ALEKS Math Practice Test Questions

ALEKS Math Practice Test Questions

Preparing for the ALEKS Math test? Try following free ALEKS Mathematics Practice questions. Reviewing practice questions is the best way to brush up your Math skills. Here, we walk you through solving 10 common ALEKS Math practice problems covering the most important math concepts on the ALEKS Math test.

These ALEKS Math practice questions are designed to be similar to those found on the real ALEKS Math test. They will assess your level of preparation and will give you a better idea of what to study on your exam.

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10 Sample ALEKS Math Practice Questions

1- If \(f(x) = 5 + x\) and \( g(x) = \ – x^2 – 1 – 2x \), then find \((g – f)(x)\)?


2- Simplify the following inequality.
\( \frac{|3+x|}{7 }≤ 5 \) _____________


3- \( tan (– \frac{π}{6}) = ? \) _____________


4- \( \frac{\sqrt{32a^5 b^3}}{\sqrt{2ab^2} } = ? \) _____________


5- The cost, in thousands of dollars, of producing x thousands of textbooks is
\( C (x) = x^2 + 10x + 30 \). The revenue, also in thousands of dollars, is \(R(x) = 4x\). Find the profit or loss if 3,000 textbooks are produced. \((profit = revenue – cost)\) ____________


6- Suppose a triangle has the dimensions indicated below:
Then Sin \(B =\) ? ____________


7- Find the slope–intercept form of the graph \(6x – 7y = – 12\) __________


8- What are the zeros of the function:\( f(x)=x^3+5x^2+6x?\) ___________


9- What is the solution of the following system of equations?\( \begin{cases}5x+y=9\\10x-7y=-18\end{cases} \) ___________


10- Find the Radius of the graph \( (x – 3)^2 + (y + 6)^2 = 16 \) __________


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Answers:

1- \(x^2 – 3x – 6\)
\((g – f)(x) = g(x) – f(x) = (– x^2 – 1 – 2x) – (5 + x) – x^2 – 1 – 2x – 5 – x = \ – x^2 – 3x – 6\)

2- \(-38≤x≤32\)
\(\frac{|3+x|}{7}≤5⇒|3+x|≤35⇒-35≤3+x≤35⇒-35-3≤x≤35-3⇒ -38≤x≤32 \)

3- \(–\frac {\sqrt{3}}{3} \)
\( tan (– \frac{π}{6}) = \ –\frac {\sqrt{3}}{3} \)

4- \(4a^2\sqrt{b} \)
\( \frac{\sqrt{32a^5 b^3}}{\sqrt{2ab^2} } = )( \frac{4a^2 \sqrt{2ab}}{b\sqrt{2a} } = 4a^2\sqrt{b} \)

5- 57,000 loss
\(c(3)=(3)^2+10(3)+30=9+30+30=69 4×3=12⇒12-69=-57⇒57,000 loss\)

6- \(\frac{4}{5}\)
\( sinB=\frac{Opposite Side}{Hypotenuse}=\frac{4}{5}\)

7- \(y=\frac{6}{7} x+\frac{12}{7}\)
\(-7y=-6x-12⇒y=\frac{-6}{-7} x-\frac{12}{-7}⇒y=\frac{6}{7} x+\frac{12}{7}\)

8- \(0, -2, -3\)
Frist factor the function: \(x (x+2)(x+3)\) To find the zeros, \(f(x)\) should be zero. \(f(x)=x (x+2)(x+3)=0\) Therefore, the zeros are: \(x=0 (x+2)=0 ⇒ x= -2 (x+3)=0 ⇒ x= -3\)

9- \((1,4)\)
\(\begin{cases}5x+y=9\\10x-7y=-18\end{cases}\)⇒ Multiply the first equation by\( -2 ⇒\) \(\begin{cases}-10x-2y=-18\\10x-7y=-18\end{cases}\) Add two equations \(⇒ –9y = –36 ⇒ y = 4\) then: \(x = 1\)

10- 4
\((x – h)^2 + (y – k)^2 = r^2 ⇒\) center: \((h,k)\) and radius: \(r\)
\((x – 3)^2 + (y + 6)^2 = 12 ⇒\) center: \((3,-6)\) and radius: \(\sqrt{16}=4\)

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