NAEP Grade 12 Algebra. Practice it free.

At the surface, the beam has traveled $0$ millimeters, so $t = 0$. Then $N(0) = 680(0.72)^0 = 680$ particles.

$500 \times 1.10 = 550$.

Slope-intercept form is $f(x) = mx + b\textsf{.}$ When $x = 0\textsf{,}$ the amount remaining is $f(0) = 48\textsf{,}$ so $b = 48\textsf{.}$

The outputs increase by $3$ each time $x$ increases by $1\textsf{,}$ so the constant rate is $3\textsf{.}$ When $x = 0\textsf{,}$ $s(0) = 7\textsf{,}$ so $s(x) = 3x + 7\textsf{.}$

Point-slope form is $y - y_1 = m(x - x_1)$. With $m = -2$ and $(x_1, y_1) = (5, -3)$, the equation is $y - (-3) = -2(x - 5)$, or $y + 3 = -2(x - 5)\textsf{.}$

Take $\ln$ of both sides: $x+3=\ln 8$. Subtract $3$ to get $x=\ln 8-3\textsf{.}$

Set $f(x) = 0$: $2(x - 6)(x - 1)(x + 10) = 0$. The constant $2 \neq 0$, so the zeros are $x = 6$, $x = 1$, and $x = -10$. Among the choices, only $-10$ is an x-coordinate of an x-intercept.

The equation is in vertex form $f(x) = a(x - h)^2 + k$ with $a = 1 > 0$, so the graph opens upward and the minimum occurs at the vertex. Here $h = 5$, so the vertex is at $x = 5$. Therefore, $f(x)$ reaches its minimum when $x = 5$.