ISAT Grade 11 Functions. Practice it free.

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

$3^{2} = 9\textsf{,}$ so $\log_{3}9 = 2\textsf{.}$

The amount after $t$ years is $A(t) = P(1 + r)^{t}\textsf{.}$ Here $P = 256\textsf{,}$ $r = 0.25\textsf{,}$ and $t = 3\textsf{,}$ so $A(3) = 256(1.25)^{3} = 500\textsf{.}$

When $t = 3$, the exponent is $3/3 = 1$. So $S(3) = 450(0.85)^1 = 382.5$ centimeters.

$2 \div 1 = 2$ half-lives. Remaining: $\left(\dfrac{1}{2}\right)^{2} = \dfrac{1}{4}$.

Set $f(x) = 0$: $6(x + 9)(x - 13) = 0$. The constant $6 \neq 0$, so the zeros are $x = -9$ and $x = 13$. Of the choices, $-9$ is an x-coordinate of an x-intercept.

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 $a_n = 10 \cdot 2^{n-1}$.