To show that these equalities are true, we will employ the definitions of Big-O and Theta notation. Here's a step-by-step explanation for each part:

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(a) $5n^2 + 6n = \Theta(n^2)$

1. Definition of $\Theta$-notation: $f(n) = \Theta(g(n))$ means that there exist constants $c_1, c_2, n_0$ such that for all $n \geq n_0$: $c_1 g(n) \leq f(n) \leq c_2 g(n)$
2. Here, $f(n) = 5n^2 + 6n$ and $g(n) = n^2$.
3. As $n$ grows large, the term $5n^2$ dominates $6n$, meaning that: $5n^2 + 6n = \Theta(n^2)$
   • For sufficiently large $n$, the linear term $6n$ becomes negligible compared to $5n^2$, so $5n^2 + 6n$ grows asymptotically like $n^2$.

Thus, $5n^2 + 6n = \Theta(n^2)$.

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(b) $n! = O(n^n)$

1. Definition of $O$-notation: $f(n) = O(g(n))$ means that there exist constants $c$ and $n_0$ such that for all $n \geq n_0$: $f(n) \leq c g(n)$
2. Here, $f(n) = n!$ and $g(n) = n^n$.
3. Stirling's approximation gives us an approximation for $n!$: $n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$
4. $\left(\frac{n}{e}\right)^n$ grows slower than $n^n$, meaning: $n! = O(n^n)$
   • This is because $n!$ grows slower than $n^n$, and hence we can find a constant $c$ such that $n! \leq c \cdot n^n$ for large $n$.

Thus, $n! = O(n^n)$.

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(c) ( \frac{n^2}{n} + 6 \cdot 2^n = \Theta \left(