## 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) $$(2 \ x \ - \ 1)$$

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

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Bundle Includes Accuplacer Math Prep Books, Workbooks, and Practice Tests

### Practice Test 1

Simulate test day with an official practice test. Then, score your test. The answers come with explanations so you can learn from your mistakes.

### Practice Test 2

Simulate test day with an official practice test. Then, score your test. The answers come with explanations so you can learn from your mistakes.

## More Articles

### TABE 11 & 12 Math in 10 Days

$24.99$13.99

### Prepare for the ACT Math Test in 7 Days

$14.99$12.99

### CLEP College Algebra Exercise Book

$18.99$14.99

### STAAR Grade 8 Math Practice Workbook

$25.99$14.99