Free Full Length ACCUPLACER Mathematics Practice Test

Full Length ACCUPLACER Mathematics Practice Test

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ACCUPLACER Mathematics Practice Test 3 (Non–Calculator) 2 Sections – 40 questions Total time for two sections: No Time Limit You may not use a calculator on this section.
Arithmetic and Elementary Algebra
1-	Which of the following is a factor of both $ x^2 \ -\ 9 \ x \ + \ 18$ and $ x^2 \ – \ 7 \ x \ +\ 6$ ?
(A)	($x \ + \ 6$)
(B)	($x \ + \ 3$)
(C)	($x \ - \ 6$)
(D)	($x \ - \ 1$)
2-	What is the area of an isosceles right triangle that has one leg that measures $ 6 $ cm?
(A)	$18$ cm$^2$
(B)	$36$ cm$^2$
(C)	$9$ cm$^2$
(D)	$20$ cm$^2$
3-	If two angles in a triangle measure $ 46 $ degrees and $72$ degrees, what is the value of the third angle?
(A)	$64$ degrees
(B)	$56$ degrees
(C)	$46$ degrees
(D)	$62$ degrees
4-	Which of the following expressions is equivalent to $ 15 \ – \ \frac{4}{3} \ x \ \leq \ 1$
(A)	$x \leq \ – \ 10.5 $
(B)	$x \geq \ 10.5 $
(C)	$x \geq \ 12 $
(D)	$x \leq \ 12 $
5-	$ \frac{1}{2\ b^2} \ + \ \frac{1}{ b} \ = \ \frac{1}{b^2} $ , then $ b = $ ?
(A)	$\frac{1}{2}$
(B)	$- \ \frac{1}{2}$
(C)	$- \ \frac{3}{2}$
(D)	$ \frac{3}{2}$
6-	$ 7^{\frac{6}{5}} \ × \ 7^{\frac{4}{5}} $=
(A)	$7^5$
(B)	$7^3$
(C)	$7^4$
(D)	$7^2$
7-	If $ a^7 \ + \ c^7 \ = \ c^7 \ + \ b^7$ , then $ a \ =$ ?
(A)	$ c$
(B)	$ b$
(C)	$ b^{4} \ – \ a^{4} $
(D)	$ b^{3} \ – \ a^{4} $
8-	What is $178.4786 $ rounded to the nearest hundredth?
(A)	$178.47$
(B)	$178.50$
(C)	$178.46$
(D)	$178.48$
9-	The equation of a line is given as: $ y \ = \ 6 \ x \ – \ 1$ . Which of the following points does not lie on the line?
(A)	$(1,2)$
(B)	$(- \ 2 ,- \ 13 )$
(C)	$(3 , 17 )$
(D)	$(1, 5 )$
10-	A soccer team played $ 210 $ games and won $ 60 $ percent of them. How many games did the team win?
(A)	$125$
(B)	$124$
(C)	$123$
(D)	$126$
11-	The sum of three numbers is $ 64 $ . If another number is added to these three numbers, the average of the four numbers is $ 28 $ . What is the fourth number?
(A)	$58$
(B)	$48$
(C)	$38$
(D)	$28$
12-	Line m passes through the point $ (2, 3) $ . Which of the following CANNOT be the equation of line m?
(A)	$y = 1 \ - \ x$
(B)	$ y = 3 $
(C)	$ x = 2 $
(D)	$ y = x \ + \ 1$
13-	If $ a = 9 $ what’s the value of $ 3 \ a^{2} \ + \ 7 \ a \ - \ 8 $?
(A)	$296$
(B)	$298$
(C)	$300$
(D)	$295$
14-	David owed $ $7216 $ . After making $ 32 $ payments of $ $115 $ each , how much did he have left to pay?
(A)	$$3,526$
(B)	$$3,525$
(C)	$$3,515$
(D)	$$3,536$
15-	$ (q^{3}) \ . \ (q^{3}) \ =$ ?
(A)	$ q^3$
(B)	$ q^9$
(C)	$ q^6$
(D)	$ q^5$
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16-	$ x^{2} \ – \ 64 \ = \ 0 $ , $ x $ could equal to:
(A)	$8$
(B)	$9$
(C)	$6$
(D)	$5$
17-	$ (x \ + \ 2) \ (x \ - \ 6) \ =$
(A)	$x^{2} \ - \ 4 \ x \ + \ 12$
(B)	$x^{2} \ + \ 4 \ x \ + \ 12$
(C)	$x^{2} \ - \ 4 \ x \ - \ 12$
(D)	$x^{2} \ + \ 4 \ x \ - \ 12$
18-	If $x$ is a positive integer divisible by $9$, and $ x \ < \ 81 $ , what is the greatest possible value of $ x $ ?
(A)	$54$
(B)	$72$
(C)	$36$
(D)	$45$
19-	If $ 7.5 \ < \ x \ ≤ \ 11.0 $ , then $ x$ cannot be equal to:
(A)	$7.5$
(B)	$11.0$
(C)	$9.2$
(D)	$10.1$
20-	If $a = 6$ , what is the value of b in this equation? $b = \ \frac{a^{2}}{3} \ + \ 4$
(A)	$14$
(B)	$8$
(C)	$16$
(D)	$13$
College–Level Mathematics
21-	Suppose a triangle has the dimensions indicated below: Then Sin $B =$ ?
(A)	$\frac{3}{4}$
(B)	$\frac{7}{3}$
(C)	$\frac{3}{6}$
(D)	$\frac{3}{7}$
22-	A number is chosen at random from $1$ to $ 50 $ . Find the probability of not selecting a composite number.
(A)	$\frac{8}{25}$
(B)	$\frac{6}{25}$
(C)	$\frac{5}{25}$
(D)	$\frac{4}{25}$
23-	The cost, in thousands of dollars, of producing $x $ thousands of textbooks is $C (x) \ = \ x^{2} \ + \ 10 \ x \ + \ 30$ . The revenue, also in thousands of dollars, is $R (x) = 3 \ x $. Find the profit or loss if $ 2,000 $ textbooks are produced. (profit $=$ revenue $–$ cost)
(A)	$$22,000$ profit
(B)	$$22,000$ loss
(C)	$$28,000$ loss
(D)	$$28,000$ profit
24-	Find the slope – intercept form of the graph $ 5 \ x \ – \ 9 \ y \ = \ – \ 16$
(A)	$\ y = \frac{5}{9} \ x \ + \ \frac{16}{9} $
(B)	$ - \ 9 \ y = - \ 5 \ x \ - \ 16 $
(C)	$y = - \ 5 \ x \ - \ 16 $
(D)	$y = 5 \ x \ - \ 16 $
25-	Solve $ e^{(x \ + \ 4 )} = 32 $
(A)	$ln \ (32) \ + \ 4$
(B)	$ln \ (32) \ + \ 2$
(C)	$ln \ (32) \ - \ 4$
(D)	$ln \ (32) \ - \ 2$
26-	If tan $\theta \ = \ \frac{8}{15} $ and sin $\theta \ > \ 0 $, then cos $\theta \ =$ ?
(A)	$\frac{17}{15} $
(B)	$\frac{15}{17} $
(C)	$\frac{8}{17} $
(D)	$\frac{17}{8} $
27-	Michael (M ) is $ 3$ years older than her friend Alex (A) who is $ 5$ years younger than her sister John (J ). If M , A and J denote their ages, which one of the following represents the given information?
(A)	$\begin{cases}M \ = \ A \ + \ 3\\ J \ = \ A \ - \ 5\end{cases}$
(B)	$\begin{cases}M = A \ - \ 3 \\ A = J \ - \ 5\end{cases}$
(C)	$\begin{cases}M = 3 \ + \ A \\ A = 5 \ - \ J\end{cases}$
(D)	$\begin{cases}M = 3 \ + \ A \\ A = J \ - \ 5\end{cases}$
28-	Solve the equation: $ log_2⁡ \ (x \ + \ 6) \ – \ log_2⁡(x \ - \ 2) \ = \ 1$
(A)	$ - \ 2$
(B)	$ - \ 10$
(C)	$2$
(D)	$10$
29-	From $9$ students in an algebra class, a group of $ 3 $ students will be chosen to work on a group project. How many different groups of $4 $ students can be chosen?
(A)	$72$
(B)	$21$
(C)	$36$
(D)	$48$
30-	What is the domain of the following function? $ f (x) \ = \ \sqrt{(x \ - \ 6 )} \ + \ 4$
(A)	$x \ ≥ - \ 6 $
(B)	$x \leq - \ 6 $
(C)	$x \leq 6 $
(D)	$x \ \geq 6 $
31-	Which of the following point is the solution of the system of equations? $ \begin{cases}12\ x \ + \ 4 \ y \ = \ 32 \\6 \ x \ - \ 2 \ y = 8\end{cases} $
(A)	$ (- \ 2, 2) $
(B)	$ ( 2, 2) $
(C)	$ ( 2, - \ 2) $
(D)	$ (- \ 2, - \ 2) $
32-	if $ f (x) \ = \ \frac{(4 \ x \ - \ 1 )}{2 } $ and $ f ^{ \ – \ 1}(x) $ , is the inverse of $ f (x) $, what is the value of $ f ^{ \ – \ 1}(3) $ ?
(A)	$\frac{5}{4}$
(B)	$\frac{3}{4}$
(C)	$\frac{7}{4}$
(D)	$\frac{9}{4}$
33-	If $ f (x) \ = \ 7 \ - \ x $ and $g (x) \ = \ – \ x^2 \ – \ 3 \ – \ 5 \ x$ , then find $(g \ + \ f) (x)$?
(A)	$– \ x^2 \ – \ 6 \ x \ + \ 4$
(B)	$– \ x^2 \ – \ 6 \ x \ - \ 4$
(C)	$– \ x^2 \ + \ 6 \ x \ - \ 4$
(D)	$– \ x^2 \ + \ 6 \ x \ + \ 4$
34-	Find the Center and Radius of the graph $ (x \ - \ 4)^{2} \ + \ (y \ + \ 7)^{2} \ = \ 18 $ ?
(A)	$( 4 , 7 )$, $3 \ \sqrt{2} $
(B)	$( 4 , 7 )$, $2 \ \sqrt{3} $
(C)	$(- \ 4 , 7 )$, $3 \ \sqrt{2} $
(D)	$(4 ,- \ 7 )$, $3 \ \sqrt{2} $
35-	Which of the following lines is parallel to the graph of $ y \ = \ 4 \ x $ ?
(A)	$4 \ x \ + \ y \ = \ 4$
(B)	$4 \ x \ - \ y \ = \ 4$
(C)	$2 \ x \ - \ y \ = \ 4$
(D)	$2 \ x \ - \ y \ = \ 2$
36-	$ \frac{\| \ 6 \ + \ x \ \|}{8} \ ≤ \ 3 $ , then $ x $ =?
(A)	$ - \ 30 \ ≤ \ x \ ≤ \ - \ 18 $
(B)	$ 30 \ ≤ \ x \ ≤ \ 18 $
(C)	$ - \ 30 \ ≤ \ x \ ≤ \ 18 $
(D)	$ - \ 18 \ ≤ \ x \ ≤ \ 18 $
37-	Simplify $ \frac{(4 \ + \ i)}{2\ - \ 3 \ i} $
(A)	$ \frac{5}{13} \ - \ i \ \frac{14}{13}$
(B)	$ \frac{5}{13} \ +\ i \ \frac{14}{13}$
(C)	$ - \ \frac{5}{13} \ +\ i \ \frac{14}{13}$
(D)	$ - \ \frac{5}{13} \ - \ i \ \frac{14}{13}$
38-	tan $( \frac{π}{3}) \ =$ ?
(A)	$- \ \sqrt3$
(B)	$- \ 3 \ \sqrt3$
(C)	$3 \ \sqrt3$
(D)	$ \sqrt3$
39-	$ \frac{\sqrt{48 \ a^7 \ b^2}}{\sqrt{4 \ a^2 \ b^2}} $ =?
(A)	$ 2 \ a\ \sqrt{3}$
(B)	$ 2 \ a^2\ \sqrt{3 \ a}$
(C)	$ 2 \ a^2\ \sqrt{3 \ b}$
(D)	$ 2 \ a\ \sqrt{3 \ b}$
40-	Find the inverse function of $ f (x) \ = \ \frac{(x \ - \ 4 )}{8} $
(A)	$4 \ (2 \ x \ - \ 1)$
(B)	$4 \ (2 \ x \ + \ 1)$
(C)	$ (2 \ x \ + \ 1)$
(D)	$ (2 \ x \ - \ 1)$

ACCUPLACER Mathematics Practice Test Answers and Explanations

1- Choice C is correct

The correct answer is ($x \ – \ 6$)
Factor each trinomial
$ x^2 \ - \ 9 \ x \ + \ 18 \ $ and $ x^2 \ – \ 7 \ x \ + \ 6 $
$x^2 \ - \ 9 \ x \ + \ 18 \Rightarrow (x \ – \ 6) \ (x \ - \ 3) $
$ x^2 \ – \ 7 \ x \ + \ 6 \Rightarrow (x \ – \ 1) \ (x \ – \ 6) $

2- Choice A is correct

The correct answer is $18$ cm$^2$
$ a=6 \Rightarrow $ area of triangle is $ = \frac{1}{2} \ ( 6\times \ 6) = \frac{36}{2} \ = 18 $ cm$^2$

3- Choice D is correct

The correct answer is $62$ degrees
$ 46^{\circ} \ + \ 72^{\circ } = 118^{\circ} $
$ 180^{\circ } \ – \ 118^{\circ} = 62^{\circ} $
The value of the third angle is $ 62^\circ $ .

4- Choice B is correct

The correct answer is $x \geq \ 10.5 $
Simplify:
$ 15 \ – \ \frac{4}{3} \ x \ \leq \ 1 \ \Rightarrow \ – \ \frac{4}{3} \ x \ \leq \ – \ 14 \ \Rightarrow \ – \ x \ \leq\ – \ 10.5 \ \Rightarrow x \geq \ 10.5 $

5- Choice A is correct

The correct answer is $\frac{1}{2}$
$ \frac{1 \ + \ 2\ b}{2 \ b^2} \ = \ \frac{1}{b^2} \Rightarrow (b\neq 0) $
$\ b^2 \ + \ 2 \ b^3 \ = \ 2 \ b^2 \ \Rightarrow$
$2 \ b^3 \ - \ \ b^2 \ = \ 0 \ \Rightarrow$
$\ b^2 \ (2 \ b \ - \ 1) \ = \ 0 \ \Rightarrow$
$2 \ b \ - \ 1 \ = \ 0 \ \Rightarrow \ b \ = \frac{1}{2} $

6- Choice D is correct

The correct answer is $5^2$
$ 7^{\frac{6}{5} } \ × \ 7^{\frac{4}{5} } = 7^ { \frac{6}{5} \ + \ \frac{4}{5} } = 7^\frac{10}{5} = 7^2 $

7- Choice B is correct

The correct answer is $b$
If $ a^7 \ + \ c^7 \ = \ c^7 \ + \ b^7 $
then: $ a^7 \ = \ b^7 \ ⇒ \ a \ = \ b $

8- Choice D is correct

The correct answer is $178.48$
Underline the hundredths place:
$ 178.4\underline{\\7}86 $
Look to the right if it is $ 7 $ or above, give it a shove.
Then, round up to $ 178.48 $

9- Choice A is correct

The correct answer is $(1, 2)$
$ y \ = \ 6 \ x \ – \ 1 $
A.$ (1, 2) \Rightarrow 2 \ = \ 6 \ – \ 1 \Rightarrow 2 \ ≠ \ 5 $
B.$ (– \ 2, \ –\ 13) \Rightarrow \ – \ 13 \ = \ – \ 12 \ – \ 1 \Rightarrow \ – \ 13 \ = \ – \ 13 $
C.$ (3, 17) \Rightarrow \ 17 \ = \ 18 \ – \ 1 \Rightarrow 17 = 17 $
D.$ (1, 5) \Rightarrow 5 \ = \ 6 \ – \ 1 \Rightarrow \ 5 \ = \ 5 $

10- Choice D is correct

The correct answer is $126$
$ 210 \ × \ \frac{60}{100} \ = \ 126 $

11- Choice B is correct

The correct answer is $35$
$ a \ + \ b \ +\ c = 64 $
$ \frac{a \ +\ b \ +\ c \ +\ d}{4} \ =\ 28 \ ⇒\ a \ + \ b \ + \ c \ + \ d \ = \ 112 \ ⇒ \ 64 \ + \ d \ = \ 112 $
$ d \ = \ 112 \ – \ 64 \ = \ 48 $

12- Choice A is correct

The correct answer is $y = 1 \ - \ x$
Solve for each equation:
$ (2, 3) $
A.$ y \ = \ 1 \ – \ x \ ⇒ \ 3 \ = \ 1 \ – \ 3 \ ⇒ \ 3 ≠ - \ 2 $
B.$ y \ = 3 \ ⇒ 3 = 3$
C.$ x = 2 \ ⇒ 2 = 2 $
D.$ y = x \ + \ 1 ⇒ 3 = 2 \ + \ 1 \ ⇒ \ 3 \ = \ 3$

13- Choice B is correct

The correct answer is $298$
If $ a \ = \ 9 $
then : $ 3 \ a^2 \ + \ 7 \ a \ - \ 8 ⇒ 3 \ (9)^2 \ + \ 7 \ (9) \ - \ 8 ⇒ 3 \ (81) \ + \ 63 \ - \ 8 = 298 $

14- Choice D is correct

The correct answer is $$3,536$
$32 \ × \ $115 \ = \ $ 3680 $
Payable amount is:
$ $7216 \ - \ $3680 \ = \ $3536 $

15- Choice C is correct

The correct answer is $q^{6}$
$ (q^3) \ . \ q^3) \ = \ q ^{3 \ + \ 3} \ = \ q^{6} $

16- Choice A is correct

The correct answer is $8$
$ x^2 \ – \ 64 \ = \ 0 \ ⇒ \ x^2 \ = \ 64 \ ⇒ \ x \ = \ 8 $

17- Choice C is correct

The correct answer is $x^{2} \ - \ 4 \ x \ - \ 12$
Use FOIL (First, Out, In, Last)
$ (x \ + \ 2) \ (x \ - \ 6) \ = \ x^2 \ - \ 6 \ x \ + \ 2 \ x \ - \ 12 = x ^2 \ - \ 4 \ x \ - \ 12 $

18- Choice B is correct

The correct answer is $72$
$ \frac{54}{9} \ = \ \frac{18}{3} = 6$ , $ \frac{72}{9} \ = \ \frac{24}{3} = 8 $ , $ \frac{36}{9} \ = \ \frac{18}{3} =4 $ , $ \frac{45}{9} \ = \ \frac{15}{3}= 5 $
$ 72 $ is prime number

19- Choice A is correct

The correct answer is $7.5$
If $7.5 \ < \ x \ ≤ \ 11.0 $, then $ x$ cannot be equal to $7.5$

20- Choice C is correct

The correct answer is $16$
If $a = \ 8 $ , then:
$b = \ \frac{6^2}{3} \ + \ 4 \ ⇒ $
$b = \ \frac{36}{3} \ + \ 4 \ ⇒ $
$b = \ 12 \ + \ 4 \ = \ 16 $

21- Choice D is correct

The correct answer is $\frac{3}{7}$
sin$ B = \ \frac{(the \ length \ of \ the \ side \ that \ is \ opposite \ that \ angle)}{(the \ length \ of \ the \ longest \ side \ of \ the \ triangle)} \ = \ \frac{3}{7} $

22- Choice A is correct

The correct answer is $\frac{8}{25}$
Set of number that are not composite between $ 1 $ and $ 50 $ : $A = \ {1, 2, 3, 5, 7, 11, 13, 17, 19, 23 , 29, 31, 37 , 41,43,47}$
$n (A) = \ 16 \ ⇒ p = \ \frac{16}{50} \ = \ \frac{8}{25} $

23- Choice B is correct

The correct answer is $$22,000$ loss
$c (2) \ = \ (2)^2 \ + \ 6\ (2) \ + \ 12 \ = \ 4 \ + \ 12 \ + \ 12 = 28$
$3 \ × \ 2 = 6 ⇒ 6 \ - \ 28 = \ - \ 22 \ ⇒ \ $22,000$ loss

24- Choice A is correct

The correct answer is$\ y = \frac{5}{9} \ x \ + \ \frac{16}{9} $
$ - \ 9 \ y = - \ 5 \ x \ - \ 16 \ ⇒$
$\ y = \frac{- \ 5}{- \ 9} \ x \ - \ \frac{16}{- \ 9} \ ⇒$
$\ y = \frac{5}{9} \ x \ + \ \frac{16}{9} $

25- Choice C is correct

The correct answer is $ln \ (32) \ - \ 4$
$ e^{x \ + \ 4} = 32 ⇒ ln \ (e^{x \ + \ 4}) = \ ln \ (32)$
$ (x \ + \ 4) \ ln \ (e) = ln \ (32) $
$ x \ + \ 4 = ln \ (32) \ ⇒ \ x \ = \ ln \ (32) \ - \ 4 $

26- Choice B is correct

The correct answer is $\frac{15}{17} $
tan $\theta \ = \ \frac{8}{15} \ ⇒$ we have following triangle, then
$ c \ = \ \sqrt{8^2 \ + \ 15^2} \ = \ \sqrt{64 \ + \ 225} \ = \ \sqrt{289} \ = \ 17 $
cos $\theta \ = \ \frac{15}{17} $

27- Choice D is correct

The correct answer is:
M $ = \ 3 \ + $ A
A $ = $ J $ – \ 5 $

28- Choice D is correct

The correct answer is $2$
METHOD ONE:
$ \log_{2}{(x \ + \ 6)} \ – \ \log_{2}{(x \ - \ 2)} = 1 $
Add $ \log_{2}{(x \ - \ 2)} $ to both sides:
$ \log_{2}{(x \ + \ 6)} \ – \ \log_{2}{(x \ - \ 2)} + \log_{2}{(x \ - \ 2)} = 1 + \log_{2}{(x \ - \ 2)} $
And simplify:
$ \log_{2}{(x \ + \ 6)} = 1 + \log_{2}{(x \ - \ 2)} $
Logarithm rule: $ a \ = \ \log_{b}{b^a} \ ⇒ \ 1 \ = \log_{2}{2^1} \ = \ \log_{2}{2} $
then: $ \log_{2}{(x \ + \ 6)} = \log_{2}{2} \ + \ \log_{2}{(x \ - \ 2)} $
Logarithm rule: $ \log_{c}{a} +\log_{c}{b} = \log_{c}{a\ b} $
then: $ \log_{2}{2} \ + \ \log_{2}{(x \ - \ 2)} \ = \ \log_{2}{2 \ (x \ - \ 2)}⇒
\log_{2}{(x \ + \ 6)} \ = \ \log_{2}{2 \ (x \ - \ 2)} $
When the logs have the same base: $ \log_{b}{(f(x))} \ = \ \log_{b}{(g(x))} \ ⇒ \ f (x) \ = \ g (x) $
$ x \ + \ 6 \ = \ 2 \ (x \ - \ 2) \ ⇒ \ x \ + \ 6 \ = \ 2 \ x \ – \ 4 \ ⇒ \ - \ x \ = \ - \ 10 \ ⇒ \ x \ = \ 10 $
METHOD TWO
We know that: $\log_{a}{b} \ - \ \log_{a}{c} \ = \ \log_{a}{\frac{b}{c}} $ and $ \log_{a}{b} \ = \ c \ ⇒ \ b \ = \ a^c $
Then: $ \log_{2}{(x \ + \ 6)} \ - \ \log_{2}{(x \ - \ 2)}=\log_{2}{\frac{x \ + \ 6}{x \ - \ 2}} \ = \ 1⇒$
$\frac{x \ + \ 6 }{x \ - \ 2} = 2^1 = 2 ⇒ x \ + \ 6 \ = \ 2 \ (x \ - \ 2)$
$ ⇒ \ x \ + \ 6 = 2 \ x \ - \ 4 ⇒ 2 \ x \ - \ x \ = \ 6 \ + \ 4 \ ⇒ \ x \ = \ 10$

29- Choice B is correct

The correct answer is $21$
$ C_4^9 \ = \ \frac{9!}{4!(9 \ - \ 3)!} \ = \ \frac {9!}{4! \ 6!} \ = \ \frac{9 \ × \ 8 \ × \ 7 \ × \ 6!}{4! \ × \ 6!} \ = \ \frac{9\ ×\ 8 \ × \ 7 }{4 \ × \ 3 \ × \ 2 \ × \ 1} \ = \ 21 $

30- Choice D is correct

The correct answer is $x \ ≥ \ 6 $
The number under the square root symbol must be zero or greater than zero therefore:
$ x \ - \ 6 \ ≥ \ 0 \ ⇒ \ x \ ≥ \ 6 $ domain of function $= \ [6 , \ + \ ∞) $

31- Choice B is correct

The correct answer is $ (2, 2) $
$ \begin{cases}12 \ x \ + \ 4 \ y = 32 \\6 \ x \ - \ 2 \ y = 8 \end{cases}\Rightarrow $ Multiplication $ (– \ 2)$ in first equation $ \begin{cases}12 \ x \ + \ 4 \ y = 32 \\- \ 12 \ x \ + \ 4 \ y = - \ 16\end{cases} $
Add two equations together $ \ ⇒8 \ y = 16 \ ⇒ \ y =2$ then: $ x = 2 $

32- Choice C is correct

The correct answer is $\frac{7}{4}$
$ f (x)=\frac{4 \ x \ - \ 1}{2} ⇒$
$y=\frac{4 \ x \ - \ 1}{2} \ ⇒$
$2 \ y \ = \ 4 \ x \ – \ 1 \ ⇒$
$2 \ y \ + \ 1 \ = \ 4\ x \ ⇒$
$ \frac{2 \ y \ + \ 1}{4} \ = \ x $
$ f^{ \ - \ 1} = \frac{2 \ y \ + \ 1}{4} \ ⇒$
$f^{ \ - \ 1} (3) \ = \ \frac{7}{4} $

33- Choice A is correct

The correct answer is $– \ x^2 \ – \ 6 \ x \ + \ 4$
$ (g \ + \ f) (x) \ = \ g (x) \ + \ f (x) \ = \ (– \ x^2 \ – \ 3 \ – \ 5 \ x) \ + \ (7 \ - \ x) $
$ – \ x^2 \ + \ 4 \ – \ 5 \ x \ – \ x \ = \ – \ x^2 \ – \ 6\ x \ + \ 4$

34- Choice D is correct

The correct answer is $(4 , - \ 7 )$, $3 \ \sqrt{2} $
$ (x \ – \ h)^2 \ + \ (y \ – \ k)^2 \ = \ r^2 \ ⇒$
center: $ (h,k)$ and radius:$ r $
$(x \ – \ 4)^2 \ + \ (y \ + \ 7 )^2 \ = \ 18 \ ⇒$
center: $(4 , - \ 7) $ and radius: $ 3 \ \sqrt{2} $

35- Choice A is correct

The correct answer is $4 \ x \ + \ y \ = \ 4$
If two lines are parallel with each other, then the slope of the two lines is the same.
Then in line $ y \ = \ 4 \ x $ , the slope is equal to $ 4$
And in the line $ 4 \ x \ + \ y \ = \ 4 \ ⇒ y \ = \ 4 \ x \ - \ 4$
the slope equal to $ 4$

36- Choice C is correct

The correct answer is $ - \ 30 \ ≤ \ x \ ≤ \ 18 $
$ \frac{| \ 6 \ + \ x \ |}{8} \ ≤ \ 3 \ ⇒ \ | \ 6 \ + \ x \ | \ ≤ \ 24 \ ⇒$
$ - \ 24 \ ≤ \ 6 \ + \ x \ ≤ \ 24 \ ⇒$
$ - \ 24\ - \ 6 \ ≤ \ x \ ≤ \ 24 \ - \ 6 \ ⇒ $
$ - \ 30 \ ≤ \ x \ ≤ \ 18 $

37- Choice B is correct

The correct answer is $ \frac{5}{13} \ + \ i \ \frac{14}{13}$
If $ z_1 \ = \ x_1 \ + \ i \ y_1$ and $ z_2 \ = \ x_2 \ + \ i \ y_2 \ ⇒$
$ \frac{z _ 1}{z _ 2} \ = \ \frac{x_1 \ x_2 \ + \ y_1 \ y_2}{x_2^2 \ + \ y_2^2} \ + \ i \ \frac{x_2 \ y_1 \ - \ x_1 \ y_2}{x_2^2 \ + \ y_2^2}$
In this problem: $ x_1 \ = \ 4, \ x_2 \ = 2, \ y_1 \ = \ 1, \ y_2 \ = - \ 3$
$\frac{4 \ +\ i}{ 2 \ - \ 3 \ i} \ = \ \frac{5}{13} \ + \ i \ \frac{14}{13} $

38- Choice D is correct

The correct answer is $\sqrt{3}$
tan $( \frac{π}{3}) \ = \sqrt{3} $

39- Choice B is correct

The correct answer is $ 2 \ a^2 \ \sqrt{3 \ a}$
$ \frac{\sqrt{48 \ a^7 \ b^2}}{\sqrt{2 \ a^2 \ b^2}} = \frac{4 \ a^2 \ b \sqrt{3}}{2 \ a \ b } \ = \ 2 \ a^2 \ \sqrt{3 \ a} $

40- Choice B is correct

The correct answer is $4 \ (2 \ x \ + \ 1)$
$ f(x)=\frac{x \ - \ 4}{8} \ ⇒ \ y \ = \ \frac{x \ - \ 4}{8} \ ⇒ \ 8 \ y \ = \ x \ – \ 4 \ ⇒ \ 8\ y \ + \ 4 \ = \ x $
$ f ^{ \ - \ 1} \ = \ 8 \ x \ + \ 4 \ = \ 4 \ ( 2 \ x \ + \ 1) $