 ## Full-Length ACCUPLACER Mathematics Practice Test

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ACCUPLACER Mathematics Practice Test 4

(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- Last Friday Jacob had $$45.12$$ . Over the weekend he received some money for cleaning the attic. He now has $$62$$ . How much money did he receive?
(A) $$16.68$$
(B) $$16.78$$
(C) $$16.88$$
(D) $$16.98$$
2- $$x^2 \ - \ 8 \ x \ + \ 15 \ =$$ ?
(A) $$(x \ – \ 3) \ (x \ - \ 5)$$
(B) $$(x \ + \ 3) \ (x \ - \ 5)$$
(C) $$(x \ + \ 3) \ (x \ + \ 5)$$
(D) $$(x \ - \ 3) \ (x \ + \ 5)$$
3- What is $$6254.74164$$ rounded to the nearest tenth?
(A) $$6254.7$$
(B) $$6254.8$$
(C) $$6254.9$$
(D) $$6254.0$$
4- In the following diagram, what is the value of $$x$$ in the following triangle? (A) $$40^\circ$$
(B) $$46^\circ$$
(C) $$60^\circ$$
(D) $$45^\circ$$
5- $$45$$ is what percent of $$120$$ ?
(A) $$37\%$$
(B) $$37.5\%$$
(C) $$38.5\%$$
(D) $$38\%$$
6- Which of the following equations has a graph that is a straight line?
(A) $$3 \ y \ + \ 6 = 4 \ x$$
(B)  $$y \ = \ 3 \ x^2 \ + \ 9$$
(C) $$3 \ y^2 \ + \ 6 = x$$
(D) $$3 \ y \ + \ 6 = x^2$$
7- Find all values for which $$2 \ x^2 \ - \ 5 \ x \ - \ 3 \ = \ 0$$
(A) $$- \ 12 , \frac{1}{2}$$
(B) $$\frac{1}{2} , 12$$
(C) $$- \ \frac{1}{2} , 3$$
(D) $$- \ \frac{3}{2} , 12$$
8- What is the distance between the points $$(2, 4)$$ and $$(- \ 4, - \ 4)$$ ?
(A) $$100$$
(B) $$10$$
(C) $$5$$
(D) $$7$$
9- Simplify $$\frac{\frac{4}{3} \ - \ \frac{2 \ x \ - \ 3}{9}}{\frac{x^2}{3} \ - \ \frac{7}{3}}$$
(A) $$\frac{- \ 2 \ x \ - \ 15}{3 \ x^2 \ - \ 21}$$
(B) $$\frac{- \ 2 \ x \ - \ 15}{3 \ x^2 \ +\ 21}$$
(C) $$\frac{- \ 2 \ x \ + \ 15}{3 \ x^2 \ + \ 21}$$
(D) $$\frac{- \ 2 \ x \ + \ 15}{3 \ x^2 \ - \ 21}$$
10- A man owed $$3254$$ on his car. After making $$42$$ payment of $$65$$ each, how much did he have left to pay?
(A) $$234$$
(B) $$236$$
(C) $$524$$
(D) $$253$$
11- Write the $$\frac{7}{120}$$ as a decimal.
(A) $$0.059$$
(B) $$0.058$$
(C) $$0.60$$
(D) $$0.62$$
12- $$\sqrt{75}$$ is between which two whole numbers?
(A) $$7$$ and $$8$$
(B) $$8$$ and $$9$$
(C) $$6$$ and $$7$$
(D) $$5$$ and $$6$$
13- Which of the following is one solution of this equation?
$$3 \ x^2 \ + \ 10 \ x \ - \ 2 \ = \ 0$$
(A) $$\frac{- \ 5 \ ± \ \sqrt{31}}{3}$$
(B) $$\frac{- \ 5 \ - \ \sqrt{31}}{3} , 0.12$$
(C) $$\frac{- \ 5 \ + \ \sqrt{31}}{3} , 0.358$$
(D) $$\frac{- \ 5 \ + \ \sqrt{31}}{3} , 0.438$$
14- $$(x \ + \ 7 ) \ (x^{2} \ - \ 4 \ x \ + \ 3) =$$ ?
(A) $$x^3 \ - \ 3 \ x^2 \ – \ 25 \ x \ + \ 21$$
(B) $$x^3 \ - \ 3 \ x^2 \ + \ 25 \ x \ + \ 21$$
(C) $$x^3 \ - \ 3 \ x^2 \ + \ 25 \ x \ - \ 21$$
(D) $$x^3 \ + \ 3 \ x^2 \ – \ 25 \ x \ + \ 21$$
15- $$(x^{5})^{\frac{2}{7}}$$
(A) $$x^{ \frac{11}{7}}$$
(B) $$x^{ \frac{10}{7}}$$
(C) $$x^{ \frac{9}{7}}$$
(D) $$x^{ \frac{8}{7}}$$
16- How many $$6 \ × \ 6$$  squares can fit inside a rectangle with a height of $$42$$ and width of $$18$$?
(A) $$21$$
(B) $$16$$
(C) $$24$$
(D) $$36$$
17- If a vehicle is driven $$42$$ miles on Monday, $$47$$ miles on Tuesday, and $$31$$ miles on Wednesday, what is the average number of miles driven each day?
(A) $$42$$ miles
(B) $$45$$ miles
(C) $$46$$ miles
(D) $$40$$ miles
18- Alex’s average (arithmetic mean) on two mathematics tests is $$10$$ . What should Liam’s score be on the next test to have an overall of $$12$$ for all the tests?
(A) $$12$$
(B) $$18$$
(C) $$16$$
(D) $$10$$
19- If $$6\ - \ 3 \ x \ ≤ \ 18$$ , what is the value of $$x \geq$$ ?
(A) $$4$$
(B) $$3$$
(C) $$-4$$
(D) $$-3$$
20- $$9^{5 } \ × \ 9^{ - \ 8} \ =$$?
(A) $$9^{ 3}$$
(B) $$9^{ - \ 3}$$
(C) $$9^{ 14}$$
(D) $$9^{ 10}$$

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College–Level Mathematics
21- cos $$2 \ \theta =$$ ?
(A) $$1\ + \ 2 \ sin^2\ \theta$$
(B) $$1\ - \ 2 \ sin^2\ \theta$$
(C) $$- \ 2 \ sin^2\ \theta$$
(D) $$2 \ sin^2\ \theta$$
22- If $$\theta$$ is an acute angle and sin $$\theta =\frac{4}{5}$$ , cos $$\theta =$$ ?
(A) $$\frac{4}{5}$$
(B) $$\frac{3}{5}$$
(C) $$\frac{2}{5}$$
(D) $$\frac{1}{5}$$
23- If the center of a circle is at the point $$(4, - \ 1)$$  and its circumference equals to $$4 \ π$$ , what is the standard form equation of the circle?
(A) $$(x \ +\ 4)^2 \ + \ (y \ + \ 1)^2 = 2$$
(B) $$(x \ - \ 4)^2 \ + \ (y \ + \ 1)^2 \ = \ 4$$
(C) $$(x \ - \ 4)^2 \ + \ (y \ + \ 1)^2 \ = \ 2$$
(D) $$(x \ - \ 4)^2 \ + \ (y \ - \ 1)^2 \ = \ 4$$
24- What is the solution of the following system of equations?
$$\begin{cases} 4 \ x \ + \ y = 8 \\ - \ 8 \ x \ - \ 4 \ y = 16\end{cases}$$
(A) $$(6, 16)$$
(B) $$(- \ 6, 16)$$
(C) $$(6, - \ 16)$$
(D) $$(6, 16)$$
25- What is the center and radius of a circle with the following equation?
$$(x \ – \ 6)^2 \ + \ (y \ + \ 4)^2 \ = \ 5$$
(A) $$(6, 4), \ \sqrt{5}$$
(B) $$(6 , - \ 4), \ \sqrt{5}$$
(C) $$(- \ 6 , - \ 4), \ \sqrt{5}$$
(D) $$(- \ 6 , 4), \ \sqrt{5}$$
26- If sin A $$=\ \frac{2}{5}$$ in a right triangle and the angle A is an acute angle, then what is cos A ?
(A) $$\frac{\sqrt{22}}{5}$$
(B) $$\frac{\sqrt{20}}{5}$$
(C) $$\frac{\sqrt{21}}{5}$$
(D) $$\frac{\sqrt{3}}{5}$$
27- If $$\log_{3}{x \ = \ 6}$$ , then $$x \ =$$ ?
(A) $$729$$
(B) $$243$$
(C) $$719$$
(D) $$216$$
28- Simplify: $$\frac{2 \sqrt{12}}{9 \sqrt{48}}$$
(A) $$\frac{2}{9}$$
(B) $$\frac{4}{9}$$
(C) $$\frac{1}{9}$$
(D) $$\frac{5}{9}$$
29- What’s the reciprocal of $$\frac{25}{x^3}$$ ?
(A) $$\frac{x^3}{25 }$$
(B) $$\frac{25}{x^3 }$$
(C) $$\frac{5}{x }$$
(D) $$\frac{5}{x ^3 }$$
30- If $$\log_{4}{x \ = \ 3}$$ , then $$x \ =$$ ?
(A) $$81$$
(B) $$16$$
(C) $$64$$
(D) $$36$$
31- What is sin  $$60^\circ$$?
(A) $$\frac{\sqrt{3}}{2}$$
(B) $$\sqrt{3}$$
(C) $$- \ \sqrt{3}$$
(D) $$- \ \frac{\sqrt{3}}{2}$$
32- Find the inverse function for ln $$(4 \ x \ - \ 3)$$
(A) $$\frac{1}{4} (e^{\ x} \ - \ 3)$$
(B) $$\frac{1}{2} (e^{\ x} \ + \ 3)$$
(C) $$\frac{1}{4} (e^{\ x} \ + \ 3)$$
(D) $$\frac{1}{2} (e^{\ x} \ - \ 3)$$
33- Solve.
$$| \ 15 \ – \ (18 \ ÷ \ | \ 3 \ + \ 6 \ |)| =$$ ?
(A) $$- \ 13$$
(B) $$- \ 12$$
(C) $$12$$
(D) $$13$$
34- Simplify $$( – \ 7 \ + \ 2 \ i) \ (9 \ + \ 4 \ i)$$ .
(A) $$- \ 10 \ i \ - \ 71$$
(B) $$- \ 10 \ i \ + \ 71$$
(C) $$10 \ i \ + \ 71$$
(D) $$10 \ i \ - \ 71$$
35- If $$f(x) \ = \ x \ – \frac{4}{5}$$ and $$f ^{ \ – \ 1}$$ is the inverse of $$f(x)$$ , what is the value of  $$f^{ \ - \ 1}(2)$$
(A) $$\frac{7}{5}$$
(B) $$\frac{16}{5}$$
(C) $$\frac{6}{5}$$
(D) $$\frac{14}{5}$$
36- Find tan $$\frac{4 \ π}{3}$$
(A) $$- \ \sqrt{3}$$
(B) $$\sqrt{3}$$
(C) $$2 \ \sqrt{3}$$
(D) $$- \ 2 \ \sqrt{3}$$

37- Solve the equation: $$\log_{5}{(x \ - \ 4)} \ – \ \log_{5}{(x \ + \ 3)}⁡ \ = \ 1$$
(A) $$\frac{9}{4}$$
(B) $$\frac{18}{4}$$
(C) $$\frac{14}{4}$$
(D) $$-\frac{19}{4}$$
38- If $$f(x) \ = 2 \ x \ + \ 6$$ and $$g(x) \ = x^2 \ + \ 3 \ x$$ , then find $$(\frac{f}{g}) (x)$$
(A) $$\frac{2 \ x \ - \ 6}{x^2 \ +\ 3 \ x}$$
(B) $$\frac{2 \ x \ + \ 6}{x^2 \ +\ 3 \ x}$$
(C) $$\frac{2 \ x \ + \ 6}{x^2 \ - \ 3 \ x}$$
(D) $$\frac{2 \ x \ - \ 6}{x^2 \ - \ 3 \ x}$$
39- The slop of a line with the equation $$y \ = \ 6 \ x \ + \ 12$$ is  …
(A) $$4$$
(B) $$\frac{5}{3}$$
(C) $$\frac{5}{6}$$
(D) $$6$$
40- Solve $$e^{(7 \ x \ + \ 1 )} \ = \ 8$$
(A) $$\frac{ln⁡(8) \ +\ 1}{7}$$
(B) $$\frac{ln⁡(8) \ +\ 1}{5}$$
(C) $$\frac{ln⁡(8) \ -\ 1}{5}$$
(D) $$\frac{ln⁡(8) \ -\ 1}{7}$$
 1- Choice C is correct The correct answer is $$16.88$$$$62 \ - \  45.12 \ = \  16.88$$ 2- Choice A is correct The correct answer is $$(x \ – \ 3) \ (x \ - \ 5)$$$$x^2 \ - \ 8 \ x \ + \ 15 \ = \ (x \ – \ 3) \ (x \ - \ 5)$$ 3- Choice A is correct The correct answer is $$6254.7$$Underline the tenth place:$$6254.\underline{\\7}4164$$Look to the right if it is $$7$$ or above, give it a shove.Then, round up to $$6254.7$$ 4- Choice D is correct The correct answer is $$45^\circ$$$$90^\circ \ + \ 45^\circ = 135^\circ$$$$180^\circ \ - \ 135^\circ = 45^\circ$$ 5- Choice B is correct The correct answer is $$37.5\%$$$$120\ × \ \frac{x}{100} \ = \ 45 \ ⇒ \ 120 \ × \ x \ = \ 4500 \ ⇒ \ x \ = \ \frac{4500}{120} =37.5\%$$ 6- Choice A is correct The correct answer is $$3 \ y \ + \ 6 = 4 \ x$$$$3 \ x \ + \ 6 = 4 \ x$$ has a graph that is a straight line 7- Choice C is correct The correct answer is $$-\frac{1}{2} , 3$$$$x_{1,2} \ = \ \frac{- \ b \ ± \ \sqrt{b^2 \ - \ 4 \ a \ c}}{2 \ a}$$ $$a \ x^2 \ + \ b \ x \ + \ c \ = \ 0$$ $$2 \ x^2 \ - \ 5 \ x \ – \ 3 \ = \ 0 \ ⇒$$ then: $$a \ = \ 2, b \ = - \ 5 \$$ and $$\ c \ = \ – \ 3$$$$x \ = \ \frac{- \ (- \ 5) \ + \ \sqrt{(5^2) \ - \ (4) \ (2) \ ( - \ 3)}}{2 .2} = 3$$$$x \ = \ \frac{- \ (- \ 5 )\ - \ \sqrt{(5^2) \ - \ (4)\ (2) \ (- \ 3)}}{2.2} \ = -\ \frac{1}{2}$$ 8- Choice B is correct The correct answer is $$10$$$$C \ = \ \sqrt{(x_A \ - \ x_B)^2 \ + \ (y_A \ - \ y_B)^2 }$$$$C \ = \ \sqrt{(2 \ - \ (- \ 4) )^2 \ + \ ( 4 \ - \ (- \ 4))^2 }$$$$C \ = \ \sqrt{(6)^2 \ + \ ( 8)^2 }$$$$C \ = \ \sqrt{36 \ + \ 64}$$ $$C \ = \ \sqrt{100} = 10$$ 9- Choice D is correct The correct answer is $$\frac{- \ 2 \ x \ + \ 15}{3 \ x^2 \ - \ 21}$$Simplify:$$\frac{\frac{4}{3} \ - \ \frac{2 \ x \ - \ 3}{9}}{\frac{x^2}{3} \ - \ \frac{7}{3}} = \frac{\frac{4}{3} \ - \ \frac{2 \ x \ - \ 3}{9}}{\frac{x^2 \ - \ 7}{3}} = \frac{ 3 \ (\frac{4}{3} \ - \ \frac{2 \ x \ - \ 3}{9 })}{x^2 \ - \ 7}$$⇒Simplify: $$\frac{4}{3} \ - \ \frac{2 \ x \ - \ 3}{9 } = \frac{ -\ 2\ x \ + \ 15 }{9}$$then: $$\frac{3(\frac{ - \ 2 \ x \ + \ 15 }{9})}{x^2 \ - \ 7} = \frac{\frac{ -\ 2 \ x \ + \ 15}{3 }}{x^2 \ - \ 7} = \frac{ - \ 2 \ x \ + \ 15}{3\ (x^2 \ - \ 7)} = \frac{-2 \ x \ +\ 15}{3 \ x^2 \ - \ 21}$$ 10- Choice C is correct The correct answer is $$254$$$$42\ × \  65 \ = \  2730$$ Payable amount is: $$3254 \ - \  2730 \ = \  524$$ 11- Choice B is correct The correct answer is $$0.058$$$$\frac{7}{120} \ = \ \frac{1}{17.14} \ =0.05833333333 \ \cong \ 0.058$$ 12- Choice B is correct The correct answer is $$8$$ and $$9$$$$\sqrt{75} \ = 8.66025$$ ...then:$$\sqrt{75 }$$ is between $$8$$  and  $$9$$ 13- Choice A is correct The correct answer is $$\frac{- \ 5 \ ± \ \sqrt{31}}{3}$$ $$x_{1,2} \ = \ \frac{- \ b \ ± \ \sqrt{b^2 \ - \ 4 \ a \ c}}{2 \ a}$$$$a \ x^2 \ + \ b \ x \ + \ c \ = \ 0$$$$3 \ x^2 \ + \ 10 \ x \ - \ 2 = 0 \ ⇒$$ then: a $$= \ 3$$, b $$= 10$$ and c $$= -\ 2$$$$x \ = \frac{- \ 10 \ + \ \sqrt{10^2 \ - \ (4) .(3 )(- \ 2 )}}{2.3} \ =\frac{- \ 5 \ - \ \sqrt{31}}{3} = -3.523$$$$x \ = \frac{- \ 10 \ + \ \sqrt{10^2 \ - \ (4) .(3) .(-\ 2)}}{2.3} \ = \frac{- \ 5 \ + \ \sqrt{31}}{3} = 0.189$$ 14- Choice D is correct The correct answer is $$x^3 \ + \ 3 \ x^2 \ – \ 25 \ x \ + \ 21$$Use FOIL (First, Out, In, Last)$$(x \ + \ 7) (x^2 \ - \ 4 \ x \ + \ 3) = x^3 \ - \ 4 \ x^2 \ + \ 3 \ x \ + \ 7 \ x^2 \ – \ 28 \ x \ + \ 21 = x^3 \ + \ 3 \ x^2 \ – \ 25 \ x \ + \ 21$$ 15- Choice B is correct The correct answer is $$x^{ \frac{10}{7}}$$$$(x^5)^{\frac{2}{7}} = x^{5 \ × \ \frac{2}{7}} \ = \ x^{ \frac{10}{7}} \ = \ x^{ \frac{10}{7}}$$ 16- Choice A is correct The correct answer is $$21$$Number of squares equal to: $$\frac{42 \ × \ 18}{6 \ × \ 6} \ = \ 7 \ × \ 3 = 21$$ 17- Choice D is correct The correct answer is $$40$$ miles$$42 \ + \ 47 \ + \ 31 = 120$$Average $$= \ \frac{120}{3} = 40$$ miles 18- Choice C is correct The correct answer is $$16$$$$\frac{a \ + \ b}{2} \ = \ 10\ ⇒ \ a \ + \ b \ = \ 20$$$$\frac{a \ + \ b \ + \ c}{3} \ = 12 \ ⇒ \ a \ + \ b \ + \ c \ = 36$$$$20 \ + \ c \ = 36 \ ⇒ \ c \ = \ 36\ – \ 20 \ = \ 16$$ 19- Choice C is correct The correct answer is $$-4$$Simplify: $$6 \ - \ 3 \ x \ ≤ \ 18 \ ⇒ - \ 3 \ x \ ≤ \ 18 \ – \ 6 \ ⇒ - \ 3 \ x \ ≤ \ 12 ⇒ x \ \geq \ -4$$ 20- Choice B is correct The correct answer is $$9^{ - \ 3}$$$$9^{ 5} \ × \ 9^{ - \ 8} \ = 9^{ 5 \ + (- \ 8)} = 9^{- \ 3}$$ 20- Choice B is correct The correct answer is $$9^{ - \ 3}$$$$9^{ 5} \ × \ 9^{ - \ 8} \ = 9^{ 5 \ + (- \ 8)} = 9^{- \ 3}$$ 21- Choice B is correct The correct answer is $$1\ - \ 2 \ sin^2\ \theta$$$$cos\ 2\ \theta = cos^2\ \theta \ - \ sin^2\ \theta = 2 \ cos^2\ \theta \ -\ 1 = 1 \ -\ 2\ sin^2\ \theta$$ 22- Choice B is correct The correct answer is $$\frac{3}{5}$$sin $$θ \ = \ \frac{4}{5} \ ⇒$$ we have following triangle, then:c $$= \ \sqrt{5^2 \ - \ 4 ^2} \ = \ \sqrt{25 \ - \ 16} \ = \ \sqrt{9} \ = \ 3$$cos $$θ \ = \ \frac{3}{5}$$ 23- Choice B is correct The correct answer is $$(x \ - \ 4)^2 \ + \ (y \ + \ 1)^2 \ = \ 2$$$$(x \ – \ h)^2 \ + \ (y \ – \ k)^2 \ = \ r^2 \ ⇒$$ center: $$(h,k)$$  and radius: $$r$$center: $$(4 , - \ 1 ) \ ⇒ \ h = 4, k = - \ 1$$circumference $$= 4 \ π ⇒$$ circumference $$= 2 \ π \ r = 4 \ π \ ⇒ \ r \ = \ 2$$$$(x \ - \ 4)^2 \ + \ (y \ + \ 1)^2 \ = \ 4$$ 24- Choice C is correct The correct answer is $$(6, - \ 16)$$$$\begin{cases}4 \ x \ + \ y \ = 8 \\ - \ 8 \ x \ - \ 4 \ y \ = 16\end{cases}$$⇒ Multiplication $$(4)$$ in first equation $$⇒ \ \begin{cases}16 \ x \ + \ 4 \ y \ = 32 \ \\ - \ 8 \ x \ - \ 4 \ y = 16\end{cases}$$Add two equations together $$⇒\ 8 \ x \ = 48 \ ⇒ x = 6$$ then: $$y = - \ 16$$ 25- Choice B is correct The correct answer is $$(6, - \ 4), \ \sqrt{5}$$$$(x \ – \ h)^2 \ + (y \ – \ k)^2 = r^2 \ ⇒$$ center: $$(h,k)$$ and radius: $$r$$$$(x \ – \ 6)^2 \ + \ (y \ + \ 4)^2 = 5 ⇒$$ center: $$(6, - \ 4)$$ and radius:$$\ \sqrt{5}$$ 26- Choice C is correct The correct answer is $$\frac{\sqrt{21}}{5}$$sin A$$= \ \frac{2}{5 } \ ⇒$$ we have following triangle, then$$c \ = \ \sqrt{5 ^{2} \ - \ 2^{2}} \ = \ \sqrt{25 \ - \ 4} \ = \ \sqrt{21}$$cos⁡ A $$= \ \frac{adjacent}{hypotenuse} \ →$$ cos A $$= \ \frac{\sqrt{21}}{5}$$ 27- Choice A is correct The correct answer is $$729$$METHOD ONE$$\log_{3}{x} \ = \ 6$$Apply logarithm rule: $$a \ = \ \log_{3}{(b^a)}$$$$6 \ = \ \log_{3}{3^6} \ = \ \log_{3}{729}$$$$\log_{3}{x} \ = \ \log_{3}{729}$$When the logs have the same base: $$\log_{b}{(f (x))} \ = \ \log_{b}{(g (x))} \ ⇒ \ f (x) \ = \ g (x)$$then: $$x \ = \ 729$$METHOD TWOWe know that: $$\log_{a}{b} \ = \ c \ ⇒ \ b \ = \ a^c \ \log_{3}{x} \ = 6 \ ⇒ \ x \ = \ 3^6 \ = \ 729$$ 28- Choice C is correct The correct answer is $$\frac{1}{9}$$$$\frac{ 2 \ \sqrt{12}}{ 9 \sqrt{48 }}$$ $$\sqrt{48} \ = 2 \ \sqrt{12 }$$$$\frac{ 2 \sqrt{12}}{9 .2 \ \sqrt{12}} \ = \ \frac{1}{9}$$ 29- Choice A is correct The correct answer is $$\frac{x^3}{25 }$$$$\frac{25}{x^3} ⇒$$ reciprocal is : $$\frac{x^3}{25 }$$ 30- Choice C is correct The correct answer is $$64$$METHOD ONE$$\log_{4}{x}⁡ = 3$$ Apply logarithm rule: $$a \ = \ log_b⁡(b^a)$$$$3 = \log_{4}{4^3} = \log_{4}{64}$$$$\log_{4}{64} = \log_{4}{x}$$When the logs have the same base: $$\log_{b}{(f (x))} \ = \ \log_{b}{(g (x))} \ ⇒ \ f (x) \ = \ g (x)$$then: $$x = 64$$METHOD TWOWe know that: $$\log_{a}{b} \ = \ c \ ⇒ \ b \ = \ a^c$$ $$\log_{4}{x}= 3 ⇒ x = 4^3 = 64$$ 31- Choice A is correct The correct answer is $$\frac{\sqrt{3}}{2}$$sin $$60^{°} \ = \ \frac{\sqrt{3}}{2}$$ 32- Choice C is correct The correct answer is $$\frac{1}{4} (e^{\ x} \ + \ 3)$$$$f (x) \ = \ ln \ (4\ x \ - \ 3)$$$$y \ = \ ln \ (4 \ x \ - \ 3 )$$Change variables $$x$$ and $$y: x \ = \ ln \ (4 \ y \ - \ 3)$$solve: $$x \ = \ ln \ (4 \ y \ - \ 3 )$$$$y \ = \ \frac{\ e^{\ x} \ + \ 3}{4 } \ = \ \frac{1}{4} (e^{\ x} \ + \ 3)$$ 33- Choice D is correct The correct answer is $$13$$$$| \ 15 \ – \ (18 \ ÷ \ | \ 3 \ + \ 6 \ |)| = | (15 \ - \ (18 \ ÷ \ | 9 |))| \ =$$$$| \ 15 \ - \ (18 \ ÷ \ 9) \ | = |15 \ - \ 2| \ = \ | \ 13 \ | \ = \ 13$$ 34- Choice A is correct The correct answer is $$- \ 10 \ i \ + \ 71$$We know that: $$i \ = \ \sqrt{\ - \ 1} \ ⇒ \ i^2 \ = \ - \ 1$$$$(- \ 7 \ + \ 2\ i) \ (9 \ + \ 4 \ i) = - \ 63 \ - \ 28 \ i \ + \ 18 \ i \ + \ 8 \ i^2 \ =$$$$- \ 63 \ + \ 10 \ i \ - \ 8 \ =- \ 10 \ i \ - \ 71$$ 35- Choice D is correct The correct answer is $$\frac{8}{5}$$$$f(x) \ = \ x \ – \frac{4}{5} \ ⇒ \ y \ = \ x \ – \ \frac{4}{5} \ ⇒ \ y \ + \ \frac{4}{5} \ = \ x$$ $$f^{ \ - \ 1} \ = \ x \ + \ \frac{4}{5}$$$$f ^{ \ – \ 1}(2) \ = 2 \ + \ \frac{4}{5} \ = \ \frac{14}{5}$$ 36- Choice B is correct The correct answer is $$\sqrt{3}$$tan$$\frac{4 \ π}{3} \ = \ \frac{sin \ \frac{4 \ π}{3}}{cos \ \frac{4 \ π}{3}} = \frac{\frac{ - \ \sqrt{3}}{2}}{- \ \frac{1}{2}} \ = \sqrt{3}$$ 37- Choice D is correct The correct answer is $$-\frac{19}{4}$$METHOD ONE$$\log_{5}{(x \ - \ 4)} – \log_{5}{(x \ + \ 3)} \ = \ 1$$Add $$\log_{5}{(x \ + \ 3)}$$ to both sides$$\log_{5}{(x \ - \ 4)} \ - \ \log_{5}{(x \ + \ 3)} \ + \ \log_{5}{(x \ + \ 3)} \ = \ 1 \ + \ \log_{5}{(x \ + \ 3)}$$Apply logarithm rule: $$a \ = \ \log_{b}{b^a} \ ⇒ \ 1 \ = \ \log_{5}{5^1} \ = \ \log_{5}{5}$$then: $$\log_{5}{(x \ - \ 4)} \ = \ \log_{5}{5} \ + \ \log_{5}{(x \ + \ 3)}$$Logarithm rule: $$\log_{c}{a} \ + \ \log_{c}{b} \ = \ \log_{c}{ab}$$then: $$\log_{5}{5} \ + \ \log_{5}{(x \ + \ 3)} = \log_{5}{(5(x \ + \ 3))}$$$$\log_{5}{(x \ - \ 4)} \ = \ \log_{5}{(5 \ (x \ + \ 3))}$$When the logs have the same base: $$\log_{b}{(f(x))} \ = \ \log_{b}{(g(x))} \ ⇒ \ f(x) \ = \ g(x)$$$$(x \ - \ 4) \ = 5 \ (x \ + \ 3)$$$$x \ =- \ \frac{19}{4}$$METHOD TWOWe know that: $$\log_{a}{b} \ - \ \log_{a}{c} \ = \ \log_{a}{\frac{b}{c}}$$ and $$\log_{a}{b} \ = \ c \ ⇒ \ b \ = \ a^c$$Then: $$\log_{5}{x \ - \ 4} \ - \ \log_{5}{x \ +\ 3} \ = \ \log_{5}{\frac{x \ - \ 4} {x \ + \ 3}} \ = \ 1 \ ⇒$$$$\frac{x \ - \ 4}{x \ + \ 3} \ = \ 5^1 \ = \ 5 \ ⇒ \ x \ - \ 4 \ = \ 5 \ (x \ + \ 3)$$$$⇒ \ x \ - \ 4\ = \ 5 \ x \ + \ 15 \ ⇒ \ 5 \ x \ - \ x \ =\ -15 \ - \ 4 \ → \ 4 \ x \ = \ -19 \ ⇒ \ x \ =- \ \frac{19}{4}$$ 38- Choice B is correct The correct answer is $$\frac{2 \ x \ + \ 6}{x^2 \ +\ 3 \ x}$$$$(\frac{f}{g}) (x) \ = \ \frac{f(x)}{g(x)} \ = \ \frac{2 \ x \ + \ 6}{x^2 \ +\ 3\ x}$$ 39- Choice D is correct The correct answer is $$6$$$$y \ = \ m \ x \ + \ b$$$$m \ =$$ slop$$y \ = \ 6 \ x \ + \ 12$$$$m \ = \ 6$$ 40- Choice D is correct The correct answer is $$\frac{ln⁡(8) \ - 1}{7}$$$$e^{7 \ x \ + \ 1 } \ = \ 8$$ If $$f(x) \ = \ g(x)$$ , then $$ln(f(x)) \ = \ ln(g(x))$$$$ln⁡(e^{7 \ x \ + \ 1 }) \ = \ ln(8)$$Apply logarithm rule: $$\log_{a}{x^b} \ = \ b \log_{a}{x}$$$$ln⁡(e^{7 \ x \ + \ 1 }) = (7 \ x \ + \ 1) \ ln(e)$$$$(7 \ x \ + \ 1) \ ln \ (e) \ = \ ln \ (8)$$$$(7 \ x \ + \ 1) \ ln \ (e) \ = \ (7 \ x \ + \ 1)$$$$(7 \ x \ + \ 1) \ = \ ln \ (8) \ ⇒ \ x \ = \ \frac{ln⁡(8) \ - 1}{7}$$

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