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
The function \(f\) is defined by \(f(r)=(r-1)(r+2)^2\). If \(f(h-5)=0\), where \(h\) is a constant, what is one possible value of \(h\) ?
▶️Answer/Explanation
Ans: 3 , 6
### Question 19:
To find one possible value of \(h\) given \(f(h-5) = 0\), we first substitute \(h-5\) for \(r\) in the function \(f(r)\) and set it equal to 0:
\[
f(r) = (r-1)(r+2)^2
\]
\[
f(h-5) = ((h-5)-1)((h-5)+2)^2 = 0
\]
\[
((h-6)(h-3)^2 = 0
\]
Now, we have a product equal to zero, which means at least one of the factors must equal zero. So, we set each factor equal to zero and solve for \(h\):
1. \((h-6) = 0\):
\[
h – 6 = 0
\]
\[
h = 6
\]
2. \((h-3)^2 = 0\):
\[
h – 3 = 0
\]
\[
h = 3
\]
So, possible value of \(h\) is \(h = \boxed{6, 3}\).
[Calc] Question Hard
\[
(x-5)^2+1=0
\]
How many distinct real solutions does the given equation have?
A) Zero
B) Exactly one
C) Exactly two
D) Infinitely many
▶️Answer/Explanation
Ans:A
To determine how many distinct real solutions the equation \((x-5)^2 + 1 = 0\) has,
\[
(x-5)^2 + 1 = 0
\]
First, isolate \((x-5)^2\) by subtracting 1 from both sides:
\[
(x-5)^2 = -1
\]
Now, observe the equation \((x-5)^2 = -1\). Since \((x-5)^2\) represents a square of a real number, it is always non-negative (i.e., it can never be less than zero). The smallest value \((x-5)^2\) can achieve is 0, and it cannot be negative.
However, the right side of the equation is -1, which is a negative number. Since a square of any real number cannot be negative, there are no real values of \(x\) that satisfy the equation \((x-5)^2 = -1\).
Therefore, the equation has no real solutions.
The correct answer is:A) Zero
[calc] Question Hard
The function \(f\) is defined by \(f(r)=(r-1)(r+2)^2\). If \(f(h-5)=0\), where \(h\) is a constant, what is one possible value of \(h\) ?
▶️Answer/Explanation
Ans: 3 , 6
### Question 19:
To find one possible value of \(h\) given \(f(h-5) = 0\), we first substitute \(h-5\) for \(r\) in the function \(f(r)\) and set it equal to 0:
\[
f(r) = (r-1)(r+2)^2
\]
\[
f(h-5) = ((h-5)-1)((h-5)+2)^2 = 0
\]
\[
((h-6)(h-3)^2 = 0
\]
Now, we have a product equal to zero, which means at least one of the factors must equal zero. So, we set each factor equal to zero and solve for \(h\):
1. \((h-6) = 0\):
\[
h – 6 = 0
\]
\[
h = 6
\]
2. \((h-3)^2 = 0\):
\[
h – 3 = 0
\]
\[
h = 3
\]
So, possible value of \(h\) is \(h = \boxed{6, 3}\).
[calc] Question Hard
\[
f(x)=\left(\frac{3}{4}\right)^x+5
\]
If the function \(f\) is graphed in the \(x y\)-plane, where \(y=f(x)\), what is the \(y\)-intercept of the graph?
A) \(\left(\frac{3}{4}, 5\right)\)
B) \(\left(\frac{3}{4}, 0\right)\)
C) \((0,5)\)
D) \((0,6)\)
▶️Answer/Explanation
Ans: D
The function given is:
\[
f(x) = \left( \frac{3}{4} \right)^x + 5
\]
To find the \(y\)-intercept, we evaluate \(f(x)\) at \(x = 0\):
\[
f(0) = \left( \frac{3}{4} \right)^0 + 5
\]
\[
f(0) = 1 + 5
\]
\[
f(0) = 6
\]
So, the \(y\)-intercept is \((0, 6)\).
Thus, the correct answer is:
\[
\boxed{(0, 6)}
\]
[Calc] Question Hard
Two numbers, \(a\) and \(b\), are each greater than zero, and the square root of \(a\) is equal to the cube root of \(b\). For what value of \(x\) is \(a^{2 x-1}\) equal to \(b\) ?
▶️Answer/Explanation
Ans:1.25 or 5/4
\[
\sqrt{a}=\sqrt[3]{b}
\]
We want to find the value of \(x\) such that \(a^{2 x-1}=b\).
Using the given relationship, we can rewrite \(a\) and \(b\) in terms of each other:
\[
\sqrt{a}=\sqrt[3]{b} \Longrightarrow a=(\sqrt[3]{b})^2
\]
Now, substitute \(a\) into the equation \(a^{2 x-1}=b\) :
\[
\left((\sqrt[3]{b})^2\right)^{2 x-1}=b
\]
\( \begin{aligned} & b^{\frac{4 x-2}{3}}=b \\ & 4 x-2=1 \\ & x=3 / 4\end{aligned} \)