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 Hard
For an exponential function \(g\), the value of \(g(x)\) decreases by \(20 \%\) for each 1 -unit increase in the value of \(x\). If \(g(2)=16\), which equation could define \(g\) ?
A. \(g(x)=16(0.8)^{x-2}\)
B. \(g(x)=16(0.8)^{x+2}\)
C. \(g(x)=16(0.2)^{x-2}\)
D. \(g(x)=16(0.2)^{x+2}\)
▶️Answer/Explanation
Ans:A
To solve this problem, we need to determine the correct equation that defines the function \( g(x) \) given the following information:
1. The value of \( g(x) \) decreases by 20% for each 1-unit increase in the value of \( x \).
2. \( g(2) = 16 \).
Decrease rate: \( 20\% = 0.2 \)
Initial value: \( g(2) = 16 \)
1. Represent the decrease of 20% using the factor \( (1 – 0.2) = 0.8 \), implying that the function value is multiplied by \( 0.8 \) for each 1-unit increase in \( x \).
2. Analyze the given options to determine the correct equation for \( g(x) \).
Option A: \( g(x) = 16(0.8)^{x-2} \)
This option matches the given condition \( g(2) = 16 \) when \( x = 2 \).
Thus, option A is correct.
Option B: \( g(x) = 16(0.8)^{x+2} \)
Incorrect because the exponent \( x+2 \) does not match \( g(2) = 16 \).
Option C: \( g(x) = 16(0.2)^{x-2} \)
Incorrect due to the incorrect base \( 0.2 \) representing an 80% decrease.
Option D: \( g(x) = 16(0.2)^{x+2} \)
Incorrect due to the incorrect base \( 0.2 \) representing an 80% decrease and \( x+2 \) not matching \( g(2) = 16 \).
[No calc] Question Hard
What is an equation of the graph shown?
A) \(y=2^x +2\)
B) \(y=2^x +1\)
C) \(y=2^x -1\)
D) \(y=2^x -2\)
▶️Answer/Explanation
D) \(y=2^x -2\)
From graph when $x=0 \rightarrow y =-1$ and $x=1 \rightarrow y =0$
To find the equation of the graph given the points \((0, -1)\) and \((1, 0)\), we need to determine which of the given equations fits these points.
The equations given are of the form \(y = 2^x + k\), where \(k\) is a constant.
For \(x = 0\):
\(y = 2^0 + k = 1 + k\)
We know that when \(x = 0\), \(y = -1\), so:
\[ 1 + k = -1 \]
\[ k = -2 \]
For \(x = 1\):
\(y = 2^1 + k = 2 + k\)
We know that when \(x = 1\), \(y = 0\), so:
\[ 2 + k = 0 \]
\[ k = -2 \]
Both points suggest that the constant \(k\) is \(-2\).
Thus, the equation that fits both points is:
\[ y = 2^x – 2 \]
Therefore, the correct answer is option D: \(y = 2^x – 2\).
[Calc] Question Hard
By examining pollen in the soil, scientists estimated that the number of Ulmus trees in an ancient population doubled every 664 years. There were n trees in the earliest known sample, where n is a constant. Which expression gives the estimated number of Ulmus trees x years after the year of the earliest known sample?
A. \(n(2)^{664x}\)
B. \(n(2)^{\frac{664}{x}}\)
C. \(n(2)^{\frac{x}{664}}\)
D. \(n(2)^{664+x}\)
▶️Answer/Explanation
Ans: C
Given that the number of Ulmus trees doubled every 664 years, we want to find an expression for the number of trees \(x\) years after the earliest known sample.
If the number of trees doubles every 664 years, then the population can be modeled by an exponential function of the form:
\[ N(x) = n \cdot 2^{\frac{x}{664}} \]
\(n\) is the initial number of trees at the earliest known sample.
\(2^{\frac{x}{664}}\) represents the number of times the population doubles in \(x\) years.
This exponential growth function reflects the doubling of the population every 664 years. Specifically:
When \(x = 664\), \(\frac{x}{664} = 1\), and \(N(x) = n \cdot 2^1 = 2n\), indicating the population has doubled after 664 years.
For any other value of \(x\), \(\frac{x}{664}\) calculates how many 664-year periods fit into \(x\) years, thus determining the number of times the population has doubled.
Therefore, the correct expression for the number of Ulmus trees \(x\) years after the earliest known sample is:
\[ \boxed{n(2)^{\frac{x}{664}}} \]
[No calc] Question Hard
What is the graph of the equation \(y=2^{-x} +1\)
▶️Answer/Explanation
A
Here is the graph of the equation \(y=2^{-x}+1\) :
- The point \((0,2)\) is plotted in red, showing that when \(x=0, y=2\).
- The green dashed line represents the horizontal asymptote \(y=1\), indicating that as \(x\) increases, the value of \(y\) approaches 1 but never actually reaches it.
- There are no real \(x\) values that make \(y=0\), as the function \(2^{-x}+1\) is always positive.
The graph confirms the behavior of the exponential decay function shifted upwards by 1 unit.
[Calc] Question Hard
The population, in millions, of Suzhou, China, can be modeled by the function \(p(t)=1.1(1.066)^t\), where \(t\) represents the number of years after 1990, and \(0 \leq t \leq 25\). Which of the following equations best models the population, in millions, of Suzhou, where \(n\) represents the number of years after 1995 , and \(0 \leq n \leq 25\) ?
A) \(r(n)=1.1(1.066)^{5 n}\)
B) \(r(n)=1.1(1.066)^{n-5}\)
C) \(r(n)=1.1(1.066)^5(1.066)^n\)
D) \(r(n)=(1.1)^5(1.066)^5(1.066)^n\)
▶️Answer/Explanation
C
To find the correct equation that models the population of Suzhou, China, in terms of \( n \), where \( n \) represents the number of years after 1995, we need to adjust the original equation \( p(t) = 1.1(1.066)^t \).
Given:
\( t \) is the number of years after 1990.
\( n \) is the number of years after 1995.
We can express \( t \) in terms of \( n \):
\[ t = n + 5 \]
substitute \( t = n + 5 \)
\[ p(t) = 1.1(1.066)^t \]
\[ p(n + 5) = 1.1(1.066)^{n + 5} \]
simplify this equation:
\[ p(n + 5) = 1.1(1.066)^{n + 5} \]
\[ p(n + 5) = 1.1(1.066)^5(1.066)^n \]
Therefore, the equation that models the population in terms of \( n \) is:
\[ r(n) = 1.1(1.066)^5(1.066)^n \]
So the correct answer is:
\[ \boxed{C) \ r(n) = 1.1(1.066)^5(1.066)^n} \]
[Calc] Question Hard
The graph of $y=f(x)+1$ is shown. Which equation defines the function $f$ ?
A) $f(x)=2^x$
B) $f(x)=3^x$
C) $f(x)=2^x+1$
D) $f(x)=3^x+2$
▶️Answer/Explanation
B
Question
Radioactive substances decay over time. The mass \(M\), in grams, of a particular radioactive substance \(d\) days after the beginning of an experiment is shown in the table below.
If this relationship is modeled by the function \(M(d) = a. 10^{bd}\), which of the following could be the values of \(a\) and \(b\) ?
- \(a = 12\) and \(b = 0.0145\)
- \(a= 12\) and \(b = -0.0145\)
- \(a = 120\) and \(b = 0.0022\)
- \(a= 120\) and \(b = -0.0022\)
▶️Answer/Explanation
Ans: D