Full Length ACCUPLACER Mathematics Practice Test

The correct answer is $$ 16.88$
$$ 62 \  - \ $  45.12 \ = \ $ 16.88$

The correct answer is $(x \ – \  3) \ (x \ - \ 5)$
$x^2 \  -  \ 8 \ x \ + \ 15 \ = \ (x \ – \ 3) \ (x \ - \ 5)$

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 $6254.7$

The correct answer is $45^\circ$
$90^\circ \ + \ 45^\circ   =  135^\circ$
$180^\circ \ - \ 135^\circ  =  45^\circ$

The correct answer is $37.5\%$
$120\ × \ \frac{x}{100} \ = \ 45 \ ⇒ \ 120 \ × \ x \ = \ 4500 \ ⇒ \ x \ = \ \frac{4500}{120} =37.5\%$

The correct answer is $3 \ y  \ +  \ 6  =  4 \ x$
$3 \ x \ + \ 6  =  4 \ x$ has a graph that is a straight line

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 $\ c \ = \ – \ 3$
$x \ = \ \frac{- \ (- \ 5) \ + \ \sqrt{(5^2) \ - \ (4) \ (2) \ ( - \ 3)}}{2 .2}  = 3$
$x \ = \ \frac{- \ (- \ 5 )\ - \ \sqrt{(5^2) \ - \ (4)\ (2) \ (- \ 3)}}{2.2} \ = -\ \frac{1}{2}$

The correct answer is $10$
$C \ = \ \sqrt{(x_A \ - \ x_B)^2 \ + \ (y_A \ - \ y_B)^2  }$
$C \ = \ \sqrt{(2 \ - \ (- \ 4) )^2 \ + \ ( 4 \ - \ (- \ 4))^2 }$
$C \ = \ \sqrt{(6)^2 \ + \ ( 8)^2 }$
$C \ = \ \sqrt{36 \ + \ 64}$ 
$C \ = \ \sqrt{100}  =  10$

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: $\frac{3(\frac{ - \ 2 \ x \ + \ 15 }{9})}{x^2 \ - \ 7} = \frac{\frac{ -\ 2 \ x \ + \ 15}{3 }}{x^2 \ - \ 7} = \frac{ - \ 2 \ x \ + \ 15}{3\ (x^2 \ - \ 7)} = \frac{-2 \ x \ +\ 15}{3 \ x^2 \ - \ 21}$

The correct answer is $$  254$
$42\ × \ $  65 \ = \ $  2730$ Payable amount is: $$ 3254 \ - \ $ 2730 \ = \ $  524$    

The correct answer is $0.058$
$\frac{7}{120} \ = \ \frac{1}{17.14} \ =0.05833333333 \ \cong  \ 0.058$

The correct answer is $8$ and $9$
$\sqrt{75} \ = 8.66025$ ...
then:
$\sqrt{75 }$ is between $8$  and  $9$

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 $= -\ 2$
$x \ = \frac{- \ 10 \ + \ \sqrt{10^2 \ - \ (4) .(3 )(- \ 2 )}}{2.3} \ =\frac{- \ 5 \ - \ \sqrt{31}}{3} = -3.523$
$x \ = \frac{- \ 10 \ + \ \sqrt{10^2 \ - \ (4) .(3) .(-\ 2)}}{2.3} \ = \frac{- \ 5 \ + \ \sqrt{31}}{3} = 0.189$

The correct answer is $x^3 \ + \ 3 \   x^2 \ – \ 25 \ x \ + \  21$
Use FOIL (First, Out, In, Last)
$(x \ + \ 7) (x^2 \ - \ 4 \ x \ + \ 3)  =  x^3 \ - \ 4 \ x^2 \ + \ 3 \ x \ + \ 7 \ x^2 \ – \ 28 \ x \ + \ 21  =   x^3 \ + \ 3 \   x^2 \ – \ 25 \ x \ + \  21$

The correct answer is $x^{ \frac{10}{7}}$
$(x^5)^{\frac{2}{7}}  =  x^{5 \ × \  \frac{2}{7}} \ = \ x^{ \frac{10}{7}} \ = \ x^{ \frac{10}{7}}$

The correct answer is $21$
Number of squares equal to: $\frac{42 \ × \ 18}{6 \ × \ 6} \ = \ 7 \ × \ 3 =  21$

The correct answer is $40$ miles
$42 \ + \ 47 \ + \ 31  =  120$
Average $= \ \frac{120}{3}  = 40$ miles

The correct answer is $16$
$\frac{a \ + \ b}{2} \ = \ 10\ ⇒ \ a \ + \ b \ = \ 20$
$\frac{a \ + \ b \ + \ c}{3} \ = 12 \ ⇒ \ a \ + \ b \ + \ c \ = 36$
$20 \ + \ c \ = 36 \ ⇒ \ c \ = \ 36\ – \ 20 \ = \ 16$

The correct answer is $-4$
Simplify: 
$6 \ - \ 3 \ x \ ≤ \ 18 \ ⇒ - \ 3 \ x \  ≤ \ 18 \ – \ 6 \ ⇒ - \ 3 \ x \ ≤ \ 12 ⇒  x \ \geq \ -4$

The correct answer is $9^{ - \ 3}$
$9^{ 5} \ × \ 9^{ -  \  8} \ =  9^{ 5 \ + (- \ 8)} =  9^{- \ 3}$

The correct answer is $9^{ - \ 3}$
$9^{ 5} \ × \ 9^{ -  \  8} \ =  9^{ 5 \ + (- \ 8)} =  9^{- \ 3}$

The correct answer is $1\  - \ 2 \ sin^2\ \theta$
$cos\ 2\ \theta = cos^2\ \theta \ - \  sin^2\ \theta = 2 \ cos^2\ \theta \ -\  1 = 1 \ -\  2\  sin^2\ \theta$

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 $θ \ = \ \frac{3}{5}$

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 $=  2 \ π \ r  =  4 \ π \ ⇒ \ r \ = \ 2$
$(x \ -  \ 4)^2 \ + \ (y \ + \ 1)^2 \ = \ 4$

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: $y = - \ 16$

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:$\ \sqrt{5}$

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 $= \ \frac{\sqrt{21}}{5}$

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 TWO
We know that: $\log_{a}{b} \ = \ c \ ⇒ \ b \ = \ a^c \ \log_{3}{x} \ = 6 \ ⇒ \ x \ = \ 3^6 \ = \ 729$

The correct answer is $\frac{1}{9}$
$\frac{ 2 \ \sqrt{12}}{ 9 \sqrt{48 }}$ 
$\sqrt{48} \ =   2 \ \sqrt{12  }$
$\frac{ 2 \sqrt{12}}{9 .2 \ \sqrt{12}} \ = \ \frac{1}{9}$

The correct answer is $\frac{x^3}{25  }$
$\frac{25}{x^3}  ⇒$ reciprocal is : $\frac{x^3}{25 }$

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 TWO
We know that: $\log_{a}{b} \ = \ c \ ⇒ \ b \ = \ a^c$ $\log_{4}{x}= 3 ⇒ x = 4^3 = 64$

The correct answer is $\frac{\sqrt{3}}{2}$
sin $60^{°} \ = \ \frac{\sqrt{3}}{2}$

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: $x \ = \ ln \ (4 \ y \ - \ 3 )$
$y \ = \ \frac{\ e^{\ x} \ + \ 3}{4 } \ = \ \frac{1}{4} (e^{\ x} \ + \ 3)$

The correct answer is $13$
$| \ 15 \ – \ (18 \ ÷ \ | \ 3 \ + \ 6 \ |)| = | (15 \ - \ (18 \ ÷ \ | 9 |))| \ =$
$| \ 15 \ - \ (18 \ ÷ \ 9) \ | = |15 \ - \ 2| \ = \ | \ 13 \ | \ = \ 13$

The correct answer is $- \ 10 \  i \ + \ 71$
We know that: $i \ = \ \sqrt{\ - \ 1} \ ⇒ \ i^2 \ = \ - \ 1$
$(- \ 7 \ + \  2\ i) \ (9 \ + \ 4 \ i)  =  - \ 63 \ - \ 28 \ i \ + \ 18 \ i \ + \ 8 \ i^2 \ =$
$- \ 63 \ + \ 10 \ i \ - \ 8 \ =-  \ 10 \ i \ - \ 71$

The correct answer is $\frac{8}{5}$
$f(x) \ = \ x \ – \frac{4}{5}  \ ⇒ \ y \ = \ x \ – \ \frac{4}{5} \ ⇒ \ y \ + \ \frac{4}{5} \ = \ x$ 
$f^{ \ - \ 1} \ = \ x \ + \ \frac{4}{5}$
$f ^{ \ – \ 1}(2) \ =  2 \ + \ \frac{4}{5} \ = \ \frac{14}{5}$

The correct answer is $\sqrt{3}$
tan$\frac{4 \ π}{3} \ = \   \frac{sin \ \frac{4 \ π}{3}}{cos \ \frac{4 \ π}{3}}  =   \frac{\frac{ - \ \sqrt{3}}{2}}{- \ \frac{1}{2}} \ =   \sqrt{3}$

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 TWO
We know that: $\log_{a}{b} \ - \ \log_{a}{c} \ = \ \log_{a}{\frac{b}{c}}$ and $\log_{a}{b} \ = \ c \ ⇒ \ b \ = \ a^c$
Then: $\log_{5}{x \ - \ 4} \ - \ \log_{5}{x \ +\ 3} \ = \ \log_{5}{\frac{x \ - \ 4} {x \ + \ 3}} \ = \ 1 \ ⇒$
$\frac{x \ - \ 4}{x \ + \ 3} \ = \ 5^1 \ = \ 5 \ ⇒ \ x \ - \ 4 \ = \ 5 \ (x \ + \ 3)$
$⇒ \ x \ - \  4\ = \ 5 \ x \ + \ 15 \ ⇒ \ 5 \ x \ - \ x \ =\ -15 \ - \ 4 \ → \ 4 \ x \ = \ -19 \ ⇒ \ x \ =-  \ \frac{19}{4}$

The correct answer is $\frac{2 \ x \ + \ 6}{x^2 \ +\ 3 \  x}$
$(\frac{f}{g})  (x) \ = \ \frac{f(x)}{g(x)} \ = \  \frac{2 \ x \ + \ 6}{x^2 \ +\ 3\  x}$

The correct answer is $6$
$y \ = \ m \ x \ + \ b$
$m \ =$ slop
$y \ = \ 6 \ x \ + \ 12$
$m \ = \ 6$

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: $\log_{a}{x^b} \ = \ b \log_{a}{x}$
$ln⁡(e^{7 \ x \ + \ 1 })  =  (7 \ x \ + \ 1) \ ln(e)$
$(7 \ x \  + \ 1) \ ln \ (e) \ = \ ln \ (8)$
$(7 \ x \ + \ 1) \ ln \ (e) \ = \ (7 \ x \ + \ 1)$
$(7 \ x \ + \ 1) \ = \ ln \ (8) \ ⇒ \ x \ = \ \frac{ln⁡(8) \ - 1}{7}$