OSTP Grade 11 CCRA Functions. Practice it free.

The amount after $t$ years is $A(t) = P(1 + r)^{t}\textsf{.}$ Here $P = 500\textsf{,}$ $r = 0.20\textsf{,}$ and $t = 3\textsf{,}$ so $A(3) = 500(1.20)^{3} = 864\textsf{.}$

When $w = 2$, the exponent is $2/2 = 1$. So $L(2) = 180(0.88)^1 = 158.4$ lumens.

$10 \div 5 = 2$ half-lives. Remaining: $\left(\dfrac{1}{2}\right)^{2} = \dfrac{1}{4}$.

Slope-intercept form is $f(x) = mx + b\textsf{.}$ When $x = 0\textsf{,}$ the cost is $f(0) = 10\textsf{,}$ so $b = 10\textsf{.}$

The temperature decreases by $3$ degrees each hour, and at noon the temperature is $72\textsf{.}$ So $w(x) = -3x + 72\textsf{.}$

First write slope-intercept form: $5 = -4(1) + b$ gives $b = 9$, so $y = -4x + 9$. Rearranging into standard form gives $4x + y = 9\textsf{.}$

Set $y = 0$: $-(x + 8)(x - 2) = 0$. The factor $-1$ is nonzero, so the zeros come from $x + 8 = 0 \Rightarrow x = -8$ and $x - 2 = 0 \Rightarrow x = 2$. Of the choices, $-8$ is an x-coordinate of an x-intercept.

Set $y = \dfrac{2x + 6}{5}\textsf{.}$ Swap: $x = \dfrac{2y + 6}{5}\textsf{.}$ Multiply by $5\textsf{,}$ subtract $6\textsf{,}$ and divide by $2\textsf{:}$ $y = \dfrac{5x - 6}{2}\textsf{.}$ So $f^{-1}(x) = \dfrac{5x - 6}{2}\textsf{.}$