Georgia Milestones Algebra II Building Functions. Practice it free.

Set $y = \dfrac{x}{2} - 5\textsf{.}$ Swap: $x = \dfrac{y}{2} - 5\textsf{.}$ Add $5$ and multiply by $2\textsf{:}$ $y = 2x + 10\textsf{.}$ So $f^{-1}(x) = 2x + 10\textsf{.}$

Subtracting $3$ outside the function decreases every output by $3$, so the graph shifts $3$ units down.

Adding $9$ outside $f$ increases every output by $9\textsf{.}$ The graph shifts up $9$ units.

A right shift of $3$ replaces $x$ with $x - 3$. At the $y$-intercept, $x = 0$, so $5(-3) + 3y = 15$. Then $-15 + 3y = 15$, $3y = 30$, and $y = 10$.

The product rule rewrites a log of a product as a sum of logs, so $\log(3\cdot 4) = \log 3 + \log 4\textsf{.}$

The common difference is $d = -3$. With $a_1 = 25$, the $15$th term is $a_{15} = 25 + 14(-3) = -17$.

Work from the outside in. Since $m(3) = 1$, we need $n(k(x)) = 3$. Since $n(3) = 3$, we need $k(x) = 3$. From the table, $k(5) = 3$, so $x = 5$.

Following the pattern:
1st term, or $n = 1\textsf{:}$ $10 \cdot 2^{0}$
2nd term, or $n = 2\textsf{:}$ $10 \cdot 2^{1}$
3rd term, or $n = 3\textsf{:}$ $10 \cdot 2^{2}$
4th term, or $n = 4\textsf{:}$ $10 \cdot 2^{3}$
Since each term has an exponent that is $1$ less than the term number, the $n$th term is $a_n = 10 \cdot 2^{n-1}$.