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: $y(x)=8\ x$