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
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
▶️Answer/Explanation
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:
\[
\boxed{\text{C) The time, in seconds, it takes a pendulum of length 12 meters to complete one cycle}}
\]
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
▶️Answer/Explanation
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)
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
▶️Answer/Explanation
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 \(f(x)\).
So the answer is:
\[ \boxed{B) \, 3} \]
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
▶️Answer/Explanation
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 \( (-1)^6 = 1 \):
\[ q(6) = 5 \times 1 \]
\[ q(6) = 5 \]
So the answer is:
\[ \boxed{C} \]
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)
▶️Answer/Explanation
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.