Home / Digital SAT Math Practice Questions – Medium : Nonlinear functions

# Digital SAT Math Practice Questions – Medium : Nonlinear functions

## SAT MAth Practice questions – all topics

• Advanced Math Weightage: 35% Questions: 13-15
• Equivalent expressions
• Nonlinear equations in one variable and systems of equations in two variables
• Nonlinear functions

## SAT MAth and English  – full syllabus practice tests

[Calc]  Question   Medium

Under certain conditions, the equation $$9.8 T^2=4 \pi^2 L$$ models the time $$T$$, in seconds, it takes for a pendulum to complete one cycle, where $$L$$ is the length, in meters, of the pendulum. Solving the equation for $$T$$ yields $$T=2 \pi \sqrt{\frac{L}{9.8}}$$. Which of the following is the best interpretation of $$T=2 \pi \sqrt{\frac{12}{9.8}}$$ in this context?
A) The time, in seconds, it takes a pendulum of $$2 \pi \sqrt{\frac{12}{9.8}}$$ length meters to complete one cycle
B) The time, in seconds, it takes a pendulum of length 9.8 meters to complete one cycle
C) The time, in seconds, it takes a pendulum of length 12 meters to complete one cycle
D) The time, in seconds, it takes a pendulum of length $$\frac{12}{9.8}$$ meters to complete one cycle

Ans:C

The given equation $$9.8 T^2 = 4 \pi^2 L$$ models the time $$T$$ it takes for a pendulum to complete one cycle, where $$L$$ is the length of the pendulum. Solving this equation for $$T$$, we get $$T = 2 \pi \sqrt{\frac{L}{9.8}}$$.

Given $$T = 2 \pi \sqrt{\frac{12}{9.8}}$$, we are looking at the time it takes for a pendulum with a length of $$12$$ meters.

So, the best interpretation is:
$\boxed{\text{C) The time, in seconds, it takes a pendulum of length 12 meters to complete one cycle}}$

[Calc]  Question   medium

The graph models the mass $$y$$, in nanograms, of cobalt-60 (Co-60) remaining in a sample after $$x$$ halflives. The half-life of Co-60 is 5.27 years.

What is the mass, in nanograms, of Co-60 remaining in the sample after 10.54 years?
A. 0.47
B. 1.25
C. 2.00
D. 2.64

Ans:B

The relationship between the remaining mass of cobalt-60 (Co-60) and the number of half-lives that have passed. The half-life of Co-60 is 5.27 years, and the mass of Co-60 decreases by half every 5.27 years.

Initial mass $$y_0 = 5$$ nanograms
Half-life $$t_{1/2} = 5.27$$ years

We need to find the mass remaining after 10.54 years. First, we determine how many half-lives have passed in 10.54 years:

$\text{Number of half-lives} = \frac{10.54 \text{ years}}{5.27 \text{ years/half-life}} = 2$

Since 10.54 years corresponds to 2 half-lives, the mass of Co-60 remaining after 2 half-lives can be calculated as follows:

$y = y_0 \times \left(\frac{1}{2}\right)^{\text{Number of half-lives}}$

Substitute the values into the equation:

$y = 5 \times \left(\frac{1}{2}\right)^2$
$y = 5 \times \frac{1}{4}$
$y = 5 \times 0.25$
$y = 1.25$

So, the mass of Co-60 remaining in the sample after 10.54 years is $$1.25$$ nanograms.

Alternative method:

Which is only in option B (because left mass is greater then 1 and less then 2)

[Calc]  Question  Medium

In the $$x y$$-plane, exactly how many $$x$$-intercepts does the graph of $$f(x)=x(x-4)^2(x-5)^3$$ have?
A) 2
B) 3
C) 5
D) 6

B

To find the $$x$$-intercepts of the graph of $$f(x) = x(x-4)^2(x-5)^3$$, we need to find the values of $$x$$ where $$f(x) = 0$$.

The $$x$$-intercepts occur where $$f(x) = 0$$, meaning the function crosses the x-axis.

$f(x) = 0$
$x(x-4)^2(x-5)^3 = 0$

For this equation to be true, at least one of the factors must be equal to zero. So, the $$x$$-intercepts occur at the values of $$x$$ that make each factor zero.

1. $$x = 0$$ (from the factor $$x$$).
2. $$x – 4 = 0 \Rightarrow x = 4$$ (from the factor $$(x – 4)^2$$).
3. $$x – 5 = 0 \Rightarrow x = 5$$ (from the factor $$(x – 5)^3$$).

Therefore, there are $$\boxed{3}$$ $$x$$-intercepts for the graph of $$f(x)$$.

$\boxed{B) \, 3}$

[Calc]  Question Medium

The function $$q$$ is defined by $$q(x)=5(-1)^x$$, where $$x$$ is an integer. What is the value of $$q(6)$$ ?
A) -30
B) -5
C) 5
D) 30

C

The function $$q$$ is defined by $$q(x) = 5(-1)^x$$, where $$x$$ is an integer. We need to find the value of $$q(6)$$.

Substitute $$x = 6$$ into the function:
$q(6) = 5(-1)^6$

Since $$(-1)^6 = 1$$:
$q(6) = 5 \times 1$
$q(6) = 5$

$\boxed{C}$

[No calc]  Question   medium

The function f is defined by 𝑓(𝑥) = (𝑥 − 1)(𝑥 +1)(𝑥 + 2). Which of the following is NOT an x-intercept of the graph 𝑦 = 𝑓(𝑥) in the xy-plane?
A. (-2, 0
B. (-1, 0)
C. (1, 0)
D. (2, 0)

Ans: D

We’ll find the roots of the function $$f(x)$$ to identify the $$x$$-intercepts. The roots are the solutions to $$f(x) = 0$$.

$$f(x) = (x – 1)(x + 1)(x + 2) = 0$$

Roots: $$x = -2, -1, 1$$

So,  options A, B, C are $$x$$-intercepts. D are not $$x$$-intercepts.

[Calc]  Question  Medium

A piece of paper is cut two times, resulting in three smaller pieces of paper of the same shape and size. Then, the three smaller pieces are stacked and cut two times to form nine even smaller pieces, each with the same shape and size. This process continues until the pieces of paper are too small to cut. Which of the following functions gives the number of pieces of paper, $$F(c)$$, that result after $$c$$ cuts, where $$c$$ is an even number?
A) $$F(c)=3^{\frac{c}{2}}$$
B) $$F(c)=3^{\frac{c}{2}+1}$$
C) $$F(c)=3^{2 c}$$
D) $$F(c)=3^{2 c+1}$$

A

After the first cut, we have 2 pieces.
After the second cut, we have 3 pieces.
After the third cut, we have $$3^2 = 9$$ pieces.
After the fourth cut, we have $$3^3 = 27$$ pieces.
And so on.

So, the number of pieces after $$c$$ cuts (where $$c$$ is an even number) follows the pattern of $$3^{\frac{c}{2}}$$, because each cut doubles the number of pieces, and $$3^{\frac{c}{2}}$$ represents the number of pieces after $$c$$ cuts.

Therefore, the correct function is:
$\boxed{A) \, F(c) = 3^{\frac{c}{2}}}$

[No calc]  Question   medium

$$h(t)=-4.9t^{2}+10t$$

The function h models the height h(t), in meters, of a football t seconds after it is kicked. What is the interpretation of h(2) =0.40 in this context?

A. The football has a maximum height of 0.40 meter.
B. The football has a maximum height of 2 meters.
C. The football has a height of 0.40 meter 2 seconds after it is kicked.
D. The football has a height of 2 meters 0.40 second after it is kicked.

Ans: C

The given function $$h(t) = -4.9t^2 + 10t$$ models the height $$h(t)$$, in meters, of a football $$t$$ seconds after it is kicked.

To interpret $$h(2) = 0.40$$, we substitute $$t = 2$$ into the function:
$h(2) = -4.9(2)^2 + 10(2)$
$h(2) = -4.9(4) + 20$
$h(2) = -19.6 + 20$
$h(2) = 0.4$

This means that $$2$$ seconds after the football is kicked, it has a height of $$0.40$$ meters.

C) The football has a height of $$0.40$$ meter $$2$$ seconds after it is kicked.

[Calc]  Question  medium

$$m(t)=10(\frac{1}{2})^{\frac{t}{5730}}$$

The given function M models the mass of the radioactive isotope carbon-14 in a sample, in picograms, t years after the initial measurement. How much time, in years, does it take for the mass
of carbon-14 in the sample to decrease to 5 picograms?

Ans: 5730

To find the time it takes for the mass of carbon-14 in the sample to decrease to $$5$$ picograms, we need to set up the equation:

$M(t) = 5$

$M(t) = 10 \left( \frac{1}{2} \right)^{\frac{t}{5730}}$

$5 = 10 \left( \frac{1}{2} \right)^{\frac{t}{5730}}$

$\frac{1}{2} = \left( \frac{1}{2} \right)^{\frac{t}{5730}}$

Since $$\left( \frac{1}{2} \right)^n = \frac{1}{2^n}$$, we can rewrite the equation as:

$2 = 2^{\frac{t}{5730}}$

$\frac{t}{5730} = 1$

$t = 5730$

So, it takes $$5730$$ years for the mass of carbon-14 in the sample to decrease to $$5$$ picograms.

[Calc]  Question  Medium

The line graph shows the number of cars a salesperson sold for each of the first six months of the year.

For how many of these months did the salesperson sell 10 or more cars?

Ans: 3

To determine how many months the salesperson sold 10 or more cars, we need to look at the number of cars sold for each month on the line graph.

From the graph:

• January: 5 cars
• February: 5 cars
• March: 7 cars
• April: 12 cars
• May: 22 cars
• June: 20 cars

The salesperson sold 10 or more cars in the following months:

• April: 12 cars
• May: 22 cars
• June: 20 cars

So, the salesperson sold 10 or more cars in 3 months.

[Calc]  Question  Medium

What is they-intercept of the graph of $$y = a(17)^x – c$$ in the xy-plane, where a and care
positive constants ?
A) (0, a)
B) (0, -c)
C) (0, -ac)
D) (0, a-c)

D) (0, a-c)

To find the $$y$$-intercept of the graph of $$y = a(17)^x – c$$, we set $$x = 0$$ because the $$y$$-intercept occurs when $$x = 0$$.

Substituting $$x = 0$$ into the equation, we get:
$y = a(17)^0 – c = a – c$

So, the $$y$$-intercept is the point $$(0, a – c)$$.

$\boxed{D) \ (0, a – c)}$

[No- Calc]  Question  Medium

y=(x-1)(x+1)(x+2)
The graph in the xy-plane of the equation above contains the point (a, b). If -1 ≤ a ≤ 1, which of the following is NOT a possible value of b ?

A. -2

B. -1

C. 0

D. 1

Ans: D

$y = (x – 1)(x + 1)(x + 2)$

$y = (x^2 – 1)(x + 2)$

$y = x^3 + 2x^2 – x – 2$

To find the possible values of $$b$$ when $$-1 \leq a \leq 1$$, we substitute each value of $$a$$ into the equation:

For $$a = -1$$:
$b = (-1)^3 + 2(-1)^2 – (-1) – 2$
$b = -1 + 2 + 1 – 2$
$b = 0$

For $$a = 0$$:
$b = 0^3 + 2(0)^2 – 0 – 2$
$b = 0 – 0 – 0 – 2$
$b = -2$

For $$a = 1$$:
$b = 1^3 + 2(1)^2 – 1 – 2$
$b = 1 + 2 – 1 – 2$
$b = 0$

So, the possible values of $$b$$ are $$0$$ and $$-2$$. Therefore, the answer is D) 1.

[Calc]  Question   Medium

During an experiment, the number of phytoplankton in a population doubled each day. There were 300 phytoplankton at the start of the experiment. Which function represents the population size, P(x), x days after the start of the experiment ?
A) $$P(x) = 2^{3oox}$$
B) $$P(x) = 300^{2x}$$
C) $$P(x) = 2(300)^x$$
D) $$P(x) = 300(2)^x$$

D) $$P(x) = 300(2)^x$$

The population of phytoplankton doubles each day starting from an initial population of 300 phytoplankton. We need to determine which function represents the population size $$P(x)$$ $$x$$ days after the start of the experiment.

1. Identify the initial population:
Initial population $$P(0) = 300$$

2. Understand the doubling process:
Each day, the population doubles. This is an exponential growth situation where the base of the exponent is 2.

3. Write the exponential function:
The general form for exponential growth is $$P(x) = P_0 \cdot 2^x$$ where $$P_0$$ is the initial population and $$x$$ is the number of days.
Here, $$P_0 = 300$$, so the function becomes $$P(x) = 300 \cdot 2^x$$.

Thus, the function that represents the population size $$P(x)$$ $$x$$ days after the start of the experiment is:
$\boxed{300(2)^x}$

[No- Calc]  Question  Medium

The graph of the function f is shown. Which of the following is a value of x for which f(x) =0 ?

A. -1

B. 0

C. 1

D. 4

Ans: C

Values of x  for f(x) = 0 is$-2,1,2.3$

[Calc]  Question  medium

The initial number of bacteria in a population is 10 thousand. The bacteria in the population are observed to double in number every 12 hours.

Which graph represents the number of bacteria y, in thousands, x hours after the initial observation?

A

The general form of the exponential growth formula is:
$N(t)=N_0 \cdot 2^{\frac{t}{T}}$
where:
$$N(t)$$ is the number of bacteria at time $$t$$,
$$N_0$$ is the initial number of bacteria,
$$T$$ is the doubling time,
$$t$$ is the time elapsed.

1. Number of Bacteria After 24 Hours:
$N(24)=10,000 \cdot 2^{\frac{24}{12}}=10,000 \cdot 2^2=10,000 \cdot 4=40,000$
2. Number of Bacteria After 36 Hours:
$N(36)=10,000 \cdot 2^{\frac{36}{12}}=10,000 \cdot 2^3=10,000 \cdot 8=80,000$
3. Number of Bacteria After 48 Hours:
$N(48)=10,000 \cdot 2^{\frac{48}{12}}=10,000 \cdot 2^4=10,000 \cdot 16=160,000$

[Calc]  Question  Medium

The function f is defined by $$f(x)=x^{2}-7$$. What is the value of f(3) ?

A. -16

B. -2

C. 2

D. 16

Ans: C

The function $$f$$ is defined by $$f(x) = x^2 – 7$$. To find the value of $$f(3)$$:

$f(3) = 3^2 – 7$
$f(3) = 9 – 7$
$f(3) = 2$

Therefore, the value of $$f(3)$$ is:$\boxed{2}$

[No- Calc]  Question   Medium

What is the graph of $$y=4-2(0.5)^x$$ ?

Ans:C

Substitute $$x=0$$ into the equation:
\begin{aligned} & y=4-2(0.5)^0 \\ & \text { Since }(0.5)^0=1 \text { : } \\ & y=4-2 \cdot 1 \\ & y=4-2 \\ & y=2 \end{aligned}

So, the point $$(0,2)$$ lies on the graph.

Set $$y=0$$ and solve for $$x$$ :
$0=4-2(0.5)^x$

Rearrange the equation:
$2(0.5)^x=4$

Divide both sides by 2 :
$(0.5)^x=2$

Recall that $$0.5=\frac{1}{2}$$, so we can write:
$\left(\frac{1}{2}\right)^x=2$

This can be rewritten as:
$2^{-x}=2$

Thus:
\begin{aligned} & -x=1 \\ & x=-1 \end{aligned}

So, the point $$(-1,0)$$ lies on the graph.

So graph C is best fit

[Calc]  Question  Medium

$$v=(1,000)(1.05)^t$$

The given equation models the value of an antique dresser $t$ years after its restoration, where $0 \leq t \leq 5$. Which of the following equations best models the value of the dresser $\mathrm{m} \underline{\text { months }}$ after its restoration, where $0 \leq m \leq 60$ ?
A) $\quad v=\left(\frac{1,000}{12}\right)(1.05)^m$
B) $\quad v=(1,000)\left(\frac{1.05}{12}\right)^m$
C) $\quad v=(1,000)(1.05)^{\frac{m}{12}}$
D) $\quad v=(1,000)(1.05)^{12 m}$

C

[Calc]  Question  Medium

The table gives some values of $x$ and the corresponding values of $f(x)$ for polynomial function $f$. Which of the following could be the graph of $f$ in the $x y$-plane, where $y=f(x)$ ?

A

B

C

D

D

[Calc]  Question  Medium

$$W(L)=0.04(1.22)^L$$

The function $W$ gives the estimated weight $W(L)$, in pounds, of a rainbow trout based on its length $L$, in inches. Which of the following is the best interpretation of the number 1.22 in this context?
A) For each increase of 1 pound in weight, the estimated length of the trout, in inches, increases by $22 \%$.
B) For each increase of 1 inch in length, the estimated weight of the trout, in pounds, increases by $22 \%$.
C) For each increase of 1 pound in weight, the estimated length of the trout increases by 1.22 inches.
D) For each increase of 1 inch in length, the estimated weight of the trout increases by 1.22 pounds.

B

[Calc]  Question  Medium

The half-life of a certain substance in an aquatic environment is about 150 years. Which of the following exponential functions best models the amount $A(t)$, in grams, of the substance that remains $t$ years after 200 grams of the substance is applied to the aquatic environment? (The halflife is the length of time needed for an amount of the substance to decrease to one-half of that amount.)
A) $A(t)=150\left(\frac{1}{2}\right)^{\frac{t}{200}}$
B) $A(t)=150\left(\frac{1}{2}\right)^{200 t}$
C) $A(t)=200\left(\frac{1}{2}\right)^{150 t}$
D) $A(t)=200\left(\frac{1}{2}\right)^{\frac{t}{150}}$

D

Question

A function $$f$$ has the property that if point ($$a$$,$$b$$) is on the graph of the equation $$y$$=$$f$$($$x$$) in the $$xy$$-plane, then the point $(a+1,\frac{1}{3}b)$ is also on the graph. Which of the following could define $$f$$?

1. $f(x)=\frac{1}{3}\left&space;(&space;\frac{1}{12}&space;\right&space;)^{x}$
2. $f(x)=12\left&space;(&space;\frac{1}{3}&space;\right&space;)^{x}$
3. $f(x)=\frac{1}{3}\left&space;(&space;12&space;\right&space;)^{x}$

B

Questions

There were no jackrabbits in Australia before 1788 when 24 jackrabbits were introduced, By 1920 the population of jackrabbits had reached 10 billion. If the population had grown exponentially, this would correspond to a $16.2 \%$ increase, on average, in the population each year. Which of the following functions best models the population $p(t)$ of jackrabbits $t$ years after 1788?
A. $p(t)=1.162(24)^t$
B. $p(t)=24(2)^{1.162 t}$
C. $p(t)=24(1.162)^t$
D. $p(t)=(24.1 .162)^t$

Ans: C

Questions

The figure above shows a graph with six regions that correspond to temperature, in degrees Fahrenheit (°F), and humidity conditions, in grams of water vapor per cubic meter of air (g/m3 ), that will result in different snow crystal shapes when the crystals are grown in a laboratory. Based on the graph, which of the following is a combination of temperature and humidity at which prisms will be formed?

1. 5°F and 0.15 g/m3
2. 15°F and 0.18 g/m3
3. 20°F and 0.02 g/m3
4. 30°F and 0.08 g/m3

Ans: C

Questions

$\frac{2}{x-2}+\frac{3}{x+5}=\frac{r x+t}{(x-2)(x+5)}$

The equation above is true for all $x>2$, where $r$ and $t$ are positive constants. What is the value of $r t$ ?
A. -20
B. 15
C. 20
D. 60

Ans: C

Questions

. $f(x)=(x+4)(x-1)(2 x-3)$

The function $f$ is defined above. Which of the following is NOT an $x$-intercept of the graph of the function in the $x y$-plane?
A. $(-4,0)$
B. $\left(-\frac{2}{3}, 0\right)$
C. $(1,0)$
D. $\left(\frac{3}{2}, 0\right)$

Ans: B

Questions

In the figure above, X is a mark on the side of a tire of a car at rest. The car, starting from rest, will experience an acceleration for some period of time. Which of the following graphs could represent the distance between the mark X and the ground after the car starts to accelerate and the tire makes its first few revolutions? 4.5

Ans: A

Question

$x^2-4 x+2=0$

Which of the following is a solution to the equation above?
A. $x=-2+\sqrt{2}$
B. $x=-2+\sqrt{6}$
C. $x=2+\sqrt{2}$
D. $x=2+\sqrt{6}$

Ans: C

Questions

$i^2+(-i)^2$

In the complex number system, what is the value of the given expression? (Note: $i=\sqrt{-1}$ )
A. -2
B. 0
C. 2
D. $2 i$

Ans: A

Questions

$f(x)=-500 x^2+25,000 x$

The revenue $f(x)$, in dollars, that a company receives from sales of a product is given by the function $f$ above, where $x$ is the unit price, in dollars, of the product. The graph of $y=f(x)$ in the $x y$-plane intersects the $x$-axis at 0 and $a$. What does $a$ represent?
A. The revenue, in dollars, when the unit price of the product is $\$ 0$B. The unit price, in dollars, of the product that will result in maximum revenue C. The unit price, in dollars, of the product that will result in a revenue of$\$0$
D. The maximum revenue, in dollars, that the company can make

Ans: C

Questions

$g(x)=2^x-2$

The function $g$ is defined by the equation above. Which of the following points in the $x y$-plane is an $x$-intercept of the graph of the equation $y=g(x)$ ?
A. $(-1, g(-1))$
B. $(0, g(0))$
C. $(1, g(1))$
D. $(2, g(2))$

Ans: C

Question

Which of the following is a solution to the equation $4 x^2+4 x-3=0$ ?
A. -1.5
B. -0.5
C. 1
D. 3

Ans: A

Questions

The graph of the function $$f$$ is shown in the $$xy$$-plane above, where $$y=f(x)$$. Which of the following functions could define $$f$$?

1. $$f(x)=(x-3)(x-1)^2(x+2)^2$$
2. $$f(x)=(x-3)^2(x-1)(x+2)$$
3. $$f(x)=(x+3)(x+1)^2(x-2)^2$$
4. $$f(x)=(x+3)^2(x+1)(x-2)$$