HSF.LE.A.2 Math Practice Questions — Sports Academy Games

Practice playeasy

An account starts with $\text{\char36}500$ and earns $20\%$ interest compounded yearly.

Choose the function model that correctly calculates the amount in the account after $3$ years.

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 $A(3) = 500(1.20)^{3} = 864\textsf{.}$

Practice playeasy

You invest $\text{\char36}500$ at $10\%$ annual interest for $1$ year. What is the total?

$500 \times 1.10 = 550$ .

Practice playeasy

The function $g$ is defined by $g(t) = 840(0.5)^{t/4}$ . The function $g$ models the mass, in milligrams, of a medicine in a patient's bloodstream, where $t$ is the number of hours after an injection. According to the model, what is the estimated mass, in milligrams, of the medicine at the time of the injection?

At the time of injection, $t = 0$ . So $g(0) = 840(0.5)^0 = 840$ milligrams.

Practice playeasy

Taxi cost, in dollars, after $x$ miles:
$\begin{array}{c|c} x & f(x) \\ \hline 1 & 6 \\ 2 & 10 \\ 3 & 14 \\ 4 & 18 \end{array}$
The cost is $f(x)=mx+2\textsf{.}$ What is $m\textsf{?}$

Slope-intercept form is $f(x) = mx + b\textsf{.}$ The cost increases by $\text{\char36}4$ per mile, so $m = 4\textsf{.}$

Practice playeasy

The table shows values of a linear function $g\textsf{.}$
$\begin{array}{c|c} x & g(x) \\ \hline 0 & 15 \\ 1 & 12 \\ 2 & 9 \\ 3 & 6 \end{array}$
Which of the following is $g(x)\textsf{?}$

The outputs decrease by $3$ each time $x$ increases by $1\textsf{,}$ so the constant rate is $-3\textsf{.}$ When $x = 0\textsf{,}$ $g(0) = 15\textsf{,}$ so $g(x) = -3x + 15\textsf{.}$

Practice playeasy

A line has slope $4$ and passes through the point $(-1, 2)$ . Which equation represents this line in point-slope form?

Point-slope form is $y - y_1 = m(x - x_1)$ . With $m = 4$ and $(x_1, y_1) = (-1, 2)$ , the equation is $y - 2 = 4(x - (-1))$ , or $y - 2 = 4(x + 1)\textsf{.}$

Practice playeasy

An $80$ -milligram sample of a radioactive isotope has a half-life of $4$ years.

Choose the function model that correctly calculates the amount remaining after $12$ years.

The remaining amount after $t$ years is $A(t) = 80\left(\dfrac{1}{2}\right)^{\textstyle\frac{t}{h}}\textsf{,}$ where $h$ is the half-life. Here $t = 12$ and $h = 4\textsf{,}$ so $A(12) = 80\left(\dfrac{1}{2}\right)^{\textstyle\frac{12}{4}} = 80\left(\dfrac{1}{2}\right)^{3} = 10\textsf{.}$

Practice playeasy

A substance has a half-life of $4$ years. What fraction remains after $12$ years?

$12 \div 4 = 3$ half-lives. Remaining: $\left(\dfrac{1}{2}\right)^{3} = \dfrac{1}{8}$ .

HSF.LE.A.2 math practice. Learn by doing.

Skills aligned to HSF.LE.A.2

Exponential functions in context (SAT)

Linear functions from tables (ACT/SAT)

Linear rule from a function table (ACT)

Slope-point line equations (ACT/SAT/PSAT 8/9/10)

Exponential functions

Exponential and log formulas

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