Answer: A. $12,579c$

Explanation:1. Understand the problem: The speeds of particles A, B, and C are $a$, $b$, and $c$ meters per second, respectively. 1) The speed of particle A is 7,900% of the speed of particle C. 2) The speed of particle C is 0.008% of the speed of particle B. We need to find an expression for $a+b$ in terms of $c$.

2. Translate percentage statements into equations: From (1): $a = \dfrac{7900}{100} \times c = 79c$. From (2): $c = \dfrac{0.008}{100} \times b = 0.00008b$.

3. Express $b$ in terms of $c$: Using the second equation, $c = 0.00008b$, solve for $b$: $b = \dfrac{c}{0.00008} = \dfrac{c}{8 \times 10^{-5}} = \dfrac{10^5}{8} c = \dfrac{100000}{8} c = 12500c$.

4. Calculate $a+b$ in terms of $c$: Substitute the expressions for $a$ (which is $79c$) and $b$ (which is $12500c$) into $a+b$: $a+b = 79c + 12500c$ $a+b = (79 + 12500)c$ $a+b = 12579c$.

5. Compare with the given options: The expression $12579c$ matches option [A].

6. Conclusion: The value of $a+b$ in terms of $c$ is $12,579c$.