Adding attenuations in dB
Part of the advantages of using dB to express gain or attenuation is that they're inherently logarithmic.
So if you're speaking in dB, when you add a 10dB attenuation to another 10dB of attenuation, what you're saying is that you're multiplying the signal by $\sqrt{0.1}$, and then by $\sqrt{0.1}$ again.  The result is 20dB of attenuation -- or multiplying the signal by $\sqrt{0.01} = 0.1$.
So if your ultimate gain is $A$, as a normal number, and you have attenuations $a_1, a_2, \cdots$ in dB, then the actual $A$ by which you multiply your signal is $$A = 10^{-\frac{1}{20}\sum a_n}$$