Free Full Length ACCUPLACER Mathematics Practice Test

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

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$^2$

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 $62^\circ$ .

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

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

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

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

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 $178.48$

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.$(1, 5) \Rightarrow 5 \ = \ 6 \ – \ 1 \Rightarrow \ 5 \ = \ 5$

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

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

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.$y  =  x \ + \ 1  ⇒  3 =  2 \ + \ 1 \ ⇒ \ 3 \ = \ 3$

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $\theta \ = \ \frac{15}{17}$

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

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: $\log_{2}{(x \ + \ 6)} \ - \ \log_{2}{(x \ - \ 2)}=\log_{2}{\frac{x \ + \ 6}{x \ - \ 2}} \ = \ 1⇒$
$\frac{x \ + \ 6 }{x \ - \ 2}  =  2^1  =  2  ⇒  x  \ + \ 6 \ = \ 2 \ (x \ - \ 2)$
$⇒ \ x \ + \ 6   =  2 \ x \ - \ 4  ⇒ 2 \ x \ - \ x \ = \ 6 \ + \ 4 \ ⇒ \ x \ = \ 10$

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

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 $= \ [6 , \ + \ ∞)$

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: $x = 2$

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

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

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: $3 \ \sqrt{2}$

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 $4$

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

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: $x_1 \ = \ 4, \  x_2 \ = 2, \  y_1 \ = \ 1, \ y_2 \ =  - \ 3$
$\frac{4 \ +\ i}{ 2 \ - \ 3  \ i} \ = \ \frac{5}{13} \ + \  i \ \frac{14}{13}$

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

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

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