OSTP Grade 11 CCRA 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 $y + 3 = -2(x - 5)\textsf{.}$

Subtract the fifth-year salary from the first: $\text{\char36}46{,}700 - \text{\char36}39{,}500 = \text{\char36}7{,}200$. Over five years there are four equal raises, so divide by $5 - 1 = 4$ to get $x = \text{\char36}1{,}800$. Add $\text{\char36}1{,}800$ each year after the first: $\text{\char36}39{,}500$, $\text{\char36}41{,}300$, $\text{\char36}43{,}100$, $\text{\char36}44{,}900$, $\text{\char36}46{,}700$. Sum all five years: $\text{\char36}215{,}500$.

The equation means $x - 14 = 20$ or $x - 14 = -20$. So $x = 34$ or $x = -6$. Both solutions are $-6$ and $34\textsf{.}$

Rewrite $|4x + 16| = |4(x + 4)| = |4|\,|x + 4| = 4|x + 4|$. Since $4|x + 4| = 52$, divide by $4$ to get $|x + 4| = 13$. The expression $x + 4$ equals $13$ or $-13$; the negative value is $-13\textsf{.}$