens digit of $7!+8! + 9!$. We can compute $7!=5040$, therefore $$7!+8!+9!\equiv 40+40\times 8 + 40\times 8\times 9 \equiv 40 + 20 + 80\equiv 40\pmod{100}$$