Free Full Length CLEP College Algebra Practice Test

The correct answer is \(\frac{3 \ x \ -\ 1}{x^2 \ - \ x}\)
\((\frac{f}{g})(x)=\) \(\frac{f(x)}{g(x)}\)\(=\)\(\frac{3 \ x \ – \ 1}{x^2 \ - \ x}\)

The correct answer is \(y=4 \ x \ - \ 17\)
The equation of a line is: \(y=m \ x \ + \ b\), where \(m\) is the slope and \(b\) is the \(y \ -\)intercept.
First find the slope:\(m=\frac{y_2 \ - \ y_1}{x_2 \ - \ x_1 }=\frac{15 \ - \ (- \ 5)}{8 \ - \ 3}=\frac{20}{5}=4\). Then, we have: \(y=4 \ x \ + \ b\)
Choose one point and plug in the values of \(x\) and y in the equation to solve for \(b\).
Let’s choose the point \((3, \ -5)\). \(y=4 \ x \ + \ b → - \ 5=4 \ (3) \ + \ b → - \ 5=12 \ + \ b → b=- \ 17\)
The equation of the line is: \(y=4 \ x \ - \ 17\)

The correct answer is \(33\)
Since \(N=6\), substitute \(6\) for \(N\) in the equation \(\frac{x \ - \ 3}{5}=N\), which gives \(\frac{x \ - \ 3}{5}=6\).
Multiplying both sides of \(\frac{x\ -\ 3}{5}=6\) by \(5\) gives \(x \ - \ 3=30\) and then adding \(3\) to both sides of
\(x \ - \ 3=30\) then, \(x=33\).

The correct answer is \(\sqrt[5]{b^3}\)
\(b^{\frac{m}{n}}\)\(=\sqrt[n]b^m\) For any positive integers \(m\) and \(n\). Thus, \(b^{\frac{3}{5}}=\sqrt[5]b^3\).

The corrcet answer is \(144^\circ\)
The sum of all angles in a quadrilateral is \(360\) degrees. Let \(x\) be the smallest angle in the quadrilateral.
Then the angles are: \(x, \ 2 \ x, \ 3 \ x, \ 4 \ x\)
\(x \ + \ 2 \ x \ + \ 3 \ x \ + \ 4 \ x=360 → 10 \ x=360 → x=36\)
The angles in the quadrilateral are: \(36^\circ, \ 72^\circ, \ 108^\circ\), and \(144^\circ\)

The corrcet answer is \(\frac{8\sqrt{π}}{π}\)
Formula for the area of a circle is: \(A=π \ r^2\).
Using \(64\) for the area of the circle we have: \(64=π \ r^2\). Let’s solve for the radius \((r)\).
\(\frac{64}{π}=r^2 → r=\sqrt{\frac{64}{π}}=\frac{8}{\sqrtπ}=\frac{8}{\sqrt{π}} \ × \frac{\sqrt{π}}{\sqrt{π}}=\frac{8 \sqrt{π}}{π}\)

The corrcet answer is \(\frac{4 \ + \ 19 \ i}{29}\)
To rewrite \(\frac{2 \ + \ 3 \ i}{5 \ - \ 2 \ i}\) in the standard form a\(+\)bi, multiply the numerator and denominator of \(\frac{2 \ + \ 3 \ i}{5 \ - \ 2 \ i}\) by the conjugate, \(5 \ + \ 2 \ i\).
This gives \((\frac{2 \ + \ 3 \ i}{5 \ - \ 2 \ i}) (\frac{5 \ + \ 2 \ i}{5 \ + \ 2 \ i})=\frac{10 \ + \ 4 \ i \ + \ 15 \ i \ + \ 6 \ i^2}{5^2 \ - \ (2 \ i)^2}\) . Since \(i^2=- \ 1\), this last fraction can be rewritten as  \(\frac{10 \ + \ 4 \ i \ + \ 15 \ i \ + \ 6 \ (- \ 1)}{25 \ - \ 4 \ (- \ 1)}=\frac{4 \ + \ 19 \ i}{29}\).

The corrcet answer is \(60\)
First find the value of \(b\), and then find \(f(3)\). Since \(f(2)=35\), substuting \(2\) for \(x\) and \(35\) for \(f(x)\) gives \(35=b \ (2)^2 \ + \ 15=4 \ b \ + \ 15\).
Solving this equation gives \(b=5\). Thus
\(f(x)=5 \ x^2 \ + \ 15\), \(f(3)=5 \ (3)^2 \ + \ 15 → f(3)=45 \ + \ 15\), \(f(3)=60\)

The corrcet answer is \(30\)
The sum of supplement angles is \(180\). Let \(x\) be that angle. Therefore, \(x \ + \ 5 \ x=180\)
\(6 \ x=180\), divide both sides by \(6\): \(x=30\)

The correct answer is \(- \ 2\)
Solving Systems of Equations by Elimination
Multiply the first equation by \((–\ 2)\), then add it to the second equation.
\(\cfrac{\begin{align}-\ 2\ (2\ x\ +\ 5\ y\ ) = 11\\ 4\ x\ -\ 2\ y\ = -\ 14\end{align}}{} \)\( \\ -\ 4\ x\ -\ 10\ y= -\ 22\\ 4\ x\ -\ 2\ y=-\ 14⇒ -\ 12\ y= -\ 36 ⇒ y= 3\)
Plug in the value of \(y\) into one of the equations and solve for \(x\).
\(2\ x\ +\ 5\ (3)= 11 ⇒ 2\ x\ +\ 15= 11 ⇒ 2\ x= -\ 4 ⇒ x= -\ 2\)

The correcte answer is \(44\)
Identify the input value. Since the function is in the form \(f(x)\) and the question asks to calculate \(f(4)\), the input value is four.
\(f(4) → \ x=4\), Using the function, input the desired \(x\) value.
Now substitute \(4\) in for every \(x\) in the function.
\(f(x)=3 \ x^2 \ - \ 4\), \(f(4)=3 \ (4)^2 \ - \ 4\), \(f(4)=48 \ - \ 4\), \(f(4)=44\)

The correct answer is \(- \ 8\)
The problem asks for the sum of the roots of the quadratic equation \(2 \ n^2 \ + \ 16 \ n \ + \ 24=0\).
Dividing each side of the equation by \(2\) gives \(n^2 \ + \ 8 \ n \ + \ 12=0\). If the roots of
\(n^2 \ + \ 8 \ n \ + \ 12=0\) are \(n_1\) and \(n_2\), then the equation can be factored as
\(n^2 \ + \ 8 \ n \ + \ 12=(n \ - \ n_1)\ (n \ - \ n_2 )=0\).
Looking at the coefficient of \(n\) on each side of \(n^2 \ + \ 8 \ n \ + \ 12=(n \ + \ 6)\ (n \ + \ 2)\) gives \(n=- \ 6\) or \(n=- \ 2\) , then, \(- \ 6 \ + \ (- \ 2)=- \ 8\)

The correct answer is \(16 \ π\)
The equation of a circle in standard form is: \((x \ - \ h)^2 \ + \ (y \ - \ k)^2=r^2\), where \(r\) is the radius of the circle.
In this circle the radius is \(4\). \(r^2=16 → r=4\). \((x \ + \ 2)^2 \ + \ (y \ - \ 4)^2=16\)
Area of a circle: \(A=π \ r^2=π \ (4)^2=16 \ π\)

The correct answer is \(6.7 \ × \ 10^5\)
\(670000=6.7 \ × \ 10^5\)

The correct answer is \(\frac{17 \ + \ 7 \ i}{26}\) 
To perform the division  \(\frac{3 \ + \ 2 \ i}{5 \ + \ i}\),
multiply the numerator and denominator of \(\frac{3 \ + \ 2 \ i}{5 \ + \ 1 \ i}\) by the conjugate of the denominator, \(5 \ - \ i\).
This gives \(\frac{(3 \ + \ 2 \ i)\ (5 \ - \ i)}{(5 \ + \ 1 \ i)\ (5 \ - \ i)}\)=\(\frac{15 \ - \ 3 \ i \ + \ 10 \ i \ - \ 2i^2}{5^2 \ - \ i^2}\).
Since \(i^2=- \ 1\), this can be simplified to \(\frac{15 \ - \ 3 \ i \ + \ 10 \ i \ + \ 2}{25 \ + \ 1}=\frac{17 \ + \ 7 \ i}{26}\)

The correct answer is \(y=(x \ - \ 3) \ (x \ - \ 4)\)
The \(x \ -\)intercepts of the parabola represented \(by y=x^2 \ - \ 7 \ x \ + \ 12\) in the \(x \ y \ - \)plane are the values of \(x\) for which \(y\) is equal to \(0\).
The factored form of the equation, \(y=(x \ - \ 3) \ (x \ - \ 4)\), shows that \(y\) equals \(0\) if and only if
\(x=3\) or \(x=4\). Thus, the factored form \(y=(x \ - \ 3) \ (x \ - \ 4)\), displays the \(x \ - \)intercepts of the parabola as the constants \(3\) and \(4\).

The correct answer is \(x \ - \ 2\)
If \(x \ - \ a\) is a factor of \(g(x)\), then \(g(a)\) must equal \(0\).
Based on the table \(g(2)=0\).
Therefore, \(x \ - \ 2\) must be a factor of \(g(x)\).

The correct answer is \(2\)
To solve this problem first solve the equation for \(c\).
\(\frac{c}{b}=2\). Multiply by \(b\) on both sides. Then: \(b \ × \frac{c}{b}=2 \ × \ b → c=2 \ b\) .
Now to calculate \(\frac{4 \ b}{c}\), substitute the value for \(c\) into the denominator and simplify.
\(\frac{4 \ b}{c}=\frac{4 \ b}{2 \ b}=\frac{4}{2}=\frac{2}{1}=2\)

The correct answer is \(27 \ x^2 \ + \ 16 \ x \ + \ 1\)
Simplify the numerator. \(\frac{x \ + \ (4 \ x)^2 \ + \ (3 \ x)^3}{x}=\frac{x \ + \ 4^2 \ x^2 \ + \ 3^3 \ x^3}{x}=\frac{x \ + \ 16 \ x^2 \ + \ 27 \ x^3}{x}\).
Pull an x out of each term in the numerator. \(\frac{x \ (1 \ + \ 16 \ x \ + \ 27 \ x^2)}{x}\).
The \(x\) in the numerator and the \(x\) in the denominator cancel: \(1 \ + \ 16 \ x \ + \ 27 \ x^2=27 \ x^2 \ + \ 16 \ x \ + \ 1\)

The correct answer is \(\frac{a}{b}=\frac{21}{11}\)
The equation \(\frac{a \ - \ b}{b}=\frac{10}{11}\) can be rewritten as \(\frac{a}{b} \ - \frac{b}{b}=\frac{10}{11}\), from which it follows that \(\frac{a}{b} \ - \ 1=\frac{10}{11}\), or  \(\frac{a}{b}=\frac{10}{11} \ + \ 1=\frac{21}{11}\).

The correct answer is \(y=\frac{1}{3} \ x \ + \ 2\)
First write the equation in slope intercept form. Add \(2 \ x\) to both sides to get \(6 \ y=2 \ x \ + \ 24\).
Now divide both sides by \(6\) to get \(y=\frac{1}{3} \ x \ + \ 4\).
The slope of this line is  \(\frac{1}{3}\), so any line that also has a slope of \(\frac{1}{3}\) would be parallel to it.
Only choice  A has a slope of \(\frac{1}{3}\).

The correct answer is \(16\)
The rate of construction company\(=\frac{30\ cm}{1\ min}=30\) cm/min
Height of the wall after \(40\) minutes \(=\frac{30 \ cm}{1 \ min} \ × \ 40\) min\(=1200 \) cm 
Let \(x\) be the height of wall, then \(\frac{3}{4} \ x=1200\) cm\( → \ x=\frac{4 \ × \ 1200}{3} → x=1600\) cm\(=16\) m

The correct answer is \(x \ ≥ \ 5 \ ∪ \ x \ ≤ \ - \ 1\)
\(x \ - \ 2 \ ≥ \ 3 → x \ ≥ \ 3 \ + \ 2 → x \ ≥ \ 5\) Or \(x \ - \ 2 \ ≤ - \ 3 → x \ ≤ \ - \ 3 \ + \ 2 → \ x \ ≤ \ - \ 1\)
Then, solution is: \(x \ ≥ \ 5 \ ∪ \ x \ ≤ \ - \ 1\)

The correct answer is \(21\)
When 5 times the number \(x\) is added to \(10\), the result is \(10 \ + \ 5 \ x\).
Since this result is equal to \(35\), the equation \(10 \ + \ 5 \ x \ = 35\) is true.
Subtracting \(10\) from each side of \(10 \ + \ 5 \ x = 35\) gives \(5 \ x=25\), and then dividing both sides by \(5\) gives \(x=5\).
Therefore, \(3\) times \(x\) added to \(6\), or \(6 \ + \ 3 \ x\), is equal to \(6 \ + \ 3 \ (5)=21\)

The correct answer is \(g=5 \ h \ + \ 4\)
Fining g in term of \(h\), simply means “solve the equation for \(g\)”.
To solve for \(g\), isolate it on one side of the equation.
Since \(g\) is on the left-hand side, just keep it there.
Subtract both sides by \(3 \ h\). \(3 \ h \ + \ g \ - \ 3 \ h=8 \ h \ + \ 4 \ - \ 3 \ h\)
And simplifying makes the equation \(g=5 \ h \ + \ 4\), which happens to be the answer.

The correct answer is \(84\)
The description \(8 \ + \ 2 \ x\) is \(16\) more than \(20\) can be written as the equation \(8 \ + \ 2 \ x=16 \ + \ 20\), which is equivalent to \(8 \ + \ 2 \ x=36\).
Subtracting \(8\) from each side of \(8 \ + \ 2 \ x \ =36\) gives \(2 \ x=28\).
Since \(6 \ x\) is \(3\) times \(2 \ x\), multiplying both sides of \(2 \ x=28\) by \(3\) gives \(6 \ x=84\)

The correct answer is \((13, \ 2)\)
From the choices provided, plugin the values of \(a\) and \(b\) into both inequalities and check.
A. \((11, \ 0) → a \ - \ b=11 \ - \ 0=11 \ > \ 10\) and \(a \ + \ b=11 \ + \ 0=11 \ < \ 14\)
B. \((13, \ 2) → a \ - \ b=13 \ - \ 2=11 \ > \ 10\) and \(a \ + \ b=13 \ + \ 2=15 \ > \ 14\)
C. \((13, \ 0) → a \ - \ b=13 \ - \ 0=13 \ > \ 10\) and \(a \ + \ b=13 \ + \ 0=13 \ < \ 14\)
D. \((12, \ 1) → a \ - \ b=12 \ - \ 1=11 \ > \ 10\) and \(a \ + \ b=12 \ + \ 1=13 \ < \ 14\)
E. \((12, \ 2) → a \ - \ b=12 \ - \ 2=10=10\) and \(a \ + \ b=12 \ + \ 2=14 \ < \ 14\)
For choice \(B, \ 15\) is not less than \(14\). Therefore, choice B does not provide the correct values of \(a\) and \(b\).

The correct answer is \(7 \ x^3 \ y^5 \ - \ x^2 \ y^3\)
\(4 \ x^2 \ y^3 \ + \ 5 \ x^3\   y^5 \ – \ (5 \ x^2\   y^3 \ – \ 2 \ x^3 \ y^5 )=4 \ x^2 \ y^3 \ - \ 5 \ x^2 \ y^3 \ + \ 5 \ x^3 \ y^5 \ + \ 2 \ x^3 \ y^5=7 \ x^3 \ y^5 \ - \ x^2 \ y^3\)

The correct answer is \(0, \ – \ 2, \ – \ 3\)
Frist factor the function: \(f(x)=x^3 \ + \ 5 \ x^2 \ + \ 6 \ x=x (x \ + \ 2)\ (x \ + \ 3)\)
To find the zeros, \(f(x)\) should be zero. \(f(x)=x \ (x \ + \ 2)\ (x \ + \ 3)=0\)
Therefore, the zeros are: \(x=0\), \((x \ + \ 2)=0 ⇒ x=- \ 2\), \((x \ + \ 3)=0 ⇒ x=- \ 3\)

The correct answer is \(2\)
\(sin^2 \ a \ + \) cos\(^2\ a=1\), then: \(x\ +\ 1=3\), \(x=2\)

The correct answer is \(\frac{y}{6}\)
Solve for \(x\) \(\sqrt{6\ x}=\sqrt{y}\). Square both sides of the equation: \((\sqrt{6 \ x})^2=(\sqrt{y})^2 → 6 \ x=y → x=\frac{y}{6}\)

The correct answer is \(61.28\)
average \(=\frac{sum \ of \ terms}{number \ of \ terms}\). The sum of the weight of all girls is: \(18\ ×\ 60 = 1080\)  kg
The sum of the weight of all boys is: \(32\ ×\ 62=1984\) kg.
The sum of the weight of all students is: \(1080\ +\ 1984 = 3064\) kg. average \(=\frac{3064}{50} = 61.28\)

The correct answer is \(9 \ x^6\)
\(y=(- \ 3 \ x^3)^2=(- \ 3)^2 (x^3)^2=9 \ x^6\)

The correct answer is \(36\)
Plug in the value of \(x\) and \(y\).
\(x=3\) and \(y=- \ 2\)
\(5 \ (x \ - \ 2 \ y) \ + \ (2 \ - \ x)^2=5 \ (3 \ - \ 2 \ (- \ 2)) \ + \ (2 \ - \ 3)^2=5 \ (3 \ + \ 4) \ + \ (- \ 1)^2 = 35 \ + \ 1=36\)

The correct answer is \(x=\frac{3}{7}\)
Substituting \(x\) for \(y\) in first equation.
\(5 \ x \ + \ 2 \ y=3\), \(5 \ x \ + \ 2 \ (x)=3\), \(7 \ x \ =3\)
Divide both side of \(7 \ x=3\) by \(3\) gives \(x=\frac{3}{7}\)

The correct answer is \(5 \ x \ y\)
There are \(5\) floors, \(x\) rooms in each floor, and \(y\) chairs per room.
If you multiply \(5\) floors by \(x\), there are \(5 \ x\) rooms in the hotel.
To get the number of chairs in the hotel, multiply \(5 \ x\) by \(y\). \(5 \ x \ y\) is the number of chairs in the hotel.

The correct answer is \(6\)
If \(β=3\ γ\), then multiplying both sides by \(12\) gives \(12 \ β=36 \ γ\).
\(α=2 \ β\), thus \(α=6 \ γ\). Multiply both sides of the equation by \(6\) gives \(6 \ α=36 \ γ\).

The correct answer is \(\sqrt{6} \ - \ 1\)
\(x_{1,2} =\frac{- \ b\ ± \sqrt{b^2 \ - \ 4 \ a\ c}}{2\ a}\) \(a\ x^2\ +\ b \ x\ +\ c=0\)
\(\ x ^2 \ + \ 2 \ x \ – \ 5=0\) ⇒ then: \(a=1, \ b=2\) and \(c= –\ 5\)
\(x =\frac{- \ 2 \ + \sqrt{2^2  \ - \ 4.1 .- \ 5}}{2.1}=\sqrt{6}- \ 1\) \(x =\frac{- \ 2 - \sqrt{2^2 \ - \ 4 .1 . \ - \ 5}}{2.1} =- \ 1 \ -\sqrt{6}\)

The correct answer is \(\frac{3}{4} \ + \ i\)
\(\frac{4 \ - \ 3 \ i}{- \ 4 \ i}× \frac{i}{i}=\frac{4 \ i \ - \ 3 \ i^2}{- \ 4 \ i^2}\).
\(i^2 \ - \ 1\), Then: \(\frac{4 \ i \ - \ 3 \ i^2}{- \ 4 \ i^2}=\frac{4 \ i \ -3 \ (- \ 1)}{- \ 4\ (- \ 1)}=\frac{4 \ i \ + \ 3}{4}=\frac{4 \ i}{4} \ + \frac{3}{4}=i \ + \frac{3}{4}\)

The correct answer is \(7 \ x^2  \ + \ x \ + \ 5\)
The sum of the two polynomials is \((4 \ x^2 \ + \ 6 \ x \ - \ 3) \ + \ (3 \ x^2 \ - \ 5 \ x \ + \ 8)\)
This can be rewritten by combining like terms: \((4 \ x^2 \ + \ 6 \ x \ - \ 3) \ + \ (3 \ x^2 \ - \ 5 \ x \ + \ 8)=(4 \ x^2 \ + \ 3 \ x^2 ) \ + \ (6 \ x \ - \ 5 \ x) \ + \ (- \ 3 \ + \ 8)=7 \ x^2 \ + \ x \ + \ 5\)

The correct answer is \(g(x)=- \ 2 \ x \ + \ 1\)
For \((1, \ - \ 1)\) check the options provided:
A. \(g(x)=2 \ x \ + \ 1 → - \ 1=2 \ (1) \ + \ 1 → - \ 1=3\) This is NOT true.
B. \(g(x)=2 \ x \ - \ 1 → - \ 1=2 \ (1) \ - \ 1=1\) This is NOT true.
C. \(g(x)=- \ 2 \ x \ +\ 1 → - \ 1=2 \ (- \ 1) \ + \ 1 → - \ 1=- \ 1\) This is true.
D. \(g(x)=x \ +\ 2 → - \ 1=1 \ + \ 2 → - \ 1=3\) This is NOT true.
E. \(g(x)=2 \ x \ + \ 2 → - \ 1=2 \ (1) \ + \ 2=1\) This is NOT true.
From the choices provided, only choice C is correct.

The correct answer is \(\frac{3 \ x}{4}\)
Simplify the expression. \(\sqrt{\frac{x^2}{2} \ + \frac{x^2}{16}}\)\(=\sqrt{\frac{8 \ x^2}{16} \ + \frac{x^2}{16}}=\)\(\sqrt{\frac{9 \ x^2}{16}}=\)\(\sqrt{\frac{9}{16} \ x^2}\)=\(\sqrt{\frac{9}{16}}×\sqrt{x^2}=\frac{3}{4} \ × \ x=\frac{3 \ x}{4}\)

The correct answer is \(- \ 4\)
To find the \(y \ -\)intercept of a line from its equation, put the equation in slope-intercept form:
\(x \ - \ 3 \ y=12\), \(- \ 3 \ y=- \ x \ + \ 12\), \(3 \ y=x \ - \ 12\), \(y=\frac{1}{3} x \ - \ 4\)
The \(y \ -\)intercept is what comes after the \(x\). Thus, the \(y \ -\)intercept of the line is \(- \ 4\).

The correct answer is \(30\)
Adding both side of \(4 \ a \ - \ 3=17\) by \(3\) gives \(4 \ a=20\)
Divide both side of \(4 \ a=20\) by \(4\) gives \(a=5\), then \(6 \ a=6 \ (5)=30\)

The correct answer is \(1\)
The easiest way to solve this one is to plug the answers into the equation.
When you do this, you will see the only time \(x=x^{(- \ 6)}\) is when \(x=1\) or \(x=0\).
Only \(x=1\) is provided in the choices.

The correct answer is \(\frac{12 \ x \ + \ 1}{x^3}\)
First find a common denominator for both of the fractions in the expression \(\frac{5}{x^2} + \frac{7 \ x \ - \ 3}{x^3}\).
of \(x^3\), we can combine like terms into a single numerator over the denominator:
\(\frac{5 \ x \ + \ 4}{x^3} \ + \frac{7 \ x \ -\ 3}{x^3} =\frac{(5 \ x \ + \ 4) \ + \ (7 \ x \ - \ 3)}{x^3} =\frac{12 \ x \ + \ 1}{x^3}\)

The correct answer is \(y=4 \ (x \ - \ 3)^2 \ - \ 3\)
Let’s find the vertex of each choice provided:
A. \(y=3 \ x^2 \ - \ 3\) The vertex is: \((0, \ - \ 3)\)
B. \(y=- \ 3 \ x^2 \ + \ 3\) The vertex is: \((0, \ 3)\)
C. \(y=x^2 \ + \ 3 \ x \ - \ 3\)
The value of \(x\) of the vertex in the equation of \(a\) quadratic in standard form is: \(x=\frac{- \ b}{2 \ a}=\frac{- \ 3}{2}\)
(The standard equation of a quadratic is: \(a \ x^2 \ + \ b \ x \ + \ c=0)\)
The value of \(x\) in the vertex is \(3\) not \(\frac{- \ 3}{2}\).
D. \(y=4 \ (x \ - \ 3)^2 \ - \ 3\)
Vertex form of \(a\) parabola equation is in form of \(y=a \ (x \ - \ h)^2 \ + \ k\) , where \((h, \ k)\) is the vertex. Then \(h=3\) and \(k=- \ 3\). (This is the answer)
E. \(y=4 \ x^2 \ + \ 3 \ x \ - \ 3\). \(x=\frac{- \ b}{2 \ a}=\frac{- \ 3}{2 \ × \ 8}=\frac{- \ 3}{16}\). The value of \(x\) in the vertex is \(3\) not \(\frac{- \ 3}{16}\).

The correct answer is \(2\ x\ -\frac{1}{3}\)
To find the average of three numbers even if they’re algebraic expressions, add them up and divide by \(3\).
Thus, the average equals: \(\frac{(4\ x\ +\ 2) \ + \ (-\ 6\ x\ -\ 5) \ + \ (8\ x\ +\ 2)}{3}=\frac{6\ x\ -\ 1}{3}=2\ x\ -\frac{1}{3}\)

The correct answer is \(-\frac{1}{2}\)
The equation of a line in slope intercept form is: \(y=m\ x\ +\ b\).
Solve for \(y\). \(4\ x\ -\ 2\ y=12 ⇒ -\ 2\ y=12\ -\ 4\ x ⇒ y=(12\ -\ 4\ x) \ ÷ \ (-\ 2) ⇒ y=2\ x\ -\ 6\).
The slope is \(2\). The slope of the line perpendicular to this line is: \(m_1\ × m_2=-\ 1 ⇒ 2\ × m_2  = -\ 1 ⇒ m_2=-\frac{1}{2}\)

The correct answer is \((1, \ 6), (2, \ 5), (−\ 5, \ 8)\)
Since the triangle ABC is reflected over the \(y\ -\)axis, then all values of \(y\)’s of the points don’t change and the sign of all \(x\)’s change.
(remember that when a point is reflected over the \(y\ -\ a\) xis,
the value of \(y\) does not change and when a point is reflected over the \(x\ -\ a\)xis,
the value of \(x\) does not change). Therefore: \((−\ 1, \ 6)\) changes to \((1, \ 6)\).
\((−\ 2, \ 5)\) changes to \((2, \ 5\)). \((5, \ 8)\) changes to \((−\ 5, \ 8)\)

The correct answer is \(39\)
The area of rectangle is:\(9\ ×\ 4=36\) cm\(^2\). The area of circle is: \(π\ r^2=π\ ×(\frac{10}{2})^2=3\ ×\ 25=75\) cm\(^2\).
Difference of areas is: \(75\ -\ 36=39\)

The correct answer is \(\frac{2}{x^3 \ +\ 2}\)
\(f(g(x))=2\ × \)\((\frac{1}{x})^3\ +\ 2=\frac{2}{x^3}+2\)

The correct answer is \(4\)
\(1269=6^4 →6^x=6^4 → x=4\)

The correct answer is \(170\) miles
Use the information provided in the question to draw the shape.
Use Pythagorean Theorem: \(a^ 2\ +\ b^2 = c^2\)
\(80^2\ +\ 150^2=c^2 ⇒ 6400\ +\ 22500 = c^2 ⇒ 28900 = c^2 ⇒ c =170\)

The correct answer is \(45\) m\(^2\)
Let \(L\) be the length of the rectangular and \(W\) be the with of the rectangular. Then, \(L=4\ W\ +\ 3\)
The perimeter of the rectangle is \(36\) meters. Therefore: \(2\ L\ +\ 2\ W=36\) \(L\ +\ W=18\)
Replace the value of L from the first equation into the second equation and solve for \(W\):
\((4\ W\ +\ 3)\ +\ W=18 → 5\ W\ +\ 3=18 → 5\ W=15 → W=3\)
The width of the rectangle is \(3\) meters and its length is: \(L=4\ W\ +\ 3=4\ (3)\ +\ 3=15\)
The area of the rectangle is: length \(×\) width \(= 3\ ×\ 15 = 45\)

The correct answer is \(5\)
Let \(x\) be the number of adult tickets and \(y\) be the number of student tickets. Then:
\(x\ +\ y=12\), \(12.50\ x\ +\ 7.50 \ y=125\)
Use elimination method to solve this system of equation.
Multiply the first equation by \(-\ 7.5\) and add it to the second equation.
\(-\ 7.5\ (x\ +\ y=12)\), \(-\ 7.5\ x\ -\ 7.5\ y=-\ 90\), \(12.50\ x\ +\ 7.50\ y=125\). \(5\ x=35\), \(x=7\)
There are \(7\) adult tickets and \(5\) student tickets.

The correct answer is \(\frac{1}{25}\)
Write the ratio of \(5\ a\) to \(2\ b\). \(\frac{5\ a}{2\ b}=\frac{1}{10}\)
Use cross multiplication and then simplify.
\(5\ a\ ×\ 10=2\ b\ ×\ 1 → 50\ a=2 \ b → a=\frac{2b}{50}=\frac{b}{25}\)
Now, find the ratio of \(a\) to \(b\).
\(\frac{a}{b}=\frac{\frac{b}{25}}{b}\)\(→\frac{b}{25}\ ÷\ b=\frac{b}{25}\ ×\frac{1}{b}=\frac{b}{25\ b}=\frac{1}{25}\)

The correct answer is \(30\)
Plug in the value of \(x\) in the equation and solve for \(y\).
\(2\ y=\frac{2\ x^2}{3}\ +\ 6 → 2\ y\) =\(\frac{2\ (9)^2}{3}\ +\ 6 → 2\ y=\) \(\frac{2\ (81)}{3}\ +\ 6 → 2\ y= 54\ +\ 6=60\)
\(2\ y = 60 → y=30\)

The correct answer is \(1.085\ (3\ p)\ +\ 6\)
Since a box of pen costs \($3\), then \(3\ p\) Represents the cost of \(p\) boxes of pen.
Multiplying this number times \(1.085\) will increase the cost by the \(8.5%\) for tax.
Then add the \($6\) shipping fee for the total: \(1.085\ (3\ p)\ +\ 6\)

The correct answer is \(y(x)=8\ x\)
Rate of change (growth or \(x\)) is \(8\) per week. \(40\ ÷\ 5=8\)
Since the plant grows at a linear rate,
then the relationship between the height \((y)\) of the plant and number of weeks of growth \((x)\) can be written as: \(y(x)=8\ x\)