Free Full Length ACT Mathematics Practice Test

The correct answer is \(36^\circ\) 
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 correct answer is \(0 \) 
Solving Systems of Equations by Elimination
Multiply the first equation by \(- \ 2\), then add it to the second equation.
\(\cfrac{\begin{align} - \ 2 \ (2 \ x \ + \ 3 \ y \ = \ 12) \\ 4 \ x \ - \ 4 \ y \ = - \ 16 \end{align}}{} \)
\(\cfrac{ \begin{align} - \ 4 \ x \ - \ 6 \ y \ = \ - \ 24 \\ 4 \ x \ - \ 4 \ y \ = - \ 16 \end{align} }{\begin{align} - \ 10\ y \ = - \ 40 \\ ⇒ y \ = \ 4 \end{align}} \)
Plug in the value of \(y\) into one of the equations and solve for \(x\).
\(2 \ x \ + \ 3 \ (4)= 12 ⇒\)
\(2 \ x \ + \ 12= 12 ⇒\)
\(2 \ x=0⇒ x= 0\)

The correct answer is \(\frac{2 \ \sqrt{6}}{5}\)
sin \(A=\frac{1}{5}⇒ \)
Since sin\(θ=\frac{opposite}{hypotenuse}\), we have the following right triangle. Then:
\(c=\sqrt{5^2\ -\ 1^2 }= \sqrt{25\ -\ 1}=\sqrt{24}=2 \ \sqrt{6}\)
cos \(A=\frac{2 \ \sqrt{6}}{5}\)

The correct answer is \(420\)
The ratio of boy to girls is \(3:4\).
Therefore, there are \(3\) boys out of \(4\) students.
To find the answer, first divide the total number of students by \(4\), then multiply the result by \(3\). 
\(560 \ ÷\ 4 = 140 ⇒ 140 \ ×\ 3 = 420\)

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

The correct answer is\(8\)
Plug in the value of each option in the inequality.
A. \(1\)    \((1\ -\ 2)^2\ +\ 1>3\ (1) \ -\ 1→2>2\)        No!
B. \(6\)    \((6\ -\ 2)^2\ +\ 1>3\ (6)\ -\ 1→17>17\)    No!
C. \(8\)    \((8\ -\ 2)^2\ +\ 1>3\ (8)\ -\ 1→37>23\)    Bingo!
D. \(3\)    \((3\ -\ 2)^2\ +\ 1>3\ (3)\ -\ 1→2>8\)        No!
E \(4\)     \((4\ -\ 2)^2\ +\ 1>3\ (4)\ -\ 1→5>11\)      No!

The correct answer is \(25\ π\)
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 \(5\).
\( r^2=25→r=5\)
\((x\ +\ 2)^2\ +\ (y\ -\ 4)^2=25\)
Area of a circle: \(A=π\ r^2=π\ (5)^2=25\ π\)

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 \(\frac{7\sqrt{π}}{π}\)
Formula for the area of a circle is: \(A=π\ r^2\) 
Using \(49\) for the area of the circle we have:\(49=π\ r^2\) Let’s solve for the radius \((r)\). 
\(\frac{49}{π}=r^2→r=\sqrt{\frac{49}{π}}=\frac{7}{\sqrt{π}}=\frac{7}{\sqrt{π}}\ ×\ \frac{\sqrt{π}}{\sqrt{π}}=\frac{7\sqrt{π}}{π}\)

The correct answer is \($750\)
Use simple interest formula:
\(I=prt\)
(\(I =\) interest, \(p =\) principal, \(r =\) rate, \(t =\) time)
\(I=(15000) \ (0.025) \ (2)=750\)

The correct answer is \(x\ ≥9 \ ∪ \ x≤ 1\)
\(x\ -\ 5\ ≥\ 4→x\ ≥4\ +\ 5→x≥9\)
Or
\(x\ -\ 5≤-\ 4→x≤-\ 4\ +\ 5→x≤ 1\)
Then, solution is: \(x\ ≥9 \ ∪ \ x≤ 1\)

The correct answer is \(48\) 
Length of the rectangle is: \(\frac{3}{5}\ ×\ 15=9\) 
perimeter of rectangle is: \(2\ ×\ (9\ +\ 15)=48\)

The correct answer is \(60\)
First, find the number.
Let \(x\) be the number.
Write the equation and solve for \(x\). 
\(120 \% \) of a number is \(90\), then:
\(1.2\ ×\ x=90 ⇒ x=90 \ ÷\ 1.2=75\)
\(80 \%\) of \(75\) is: \(0.7 \ ×\ 75 = 60\)

The correct answer is \(40\%\) 
The population is increased by \(12\%\) and \(25\%\).
\(12\%\) increase changes the population to \(112\%\) of original population.
For the second increase, multiply the result by \(125\%\).
\((1.12) \ ×\ (1.25) = 1.40= 140\%\)
\(40\) percent of the population is increased after two years.

The correct answer is \(40\%\) 
The population is increased by \(12\%\) and \(25\%\).
\(12\%\) increase changes the population to \(112\%\) of original population.
For the second increase, multiply the result by \(125\%\).
\((1.12) \ ×\ (1.25) = 1.40= 140\%\)
\(40\) percent of the population is increased after two years.

The correct answer is \(\frac{4}{19}\)
The probability of choosing a Hearts is \(\frac{12}{57} = \frac{4}{19}\)

The correct answer is \(9\ x^3\ y^5\ -\ 5 \ x^2\ y^3\)
\(4\ x^2\ \ y^3\ + \ 5\ x^3\ y^5\ –\ 9\ x^2\ y^3\ +\ 4\ x^3\ y^5\ =\)
\(4\ x^2\ y^3\ -\ 9\ x^2\ y^3\ +\ 5\ x^3\ y^5\ + \ 4 \ x^3 \ y^5\ =\)
\(9\ x^3\ y^5\ -\ 5 \ x^2\ y^3\)

The correct answer is \(142\) degree 
The angle \(x\) and \(35\) are complementary angles.
Therefore: \(x\ +\ 38=180\) 
\(180^\circ\ - \ 38^\circ=142^\circ\)

The correct answer is \(\frac{3}{5}\)
tan ⁡\(=\frac{opposite}{djacent}\), and tan⁡ \(x=\frac{6}{8}\), therefore, the opposite side of the angle \(x\) is \(6\) and the adjacent side is \(8\).
Let’s draw the triangle.
Using Pythagorean theorem, we have:
\(a^2\ +\ b^2=c^2→6^2\ +\ 8^2=c^2→36\ +\ 64=c^2→c=10\)
sin ⁡\(=\frac{opposite}{hypotnuse}=\frac{6}{10} = \frac{3}{5}\)

The correct answer is \(36\)
The sum of supplement angles is \(180\). Let \(x\) be that angle.
Therefore, \( x \ +\ 4\ x\ = 180\)
\(5\ x\ = 180\), divide both sides by \(5\):
\(x = 36\)

The correct answer is \(4\)
sin\(^2 \ a\ +\) cos\(^2 \ a=1\), then: \(x\ +\ 1=5, x=4\)

The correct answer is \(24\%\)
The question is this: \(474.24 \) is what percent of \(624\)?
Use percent formula: part \(= \frac{percent}{100} \ ×\) whole
\(474.24 = \frac{percent}{100} \ ×\ 624 ⇒ 474.24= \frac{percent \ ×\ 624}{100} ⇒47424 =\) percent \(×\ 624 ⇒\) 
percent \(= \frac{47424}{624} = 83.39\)
\(47424\) is \(76 \%\) of \(624\). Therefore, the discount is: \(100\% \ –\ 76\% = 24\%\)

The correct answer is \(\frac{\sqrt{5} }{2 \ \sqrt{3}} \ x\)
Simplify the expression.
\(\sqrt{\frac{x^2}{3}\ +\ \frac{x^2}{12}}=\sqrt{\frac{4 \ x^2}{12}\ +\ \frac{x^2}{12}}=\sqrt{\frac{5x^2}{12}}=\sqrt{\frac{5}{12} \ x^2 }=
\sqrt{\frac{5}{12}} \ ×\ \sqrt{x^2} = \frac{\sqrt{5}}{\sqrt{12}}\ ×\ \sqrt{x^2}= \frac{\sqrt{5}}{2 \ \sqrt{3}}\ ×\ \sqrt{x^2}=\frac{\sqrt{5} }{2 \ \sqrt{3}} \ x\)

The correct answer is \( 45,000\) 
Three times of \(18,000\) is \(54,000\). One sixth of them cancelled their tickets.
One sixth of \(54,000\) equals \(9,000\ (\frac{1}{6} \ ×\ 54000 = 9000)\). 
\(45,000 \ (54000 \ –\ 9000 = 45000)\) fans are attending this week

The correct answer is \(27.5\)
Let \(x\) be the smallest number.
Then, these are the numbers:
\(x, x\ + \ 1, x \ + \ 2, x \ + \ 3, x \ + \ 4, x \ + \ 5\)
average \(= \frac{sum \ of \ terms}{number \ of \ terms} ⇒\)
\(30 = \frac{x \ + \ (x \ + \ 1) \ + \ (x \ + \ 2) \ + \ (x \ + \ 3) \ + \ (x \ + \ 4)\ + \ (x \ + \ 5)}{6}⇒\)
\(30=\frac{6 \ x \ + \ 15}{6} ⇒\) 
\(180 = 6 \ x \ + \ 15 ⇒\)
\(165 = 6 \ x ⇒ x=27.5\)

The correct answer is \(− \ \frac{1}{2}\)
The equation of a line in slope intercept form is: \(y=m \ x \ + \ b\)
Solve for \(y\).
\(4 \ x \ - \ 2 \ y=16 ⇒\)
\(- \ 2 \ y=16 \ - \ 4 \ x ⇒\)
\(y=(16 \ - \ 4 \ x) \ ÷ \ (- \ 2) ⇒\)
\(y=2 \ x \ - \ 8\)
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 \(30\)
Plug in the value of \(x\) and \(y\). 
\(x=2\) and \(y=-\ 2\)
\(5\ (x\ -\ 2\ y)\ +\ (2 \ - \ x)^2=5\ (2\ -\ 2(-\ 2))\ +\ (2\ -\ 2)^2=5\ (2\ +\ 4)\ +\ (0)^2 = 30\ +\ 0=30\)

The correct answer is \(3.2\ ×\ 10^5\)
\(320000=3.2\ ×\ 10^5\)

The correct answer is \(\frac{y}{8}\)
Solve for \(x\).
\(\sqrt{8 \ x}=\sqrt{y}\)
Square both sides of the equation:
\((\sqrt{8 \ x})^2=(\sqrt{y})^2\)  ,  \(8\ x=y \)  ,   \( x=\frac{y}{8}\)

The correct answer is Length of AD equal to length BC
In any rectangle opposite sides are equal.

The correct answer is \(2\) cm
Formula for the Surface area of a cylinder is:
S A \(=2 \ π \ r^2 \ + \ 2 \ π \ r \ h→\)
\(48 \ π=2 \ π \ r^2 \ + \ 2 \ π \ r \ (10)→\)
\(r^2 \ + \ 10 \ r \ - \ 24=0 \)
Factorize and solve for \(r\).
\((r \ + \ 12)(r \ - \ 2)=0→r=2\) or \(r= \ - \ 12\) (unacceptable)

The correct answer is \(8, 4\)
\((x\ +\ 2)\ (x\ +\ p)=x^2\ +\ (2\ +\ p)x\ +\ 2\ p→2\ +\ p=6→p=4\) and
\(r=2\ p=8\)

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 \(y-\)axis, the value of \(y\) does not change and when a point is reflected over the \(x-\)axis, 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 \(170\) 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 \(\frac{1}{10}\)
Write the ratio of \(2 \ a\) to \(3 \ b\).
\(\frac{2 \ a}{3 \ b}=\frac{1}{15}\)
Use cross multiplication and then simplify.
\(2 \ a \ × \ 15=3 \ b \ ×\ 1→30 \ a=3 \ b→a=\frac{3 \ b}{30}=\frac{b}{10}\)
Now, find the ratio of \(a\) to \(b\).
\(\frac{a}{b}=\frac{\frac{b}{10}}{b}→\frac{b}{10} \ ÷\ b=\frac{b}{10} \ × \ \frac{1}{b}=\frac{b}{10 \ b}=\frac{1}{10}\)

The correct answer is \($4.5\)
Let \(x\) be the cost of one-kilogram orange, then: \(4\ x\ + \ (2 \ × \ 4.2)=25.4→\)
\(4\ x\ +\ 8.4=25.4→4\ x=25.4\ -\ 8.4→4\ x=18→x=\frac{18}{4}=$4.5\)

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

The correct answer is \(\begin{bmatrix}-\ 4&5\\ 4&-\ 8\end{bmatrix}\)
First, find \(2\) A.
A\(=\begin{bmatrix}-\ 1 & 3\\ 1 & -\ 3\end{bmatrix}\)
\(2\)A\(=2\ ×\ \begin{bmatrix}-\ 1&3\\ 1 & -\ 3\end{bmatrix}=\begin{bmatrix} -\ 2& 6\\ 2 & -\ 6\end {bmatrix}\)
Now, solve for \(2\) A \(-\) B.
\(2\) A\(-\)B\(=\begin{bmatrix} -\ 2&6 \\ 2& -\ 6\end {bmatrix}\ -\ \begin{bmatrix} 2&1\\ -\ 2 & 2\end{bmatrix}=\begin{bmatrix}-\ 2 \ -\ 2 & 6 \ -\ 1\\ 2\ -\ (-\ 2) & -\ 6\ -\ 2\end{bmatrix}=\begin{bmatrix}-\ 4&5\\ 4&-\ 8\end{bmatrix}\)

The correct answer is \(x ^ {\frac{15}{7}}\)
\((x^5)^{\frac{3}{7}} = x^{5 \ ×\ \frac{3}{7}} =x ^ {\frac{15}{7}}\)

The corrcet answer is \(3,375\)
If the length of the box is \(45\), then the width of the box is one third of it, \(15\), and the height of the box is \(5\) (one third of the width).
The volume of the box is:
\(V = lwh = (45)\ (15)\ (5) = 3,375\)

The correct answer is \(2\)
The amplitude in the graph of the equation \(y=a\) cos \(b\ x\) is \(a\). (\(a\) and \(b\) are constant)
In the equation \(y\ -\ 1=2\) cos \(3\ x\), the amplitude is \(2\).

The correct answer is \(\frac{\sqrt{3}}{2}\)
The relationship among all sides of right triangle \(30^\circ \ - \ 60^\circ \ - \ 90^\circ \) is provided in the following triangle: 
cos of \(30^\circ \) equals to:\(\frac{adjacent}{hypotenuse}=\frac{x \ \sqrt{3}}{2\ x}=\frac{\sqrt{3}}{2}\)

The correct answer is \(43\)
The area of rectangle is: \(8 \ × \ 4=32\) cm\(^2\)
The area of circle is: \(π \ r^2=π \ × \ ( \frac{10}{2})^2=3 \ × \ 25=75\) cm\(^2\)
Difference of areas is: \(75 \ - \ 32=43\)

The correct answer is \(62.84 \) kg
average\( = \frac{sum \ of\ terms }{number \ of\ terms}\)
The sum of the weight of all girls is: \(18 \ ×\ 60 = 900\) kg
The sum of the weight of all boys is: \(37 \ ×\ 64 = 2368\) kg
The sum of the weight of all students is: \(900 \ +\ 2368 = 3268\) kg
average \(= \frac{3268}{52} = 62.84\)

The correct answer is January and February
First find the number of pants sold in each month.
A. January: \(110\), February: \(88\), March: \(90\), April: \(70\), May: \(85\), June: \(65\)
Check each option provided.
January and February, 
\((\frac{110 \ - \ 88}{110}) \ × \ 100=\frac{22}{110} \ × \ 100=20\%\)
B. February and March, there is an increase from February to March.
C. March and April
\((\frac{90 \ - \ 70}{90}) \ × \ 100=\frac{20}{90} \ × \ 100=22.22\%\)
D. April and May: there is an increase from April to May
May and June
\((\frac{85 \ - \ 65}{85}) \ × \ 100=\frac{20}{85} \ × \ 100=23.53\%\)

The correct answer is \(147.5, 30\)
Let’s order number of shirts sold per month:
\(130,140,145,150,160,170\)
median is: \(\frac{145 \ + \ 150}{2}=147.5\)
Let’s list the number of shoes sold per month:
\(20,25,25,35,35,40\)
mean is: \(\frac{20 \ + \ 25 \ + \ 25 \ + \ 35 \ + \ 35 \ + \ 40}{6}=\frac{180}{6}=30\)

The correct answer is \(50\)
Let \(x\) be the number of shoes need to be added in April. Then:
\(\frac{70}{20 \ + \ x}=(\frac{5}{17}) \ (\frac{85}{25}) →\frac{70}{20 \ + \ x}=\frac{425}{425}=1→\)
\(70=20 \ + \ x→x=50\)

The correct answer is \(5\)
\(3125=5^5→5^{ \ x}=5^5→x=5\)

The correct answer is \(18\)
Plug in the value of \(x\) in the equation and solve for \(y\).
\(3 \ y=\frac{2 \ x^3}{5} \ + \ 4→\)
\(3 \ y =\frac{2(5)^3}{5} \ + \ 4→\)
\(3 \ y=\frac{2 \ (125)}{5} \ + \ 4→\)
\(3 \ y= 50 \ + \ 4=54\)
\(3 \ y = 54→y=18\)

The correct answer is \(\frac{15 \ \sqrt{3}}{2}\)
Based on triangle similarity theorem:
\(\frac{a}{a\ +\ b}=\frac{c}{4}→c=\frac{4\ a}{a\ +\ b}=\frac{4\sqrt3}{4\sqrt3}=1→\) 
area of shaded region is:
\((\frac{c\ +\ 4}{2})\ (b)=\frac{5}{2}\ ×\ 3\sqrt{3}=\frac{15 \ \sqrt{3}}{2}\)

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 \(110 \ x \ + \ 16,000 \ ≤ \ 23,000\)
Let \(x\) be the number of shoes the team can purchase.
Therefore, the team can purchase \(110 \ x\).
The team had \($23,000\) and spent \($16000\).
Now the team can spend on new shoes \($7000\) at most.
Now, write the inequality: \(110 \ x \ + \ 16,000 \ ≤ \ 23,000\)

The correct answer is \(\frac{3}{x^2} \ + \ 5\)
\(f(g(x))=3 \ × \ (- \ \frac{1}{x})^2 \ + \ 5=\frac{3}{x^2} \ + \ 5\)

The correct answer is \(64\) m\(^2\)
Let \(L\) be the length of the rectangular and \(W\) be the with of the rectangular. Then,
\(L=3 \ W \ + \ 4\)
The perimeter of the rectangle is \(40\) meters.
Therefore: \(2 \ L \ + \ 2 \ W=40\)
\(L \ + \ W=20\)
Replace the value of \(L\) from the first equation into the second equation and solve for \(W\):
\((3 \ W \ + \ 4) \ + \ W=20→4 \ W \ + \ 4=20→4 \ W=16→W=4\)
The width of the rectangle is \(4\) meters and its length is:
\(L=3 \ W \ + \ 4=3 \ (4) \ + \ 4=16\)
The area of the rectangle is:
length \(×\) width \(= 4 \ × \ 16 = 64\)

The correct answer is \(120\)
Here is the list of all prime numbers between \(20\) and \(40\):
\(23, 29, 31, 37\)
The sum of all prime numbers between \(20\) and \(40\) is:
\(23 \ + \ 29 \ + \ 31 \ + \ 37 = 120\)

The correct answer is \(i\ +\frac{ 3}{4}\)
To simplify the fraction, multiply both numerator and denominator by \(i\). 
\(\frac{4\ -\ 3\ i}{-\ 4\ i}\ ×\ \frac{i}{i}=\frac{4\ i\ -\ 3\ i^2} { -\ 4\ i^2} \)
\(i^2\ -\ 1\), 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 \(0, -\ 2 , -\  3\)
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 \(24\)
Let \(x\) be the length of AB, then:
\(12=\frac{x×3}{2}→x=8\)
The length of AC \(=\sqrt{6^2\ +\ 8^2 }=\sqrt{100}=10\)
The perimeter of ∆ABC\(=6\ +\ 8\ +\ 10=24\)

The correct answer is \(12.5\) feet
Use formula of rectangle prism volume.
\(V =\) (length) (width) (height) \(⇒ 5000 = (20) \ (20) \) (height) \(⇒\) height \(= 5000 \ ÷ \ 400 = 12.5\)

The correct answer is \( 25 \ x^8\)
\(y=(-\ 5 \ x^4)^2=(-\ 5^2) \  (x^4)^2= 25 \ x^8\)