To determine the big-O estimates for the provided functions, let's break down each part systematically.

(a) $3n + n^3 + 4$

1. Big-O considers the term with the largest growth rate as $n \to \infty$.
2. The dominant term here is $n^3$ because it grows faster than $n$ or any constant.
3. So, $3n + n^3 + 4 \in O(n^3)$.

(b) $1 + 2 + 3 + \dots + n$

1. The sum of the first $n$ integers is given by: $1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$
2. Simplify to $\frac{n^2}{2} + \frac{n}{2}$.
3. The dominant term is $n^2$, so: $1 + 2 + 3 + \dots + n \in O(n^2)$

(c) $\log_{10}(2^n) + 10^{10} n^2$

1. The term $\log_{10}(2^n)$ simplifies as: $\log_{10}(2^n) = n \cdot \log_{10}(2)$ So, this term grows linearly as $n$.
2. The term $10^{10} n^2$ grows as $n^2$.
3. The dominant term here is $n^2$ because it grows faster than $n$ as $n \to \infty$.
4. Thus: $\log_{10}(2^n) + 10^{10} n^2 \in O(n^2)$

Final Big-O Estimates:

(a) $O(n^3)$
(b) $O(n^2)$
(c) $O(n^2)$

Would you like a detailed explanation of any of these steps? Here are 5 follow-up questions you might find helpful:

1. What are the rules for simplifying Big-O estimates?
2. How does logarithmic growth compare to polynomial growth in Big-O terms?
3. Why is $O(n^3)$ considered simpler than expressing all terms of a function?
4. Can constants ever affect the Big-O notation?
5. What practical implications does Big-O have for algorithm design?

Tip: Always focus on the highest-order term for Big-O estimation—it determines the asymptotic growth.