CLEP College Algebra FREE Sample Practice Questions

3- A
Subtracting \(2x\) and adding 5 to both sides of \(2x – 5 ≥ 3x – 1\) gives \(-4 ≥ x\). Therefore, \(x\) is a solution to \(2x – 5 ≥ 3x – 1\) if and only if \(x\) is less than or equal to \(-4\) and x is NOT a solution to\( 2x – 5 ≥ 3x – 1\) if and only if \(x\) is greater than \(-4\). Of the choices given, only \(-2\) is greater than \9-4\) and, therefore, cannot be a value of \(x\). For additional educational resources, .

4- D
Given the two equations, substitute the numerical value of a into the second equation to solve for \(x\). \(a=\sqrt{3}, 4a=\sqrt{4x}\)
Substituting the numerical value for an into the equation with x is as follows.
\(4(\sqrt{3})=\sqrt{4x}\),From here, distribute the 4. \(4\sqrt{3}=\sqrt{4x}\)
Now square both sides of the equation. \((4\sqrt{3})^2=(\sqrt{4x})^2\)
Remember to square both terms within the parentheses. Also, recall that squaring a square root sign cancels them out. \(4^2 \sqrt{3}^2=4x, 16(3)=4x, 48=4x, x=12\) For additional educational resources, .

5- C
First square both sides of the equation to get \(4m-3=m^2\)
Subtracting both sides by \(4m-3\) gives us the equation \(m^2-4m+3=0\)
Here you can solve the quadratic equation by factoring to get \((m-1)(m-3)=0\)
For the phrase \((m-1)(m-3)\) to equal zero, \(m=1\) or \(m=3\) For additional educational resources, .

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6- C
Let \(x\) be the number. Write the equation and solve for \(x. (28-x)÷x= 3\)
Multiply both sides by \(x\). \((28-x)= 3x\), then add x both sides. \(28=4x\), now divide both sides by 4.\( x=7\) For additional educational resources, .

7- B
The sum of supplement angles is 180. Let \(x\) be that angle. Therefore, \(x+9x=180\)
\(10x=180\), divide both sides by 10: \(x=18\) For additional educational resources, .

8- B
tan\(θ=\frac{opposite}{adjacent}\)
tan\(θ=\frac{5}{12}⇒\) we have the following right triangle. Then
\(c=\sqrt{5^2+12^2 }=\sqrt{25+144}=\sqrt{169}=13\)
cos\(θ=\frac{adjacent}{hypotenuse}=\frac{12}{13}\)
\(\img{https://appmanager.effortlessmath.com/public/images/questions/ppp.png
}\) For additional educational resources, .

9- D
The amplitude in the graph of the equation \(y=\)acosb\(x\) is a. (a and b are constant)
In the equation \(y=\)cos\(x\), the amplitude is 2 and the period of the graph is \(2π\).
The only choice that has two times the amplitude of graph \(y =\) cos \(x\) is \(y=2+2\) cos \(x\)
They both have an amplitude of 2 and a period of \(2π\). For additional educational resources, .

10- D
Substituting 6 for \(x\) and 14 for \(y\) in \(y = nx+2\) gives \(14=(n)(6)+2\),
which gives \(n=2\). Hence, \(y=2x+2\). Therefore, when \(x = 10\), the value of \(y\) is
\(y=(2)(10)+2 = 22\). For additional educational resources, .

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