Full Length PSAT Math Practice Test

The correct answer is \(0.05 \ x \ + \ 4,500\)
\(x\) is the number of all John’s sales per month and \(5\%\) of it is:
\(5\% \ × \ x=0.05 \ x\)
John’s monthly revenue: \(0.05 \ x \ + \ 4,500\)

The correct answer is \(37\)
\(x^2=121→x=11\) (positive value) Or \(x=  - \ 11\) (negative value)
Since \(x\) is positive, then:
\(f(121)=f(11^2 )=3 \ (11) \ + \ 4=33 \ + \ 4=37\)

The correct answer is \(- \ 5\)
\(2 \ x^2 \ - \ 11 \ x \ + \ 8= \ - \ 3 \ x \ + \ 18→\)
\(2 \ x^2 \ - \ 11 \ x \ + \ 3 \ x \ + \ 8 \ - \ 18=0→\)
\(2 \ x^2 \ - \ 8 \ x \ - \ 10=0→\)
\(2 \ (x^2 \ - \ 4 \ x \ - \ 5)=0→\) Divide both sides by \(2\).
Then:
\(x^2 \ - \ 4 \ x \ - \ 5=0\), Find the factors of the quadratic equation.
\(→(x \ - \ 5) \ (x \ + \ 1)=0→x=5\) or \(x= \ - \ 1\)
\(a \ > \ b\), then: \(a=5\) and \(b= \ - \ 1\)
\(\frac{a}{b}=\frac{5}{- \ 1}= \ - \ 5\)

The correct answer is \((9,3)\)
First, find the equation of the line.
All lines through the origin are of the form \(y=m \ x\), 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 \(x \ < \ 3\)
\(2 \ x \ + \ 4 \ > \ 11 \ x \ - \ 12.5 \ - \ 3.5 \ x→\) Combine like terms:
\(2 \ x \ + \ 4 \ > \ 7.5 \ x \ - \ 12.5→\) Subtract \(2 \ x\) from both sides: \(4>5.5x-12.5\)
Add \(12.5\) both sides of the inequality.
\(16.5 \ > \ 5.5 \ x\), Divide both sides by \(5.5\).
\(\frac{16.5}{5.5} \ > \ x→x \ < \ 3\)

The correct answer is \(11.5\)
\(3 \ a=4 \ b→b=\frac{3 \ a}{4}\) and \(3 \ a=5 \ c→c=\frac{3 \ a}{5}\)
\(a \ + \ 2 \ b \ + \ 15 \ c=a \ + \ (2 \ × \ \frac{3 \ a}{4}) \ + \ (15 \ × \ \frac{3 \ a}{5})=a \ + \ 1.5 \ a \ + \ 9 \ a=11.5 \ a\)
The value of \(a \ + \ 2 \ b \ + \ 15 \ c\) is \(11.5\) times the value of \(a\).

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 \(\frac{11}{30}\)
Of the \(30\) employees, there are \(5\) females under age \(45\) and \(6\) males age \(45\) or older.
Therefore, the probability that the person selected will be either a female under age \(45\) or a male age \(45\) or older is: \(\frac{5}{30} \ + \ \frac{6}{30}=\frac{11}{30}\)

The correct answer is \(4\)
Plug in the values of x and y of the point \((2, 12)\) in the equation of the parabola. Then:
\(12=a \ (2)^2 \ + \ 5 \ (2) \ + \ 10→\)
\(12=4 \ a \ + \ 10 \ + \ 10→\)
\(12=4 \ a \ + \ 20\)
\(→4 \ a=12 \ - \ 20=- \ 8→\)
\(a=\frac{- \ 8}{4}=- \ 2→\)
\(a^2=(- \ 2)^2=4\)

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:
\(y=m \ x \ + \ b\)
Where, \(m=\) slope and \(y=y-\)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 tall when 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 \(x=22, y=48\)
\(\begin{cases}\frac{ - \ x}{2} \ + \ \frac{y}{4} = 1 \\ \frac{ - \ 5 \ y}{6} \ + \ 2 \ x =4 \end{cases} \rightarrow\) Multiply the top equation by \(4\). Then,
\(\begin{cases}- \ 2 \ x \ + \ y =4\\\frac{- \ 5 \ y}{6} \ + \ 2 \ x =4\end{cases} \rightarrow\) Add two equations.
\(\frac{1}{6} \ y=8→y=48\), plug in the value of \(y\) into the first equation \(→x=22\)

The correct answer is \(4.8\)
Two triangles \(\triangle\)BAE and \(\triangle\)BCD are similar. Then:
\(\frac{AE}{CD}=\frac{AB}{BC}→\frac{4}{6}=\frac{x}{12}→\)
\(48 \ - \ 4 \ x=6 \ x→\)
\(10 \ x=48→x=4.8\)

The correct answer is \(\frac{25}{16}\)
First, simplify the numerator and the denominator.
\(\frac{(10 \ (x) \ (y^2)^2}{(8 \ x \ y^2 )^2} =\frac{100 \ x^2 \ y^4}{64 \ x^2 \ y^4 )}\)
Remove \(x^2 \ y^4\) from both numerator and denominator.
\(\frac{100 \ x^2 \ y^4}{64 \ x^2 \ y^4 )}=\frac{100}{64}=\frac{25}{16}\)

The correct answer is \(50\)
\(\frac{y}{5}=x \ - \ \frac{2}{5} \ x \ + \ 10\), Multiply both sides of the equation by \(5\). Then:
\(5 \ × \ \frac{y}{5}=5 \ × \ (x \ - \ \frac{2}{5} \ x \ + \ 10)→\)
\(y=5 \ x \ - \ 2 \ x \ + \ 50→y=3 \ x \ + \ 50\)
Now, subtract \(3 \ x\) from both sides of the equation. Then:
\(y \ - \ 3 \ x=50\)

The correct answer is \(\frac{10}{3}\)
First, factorize the numerator and simplify.
\(\frac{x^2 \ - \ 9}{x \ + \ 3} \ + \ 2 \ (x \ + \ 4)=15→\)
\(\frac{(x \ - \ 3) \ (x \ + \ 3)}{(x \ + \ 3)} \ + \ 2 \ x \ + \ 8=15\)
Divide both sides of the fraction by \((x \ + \ 3)\). Then:
\(x \ - \ 3 \ + \ 2 \ x \ + \ 8=15→3 \ x \ + \ 5=15\)
Subtract \(5\) from both sides of the equation. Then:
\(→3 \ x=15 \ - \ 5=10→x=\frac{10}{3}\)

The correct answer is \(\frac{10}{3}\)
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 \(\frac{10}{3}\)
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 \(84\)
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 \(10\)
\(\frac{2}{5} \ × \ 25=\frac{50}{5}=10\)

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 D \((y=x)\) has slope of \(1\).

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 \(35\)
Choices A, C and D are incorrect because \(80\%\) of each of the numbers is a non-whole number.
A. \(49\),      \(80\%\) of \(49 = 0.80 \ × \ 49=39.2 \)
B. \(35\),      \(80\%\) of \(35=0.80 \ × \ 35=28\)
C. \(12\),      \(80\%\) of \(12=0.80 \ × \ 12=9.6\)
D. \(32\),      \(80\%\) of \(32=0.80 \ × \ 32=25.6\)
Only choice B gives a whole number.

The correct answer is \(25\)
The capacity of a red box is \(20\%\) bigger than the capacity of a blue box and it can hold \(30\) books.
Therefore, we want to find a number that \(20\%\) bigger than that number is \(30\).
Let \(x\) be that number. Then:
\(1.20 \ × \ x=30\), Divide both sides of the equation by \(1.2\). Then:
\(x=\frac{30}{1.20}=25\)

The correct answer is \(− \ 5\)
The smallest number is \(- \ 15\).
To find the largest possible value of one of the other five integers, we need to choose the smallest possible integers for four of them.
Let \(x\) be the largest number. Then:
\(- \ 70=(- \ 15) \ + \ (- \ 14) \ + \ (- \ 13) \ + \ (- \ 12) \ +(- \ 11) \ + \ x→- \ 70=- \ 65 \ + \ x\)
\(→x=- \ 70 \ + \ 65=- \ 5\)

The correct answer is \(x \ < \ \sqrt{x^2 \ + \ 1} \ < \ \sqrt{x^2} \ + \ 1\)
Let \(x\) be equal to \(0.5\), then: \(x=0.5 \)
\(\sqrt{x^2 \ + \ 1}=\sqrt{0.5^2 \ + \ 1}=\sqrt{1.25}≈1.12\)
\(\sqrt{x^2 } \ + \ 1=\sqrt{0.5^2} \ + \ 1=0.5 \ + \ 1=1.5\)
Then, option A is correct.
\(x \ < \ \sqrt{x^2 \ + \ 1} \ < \ \sqrt{x^2} \ + \ 1\)

The correct answer is \(𝑎=𝑐\)
Let’s find the mean (average), mode and median of the number of cities for each type of pollution.
Number of cities for each type of pollution: \(6, 3, 4, 9, 8\)
average (mean) \(= \frac{sum \ of \ terms}{number \ of \ terms}= \frac{6 \ + \ 3 \ + \ 4 \ + \ 9 \ + \ 8}{5}=\frac{30}{5}=6 \)
Median is the number in the middle.
To find median, first list numbers in order from smallest to largest.
\(3, 4, 6, 8, 9\)
Median of the data is \(6\).
Mode is the number which appears most often in a set of numbers.
Therefore, there is no mode in the set of numbers.
Median \(=\) Mean, then, \(𝑎=𝑐 \)

The correct answer is \(60\%, 40\%, 90\%\)
Percent of cities in the type of pollution A: \(\frac{6}{10} \ × \ 100=60\%\)
Percent of cities in the type of pollution C: \(\frac{4}{10} \ × \ 100=40\%\)
Percent of cities in the type of pollution E: \(\frac{9}{10} \ × \ 100=90\%\)

The correct answer is \(2\)
Let \(x\) be the number of cities need to be added to type of pollutions B. Then:
\(\frac{x \ + \ 3}{8}=0.625→x \ + \ 3=8 \ × \ 0.625→x \ + \ 3=5→x=2\)

The correct answer is \(12\)
The ratio of boy to girls is \(4:7\).
Therefore, there are \(4\) boys out of \(11\) students.
To find the answer, first divide the total number of students by \(11\), then multiply the result by \(4\).
\(44 \ ÷ \ 11 = 4 ⇒ 4 \ × \ 4 = 16\)
There are \(16\) boys and \(28 \ (44 \ – \ 16)\) girls.
So, \(12\) more boys should be enrolled to make the ratio \(1:1\)

The correct answer is \(\frac{1}{2}\)
AB \(=5\) And BC \(=12\)
AC \(=\sqrt{12^2 \ + \ 5^2 }=\sqrt{144 \ + \ 25}=\sqrt{169}=13\)
Perimeter \(=5 \ + \ 12 \ + \ 13=30 \)
Area \(=\frac{5 \ × \ 12}{2}=5 \ × \ 6=30\)
In this case, the ratio of the perimeter of the triangle to its area is: \(\frac{30}{30}=1\)
If the sides AB and BC become twice longer, then:
AB \(=24\) And AC \(=10\)
BC \(=\sqrt{24^2 \ + \ 10^2}=\sqrt{576 \ + \ 100}=\sqrt{676}=26\)
Perimeter \(=26 \ + \ 24 \ + \ 10=60 \)
Area \(=\frac{10 \ × \ 24}{2}=10 \ × \ 12=120\)
In this case the ratio of the perimeter of the triangle to its area is: \(\frac{60}{120}=\frac{1}{2}\)

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 \(0.97\)
Ratio of women to men in city A: \(\frac{570}{600}=0.95\)
Ratio of women to men in city B: \(\frac{291}{300}=0.97\)
Ratio of women to men in city C: \(\frac{665}{700}=0.95\)
Ratio of women to men in city D: \(\frac{528}{550}=0.96\)
Choice B provides the maximum ratio of women to men in the four cities.

The correct answer is \(1.05\)
Percentage of men in city A \(= \frac{600}{1170} \ × \ 100=51.28\%\)
Percentage of women in city C \(= \frac{665}{1365} \ × \ 100=48.72\%\)
Percentage of men in city A to percentage of women in city C \(= \frac{51.28}{48.72}=1.05 \)

The correct answer is \(132\)
Let the number of women should be added to city D be \(x\), then:
\(\frac{528 \ + \ x}{550}=1.2→\)
\(528 \ + \ x=550 \ × \ 1.2=660→\)
\(x=132\)

The correct answer is \(5, 10\)
The perimeter of the rectangle is: \(2 \ x \ + \ 2 \ y=30→\)
\(x \ + \ y=15→\)
\(x=15 \ - \ y\)
The area of the rectangle is: \(x \ × \ y=50→\)
\((15 \ - \ y) \ (y)=50→\)
\(y^2 \ - \ 15 \ y \ + \ 50=0\)
Solve the quadratic equation by factoring method.
\((y \ - \ 5) \ (y \ - \ 10)=0→y=5\) (Unacceptable, because \(y\) must be greater than \(5\)) or \(y=10\)
If \(y=10 →x \ × \ y=50→x \ × \ 10=50→x=5\)

The correct answer is \(80 \ - \ 6 \ x \)
The amount of petrol consumed after \(x\) hours is: \(6 \ × \ x=6 \ x\)
Petrol remaining after \(x\) hours driving: \(80 \ - \ 6 \ x \)

The correct answer is \(86\)
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 \(14.5\)
average (mean) \(= \frac{sum \ of \ terms}{number \ of \ terms}=\)
\(\frac{9 \ + \ 12 \ + \ 15 \ + \ 16 \ + \ 19 \ + \ 16 \ + \ 14.5}{7}=14.5\)

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 \(x=− \ 2, y=3\)
\(\begin{cases}x \ + \ 4 \ y =10\\ 5 \ x \ + \ 10 \ y =20\end{cases}\rightarrow\) Multiply the top equation by \(- \ 5\) then,
\(\begin{cases}- \ 5 \ x \ - \ 20 \ y =- \ 50\\ 5 \ x \ + \ 10 \ y =20\end{cases}\rightarrow\) Add two equations
\(- \ 10 \ y=- \ 30→y=3\), plug in the value of \(y\) into the first equation
\(x \ + \ 4 \ y=10→x \ + \ 4 \ (3)=10→x \ + \ 12=10\)
Subtract \(12\) from both sides of the equation. Then:
\(x \ + \ 12=10→x=- \ 2\)

The correct answer is \(\frac{𝑎}{\sqrt{𝑎^2 \ + \ 𝑏^2}}\)
Cos ⁡\(β=\frac{Adjacent \ side}{hypotenuse}\)
To find the hypotenuse, we need to use Pythagorean theorem.
\(a^2 \ + \ b^2=c^2→c=\sqrt{a^2 \ + \ b^2 }\)
cos \((β)=\frac{a}{c}=\frac{a}{\sqrt{a^2 \ + \ b^2}}\)

The correct answer is \(\frac{10}{3} \ < \ x \ < \ 10 \)
\(|\frac{x}{2} \ - \ 2 \ x \ + \ 10| \ < \ 5→|- \ \frac{3}{2} \ x \ + \ 10| \ < \ 5→- \ 5< \ - \ \frac{3}{2} \ x \ + \ 10 \ < \ 5\)
Subtract \(10\) from all sides of the inequality.
\(→- \ 5 \ - \ 10 \ < \ - \ \frac{3}{2} \ x \ + \ 10 \ - \ 10 \ < \ 5 \ - \ 10→- \ 15 \ < \ - \ \frac{3}{2} \ x \ < \ - \ 5\)
Multiply all sides by \(2\).
\(→2 \ × \ (- \ 15) \ < \ 2 \ × \ (- \ \frac{3 \ x}{2}) \ < \ 2 \ × \ (- \ 5)→- \ 30 \ < \ - \ 3 \ x \ < \ - \ 10\)
Divide all sides by \(- \ 3\).
(Remember that when you divide all sides of an inequality by a negative number, the inequality sing will be swapped. \(<\) becomes \(>\))
\(→\frac{- \ 30}{- \ 3} \ > \ \frac{- \ 3 \ x}{- \ 3} \ > \ \frac{- \ 10}{- \ 3}\)
\(→10 \ > \ x \ > \ \frac{10}{3}→\frac{10}{3} \ < \ x \ < \ 10\)

The correct answer is \(5\)
\(x\) is directly proportional to the square of \(y\). Then:
\(x=c \ y^2\)
\(12=c \ (2)^2→12=4 \ c→c=\frac{12}{4}=3\)
The relationship between \(x\) and \(y\) is:
\(x=3 \ y^2\)
\(x=75\)
\(75=3 \ y^2→y^2=\frac{75}{3}=25→y=5\)

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 \(1\)
The intersection of two functions is the point with \(2\) for \(x\).
Then:
\(f(2)=g(2)\) and \(g(2)=(2 \ × \ (2)) \ - \ 3=4 \ - \ 3=1\)
Then, \(f(2)=1→a \ (2)^2 \ + \ b \ (2) \ + \ c=1→4 \ a \ + \ 2 \ b \ + \ c=1\) (i)
The value of \(x\) in the vertex of the parabola is: \(x=- \ \frac{b}{2 \ a}→- \ 2=- \ \frac{b}{2 \ a}→b=4 \ a\) (ii)
In the point \((- \ 2, 5)\), the value of the \(f(x)\) is \(5\).
\(f(- \ 2)=5→a \ (- \ 2)^2 \ + \ b \ (- \ 2) \ + \ c=5→4 \ a \ - \ 2 \ b \ + \ c=5\) (iii)
Using the first two equations:
\(\begin{cases}4 \ a \ + \ 2 \ b \ + \ c=1\\4 \ a \ - \ 2 \ b \ + \ c=5\end{cases}\)
Equation \(1\) minus equation \(2\) is:
(i) \(-\) (iii) \(→4 \ b=- \ 4→b=- \ 1\) (iv)
Plug in the value of b in the second equation:
\(b=4 \ a →a=\frac{b}{4}=- \ \frac{1}{4}\)
Plug in the values of a and be in the first equation. Then:
\(→4(- \ \frac{1}{4}) \ + \ 2 \ (- \ 1) \ + \ c=1→- \ 1 \ - \ 2 \ + \ c=1→c=1 \ + \ 3→c=4\)
The product of \(a, \ b\) and \(c=(- \ \frac{1}{4}) \ × \ (- \ 1) \ × \ 4=1\)

The correct answer is \(90\)
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 latter is the hypotenuse.
Therefore, the latter is \(90\) feet.

The correct answer is \(2\)
Let \(x\) be the length of an edge of cube, then the volume of a cube is: \(V=x^3\)
The surface area of cube is: \(SA=6 \ x^2\)
The volume of cube \(A\) is \(\frac{1}{3}\) of its surface area. Then:
\(x^3=\frac{6 \ x^2}{3}→x^3=2 \ x^2\), divide both side of the equation by \(x^2\). Then:
\(\frac{x^3}{x^2} =\frac{3 \ x^2}{x^2} →x=2\)

The correct answer is \(5\)
\(3 \ x \ + \ 6 \ y=\frac{- \ 3 \ y^2 \ + \ 15}{x}\), Multiply both sides by \(x\).
\(x \ × \ (3 \ x \ + \ 6 \ y)=x \ × \ (\frac{- \ 3 \ y^2 \ + \ 15}{x})→3 \ x^2 \ + \ 6 \ x \ y=- \ 3 \ y^2 \ + \ 15\)
\(→3 \ x^2 \ + \ 6 \ x \ y \ + \ 3 \ y^2=15→3 \ × \ (x^2 \ + \ 2 \ x \ y \ + \ y^2 )=15→x^2 \ + \ 2 \ x \ y \ + \ y^2=\frac{15}{3}\)
\(x^2 \ + \ 2 \ x \ y \ + \ y^2=(x \ + \ y)^2\), Then:
\(→(x \ + \ y)^2=5\)