1 Choice A is correct
The correct answer is \(10.5\) \(5 \ x \  \ 8=4.5→5 \ x=4.5 \ + \ 8=12.5→x=\frac{12.5}{3}=2.5 \) Then; \(3 \ x \ + \ 3=3 \ (2.5) \ + \ 3=7.5 \ + \ 3=10.5\)

2 Choice B is correct
The correct answer is \(9 \ π‘^4 \ + \ 6 \ π‘^2 \ − \ 5\) \(f(x)=x^2 \ + \ 2 \ x \  \ 5\) \(f(3 \ t^2 )=(3 \ t^2 )^2 \ + \ 2 \ (3 \ t^2 ) \  \ 5=9 \ t^4 \ + \ 6 \ t^2 \  \ 5\)

3 Choice D is correct
The correct answer is \(9\) \(x \ p \ + \ 2 \ y \ q=26→x \ p=26 \  \ 2 \ y \ q\) \((1)\) \(x \ p \ + \ y \ q=17\) \((2)\) \((1)\) in \((2) \ \ →26 \  \ 2 \ y \ q \ + \ y \ q=17→26 \  \ y \ q=17→y \ q=26 \  \ 17=9\)

4 Choice D is correct
The correct answer is \($810\) Let \(x\) be all expenses, then \(\frac{22}{100} \ x=$660 →x=\frac{100 \ × \ $660}{22}=$3,000\) He spent for his rent: \(\frac{27}{100} \ × \ $3,000=$810\)

5 Choice A is correct
The correct answer is \(− \ 108\) \(12 \ x^2 \ + \ n=a \ (x^2 \ + \ 3) \ (x^2 \  \ 3)=a \ x^4 \  \ 9 \ a→a=12\) And \(n= \ 9 \ a= \ 9 \ × \ 12= \ 108\)

6 Choice C is correct
The correct answer is \(25\) \(\frac{1}{8}=0.125→C=5\) \(\frac{1}{20}=0.05→D=5→C \ × \ D=5 \ ×\ 5=25\)

7 Choice C is correct
The correct answer is \( \ 5\) \(y\) is the intersection of the three circles? Therefore, it must be odd (from circle A), negative (from circle B), and multiple of \(5\) (from circle C). From the options, only \( \ 5\) is odd, negative and multiple of \(5\).

8 Choice D is correct
The correct answer is \(369\) let \(x\) be total number of cards in the box, then number of red cards is: \(x \  \ 246\) The probability of choosing a red card is one third. Then: probability \(=\frac{1}{3}=\frac{x \  \ 132}{x}\) Use cross multiplication to solve for \(x\). \(x \ × \ 1=3 \ (x \  \ 246)→x=3 \ x \  \ 738→2 \ x=738→x=369\)

9 Choice C is correct
The correct answer is II and III Plug in the values of \(x\) in each equation and check. I. \(( \ 2)^2 \  \ 2 \ + \ 6=4 \  \ 2 \ + \ 6=8≠0\) \((3)^2 \  \ 3 \ + \ 6=3 \  \ 3 \ + \ 6=12≠0\) II. \(2 \ ( \ 2)^2 \  \ 2 \ ( \ 2)=8 \ + \ 4=12→12=12\) \(2 \ (3)^2 \  \ 2 \ (3)=18 \  \ 6=12→12=12\) III. \(5 \ ( \ 2)^2 \  \ 5 \ ( \ 2) \  \ 30=20 \ + \ 10 \  \ 30=0\) \(5 \ (3)^2 \  \ 5 \ (3) \  \ 30=45 \  \ 15 \  \ 30=0\) Equations II and III are correct.

10 Choice C is correct
The correct answer is \(9 \ π\) Let P be circumference of circle A, then; \(2 \ π \ r_{A}=18 \ π→r_{A}=9\) \(r_{A}=3 \ r_{B}→r_{B}=\frac{9}{3}=3→\) Area of circle B is; \(π \ r_{B}^2=9 \ π\)

11 Choice B is correct
The correct answer is \(\frac{1}{6}\) Number of biology book: \(35\) Total number of books; \(35 \ + \ 95 \ + \ 80=210\) the ratio of the number of biology books to the total number of books is: \(\frac{35}{210}=\frac{1}{6}\)

12 Choice B is correct
The correct answer is \(67^\circ\) \(α=180^\circ \  \ 112^\circ=68^\circ\) \(β=180^\circ \  \ 135^\circ=45^\circ\) \(x \ + \ α \ + \ β=180^\circ→x=180^\circ \  \ 68^\circ \  \ 45^\circ=67^\circ\)

13 Choice D is correct
The correct answer is \(π(x)=\sqrt{x } \ + \ 4 \) A. \(f(x)=x^2 \  \ 5\) if \(x=1→f(1)=(1)^2 \  \ 5=1 \  \ 5= \ 4≠5 \) B. \(f(x)=x^2 \  \ 1\) if \(x=1→f(1)=(1)^2 \  \ 1=1 \  \ 1=0≠5\) C. \(f(x)=\sqrt{x \ + \ 2}\) if \(x=1→f(1)=\sqrt{1 \ + \ 2}=\sqrt{3}≠5\) D. \(f(x)=\sqrt{x} \ + \ 4\) if \(x=1→f(1)=\sqrt{1} \ + \ 4=5\) Choice D is correct.

14 Choice C is correct
The correct answer is \(150\) Let \(b\) be the amount of time Alec can do the job, then, \(\frac{1}{a} \ + \ \frac{1}{b}=\frac{1}{100}→\frac{1}{300} \ + \ \frac{1}{b}=\frac{1}{100}→\frac{1}{b}=\frac{1}{100} \  \ \frac{1}{300}=\frac{2}{300}=\frac{1}{150}\) Then: \(b=150\) minutes

15 Choice C is correct
The correct answer is \(66\) In the equilateral triangle if \(x\) is length of one side of triangle, then the perimeter of the triangle is \(3 \ x\). Then \(3 \ x=33→x=11\) and radius of the circle is: \(x=11\) Then, the perimeter of the circle is: \(2 \ π \ r=2 \ π \ (11)=22 \ π\) \(π=3→22 \ π=22 \ × \ 3=66\)

16 Choice C is correct
The correct answer is \(472\) \(\frac{12}{100} \ x=72→x=\frac{72 \ × \ 100}{12}=600\) \(\frac{1}{8} \ y=16→y=8 \ × \ 16=128\) \(→x \  \ y=600 \  \ 128=472\)

17 Choice J is correct
The correct answer is \(3 \ \frac{1}{2}\) One degree equals \(\frac{π}{180 }\). The angle \(α\) in radians is equal to the angle α in degrees times \(π\) constant divided by \(180\) degrees. Then: \(1\) degree \(= \frac{π}{180}→630\) degrees \(=\frac{630 \ π}{180}=3.5 \ π\) \(3.5 \ π=x \ π→x=3.5\)

17 Choice J is correct
The correct answer is \(3 \ \frac{1}{2}\) One degree equals \(\frac{π}{180 }\). The angle \(α\) in radians is equal to the angle α in degrees times \(π\) constant divided by \(180\) degrees. Then: \(1\) degree \(= \frac{π}{180}→630\) degrees \(=\frac{630 \ π}{180}=3.5 \ π\) \(3.5 \ π=x \ π→x=3.5\)

19 Choice A is correct
The correct answer is \(6.5\) \( \ 3 \ a \ + \ 5 \ a \ + \ 7 \ a=45→9 \ a=45→a=\frac{45}{9}=5\) Then; \(\frac{3 \ a \  \ 2}{2}=\frac{3 \ (5) \  \ 2}{2}=\frac{15 \  \ 2}{2}=6.5\)

20 Choice C is correct
The correct answer is \(15\) All integers from \(11\) to \(19\) are: \(11, 12, 13, 14, 15, 16, 17, 18, 19\) The mean of these integers is: \(\frac{11 \ + \ 12 \ + \ 13 \ + \ 14 \ + \ 15 \ + \ 16 \ + \ 17 \ + \ 18 \ + \ 19}{9}=\frac{135}{9}=15\)

21 Choice A is correct
The correct answer is \(11\) \( \ 12 \  \ 5 \  \  \ 8 \ + \ 2= \ 17 \  \  \ 6=17 \  \ 6=11\)

22 Choice D is correct
The correct answer is \(13\) Based on the table provided: \(g( \ 2)=g(x= \ 2)=3\) \(g(3)=g(x=3)= \ 2\) \(3 \ g( \ 2) \  \ 2 \ g(3)=3 \ (3) \  \ 2 \ ( \ 2)=9 \ + \ 4=13\)

23 Choice B is correct
The correct answer is \(12\) let \(x\) be the number of gallons of water the container holds when it is full. Then; \(\frac{7}{24} \ x=3.5→x=\frac{24 \ × \ 3.5}{7}=12\)

24 Choice A is correct
The correct answer is \(27\) The quadrilateral is a trapezoid. Use the formula of the area of trapezoids. Area \(=\frac{1}{2} \ h \ (b_{1} \ + \ b_{2} )\) You can find the height of the trapezoid by finding the difference of the values of \(y\) for the points A and D. (or points B and C) \(h=8 \  \ 2=6\) AB \(=\sqrt{(x_{1} \  \ x_{2} )^2 \ + \ (y_{1} \  \ y_{2})^2 }=\sqrt{(6 \  \ 3)^2 \ + \ (8 \  \ 8)^2 }=\sqrt{9 \ + \ 0}=3\) CD \(=\sqrt{(x_{1} \  \ x_{2} )^2 \ + \ (y_{1} \  \ y_{2})^2 }=\sqrt{(8 \  \ 2)^2 \ + \ (2 \  \ 2)^2 }=\sqrt{36 \ + \ 0}=6\) Area of the trapezoid is: \(\frac{1}{2} \ h \ (b_{1} \ + \ b_{2} )=\frac{1}{2} \ (6) \ (3 \ + \ 6)=27\)

25 Choice D is correct
The correct answer is \(2 \ a \  \ 2\) Choose a random number for a and check the options. Let \(a\) be equal to \(15\) which is divisible by \(5\), then: A. \(a \  \ 1=15 \  \ 1=14\) is not divisible by \(4\) B. \(a \ + \ 1=15 \ + \ 1=16\) is divisible by \(4\) but if \(a=5→a \ + \ 1=5 \ + \ 1=6\) is not divisible by \(4\) C. \(2 \ a=2 \ × \ 15=30\) is not divisible by \(4\) D. \(2 \ a \  \ 2=(2 \ × \ 15) \  \ 2=28\) is divisible by \(4\)

26 Choice C is correct
The correct answer is \(4\) \((3^a )^b=81→3^{a \ b}=81\) \(81=3^4→3^{a \ b}=3^4\) \(→a \ b=4\)

27 Choice A is correct
The correct answer is January and February First find the number of pants sold in each month. January: \(110\), February: \(88\), March: \(90\), April: \(70\), May: \(85\), June: \(65\) Check each option provided. A. 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

28 Choice D is correct
The correct answer is \(147.5, 30\) First, order the number of shirts sold each month: \(130,140,145,150,160,170\) median is: \(\frac{145 \ + \ 150}{2}=147.5\) Put the number of shoes sold per month in order: \(20,25,25,35,35,40\) mean is: \(\frac{20 \ + \ 25 \ + \ 25 \ + \ 35 \ + \ 35 \ + \ 40}{6}=\frac{180}{6}=30\)

29 Choice D is correct
The correct answer is \(50\) The ratio of number of pants to number of shoes in May equals \(\frac{85}{25}\). Fiveseventeenth of this ratio is \((\frac{5}{17}) \ (\frac{85}{25})\). Now, Let \(x\) be the number of shoes needed to be added in April. \(\frac{70}{20 \ + \ x}=(\frac{5}{17})\ (\frac{85}{25})→\frac{70}{20 \ + \ x}=\frac{425}{425}=1→70=20 \ + \ x→x=50\)

30 Choice C is correct
The correct answer is \(\frac{5}{2}\) The value of \(y\) in the \(x\)intercept of a line is zero. Then: \(y=0→2 \ x \  \ 2 \ (0)=5→2 \ x=5→x=\frac{5}{2}\) then, \(x\)intercept of the line is \(\frac{5}{2}\)

31 Choice A is correct
The correct answer is \(4\) cm The sum of the lengths of any two sides of triangle is greater than the length of the third side, therefore the greatest possible value of the biggest side equal to \(4\) cm. \(4 \ < \ 6\)

32 Choice B is correct
The correct answer is \(2\) \((x \  \ 2)^3=27→\) Find the third root of both sides. Then: \(x \  \ 2=3→x=5\) \(→(x \  \ 4) \ (x \  \ 3)=(5 \  \ 4) \ (5 \  \ 3)=(1) \ (2)=2\)

33 Choice B is correct
The correct answer is \(31,752\) Number of Mathematics book: \(0.3 \ × \ 840=252\) Number of English book: \(0.15 \ × \ 840=126\) Product of number of Mathematics and number of English books: \(252 \ × \ 126=31,752\)

34 Choice D is correct
The correct answer is \(108^\circ, 45^\circ\) The angle \(α\) is: \(0.3 \ × \ 360=108^\circ\) The angle \(β\) is: \(0.15 \ × \ 360=54^\circ\)

35 Choice B is correct
The correct answer is \(120\) According to the chart, \(50\%\) of the books are in the Mathematics and Chemistry sections. Therefore, there are \(420\) books in these two sections. \(0.50 \ × \ 840 = 420\) \(\gamma \ + \ \alpha=420\), and \(\gamma=\frac{2}{5} \ \alpha\) Replace \(\gamma\) by \(\frac{2}{5} \ \alpha\) in the first equation. \(\gamma \ + \ \alpha=420→\frac{2}{5} \ \alpha \ + \ \alpha=420→\frac{7}{5} \ \alpha=420→\) multiply both sides by \(\frac{5}{7}\) \((\frac{5}{7}) \ \frac{7}{5} \ \alpha=420 \ × \ (\frac{5}{7})→\alpha=\frac{420 \ × \ 5}{7}=300\) \(\alpha=300→\gamma=\frac{2}{5} \ \alpha→\gamma=\frac{2}{5} \ × \ 300=120\) There are \(120\) books in the Chemistry section.

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

37 Choice C is correct
The correct answer is \(\frac{100 \ x \ + \ 800}{x}\) The amount of money for \(x\) bookshelf is: \(100 \ x\) Then, the total cost of all bookshelves is equal to: \(100 \ x \ + \ 800\) The total cost, in dollar, per bookshelf is: \(\frac{Total \ cost}{number \ of \ items}=\frac{100 \ x \ + \ 800}{x}\)

38 Choice C is correct
The correct answer is \(0\) \(\sqrt{x}=4→x=16\) then; \(\sqrt{x} \  \ 7=\sqrt{16} \  \ 7=4 \  \ 7= \ 3\) and \(\sqrt{x \  \ 7}=\sqrt{16 \  \ 7}=\sqrt{9}=3\) Then: \((\sqrt{x \  \ 7}) \ + \ (\sqrt{x \  \ 7})=3 \ + \ ( \ 3)=0\)

39 Choice B is correct
The correct answer is \(25\) The angles on a straight line add up to \(180\) degrees. Then: \(x \ + \ 25 \ + \ y \ + \ 2 \ x \ + \ y=180\) Then, \(3 \ x \ + \ 2 \ y=180 \  \ 25→3 \ (35) \ + \ 2 \ y=155\) \(→2 \ y=155 \  \ 105=50→y=25\)

40 Choice B is correct
The correct answer is \(10 \ π\) The distance of A to B on the coordinate plane is: \(\sqrt{(x_{1} \  \ x_{2} )^2 \ + \ (y_{1} \  \ y_{2} )^2 }= \sqrt{(10 \  \ 4)^2 \ + \ (11 \  \ 3)^2 }=\sqrt{6^2 \ + \ 8^2}\) \(=\sqrt{36 \ + \ 64}=\sqrt{100}=10 \) The diameter of the circle is \(10\) and the radius of the circle is \(5\). Then: the circumference of the circle is: \(2 \ π \ r=2 \ π \ (5)=10 \ π\)

41 Choice C is correct
The correct answer is \(37\) Square root of \(16\) is \(\sqrt{16}=4 \ < \ 6\) Square root of \(25\) is \(\sqrt{25}=5\ < \ 6\) Square root of \(37\) is \(\sqrt{37}=\sqrt{36 \ + \ 1} \ > \ \sqrt{36}=6\) Square root of \(49\) is \(\sqrt{49}=7 \ > \ 6\) Since, \(\sqrt{37} \ < \ \sqrt{49}\), then the answer is C.

42 Choice C is correct
The correct answer is \(46\) cm The area of the trapezoid is: Area \(=\frac{1}{2} \ h \ (b_{1} \ + \ b_{2} )=\frac{1}{2} \ (x) \ (13 \ + \ 8)=126\) \(→10.5 \ x=126→x=12\) \(y=\sqrt{5^2 \ + \ 12^2}=\sqrt{25 \ + \ 144}=\sqrt{169}=13\) The perimeter of the trapezoid is: \(12 \ + \ 13 \ + \ 8 \ + \ 13=46\)

43 Choice A is correct
The correct answer is \(x \ ≥ \ 5 \ ∪ \ x \ ≤ \ − \ 1\) \(x \  \ 2 \ ≥ \ 3\) Then: \(x \  \ 2 \ ≥ \ 3→x \ ≥ \ 3 \ + \ 2→x \ ≥ \ 5\) Or \(x \  \ 2 \ ≤ \  \ 3→x \ ≤ \  \ 3 \ + \ 2→x \ ≤ \  \ 1\) Then, the solution is: \(x \ ≥ \ 5 \ ∪ \ x \ ≤ \ − \ 1\)

44 Choice B is correct
The correct answer is \(50\) Since, E is the midpoint of AB, then the area of all triangles DAE, DEF, CFE and CBE are equal. Let \(x\) be the area of one of the triangle, Then: \(4 \ x=100→x=25\) The area of DEC \(=2 \ x=2 \ (25)=50 \)

45 Choice A is correct
The correct answer is \(− \ 8 \ < \ x \ < \ − \ 5\) \(13 \ < \  \ 3 \ x \  \ 2 \ < \ 22→\) Add \(2\) to all sides. \(13 \ + \ 2 \ < \  \ 3 \ x \  \ 2 \ + \ 2 \ < \ 22 \ + \ 2\) \(→15 \ < \  \ 3 \ x \ < \ 24→\) 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 \(>\)) \(\frac{15}{ \ 3} \ > \ \frac{ \ 3 \ x}{ \ 3} \ > \ \frac{24}{ \ 3}\) \( \ 8 \ < \ x \ < \  \ 5\)

46 Choice C is correct
The correct answer is \(6.93\) Based on triangle similarity theorem: \(\frac{a}{a \ + \ b}=\frac{c}{3}→c=\frac{3 \ a}{a \ + \ b}=\frac{3 \ \sqrt{3}}{3 \ \sqrt{3}}=1→\) area of the shaded region is: \((\frac{c \ + \ 3}{2}) \ (b)=4 \ \sqrt{3}\) Round \(4 \ \sqrt{3}\) to the nearest hundredths place gives \(6.93\).

47 Choice B is correct
The correct answer is \(0.94\) sinβ‘\((A)=\frac{opposite}{hypotenuse}=\frac{1}{3}⇒\) We have the following triangle, then: \(c=\sqrt{3^2 \  \ 1^2}=\sqrt{9 \  \ 1}=\sqrt{8}\) cosβ‘\((A)=\frac{\sqrt{8}}{3}\) Rounding the answer to the nearest hundredths, gives \(0.94\)

48 Choice C is correct
The correct answer is \(80\) One liter \(=1000\) cm\(^3→ 6\) liters \(= 6,000\) cm\(^3\) \(6,000=15 \ × \ 5 \ × \ h→h=\frac{6,000}{75}=80\) cm

49 Choice C is correct
The correct answer is \(12\) Based on corresponding members of each matrix, write two equations: \(\begin{cases}2 \ x=x \ + \ 3 \ y \  \ 5\\4 \ x = 2 \ y \  \ 10\end{cases} \rightarrow \begin{cases}x \  \ 3 \ y=  \ 5\\4 \ x \  \ 2 \ y =10\end{cases}\) Multiply first equation by \(( \ 4)\), then: \(\begin{cases} \ 4 \ x \ + \ 12 \ y =20\\4 \ x \  \ 2 \ y =10\end{cases}\) Add two equations: \(→10 \ y=30→y=3→x=4→ x \ × \ y=12\)
