Free Full Length CLEP College Algebra Practice Test

The correct answer is \(20\)
Add \(8\) both sides of the equation \(8 \ x \ - \ 8 = 24\) gives \(8 \ x = 24 \ + \ 8 = 32\).
Dividing each side of the equation \(8 \ x = 32\) by \(8\) gives \(x = 4\).
Substituting \(4\) for \(x\) in the expression \(6 \ x \ - \ 4\) gives \(6 \ (4) \ - \ 4 = 20\).

The correct answer is \((4, -\ 4)\)
Method \(1\): Plugin the values of \(x\) and \(y\) provided in the options into both equations.
A. \((4, \ 3)\)       \(x \ + \ y =0 → 4 \ + \ 3 ≠0\)
B. \((5, \ 4)\)       \(x \ + \ y =0 → 5 \ + \ 4 ≠0\)
C. \((4,- \ 4)\)        \(x \ + \ y = 0 → 4 \ + \ (- \ 4) =0\)
D. \((4, \ - \ 6)\)     \(x \ + \ y =0 → 4 \ + \ (- \ 6) ≠0\)
E. \((4, \ 6)\)
Method \(2\): Multiplying each side of \(x \ + \ y=0\) by \(2\) gives \(2 \ x \ + \ 2 \ y=0\).
Then, adding the corresponding side of \(2 \ x \ + \ 2 \ y=0\) and \(4 \ x \ - \ 2 \ y=24\) gives  \(6 \ x=24\).
Dividing each side of   \(6 \ x=24\) by \(6\) gives \(x  =4\).
Finally, substituting \(4\) for \(x\) in \(x \ + \ y =0\), or \(y = - \ 4\).
Therefore, the solution to the given system of equations is \((4, \ - \ 4)\).

The correct answer is \(28\ x \ + \ 6 \)
If \(f(x) = 3 \ x \ + \ 4 \ (x \ + \ 1) \ + \ 2\), then find \(f(4 \ x)\) by substituting \(4 \ x\) for every \(x\) in the function. This gives: \(f(4 \ x)= 3 \ (4 \ x) \ + \ 4 \ (4 \ x \ + \ 1) \ + \ 2\),
It simplifies to: \(f \ (4 \ x)= 3 \ (4 \ x) \ + \ 4 \ (4 \ x \ + \ 1) \ + \ 2=12 \ x \ + \ 16 \ x \ + \ 4 \ + \ 2 =28 \ x \ + \ 6\)

The correct answer is \((9, \ 3)\)
First, find the equation of the line. All lines through the origin are of the form \(y=mx\), so the equation is \(y=\frac{1}{3} \ x\).
Of the given choices, only choice \(C \ (9, \ 3)\), satisfies this equation:
\(y=\frac{1}{3}\) \(x →3= \frac{1}{3} \ (9)=3\)

The correct answer is \(n^2 \ + \ 2 \ n \ + \ 10\)
\((3 \ n^2 \ + \ 2 \ n \ + \ 6) \ - \ (2 \ n^2 \ - \ 4)\)
Add like terms together: \(3 \ n^2 \ - \ 2 \ n^2 = n^2\)
\(2 \ n\) doesn’t have like terms. \(6 \ - \ ( \ -\ 4) =10\)
Combine these terms into one expression to find the answer: \(n^2 \ + \ 2 \ n \ + \ 10\)

The correct answer is \(23, \  26\)
You can find the possible values of \(a\) and \(b\) in \((a \ x \ + \ 4) \ (b \ x \ + \ 3)\)
by using the given equation \(a \ + \ b =7\) and finding another equation that relates the variables \(a\) and \(b\).
Since\((a \ x \ + \ 4)\) \((b \ x \ + \ 3)=10 \ x^2 \ + \ c \ x \ + \ 12\), expand the left side of the equation to obtain
\(a\ b\ x^2 \ + \ 4 \ b \ x \ + \ 3 \ a \ x \ + \ 12 =10 \ x^2 \ + \ c \ x \ + \ 12\)
Since ab is the coefficient of \(x^2\) on the left side of the equation and \(10\) is the coefficient of \(x^2\) on the right side of the equation,
it must be true that \(a \ b=10\)
The coefficient of \(x\) on the left side is \(4 \ b \ + \ 3 \ a\) and the coefficient of \(x\) in the right side is \(c\).
Then: \(4 \ b \ + \ 3 \ a = c\), \(a \ + \ b =7\), then: \(a =7- \ b\)
Now, plug in the value of \(a\) in the equation \(ab =10\). Then:
\(ab=10 → (7 \ - \ b) \ b =10 → 7 \ b \ - \ b^2 = 10\)
Add \(-7 \ b \ + \ b^2\) both sides. Then: \(b^2 \ - \ 7 \ b \ + \ 10 = 0\)
Solve for \(b\) using the factoring method. \(b^2 \ - \ 7 \ b \ + \ 10=0 → (b \ - \ 5) \ (b \ - \ 2)=0\)
Thus, either \(b=2\) and \(a\) \(=5\), or \(b = 5\) and \(a = 2\). If \(b=2\) and \(a=5\), then
\(4 \ b \ + \ 3 \ a =c → 4 \ (2) \ + \ 3 \ (5) =c → c =23\)
If \(5 = 2\) and \(a = 2\), then, \(4 \ b \ + \ 3 \ a = c → 4 \ (5) \ + \ 3 \ (2) = c → c = 26\)
Therefore, the two possible values for \(c\) are \(23\) and \(26\).

The correct answer is \(\frac{(x \ - \ 5) \ (x \ + \ 4)}{((x \ - \ 5) \ + \ (x \ + \ 4)}\)
To rewrite \(\frac{1}{\frac{1}{x \ - \ 5} \ + \ \frac{1}{x \ + \ 4}}\), first simplify \(\frac{1}{x \ - \ 5} \ + \ \frac{1}{x \ + \ 4}\).
\(\frac{1}{(x \ - \ 5)}\)\(+\)\(\frac{1}{(x \ + \ 4)}\)\(=\)\(\frac{1(x \ + \ 4)}{((x \ - \ 5) \ (x \ + \ 4)}\) \(+\frac{1(x \ - \ 5)}{(x \ + \ 4) \ (x \ - \ 5)}\)\(=\)\(\frac{(x \ + \ 4) \ + \ (x \ - \ 5)}{ \ (x \ + \ 4) \ (x \ - \ 5)}\) 
Then:
\(\frac{1}{\frac{1}{(x \ - \ 5)} \ + \ \frac{1}{(x \ + \ 4)}}\)\(=\)\(\frac{1}{\frac{(x \ + \ 4) \ + \ (x \ - \ 5)}{(x \ + \ 4) \ (x \ - \ 5)}}\)\(=\)\(\frac{(x \ - \ 5)(x \ + \ 4)}{(x \ - \ 5) \ + \ (x \ + \ 4)}\). (Remember, \(\frac{1}{\frac{1}{x}}\)\(= x\))
This result is equivalent to the expression in choice A.

The correct answer is \(a \ > \ b\)
Since \((0, \ 0)\) is a solution to the system of inequalities, substitutin \(0\) for \(x\) and \(0\) for \(y\) in the given system must result in two true inequalities.
After this substitution, \(y \ < \ a \ - \ x\) becomes \(0 \ < \ a\), and \(y \ > \ x \ + \ b\) becomes \(0 \ > \ b\).
Hence, \(a\) is positive and \(b\) is negative.
Therefore, \(a \ > \ b\).

The correct answer is \((6, \ 6)\)
First find the slope of the line using the slope formula. \(m =\frac{y_2 \ - \ y_1}{x_2 \ - \ x _1}\)
Substituting in the known information. \((x_1, \ y_1 )=(2, \ 4)\), \((x_2, \ y_2 )=(4, \ 5)\)
\(m =\frac{5 \ - \ 4}{4 \ - \ 2}=\frac{1}{2}\). Now the slope to find the equation of the line passing through these points.
\(y =m\ x \ + \ b\) Choose one of the points and plug in the values of \(x\) and \(y\) in the equation to solve for \(b\). Let’s choose point \((4, \ 5)\). Then:
\(y =m\ x \ + \ b → 5 \ = \frac{1}{2}\)\((4) \ + \ b → 5=2 \ + \ b → b=5 \ - \ 2=3\)
The equation of the line is: \(y \ =\frac{1}{2} \ x \ + \ 3\)
Now, plug in the points provided in the choices into the equation of the line.
A. \((9, \ 9)\) \(y \ =\frac{1}{2} \ x \ + \ 3 → 9 =\frac{1}{2} \ (9) \ + \ 3 → 9 =7.5\) This is NOT true.
B. \((9, \ 6)\) \(y =\frac{1}{2} \ x \ + \ 3 → 6 =\frac{1}{2} \ (9) \ + \ 3 → 6 =7.5\) This is NOT true.
C. \((6, \ 9)\) \(y =\frac{1}{2} \ x \ + \ 3 → 9 =\frac{1}{2} \ (6) \ + \ 3 → 9 =6\) This is NOT true.
D. \((6, \ 6)\) \(y =\frac{1}{2} \  x \ + \ 3 → 6 = \frac{1}{2} \ (6) \ + \ 3 → 6 =6\) This is true!
E. \((6, \ - 6)\) \(y=\frac{1}{2} \  x \ + \ 3 → - \ 6 =\frac{1}{2} \ (6) \ + \ 3 → -6=6\) This is NOT true.
Therefore, the only point from the choices that lies on the line is \((6, \ 6)\).

The correct answer is \(10\)
The input value is \(5\). Then: \(x =5\)
\(f(x) = x^2 \ - \ 3 \ x → f(5)=5^2 \ - \ 3 \ (5)=25 \ - \ 15=10\)

The correct answer is The \(y-\)intercept represents the starting height of \(6\) inches
To solve this problem, first recall the equation of a line: \(y=mx \ + \ b\)
Where \(m=\) slope \(y=y-\)intercept
Remember that slope is the rate of change that occurs in a function and that the \(y-\)intercept is the \(y\) value corresponding to \(x=0\).
Since the height of John’s plant is \(6\) inches tallwhen he gets it. Time (or \(x\)) is zero.
The plant grows \(4\) inches per year. Therefore, the rate of change of the plant’s height is \(4\).
The \(y-\)intercept represents the starting height of the plant which is \(6\) inches.

The correct answer is \(-2\)
Multiplying each side of \(\frac{4}{x}=\frac{12}{x \ - \ 8}\) by \(x\) \((x \ - \ 8)\) gives \(4 \ (x \ - \ 8)=12 \ (x)\),
distributing the \(4\) over the values within the parentheses yields \(x \ - \ 8=3\  x\) or \(x =- \ 4\).
Therefore, the value of  \(\frac{x}{2}=\frac{-4}{2} =- \ 2.\)

The correct answer is \(x^2 \ + \ (y \ - \ 4)^2 =\frac{61}{9}\)
The equation of a circle can be written as \((x \ - \ h)^2 \ + \ (y \ - \ k)^2 =r^2\) 
where \((h, \ k)\) are the coordinates of the center of the circle and \(r\) is the radius of the circle.
Since the coordinates of the center of the circle are \((0, \ 4)\), the equation is \(x^2 \ + \ (y \ - \ 4)^2=r^2\), where \(r\) is the radius.
The radius of the circle is the distance from the center \((0, \ 4)\), to the given endpoint of a radius, \((\frac{5}{3}, \ 6)\). By the distance formula,
\(r^2 =(\frac{5}{3} \ - \ 0)^2 \ + \ (6 \ - \ 4)^2=\frac{61}{9}\)
Therefore, an equation of the given circle is \(x^2 \ + \ (y \ - \ 4)^2=\frac{61}{9}\)

The correct answer is \(2 \ x^2 \ - \ 4 \ x \ + \ 4\)
In order to figure out what the equation of the graph is, fist find the vertex.
From the graph we can determine that the vertex is at \((1, \ 2)\).
We can use vertex form to solve for the equation of this graph.
Recall vertex form, \(y =a \ (x \ - \ h)^2 \ + \ k\), where h is the \(x\) coordinate of the vertex, and \(k\) is the \(y\) coordinate of the vertex.
Plugging in our values, you get \(y=a \ (x \ - \ 1)^2 \ + \ 2\)
To solve for \(a\), we need to pick \(a\) point on the graph and plug it into the equation.
Let’s pick \((-\ 1, \ 10)\). \(10=a \ (-\ 1 \  -\ 1)^2 \ + \ 2 → 10=a \ (-\ 2)^2 \ + \ 2 → 10=4 \ a \ + \ 2\)
\(8=4 \ a → a=2\) Now the equation is : \(y=2 \ (x \ - \ 1)^2 \ + \ 2\)
Let’s expand this, \(y=2 \ (x^2 \ - \ 2 \ x \ + \ 1) \ + \ 2 → y=2 \ x^2 \ - \ 4 \ x \ + \ 2 \ + \ 2 → y=2 \ x^2 \ - \ 4 \ x \ + \ 4\)
The equation in Choice C is the same.

The correct answer is \(7 \ ≤ \ x \ ≤ \ 13\)
\(|\ x \ - \ 10\ | \ ≤ \ 3 → \ -3 \ ≤ \ x \ - \ 10 \ ≤ \ 3 → - \ 3 \ + \ 10 \ ≤ \ x \ - \ 10 \ + \ 10 \ ≤ \ 3 \ + \ 10 → 7 \ ≤ \ x \ ≤ \ 13\)

The correct answer is {\(5, \ 11\)}
The union of \(A\) and \(B\) is: \(A \ ∪ \ B\)\(=\){\(1, \ 2, \ 3, \ 4, \ 5, \ 6, \ 11, \ 15\)}
The intersection of \((A \ ∪ \ B)\) and C is: \((A \ ∪ \ B) \ ∩ \ C=\) {\(5, \ 11\)}

The correct answer is \(7\)
Adding \(6\) to each side of the inequality \(4 \ n \ - \ 3 \ ≥ \ 1\) yields the inequality \(4 \ n \ + \ 3 \ ≥ \ 7\).
Therefore, the least possible value of \(4 \ n \ + \ 3\) is \(7\).

The correct answer is \(60\) feet
The relationship among all sides of special right triangle
\(30^\circ \ - \ 60^\circ \ - \ 90^\circ\) is provided in this triangle:
In this triangle, the opposite side of \(30^\circ\) angle is half of the hypotenuse.
Draw the shape of this question:
The ladder is the hypotenuse. Therefore, the ladder is \(60\) ft.

The correct answer is \(9\)
The union of \(A\) and \(B\) is: \(A \ ∪ \ B=\) {\(1, \ 2, \ 4, \ 8, \ 12, \ 16, \ 24, \ 32, \ 48\)}. There are \(9\) elements in \(A \ ∪ \ B\).

The correct answer is \(8\)
\(x^2 \ + \ 6 \ x \ + \ r =(x \ + \ 2) \ (x \ + \ p)=x^2 \ + \ (2 \ + \ p) \ x \ + \ 2 \ p\)

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

The correct answer is \(80\)
he speed of car A is \(56\) mph and the speed of car B is \(64\) mph.
When both cars drive in a straight line toward each other,
the distance between the cars decreases at the rate of \(120\) miles per hour: \(56 \ + \ 64=120\)
\(40\) minutes is two third of an hour. Therefore, they will be \(80\) miles apart \(40\) minutes before they meet.
\(\frac{2}{3} \ × \ 120=80\)

The correct answer is \(-\  \frac{8}{7}\)
Since \(f(x)\) is linear function with a negative slop, then when \(x=- \ 2, \ f(x)\) is maximum and when \(x=3, \ f(x)\) is minimum.
Then the ratio of the minimum value to the maximum value of the function is: \(\frac{f(3)}{f(-\ 2)}\)\(=\)\(\frac{ - \ 3 \ (3) \ + \ 1}{-\ 3 \ ( \ - \ 2) \ + \ 1}\)\(=\)\(\frac{\ - \ 8}{7}=-\  \frac{ 8}{7}\)

The correct answer is \(2\)
There can be \(0, \ 1\), or \(2\) solutions to a quadratic equation.
In standard form, a quadratic equation is written as: \(a \ x^2 \ + \ b \ x \ + \ c=0\)
For the quadratic equation, the expression \(b^2 \ - \ 4 \ a \ c\) is called discriminant.
If discriminant is positive, there are \(2\) distinct solutions for the quadratic equation.
If discriminant is \(0\), there is one solution for the quadratic equation and if it is negative the equation does not have any solutions.
To find number of solutions for \(x^2 \ = \ 4 \ x \ - \ 3\), first, rewrite it \(a \ s \ x^2 \ - \ 4 \ x \ + \ 3=0\).
Find the value of the discriminant. \(b^2 \ - \ 4 \ a \ c =(- \ 4)^2 \ -4 \ (1) \ (3)=16 \ - \ 12 =4\)
Since the discriminant is positive, the quadratic equation has two distinct solutions.

The correct answer is no value of \(x\)
If the value of \(| \ x \ - \ 3 \ | \ + \ 3\) is equal to \(0\), then \(| \ x \ - \ 3 \ | \ + \ 3=0\).
Subtracting \(3\) from both sides of this equation gives \(| \ x \ - \ 3 \ |=- \ 3\).
The expression \(| \ x \ - \ 3 \ |\) on the left side of the equation is the absolute value of \(x \ - \ 3\),
and the absolute value can never be a negative number.
Thus \(| \ x \ - \ 3 \ |=- \ 3\) has no solution.
Therefore, there are no values for \(x\) for which the value of \(| \ x \ - \ 3 \ | \ + \ 3\) is equal to \(0\).

The correct answer is I and III only
I. \(| \ a \ | \ < \ 1 → \ - \ 1 \ < \ a \ < \ 1\)
Multiply all sides by \(b\). Since,\(b \ > \ 0 → \ - \ b \ < \ b\ a \ < \ b\) (it is true!)
II. Since, \( \ - \ 1 \ < \ a \ < \ 1 \) ,and \(a \ < \ 0 \ → \ -\ a \ > \ a^2 \ > \ a\) (plug in \(-\frac {1}{2}\), and check!) (It’s false)
III. \(- \ 1 \ < \ a \ < \ 1\),multiply all sides by \(2\),then: \(- \ 2 \ < \ 2 \ a \ < \ 2\)
Subtract \(3\) from all sides. Then:
\(- \ 2 \ - \ 3 \ < \ 2 \ a \ - \ 3 \ < \ 2 \ - \ 3 → \ - \ 5 \ < \ 2 \ a \ - \ 3 \ < \ - \ 1\) (It is true!)

The correct answer is \(88\)
In the figure angle A is labeled \((3 \ x \ - \ 2)\) and it measures \(37\). Thus, \(3 \ x \ - \ 2=37\) and \(3 \ x =39\) or \(x =13\).
That means that angle B, which is labeled \((5 \ x)\), must measure \(5 \ × \ 13=65\).
Since the three angles of a triangle must add up to \(180, \ 37 \ + \ 65 \ + \ y \ - \ 8=180\), then:
\(y \ + \ 94=108 → y =180 \ - \ 94=86\)

The correct answer is \(20\)
Five years ago, Amy was three times as old as Mike. Mike is \(10\) years now. Therefore, \(5\) years ago Mike was \(5\) years.
Five years ago, Amy was: \(A=3 \ × \ 5=15\)
Now Amy is \(20\) years old: \(15 \ + \ 5 =20\)

The correct answer is \(600\) ml
\(4\%\) of the volume of the solution is alcohol. Let \(x\) be the volume of the solution.
Then: \(4\%\) of \(x=24\) ml ⇒ \(0.04 \ x=24 ⇒ \ x=24 \ ÷ \ 0.04=600\)

The correct answer is \(4\) and \(5\)
A. \(1\) and \(2\). \(13 = 1\) and \(23 = 8\), \(112\) is not between these two numbers.
B. \(2\) and \(3\). \(23 = 8\) and \(33 = 27\), \(112\) is not between these two numbers.
C. \(3\) and \(4\). \(33 = 27\) and \(43 = 64\), \(112\) is not between these two numbers.
D. \(4\) and \(5. 43 = 64\) and \(53 = 125\), \(112\) is between these two numbers.
E. \(5\) and \(6. 53 = 125\) and \(63 = 216, 112\) is not between these two numbers.

The correct answer is \(4\)
Solve for \(x\). \(\frac{3 \ x}{16}=\)\(\frac{x \ - \ 1}{4}\). Multiply the second fraction by \(4\). \(\frac{3 \ x}{16}=\)\(\frac{4 \ (x \ - \ 1)}{4 \ × \ 4}\)
Tow denominators are equal. Therefore, the numerators must be equal.
\(3 \ x =4 \ x \ - \ 4\),     \(0=x \ - \ 4\)       \(4=x\)

The correct answer is \(2\)
\((x \ - \ 2)^3=27 → x \ - \ 2 =3 → x=5\)
\(→ (x \ - \ 4) \ (x \ - \ 3)=(5 \ - \ 4) \ (5 \ - \ 3)=(1) \ ( 2)=2\)

The correct answer is \(- \ 60 \ + \ 2 \ i\)
We know that: \(i=\sqrt{-\ 1} ⇒ i^2=- \ 1\)
\((- \ 5 \ + \ 9 \ i) \ (3 \ + \ 5 \ i)=- \ 15 \ - \ 25 \ i \ + \ 27 \ i \ + \ 45 \ i^2=- \ 15 \ + \ 2 \ i \ - \ 45=- \ 60 \ + \ 2 \ i\)

The correct answer is \(22\)
Substituting \(6\) for \(x\) and \(14\) for \(y\) in \(y=n \ x \ + \ 2\) gives \(14=(n) \ (6) \ + \ 2\),
which gives \(n=2\). Hence, \(y=2 \ x \ + \ 2\). Therefore, when \(x=10\), the value of y is
\(y =(2) \ (10) \ + \ 2=22\).

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

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

The correct answer is \(1, \ 3\)
First square both sides of the equation to get \(4 \ m \ - \ 3=m^2\)
Subtracting both sides by \(4 \ m \ - \ 3\) gives us the equation \(m^2 \ - \ 4 \ m \ + \ 3=0\)
Here you can solve the quadratic equation by factoring to get \((m \ - \ 1) \ (m \ - \ 3)=0\)
For the expression \((m \ - \ 1) \ (m \ - \ 3)\) to equal zero, \(m=1\) or \(m=3\)

The correct answer is \(\frac{5}{4}\)
To solve the equation for y, multiply both sides of the equation by the reciprocal of 
\(\frac{6}{5}\) , which is \(\frac{5}{6}\) , this gives  \((\frac{5}{6}) \ × \frac{6}{5} \  y=\frac{3}{2} \ × \ (\frac{5}{6})\), which simplifies to \(y=\frac{15}{12}=\frac{5}{4}\).

The correct answer is \(45\)
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 \(I \ > \ 2000 \ x \ + \ 24000\)
Let \(x\) be the number of years. Therefore, \($2,000\) per year equals \(2000 \ x\).
starting from \($24,000\) annual salary means you should add that amount to \(2000 \ x\).
Income more than that is: \(I \ > \ 2000 \ x \ + \ 24000\)

A zero of a function corresponds to an \(x \ -\)intercept of the graph of the function in the \(xy \ - \) plane.
Therefore, the graph of the function \(g(x)\), which has three distinct zeros, must have three \(x \ - \)intercepts.
Only the graph in choice C has three \(x \ - \)intercepts.

The correct answer is \(225\)
\(0.6 \ x =(0.3) \ × \ 20 → x=10 → (x \ + \ 5)^2=(15)^2 =225\)

The correct answer is \(y=x\)
The slop of line A is: \(m=\frac{y_2 \ - \ y_1}{x_2 \ - \ x_1}=\frac{3 \ - \ 2}{4 \ - \ 3}=1\)
Parallel lines have the same slope and only choice E \((y=x)\) has slope of \(1\).

The correct answer is \(x\) is divided by \(3\)
\(\frac{8\ y \ +\ \frac{r}{r\ +\ 1}}{\frac{6}{\frac{z}{3}}}=\frac{8\ y\  +\ \frac{r}{r\ +\ 1}}{\frac{3\ ×\ 6}{z}}=\frac{8\ y \ +\ \frac{r}{r\ +\ 1}}{3\ ×\ {\frac{6}{z}}}=\frac{1}{3}\ ×\ \frac{8\ y \ +\ \frac{r}{r\ +\ 1}}{\frac{6}{{z}}}=\frac{x}{3}\)

The correct answer is \(50\) miles
Use the information provided in the question to draw the shape.
Use Pythagorean Theorem: \(a^2 \ + \ b^2=c^2\)
\(40^2 \ + \ 30^2=c^2 ⇒ 1600 \ + \ 900=c^2 ⇒ 2500=c^2 ⇒ c=50\)

The correct answer is \(6\) 
The four-term polynomial expression can be factored completely, by grouping, as follows: \((x^3 \ - \ 6 \ x^2 ) \ + \ (3 \ x \ - \ 18)=0\)
\(x^2 \ (x \ - \ 6) \ + \ 3 \ (x \ - \ 6)=0\) \((x \ - \ 6) \ (x^2 \ + \ 3)=0\)
By the zero-product property, set each factor of the polynomial equal to \(0\) and solve each resulting equation for \(x\).
This gives \(x=6\) or \(x=± \ i \sqrt{3}\), respectively.
Because the equation the question asks for the real value of \(x\) that satisfies the equation, The correct answer is \(6\).

The correct answer is \(p^3\)
To solve for f\((3 \ g(P))\), first, find \(3 \ g(p)\). \(g(x)=\log_3 x \ → g(p)=\log_3 \ p \ →3 \ g(p)=3 \log_3 p=\log_3 \ p^3\)
Now, find \(f(3 \ g(p))\): \(f(x)=  3^x → f \ (\log_3 \ p^3 )=\) \(3^{\log_3 \ p^3} \)
Logarithms and exponentials with the same base cancel each other.
This is true because logarithms and exponentials are inverse operations. Then: \(f(\log_3 \ p^3 )=3^{\log_3 p^3}=p^3\)

The correct answer is \(c=0.35 \ (60 \ h)\)
\($0.35\) per minute to use car. This per-minute rate can be converted to the hourly rate using the conversion \(1\) hour \(= 60\) minutes, as shown below.
\(\frac{0.35}{minute} \ × \frac{60 \ minutes}{1 \ hours}=\frac{$(0.35 \ × \ 60)}{hour}\)
Thus, the car costs \($(0.35 \ × \ 60)\) per hour.
Therefore, the cost c, in dollars, for h hours of use is \(c=(0.35 \ × \ 60) \ h\),
Which is equivalent to \(c =0.35 \ (60 \ h)\)

The correct answer is \(100\)
The best way to deal with changing averages is to use the sum.
Use the old average to figure out the total of the first \(4\) scores: Sum of first \(4\) scores: \((4) \ (90)=360\)
Use the new average to figure out the total she needs after the \(5^{th}\) score: Sum of 5 score: \((5) \ (92)=460\)
To get her sum from \(360\) to \(460\), Mary needs to score \(460 \ - \ 360=100\).

The correct answer is \(2\)
To solve a quadratic equation, put it in the \(a\ x^2 \ + \ b \ x \ + \ c=0\) form,
factor the left side, and set each factor equal to \(0\) separately to get the two solutions.
To solve \(x^2=5 \ x \ - \ 4\) , first, rewrite it as \(x^2 \ - \ 5 \ x \ + \ 4=0\).
Then factor the left side: \(x^2 \ - \ 5 \ x \ + \ 4=0\) , \((x \ - \ 4) \ (x \ - \ 1)=0\)
\(x=1\) Or \(x=4\), There are two solutions for the equation.

The correct answer is \(105\)
\(y=4 \ a \ b \ + \ 3 \ b^3\). Plug in the values of a and b in the equation: \(a=2\) and \(b=3\)
\(y=4 \ (2) \ (3) \ + \ 3 \ (3)^3=24 \ + \ 3 \ (27)=24 \ + \ 81=105\)

The correct answer is \(– \ x^2 \ – \ 3 \ x \ – 6\)
\((g \ – \ f)(x)=g(x) \ – \ f(x)=(– \ x^2 \ – \ 1 \ – \ 2 \ x) \ – \ (5 \ + \ x)
\ – \ x^2 \ – \ 1 \ – 2 \ x \ – \ 5 \ – \ x=– \ x^2 \ – \ 3 \ x \ – 6\)

The correct answer is \(\frac{100 \ x}{y}\)
Let the number be A. Then: \(x =y\% \ × \ A\). Solve for \(A. x=\frac{y}{100} \ × \ A\)
Multiply both sides by \(\frac{100}{y}\): \(x \ × \frac{100}{y}=\frac{y}{100} \ × \frac{100}{y} \ × \ A → A=\frac{100\ x}{y}\)

The correct answer is \(6\sqrt{2}\)
The line passes through the origin, \((6, \ m)\) and \((m, \ 12)\).
Any two of these points can be used to find the slope of the line.
Since the line passes through \((0, \ 0)\) and \((6, \ m)\), the slope of the line is equal to \(\frac{m \ - \ 0}{6 \ - \ 0}=\frac{m}{6}\).
Similarly, since the line passes through \((0, \ 0)\) and \((m, \ 12)\), the slope of the line is equal to \(\frac{12 \ - \ 0}{m \ - \ 0}=\frac{12}{m}\).
Since each expression gives the slope of the same line, it must be true that \(\frac{m}{6}=\frac{12}{m}\)
Using cross multiplication gives
\(\frac{m}{6}= \frac{12}{m} → m^2 = \ 72 → m=± \ \sqrt{72}=± \ \sqrt{36 \ × \ 2}=± \  \sqrt{36} \ × \sqrt{2}=± \ 6 \ \sqrt{2}\)

The correct answer is \(6\)
 It is given that \(g(5)=4\). Therefore, to find the value of \(f(g(5))\), then \(f(g(5))=f(4)=6\)

The correct answer is \((–1, \ 3)\)
Plug in each pair of number in the equation:
A. \((2, \ 1)\): \(2 \ (2) \ + \ 4 \ (1)=8\) Nope!
B. \((– \ 1, \ 3)\): \(2 \ (– \ 1) \ + \ 4 \ (3)=10\) Bingo!
C. \((– \ 2, \ 2)\): \(2 \ (– \ 2) \ + \ 4 \ (2)=4\) Nope!
D. \((2, \ 2)\): \(2 \ (2) \ + \ 4 \ (2)=12\) Nope!
E. \((2, \ 8)\): \(2 \ (2) \ + \ 4 \ (8)=36\) Nope!

The correct answer is \(29\)
Here we can substitute \(8\) for \(x\) in the equation. Thus, \(y \ - \ 3=2 \ (8 \ + \ 5)\), \(y \ - \ 3=26\)
Adding \(3\) to both side of the equation: \(y=26 \ + \ 3\),  \(y=29\)

The correct answer is I and III only
Let’s review the options: \(I. | \ a \ | \ < \ 1 → - \ 1 \ < \ a \ < \ 1\)
Multiply all sides by \(b\). Since, \(b \ > \ 0 \ → \ - \ b \ < \ b \ a \ < \ b\)
II. Since, \(- \ 1 \ < \ a \ < \ 1\),and \(a \ < \ 0 → - \ a \ > \ a^2 \ > \ a\) (plug in \(- \frac{1}{2}\), and check!)
III. \(- \ 1 \ < \ a \ < \ 1\),multiply all sides by \(2\),then:
\(- \ 2 \ < \ 2 \ a \ < \ 2\),subtract \(3\) from all sides,then:
\(- \ 2 \ - \ 3 \ < \ 2 \ a \ - \ 3 \ < \ 2 \ - \ 3 → - \ 5 \ < \ 2 \ a \ - \ 3 \ < \ - \ 1\)
I and III are correct.

The correct answer is \(a \ > \ 1\)
he equation can be rewritten as
\(c \ - \ d =a \ c →\) (divide both sides byc) \(1 \ - \frac{d}{c}=a\), since \(c \ < \ 0\) and \(d \ > \ 0\), the value of \(- \frac{d}{c}\) is positive.
Therefore, \(1\) plus a positive number is positive. a must be greater than \(1. \ a \ > \ 1\)

The correct answer is \(0\)
\(g(x)=- \ 2\), then \(f(g(x) )\) \(=f(-\  2)=2\) \((\ -\  2)^3 \ + \ 5 \ (\ - \ 2)^2 \ + \ 2 \ (\ -\  2)=- \ 16 \ + \ 20 \ - \ 4=0\)