- Find the indicated root $\sqrt[3]{-64}$
- $-4$
- $-16 $
- $ 16$
- $4 $
- Cannot be evaluated
- $3^{3x}=9^{x-1}$
- $2$
- $-2 $
- $\frac{1}{2} $
- $\frac{1}{3} $
- $3 $
- $3\sqrt{20}+6\sqrt{45}$
- $24\sqrt{5}$
- $9\sqrt{65} $
- $3\sqrt{5} $
- $20\sqrt{5} $
- $18\sqrt{5} $
- $2a^{2}-3(b-c)^{2}$
- $[\sqrt{2}a-\sqrt{3}b+\sqrt{3}c][\sqrt{2}a+\sqrt{3}b-\sqrt{3}c]$
- $[\sqrt{2}a-\sqrt{3}b+\sqrt{3}c][\sqrt{2}a-\sqrt{3}b+\sqrt{3}c]$
- $[\sqrt{2}a-\sqrt{3}b.\sqrt{3}c][\sqrt{2}a+\sqrt{3}b.\sqrt{3}c]$
- $[2a-3b+3c][2a+3b-3c]$
- $[\sqrt{2}a+\sqrt{3}b+\sqrt{3}c][\sqrt{2}a+\sqrt{3}b+\sqrt{3}c]$
- $(2\sqrt{5}+5\sqrt{2})(5\sqrt{5}-2\sqrt{2})$
- $30-21\sqrt{10}$
- $30+21\sqrt{10} $
- $-30+21\sqrt{10} $
- $-10 $
- $+10 $
- $x^{3}-36x=0$
- $-6$ and $6$
- $-3$, $0$ and $3$
- $-6$, $0$ and $6$
- $36 $
- $6 $
- $g(x)=-2x^{2}+5x-1$
$g(-5)$
- $log_{25}(\frac{1}{125})=$
- $\frac{-3}{2}$
- $\frac{2}{3} $
- $\frac{3}{2} $
- $ -3$
- $ 2$
- If $log_{b}(5)=0.81$ and $log_{b}(3)=0.13$ find $log_{b}(15)$
- $0.1053$
- $0.94 $
- $0.10 $
- $0.9 $
- $0.95 $
- If $log_{5}N=2$, find $N$
- $\frac{5}{2}$
- $10 $
- $\frac{2}{5} $
- $log_{2}5 $
- $ 25$
This site prepare students for math exams, SAT, GRE, GMAT,CLEP practice test, CLEP college Algebra,CLEP precalculus,CLEP Mathematics
Tuesday, June 28, 2011
CLEP College Algebra Practice Questions - 5
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment